[{"content":" In a BS 7910 fracture assessment most attention goes to flaw size, stress and toughness — but one thing hidden in the material is easy to miss: when this steel yields, does it pass smoothly into plasticity, or does it \u0026ldquo;catch, then let go\u0026rdquo; first? This small metallurgical detail grows a cliff in the failure assessment line near $L_r=1$ — get it wrong and you badly overestimate the safety margin. This article explains what yielding is, how continuous and discontinuous yielding differ, and how Option 1 and Option 2 each handle them.\nIf you are not yet comfortable with the horizontal axis $L_r$, the vertical axis $K_r$ and the dividing failure assessment line of the Failure Assessment Diagram (FAD), read the BS 7910 Fracture Assessment — A Concise Guide first; for an overview of how to choose among the three assessment options, see Clause 7\u0026rsquo;s Three Assessment Options: How to Choose Between Option 1 / 2 / 3 . This article covers one thing only: how the material\u0026rsquo;s yielding behaviour changes the shape of that FAL curve.\n1. First, yielding: the step from \u0026ldquo;elastic\u0026rdquo; to \u0026ldquo;plastic\u0026rdquo; Pull a metal bar and at first it acts like a spring — remove the load and it springs back. This is elastic deformation. But once the load passes a threshold, the bar keeps a permanent stretch that does not recover on unloading. That is yielding: the material crosses from recoverable elastic behaviour into unrecoverable plastic deformation. That threshold stress is the yield strength $\\sigma_Y$.\nExactly which value you take for the yield strength depends on how \u0026ldquo;cleanly\u0026rdquo; the material yields:\nSome materials show a clear upper yield point $R_{eH}$ (the highest stress reached before the dislocations break free) and a lower yield strength $R_{eL}$ (the value the stress falls back to and settles at). For these, engineering practice takes the lower yield strength $R_{eL}$ as the yield strength — it is the more stable, safer value. Other materials yield \u0026ldquo;gradually\u0026rdquo;, with no clear yield point at all. For these a value is defined by convention: the stress that produces $0.2\\%$ permanent plastic strain, written $0.2\\%$ proof strength $R_{p0.2}$, is used as the nominal yield strength (BS 7910:2019, §7.1.3). Those two ways of \u0026ldquo;yielding cleanly or not\u0026rdquo; are exactly the subject of this article — continuous versus discontinuous yielding.\n2. Two yielding behaviours: continuous vs discontinuous Draw the stress-strain curves of the two material types together and the difference is immediate:\nFigure 1: Stress-strain curves for the two yielding behaviours (the elastic part is schematic, not to true-modulus scale, so the yield details stay visible). Left: continuous yielding — past the yield region the curve rises smoothly with no plateau, and the yield strength is set by the 0.2% proof strength R_p0.2. Right: discontinuous yielding — the stress first peaks at the upper yield point R_eH, drops back to the lower yield strength R_eL, then runs along a Lüders plateau where the stress barely rises while the strain grows a lot (length Δε); strain hardening only begins once the plateau ends.\nContinuous yielding: the stress-strain curve passes smoothly, and past the yield point the stress keeps rising with strain (strain hardening) without any pause. Discontinuous yielding: after peaking at the upper yield point the stress falls back to the lower yield strength, and then a nearly flat plateau appears — the stress barely rises while the strain runs on. This plateau is the yield plateau or Lüders plateau, the large strain produced along it is the Lüders strain $\\Delta\\varepsilon$, and visible slanted slip bands — Lüders bands — appear on the surface. Why is there a plateau? In microscopic terms: in low-carbon steel the carbon and nitrogen atoms gather around the dislocations and \u0026ldquo;pin\u0026rdquo; them like small locks (this cloud of atoms is the Cottrell atmosphere). To start deforming you must pull the dislocations free of the locks with a large force — this gives the raised upper yield point. But once free, the dislocations slide more easily through the lattice, so they move in large groups: macroscopically the load need not rise while the bar keeps stretching — the yield plateau. Only after most of these dislocations have run does the material enter the strain-hardening stage where force must rise again.\nA note on terms: the rise of stress with strain after yield is called strain hardening in fracture mechanics (also \u0026ldquo;work hardening\u0026rdquo; in manufacturing); the strain-hardening exponent $n$ used later in Option 2 measures how strong it is.\n3. Which materials are which? Yielding type is set mainly by composition, heat treatment and processing history. Roughly:\nMaterials that tend to yield discontinuously (with a plateau):\nAs-rolled or normalized low-carbon and C-Mn steels, especially rolled steel plate — the most typical group; carbon steels sensitive to strain aging, whose plateau is especially marked when they operate in the relevant temperature range. Materials that tend to yield continuously (no plateau, smooth transition):\naustenitic stainless steels and aluminium alloys (their lattice does not readily form the Cottrell \u0026ldquo;locks\u0026rdquo;); quenched-and-tempered high-strength steels; cold-worked (cold-drawn, cold-rolled) steels — the dislocations were already pulled free during processing, so the yield plateau is used up in advance. This is only an empirical tendency, not a rule. In §7.1.3.6, Table 7.4, BS 7910 gives detailed guidance on whether to assume a yield plateau in rolled steel plate, by yield strength range, process route, composition and heat treatment — when you have a specific grade and process, use that table (BS 7910:2019, §7.1.3.6, Table 7.4).\n4. Not sure which? Be safe — assume \u0026ldquo;discontinuous\u0026rdquo; In real work you cannot always get a full stress-strain curve for a given steel, nor be certain whether it has a plateau. What then?\nThe spirit of BS 7910 is: when data are insufficient and you cannot decide, the safe course is to [assume discontinuous yielding] (that is, assume a Lüders plateau exists). The standard even gives, for materials with $\\sigma_Y \u003c 1000\\ \\mathrm{N/mm^2}$, a direct estimate of the plateau length (BS 7910:2019, §7.1.3.6, Eq. (7.8)):\n$$\\Delta\\varepsilon = 0.0375\\,(1 - 0.001\\,\\sigma_Y)$$Why is \u0026ldquo;discontinuous\u0026rdquo; the conservative assumption? As Section 5 below shows, discontinuous yielding makes the failure assessment line drop like a cliff at $L_r=1$, pushing the allowed fracture ratio $K_r$ in that region very low. Assuming a plateau tightens the acceptance standard on purpose; conversely, treating a steel that actually has a plateau as continuous yielding overestimates the safety margin near $L_r\\approx1$ and gives an unsafe result. Safety comes first — when unsure, it is better to assume the cliff is there.\nWhen you have a specific grade and process: use Table 7.4 If you know the material\u0026rsquo;s yield strength range, process route, composition and heat treatment, BS 7910 gives a decision table for rolled steel plate (BS 7910:2019, §7.1.3.6, Table 7.4):\nYield strength range (N/mm²) Process route Composition Heat treatment Assume yield plateau (discontinuous)? $R_{eH}\\le350$ As-rolled Conventional steels (e.g. BS EN 10025-2, no microalloy additions) — Yes $R_{eH}\\le350$ As-rolled Mo, Cr, Nb, Al or Ti present — (No) $R_{eH}\\le350$ Normalized BS EN 10025-2 type, no microalloy Conventional normalizing Yes $R_{eH}\\le350$ Normalized BS EN 10025-3 type, with microalloy Conventional normalizing Yes $R_{eH}\\le350$ Controlled rolled BS EN 10025-3 / 10025-4 — Yes $R_{eH}\u003e350$ Controlled rolled BS EN 10025-3 / 10025-4 Light TMCR ($R_{eH}\u003c400$) Yes $R_{eH}\u003e350$ Controlled rolled BS EN 10025-3 / 10025-4 Heavy TMCR ($R_{eH}\u003e400$) (Yes) $R_{eH}\\le500$ Quenched \u0026amp; tempered Mo or B present, with microalloy Cr/V/Nb/Ti Heavy tempering (favours plateau) Yes $R_{eH}\\le500$ Quenched \u0026amp; tempered Mo or B present, with microalloy Light tempering (favours no plateau) (Yes) $R_{eH}\\le500$ Quenched \u0026amp; tempered No Mo/B, but microalloy present (V especially strong) Heavy tempering (Yes) $R_{eH}\\le500$ Quenched \u0026amp; tempered No Mo/B, but microalloy present Light tempering (No) $R_{p0.2}$ or $R_{eH}\u003e500$ Quenched \u0026amp; tempered Mo or B present, with microalloy Tempered to $R_{p0.2}\u003c\\sim690$ (No) $R_{p0.2}$ or $R_{eH}\u003e500$ Quenched \u0026amp; tempered Mo or B present, with microalloy Tempered to $R_{p0.2}\u003e\\sim690$ No $R_{p0.2}$ or $R_{eH}\u003e500$ Quenched \u0026amp; tempered No Mo/B, but microalloy present Tempered to $R_{p0.2}\u003c\\sim690$ Yes $R_{p0.2}$ or $R_{eH}\u003e500$ Quenched \u0026amp; tempered No Mo/B, but microalloy present Tempered to $R_{p0.2}\\ge\\sim690$ (No) $R_{p0.2}\\le1000$ As-quenched All compositions — No A bracketed \u0026ldquo;(Yes)/(No)\u0026rdquo; in the table means the call is uncertain; then run a sensitivity analysis on whether the presence or absence of a plateau affects the assessment. Yield strength in the table is defined by the upper yield strength $R_{eH}$ (to harmonize with the relevant standards). TMCR means thermo-mechanical controlled rolling. Source: BS 7910:2019, §7.1.3.6, Table 7.4.\nNow to the main point: how Option 1 and Option 2 each draw the two yielding behaviours into the FAL.\nBefore starting, keep in mind one thing that runs through the whole article: the assessment point $(L_r,\\,K_r)$ is computed the same way regardless of yielding type; yielding type only changes the shape of the FAL curve $f(L_r)$. The horizontal $L_r=\\sigma_{ref}/\\sigma_Y$ and the vertical $K_r=(K_I^{\\,p}+K_I^{\\,s})/K_{mat}+\\rho$ are computed as usual (BS 7910:2019, §7.3.6/§7.3.7).\n5. Option 1\u0026rsquo;s way: swap the equations Option 1 needs the least data (only $\\sigma_Y$, $\\sigma_U$, $E$), and its way of handling the two cases is direct — it keeps two sets of equations and switches by material.\nFigure 2: The two failure assessment lines Option 1 gives for the two yielding behaviours (representative structural steel σ_Y=350, σ_U=500 MPa, E=210 GPa). Blue: continuous yielding (Eq. 7.26–7.28). Red: discontinuous yielding (Eq. 7.29–7.32) — note it runs slightly above the blue line for Lr\u0026lt;1, yet drops vertically at Lr=1 from 0.816 to 0.253 and stays below the blue line afterwards. Both curves share the same plastic-collapse cut-off L_r,max at the right.\nContinuous yielding uses this set (BS 7910:2019, §7.3.3, Eq. (7.26)–(7.28)):\n$$f(L_r) = \\left(1 + \\tfrac{1}{2}L_r^2\\right)^{-0.5}\\left[\\,0.3 + 0.7\\exp\\!\\left(-\\mu L_r^6\\right)\\right] \\quad (L_r \\le 1)$$$$\\mu = \\min\\!\\left(0.001\\tfrac{E}{\\sigma_Y},\\; 0.6\\right), \\qquad N = 0.3\\left(1 - \\tfrac{\\sigma_Y}{\\sigma_U}\\right)$$Discontinuous yielding switches to another set (BS 7910:2019, §7.3.3, Eq. (7.29)–(7.33)). There are two key differences:\nThe $L_r\u003c1$ branch becomes $f(L_r) = \\left(1 + \\tfrac{1}{2}L_r^2\\right)^{-0.5}$ — the square-bracket term is gone (and with it $\\mu$). So this branch sits slightly higher than the continuous curve. At $L_r=1$, $f$ drops vertically from its pre-plateau value to a lower one: $$f(L_r{=}1) = \\left(\\lambda + \\tfrac{1}{2\\lambda}\\right)^{-0.5}, \\qquad \\lambda = 1 + \\frac{E\\,\\Delta\\varepsilon}{R_{eL}}$$Why the vertical drop? At $L_r=1$ the reference stress just equals the yield strength, and the material runs a whole strain increment $\\Delta\\varepsilon$ along the yield plateau at nearly constant stress. The plastic deformation capacity of the flawed section is released at once, so the allowed $K_r$ drops like a cliff from its pre-plateau value to the post-plateau value. The parameter $\\lambda$ measures the \u0026ldquo;total strain at the end of the plateau\u0026rdquo; relative to the \u0026ldquo;initial elastic yield strain\u0026rdquo; — the longer the plateau (the larger $\\Delta\\varepsilon$), the larger $\\lambda$ and the deeper the drop (in Figure 2, $\\lambda=15.6$ and $f$ falls straight from $0.816$ to $0.253$). After the drop, the $1","permalink":"https://mechcalc.net/blog/en/posts/bs7910-yielding-behaviour-fad/","summary":"\u003cblockquote\u003e\n\u003cp\u003eIn a BS 7910 fracture assessment most attention goes to flaw size, stress and toughness — but one thing hidden in the material is easy to miss: when this steel yields, does it pass smoothly into plasticity, or does it \u0026ldquo;catch, then let go\u0026rdquo; first? This small metallurgical detail grows a \u003cstrong\u003ecliff\u003c/strong\u003e in the failure assessment line near $L_r=1$ — get it wrong and you badly overestimate the safety margin. This article explains what yielding is, how continuous and discontinuous yielding differ, and how Option 1 and Option 2 each handle them.\u003c/p\u003e","title":"Continuous or Discontinuous Yielding? The Two Ways BS 7910 Draws Option 1 and Option 2"},{"content":" In a BS 7910 fracture assessment, the horizontal axis $L_r$, the vertical axis $K_r$, and the failure assessment line (FAL) that separates \u0026ldquo;safe\u0026rdquo; from \u0026ldquo;unsafe\u0026rdquo; — how the assessment point is computed and how the verdict is read — are covered thoroughly in the BS 7910 Fracture Assessment — A Concise Guide . This article covers one thing only: Clause 7 gives three ways to draw that FAL curve (Option 1 / 2 / 3). They step up in the material data they need, in computational accuracy and in conservatism; understanding their differences is a key step to doing a fracture assessment \u0026ldquo;correctly and economically\u0026rdquo;.\n1. One FAD, three ways to draw it First, an easily-missed fact: the three options change only how the failure assessment line $f(L_r)$ is drawn; the assessment point $(L_r,\\,K_r)$ is computed identically in all three.\nThat is, whichever option you choose:\nthe horizontal load ratio $L_r = \\dfrac{P}{P_L(a,\\sigma_Y)} = \\dfrac{\\sigma_{ref}}{\\sigma_Y}$ is still computed from the primary load only (BS 7910:2019, §7.3.7, Eq. (7.40)); the vertical fracture ratio $K_r$ still assembles the primary stress $K_I^{\\,p}$, the secondary stress $K_I^{\\,s}$ and the plasticity interaction correction as below (BS 7910:2019, §7.3.6, Eq. (7.39)): $$K_r = \\frac{K_I^{\\,p} + K_I^{\\,s}}{K_{mat}} + \\rho$$The only difference between the three options is what that FAL curve $f(L_r)$ looks like. In other words, choosing an option means choosing \u0026ldquo;how much material data to trace the boundary with\u0026rdquo; — the more data, the closer the boundary follows the material\u0026rsquo;s real behaviour.\nBefore comparing them, one hard boundary shared by all three must be set up first.\nThe plastic-collapse cut-off $L_{r,\\max}$: the right wall for all three curves Whichever option, the horizontal axis has an upper limit $L_{r,\\max}$. Once $L_r$ reaches it, the structure fails by overall plastic flow of the flawed section no matter how high the toughness, and $f(L_r)$ is taken as zero (BS 7910:2019, §7.3.2, Eq. (7.25)):\n$$L_{r,\\max} = \\frac{\\sigma_Y + \\sigma_U}{2\\sigma_Y}$$The numerator $(\\sigma_Y+\\sigma_U)/2$ is the flow stress $\\sigma_f$ (the mean of yield and tensile strength), representing the average stress level as the material goes fully plastic. So $L_{r,\\max}$ is essentially \u0026ldquo;the flow stress relative to the yield strength\u0026rdquo;, computed with mean (not lower-bound) tensile properties. This vertical wall applies to all three options alike.\n2. Option 1: yield and tensile strength only Option 1 is the option with the lowest data need and the most common in practice: it does not need a full stress-strain curve — the whole FAL can be set from the yield strength $\\sigma_Y$, tensile strength $\\sigma_U$ and elastic modulus $E$ alone (BS 7910:2019, §7.3.3). It is a general approximate lower bound the standard fitted from data on many common structural steels.\nContinuous-yielding materials (no yield plateau) $$f(L_r) = \\left(1 + \\tfrac{1}{2}L_r^2\\right)^{-0.5}\\left[\\,0.3 + 0.7\\exp\\!\\left(-\\mu L_r^6\\right)\\right] \\quad (L_r \\le 1)$$$$f(L_r) = f(L_r{=}1)\\;L_r^{\\,(N-1)/(2N)} \\quad (1 \u003c L_r \u003c L_{r,\\max})$$The two material parameters are (BS 7910:2019, §7.3.3, Eq. (7.26), (7.27)):\n$$\\mu = \\min\\!\\left(0.001\\frac{E}{\\sigma_Y},\\; 0.6\\right), \\qquad N = 0.3\\left(1 - \\frac{\\sigma_Y}{\\sigma_U}\\right)$$Reading: the $(1+\\tfrac{1}{2}L_r^2)^{-0.5}$ outside the bracket comes from a small-scale-yielding correction, and the exponential decay inside the bracket captures the rapid rise of driving force as yield is approached; the lower the yield-to-tensile ratio $\\sigma_Y/\\sigma_U$ (the stronger the strain hardening), the larger $N$, and the more gently the curve drops in the $L_r\u003e1$ branch.\nDiscontinuous-yielding materials (with a yield plateau) Many low-carbon and C-Mn steels have a yield plateau (Lüders plateau) at the lower yield strength — the stress barely rises while the strain suddenly grows. The FAL for these materials switches to another set of equations, with a vertical drop at $L_r = 1$ (BS 7910:2019, §7.3.3, Eq. (7.29)–(7.33)):\n$$\\lambda = 1 + \\frac{E\\,\\Delta\\varepsilon}{R_{eL}}$$Why the vertical drop? At $L_r=1$ (the reference stress just reaches yield) the material runs a strain increment $\\Delta\\varepsilon$ along the yield plateau at nearly constant stress; the plastic deformation capacity of the flawed section is released abruptly, so the allowed $K_r$ drops like a cliff from its pre-plateau value to a lower post-plateau value. The parameter $\\lambda$ measures the \u0026ldquo;total strain at the end of the yield plateau\u0026rdquo; relative to the \u0026ldquo;initial elastic yield strain\u0026rdquo; — the longer the plateau (the larger $\\Delta\\varepsilon$), the larger $\\lambda$ and the deeper the drop.\nDo not use the wrong yielding type: for a steel with a yield plateau, using the continuous-yielding equation (7.26) overestimates the safety margin near $L_r \\approx 1$ — an unsafe error. For deciding whether a material has a yield plateau, see §7.1.3.6. (See also the dedicated article, Continuous or Discontinuous Yielding? )\n3. Option 2: the true stress-true strain curve, the general option Option 2 builds the FAL from the mean uniaxial true stress-true strain curve at the assessment temperature; it is suitable for all metals regardless of stress-strain behaviour, so it is more general than Option 1 and its FAL follows the material more closely (BS 7910:2019, §7.3.4, Eq. (7.34)):\n$$f(L_r) = \\left(\\frac{E\\,\\varepsilon_{ref}}{L_r\\,\\sigma_Y} + \\frac{L_r^3\\,\\sigma_Y}{2\\,E\\,\\varepsilon_{ref}}\\right)^{-0.5} \\quad (L_r \u003c L_{r,\\max})$$Here $\\varepsilon_{ref}$ is the true strain at the true stress $\\sigma_{ref} = L_r\\sigma_Y$. A handy reading: the denominator $L_r\\sigma_Y$ of the first term is exactly the reference stress $\\sigma_{ref}$, so the first term $E\\varepsilon_{ref}/\\sigma_{ref}$ is the inverse of \u0026ldquo;the elastic-strain estimate $\\sigma_{ref}/E$ over the true reference strain $\\varepsilon_{ref}$\u0026rdquo;. As $L_r$ rises and the material goes plastic, $\\varepsilon_{ref}$ departs from the elastic value, the first term grows, and $f(L_r)$ falls — this is how the material\u0026rsquo;s real plastic deformation capacity is written into the FAL.\nTo use Option 2, you must first build the true stress-true strain curve from tensile data (or a Ramberg-Osgood fit, Eq. (7.4)–(7.7)), especially the knee below 1% strain — the FAD shape near the knee at $L_r\\approx1$ depends heavily on the quality of that data. The full step-by-step procedure for building the FAL point by point is a separate topic.\nTrue stress ≠ engineering stress: Option 2 requires true stress-true strain input. A tensile test gives engineering stress-engineering strain directly; feeding it in without converting to true values in the plastic range causes a clear error.\n4. Option 3: directly from elastic-plastic $J$ — the most accurate but not general Option 3 targets a specific material, geometry and loading type: for the flawed structure it runs both an elastic analysis and an elastic-plastic analysis (usually by finite element), and defines the FAL from the ratio of their $J$-integrals (BS 7910:2019, §7.3.5, Eq. (7.36)):\n$$f(L_r) = \\sqrt{\\frac{J_e}{J}} \\quad (L_r \u003c L_{r,\\max})$$where $J_e$ is the $J$ from an elastic analysis at a given $L_r$ load and $J$ is the $J$ from an elastic-plastic analysis at the same load. Reading: the ratio $J_e/J$ measures \u0026ldquo;how much a purely elastic calculation would underestimate the true driving force\u0026rdquo; — the more plasticity, the more $J \\gg J_e$, the smaller $\\sqrt{J_e/J}$, and the lower the FAL.\nOption 3 is the most accurate but also the most costly: it is not general — one curve is valid only for the specific configuration it was derived for — so it is used only in specific critical assessments as an alternative to Options 1 and 2.\n5. Comparing the three: how data, accuracy and conservatism step up Put the three FALs on one failure assessment diagram and the pattern is clear at a glance: \u0026ldquo;more data → curve further out → larger acceptable region\u0026rdquo;:\nFigure 1: The three Clause 7 assessment options\u0026rsquo; failure assessment lines compared (representative structural steel σ_Y=350, σ_U=500 MPa, E=210 GPa). Option 1 (blue, needs only σ_Y/σ_U/E) and Option 2 (green, from the true stress-true strain curve) are computed exactly from the code equations; Option 3 (grey dashed, elastic-plastic J) is schematic — it must be found by finite element J for a specific configuration and is not a general curve. All three share the same plastic-collapse cut-off L_r,max at the right. The higher the data need, the closer the FAL follows the material\u0026rsquo;s real behaviour, the less conservative the margin, and the larger the acceptable flaw.\nThe trade-off in one line:\nOption Data needed Cost Character Option 1 Only $\\sigma_Y$, $\\sigma_U$, $E$ Lowest (plug into the general formula) General approximate lower bound, least data, the engineering default starting point Option 2 True stress-true strain curve (especially the \u0026lt;1% knee) Medium (built point by point) Suitable for all metals, follows the material\u0026rsquo;s real hardening behaviour Option 3 Elastic-plastic $J$-integral solution (usually FEA) Highest (numerical analysis) Most accurate, valid only for a specific configuration, not general In general, conservatism runs Option 1 \u0026gt; Option 2 \u0026gt; Option 3 — the further along, the closer to reality and the less margin spent.\nAn exception you must know: low-strain-hardening materials \u0026ldquo;Higher option = less conservative\u0026rdquo; is not absolute. For some low-strain-hardening (very weak work-hardening) materials, the Option 2 curve can actually fall inside the general Option 1 curve (BS 7910:2019, §7.4.2).\nThe reason: Option 1 is an approximate lower bound the standard fitted from common structural steels. If a material\u0026rsquo;s stress-strain curve flattens quickly after yield, then just past yield its plastic deformation (and elastic-plastic $J$) grows far faster than for ordinary materials, so $f(L_r)$ drops sharply and falls inside the Option 1 curve. Here Option 1 is on the unsafe side, and Option 2, based on the real stress-strain curve, must be used to envelop the risk faithfully. So moving up an option is not only about \u0026ldquo;spending less margin\u0026rdquo; — sometimes it is about \u0026ldquo;correcting the distortion of the general curve for a specific material\u0026rdquo;.\nHow to choose: by scenario and data completeness Which option to choose depends on the completeness of the data at hand and the urgency of the assessment:\nRoutine quick assessment, limited material data: start from Option 1. It needs the least data and is the safest; in most cases, if it passes there is no need to go further. Option 1 fails but the margin is worth pursuing, and you have (or can afford) the material\u0026rsquo;s true stress-strain curve: move up to Option 2 and use the real hardening behaviour to recover the margin Option 1 \u0026ldquo;buried\u0026rdquo;. Margin is critical, the configuration is important, and numerical analysis is justified: use Option 3, run an elastic-plastic $J$ analysis for the specific configuration, and make the assessment as realistic as possible. For a low-strain-hardening material: even if Option 1 passes, check with Option 2 — because Option 1 may be non-conservative here (see above). 6. Three common pitfalls Pitfall 1: using the wrong yielding type. For a carbon steel with a yield plateau, not using the discontinuous-yielding equations with their drop segment badly overestimates the load-bearing safety margin in the $L_r\\ge1$ region (§7.4.2). Pitfall 2: treating \u0026ldquo;higher option\u0026rdquo; as \u0026ldquo;safer\u0026rdquo;. Moving up an option only means the calculation follows reality more closely and removes some conservative assumptions of the lower option (saving margin) — it adds no extra safety. The BS 7910 method has no built-in safety factor of its own — the assessment point lying inside the curve only means \u0026ldquo;in theory it just does not fail\u0026rdquo;. The real safety margin must be confirmed by the user with a sensitivity analysis on the flaw size, stress, toughness and other inputs (§7.1.12). Pitfall 3: putting $\\rho$ inside the fraction. The plasticity interaction correction $\\rho$ is a separate term added outside the fraction, $K_r = \\dfrac{K_I^{\\,p}+K_I^{\\,s}}{K_{mat}} + \\rho$ — never inside the numerator bracket or the denominator. The standard also gives an equivalent multiplier form $K_r = \\dfrac{K_I^{\\,p}+V\\,K_I^{\\,s}}{K_{mat}}$ (Eq. (7.38)), but once you use the additive term $\\rho$ it must be fully separate (BS 7910:2019, §7.3.6). Note: all three FALs can be used for ductile-tearing analysis (§7.3.8) — assume a set of tearing amounts, compute a string of assessment points with the corresponding enhanced toughness, and join them into a locus; where the locus is tangent to the chosen FAL is the instability point. Changing the option only changes the curve being compared against; the tearing analysis procedure is unchanged.\nTry the three options online with MechCalc Once the differences are clear, hands-on is quickest. In the online BS 7910 fracture assessment calculator, the FAD Option drop-down switches between Option 1 / 2 / 3: Option 1 needs only yield/tensile strength, Option 2 provides Ramberg-Osgood parameters, and Option 3 lets you enter the elastic $J_e$ and elastic-plastic $J$ from your own analysis. Switch options and watch in real time how the FAL curve and assessment point move, and how the reserve factors tighten or loosen.\n🧮 在线计算器：BS 7910 Fracture Assessment Calculator — Switch between Option 1 / 2 / 3 in the \u0026#39;FAD Option\u0026#39; drop-down and compare in real time the margin at the same assessment point under the three FALs.\nThe figures here are original MechCalc illustrations for teaching only; they do not replace the BS 7910 text. The Option 3 curve in the figure is schematic; in practice it must be determined by finite element J for a specific configuration. For engineering assessment use the current BS 7910:2019+A1:2020.\n","permalink":"https://mechcalc.net/blog/en/posts/bs7910-clause7-fad-options/","summary":"\u003cblockquote\u003e\n\u003cp\u003eIn a BS 7910 fracture assessment, the horizontal axis $L_r$, the vertical axis $K_r$, and the \u003cstrong\u003efailure assessment line (FAL)\u003c/strong\u003e that separates \u0026ldquo;safe\u0026rdquo; from \u0026ldquo;unsafe\u0026rdquo; — how the assessment point is computed and how the verdict is read — are covered thoroughly in the \u003ca href=\"/blog/en/posts/bs7910-fracture-assessment-tutorial/\"\u003eBS 7910 Fracture Assessment — A Concise Guide\u003c/a\u003e\n. This article covers one thing only: \u003cstrong\u003eClause 7 gives three ways to draw that FAL curve (Option 1 / 2 / 3)\u003c/strong\u003e. They step up in the material data they need, in computational accuracy and in conservatism; understanding their differences is a key step to doing a fracture assessment \u0026ldquo;correctly and economically\u0026rdquo;.\u003c/p\u003e","title":"Clause 7's Three Assessment Options: How to Choose Between Option 1 / 2 / 3, and How They Differ"},{"content":" A fracture assessment watches two things at once — how close to brittle fracture, and how close to plastic collapse. Annex M (the stress intensity factor) covers the first and gives the vertical axis $K_r$ of the Failure Assessment Diagram (FAD); this guide\u0026rsquo;s reference stress $\\sigma_{ref}$ (Annex P) covers the second and gives the horizontal axis $L_r$. Together they complete the Clause 7 fracture assessment .\nPrologue: the horizontal axis measures \u0026ldquo;how close to plastic collapse\u0026rdquo; A cracked structure can fail along two paths: brittle fracture, where the crack-tip driving force exceeds the toughness, and plastic collapse, where the cracked section as a whole yields and loses its capacity. The first is measured by $K_I$ (vertical axis); the second relies on the reference stress $\\sigma_{ref}$ (horizontal axis).\n1. What the reference stress is The reference stress $\\sigma_{ref}$ is an equivalent uniform stress obtained by normalising the complex geometry and loading of a cracked structure. To be clear from the start:\nIt is not the real stress at the crack tip (that is what $K_I$ describes); It is a yardstick for \u0026ldquo;how close to plastic collapse\u0026rdquo;: when $\\sigma_{ref}$ reaches the material yield strength $\\sigma_Y$, the remaining load-bearing section around the flaw (the ligament) yields as a whole — the onset of plastic collapse. (BS 7910:2019, §P.2) 2. The core idea of the reference-stress method A real cracked structure is an elastic-plastic problem, and computing it exactly means a complex $J$-integral (non-linear finite elements). The reference-stress method (Ainsworth, 1984) bypasses this:\nUse the material\u0026rsquo;s uniaxial tensile (yield) properties and the structure\u0026rsquo;s plastic limit load to estimate the elastic-plastic behaviour of the cracked structure by equivalence.\nConcretely, a closed-form formula gives the structure\u0026rsquo;s plastic limit load $P_L$ (or the equivalent $\\sigma_{ref}$) directly, which then drives the plasticity correction — saving the precise non-linear analysis.\n$\\sigma_{ref}$ and the limit load $P_L$ are really two languages for the same thing (stress language vs load language), linked by the load-bearing area:\n$$L_r = \\frac{P}{P_L} = \\frac{\\sigma_{ref}}{\\sigma_Y}$$\rFigure 1: Plate/shell geometry and the load-bearing section. The reference-stress method treats \u0026lsquo;a cracked section under complex loading\u0026rsquo; as \u0026lsquo;a uniform stress σ_ref acting on the load-bearing section\u0026rsquo; — σ_ref reaching the yield strength corresponds to the plastic limit load P_L of the cracked section.\n3. How $\\sigma_{ref}$ gives the horizontal axis $L_r$ Once $\\sigma_{ref}$ is known, the FAD horizontal axis is the load ratio:\n$$L_r = \\frac{\\sigma_{ref}}{\\sigma_Y}$$ $L_r \u003c 1$: the load is well short of plastic collapse, the ligament essentially elastic; $L_r = 1$: the cracked section just reaches yield (near the \u0026ldquo;knee\u0026rdquo; of the FAL); $L_r \\ge L_{r,max}$: whether or not it fractures, the structure has failed by overall plastic flow. (BS 7910:2019, §7.3.2) Note: only primary stress enters $L_r$. Secondary stress (residual, thermal) is self-balancing and relaxes with plasticity; it does not drive overall collapse, so it stays out of the horizontal axis and enters only the vertical axis $K_r$.\nFigure 2: Primary vs secondary stress. Primary stress (Pm/Pb) is sustained by external loads and enters both K_r and L_r; secondary stress (Qm/Qb) is self-balancing and enters K_r only, not L_r. σ_ref and L_r are computed from primary stress alone.\n4. The general Annex P framework Most Annex P solutions (plate and shell types) write $\\sigma_{ref}$ as a combination of membrane stress $P_m$ and bending stress $P_b$. The signature form is for a through-thickness crack in a plate (BS 7910:2019, §P.4):\n$$\\sigma_{ref} = \\frac{P_b + \\sqrt{P_b^2 + 9P_m^2\\,(1-\\alpha)^2}}{3\\,(1-\\alpha)^2}$$Many other geometry solutions are variations of it. Here $\\alpha$ (written $a''$ in some solutions) is the equivalent flaw parameter, representing how much the flaw weakens the load-bearing section:\n$\\alpha=0$: no flaw, $\\sigma_{ref}=\\sigma_Y$ (the denominator $(1-\\alpha)^2=1$; for pure membrane it reduces to $\\sigma_{ref}=P_m$); larger $\\alpha$: less ligament, more capacity lost, the denominator $(1-\\alpha)^2$ shrinks fast and $\\sigma_{ref}$ rises sharply. Physically, the denominator $(1-\\alpha)^2$ is exactly the non-linear degradation of net-section capacity: the deeper the crack, the higher the equivalent stress the remaining ligament needs to carry the same load.\nConstraint and boundary conditions also affect $\\sigma_{ref}$: fixed-grip (restrained) vs pin-jointed (free rotation), plane stress vs plane strain — each gives a different solution; in general, more constraint means a higher limit load and a lower $\\sigma_{ref}$. (BS 7910:2019, §P.5)\n5. The plastic cut-off $L_{r,max}$ and the flow stress The FAD horizontal axis has a hard cut-off line: for $L_r \\ge L_{r,max}$ the flaw is judged to fail by plastic collapse, regardless of toughness (BS 7910:2019, §7.3.2):\n$$L_{r,max} = \\frac{\\sigma_Y + \\sigma_U}{2\\sigma_Y}$$The numerator $(\\sigma_Y+\\sigma_U)/2$ is the flow stress $\\sigma_f$ — the average stress level between general yield and tensile failure. Limit-load analysis treats the material as \u0026ldquo;elastic-perfectly-plastic\u0026rdquo;, taking the flow stress once plastic. $L_{r,max}$ is the ratio of flow stress to yield strength, usually between 1.1 and 1.4 (not 1).\nFigure 3: The Failure Assessment Diagram (FAD). The horizontal axis L_r comes from σ_ref/σ_Y; the vertical dashed line L_r,max on the right is the plastic cut-off — the further right the assessment point, the closer to plastic collapse, and beyond L_r,max it fails. Inside the green safe region and below the curve is acceptable.\n6. Solutions for different geometries: local vs global collapse Annex P §P.4–P.14 give reference-stress / limit-load solutions by geometry family (plate through/surface/embedded/corner, cylinder axial/circumferential, sphere, round bar/bolt, welded mismatch, etc.), covering pressurised thick-wall cylinders and the Folias bulging of axial cracks in a cylinder. The method is \u0026ldquo;match to the right entry\u0026rdquo;: fix the flaw geometry → find the section → check the load type and applicable range → apply the formula.\nOne key distinction: local collapse vs global (net-section) collapse:\nLocal collapse: the remaining ligament $(B-a)$ at the flaw yields first; the local solution gives a higher $\\sigma_{ref}$ (more conservative). Global collapse: the mean stress over the whole cracked section reaches yield; the global solution is closer to reality. For surface / embedded flaws both solutions are given; which mode arrives first is judged from the plate width $W$ versus the flaw characteristic size $2(c+B)$: a wide plate with a small flaw tends local, a large flaw or a narrow plate tends global. A through-thickness flaw has no remaining ligament and is always global collapse. (BS 7910:2019, §P.3 / §P.5)\n7. Calculation steps From \u0026ldquo;geometry + load + crack size + material strength\u0026rdquo; to \u0026ldquo;computed $\\sigma_{ref}$, $L_r$, $L_{r,max}$\u0026rdquo; (BS 7910:2019, §P and §7.2):\nPrepare inputs: geometry ($W$, $B$, cylinder diameters), flaw ($a$, $c$, position), primary loads (tension / moment / pressure), material strengths $\\sigma_Y$, $\\sigma_U$. (mandatory) Judge the collapse mode: use the plate-width rule to decide local or global collapse, which fixes the solution to use. (mandatory) Select the Annex P solution: match the flaw geometry / position / load to the right §P.4–P.14 entry. (mandatory) Convert to membrane / bending stress: turn the actual loads into $P_m=F/(WB)$, $P_b=6M^b/(WB^2)$, etc. (as needed) Compute $\\sigma_{ref}$: substitute $P_m$, $P_b$ and the flaw parameter $\\alpha$. (mandatory) Compute the load ratio $L_r=\\sigma_{ref}/\\sigma_Y$. (mandatory) Compute the plastic cut-off $L_{r,max}=(\\sigma_Y+\\sigma_U)/(2\\sigma_Y)$; if $L_r\\ge L_{r,max}$ it is unacceptable outright. (mandatory) Into the FAD: plot the assessment point with the horizontal coordinate $L_r$ and the vertical coordinate $K_r$ from Annex M , and judge. (mandatory) 8. Common pitfalls Treating $\\sigma_{ref}$ as the crack-tip stress: $\\sigma_{ref}$ is the equivalent uniform stress over the cracked section and measures plastic collapse; the real crack-tip stress is described by $K_I$. They belong to the horizontal and vertical axes respectively. Thinking a larger $\\sigma_{ref}$ is safer: the opposite — a larger $\\sigma_{ref}$ means a larger $L_r$, hence closer to collapse and less safe. The local solution gives a larger, more conservative $\\sigma_{ref}$. Counting secondary stress in $L_r$: only primary stress enters $L_r$; residual / thermal stress enters $K_r$ only. Thinking $L_{r,max}=1$: $L_{r,max}=(\\sigma_Y+\\sigma_U)/(2\\sigma_Y)\u003e1$ (about 1.1–1.4); $L_r=1$ is only the knee, not the cut-off. Ignoring local vs global: for surface / embedded flaws both solutions must be checked, taking whichever mode arrives first by the plate-width rule; choosing wrong is non-conservative. Compute $\\sigma_{ref}$ online with MechCalc Once you understand the principle, the fastest way to learn is to try it. Use the online BS 7910 Annex P calculator: choose the flaw geometry and solution, enter the crack size and primary stresses, and it computes $\\sigma_{ref}$, $L_r$ and the plastic cut-off $L_{r,max}$ per Annex P, with a white-box derivation; it covers 33 geometries (plates, thin/thick cylinders, spheres, round bars/bolts).\n🧮 在线计算器：BS 7910 Annex P — Reference Stress Calculator — Choose the geometry solution, enter a/c/B/W and Pm/Pb, and get σ_ref, L_r and L_r,max.\nThe figures in this article are original MechCalc diagrams for teaching only and do not replace the BS 7910 standard. For engineering assessment, refer to the current BS 7910:2019+A1:2020.\n","permalink":"https://mechcalc.net/blog/en/posts/bs7910-annex-p-ref-stress-tutorial/","summary":"\u003cblockquote\u003e\n\u003cp\u003eA fracture assessment watches two things at once — how close to \u003cstrong\u003ebrittle fracture\u003c/strong\u003e, and how close to \u003cstrong\u003eplastic collapse\u003c/strong\u003e. \u003ca href=\"https://mechcalc.net/blog/en/posts/bs7910-annex-m-ki-tutorial/\"\u003eAnnex M (the stress intensity factor)\u003c/a\u003e\n covers the first and gives the vertical axis $K_r$ of the Failure Assessment Diagram (FAD); this guide\u0026rsquo;s \u003cstrong\u003ereference stress $\\sigma_{ref}$\u003c/strong\u003e (Annex P) covers the second and gives the horizontal axis $L_r$. Together they complete the \u003ca href=\"https://mechcalc.net/blog/en/posts/bs7910-fracture-assessment-tutorial/\"\u003eClause 7 fracture assessment\u003c/a\u003e\n.\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003ch2 id=\"prologue-the-horizontal-axis-measures-how-close-to-plastic-collapse\"\u003ePrologue: the horizontal axis measures \u0026ldquo;how close to plastic collapse\u0026rdquo;\u003c/h2\u003e\n\u003cp\u003eA cracked structure can fail along two paths: \u003cstrong\u003ebrittle fracture\u003c/strong\u003e, where the crack-tip driving force exceeds the toughness, and \u003cstrong\u003eplastic collapse\u003c/strong\u003e, where the cracked section as a whole yields and loses its capacity. The first is measured by $K_I$ (vertical axis); the second relies on the \u003cstrong\u003ereference stress $\\sigma_{ref}$\u003c/strong\u003e (horizontal axis).\u003c/p\u003e","title":"A Concise Guide to BS 7910 Annex P"},{"content":" The first step of a fracture assessment is to compute the stress intensity factor $K_I$ at the crack tip. This guide explains what $K_I$ is and how BS 7910 Annex M computes it — from the general master formula and its correction factors to the common semi-elliptical surface-crack solution and the weld-toe correction. It feeds the vertical axis $K_r$ of the Failure Assessment Diagram (FAD), and together with Annex P (reference stress) for the horizontal axis $L_r$ completes the Clause 7 fracture assessment .\nPrologue: at the crack tip, the stress is \u0026ldquo;infinite\u0026rdquo; The first question in a fracture assessment is: how strong is this crack\u0026rsquo;s driving force?\nOrdinary strength design uses \u0026ldquo;working stress \u0026lt; allowable stress\u0026rdquo;. But once a crack is present, linear-elastic theory says the stress at the crack tip goes to infinity — there is no finite stress value left to compare with an allowable stress. Fracture mechanics uses a new measuring stick: the stress intensity factor $K_I$.\n1. Why $K_I$, and not ordinary stress The stress field near a crack tip has a general form (linear-elastic fracture mechanics):\n$$\\sigma_{ij} = \\frac{K_I}{\\sqrt{2\\pi r}}\\, f_{ij}(\\theta) + \\cdots$$where $r$ is the distance to the tip and $\\theta$ is the angle. As $r \\to 0$ the stress diverges like $1/\\sqrt{r}$ — this is the crack-tip stress singularity.\nThe key point: although the tip stress is infinite, the strength of the whole singular field is set by a single coefficient, and that coefficient is $K_I$. So:\nOrdinary stress is infinite at the tip and cannot be used directly; $K_I$ is a finite, computable, measurable quantity. Once $K_I$ is known, the stress/strain/displacement field near the tip is fully determined (same shape, only the magnitude differs). Different geometries, crack sizes, and loads can all be measured with the same $K_I$. $K_I$ has units of $\\mathrm{MPa}\\sqrt{\\mathrm m}$. Do not confuse it with the dimensionless stress concentration factor $k_t$, which describes stress magnification at a smooth geometric feature with no crack — a different thing entirely.\n2. $K_I$ and fracture toughness: the brittle-fracture criterion $K_I$ is the crack\u0026rsquo;s driving force; the material\u0026rsquo;s resistance to crack extension is its fracture toughness $K_{mat}$ (measured by test, varying with temperature/constraint/thickness). Comparing the two gives the brittle-fracture criterion:\n$$K_I \\ge K_{mat} \\quad\\Rightarrow\\quad \\text{unstable crack extension (brittle fracture)}$$In the FAD, the vertical axis is exactly the fracture ratio $K_r = K_I/K_{mat}$, measuring \u0026ldquo;how close to brittle fracture\u0026rdquo;. So the $K_I$ computed by Annex M is the raw material fed to the FAD\u0026rsquo;s vertical axis.\n3. The general Annex M framework Annex M fits every geometry\u0026rsquo;s $K_I$ solution into one master formula (BS 7910:2019, §M.1, Eq. M.1):\n$$K_I = (Y\\sigma)\\sqrt{\\pi a}$$Three ingredients: $\\sqrt{\\pi a}$ is the square-root term of the characteristic crack size $a$; $\\sigma$ is the driving stress; $Y$ is the collective name for the geometric correction factors that carry an \u0026ldquo;ideal infinite plate\u0026rdquo; over to a \u0026ldquo;real finite component\u0026rdquo;. All of Annex M\u0026rsquo;s work is writing the specific $Y\\sigma$ for each geometry.\nFor primary stress, $Y\\sigma$ expands as (BS 7910:2019, §M.1, Eq. M.4):\n$$(Y\\sigma)_p = M f_w\\left[k_{tm}M_{km}M_m\\,P_m + k_{tb}M_{kb}M_b\\big(P_b+(k_m-1)P_m\\big)\\right]$$It looks fierce, but it is simply a membrane-stress term plus a bending-stress term, each magnified by a string of correction factors. Secondary stress is far simpler (BS 7910:2019, §M.1, Eq. M.5): $(Y\\sigma)_s = M_m Q_m + M_b Q_b$.\nThe physical meaning of each correction factor:\nSymbol Name Physical meaning When = 1 $M_m$ / $M_b$ membrane / bending magnification factor geometric magnification of $K_I$ by crack shape and position given by each geometry solution $M$ bulging factor extra magnification from the shell wall bulging out at the crack under internal pressure flat plate / no bulging $f_w$ finite-width correction net-section reduction and stress concentration when the crack takes up much of the section crack relatively small $M_{km}$ / $M_{kb}$ weld-toe magnification factor extra magnification when the crack sits in the local stress concentration at a weld toe not at a weld toe $k_{tm}$ / $k_{tb}$ structural stress concentration factor stress concentration at a gross structural discontinuity (nozzle, joint) no structural discontinuity $k_m$ misalignment factor axial misalignment / angular distortion turns membrane stress into extra bending $(k_m-1)P_m$ no misalignment Here $P_m$, $P_b$ are primary stresses (from external loads: pressure, tension, moment) and $Q_m$, $Q_b$ are secondary stresses (self-balancing: weld residual stress, thermal stress). The two play different roles in the FAD:\nFigure 1: Primary vs secondary stress. Primary stress (Pm/Pb) is sustained by external loads and does not self-balance, so it enters both K_r and L_r. Secondary stress (Qm/Qb, e.g. weld residual stress) is self-balancing and relaxes with plastic flow, so it enters K_r only (with a plasticity-interaction correction), not L_r.\nPrimary $K_I^P$ and secondary $K_I^S$ add into the vertical axis: $K_r=(K_I^P+K_I^S)/K_{mat}+\\rho$. When $K_I^S\u003c0$ (compressive, favourable secondary stress), set both $K_I^S$ and $\\rho$ to zero (a conservative treatment).\n4. The most common solution: the semi-elliptical surface crack In practice most surface flaws (weld-toe cracks, fatigue cracks, corrosion pits) are characterised as semi-elliptical surface cracks and use the Newman–Raju solution (BS 7910:2019, §M.4.1) — the workhorse of Annex M.\nFigure 2: Size definitions for the three basic flaws. Top left, a surface crack (depth a, total length 2c, wall thickness B); top right, an embedded crack (height 2a, length 2c, distance p to the near surface); bottom, a through-thickness crack (total length 2a). A semi-elliptical surface crack is described by depth a and half-length c; the aspect ratio a/c fixes how round or flat it is.\nThe membrane magnification factor under tension (BS 7910:2019, §M.4.1, Eq. M.10):\n$$M_m = \\left[M_1 + M_2\\left(\\tfrac{a}{B}\\right)^2 + M_3\\left(\\tfrac{a}{B}\\right)^4\\right]\\frac{g\\,f_\\theta}{\\Phi}$$The bracket is a polynomial in crack depth $a/B$ (the coefficients $M_1,M_2,M_3$ depend only on the aspect ratio $a/c$, with separate sets for $a/2c\\le 0.5$ and $\u003e0.5$); $\\Phi$ is the elliptic shape factor $\\Phi=\\big[1+1.464(a/c)^{1.65}\\big]^{0.5}$; $g$ and $f_\\theta$ describe the variation along the elliptical front. Bending does not start from scratch: it takes $M_b=H\\cdot M_m$ (BS 7910:2019, §M.4.1, Eq. M.12), where $H$ folds in the through-thickness decay of bending stress.\nEvaluation points: deepest point A and surface point C $K_I$ varies along the semi-elliptical front, so two special points are evaluated:\nDeepest point A ($\\theta=\\pi/2$): the deepest part of the crack, deciding \u0026ldquo;will it break through the wall\u0026rdquo;. Surface point C ($\\theta=0$): where the crack meets the free surface, deciding \u0026ldquo;will it grow longer along the surface\u0026rdquo;. Both points must be evaluated for a surface crack (BS 7910:2019, §7.2.12); take the more critical one into the FAD. Which point becomes critical first depends on the aspect ratio and the load type (tension / bending).\nThe finite-width correction for a surface crack is $f_w=\\big\\{\\sec\\big[(\\pi c/W)(a/B)^{0.5}\\big]\\big\\}^{0.5}$: the smaller the crack relative to the plate width, the closer $f_w\\to 1$; the longer and deeper it is, the larger $f_w$. The applicable limit is roughly $2c/W\\le 0.8$.\n5. Welded joints: the weld-toe magnification factor $M_k$ If the crack sits at a weld toe (where the weld meets the parent material), there is a strong local stress concentration and $K_I$ must be multiplied by an extra weld-toe magnification factor $M_k$ (BS 7910:2019, §M.11). $M_k$ starts high at the surface and decays quickly with depth, returning to 1 at about 30% of the wall thickness:\nFigure 3: The weld-toe magnification factor Mk versus crack depth. For a very shallow crack (near the weld-toe surface concentration) Mk is clearly above 1; as the crack deepens out of the local weld-toe field, Mk falls quickly to 1 (the parent-material field). Membrane and bending each have their own Mkm/Mkb curve.\nUse $M_k$ only when the crack is at a weld toe; otherwise take 1. Likewise: $f_w$ is used only when the crack takes up a large part of the section, $k_m$ only when there is misalignment, and the bulging factor $M$ only for axial cracks in a pressurised shell — adding corrections at will ruins the result.\n6. Solutions for different geometries (overview) Annex M gives solutions by geometry family; choosing a solution is \u0026ldquo;matching to the right entry\u0026rdquo; (BS 7910:2019, Annex M):\nGeometry Crack types covered Key feature Flat plate through / edge / semi-elliptical surface / embedded / corner $M_m/M_b$ polynomials + $f_w$ Cylinder, axial through / internal \u0026amp; external surface / embedded bulging factor $M$ under internal pressure Cylinder, circumferential through / internal surface global moment converted to equivalent membrane stress Sphere equatorial through bulging + influence coefficients Round bar / bolt surface / circumferential round section + thread geometry Welded joint weld toe / weld root weld-toe magnification factor $M_k$ Four questions to pick a solution: curved shell or flat plate (bulging or not)? surface or through (two points to evaluate or not)? finite-length or extended (3-D semi-ellipse vs 2-D)? at a weld toe (need $M_k$ or not)?\n⚠️ Two rules: (1) no extrapolation — every solution has an applicable range of $a/B$, $a/c$, $2c/W$; going outside it distorts the result badly, so use the weight-function polynomial solution or finite elements instead; (2) do not count bulging twice — either \u0026ldquo;flat-plate solution times $M$\u0026rdquo; or \u0026ldquo;a cylinder solution that already includes bulging\u0026rdquo;, never both.\n7. Calculation steps From \u0026ldquo;geometry + load + crack size\u0026rdquo; to \u0026ldquo;computed $K_I$\u0026rdquo; (BS 7910:2019, §M and §7.2):\nCharacterise the flaw: idealise the measured flaw into a standard shape (surface / embedded / through / corner) and measure $a$, $c$, $B$, $W$. (mandatory) Resolve the stress: split it into membrane $P_m$ + bending $P_b$ (fit a polynomial for a non-linear profile); separate primary $P$ from secondary $Q$. (mandatory) Select the geometry solution: match the flaw geometry to the right Annex M entry and check its applicable range. (mandatory) Compute the correction factors: as needed, $M_m/M_b$, $f_w$ (only if the crack exceeds ~10% of the section), $M_k$ (weld toe), $M$ (pressurised shell), $k_m$ (misalignment). (as needed) Assemble $K_I$: multiply term by term per Eq. M.4 for $K_I^P$, and per Eq. M.5 for $K_I^S$. (mandatory) Evaluate at the right point: for a surface crack compute the deepest point A and the surface point C, taking the more critical. (mandatory) Into the FAD: use $K_r=(K_I^P+K_I^S)/K_{mat}+\\rho$ as the vertical coordinate, with the $L_r$ from Annex P to complete the assessment. (mandatory) For stress that is non-linear through the thickness (such as a weld residual-stress profile), use a 5th-order polynomial + weight-function solution (BS 7910:2019, §M.4.2 / §M.4.4): $K_I=\\sqrt{\\pi a}\\sum_{i=0}^{5}\\sigma_i(a/B)^i f_i$, with each $f_i$ read from a table by $a/B$ and $2c/a$.\n8. Common pitfalls Confusing $K_I$ with the stress concentration factor $k_t$: $K_I$ has units ($\\mathrm{MPa}\\sqrt{\\mathrm m}$) and describes the crack-tip singular field; $k_t$ is dimensionless and describes a smooth geometric feature — different things. Evaluating only one point: a surface crack must have both the deepest point A and the surface point C computed, taking the more critical. Counting secondary stress in $L_r$: residual / thermal stress enters $K_r$ only, never $L_r$. Adding corrections at will: use $f_w$ only when the crack takes up enough of the section, $M_k$ only at a weld toe, $M$ only for a pressurised shell, $k_m$ only for misalignment; otherwise take 1. Extrapolating out of range: each solution has a valid domain; outside it, switch to a weight-function or finite-element solution rather than forcing the formula. Compute $K_I$ online with MechCalc Once you understand the principle, the fastest way to learn is to try it. Use the online BS 7910 Annex M calculator: choose the flaw geometry, enter the crack size and stresses, and it assembles the correction factors per Annex M, computes $K_I$ at the deepest and surface points, and shows a white-box derivation; it supports polynomial stress (residual-stress profiles) and the weld-toe factor $M_k$.\n🧮 在线计算器：BS 7910 Annex M — Stress Intensity Factor Calculator — Choose the geometry, enter a/c/B/W and Pm/Pb, and get K_I at the deepest point A and the surface point C.\nThe figures in this article are original MechCalc diagrams for teaching only and do not replace the BS 7910 standard. For engineering assessment, refer to the current BS 7910:2019+A1:2020.\n","permalink":"https://mechcalc.net/blog/en/posts/bs7910-annex-m-ki-tutorial/","summary":"\u003cblockquote\u003e\n\u003cp\u003eThe first step of a fracture assessment is to compute the \u003cstrong\u003estress intensity factor $K_I$\u003c/strong\u003e at the crack tip. This guide explains what $K_I$ is and how BS 7910 Annex M computes it — from the general master formula and its correction factors to the common semi-elliptical surface-crack solution and the weld-toe correction. It feeds the vertical axis $K_r$ of the Failure Assessment Diagram (FAD), and together with \u003ca href=\"https://mechcalc.net/blog/en/posts/bs7910-annex-p-ref-stress-tutorial/\"\u003eAnnex P (reference stress)\u003c/a\u003e\n for the horizontal axis $L_r$ completes the \u003ca href=\"https://mechcalc.net/blog/en/posts/bs7910-fracture-assessment-tutorial/\"\u003eClause 7 fracture assessment\u003c/a\u003e\n.\u003c/p\u003e","title":"A Concise Guide to BS 7910 Annex M"},{"content":"In a fitness-for-service (FFS) assessment of a welded pressure-bearing structure, there is a class of stress that comes neither from external load nor from a residual field, but from imperfect fabrication and fit-up — the weld joint is \u0026ldquo;not aligned\u0026rdquo;. BS 7910:2019 Annex D handles exactly this: when two plates or shells to be welded together have axial misalignment or angular distortion, the load path of a tensile load is forced to bend, and a layer of local bending stress $\\sigma_s$ appears at the weld.\nAnnex D is an informative annex — the 2019 edition made all of Annexes A–U informative. It is not a standalone assessment method but a plug-in that supplies stress data to the main assessment: compute $\\sigma_s$, then feed it into the fracture assessment (FAD) and the fatigue assessment. In all, two tables cover 10 standardized configurations: Table D.1 with 7 butt-joint types (a–g) + Table D.2 with 3 cruciform/T-joint types (a–c). Below, the principle first, then a figure and algorithm for each type — none left out.\n1. Basic principle 1.1 Ideal vs real: why does misalignment \u0026ldquo;create\u0026rdquo; stress? Ideal weld: the neutral axes of the two plates lie on one straight line, the tensile force $P_m$ passes along it, and the section carries only a uniform membrane stress.\nReal weld: fabrication is not perfect and the two plates are \u0026ldquo;not aligned\u0026rdquo;. The tensile forces at the two ends are equal, opposite and pulling outward ($\\sum F=0$), but their lines of action are offset by $e$ → they form a couple $M\\approx P_m e$ → adding a layer of bending stress at the weld toe (the boundary between weld and parent metal, where cracks most love to start) (BS 7910:2019, Annex D, D.1).\nFigure 1: Why misalignment creates stress. Left — an ideal joint with the two neutral axes collinear; the end membrane forces are equal and opposite with coincident lines of action, and the section carries only uniform membrane stress. Right — a real joint offset by e; the end membrane forces are still equal and opposite (axial force balance) but their lines of action are offset by e, forming a couple M≈Pm·e that forces an induced bending stress σs at the weld toe (per BS 7910:2019, Annex D, D.1).\n💡 The physics in one line: membrane stress (tension) passing through a \u0026ldquo;skewed joint\u0026rdquo; is partly converted into bending stress. This is the most typical engineering conversion between membrane and bending stress, and the product is a local bending stress at the weld toe.\n1.2 Two basic misalignments + one special geometry Type Symbol Meaning Axial misalignment (eccentricity) $e$ the two neutral axes are parallel but offset by $e$ Angular misalignment $\\alpha$ (peaking $y$) the two plates are joined at an angle $\\alpha$, raising a ridge at the weld, with peaking $y$ Ovality (special geometry) $D_{max}-D_{min}$ a pressure vessel or pipe is out of round and is \u0026ldquo;rounded out\u0026rdquo; under internal pressure, giving extra bending at the weld (BS 7910:2019, Annex D, D.0 + Table D.1)\n1.3 Three applicability boundaries (easiest to get wrong) Directionality — $\\sigma_s$ arises only under the membrane stress component $P_m$ perpendicular to the misalignment line; it does not arise from: ① a continuous weld loaded along its length; ② a plate in pure bending only. Exception: for a section or pipe under overall bending, its membrane stress component still re-generates extra bending through the misalignment. Sign — when several misalignments coexist, the total induced bending is the sum of each type; the same section has tension (+) / compression (−) at different through-thickness positions; the axial and angular components may add with the same sign, or cancel with opposite sign, so they must be added algebraically. Restraint — the amount of bending also depends on whether the joint can rotate freely under the induced moment (which depends on the load, boundary conditions, section shape and local reinforcement). Quantifying restraint usually needs finite element analysis; unless it can be shown that restraint reduces the effect, always compute as \u0026ldquo;unrestrained\u0026rdquo; (most conservative). 💡 Three shared conventions (remember before reading the formulas)\nApplies to butt and fillet joints; except for the fillet-weld root of the cruciform case (Table D.2 case c), all formulas give $\\sigma_s$ at the weld toe. The formulas assume the shape deviation is measured under no load; if measured under load (e.g. a geometry gauge on a thin-wall pipe in service), correct it back to the no-load value. When assessing an embedded crack: the misalignment bending stress varies linearly through the thickness and is zero at the neutral axis; take the value at the crack depth from this distribution. 2. The core products: bending stress σs and magnification factor km 2.1 Two ways to use it + the definition of km Every Annex D formula first gives the dimensionless ratio $\\sigma_s/P_m$. With that ratio there are two routes: Route ① take $\\sigma_s$ directly as the bending component into the fracture assessment; Route ② convert it to a magnification factor $k_m$:\n$$k_m = \\frac{P_m+\\sigma_s}{P_m}=1+\\frac{\\sigma_s}{P_m} \\qquad k_m = 1+\\frac{\\Delta\\sigma_s}{\\Delta P_m}\\ \\text{(stress-range form)}$$(BS 7910:2019, Annex D, Eq. D.1, D.2)\nFigure 2: Stress decomposition at the weld section. Total stress = uniform membrane stress Pm + misalignment bending stress σs (linear through the thickness, zero at the neutral axis); combined, the weld-toe surface reaches a peak of Pm+σs, so the magnification factor km = (Pm+σs)/Pm = 1 + σs/Pm (per BS 7910:2019, Annex D, Eq. D.1).\n2.2 Superposition of several misalignments When axial and angular misalignment coexist, the magnification factors add linearly:\n$$k_m = 1 + (k_m-1)_{\\text{axial}} + (k_m-1)_{\\text{angular}}$$(BS 7910:2019, Annex D, Eq. D.3; when superposing, mind the sign convention of §1.3 point 2 — same-sign addition may be more dangerous, opposite-sign may cancel)\n2.3 A starting anchor: km = 1 + 3e/B 💡 One formula to get you started: equal-thickness plate butt joint, axial misalignment From Table D.1 case a, with unrestrained $\\kappa=6$ and remote loading $l_1=l_2$, it simplifies to: $$\\frac{\\sigma_s}{P_m}=\\frac{3e}{B}\\quad\\Longrightarrow\\quad k_m=1+\\frac{3e}{B}$$ The physics is clear at a glance: the larger the offset $e$ and the thinner the plate (smaller $B$), the stronger the magnification. Example: a 10 mm plate with a 2 mm offset gives $k_m=1.6$ — the membrane stress is magnified by 60%.\n2.4 Primary or secondary stress? The conservative choice is to treat it as primary stress (affecting both the vertical axis $K_r$ and the horizontal axis $L_r$); research shows that for girth-weld scenarios it can often be treated as secondary stress (affecting $K_r$ only, not $L_r$). Take primary first for safety, and relax to secondary when justified — a classic engineering case of the primary/secondary distinction.\n2.5 Through-wall cracks: possibly over-conservative The crack itself releases local bending — the longer the crack, the more it releases — so applying Annex D directly to a through-wall crack can be over-conservative; a dedicated analysis may be needed (BS 7910:2019, Annex D, D.1).\n3. Table D.1: the 7 butt-joint types in detail The common algorithm skeleton (each type follows these five steps): ① measure the geometry ($e$ or $\\alpha,y$; and $B,l$ etc.) → ② choose the formula and parameters by configuration ($\\kappa$, $n$, end condition) → ③ compute $\\sigma_s/P_m$ → ④ get $\\sigma_s=(\\sigma_s/P_m)\\,P_m$, $k_m=1+\\sigma_s/P_m$ → ⑤ feed into the assessment (see Section 5).\na) Equal-thickness plate, axial misalignment Figure 3a: Equal-thickness plate with axial misalignment e (per BS 7910:2019, Annex D, Table D.1 case a).\n$$\\frac{\\sigma_s}{P_m}=\\kappa\\,\\frac{e\\,l_1}{B(l_1+l_2)}$$Algorithm: ① measure the offset $e$, plate thickness $B$, and the two spans $l_1,l_2$; ② take the restraint factor $\\kappa$ — unrestrained take $\\kappa=6$ (most conservative); ③ for remote loading set $l_1=l_2$, and the formula simplifies to $\\sigma_s/P_m=3e/B$; ④ get $\\sigma_s$ and $k_m=1+3e/B$.\n(BS 7910:2019, Annex D, Table D.1 case a)\nb) Unequal-thickness plate, axial misalignment ($B_2\u003eB_1$) Figure 3b: Unequal-thickness plate with axial misalignment, B1 the thinner plate, B2 the thicker (per BS 7910:2019, Annex D, Table D.1 case b).\n$$\\frac{\\sigma_s}{P_m}=\\frac{6e}{B_1}\\left(\\frac{B_1^{\\,n}}{B_1^{\\,n}+B_2^{\\,n}}\\right)$$Algorithm: ① use the thinner plate $B_1$ as the reference; ② the thickness-ratio term $B_1^{\\,n}/(B_1^{\\,n}+B_2^{\\,n})$ distributes the induced bending between the thin and thick plates by stiffness (the thick plate takes more, the thin plate\u0026rsquo;s weld-toe stress is reduced); ③ take $n=1.5$ (supported by tests); ④ applies to remote loading, unrestrained joints.\n(BS 7910:2019, Annex D, Table D.1 case b)\nc) Cylinder/pipe longitudinal weld, axial misalignment ($B_2\\ge B_1$) Figure 3c: Longitudinal weld of a cylinder or pipe with axial misalignment; on the cross-section there is a radial offset e at the longitudinal weld (per BS 7910:2019, Annex D, Table D.1 case c).\n$$\\frac{\\sigma_s}{P_m}=\\frac{6e}{B_1(1-\\nu^2)}\\left[\\frac{1}{1+(B_2/B_1)^{0.6}}\\right]$$Algorithm: on top of the plate form, add two shell corrections — ① $(1-\\nu^2)$ reflects the biaxial restraint of the shell wall (raising stiffness); ② the thickness ratio uses a power of 0.6. The rest of the flow is as in cases a, b.\n(BS 7910:2019, Annex D, Table D.1 case c)\nd) Cylinder/pipe girth weld + spherical-shell weld, axial misalignment ($B_2\\ge B_1$) Figure 3d: Girth weld of a cylinder or pipe, and spherical-shell weld, with axial misalignment; axial section, with cylinder inner radius Ri (per BS 7910:2019, Annex D, Table D.1 case d).\nThe girth/spherical-shell weld uses a piecewise formula:\n$$\\sigma_s/P_m\u003c1:\\quad \\frac{\\sigma_s}{P_m}=\\frac{6e}{B_1(1-\\nu^2)}\\left[\\frac{1}{1+(B_2/B_1)^{1.5}}\\right]$$$$\\sigma_s/P_m\\ge1:\\quad \\frac{\\sigma_s}{P_m}=\\frac{2.6e}{B_1}\\left[\\frac{1}{1+0.7(B_2/B_1)^{1.4}}\\right]$$Algorithm: ① first try the first equation; ② if the result is $\u003c1$ that is the final value; ③ if $\\ge1$, recompute with the second equation (the two switch depending on whether $\\sigma_s/P_m$ reaches 1).\n(BS 7910:2019, Annex D, Table D.1 case d)\ne) Plate, angular misalignment ($\\alpha$ in radians) Figure 3e: Plate angular misalignment; α is the angle between the extension of the left plate\u0026rsquo;s neutral axis and the right plate\u0026rsquo;s neutral axis, with peaking y (per BS 7910:2019, Annex D, Table D.1 case e).\nFixed ends:\n$$\\frac{\\sigma_s}{P_m}=\\frac{3y}{B}\\left[\\frac{\\tanh(\\beta/2)}{\\beta/2}\\right]=\\frac{3\\alpha}{4}\\frac{2l}{B}\\left[\\frac{\\tanh(\\beta/2)}{\\beta/2}\\right]$$Pinned ends:\n$$\\frac{\\sigma_s}{P_m}=\\frac{6y}{B}\\left[\\frac{\\tanh\\beta}{\\beta}\\right]=\\frac{3\\alpha}{2}\\frac{2l}{B}\\left[\\frac{\\tanh\\beta}{\\beta}\\right],\\qquad \\beta=\\frac{2l}{B}\\sqrt{\\frac{3\\,\\sigma_{max,m}}{E}}$$Algorithm: ① measure the angular distortion (peaking $y$ or angle $\\alpha$, with $2l$ the span); ② choose the end condition (fixed or pinned) and take the matching equation; ③ compute $\\beta$ (needs the maximum membrane stress $\\sigma_{max,m}$ and the elastic modulus $E$); ④ the tanh correction — the bracket term is always $\\le 1$, reflecting how tension \u0026ldquo;straightens\u0026rdquo; the angular distortion: under tension, ignoring it is conservative, and it can be ignored for $2l/B\u003c10$; but under compression tanh becomes tan and cannot be ignored.\n(BS 7910:2019, Annex D, Table D.1 case e)\nf) Cylinder/vessel longitudinal or girth weld, angular misalignment Figure 3f: Longitudinal or girth weld of a cylinder or vessel with angular misalignment; a fold in the curved wall, with inner radius Ri (per BS 7910:2019, Annex D, Table D.1 case f).\nFixed ends:\n$$\\frac{\\sigma_s}{P_m}=\\frac{3d}{B(1-\\nu^2)}\\left[\\frac{\\tanh(\\beta/2)}{\\beta/2}\\right]$$Pinned ends:\n$$\\frac{\\sigma_s}{P_m}=\\frac{6d}{B(1-\\nu^2)}\\left[\\frac{\\tanh\\beta}{\\beta}\\right],\\qquad \\beta=\\frac{2l}{B}\\sqrt{\\frac{3(1-\\nu^2)\\,\\sigma_{max,m}}{E}}$$Algorithm: as in case e, but ① use the deviation-from-round $d$ (for ideal geometry $d=y/2$ or $\\alpha l/2$); ② add the shell restraint $(1-\\nu^2)$ (which also enters $\\beta$).\n(BS 7910:2019, Annex D, Table D.1 case f)\ng) Pressure pipe/vessel, ovality ($\\theta$ in degrees) Figure 3g: Ovality of a pressure pipe or vessel, Dmax≠Dmin; internal pressure \u0026lsquo;rounds out\u0026rsquo; the out-of-round section, producing extra bending stress at the weld (per BS 7910:2019, Annex D, Table D.1 case g).\nFull form:\n$$\\frac{\\sigma_s}{P_m}=\\frac{1.5(D_{max}-D_{min})\\cos 2\\theta}{B\\left\\{1+0.5\\left[\\dfrac{p_m(1-\\nu^2)}{E}\\right]\\left(\\dfrac{D}{B}\\right)^3\\right\\}}$$Conservative form:\n$$\\frac{\\sigma_s}{P_m}=\\frac{1.5(D_{max}-D_{min})}{B}$$Algorithm: ① measure the ovality $D_{max}-D_{min}$, the weld position angle $\\theta$, and the internal pressure $p_m$; ② for a quick estimate use the conservative form; ③ for a precise value use the full form (which includes $\\theta$ and the favourable reshaping by internal pressure \u0026ldquo;rounding out\u0026rdquo; the vessel — the denominator $\\{\\dots\\}\u003e1$, so $\\sigma_s$ is reduced); ④ under fatigue loading with varying $p_m$, take the mean value over the period.\n(BS 7910:2019, Annex D, Table D.1 case g)\n4. Table D.2: the 3 cruciform / T-joint types in detail The restraint parameter $\\kappa$ in Table D.2 is not a single constant — it takes different values with the joint\u0026rsquo;s restraint and support (the original table gives values for several support sketches); representative values are listed below. The first two types concern fatigue failure at the parent-plate weld toe.\na) Butt or fillet weld, axial misalignment ($l_1\\le l_2$) Figure 4a: Cruciform or T-joint butt or fillet weld with axial misalignment (per BS 7910:2019, Annex D, Table D.2 case a).\n$$\\frac{\\sigma_s}{P_m}=\\frac{\\kappa\\,e\\,l_1}{B(l_1+l_2)}$$Algorithm: ① measure $e,B,l_1,l_2$; ② choose $\\kappa$ by the actual restraint and support (the table\u0026rsquo;s representative values are $6.0 / 6.75 / 3.0 / 2.95$); ③ unrestrained with remote loading take $\\kappa=6,\\ l_1=l_2$; ④ get $\\sigma_s$, $k_m$. Concerns fatigue failure at the parent-plate weld toe.\n(BS 7910:2019, Annex D, Table D.2 case a)\nb) Butt or fillet weld, angular misalignment Figure 4b: Cruciform or T-joint butt or fillet weld with angular misalignment (per BS 7910:2019, Annex D, Table D.2 case b).\n$$\\frac{\\sigma_s}{P_m}=\\frac{\\kappa\\,\\alpha\\,l_1 l_2}{B(l_1+l_2)}$$Algorithm: ① measure the angle $\\alpha$ (radians), $B,l_1,l_2$; ② choose $\\kappa$ by restraint (representative values $6.0 / 3.0 / 0.04 / 0.02$ — the restraint varies enormously, so take the value by the actual support, or the result may be off by two orders of magnitude); ③ get $\\sigma_s$, $k_m$.\n(BS 7910:2019, Annex D, Table D.2 case b)\nc) Fillet weld, axial misalignment (failure through the weld throat/root) Figure 4c: Fillet weld with axial misalignment, the failure plane being the weld-throat inclined plane; σs/σw = e/(B+h), h the weld leg length (per BS 7910:2019, Annex D, Table D.2 case c).\n$$\\frac{\\sigma_s}{\\sigma_w}=\\frac{e}{B+h}$$Algorithm: ① note that this is referenced to the weld-throat stress $\\sigma_w$ (not $P_m$); ② $h$ is the weld leg length; ③ concerns fatigue failure of the root through the weld throat.\n⚠️ Two red lines\nHere the denominator is $B+h$ and the reference stress is $\\sigma_w$ — do not mix it with the earlier formulas using $P_m$. This case cannot be used to compute the stress intensity factor of a weld-root flaw (BS 7910:2019, Annex D, Table D.2 case c). 5. After computing σs, which step of the assessment does it go into? Annex D is not isolated — it supplies data to the main assessment (BS 7910:2019, 6.4.4).\n5.1 Fracture assessment (FAD) The misalignment bending stress $\\sigma_s$, as a bending component, enters:\nthe stress intensity factor $K_I$ calculation (Annex M); the reference stress $\\sigma_{ref}$ calculation (Annex P). Together they set the assessment point\u0026rsquo;s vertical coordinate $K_r$ and horizontal coordinate $L_r$ on the FAD; the primary/secondary choice is in §2.4.\n5.2 Fatigue assessment (Clause 8) use $k_m$ to magnify the membrane stress range, adding the bending stress range $\\Delta\\sigma_s=(k_m-1)\\Delta\\sigma_m$ into the bending stress range (BS 7910:2019, 8.4); when misalignment is assessed on its own, judge with the $k_m$ acceptance limits of §8.8.1 and Table 8.10 (BS 7910:2019, 8.8.1, Table 8.10). 5.3 Value for an embedded crack The misalignment bending stress varies linearly through the thickness and is zero at the neutral axis — when assessing an embedded flaw, take the value at the crack depth from this distribution (BS 7910:2019, Annex D, D.1).\n6. Engineering pitfalls (recap) Directionality: $\\sigma_s$ arises only under the membrane stress perpendicular to the misalignment line. Do not add this layer for a longitudinal continuous weld loaded lengthwise, or a pure-bending joint (§1.3 point 1). Compute restraint as unrestrained: unless restraint can be shown to reduce the effect, take $\\kappa=6$ for plates (§1.3 point 3). Add signs algebraically: when axial + angular coexist they may add with the same sign (more dangerous) or cancel with opposite sign; add algebraically per Eq. D.3 (§2.2). The tanh correction is counter-intuitive: it can be ignored under tension (conservative); under compression it becomes tan and cannot be ignored (cases e, f). Cruciform $\\kappa$ is not a fixed value: choose by the actual restraint, do not use the plate\u0026rsquo;s 6 (Table D.2 a, b). Fillet-weld case c uses $\\sigma_w$ and cannot be used for the root SIF: do not mix it with $P_m$ or the weld-root stress intensity factor. Through-wall cracks may be over-conservative (§2.5). Appendix: symbol quick reference Symbol Meaning Unit $B$ / $B_1,B_2$ section thickness / thickness of the two butt plates mm $D$ / $D_{max},D_{min}$ mean diameter / maximum and minimum diameter mm $d$ deviation from round caused by angular distortion mm $E$ elastic modulus N/mm² $e$ axial misalignment (eccentricity / centre-line offset) mm $h$ weld leg length mm $k_m$ misalignment stress magnification factor — $l$ / $l_1,l_2$ distance from the misaligned joint to the load point or to the ends of the angular-distortion region (shortest is $l_1$) / plate length mm $n$ factor for the bending-stress calculation of an unequal-thickness plate with axial misalignment — $P_m$ / $p_m$ primary membrane stress / maximum pressure for the ovality calculation N/mm² $y$ peaking caused by angular distortion mm $\\alpha$ angular change at the misalignment radians $\\beta$ factor for the angular-distortion bending-stress calculation — $\\Delta P_m$ / $\\Delta\\sigma_s$ primary membrane stress range / misalignment bending stress range N/mm² $\\theta$ angle between the weld and the point of maximum-diameter observation ° $\\kappa$ restraint parameter of the butt / cruciform joint — $\\nu$ Poisson\u0026rsquo;s ratio — $\\sigma_{max,m}$ membrane component of the maximum applied tensile stress N/mm² $\\sigma_s$ maximum induced bending stress due to misalignment (same sign as $P_m$) N/mm² $\\sigma_w$ applied stress on the weld throat N/mm² (BS 7910:2019, Annex D, D.0)\nMisalignment stress does not give a verdict by itself; it ultimately feeds into the fracture assessment. Take the computed $\\sigma_s$ as the bending component, or multiply $k_m$ into the membrane stress, then run the Clause 7 Failure Assessment Diagram to reach the judgment of \u0026ldquo;whether the cracked structure may remain in service\u0026rdquo;.\n🧮 在线计算器：BS 7910 Annex D Misalignment Stress Magnification k_m Calculator — Choose the misalignment geometry family, enter the geometry, and compute σs/Pm and the magnification factor km to feed into the Annex M stress intensity factor and the Clause 7 fracture assessment.\nFigure note: Figures 1–4(c) are original MechCalc illustrations drawn from the BS 7910:2019 Annex D concepts and formulas, not copies of the standard\u0026rsquo;s figures.\n📖 Reference:\nBS 7910:2019+A1:2020, Guide to methods for assessing the acceptability of flaws in metallic structures, Annex D — Stress due to misalignment ","permalink":"https://mechcalc.net/blog/en/posts/bs7910-annex-d-misalignment/","summary":"\u003cp\u003eIn a fitness-for-service (FFS) assessment of a welded pressure-bearing structure, there is a class of stress that comes neither from external load nor from a residual field, but from \u003cstrong\u003eimperfect fabrication and fit-up\u003c/strong\u003e — the weld joint is \u0026ldquo;not aligned\u0026rdquo;. BS 7910:2019 Annex D handles exactly this: when two plates or shells to be welded together have axial misalignment or angular distortion, the load path of a tensile load is forced to bend, and a layer of local bending stress $\\sigma_s$ appears at the weld.\u003c/p\u003e","title":"BS 7910 Annex D: How a Misaligned Weld Forces a Layer of Bending Stress"},{"content":"This is the second worked example in the [[FITNET|FITNET]] FAD example collection (§13.2.6, SSTP10). Its focus differs from the [[bs7910-a533b-residual-stress-fad|first A533B example]]: this time it is a welded stainless-steel wide plate with a through-thickness crack, assessed for ductile tearing (the crack grows stably as the load rises) on the FAD. We follow the [[bs7910-a533b-residual-kis-annexm|usual routine]] — run it in mechCalc\u0026rsquo;s BS 7910 Clause 7 fracture assessment calculator, read the chart, and cross-check point by point against the FITNET literature.\nSource: [[FITNET|FITNET]] FFS Procedure MK7 (2006), Vol. II §13.2.6 (SSTP10 stainless-steel wide plate, through-thickness crack, ductile tearing). The test is a welded wide plate with a through crack under monotonic tension, recording the full ductile-tearing history from initiation (4 MN) to instability (10.83 MN).\n🧮 在线计算器：BS 7910 Clause 7 Fracture Assessment Calculator — The engine used here: it combines Annex M (K_I) \u0026#43; Annex P (σ_ref) \u0026#43; the Option 1 FAD, and can be re-run online.\n1. What this problem assesses A stainless-steel wide plate, 820 mm wide and 61.2 mm thick, has a through-thickness crack at the weld centre (273 mm total length after fatigue pre-cracking). Under monotonic tension the crack does not snap brittlely; it undergoes ductile tearing: as the load rises, the crack grows a little (Δa) at a time, until instability. FITNET records three key points — initiation (4 MN), mid-tearing (10.35 MN), and instability (10.83 MN).\nThe object being assessed is whether the plate containing this through crack can keep carrying load. The FAD assessment plots each load point as a coordinate $(L_r,\\ K_r)$ and checks whether it lies inside or outside the failure assessment line (FAL).\nFigure 1: Geometry and loading of the SSTP10 stainless-steel wide plate. A plate 820 mm wide and 61.2 mm thick, with a transverse through-thickness crack at the weld centre (full length 2a=273 mm, red); monotonic tension is applied at the top and bottom ends, giving a membrane stress σ_m=P/(W·B). The side view on the right shows the crack penetrating the full plate thickness (through-wall) — hence there is no deepest point, and assessment is at a single front.\n2. Principle and inputs Assessment-point coordinates (Option 1):\n$$ K_r = \\frac{K_I^P + K_I^S}{K_{mat}} + \\rho, \\qquad L_r = \\frac{\\sigma_{ref}}{\\sigma_Y} $$ Primary stress: monotonic membrane load $\\sigma_m = P/(W\\cdot B)$ ($W$ width, $B$ thickness). $K_I^P$: the stress intensity factor of the through crack (BS 7910 Annex M.3, with finite-width correction), computed by mechCalc. $\\sigma_{ref}$ / $L_r$: Annex P net-section plastic-collapse reference stress. Toughness $K_{mat}$: for ductile tearing the toughness rises with the crack extension $\\Delta a$, given by the J-R resistance curve (Table 13.11): $$ J\\,[\\mathrm{kJ/m^2}] = 317.31\\,\\Delta a^{0.6475}\\ (\\Delta a\\ \\text{in mm}), \\qquad K_{mat} = \\sqrt{\\dfrac{J\\,E'}{1000}},\\ \\ E' = \\dfrac{E}{1-\\nu^2} $$ Parameter Value Note Crack type through-thickness weld centre Crack half-length $a$ 136.5 mm full length $2a = 273$ mm Plate thickness $B$ / width $W$ 61.2 / 820 mm Base material $\\sigma_Y$ / $\\sigma_U$ 234 / 594 MPa stainless-steel base-metal properties Young\u0026rsquo;s modulus $E$ / $\\nu$ 173 GPa / 0.3 $E' = 190{,}110$ MPa Primary membrane stress $\\sigma_m$ 79.7 / 206.2 / 215.8 MPa for 4 / 10.35 / 10.83 MN $K_{mat}$ (from J-R) 145.9 / 325.2 / 532.9 MPa·m$^{0.5}$ for $\\Delta a$ = 0.2 / 2.38 / 10.94 mm Why base-material properties: when FITNET assesses with base-material properties all points lie above the FAL (the conservative, safe side); this is the representative basis for this test, and the one compared against here.\n3. How to enter it in the calculator (using the 4 MN initiation point) Assessment Option (FAD): choose Option 1. 1. Crack / Flaw Type: set geometry to \u0026ldquo;Flat Plate\u0026rdquo; and flaw type to \u0026ldquo;through-thickness\u0026rdquo;; enter half-length $a=136.5$. 2. Component Geometry: thickness 61.2, width 820. 3. Stress: primary membrane $P_m=79.71$, bending $P_b=0$; secondary stress source \u0026ldquo;none\u0026rdquo; (see §5). 5. Material: yield 234, tensile 594, modulus 173. 6. Fracture Toughness: source \u0026ldquo;Direct $K_{mat}$\u0026rdquo;, enter 145.9 (= the J-R curve value at $\\Delta a=0.2$). Click Run Calculation. Figure 2: SSTP10 geometry in the BS 7910 Clause 7 calculator. A through-thickness crack of full length 2a=273 mm (red) on an 820 mm wide plate; crack front A spans the entire thickness — a through crack has no \u0026lsquo;deepest point\u0026rsquo; and is assessed at a single front.\n4. Results Figure 3: FAD result for the 4 MN initiation point. The assessment point (L_r, K_r)=(0.5107, 0.3844) lies inside the failure assessment line (FAL), in the green acceptable region — verdict ACCEPTABLE. Toughness reserve F_Kr=2.429, load reserve F_load=1.768; K_r utilisation 41.2%, L_r utilisation 28.9%.\nThe initiation point (4 MN) assembled by the engine per Clause 7:\nQuantity Value Source Primary stress intensity factor $K_I^P$ 56.08 MPa·m$^{0.5}$ Annex M.3 through-crack solution Reference stress $\\sigma_{ref}$ 119.5 MPa Annex P limit load Horizontal $L_r$ 0.5107 $\\sigma_{ref}/\\sigma_Y$ Vertical $K_r$ 0.3844 $(K_I^P+0)/145.9 + 0$ Verdict acceptable (large margin) $K_r \\le f(L_r)$ and $L_r \u003c L_{r,max}$ 5. Cross-check against FITNET: L_r matches digit-for-digit, K_r is cross-method Putting the three load points (base-material properties) side by side:\nLoad (MN) $\\Delta a$ (mm) $K_{mat}$ (J-R) mechCalc $L_r$ FITNET $L_r$ mechCalc $K_r$ FITNET $K_r$ 4.0 (initiation) 0.2 145.9 0.511 0.51 0.384 1.15 10.35 2.38 325.2 1.321 1.33 0.446 0.77 10.83 (instability) 10.94 532.9 1.383 1.44 0.285 0.49 The horizontal coordinate $L_r$ matches FITNET almost digit-for-digit (0.511 vs 0.51, 1.321 vs 1.33, 1.383 vs 1.44, only −4% at the instability point). $L_r$ is determined only by the primary load and geometry through the Annex P limit load, so this cleanly cross-validates mechCalc\u0026rsquo;s reference-stress / limit-load implementation for the through crack, and also reproduces $K_I^P$ (56.08 MPa·m$^{0.5}$).\nThe vertical coordinate $K_r$ is lower than FITNET, which is an expected \u0026ldquo;cross-method\u0026rdquo; difference, not an error. Two reasons:\nThe welding residual stress is not quantified. This plate carries near-yield weld-repair residual stress (tensile at the plate centre), which raises $K_r$ through $K_I^S$ — it is the main driver pushing the initiation point up to $K_r=1.15$. But the literature gives no value or profile for this residual stress (unlike the A533B example, which explicitly gave $K_I^S=46/5$), so mechCalc cannot feed it in; this post computes only the primary (membrane) side. A rough estimate is that $K_I^S \\gtrsim 110$ MPa·m$^{0.5}$ would be needed to reach 1.15. Ductile-tearing $K_r$ uses the J route. Along the tearing locus FITNET\u0026rsquo;s $K_r$ couples the J-R curve with residual stress, which differs from the simplified \u0026ldquo;primary-side $K_I^P$ ÷ J-R toughness\u0026rdquo; basis used here. Methodological note (DIFF is not FAIL): per the [[bs7910-a533b-residual-kis-annexm|verification discipline]], a difference between the same physical quantity computed by different methods is judged DIFF. Here $L_r$ is same-method (a digit-for-digit match, a positive validation); $K_r$ is cross-method and missing the residual-stress input, so only $L_r$ is compared. This agrees with the judgement on SSTP10 in the FITNET example collection.\n6. An honest boundary The [[bs7910-a533b-residual-kis-annexm|A533B example]] could match the residual $K_I^S$ point by point because the literature gave measured residual values; for SSTP10 the literature does not quantify the residual stress, so mechCalc can only validate down to the $L_r$ level — validate firmly what can be validated, and state plainly what cannot be reproduced. This is exactly the right attitude when cross-checking a calculator against the literature: the $L_r$ line (reference stress / limit load) stands up to digit-for-digit checking, and the gap in $K_r$ is localised to the missing residual-stress input, not to the engine itself.\n🧮 在线计算器：BS 7910 Clause 7 Fracture Assessment Calculator — Re-run the initiation point yourself: Option 1, through crack a=136.5, B=61.2, W=820, P_m=79.71, base material σ_Y=234/σ_U=594/E=173, K_mat=145.9, to obtain L_r=0.511, K_r=0.384, ACCEPTABLE.\n","permalink":"https://mechcalc.net/blog/en/posts/bs7910-sstp10-fad-walkthrough/","summary":"\u003cp\u003eThis is the second worked example in the [[FITNET|FITNET]] FAD example collection (§13.2.6, SSTP10). Its focus differs from the [[bs7910-a533b-residual-stress-fad|first A533B example]]: this time it is a welded stainless-steel wide plate with a \u003cstrong\u003ethrough-thickness crack\u003c/strong\u003e, assessed for \u003cstrong\u003eductile tearing\u003c/strong\u003e (the crack grows stably as the load rises) on the FAD. We follow the [[bs7910-a533b-residual-kis-annexm|usual routine]] — run it in mechCalc\u0026rsquo;s BS 7910 Clause 7 fracture assessment calculator, read the chart, and cross-check point by point against the FITNET literature.\u003c/p\u003e","title":"Re-running FITNET SSTP10 with MechCalc: FAD Assessment of a Through-Thickness Crack and an L_r Cross-Check"},{"content":"In the FAD assessment of the [[bs7910-a533b-residual-stress-fad|four A533B welded-plate problems]], the residual stress intensity factor for the as-welded case, $K_I^S \\approx 46\\ \\mathrm{MPa\\cdot m^{0.5}}$, has always been entered directly: FITNET obtained it by integrating the measured residual stress profile, and we simply fed that ready-made number into the vertical coordinate $K_r$.\nA natural follow-up question: how does that 46 actually emerge from a residual stress curve, and can mechCalc compute it on its own?\nIt can. This is exactly what the BS 7910 Annex M.4.2 solution (finite-plate surface flaw, polynomial stress) does: given a stress polynomial $\\sigma(x)$ distributed through the wall thickness, it integrates that stress along the crack front with a weight function to obtain the stress intensity factor of the crack. In this post we run that single step with the Annex M calculator in mechCalc, integrating the as-welded residual profile into the deepest-point SIF, and then compare it with the 46 from FITNET. This is a pure single-point stress intensity factor calculation, independent of any FAD assessment.\nPrinciple: polynomial stress profile → SIF The flaw being assessed is a semi-elliptical surface crack in the weld region (depth $a$, surface half-length $c$). The secondary stress (welding residual) is not constant through the wall thickness but follows a curve, which BS 7910 describes with a polynomial of degree up to five (the coordinate $x$ is measured from the cracked surface and normalised by the wall thickness $B$):\n$$ \\sigma(x) = \\sum_{n=0}^{5} \\sigma_n \\left(\\frac{x}{B}\\right)^n $$The deepest-point $K_I$ is assembled per Eq. M.13 of Annex M.4.2.2: each stress order is paired with a Fett geometry function $f_i^d$ (read from Table M.1, with bilinear interpolation over $a/B$ and $a/2c$), and the sum is multiplied by $\\sqrt{\\pi a}$:\n$$ K_I = \\sqrt{\\pi a}\\,\\sum_{i=0}^{5} \\sigma_i \\left(\\frac{a}{B}\\right)^{i} f_i^{\\,d}\\!\\left(\\frac{a}{B},\\ \\frac{a}{2c}\\right) $$ Source: BS 7910:2019, Annex M.4.2.2, Eq. M.13, Table M.1 (deepest point) / Table M.2 (surface point); geometry functions taken from Fett \u0026amp; Munz (1997) [M.10] and Fett, Munz \u0026amp; Neumann (1990) [M.11].\nThe key point: this solution\u0026rsquo;s polynomial convention is precisely $\\sigma_n(x/B)^n$ (normalised by wall thickness, $x$ measured from the cracked surface). The A533B measured residual profile is already given in terms of $x/t$ (with $t=B$), so the coefficients map term by term with no change of variable needed.\nInputs: the A533B as-welded residual profile The as-welded (LLAW / HLAW) measured residual stress profile given in §2.2 of the [[bs7910-a533b-residual-stress-fad|overview post]] (transverse to the weld, for the $x/t \\le 0.5$ range, which covers $a/t=0.273$):\n$$ \\sigma^*(x/t) = -108.35 + 3543.6\\,(x/t) - 9871.5\\,(x/t)^2 + 2930.3\\,(x/t)^3 \\quad [\\mathrm{MPa}] $$Mapping each term onto the Annex M.4.2 polynomial coefficients, and taking the crack and plate geometry of the LLAW as-welded specimen:\nParameter Value Note Crack depth $a$ 19.4 mm LLAW as-welded pre-crack depth Crack surface half-length $c$ 87.5 mm Full length $2c = 175$ mm Plate thickness $B = t$ 71 mm $a/B = 0.273$, within the Fett table grid Section width $W$ 600 mm Finite-width correction $\\sigma_0$ −108.35 MPa Surface ($x=0$) residual stress, compressive $\\sigma_1$ 3543.6 MPa First-order term $\\sigma_2$ −9871.5 MPa Second-order term $\\sigma_3$ 2930.3 MPa Third-order term $\\sigma_4,\\ \\sigma_5$ 0 The profile is a cubic polynomial Entering it in the calculator Open the BS 7910 Annex M stress intensity factor calculator in mechCalc and set it up in three steps:\nGeometry Group: choose Flat Plate. Flaw Type: choose M.4.2 · Finite Surface Polynomial [Fett] — this is the finite surface flaw solution under a polynomial stress profile. Enter the geometry and polynomial coefficients: crack depth 19.4, surface half-length 87.5; plate thickness 71, section width 600; polynomial $\\sigma_0..\\sigma_3$ as in the table above ($\\sigma_4=\\sigma_5=0$). Figure 1: Selecting M.4.2 (finite-plate surface flaw, polynomial stress) in the Annex M calculator and entering the A533B as-welded residual stress polynomial. The \u0026lsquo;Flaw Cross-section Proportional Geometry\u0026rsquo; panel on the right draws the crack to scale: a semi-elliptical surface crack (red) of depth a=19.4 mm and full length 2c=175 mm on a W=600 mm wide plate, with the deepest point D and surface point S marked.\nThe calculator also draws this residual stress distribution through the wall thickness in real time and to scale — this is exactly the quantity being integrated:\nFigure 2: Residual stress distribution through the wall thickness (orange line, polynomial σ₀~σ₅). The surface (x=0) is in compression at −108 MPa, turns rapidly to tension with depth, and peaks in the mid-thickness region; the dashed red vertical line marks the crack-front position at a=19.4 mm. The dashed blue line is the linearised equivalent (membrane + bending). The deepest-point K_I is the result of integrating this curve over [0, a] along the crack front with the weight function.\nResults Click Run Calculation to obtain the SIF at the two crack-front points:\nFigure 3: Annex M.4.2 results. Deepest point D: K_I = 44.843 MPa·m^0.5 (geometry function f₀ᵈ=1.1429); surface point S: K_I ≈ −0.028 ≈ 0. Below are the intermediate quantities a/B=0.2732, a/c=0.2217, 2c/a=9.0206, with the source cited as BS 7910 Annex M.4.2.2 Table M.1/M.2.\nCrack-front point $K_I$ [MPa·m$^{0.5}$] Note Deepest point D 44.843 The residual $K_I^S$ that enters the FAD Surface point S −0.028 (≈ 0) Near-surface compression, net contribution almost zero Why is the surface point almost zero? The surface-point geometry function weights the near-surface region most heavily, and there the residual stress is compressive (−108 MPa at $x=0$); compressive stress contributes negatively to the opening-mode SIF and just cancels the small amount of tension in the shallow layer, leaving net $K_I \\approx 0$. The deepest point, by contrast, sees the large band of tension in the mid-thickness region, and so integrates out to 44.8 MPa·m$^{0.5}$ — this is the residual $K_I^S$ the assessment needs.\nWhitebox: term-by-term expansion of Eq. M.13 The Whitebox steps in the calculator lay out the deepest-point summation in full ($a=0.0194$ m):\n$$ K_I = \\sqrt{\\pi a}\\,\\big[\\underbrace{-108.35{\\times}1.0000{\\times}1.1429}_{\\sigma_0\\text{ term}} + \\underbrace{3543.6{\\times}0.2732{\\times}0.6585}_{\\sigma_1\\text{ term}} + \\underbrace{-9871.5{\\times}0.0747{\\times}0.4820}_{\\sigma_2\\text{ term}} + \\underbrace{2930.3{\\times}0.0204{\\times}0.3873}_{\\sigma_3\\text{ term}}\\big] = 44.84 $$Here the Fett geometry functions (Table M.1, deepest point), obtained by bilinear interpolation at $a/B=0.273$ and $2c/a=9.02$, are $f_0^d{=}1.1429,\\ f_1^d{=}0.6585,\\ f_2^d{=}0.4820,\\ f_3^d{=}0.3873$. Every step traces back to a table in the standard — no black box.\nCross-check against FITNET: is mechCalc accurate? Placing the residual $K_I^S$ that mechCalc computed independently alongside the value reported by FITNET:\nQuantity mechCalc (Annex M.4.2) FITNET Difference Deepest-point residual $K_I^S$ [MPa·m$^{0.5}$] 44.84 ≈ 46 −2.5% A 2.5% difference — for a cross-method check, that is very good agreement. Both sides do the same thing, \u0026ldquo;integrate the same measured residual profile along the crack front with a weight function\u0026rdquo;, but they use different weight function solutions: mechCalc uses the Fett tabulated geometry functions of BS 7910 Annex M.4.2, while FITNET uses the weight function solution built into its own procedure. For the same physical quantity computed by different methods, a difference of a few percent is normal, and here mechCalc comes out slightly lower (closer to the data, not overshooting).\nWhat matters here is this: that \u0026ldquo;given out of nowhere\u0026rdquo; $K_I^S=46$ in the four-problem FAD assessment can be reproduced independently by mechCalc, from the raw residual stress profile, to within −2.5%, using its own Annex M.4.2 solution. In other words, residual stress — the most error-prone intermediate step — does not require mechCalc to take a fed value from the literature; it closes the loop on its own, and the result is confirmed by a third-party source.\nMethodological note (DIFF is not FAIL): this comparison is the \u0026ldquo;self-generated\u0026rdquo; path check for $K_I^S$ in the [[bs7910-a533b-residual-stress-fad|four residual stress problems]] — a −2.5% difference between two weight function solutions for the same physical quantity is a normal methodological difference, already quantified and explained, and mechCalc holds up. For methodological cleanliness, the four FAD problems still consistently use the direct $K_I^S$ value from the same FITNET source.\nSource: the [[FITNET|FITNET]] project\u0026rsquo;s Case Studies for Fracture, case \u0026ldquo;A533B-1 Steel Residual Stress Experiments\u0026rdquo;; original test report C C France, J K Sharples \u0026amp; C Wignall, AEA Technology Report AEAT-4236 (SINTAP/TASK4/AEAT18, 1998). The SIF solution follows BS 7910:2019+A1:2020, Annex M.4.2.2 (Eq. M.13, Table M.1/M.2).\n🧮 在线计算器：BS 7910 Annex M Stress Intensity Factor Calculator — Run it yourself: set Geometry Group to Flat Plate, Flaw Type to M.4.2 (Finite Surface Polynomial), enter a=19.4, c=87.5, B=71, W=600, and the polynomial σ₀=−108.35, σ₁=3543.6, σ₂=−9871.5, σ₃=2930.3 to obtain the deepest-point K_I=44.84 MPa·m^0.5.\n","permalink":"https://mechcalc.net/blog/en/posts/bs7910-a533b-residual-kis-annexm/","summary":"\u003cp\u003eIn the FAD assessment of the [[bs7910-a533b-residual-stress-fad|four A533B welded-plate problems]], the residual stress intensity factor for the as-welded case, $K_I^S \\approx 46\\ \\mathrm{MPa\\cdot m^{0.5}}$, has always been \u003cstrong\u003eentered directly\u003c/strong\u003e: FITNET obtained it by integrating the measured residual stress profile, and we simply fed that ready-made number into the vertical coordinate $K_r$.\u003c/p\u003e\n\u003cblockquote\u003e\n\u003cp\u003eA natural follow-up question: \u003cstrong\u003ehow does that 46 actually emerge from a residual stress curve, and can mechCalc compute it on its own?\u003c/strong\u003e\u003c/p\u003e","title":"Where Does the Welding Residual Stress Intensity Factor Come From? Integrating an A533B Residual Profile into a SIF with BS 7910 Annex M.4.2"},{"content":"This is the finale of the four problems on the A533B-1 welded plate. It is the counterpart to Problem 3, [[bs7910-a533b-hlaw-fad-walkthrough|HLAW]], and it pushes the power of PWHT in this test series into its most visible form.\nThe shared background and method for all four problems are covered in the overview, [[bs7910-a533b-residual-stress-fad|Where does residual stress push the assessment point?]].\nWhat this problem asks HLHT = High-$L_r$ + Heat-Treated: PWHT applied, assessment temperature −30 ℃, high load ratio. It shares the temperature and regime of HLAW, and the difference is still that one thing — post-weld heat treatment. But at −30 ℃, PWHT delivers a double dividend:\nRelaxed residuals: the residual $K_I^S$ drops from 46 in the as-welded state to 5 MPa·m$^{0.5}$; Restored toughness: −30 ℃ is already close to the ductile-brittle transition region, and PWHT lifts the weld toughness off the lower shelf, raising it by an order of magnitude — $K_{mat}$ jumps from 62 in HLAW to 321 MPa·m$^{0.5}$. These two effects together pull the vertical coordinate $K_r$ down sharply. But does that overturn the verdict? That is exactly what makes this finale worth watching.\nPrinciple: the vertical coordinate collapses, yet the horizontal one may still cross the line The assessment is still BS 7910:2019 §7.3.3 Option 1. Both terms in the numerator of $K_r$ shrink (smaller residual, larger $K_{mat}$), so the vertical coordinate must fall a great deal. But remember the FAD is a two-dimensional criterion: besides keeping the vertical $K_r$ below the failure assessment line, the horizontal axis must also satisfy $L_r \u003c L_{r,max}$ (the plastic cut-off, §7.3.2):\n$$ L_{r,max} = \\frac{\\sigma_Y + \\sigma_U}{2\\sigma_Y} \\approx 1.15 \\quad (\\sigma_Y=520,\\ \\sigma_U=677) $$No amount of toughness can save a section that has already reached net-section yielding ($L_r \\ge L_{r,max}$).\nInputs at a glance Parameter Value Difference from HLAW Crack depth $a_0$ / half-length $c_0$ 19.0 / 87 mm same Plate thickness $B=t$ / width $W$ 70 / 600 mm same Primary membrane $P_m$ / bending $P_b$ 0 / 960 MPa slightly lower (fracture load 4.83 MN) Yield $\\sigma_Y$ / tensile $\\sigma_U$ 520 / 677 MPa same Fracture toughness $K_{mat}$ 321 MPa·m$^{0.5}$ 62 → 321 (order-of-magnitude rise) Residual $K_I^S$ (direct input) 5 MPa·m$^{0.5}$ 46 → 5 (relaxed) Entering it in the calculator The steps are the same as the first three problems (pick Option 1 at the top, then fill the cards). Relative to HLAW, three things change here: primary bending $P_b=960$, fracture toughness $K_{mat}=321$, and the secondary $K_I^S$ entered directly as 5. The $\\rho$ term is still computed automatically by Annex R.\nResults Figure 1: HLHT results. K_r is only 0.60 (the high toughness K_mat=321 presses the vertical coordinate far down), but L_r=1.71 still crosses the plastic cut-off (about 1.15), so the assessment point lands to the right of the cut-off line — the NOT ACCEPTABLE verdict is driven by plastic collapse, not fracture.\nQuantity HLHT (HLAW for comparison) Primary $K_I^P$ 187.5 MPa·m$^{0.5}$ 198.2 Secondary $K_I^S$ 5.00 MPa·m$^{0.5}$ 46.00 Fracture toughness $K_{mat}$ 321 MPa·m$^{0.5}$ 62 Reference stress $\\sigma_{ref}$ 886.7 MPa 937.5 Horizontal $L_r$ 1.71 1.80 Vertical $K_r$ 0.60 3.94 Verdict not acceptable (plastic collapse) not acceptable (plastic collapse) Reading the result First, the power of PWHT: $K_r$ plunges from 3.94 in HLAW to 0.60. Almost all of that drop comes from $K_{mat}$ rising from 62 to 321 — the denominator of the vertical coordinate is five times larger — while the residual term shrinks from 46 to 5. Look at the vertical axis alone and HLHT seems perfectly safe.\nBut the FAD is a two-dimensional criterion. The horizontal $L_r = 1.71$ still crosses the plastic cut-off $L_{r,max}\\approx 1.15$, and the assessment point lands to the right of the vertical cut-off line — the flawed structure is judged not acceptable because of plastic collapse, consistent with the test fracture. No amount of toughness can stop a net section that has already yielded fully from failing. In this problem the governing mode of the verdict has switched completely from fracture to plastic collapse.\nFinally, that thought-provoking twist: HLHT has smaller residuals and five times the toughness, yet its failure load (4.83 MN) is slightly lower than HLAW\u0026rsquo;s (5.10 MN). Far from being a contradiction, this confirms the conclusion of Problem 3 — in the high-$L_r$ regime the relative influence of residual stress has already been \u0026ldquo;diluted\u0026rdquo; by plasticity, and the governing factors return to toughness and net-section yield strength. The two specimens have the same yield/tensile strengths, so their net-section load capacities are already close; further increasing toughness is no longer the bottleneck here, so the failure loads naturally land in the same range and trade places by a small margin.\nCross-check against FITNET: is mechCalc accurate? The finale is likewise compared term by term against the original FITNET text:\nQuantity mechCalc FITNET original Difference Primary $K_I^P$ [MPa·m$^{0.5}$] 187.5 ≈187 ≈0 Secondary $K_I^S$ [MPa·m$^{0.5}$] 5.0 5 same source Load ratio $L_r$ 1.71 1.63 +4.6% Fracture ratio $K_r$ 0.60 0.57 +5.2% As with Problem 3, the difference in the high-$L_r$ regime is about 5%, arising from the method difference in the reference-stress solution (the Annex P closed-form expression vs 3D finite-element analysis); the direction is consistent and mechCalc is slightly conservative. The verdict (plastic collapse, not acceptable) is identical on both sides.\nSource: the [[FITNET|FITNET]] project\u0026rsquo;s Case Studies for Fracture, case \u0026ldquo;A533B-1 Steel Residual Stress Experiments\u0026rdquo;; original test report C C France, J K Sharples \u0026amp; C Wignall, AEA Technology Report AEAT-4236 (SINTAP/TASK4/AEAT18, 1998).\nTying the four problems together Looking at the four problems together, this set of A533B-1 tests yields a complete picture:\nLow-$L_r$ brittle-fracture regime (LLAW/LLHT): residual stress is the lead actor. As-welded residuals can contribute a $K_I^S$ on the order of a primary stress, pushing the assessment point markedly higher; relaxing the residuals with PWHT pulls it back down substantially. High-$L_r$ plastic regime (HLAW/HLHT): residuals are diluted by plasticity, and the lead role passes to toughness and net-section strength; the verdict is governed by plastic collapse ($L_r \\ge L_{r,max}$). The value of PWHT is twofold: it both relaxes the residuals and (in the transition region) restores toughness — but it cannot save a section that has already reached net-section yielding. And mechCalc\u0026rsquo;s BS 7910 Clause 7 fracture-assessment engine independently reproduces the published assessment results for all four problems ($L_r$ and $K_r$ both agreeing within ±5%), showing that it assembles the residual-stress step — the most error-prone part — correctly, and is fit for engineering assessment.\n🧮 在线计算器：BS 7910 Clause 7 Fracture Assessment Calculator — Rerun this finale: P_b=960, K_mat=321, K_I^S entered directly as 5, giving K_r=0.60 and L_r=1.71 (still beyond the cut-off).\n","permalink":"https://mechcalc.net/blog/en/posts/bs7910-a533b-hlht-fad-walkthrough/","summary":"\u003cp\u003eThis is the finale of the \u003cstrong\u003efour problems\u003c/strong\u003e on the A533B-1 welded plate. It is the counterpart to Problem 3, [[bs7910-a533b-hlaw-fad-walkthrough|HLAW]], and it pushes the power of PWHT in this test series into its most visible form.\u003c/p\u003e\n\u003cblockquote\u003e\n\u003cp\u003eThe shared background and method for all four problems are covered in the overview, [[bs7910-a533b-residual-stress-fad|Where does residual stress push the assessment point?]].\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003ch2 id=\"what-this-problem-asks\"\u003eWhat this problem asks\u003c/h2\u003e\n\u003cp\u003eHLHT = \u003cstrong\u003eH\u003c/strong\u003eigh-$L_r$ + \u003cstrong\u003eH\u003c/strong\u003eeat-\u003cstrong\u003eT\u003c/strong\u003ereated: \u003cstrong\u003ePWHT applied, assessment temperature −30 ℃, high load ratio\u003c/strong\u003e. It shares the temperature and regime of HLAW, and the difference is still that one thing — post-weld heat treatment. But at −30 ℃, PWHT delivers a \u003cstrong\u003edouble dividend\u003c/strong\u003e:\u003c/p\u003e","title":"Problem 4 — HLHT: the double dividend of PWHT, and a thought-provoking twist (the A533B high-load-ratio finale)"},{"content":"This is the third of four problems on the welded A533B-1 plate. The first two both sat in the low load-ratio brittle-fracture regime and compared residual stress; this one shifts the battlefield — to the high load-ratio, large-plasticity regime.\nFor the shared background and method behind all four problems, see the overview post [[bs7910-a533b-residual-stress-fad|Where does residual stress push the assessment point?]].\nWhat this problem asks HLAW = High-$L_r$ + as-Welded: as-welded condition, assessment temperature raised to −30 ℃, load increased into the high load-ratio regime. The warmer temperature lifts the fracture toughness somewhat off the lower shelf ($K_{mat}=62\\ \\mathrm{MPa\\cdot m^{0.5}}$), while the load is raised to a failure load of 5.10 MN. The residual stress stays the as-welded value ($K_I^S=46$).\nThe question becomes: once plasticity has developed extensively in the high-$L_r$ regime, does that residual stress — so dominant in the low-temperature regime — still call the shots?\nPrinciple: once $L_r$ crosses the cut-off, the assessment is governed by plastic collapse The assessment is still BS 7910:2019 §7.3.3 Option 1, but this problem triggers the FAD\u0026rsquo;s plastic cut-off mechanism. BS 7910:2019 §7.3.2 defines a cut-off value to prevent the net section from collapsing plastically before fracture occurs:\n$$ L_{r,max} = \\frac{\\sigma_Y + \\sigma_U}{2\\sigma_Y} $$When the assessment point\u0026rsquo;s abscissa reaches $L_r \\ge L_{r,max}$, the failure assessment line takes $f(L_r)=0$ (§7.3.3 Eq. 7.28) — meaning that however low the ordinate, the flawed structure is not acceptable because of plastic collapse / net-section yielding. Here $\\sigma_Y=520$ and $\\sigma_U=677$, which gives $L_{r,max}\\approx 1.15$.\nIn addition, the plasticity interaction $\\rho$ from BS 7910 Annex R is defined only within $L_r \\le L_{r,max}$; past the cut-off, $\\rho$ is taken as 0.\nInputs at a glance Parameter Value Note Crack depth $a_0$ / half-length $c_0$ 19.0 / 87 mm Full length $2c_0=174$ mm Thickness $B=t$ / width $W$ 70 / 600 mm Primary membrane $P_m$ / bending $P_b$ 0 / 1015 MPa Pure bending (elastic conversion of the 5.10 MN fracture load) Yield $\\sigma_Y$ / tensile $\\sigma_U$ 520 / 677 MPa $L_{r,max}\\approx 1.15$ Fracture toughness $K_{mat}$ 62 MPa·m$^{0.5}$ As-welded weld @ −30 ℃ Residual $K_I^S$ (direct input) 46 MPa·m$^{0.5}$ Same as LLAW (as-welded) Entering it in the calculator The steps match the first two problems (pick Option 1 at the top, then fill in the cards). Here you swap in the HLAW values: thickness 70, primary bending $P_b=1015$, yield 520, tensile 677, $K_{mat}=62$, and the secondary $K_I^S$ entered directly as 46. Leave $\\rho$ on the Annex R auto-calculation (the engine sets it to 0 automatically once the cut-off is crossed).\nResults Figure 1: HLAW results. L_r=1.80 already exceeds the plastic cut-off value (about 1.15), so the assessment point lands to the right of the cut-off line and is judged NOT ACCEPTABLE — the message reads the section is fully plastic before fracture governs. Note ρ=0 and f(L_r)=0.\nQuantity Value Note Primary $K_I^P$ 198.22 MPa·m$^{0.5}$ Markedly larger under the high load Secondary $K_I^S$ 46.00 MPa·m$^{0.5}$ As-welded residual Reference stress $\\sigma_{ref}$ 937.5 MPa Close to yield Plasticity interaction $\\rho$ 0 (past cut-off) §7.3.2 Abscissa $L_r$ 1.80 $\u003e L_{r,max}\\approx 1.15$ Ordinate $K_r$ 3.94 Verdict Not acceptable (governed by plastic collapse) $L_r \\ge L_{r,max}$ Reading the result The abscissa $L_r = 1.80$ is far beyond the plastic cut-off value $L_{r,max}\\approx 1.15$. On the FAD, the assessment point falls to the right of the cut-off vertical line — the side controlled by plastic collapse / net-section yielding — so the flawed structure is judged not acceptable, consistent with the test fracture.\nMore telling still is that the governing factor has changed. The ordinate $K_r$ reaches 3.94, but this time the main thing driving it up is no longer residual stress: in the large-plasticity regime, secondary stresses such as residual stress are \u0026ldquo;diluted\u0026rdquo; by plastic deformation (this is also the physical reason $\\rho$ drops to zero past the cut-off and stops amplifying secondary stress). The real weak point is the low fracture toughness — even at −30 ℃, the as-welded weld only reaches $K_{mat}=62$.\nIn other words: the low-$L_r$ regime (Problems 1 and 2) is a contest over residual stress, whereas the high-$L_r$ regime is a contest over toughness and net-section strength. For one and the same set of specimens, where the operating point sits on the FAD decides who the weak link is — and that is exactly the value of the failure assessment diagram, which unifies brittle fracture and plastic collapse on a single chart.\nCross-check against FITNET: is mechCalc accurate? Setting this high-$L_r$ assessment side by side with the FITNET source, item by item:\nQuantity mechCalc FITNET source Difference Primary $K_I^P$ [MPa·m$^{0.5}$] 198.2 ≈198 ≈0 Secondary $K_I^S$ [MPa·m$^{0.5}$] 46.0 46 Same source Load ratio $L_r$ 1.80 1.72 +4.8% Fracture ratio $K_r$ 3.94 3.79 +3.9% In the high-$L_r$ regime the gap widens to about 4–5% — but this is not error, it is a methodological difference. For the reference stress in the large-plasticity regime, mechCalc uses the closed-form expression from BS 7910 Annex P, whereas the FITNET source switches to 3D finite-element analysis under strength mismatch; two ways of solving the same physical quantity will naturally differ by a few percent, and here they differ in the same direction, with mechCalc on the slightly more conservative side (its assessment point sits further out, giving a safer verdict). Most importantly, the verdicts agree completely: $L_r$ crosses the cut-off, plastic collapse, not acceptable. The difference therefore sits at the level of \u0026ldquo;solution accuracy\u0026rdquo; and does not affect the engineering conclusion — mechCalc\u0026rsquo;s accuracy and conservatism both hold up.\nSource: the [[FITNET|FITNET]] project\u0026rsquo;s Case Studies for Fracture, case \u0026ldquo;A533B-1 Steel Residual Stress Experiments\u0026rdquo;; original test report C C France, J K Sharples \u0026amp; C Wignall, AEA Technology Report AEAT-4236 (SINTAP/TASK4/AEAT18, 1998).\nThe closing problem: same −30 ℃ and high-$L_r$ regime, but with PWHT applied — relaxing the residual stress and raising the toughness by an order of magnitude ($K_{mat}$ jumps from 62 to 321). $K_r$ will collapse, but will the conclusion flip? See [[bs7910-a533b-hlht-fad-walkthrough|Problem 4, HLHT]].\n🧮 在线计算器：BS 7910 Clause 7 Fracture Assessment Calculator — Rerun this problem: P_b=1015, σ_Y=520, σ_U=677, K_mat=62, K_I^S entered directly as 46, and you get L_r=1.80 (past the cut-off), K_r=3.94.\n","permalink":"https://mechcalc.net/blog/en/posts/bs7910-a533b-hlaw-fad-walkthrough/","summary":"\u003cp\u003eThis is the third of \u003cstrong\u003efour problems\u003c/strong\u003e on the welded A533B-1 plate. The first two both sat in the \u003cstrong\u003elow load-ratio\u003c/strong\u003e brittle-fracture regime and compared residual stress; this one shifts the battlefield — to the \u003cstrong\u003ehigh load-ratio, large-plasticity\u003c/strong\u003e regime.\u003c/p\u003e\n\u003cblockquote\u003e\n\u003cp\u003eFor the shared background and method behind all four problems, see the overview post [[bs7910-a533b-residual-stress-fad|Where does residual stress push the assessment point?]].\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003ch2 id=\"what-this-problem-asks\"\u003eWhat this problem asks\u003c/h2\u003e\n\u003cp\u003eHLAW = \u003cstrong\u003eH\u003c/strong\u003eigh-$L_r$ + as-\u003cstrong\u003eW\u003c/strong\u003eelded: \u003cstrong\u003eas-welded condition, assessment temperature raised to −30 ℃, load increased into the high load-ratio regime.\u003c/strong\u003e The warmer temperature lifts the fracture toughness somewhat off the lower shelf ($K_{mat}=62\\ \\mathrm{MPa\\cdot m^{0.5}}$), while the load is raised to a failure load of \u003cstrong\u003e5.10 MN\u003c/strong\u003e. The residual stress stays the as-welded value ($K_I^S=46$).\u003c/p\u003e","title":"Problem 3 — HLAW: into the high load-ratio regime, where plasticity dilutes residual stress (A533B as-welded, −30 ℃)"},{"content":"This is the second of the four problems on the A533B-1 welded plate, and the companion to Problem 1 [[bs7910-a533b-llaw-fad-walkthrough|LLAW]]. The two problems were designed for a single-variable comparison:\nThe shared background and method for all four problems are in the overview [[bs7910-a533b-residual-stress-fad|Where do residual stresses push the assessment point?]].\nWhat this problem asks LLHT = Low-$L_r$ + Heat-Treated: post-weld heat treatment (PWHT) applied, assessment temperature −120 ℃, low load ratio. It is at the same temperature as LLAW, in the same region, and uses the same set of measured residual-stress profiles — the only variable is PWHT. The heat treatment relaxes the welding residual stress by at least an order of magnitude, so the residual $K_I^S$ drops from 46 in the as-welded state to 5 MPa·m$^{0.5}$. The load capacity rises accordingly, from LLAW\u0026rsquo;s 1.27 MN to 2.19 MN (about 1.7×).\nThe question this problem answers is the comparison question: if you relax the residual stress, how far does the assessment point fall back?\nPrinciple: same yardstick, only the residual term changes The assessment still follows BS 7910:2019 §7.3.3 Option 1, with coordinate definitions identical to Problem 1:\n$$ K_r = \\frac{K_I^P + K_I^S}{K_{mat}} + \\rho, \\qquad L_r = \\frac{\\sigma_{ref}}{\\sigma_Y} $$Residual stress is a secondary stress, so it enters $K_r$ only, not $L_r$ (§7.3.6). The only thing this comparison problem actually changes is the $K_I^S$ term in the $K_r$ numerator: from 46 to 5.\nInputs at a glance Parameter Value Difference vs. LLAW Crack depth $a_0$ / half-length $c_0$ 18.6 / 87 mm almost the same Plate thickness $B=t$ / width $W$ 71 / 600 mm the same Primary membrane $P_m$ / bending $P_b$ 0 / 424 MPa higher load (fracture load 2.19 MN) Yield $\\sigma_Y$ / tensile $\\sigma_U$ 596 / 772 MPa similar Fracture toughness $K_{mat}$ 46 MPa·m$^{0.5}$ PWHT weld @ −120 ℃ Residual $K_I^S$ (direct input) 5 MPa·m$^{0.5}$ 46 → 5 (relaxed by an order of magnitude) Entering it in the calculator The input steps are the same as Problem 1 (first pick Option 1 under Assessment Option (FAD) at the top, then fill in geometry, stress, material and toughness from the cards). For this problem you only need to swap three values to the LLHT ones: primary bending $P_b=424$, fracture toughness $K_{mat}=46$, and secondary $K_I^S$ entered directly as 5. The plasticity interaction $\\rho$ still uses the default Annex R auto-calculation.\nResults Figure 1: LLHT results. Assessment point (L_r, K_r)=(0.648, 1.92). Note the lower right: the primary K_I^p=82.97 is actually higher than LLAW\u0026rsquo;s 48.13, but the secondary K_I^s is only 5.00 (LLAW was 46), so the assessment point lands lower.\nQuantity LLHT (LLAW for comparison) Primary $K_I^P$ 82.97 MPa·m$^{0.5}$ 48.13 Secondary $K_I^S$ 5.00 MPa·m$^{0.5}$ 46.00 Reference stress $\\sigma_{ref}$ 386.0 MPa 226.5 Plasticity interaction $\\rho$ 0.010 0.046 Horizontal axis $L_r$ 0.648 0.367 Vertical axis $K_r$ 1.92 2.59 Verdict not acceptable not acceptable Reading the result: the counter-intuitive pair to watch The most thought-provoking thing in this problem is a counter-intuitive result:\nLLHT\u0026rsquo;s primary stress intensity factor $K_I^P = 82.97$ is nearly double LLAW\u0026rsquo;s 48.13 (because LLHT fractured under the higher load of 2.19 MN); yet its assessment point $K_r = 1.92$ is lower than LLAW\u0026rsquo;s 2.59. The whole difference is in the secondary term: the as-welded residual $K_I^S = 46$, but after PWHT only 5 remains. Between that drop and the rise on the primary side, PWHT pulls the vertical coordinate from 2.59 back down to 1.92. This turns \u0026ldquo;residual stress is harmful\u0026rdquo; from a qualitative statement into a quantifiable increment on the FAD vertical axis — in the brittle-fracture-governed low-$L_r$ region, as-welded residual stress can contribute a $K_I^S$ on the same order as the primary stress.\nBoth are judged not acceptable (both did fracture in the test), but PWHT lets LLHT carry 1.7× the load before fracturing — which is the direct engineering basis for insisting on post-weld heat treatment of important welds.\nCross-check against FITNET: is mechCalc accurate? As before, we line up this problem\u0026rsquo;s assessment against the FITNET original term by term to check mechCalc\u0026rsquo;s accuracy:\nQuantity mechCalc FITNET original Difference Primary $K_I^P$ [MPa·m$^{0.5}$] 82.97 ≈83 ≈0 Secondary $K_I^S$ [MPa·m$^{0.5}$] 5.0 5 same source Load ratio $L_r$ 0.648 0.64 +1.2% Fracture ratio $K_r$ 1.92 1.89 +1.7% In the low-$L_r$ region the two implementations again agree closely: $L_r$ differs by 1.2% and $K_r$ by 1.7%, both within ±2%. In this comparison problem mechCalc reproduces the trend (\u0026ldquo;PWHT pulls $K_r$ from 2.59 back to 1.92\u0026rdquo;) and also matches the FITNET original numerically — one more piece of evidence for its accuracy.\nSource: the [[FITNET|FITNET]] project\u0026rsquo;s Case Studies for Fracture, case \u0026ldquo;A533B-1 Steel Residual Stress Experiments\u0026rdquo;; original test report C C France, J K Sharples \u0026amp; C Wignall, AEA Technology Report AEAT-4236 (SINTAP/TASK4/AEAT18, 1998). The residual $K_I^S$ is from the same source on both sides; the primary SIF and limit-load solutions come from different sources (FITNET uses Sharples \u0026amp; Clayton / Sattari-Far, mechCalc uses BS 7910 Annex M / Annex P).\nNext: raise the temperature to −30 ℃ and increase the load into the high-$L_r$ (large-plasticity) region — is residual stress still that important? See [[bs7910-a533b-hlaw-fad-walkthrough|Problem 3 HLAW]].\n🧮 在线计算器：BS 7910 Clause 7 Fracture Assessment Calculator — Re-run this comparison problem: starting from the Problem 1 inputs, change P_b to 424, K_mat to 46, and enter K_I^S directly as 5, to get L_r=0.648, K_r=1.92.\n","permalink":"https://mechcalc.net/blog/en/posts/bs7910-a533b-llht-fad-walkthrough/","summary":"\u003cp\u003eThis is the second of the \u003cstrong\u003efour problems\u003c/strong\u003e on the A533B-1 welded plate, and the companion to Problem 1 [[bs7910-a533b-llaw-fad-walkthrough|LLAW]]. The two problems were designed for a \u003cstrong\u003esingle-variable comparison\u003c/strong\u003e:\u003c/p\u003e\n\u003cblockquote\u003e\n\u003cp\u003eThe shared background and method for all four problems are in the overview [[bs7910-a533b-residual-stress-fad|Where do residual stresses push the assessment point?]].\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003ch2 id=\"what-this-problem-asks\"\u003eWhat this problem asks\u003c/h2\u003e\n\u003cp\u003eLLHT = \u003cstrong\u003eL\u003c/strong\u003eow-$L_r$ + \u003cstrong\u003eH\u003c/strong\u003eeat-\u003cstrong\u003eT\u003c/strong\u003ereated: \u003cstrong\u003epost-weld heat treatment (PWHT) applied, assessment temperature −120 ℃, low load ratio\u003c/strong\u003e. It is at the same temperature as LLAW, in the same region, and uses the same set of measured residual-stress profiles — \u003cstrong\u003ethe only variable is PWHT\u003c/strong\u003e. The heat treatment relaxes the welding residual stress by at least an order of magnitude, so the residual $K_I^S$ drops from 46 in the as-welded state to \u003cstrong\u003e5 MPa·m$^{0.5}$\u003c/strong\u003e. The load capacity rises accordingly, from LLAW\u0026rsquo;s 1.27 MN to \u003cstrong\u003e2.19 MN\u003c/strong\u003e (about 1.7×).\u003c/p\u003e","title":"Problem 2 — LLHT: PWHT Relaxes Residual Stress by an Order of Magnitude, Same Temperature and Region for a Head-to-Head"},{"content":"This is the first of four problems on the A533B-1 welded plate. The shared background, common method, and residual-stress profile for all four are set out in the overview post:\nFor the overview and method, see [[bs7910-a533b-residual-stress-fad|Where does residual stress push the assessment point? — A FAD recomputation of the A533B-1 large welded-plate fracture tests]]. This post focuses on one specimen and walks it through the calculator from input to reading the chart.\nWhat this problem asks LLAW = Low-$L_r$ + as-Welded: as-welded (no post-weld heat treatment), assessment temperature −120 ℃, low load ratio. Of the four specimens, it fractured at the lowest load (1.27 MN), with crack arrest afterwards. The low temperature drops the fracture toughness onto the lower shelf ($K_{mat}=37\\ \\mathrm{MPa\\cdot m^{0.5}}$), while the full set of unrelaxed welding residual stresses still bears directly on the crack.\nThe aim here is to turn a qualitative statement — \u0026ldquo;welding residual stress is very damaging\u0026rdquo; — into a specific vertical-axis increment on the Failure Assessment Diagram (FAD).\nThe principle in brief: residual stress enters only $K_r$, not $L_r$ All four problems use the Option 1 (standard) Failure Assessment Diagram of BS 7910:2019 §7.3.3. The assessment-point coordinates:\n$$ K_r = \\frac{K_I^P + K_I^S}{K_{mat}} + \\rho, \\qquad L_r = \\frac{\\sigma_{ref}}{\\sigma_Y} $$The Failure Assessment Line (FAL):\n$$ f(L_r) = \\begin{cases} \\left(1 + \\tfrac{1}{2}L_r^2\\right)^{-1/2}\\left[0.3 + 0.7\\,e^{-\\mu L_r^6}\\right], \u0026 L_r \\le 1 \\\\ f(L_r{=}1)\\,L_r^{(N-1)/(2N)}, \u0026 1 \u003c L_r \u003c L_{r,\\max} \\\\ 0, \u0026 L_r \\ge L_{r,\\max} \\end{cases} $$$$ \\mu = \\min\\!\\left(0.001\\tfrac{E}{\\sigma_Y},\\ 0.6\\right), \\quad N = 0.3\\left(1 - \\tfrac{\\sigma_Y}{\\sigma_U}\\right), \\quad L_{r,\\max} = \\frac{\\sigma_Y + \\sigma_U}{2\\sigma_Y} $$The three branches correspond to BS 7910:2019 §7.3.3 Eq. 7.26 ($L_r \\le 1$, the brittle-fracture-governed branch), Eq. 7.27 ($1 \u003c L_r \u003c L_{r,\\max}$, the elastic-plastic transition branch, where the curve bends down with the strain-hardening exponent $N$), and Eq. 7.28 ($L_r \\ge L_{r,\\max}$, plastic cut-off, where the curve drops to zero); the cut-off point $L_{r,\\max}$ is given by Eq. 7.25. For this problem the assessment point sits at $L_r=0.367$, within the first branch, so the verdict below uses only Eq. 7.26.\nThe key engineering division of labour (BS 7910:2019, §7.3.6 / §7.3.7): welding residual stress is a secondary stress (a self-balancing field). It raises the vertical axis $K_r$ only and does not enter the horizontal axis $L_r$ — $L_r$ comes solely from the primary stress (here the bending stress of four-point bending), via the Annex P reference-stress solution. So the residual-stress contribution is simply the $K_I^S$ obtained by integrating the measured profile against the crack front with a weight function, entered directly and added into the numerator of $K_r$. For this problem the as-welded residual $K_I^S \\approx 46\\ \\mathrm{MPa\\cdot m^{0.5}}$.\nInputs at a glance Parameter Value Note Flaw type surface semi-elliptical (surface) machined in the weld region Crack depth $a_0$ / half-length $c_0$ 19.4 / 87.5 mm full length $2c_0=175$ mm Plate thickness $B=t$ / width $W$ 71 / 600 mm Primary membrane $P_m$ / bending $P_b$ 0 / 245 MPa pure bending (converted from the 1.27 MN fracture load) Yield $\\sigma_Y$ / tensile $\\sigma_U$ 618 / 791 MPa base-metal properties Elastic modulus $E$ 210 GPa Fracture toughness $K_{mat}$ 37 MPa·m$^{0.5}$ as-welded weld @ −120 ℃, lower shelf Residual $K_I^S$ (direct entry) 46 MPa·m$^{0.5}$ weight-function integration of the measured profile Entering it in the calculator mechCalc\u0026rsquo;s Fracture Assessment Calculator groups the inputs into cards, and the very first card is the assessment option — choose the method first, then fill in the data:\nAssessment Option (FAD): pick Option 1 (Section 7.3.3) (it is the default). 1. Crack / Flaw Type: set geometry to Flat Plate and flaw type to Surface semi-elliptical; enter crack depth 19.4 and half-length 87.5. 2. Component Geometry: wall thickness 71, width 600. 3. Stress \u0026amp; Polynomial Coefficients: primary membrane stress $P_m=0$, primary bending $P_b=245$; for the secondary-load source choose Enter $K_I^S$ directly and enter $K_I^S=46$. 5. Material: yield 618, tensile 791, modulus 210. 6. Fracture Toughness: source Direct $K_{mat}$, enter 37. 7. Analysis Type \u0026amp; Plasticity Interaction: leave the plasticity-interaction $\\rho$ source at the default Annex R (auto) — when a secondary stress is present, $\\rho$ must be computed and cannot be set to zero, otherwise $K_r$ would be non-conservative. Figure 1: LLAW inputs in the BS 7910 Clause 7 Fracture Assessment Calculator. The top “Assessment Option (FAD)” already selects Option 1; the primary stress is bending only at 245 MPa, and the welding residual stress is entered directly through the secondary channel as K_I^S=46.\nOnce everything is filled in, click Run Calculation.\nResults Figure 2: LLAW results. The assessment point (L_r, K_r)=(0.367, 2.59) lies far beyond the Failure Assessment Line, giving a verdict of NOT ACCEPTABLE. The three panels at lower right give the full set of intermediate quantities: K_I^p=48.13, K_I^s=46, σ_ref=226.5, ρ=0.046, f(L_r)=0.373.\nThe assessment point and intermediate quantities the engine assembles via the Clause 7 procedure:\nQuantity Value Source Primary stress intensity factor $K_I^P$ 48.13 MPa·m$^{0.5}$ Annex M semi-elliptical surface-flaw solution Secondary stress intensity factor $K_I^S$ 46.00 MPa·m$^{0.5}$ direct entry (integration of the measured profile) Reference stress $\\sigma_{ref}$ 226.5 MPa Annex P limit load Plasticity interaction $\\rho$ 0.046 Annex R Horizontal coordinate $L_r$ 0.367 $\\sigma_{ref}/\\sigma_Y$ Vertical coordinate $K_r$ 2.59 $(K_I^P+K_I^S)/K_{mat}+\\rho$ FAL height $f(L_r)$ 0.373 Eq. 7.26 Verdict not acceptable (brittle fracture governing) $K_r \\gg f(L_r)$ Reading the result Break $K_r$ apart and it becomes obvious: the numerator $K_I^P + K_I^S = 48.13 + 46 = 94.13$, of which the residual-stress term alone accounts for nearly half. If this were a piece of base metal with no residual stress ($K_I^S=0$), $K_r$ would land near $48.13/37 \\approx 1.30$; the welding residual stress pushes it to 2.59 — almost double.\nThe assessment point $(0.367,\\ 2.59)$ lies far beyond the Failure Assessment Line (at the same horizontal coordinate the FAL is only $f(L_r)=0.373$), so the flawed structure is judged not acceptable. This matches the test exactly: LLAW fractured at the lowest load. The main force driving it past the critical line is precisely that residual stress intensity factor of 46 MPa·m$^{0.5}$.\nCross-check against FITNET: is mechCalc accurate? The value of this problem is not only \u0026ldquo;getting a number\u0026rdquo; but checking mechCalc\u0026rsquo;s accuracy against the original literature result. Laying the two assessments side by side, term by term:\nQuantity mechCalc FITNET original Difference Primary $K_I^P$ [MPa·m$^{0.5}$] 48.13 ≈48 ≈0 Secondary $K_I^S$ [MPa·m$^{0.5}$] 46.0 46 same source Load ratio $L_r$ 0.367 0.36 +1.9% Fracture ratio $K_r$ 2.59 2.58 +0.4% In the low-$L_r$, brittle-fracture-governed region — the very range that matters most to this test series — the two implementations agree almost digit for digit: $K_r$ differs by only 0.4% and $L_r$ by 1.9%. Bear in mind that the primary SIF and limit load on each side come from different analytical solutions (FITNET uses Sharples \u0026amp; Clayton / Sattari-Far, mechCalc uses BS 7910 Annex M / Annex P), and only the residual $K_I^S$ is taken from the same measured profile. Agreement this close is enough to show that mechCalc assembles correctly and gives trustworthy numbers at the residual-stress step — the one most prone to error.\nSource: the [[FITNET|FITNET]] project\u0026rsquo;s Case Studies for Fracture, case \u0026ldquo;A533B-1 Steel Residual Stress Experiments\u0026rdquo;; original test report C C France, J K Sharples \u0026amp; C Wignall, AEA Technology Report AEAT-4236 (SINTAP/TASK4/AEAT18, 1998).\nNext problem preview: same temperature, same region, same measured residual profile — the only variable is that post-weld heat treatment (PWHT) was applied, dropping the residual $K_I^S$ from 46 to 5. How far does the assessment point fall back? See [[bs7910-a533b-llht-fad-walkthrough|Problem 2 LLHT]].\n🧮 在线计算器：BS 7910 Clause 7 Fracture Assessment Calculator — Re-run this problem yourself: pick Option 1, enter the inputs from the table above, enter the secondary stress directly as K_I^S=46, and you get L_r=0.367, K_r=2.59.\n","permalink":"https://mechcalc.net/blog/en/posts/bs7910-a533b-llaw-fad-walkthrough/","summary":"\u003cp\u003eThis is the first of \u003cstrong\u003efour problems\u003c/strong\u003e on the A533B-1 welded plate. The shared background, common method, and residual-stress profile for all four are set out in the overview post:\u003c/p\u003e\n\u003cblockquote\u003e\n\u003cp\u003eFor the overview and method, see [[bs7910-a533b-residual-stress-fad|Where does residual stress push the assessment point? — A FAD recomputation of the A533B-1 large welded-plate fracture tests]]. This post focuses on one specimen and walks it through the calculator \u003cstrong\u003efrom input to reading the chart\u003c/strong\u003e.\u003c/p\u003e","title":"Problem 1 LLAW: How far does residual stress push the assessment point past the FAL? — An A533B as-welded, low-temperature FAD walkthrough"},{"content":"In the fracture-mechanics and fitness-for-service (FFS) literature, FITNET is a name you cannot avoid. Many large-component fracture-test examples, and many assessment results cited for cross-checking, are marked \u0026ldquo;from FITNET\u0026rdquo;. What is it, where did it come from, and why is it authoritative? This article sets it out from public sources.\n1. In one line: what FITNET is FITNET (European Fitness-for-Service Network) is an EU-funded research collaboration network. Its goal is one sentence: to establish a unified, validated fitness-for-service procedure for flawed metal structures (welded and non-welded) — the later FITNET FFS Procedure.\nThe core question a fitness-for-service assessment answers is: an in-service structure with a discovered flaw (crack, wall thinning, damage) — can it still remain safely in service? This is exactly the question that the mechCalc series of fracture assessment calculators addresses.\n2. Background: why Europe wanted to \u0026ldquo;unify\u0026rdquo; By the end of the last century, European countries each had their own approach to flaw assessment: the UK electricity industry\u0026rsquo;s R6 procedure, the BS 7910 lineage from the same root, various industry-specific practices… many methods, not fully compatible with each other. As a result, cross-border engineering collaboration, design certification and accident assessment all had to convert back and forth among several procedures — costly, and inconsistent in basis.\nMeanwhile, across the Atlantic, the US had already organized the oil-gas and pressure-equipment assessment experience into API 579 (today\u0026rsquo;s API 579-1 / ASME FFS-1). Europe needed a unified procedure of its own that integrated the countries\u0026rsquo; existing methods and stood up to example verification — that was the motivation for launching FITNET.\n3. Lineage: from SINTAP to FITNET The unification effort did not start from scratch. As early as 1996–1999, the SINTAP (Structural INTegrity Assessment Procedures for European Industry) project, funded by the EU\u0026rsquo;s Fourth Framework Programme (FP4), organized the scattered fracture assessment methods — with R6\u0026rsquo;s Failure Assessment Diagram (FAD) as the skeleton — into a unified procedure, completed in 1999.\nFITNET is the continuation and extension of SINTAP: on the basis of SINTAP\u0026rsquo;s fracture assessment method it further brought in fatigue, creep and corrosion, and did extensive example verification. So discussing the FITNET fracture module is essentially discussing the FAD method of the R6 / SINTAP lineage.\n4. Project overview (public sources) Item Content Full name European Fitness-for-Service Network Funding A \u0026ldquo;Thematic Network\u0026rdquo; of the EU\u0026rsquo;s Fifth Framework Programme (FP5), contract number G1RT-CT-2001-05071 Start February 2002, lasting about four years Lead institution Germany\u0026rsquo;s GKSS Research Centre (Geesthacht, today part of Helmholtz-Zentrum Hereon), project coordinator Mustafa Koçak Scale About 50 institutions from 17 European countries, with additional support from institutions in the US, Japan and Korea Main output The FITNET FFS Procedure, submitted to the European Committee for Standardization (CEN) to seek adoption as a European standard Official website eurofitnet.org This is a typical European cross-border industry-academia collaboration: universities, research institutes and industry contributing together, gathering the countries\u0026rsquo; experience into one volume and cross-checking it repeatedly with unified examples.\n5. The four modules of the FITNET FFS Procedure The FITNET FFS Procedure splits \u0026ldquo;whether a flawed structure can still be used\u0026rdquo; into four assessment modules by failure mechanism:\nFracture — brittle fracture and plastic instability, the core tool being the Failure Assessment Diagram (FAD); Fatigue — crack initiation and growth life under cyclic loading; Creep — damage accumulation under high-temperature long-term loading; Corrosion — wall thinning and environmental cracking caused by the environment. A real assessment chooses the matching module by flaw type, load case, material and service environment, combining several modules where necessary. This blog currently focuses on the fracture module — the most common in engineering and the most demanding of skill.\n6. Why the fracture module is highly comparable to BS 7910 This is the section readers of this blog care about most.\nFITNET\u0026rsquo;s fracture module directly reuses SINTAP\u0026rsquo;s method, and SINTAP\u0026rsquo;s skeleton shares a root with the UK\u0026rsquo;s R6. The assessment logic of all three is the same: put the crack-tip brittle-fracture driving force (vertical axis $K_r$, fracture ratio) and the plastic-instability driving force (horizontal axis $L_r$, load ratio) on one Failure Assessment Diagram —\n$$ K_r = \\frac{K_I}{K_{mat}}, \\qquad L_r = \\frac{\\sigma_{ref}}{\\sigma_Y} $$An assessment point inside the failure assessment line (FAL) is acceptable; on or outside the line, not acceptable.\nPrecisely because R6 / SINTAP / BS 7910 / FITNET share one lineage, FITNET\u0026rsquo;s fracture assessment results are highly comparable to BS 7910:2019 Clause 7. This is why, when we verify our BS 7910 fracture assessment, we use FITNET\u0026rsquo;s public examples as an independent literature cross-anchor — if two independent implementations agree point by point on the same large-component tests, they confirm each other\u0026rsquo;s assembly.\n🧮 在线计算器：BS 7910 Clause 7 Fracture Assessment Calculator — The FAD engine that combines Annex M (K_I) \u0026#43; Annex P (σ_ref) \u0026#43; the Option 1 FAD; re-run FITNET public examples online.\n7. Status and historical place FITNET was the first to systematically unify, validate and compile Europe\u0026rsquo;s scattered fitness-for-service methods, with a deep influence on European and international structural integrity assessment ever since — many assessment procedures in today\u0026rsquo;s textbooks, standards and engineering reports trace back to this R6 → SINTAP → FITNET line.\nIt should be noted that the formal FITNET FFS Procedure document is no longer publicly issued. So when citing its examples in this blog, we simply note \u0026ldquo;from the FITNET project\u0026rsquo;s Case Studies for Fracture\u0026rdquo; and do not give specific section and table numbers.\nRelated reading A worked case: Where does residual stress push the assessment point? — Recalculating the FAD of the A533B-1 large welded-plate fracture test — using a set of FITNET public examples, recalculated independently and compared point by point with the literature. References (public sources):\nM. Koçak, \u0026ldquo;FITNET Fitness-for-Service Procedure: An Overview\u0026rdquo;, Welding in the World, 2007 (link.springer.com/article/10.1007/BF03266577 ); \u0026ldquo;FITNET FFS procedure: A unified European procedure for structural integrity assessment\u0026rdquo;, Engineering Fracture Mechanics, 2008 (sciencedirect.com/science/article/abs/pii/S1350630708000381 ); The FITNET Thematic Network official website eurofitnet.org (with the SINTAP lineage). ","permalink":"https://mechcalc.net/blog/en/posts/fitnet-ffs-overview/","summary":"\u003cp\u003eIn the fracture-mechanics and fitness-for-service (FFS) literature, \u003cstrong\u003eFITNET\u003c/strong\u003e is a name you cannot avoid. Many large-component fracture-test examples, and many assessment results cited for cross-checking, are marked \u0026ldquo;from FITNET\u0026rdquo;. What is it, where did it come from, and why is it authoritative? This article sets it out from public sources.\u003c/p\u003e\n\u003chr\u003e\n\u003ch2 id=\"1-in-one-line-what-fitnet-is\"\u003e1. In one line: what FITNET is\u003c/h2\u003e\n\u003cp\u003e\u003cstrong\u003eFITNET\u003c/strong\u003e (European Fitness-for-Service Network) is an EU-funded research collaboration network. Its goal is one sentence: to establish a \u003cstrong\u003eunified, validated fitness-for-service procedure\u003c/strong\u003e for flawed metal structures (welded and non-welded) — the later \u003cstrong\u003eFITNET FFS Procedure\u003c/strong\u003e.\u003c/p\u003e","title":"FITNET: The Origins of Europe's Unified Fitness-for-Service Procedure"},{"content":"When you run a fitness-for-service (FFS) assessment on a welded pressure component, welding residual stress is almost always one of the hurdles you cannot step around. It is a textbook secondary stress (a self-balancing field): it plays no part in static equilibrium, yet it genuinely raises the driving force at the crack tip. The question engineers have asked for years is this: on the Failure Assessment Diagram (FAD), where does residual stress actually push the assessment point — and how much can post-weld heat treatment (PWHT) pull it back?\nThe most convincing way to answer that is not a derivation, but a set of large-component tests with real fracture loads and measured residual-stress profiles. That is exactly what this post works through — a set of large welded A533B-1 steel plates tested in four-point bending to fracture. Below we first set out the background and the shared method, then break the four specimens into four problems and assess them one by one, re-running each independently with mechCalc\u0026rsquo;s BS 7910 Clause 7 fracture assessment and cross-checking point by point against the original literature.\nSource: the [[FITNET|FITNET]] project\u0026rsquo;s Case Studies for Fracture, case \u0026ldquo;A533B-1 Steel Residual Stress Experiments\u0026rdquo;; original test report C C France, J K Sharples \u0026amp; C Wignall, AEA Technology Report AEAT-4236 (SINTAP/TASK4/AEAT18, 1998). FITNET is the unified European structural-integrity assessment procedure; its fracture module shares a common lineage with BS 7910:2019 (the R6/SINTAP family), which makes it a good independent literature anchor for a BS 7910 fracture assessment.\n🧮 在线计算器：BS 7910 Clause 7 Fracture Assessment Calculator — The Failure Assessment Diagram (FAD) calculation used here: it combines Annex M (K_I) \u0026#43; Annex P (σ_ref) \u0026#43; Option 1 FAD, and lets you re-run all four problems online.\n1. Background: why run a dedicated set of \u0026ldquo;residual stress\u0026rdquo; tests A533B-1 is a typical low-alloy steel for nuclear reactor pressure vessels (RPVs). Once a buried or surface flaw turns up in a weld in this kind of pressure equipment, the assessment must account for welding residual stress as it really is. But residual stress is both hard to measure and changes a great deal with post-weld heat treatment, so for a long time the field lacked a basis calibrated against full-scale components.\nTo fill that gap, the European SINTAP project of the 1990s (later folded into the [[FITNET|FITNET]] FFS procedure) designed a set of matched tests with a clear aim: drive the FAD from a measured residual-stress profile (not a code envelope curve); quantify the real effect of residual stress through as-welded / PWHT pairs; cover both low and high load-ratio conditions; and check how conservative the assessment is against the real fracture loads.\nThe tests were built as two pairs, four specimens in all, with names that encode the condition directly — and these are also the subjects of the four problems below:\nSpecimen Weld condition Test temperature $L_r$ condition Fracture load Fracture behaviour LLAW As-Welded −120 ℃ low $L_r$ 1.27 MN brittle fracture then arrest (ran about 40 mm in depth, then stopped) LLHT PWHT −120 ℃ low $L_r$ 2.19 MN full brittle fracture; load capacity about 1.7× that of LLAW HLAW As-Welded −30 ℃ high $L_r$ 5.10 MN clear yielding before fracture + about 5 mm of ductile tearing HLHT PWHT −30 ℃ high $L_r$ 4.83 MN clear yielding before fracture + about 4 mm of ductile tearing Naming rule: Low / High $L_r$ + As-Welded / Heat-Treated.\nTwo counter-intuitive effects are already buried in here. In the low-temperature pair, the as-welded LLAW actually broke at a lower load (direct evidence of how harmful residual stress is); in the high-temperature pair, the as-welded HLAW failed at a slightly higher load than the PWHT HLHT. The four problems below pin both of these down on the FAD.\n2. Shared method and inputs 2.1 Specimen and loading Each specimen is a 600 mm × 600 mm A533B-1 steel plate, 70–72 mm thick, with a double-V groove butt weld down the middle and a semi-elliptical surface crack machined into the weld region. The primary load is applied slowly and monotonically by four-point bending, producing a through-wall bending-stress field at the cracked section, converted as:\n$$ \\sigma_p = \\frac{6M}{W t^2}, \\qquad M = 0.0975 P $$where $P$ is the fracture load, $M$ is the bending moment, $W = 0.6$ m is the plate width, and $t$ is the plate thickness. The object assessed is the semi-elliptical surface crack in the weld region, evaluated at the deepest point (where both the primary and secondary $K_I$ exceed those at the surface point, consistent with where fracture started in the test). Shared constants: $W = 600$ mm, $E = 210$ GPa, $\\nu = 0.3$; every assessment uses the parent-material properties row (the same basis as the literature\u0026rsquo;s analytical $L_r$ column).\n2.2 Measured residual-stress profile → residual $K_I^S$ The secondary stress comes from welding itself — a through-wall self-balancing residual-stress field, tensile near the surface and compressive in the core. It was not estimated; it was measured layer by layer with the block removal and splitting technique combined with strain gauges. The measured through-wall residual stress (transverse to the weld, for $x/t \\le 0.5$) is fitted as:\n$$ \\text{As-welded (LLAW, HLAW):}\\quad \\sigma^* = -108.35 + 3543.6 (x/t) - 9871.5 (x/t)^2 + 2930.3 (x/t)^3 \\ \\text{MPa} $$$$ \\text{PWHT (LLHT, HLHT):}\\quad \\sigma^* = -28.032 + 609.14 (x/t) - 1590.8 (x/t)^2 \\ \\text{MPa} $$Integrating the residual profile over the crack front with a weight function gives the deepest-point residual stress intensity factor that enters the FAD: as-welded $K_I^S \\approx 46\\ \\mathrm{MPa\\cdot m^{0.5}}$, PWHT $K_I^S \\approx 5\\ \\mathrm{MPa\\cdot m^{0.5}}$ — an order-of-magnitude gap, and that is the quantitative reason the as-welded and PWHT assessment points diverge. PWHT lowers the residual stress by at least an order of magnitude relative to the as-welded state, and that is the physical root of the dramatic differences seen later.\n[!tip] Where does the residual-stress SIF come from?\nFor a given welding residual-stress profile, you can use mechCalc\u0026rsquo;s Annex M stress intensity factor calculator, pick the M.4.2 solution (finite-plate surface flaw, polynomial stress), enter $\\sigma_0..\\sigma_5$, and integrate out the residual $K_I^S$ that fits your own measured data. We checked this with the as-welded profile from this case: Annex M.4.2 gives $44.84\\ \\mathrm{MPa\\cdot m^{0.5}}$, within 2.5% of the value quoted in FITNET ($46\\ \\mathrm{MPa\\cdot m^{0.5}}$). The procedure, with a term-by-term white-box, is in [[bs7910-a533b-residual-kis-annexm|How is the residual stress intensity factor computed? — integrating a welding residual profile into an SIF with Annex M.4.2]].\n2.3 FAD and assessment-point coordinates All four problems use the BS 7910 Option 1 (standard) Failure Assessment Diagram, with the assessment-point coordinates matched to the Failure Assessment Line (FAL) definition:\n$$ K_r = \\frac{K_I^P + K_I^S}{K_{mat}} + \\rho, \\qquad L_r = \\frac{\\sigma_{ref}}{\\sigma_Y} $$$$ f(L_r) = \\begin{cases} \\left(1 + \\tfrac{1}{2}L_r^2\\right)^{-1/2}\\left[0.3 + 0.7\\,e^{-\\mu L_r^6}\\right], \u0026 L_r \\le 1 \\\\ f(L_r{=}1)\\,L_r^{(N-1)/(2N)}, \u0026 1 \u003c L_r \u003c L_{r,\\max} \\\\ 0, \u0026 L_r \\ge L_{r,\\max} \\end{cases} $$$$ \\mu = \\min\\left(0.001\\,E/\\sigma_Y,\\ 0.6\\right), \\quad N = 0.3\\left(1 - \\tfrac{\\sigma_Y}{\\sigma_U}\\right), \\quad L_{r,\\max} = \\frac{\\sigma_Y + \\sigma_U}{2\\sigma_Y} $$Source: BS 7910:2019, §7.3.3\nThese three branches happen to split the four problems into two groups: two low-load problems (LLAW, LLHT, $L_r=0.367/0.648$) land in the first branch, where the verdict is fracture-driven; two high-load problems (HLAW, HLHT, $L_r=1.80/1.71$) cross $L_{r,\\max}\\approx 1.15$ and fall into the third branch.\nOne key point: residual stress is a secondary stress, so it enters $K_r$ (the vertical axis) only, never $L_r$ (the horizontal axis) — $L_r$ comes solely from the primary stress via the reference stress from Annex P . So the $K_I^S$ from the weight-function integration of the measured profile can simply be fed in directly (which both follows the code\u0026rsquo;s division of labour and sidesteps the awkward question of how to linearise a non-linear secondary stress).\n2.4 How mechCalc assembles it mechCalc\u0026rsquo;s BS 7910 Clause 7 fracture assessment assembles the already-verified Annex M ($K_I$), Annex P ($\\sigma_{ref}$), Annex R ($\\rho$), and Option 1 FAD following the Clause 7 procedure.\nThe input mapping is the same for all four problems:\ncrack_type = surface (semi-elliptical surface crack), with $a_0$, $2c_0$, $B=t$, $W=600$ per problem; primary stress $P_b = \\sigma_p$ (pure bending, $P_m = 0$); secondary stress kis_source = direct, feeding $K_I^S$ in directly (as-welded 46 / PWHT 5); reference stress $\\sigma_{ref}$ and $L_r$ from the Annex P semi-elliptical surface-crack limit load (parent-material properties); plasticity interaction with $\\rho$ computed by Annex R (when secondary stress is present you must compute $\\rho$; you cannot set it to zero, or $K_r$ becomes non-conservative); FAD Option 1 assessment. Note: both sides take the residual $K_I^S$ from the same set of measured profiles (46 / 5), so what the cross-check really tests is each implementation\u0026rsquo;s own primary-SIF solution, reference-stress solution, and FAD assembly. The primary SIF and limit-load solutions come from different sources (the literature uses Sharples \u0026amp; Clayton / Sattari-Far, mechCalc uses BS 7910 Annex M / Annex P), so they are different solutions for the same physical quantity, and a difference of a few percent is normal.\n3. The four problems: assessed one by one Each cross-check focuses on the FAD coordinates $L_r$ and $K_r$. The mechCalc column gives the full set of intermediate quantities; the FITNET column takes the published parent-material-property row and the analytical limit-load $L_r$.\n3.1 Problem 1 — LLAW (as-welded, −120 ℃, low $L_r$) Brittle-fracture dominated, with large residual stress. In the test, LLAW broke at just 1.27 MN — the lowest of the four — and then arrested.\nInputs\nParameter Value Crack depth $a_0$ / full length $2c_0$ 19.4 / 175 mm Plate thickness $t = B$ 71 mm Bending stress $\\sigma_p$ (fracture load 1.27 MN) 245 MPa ($P_m=0$) Parent $\\sigma_Y$ / $\\sigma_U$ 618 / 791 MPa Fracture toughness $K_{mat}$ (as-welded weld, −120 ℃) 37 MPa·m$^{0.5}$ Residual $K_I^S$ (direct) 46 MPa·m$^{0.5}$ Results (FITNET ↔ mechCalc)\nQuantity mechCalc FITNET Difference $K_I^P$ / $K_I^S$ [MPa·m$^{0.5}$] 48.13 / 46 ≈48 (Sharples \u0026amp; Clayton) / 46 — $\\rho$ (Annex R) 0.046 FITNET simplified procedure — $\\sigma_{ref}$ [MPa] 226.5 — (Sattari-Far) — $L_r$ 0.367 0.36 +1.9% $K_r$ 2.59 2.58 +0.4% Reading the result: what pushes the assessment point up to $K_r = 2.59$ is mainly the 46 MPa·m$^{0.5}$ residual $K_I^S$ (nearly half of $K_I^P + K_I^S$). The point sits far outside the FAL, consistent with the fact that the specimen really did fracture; the two implementations of $K_r$ agree almost digit for digit.\n📖 [[bs7910-a533b-llaw-fad-walkthrough|Step-by-step calculation in mechCalc (illustrated walkthrough)]]\n3.2 Problem 2 — LLHT (PWHT, −120 ℃, low $L_r$) Same temperature and same region as LLAW; the only variable is that PWHT was applied — the residual stress is relaxed by an order of magnitude. Load capacity rises to 2.19 MN (about 1.7× that of LLAW).\nInputs\nParameter Value Crack depth $a_0$ / full length $2c_0$ 18.6 / 174 mm Plate thickness $t = B$ 71 mm Bending stress $\\sigma_p$ (fracture load 2.19 MN) 424 MPa ($P_m=0$) Parent $\\sigma_Y$ / $\\sigma_U$ 596 / 772 MPa Fracture toughness $K_{mat}$ (PWHT weld, −120 ℃) 46 MPa·m$^{0.5}$ Residual $K_I^S$ (direct) 5 MPa·m$^{0.5}$ Results (FITNET ↔ mechCalc)\nQuantity mechCalc FITNET Difference $K_I^P$ / $K_I^S$ [MPa·m$^{0.5}$] 82.97 / 5 ≈83 / 5 — $\\rho$ (Annex R) 0.010 FITNET simplified procedure — $\\sigma_{ref}$ [MPa] 386.0 — (Sattari-Far) — $L_r$ 0.648 0.64 +1.2% $K_r$ 1.92 1.89 +1.7% Reading the result: note one contrast — LLHT\u0026rsquo;s primary SIF ($K_I^P = 83$) is in fact higher than LLAW\u0026rsquo;s (48), yet its assessment point $K_r = 1.92$ is lower than LLAW\u0026rsquo;s 2.59. The whole difference is in the residual $K_I^S$: 46 as-welded, only 5 after PWHT. That turns \u0026ldquo;residual stress is harmful\u0026rdquo; from a qualitative statement into a concrete increment on the FAD\u0026rsquo;s vertical axis.\n📖 [[bs7910-a533b-llht-fad-walkthrough|Step-by-step calculation in mechCalc (illustrated walkthrough)]]\n3.3 Problem 3 — HLAW (as-welded, −30 ℃, high $L_r$) Warmed up to −30 ℃ and loaded harder into the high-$L_r$ (large-plasticity) region, still as-welded. Failure load 5.10 MN.\nInputs\nParameter Value Crack depth $a_0$ / full length $2c_0$ 19.0 / 174 mm Plate thickness $t = B$ 70 mm Bending stress $\\sigma_p$ (fracture load 5.10 MN, elastic conversion) 1015 MPa ($P_m=0$) Parent $\\sigma_Y$ / $\\sigma_U$ 520 / 677 MPa Fracture toughness $K_{mat}$ (as-welded weld, −30 ℃) 62 MPa·m$^{0.5}$ Residual $K_I^S$ (direct) 46 MPa·m$^{0.5}$ Results (FITNET ↔ mechCalc)\nQuantity mechCalc FITNET Difference $K_I^P$ / $K_I^S$ [MPa·m$^{0.5}$] 198.2 / 46 ≈198 / 46 — $\\rho$ (Annex R) 0 ($L_r \u003e L_{r,max}$ cut-off) FITNET simplified procedure — $\\sigma_{ref}$ [MPa] 937.5 — (Sattari-Far / ADINA) — $L_r$ 1.80 1.72 +4.8% $K_r$ 3.94 3.79 +3.9% Reading the result: $L_r = 1.80 \u003e L_{r,max}\\approx1.15$, so the verdict is plastic collapse. $K_r$ reaches 3.94, but this time the main driver is not the residual stress (in the high-$L_r$ region residual stress is \u0026ldquo;diluted\u0026rdquo; by plasticity) — it is the low toughness ($K_{mat} = 62$). The roughly 4–5% deviation at high $L_r$ comes from the difference in the reference-stress solution (mechCalc uses the Annex P closed form, the literature uses 3D finite-element analysis under mismatch); the direction is the same, and mechCalc is slightly more conservative.\n📖 [[bs7910-a533b-hlaw-fad-walkthrough|Step-by-step calculation in mechCalc (illustrated walkthrough)]]\n3.4 Problem 4 — HLHT (PWHT, −30 ℃, high $L_r$) Same temperature and region as HLAW, but with PWHT applied. The thing to watch is the double benefit of PWHT at −30 ℃: it both lowers the residual stress and brings the toughness back up by an order of magnitude ($K_{mat}$ jumps from 62 to 321).\nInputs\nParameter Value Crack depth $a_0$ / full length $2c_0$ 19.0 / 174 mm Plate thickness $t = B$ 70 mm Bending stress $\\sigma_p$ (fracture load 4.83 MN, elastic conversion) 960 MPa ($P_m=0$) Parent $\\sigma_Y$ / $\\sigma_U$ 520 / 677 MPa Fracture toughness $K_{mat}$ (PWHT weld, −30 ℃, off the lower shelf) 321 MPa·m$^{0.5}$ Residual $K_I^S$ (direct) 5 MPa·m$^{0.5}$ Results (FITNET ↔ mechCalc)\nQuantity mechCalc FITNET Difference $K_I^P$ / $K_I^S$ [MPa·m$^{0.5}$] 187.5 / 5 ≈187 / 5 — $\\rho$ (Annex R) 0 ($L_r \u003e L_{r,max}$ cut-off) FITNET simplified procedure — $\\sigma_{ref}$ [MPa] 886.7 — (Sattari-Far / ADINA) — $L_r$ 1.71 1.63 +4.6% $K_r$ 0.60 0.57 +5.2% Reading the result: same plastic-collapse verdict ($L_r \u003e L_{r,max}$), but $K_r$ drops from HLAW\u0026rsquo;s 3.94 to 0.60 — almost entirely because $K_{mat}$ rises from 62 to 321. In the high-$L_r$ region the governing factors swing back to toughness and net-section yielding.\n📖 [[bs7910-a533b-hlht-fad-walkthrough|Step-by-step calculation in mechCalc (illustrated walkthrough)]]\n4. The four problems together, and conclusions Laying the four sets of FAD coordinates side by side:\nSpecimen mechCalc $L_r$ FITNET $L_r$ $\\Delta L_r$ mechCalc $K_r$ FITNET $K_r$ $\\Delta K_r$ LLAW 0.367 0.36 +1.9% 2.59 2.58 +0.4% LLHT 0.648 0.64 +1.2% 1.92 1.89 +1.7% HLAW 1.80 1.72 +4.8% 3.94 3.79 +3.9% HLHT 1.71 1.63 +4.6% 0.60 0.57 +5.2% A few conclusions:\nThe two independent implementations agree closely. At low $L_r$ the $K_r$ values are almost digit for digit (LLAW 2.59 vs 2.58), with the $L_r$ error under 2%; at high $L_r$ the error is still only about 4–5%. Given that the primary-SIF and limit-load solutions come from different sources, this level of agreement is enough for the two calculations to confirm each other\u0026rsquo;s Clause 7 assembly.\nThe effect of residual stress is reproduced quantitatively (Problems 1 and 2). The as-welded LLAW $K_r$ (2.59) is clearly higher than the PWHT LLHT (1.92), even though LLHT actually has the higher primary SIF — what pushes the as-welded point up is the 46 MPa·m$^{0.5}$ residual $K_I^S$.\nThe benefit of PWHT is twofold (Problems 3 and 4 compared): it both relaxes the residual stress ($K_I^S$ 46 → 5) and restores toughness ($K_{mat}$ 62 → 321). The two paths stack, so the assessment point drops markedly in both the low- and high-$L_r$ regions — which is precisely the quantitative justification for insisting on PWHT for important welds in practice.\nThe difference at high $L_r$ has a clear methodological source. For $L_r \u003e 1$, mechCalc uses an analytical Annex P reference stress while the literature computes the limit load by 3D finite-element analysis; together with the non-linearity of the plastic zone, a deviation of about 4–5% is normal and in the same direction (mechCalc slightly more conservative).\nThe conservatism of the assessment is reasonable. All four assessment points fall outside the FAL, consistent with the fact that the specimens really did fracture; the assessment correctly returns \u0026ldquo;this condition is not acceptable\u0026rdquo;, with margin to spare. For the safety assessment of a flawed structure, this kind of right-direction result with a conservative margin is exactly what you want.\nIn one sentence: this set of A533B-1 tests answers \u0026ldquo;where does residual stress push the FAD assessment point\u0026rdquo; very concretely — in the brittle-fracture-dominated low-$L_r$ region, as-welded residual stress can contribute a $K_I^S$ on the order of the primary stress, lifting the assessment point markedly; PWHT pulls it back through the combination of relaxing the residual stress and restoring toughness. And mechCalc\u0026rsquo;s BS 7910 Clause 7 calculation independently reproduces the literature results for all four problems, which shows that it assembles correctly at the step most prone to error — residual stress — and is fit for use in engineering assessment.\nReproducible re-run: the mechCalc driver script .tempTestScript/golden-verify-fitnet-fad/verify_ex1_residual.py (drives the BS 7910 Clause 7 calculation from the literature inputs, with the residual $K_I^S$ fed in directly).\nReferences: the FITNET project\u0026rsquo;s Case Studies for Fracture (A533B-1 Steel Residual Stress Experiments case); BS 7910:2019+A1:2020, Clause 7 (§7.3.3 / §7.3.6 / §7.3.7), Annex M / Annex P / Annex R; C C France, J K Sharples \u0026amp; C Wignall, AEAT-4236, SINTAP/TASK4/AEAT18 (1998).\n","permalink":"https://mechcalc.net/blog/en/posts/bs7910-a533b-residual-stress-fad/","summary":"\u003cp\u003eWhen you run a fitness-for-service (FFS) assessment on a welded pressure component, welding residual stress is almost always one of the hurdles you cannot step around. It is a textbook secondary stress (a self-balancing field): it plays no part in static equilibrium, yet it genuinely raises the driving force at the crack tip. The question engineers have asked for years is this: \u003cstrong\u003eon the Failure Assessment Diagram (FAD), where does residual stress actually push the assessment point — and how much can post-weld heat treatment (PWHT) pull it back?\u003c/strong\u003e\u003c/p\u003e","title":"BS 7910 FAD Assessment: What Residual Stress Does, Seen Through a FITNET Case"},{"content":"Pressure vessels, pipelines, and structures develop flaws during service — cracks, corrosion, wall thinning. When an inspection finds such a flaw, engineers face one key question:\nCan this equipment keep running? If so, for how long?\nThis is exactly what Fitness-for-Service (FFS) assessment answers.\nThe Limit of Traditional Rules Traditional design codes (such as ASME or GB 150) are written for new, flaw-free equipment. When a flaw is found, the codes often say: \u0026ldquo;out of tolerance — repair or retire.\u0026rdquo;\nBut in practice, not every flaw that breaks a manufacturing tolerance will cause a failure. An overly conservative call creates two problems:\nUnnecessary shutdowns — lost production time and repair costs; False confidence — the repair itself may introduce new flaws. FFS assessment replaces rule-of-thumb conservatism with fracture mechanics and materials science, giving engineers a documented, defensible verdict on whether a flawed component is still safe.\nThe Core Tool: The Failure Assessment Diagram (FAD) The most widely used FFS tool is the Failure Assessment Diagram (FAD).\nThe FAD has two axes, each measuring a different failure mode:\nVertical axis $K_r$ (fracture ratio): crack driving force divided by material fracture toughness — measures the risk of brittle fracture; Horizontal axis $L_r$ (load ratio): applied load divided by the limit load — measures the risk of plastic collapse. Plot the assessment point $(L_r, K_r)$ for the flawed structure on the diagram:\nPoint inside the Failure Assessment Line (FAL): flaw is acceptable — the component can keep running; Point on or outside the FAL: flaw is not acceptable — repair or retire. $$ K_r = \\frac{K_I^P + K_I^S}{K_{mat}} + \\rho \\leq f(L_r) $$Here $K_I^P$ is the primary stress intensity factor, $K_I^S$ is the secondary stress contribution (e.g. residual stress), $K_{mat}$ is the material fracture toughness, and $\\rho$ is the plasticity correction term.\nMain Assessment Standards Three standards are widely used around the world for FFS:\nStandard Typical use BS 7910 Welded structures, pressure vessels (UK / international) API 579-1 / ASME FFS-1 Refinery and process pressure equipment (USA / global) ASME XI Nuclear power plant pressure boundaries (USA nuclear) All three share the same FAD framework in principle, but differ in option levels, formula details, and scope.\nWhat MechCalc Is Building MechCalc turns these assessment methods into open, step-by-step online calculators:\nEvery calculation step is shown with the standard clause and formula reference; Full PDF calculation reports can be exported; Three-language interface (Chinese / English / German) for international engineering teams. This blog covers the theory, standard interpretation, and engineering examples behind the tools — so engineers don\u0026rsquo;t just use a tool, they understand the physics behind it.\n","permalink":"https://mechcalc.net/blog/en/posts/ffs-intro/","summary":"\u003cp\u003ePressure vessels, pipelines, and structures develop flaws during service — cracks, corrosion, wall thinning.\nWhen an inspection finds such a flaw, engineers face one key question:\u003c/p\u003e\n\u003cblockquote\u003e\n\u003cp\u003e\u003cstrong\u003eCan this equipment keep running? If so, for how long?\u003c/strong\u003e\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003cp\u003eThis is exactly what \u003cstrong\u003eFitness-for-Service (FFS) assessment\u003c/strong\u003e answers.\u003c/p\u003e\n\u003chr\u003e\n\u003ch2 id=\"the-limit-of-traditional-rules\"\u003eThe Limit of Traditional Rules\u003c/h2\u003e\n\u003cp\u003eTraditional design codes (such as ASME or GB 150) are written for new, flaw-free equipment.\nWhen a flaw is found, the codes often say: \u0026ldquo;out of tolerance — repair or retire.\u0026rdquo;\u003c/p\u003e","title":"What Is Fitness-for-Service (FFS) Assessment?"},{"content":"In the structural integrity assessment of pressure vessels, oil and gas pipelines, offshore steel structures and nuclear pressure-boundary components, the stress intensity factor (SIF, written $K_I$) is the central quantity of Linear Elastic Fracture Mechanics (LEFM). It measures precisely how strong the singular stress field is at the crack tip.\nThe flaw assessment standard BS 7910:2019+A1:2020, published by The Welding Institute (TWI), sets out in its Annex M a classic, high-accuracy and self-consistent system of analytical SIF solutions for the eight common geometry groups met in engineering. This article explains its physical basis, the code calculation chain and its engineering use, and introduces an interactive online calculator built entirely in pure Python.\n1. Physical basis: a general SIF superposition model Annex M uses a general SIF assembly model based on superposition (BS 7910:2019, Annex M, Eq. M.1):\n$$ K_I = Y\\sigma \\sqrt{\\pi a} $$To cover primary and secondary (e.g. residual) stress together with macro and local stress-concentration effects, the standard expands this into the form (BS 7910:2019, Annex M, Clause M.2.1):\n$$ K_I = \\left[ M \\cdot f_w \\left( k_{tm} \\cdot M_{km} \\cdot M_m \\cdot P_m + k_{tb} \\cdot M_{kb} \\cdot M_b \\cdot (P_b + (k_m - 1) P_m) \\right) + (M_m Q_m + M_b Q_b) \\right] \\sqrt{\\frac{\\pi a}{Q}} $$The physical multiplier chain: $P_m, P_b$ and $Q_m, Q_b$: primary membrane/bending and secondary membrane/bending stress (Clause M.2.1). $M_m, M_b$: unbounded-plate geometry correction factors (membrane \u0026amp; bending). $f_w$: the finite-width correction factor, e.g. the secant width correction for a through-wall crack (Eq. M.8). $Q$: the crack shape parameter ($Q = \\Phi^2$), which corrects a straight-fronted crack to a semi- or full-elliptical singular front (Eq. M.11). $M$: the thin-shell bulging factor (also the Folias factor), which compensates for the shell-tension release in cylinders/pipes and curved shells (Clause M.6). $k_m, k_{tm}, k_{tb}$: the manufacturing misalignment and macro stress-concentration correction chain. $M_{km}, M_{kb}$: the local weld-toe notch stress-concentration factors, which capture the local singularity of the weld-toe geometry (Clause M.4.1.2). 2. The classic Newman-Raju semi-elliptical solution for plates (M.4) For the most representative case in engineering — a semi-elliptical surface crack in a plate — BS 7910 Annex M.4 fully adopts the Newman-Raju (1986) polynomial fit. This solution is a milestone in fracture mechanics: it was the first to evaluate, in parallel, both the deepest point of the crack front (point A) and the surface free end (point C):\n2.1 Shape parameter and integral factor The crack shape parameter $Q$ and the elliptic-integral value $\\Phi$ are related by (Annex M.4.1.2.2, Eq. M.11):\nFor $a \\le c$: $$ \\Phi = \\sqrt{1 + 1.464 \\left(\\frac{a}{c}\\right)^{1.65}} $$For $a \u003e c$ (a long, narrow crack): $$ \\Phi = \\sqrt{1 + 1.464 \\left(\\frac{c}{a}\\right)^{1.65}} $$and the three-dimensional shape parameter is $Q = \\Phi^2$.\n2.2 Splitting the geometry factor between the two front points Newman-Raju gives a polynomial split of the membrane geometry factor $M_m$ between the deepest and surface points (Annex M.4.1.2, Eq. M.12):\nDeepest front point ($\\theta = \\pi/2$): $$ M_{m,dp} = \\frac{M_1 + M_2(a/B)^2 + M_3(a/B)^4}{\\Phi} $$ Surface front point ($\\theta = 0$): $$ M_{m,sf} = \\frac{\\left( M_1 + M_2(a/B)^2 + M_3(a/B)^4 \\right) \\cdot g \\cdot \\sqrt{a/c}}{\\Phi} $$ This split shows clearly that: for a shallow, long flaw, the surface point — under the boundary effect of surface free-energy release — can, at some ratio, overtake the deepest point in SIF and so control surface growth; while for a deep flaw, the deepest point, under high triaxial constraint, carries the highest brittle-fracture risk.\n3. The thin-shell Folias bulging factor and cylinder correction (M.6) In cylinders/pipes or curved shells, as a crack grows axially or circumferentially, internal fluid pressure or tensile load makes the shell at the cracked ligament bulge radially outward. This bulging produces a strong local bending that greatly magnifies the crack-tip SIF.\n3.1 Axial through-wall correction: The bulging magnification factor $M_T$ is set through the dimensionless shell-curvature parameter $\\lambda$ (Clause M.6.1, Eq. M.24): $$ \\lambda = \\frac{a}{\\sqrt{r_m B}} $$ $$ M_T = \\sqrt{1 + 1.61 \\lambda^2} $$ where $r_m = r + B/2$ is the mean radius of the cylinder or curved shell (Clause M.6.1).\n3.2 Surface semi-elliptical correction: For a surface semi-elliptical flaw the bulging penalty is partly reduced by the restraining stiffness of the rest of the wall, and the correction $M_s$ is reshaped as (Clause M.6.2): $$ M_s = \\frac{1 - a/(B M_T)}{1 - a/B} $$ Through an adaptive damping term this makes $M_s \\to 1.0$ for a shallow crack, and $M_s \\to M_T$ when the crack is very deep and the ligament is near instability — physically consistent.\n4. Weld-toe local magnification, Maddox/Fett correction (M.4.1.2 / M.4.2) When a flaw sits at a weld toe — for example at a cruciform joint — the highly singular concentration field of the weld toe has a strong local effect on the crack tip. BS 7910 Clause M.4.1.2 introduces the Maddox local weld-toe magnification factors $M_{km}$ and $M_{kb}$:\nthe local stress dominates only within a very thin layer at the weld toe (typically within $10\\%$–$15\\%$ of the plate thickness $B$); as the crack extends through the thickness ($a/B \u003e 0.15$), the weld-toe effect falls off sharply with depth and $M_{k} \\to 1.0$, approaching the parent-material solution. This calculator implements the depth-decay formula 1:1, avoiding the over-conservative design that a single blanket multiplier on the whole section would cause.\n5. A pure-Python interactive whitebox calculation platform So that structural engineers can leave behind manual look-up of the code curves, the whole engine is written in pure Python and deployed as:\n🧮 在线计算器：BS 7910 Annex M Stress Intensity Factor Calculator — Enter the geometry (a, c, B, W, r, p) and stresses, and read K_I with the whitebox step-by-step derivation.\nThree highlights of the full-stack module: Canvas geometry preview to scale: the web front end renders the metal section and elliptical crack to real physical proportions from your live geometry inputs ($a, c, B, W, r, p$), across 34 geometries (plate, embedded, cylinder inner/outer surface, round bar and more), and marks the deepest point A and surface point C on the canvas — for clear physical transparency.\nPure-Python analytical solving, microsecond response: the engine is rewritten 100% in pure Python, dropping external DLL wrapping and pythonnet interop overhead. Back-end compute is in the microsecond range, and it passes a full green regression against 68 golden examples to a 0.01% tolerance.\nWhitebox LaTeX step-by-step derivation: rather than only a final number, the right panel shows the full LaTeX step outline matching the code clauses 1:1, with the numerical substitution of every intermediate quantity (such as $\\Phi, Q, f_w, M_{m,dp}$).\nThrough this system, engineers can see directly how geometry, misalignment and residual stress weight the change of the $K_I$ crack-tip field — moving from \u0026ldquo;memorizing formulas\u0026rdquo; to \u0026ldquo;controlling the physics\u0026rdquo;.\n","permalink":"https://mechcalc.net/blog/en/posts/bs7910-annex-m-ki-theory/","summary":"\u003cp\u003eIn the structural integrity assessment of pressure vessels, oil and gas pipelines, offshore steel structures and nuclear pressure-boundary components, the \u003cstrong\u003estress intensity factor (SIF, written $K_I$)\u003c/strong\u003e is the central quantity of Linear Elastic Fracture Mechanics (LEFM). It measures precisely how strong the singular stress field is at the crack tip.\u003c/p\u003e\n\u003cp\u003eThe flaw assessment standard \u003cstrong\u003eBS 7910:2019+A1:2020\u003c/strong\u003e, published by The Welding Institute (TWI), sets out in its Annex M a classic, high-accuracy and self-consistent system of analytical SIF solutions for the eight common geometry groups met in engineering. This article explains its physical basis, the code calculation chain and its engineering use, and introduces an interactive online calculator built entirely in pure Python.\u003c/p\u003e","title":"BS 7910 Annex M: Analytical Calculation of the Stress Intensity Factor (K_I) — Theory and Practice"},{"content":" The fracture assessment of a cracked structure watches two competing failure modes. This guide uses a single Failure Assessment Diagram (FAD) to weave together the assessment principle, the calculation chain, and the BS 7910 assessment steps into one clear storyline. By the end you should be able to see how an assessment point is computed and how it is judged acceptable or not.\nPrologue: after a crack is found When an in-service pressure vessel, pipe, or welded structure is inspected, crack-like flaws are often found. The question the engineer must answer is not \u0026ldquo;can this crack be used\u0026rdquo; — a crack is not something you \u0026ldquo;use\u0026rdquo; — but rather: can the structure that contains this flaw still operate safely under the current loads?\nThis is exactly what a fitness-for-service (FFS) assessment addresses. Clause 7 of BS 7910:2019 covers its most central piece — the fracture assessment: deciding whether a flaw of known size is acceptable under the given loads and material.\nOrdinary strength design (\u0026ldquo;working stress \u0026lt; allowable stress\u0026rdquo;) breaks down here, because in linear-elastic theory the stress at a crack tip goes to infinity and cannot be compared with an allowable stress at all. Fracture mechanics uses a different language to answer the question.\n1. A tug-of-war between two failure modes A cracked structure can fail along two very different paths (BS 7910:2019, §7.1.1):\nBrittle fracture: the driving force at the crack tip exceeds the material\u0026rsquo;s fracture toughness, and the crack grows suddenly and rapidly, with almost no warning. Low temperature, high strength, and thick sections make this especially dangerous. Plastic collapse: the remaining load-bearing section around the flaw yields completely and loses its capacity. Tough, ductile materials tend toward this mode. Real structures usually sit somewhere between the two: the more brittle the material, the closer to the first mode; the tougher, the closer to the second; and in the ductile-to-brittle transition region both risks coexist. A single criterion cannot cover this competition — looking only at toughness misses collapse, looking only at strength misses fracture.\nBS 7910 solves this with a single two-dimensional chart that captures both failure modes at once: the Failure Assessment Diagram (FAD).\n2. The whole picture in one diagram: the FAD Start with the overview; every section that follows simply explains this one chart in more depth:\nFigure 1: The BS 7910 fracture assessment at a glance. On the left is the calculation chain — three verified building blocks K_I (Annex M), σ_ref (Annex P), and K_mat (Annex J) assemble the horizontal coordinate L_r and the vertical coordinate K_r, giving an assessment point (L_r, K_r). On the right is the FAD — a point inside the failure assessment line FAL (green SAFE region) means the flaw is acceptable; on or outside the line (red UNSAFE) means it is not.\nEach axis measures one failure mode, and the closer the assessment point is to the origin, the safer it is:\nVertical axis $K_r$ (fracture ratio): how close you are to brittle fracture. Horizontal axis $L_r$ (load ratio): how close you are to plastic collapse. Failure Assessment Line (FAL) (blue): a curve $K_r = f(L_r)$ that fuses the competition between the two failures into a single boundary. A point inside the curve → the flaw is acceptable; on or outside → not acceptable. 3. The fracture-mechanics principle Vertical axis $K_r$: how close to brittle fracture $$K_r = \\frac{K_I}{K_{mat}}$$ $K_I$ is the stress intensity factor, a measure of the strength of the singular stress field at the crack tip — it is the crack\u0026rsquo;s driving force. $K_{mat}$ is the material\u0026rsquo;s fracture toughness, its resistance to crack extension — the resistance. So $K_r$ is \u0026ldquo;driving force / resistance\u0026rdquo;. $K_r = 1$ means the driving force just equals the resistance, i.e. the point of brittle fracture. (BS 7910:2019, §7.1.1) Horizontal axis $L_r$: how close to plastic collapse $$L_r = \\frac{\\sigma_{ref}}{\\sigma_Y}$$ $\\sigma_{ref}$ is the reference stress, representing the \u0026ldquo;equivalent stress\u0026rdquo; level carried by the cracked section. $\\sigma_Y$ is the material\u0026rsquo;s yield strength. $L_r = 1$ means the cracked section as a whole reaches yield and begins to collapse. Note that it is the cracked net section that fails first, not the entire structure yielding together — this is the difference between \u0026ldquo;net-section yielding\u0026rdquo; and \u0026ldquo;plastic collapse\u0026rdquo;. (BS 7910:2019, §7.1.1) The FAL: fusing two failures into one line The FAL is not a simple join of the two axes, but a curve that drops as $L_r$ rises:\nFigure 2: The FAD explained. The blue FAL falls as L_r increases; together with the axes and the vertical cut-off line L_r,max it bounds the green safe region (SAFE). The solid green dot is the assessment point (L_r, K_r); the dashed ray from the origin through the point extends to its intersection with the FAL (open dot), and the ratio between them is the load reserve factor — how much the load can still be scaled up along a proportional loading path.\nWhy does the curve drop? As $L_r$ rises and the crack tip enters large-scale plasticity, the plasticity blunts the tip and absorbs energy, so the material\u0026rsquo;s apparent resistance to brittle fracture actually falls; at the same time the structure moves closer to overall collapse. So at high $L_r$ the allowable $K_r$ must be pulled down. This falling curve is the balance line where the two risks — brittle fracture and plastic collapse — trade off against each other. (BS 7910:2019, §7.1.1)\nOption 1 curve equation ($L_r \\le 1$ branch; BS 7910:2019, §7.3, Option 1):\n$$f(L_r) = \\left(1 + \\tfrac{1}{2}L_r^2\\right)^{-1/2}\\left[0.3 + 0.7\\,e^{-\\mu L_r^6}\\right], \\quad \\mu = \\min\\!\\left(0.001\\,\\frac{E}{\\sigma_Y},\\ 0.6\\right)$$4. The calculation: three steps to one assessment point Plotting the assessment point $(L_r, K_r)$ is essentially walking through the calculation chain on the left of Figure 1.\nStep 1: stress intensity factor $K_I$ (Annex M) $K_I$ describes the strength of the elastic stress field at the crack tip. Annex M of BS 7910 gives general expressions for standard geometries (plate, cylinder, sphere, …) containing flaws, with the skeleton:\n$$K_I = \\left(P_m M_m + P_b M_b\\right) M_k\\, f_w \\sqrt{\\frac{\\pi a}{Q}}$$The physical meaning of each correction factor:\nSymbol Name Physical meaning $M_m$ membrane correction factor geometric correction for the membrane-stress contribution of a surface/embedded crack (Newman–Raju polynomials) $M_b$ bending correction factor contribution of bending stress to crack-tip $K$ $M_k$ weld-toe magnification factor extra magnification from local stress concentration at the weld toe (= 1 with no weld) $f_w$ finite-width correction boundary effect for a plate of finite width (≈ 1 for wide plates) $Q$ flaw shape parameter $= 1 + 1.464\\,(a/c)^{1.65}$, an elliptic-integral approximation combining the crack aspect ratio Before computing, the real flaw is first characterised (idealised) into a standard geometry — an enclosing regular shape stands in for the irregular measured flaw (BS 7910:2019, §7.1.2):\nFigure 3: Size definitions for the three basic flaws. Top left, a surface crack (depth a, total length 2c, wall thickness B); top right, an embedded crack (height 2a, length 2c, distance p to the near surface); bottom, a through-thickness crack (total length 2a). Each K_I formula in Annex M corresponds to one flaw geometry.\nStep 2: reference stress $\\sigma_{ref}$ (Annex P) The key idea of the reference-stress method is to map the complex plastic behaviour of a cracked structure back onto an equivalent elastic problem. This is done by defining a fictitious reference stress $\\sigma_{ref}$ tied to the limit load $P_L$ of the cracked section — in essence $\\sigma_{ref}$ and $P_L$ are two languages for the same thing (one in stress units, one in load units), related through the load-bearing area: $\\sigma_{ref} \\times A \\approx P_L$.\nOnce $\\sigma_{ref}$ is known, the horizontal coordinate is $L_r = \\sigma_{ref} / \\sigma_Y$. How $\\sigma_{ref}$ is taken depends on the flaw type and stress distribution, and is looked up from the relevant Annex P solutions. (BS 7910:2019, Annex P)\nStep 3: assembling $K_r$ and $L_r$ A real assessment usually also involves secondary stresses such as weld residual stress or thermal stress. The full assembly of the vertical coordinate is (BS 7910:2019, §7.3, Eq. 7.39):\n$$K_r = \\frac{K_I^P + K_I^S}{K_{mat}} + \\rho$$ $K_I^P$: the SIF from primary stress (produced by external loads such as pressure or mechanical force), counted in full; $K_I^S$: the SIF from secondary stress (self-balancing, such as weld residual or thermal stress); $\\rho$: the plasticity interaction correction between primary and secondary stress (BS 7910:2019, Annex R). Why primary and secondary stresses are treated differently is the most important — and most confusing — point in Clause 7:\nFigure 4: Primary vs secondary stress. Primary stress (Pm/Pb) is sustained by external loads and does not self-balance, so it enters both K_r and L_r. Secondary stress (Qm/Qb, e.g. weld residual stress) is self-balancing and relaxes with plastic flow, so it enters K_r only (with a plasticity-interaction correction), not L_r.\nThe reason: $L_r$ measures \u0026ldquo;how far from overall plastic collapse\u0026rdquo;, and overall collapse is driven by the externally sustained primary load; secondary stress is self-balancing and relaxes with plastic flow, so it cannot push the structure into collapse on its own. Therefore:\nOnly primary stress enters $L_r$: $\\;L_r = \\sigma_{ref}/\\sigma_Y$ (horizontal axis, collapse); Both primary and secondary stress enter $K_r$ (vertical axis, fracture), with secondary stress corrected through $\\rho$. ⚠️ Key rule: when $K_I^S \u003c 0$ (compressive, favourable secondary stress), set both $K_I^S$ and $\\rho$ to zero (a conservative treatment). And $\\rho$ is a separate term added outside the fraction — never put it in the denominator.\nWhen $K_I^S = 0$ and $\\rho = 0$, the expression reduces exactly to $K_r = K_I / K_{mat}$ — the basic form of the vertical axis from Section 3.\nThe plastic cut-off $L_{r,max}$ The FAD is cut off hard at a certain horizontal coordinate: for $L_r \\ge L_{r,max}$, $f(L_r) = 0$, and the flaw is judged to fail no matter how good the toughness. The cut-off value is (BS 7910:2019, §7.3):\n$$L_{r,max} = \\frac{\\sigma_Y + \\sigma_U}{2\\sigma_Y}$$The numerator $(\\sigma_Y + \\sigma_U)/2$ is the flow stress $\\sigma_f$ (the mean of yield and tensile strength), representing the average stress level at which the material reaches full plasticity. Beyond this vertical line the structure fails by overall plastic flow, regardless of crack toughness.\nFAD options: Option 1 / 2 / 3 The FAL curve has three levels of increasing accuracy (BS 7910:2019, §7.3):\nOption Data needed Character Option 1 only $\\sigma_Y$, $\\sigma_U$ the general default, most conservative, least data-hungry (the formula in Section 3) Option 2 the material stress–strain curve built point by point from the real hardening behaviour; usually a wider safe region Option 3 a finite-element $J$-integral solution the most accurate, building the FAD directly from $J$; the most work You can start from Option 1; only when Option 1 fails and the extra margin is worth chasing do you move up a level.\n5. The BS 7910 fracture assessment steps Clause 7 lays out a complete assessment procedure (BS 7910:2019, §7.2). It can be grouped into four stages:\nStage 1 · Preparation (gather inputs)\nDefine loads and stresses: identify the worst load case and split stresses into primary / secondary. (mandatory) Determine the fracture toughness $K_{mat}$: from tests or estimates; when data are missing, convert from Charpy energy via Annex J. (mandatory) Determine tensile properties: take $\\sigma_Y$, $\\sigma_U$, $E$ of the weakest material near the flaw. (mandatory) Characterise the flaw: idealise the measured flaw into a standard geometry (Figure 3), fixing $a$, $2c$, etc. (mandatory) Stage 2 · Choose the framework\nSelect and compute the FAD: choose Option 1 / 2 / 3 per §7.3, and compute the FAL curve $f(L_r)$ and the cut-off $L_{r,max}$. (mandatory) Choose the analysis type: an initiation analysis (single-point check) or a ductile-tearing analysis (point-locus check). (mandatory) Stage 3 · Compute the point\nCompute $L_r$: $L_r = \\sigma_{ref}/\\sigma_Y$ (primary stress only). (mandatory) Compute $K_r$: assemble primary + secondary stress and $\\rho$ per Eq. 7.39. (mandatory) Estimate in-service growth: if the service life includes fatigue (Clause 8), creep (Clause 9), or environmental cracking (Clause 10), compute the growth $\\Delta a$ to obtain a series of assessment points that move with size. (as needed) Stage 4 · Judge\nPlot on the FAD: a single point for initiation, a locus for tearing. (mandatory) Compare and judge: the position of the point/locus relative to the FAL decides PASS / FAIL. (mandatory) Refine if it fails: move to a higher Option, apply a constraint correction (Annex N), a weld strength-mismatch correction (Annex I), flaw re-characterisation (Annex E), and so on — each is a physically grounded argument, not a way to \u0026ldquo;game\u0026rdquo; the result. (as needed) Report: prepare a traceable assessment report per Annex H. (mandatory) 6. The criterion: how PASS / FAIL is judged For the most common initiation analysis, the criterion is simply the position of the assessment point relative to the FAL (BS 7910:2019, §7.2 / §7.3):\nIf $K_r \\le f(L_r)$ and $L_r \u003c L_{r,max}$ → PASS (flaw acceptable); If $K_r \u003e f(L_r)$ or $L_r \\ge L_{r,max}$ → FAIL (flaw not acceptable). Telling the two failure modes apart on the diagram The point crosses the line in the upper-left ($L_r$ small, $K_r$ high) → brittle fracture dominates, toughness is short; The point crosses in the lower-right ($K_r$ small, $L_r$ approaching $L_{r,max}$) → plastic collapse dominates, capacity is short; Crossing near the knee of the curve → the two risks are comparable. Reserve factors: how much margin is left The standard itself carries no built-in safety factor, so a point sitting \u0026ldquo;just on the FAL\u0026rdquo; only means \u0026ldquo;just not failing\u0026rdquo;, with zero margin. Two reserve factors quantify the margin:\nToughness reserve factor $F_{K_r}$: the margin from the assessment point to the FAL measured along the vertical axis; Load reserve factor $F_{load}$: the margin measured along the proportional-loading ray from the origin through the point (the dashed ray in Figure 2), answering \u0026ldquo;by what factor can the load still be scaled before it touches the line\u0026rdquo;. Because there is no built-in safety factor, a sensitivity analysis is essential: perturb the flaw size, toughness, and stress each a little and check whether the point stays comfortably inside the safe region — this is the real core of engineering risk judgement.\n7. A worked example (Option 1, primary stress only) Take a surface crack in a plate under primary stress only ($K_I^S = 0$, $\\rho = 0$):\nParameter Value Parameter Value membrane stress $P_m$ 100 MPa yield strength $\\sigma_Y$ 350 MPa bending stress $P_b$ 50 MPa tensile strength $\\sigma_U$ 500 MPa crack depth $a$ 5 mm Young\u0026rsquo;s modulus $E$ 210 GPa crack half-length $c$ 10 mm fracture toughness $K_{mat}$ 90 MPa·m$^{1/2}$ plate thickness $B$ 25 mm plate width $W$ 200 mm (no weld, $M_k=1$) Step 1, $K_I$: $a/c = 0.5$, $a/B = 0.2$, $Q = 1 + 1.464 \\times 0.5^{1.65} = 1.466$; giving $K_I \\approx 15.85$ MPa·m$^{1/2}$.\nStep 2, $\\sigma_{ref}$: from the Annex P surface-crack solution, $\\sigma_{ref} \\approx 126.5$ MPa, so $L_r = 126.5/350 = 0.361$; the cut-off is $L_{r,max} = (350+500)/(2\\times350) = 1.214$.\nStep 3, assessment: $K_r = 15.85/90 = 0.176$; with $\\mu = 0.6$, $f(0.361) \\approx 0.968$.\nThe assessment point $(0.361,\\ 0.176)$ lies well inside the curve limit $(0.361,\\ 0.968)$, with a vertical margin of $0.968 - 0.176 = 0.792$.\nResult: ✅ PASS, with ample reserve.\n8. Common pitfalls \u0026ldquo;Can the crack be used\u0026rdquo; is the wrong question. What is assessed is whether the cracked structure can stay in service and whether the flaw is acceptable — not the crack itself. Counting secondary stress in $L_r$. Secondary stress (residual, thermal) enters $K_r$ only, never $L_r$. When unsure whether a stress is primary or secondary, treat it as primary (more conservative). Calling a stress \u0026ldquo;primary\u0026rdquo; just because it is large. The test is \u0026ldquo;does it drive overall plastic collapse\u0026rdquo;, not the magnitude: residual stress, even near yield in magnitude, is still treated as secondary. Putting $\\rho$ in the denominator. $\\rho$ is a separate term added outside the fraction in $K_r = \\dfrac{K_I^P + K_I^S}{K_{mat}} + \\rho$. Mixing up which tensile properties to use. A common convention: use lower-bound properties for $K_r$/$L_r$, and mean (flow-stress) properties for the FAL and $L_{r,max}$. Treating \u0026ldquo;on the line\u0026rdquo; as safe. The standard has no built-in safety factor, so being on the line means zero margin — a sensitivity analysis is a must. Run it online with MechCalc Once you understand the principle, the fastest way to learn is to try it. Use the online BS 7910 fracture assessment calculator: enter the example above (or your own flaw) and it runs Annex M ($K_I$) + Annex P ($\\sigma_{ref}$) + the FAD assessment in one click, returning a PASS/FAIL verdict, reserve factors, and a live FAD chart; you can optionally include secondary stress $K_I^S$, the Annex R plasticity interaction $\\rho$, an Annex J toughness estimate, and Option 2/3.\n🧮 在线计算器：BS 7910 Fracture Assessment Calculator — Choose Option 1, enter the inputs from Section 7, and reproduce the PASS result L_r=0.361, K_r=0.176, f(L_r)=0.968.\nThe figures in this article are original MechCalc diagrams for teaching only and do not replace the BS 7910 standard. For engineering assessment, refer to the current BS 7910:2019+A1:2020.\n","permalink":"https://mechcalc.net/blog/en/posts/bs7910-fracture-assessment-tutorial/","summary":"\u003cblockquote\u003e\n\u003cp\u003eThe fracture assessment of a cracked structure watches two competing failure modes. This guide uses a single \u003cstrong\u003eFailure Assessment Diagram (FAD)\u003c/strong\u003e to weave together the assessment principle, the calculation chain, and the BS 7910 assessment steps into one clear storyline. By the end you should be able to see how an assessment point is computed and how it is judged acceptable or not.\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003ch2 id=\"prologue-after-a-crack-is-found\"\u003ePrologue: after a crack is found\u003c/h2\u003e\n\u003cp\u003eWhen an in-service pressure vessel, pipe, or welded structure is inspected, crack-like flaws are often found. The question the engineer must answer is \u003cstrong\u003enot \u0026ldquo;can this crack be used\u0026rdquo; — a crack is not something you \u0026ldquo;use\u0026rdquo; — but rather: can the structure that contains this flaw still operate safely under the current loads?\u003c/strong\u003e\u003c/p\u003e","title":"A Concise Guide to BS 7910 Fracture Assessment"},{"content":" 🧮 Go straight to the calculator: want to skip the derivation? Open the BS 7910 fracture toughness estimator , load a standard example, and verify in one click.\n1. Why estimate K_mat from CVN? In a fitness-for-service / engineering critical assessment (FFS/ECA), fracture toughness $K_{mat}$ is the core parameter for judging whether a crack will cause brittle fracture. Yet in many real engineering situations:\nthe component has been in service for years and no fresh toughness specimen can be taken; the code required only a Charpy impact test, so the historical data hold CVN values only; a full-size fracture toughness test is extremely costly and not feasible in practice. Here, indirect conversion is the only viable route. BS 7910:2019 Annex J provides two well-validated conversion methods for exactly this purpose, for ferritic steels.\n⚠️ This method applies to ferritic steels only (yield strength $\\sigma_{ys} \\leq 690$ MPa); it does not apply to austenitic stainless steels (which have no ductile-brittle transition).\n2. Overview of the two conversion routes Route Code clause Applicability Character A: lower-shelf equation BS 7910:2019, Annex J.2.1 3 J \u0026lt; $C_v$ ≤ 27 J, fracture-surface crystallinity ≥ 80% Conservative, for the very-low-temperature brittle region B: Master Curve BS 7910:2019, Annex J.2.2 Ductile-brittle transition region Statistical-physical model, more accurate 3. Route A: the lower-shelf equation (Annex J.2.1) 3.1 Conditions of use The lower-shelf equation requires the material to be \u0026ldquo;near the lower shelf\u0026rdquo;, shown by:\na Charpy impact energy $C_v$ between 3 J and 27 J; a cleavage (brittle) area fraction (crystallinity) on the fracture surface of ≥ 80%. When $C_v \u003e 27$ J the material has entered the transition region, where the J.2.1 equation overestimates toughness; the Master Curve method must be used instead.\n3.2 The equation $$K_{mat} = \\left[(12\\sqrt{C_v} - 20) \\cdot \\left(\\frac{25}{B}\\right)^{0.25}\\right] + 20 \\quad [\\text{MPa}\\sqrt{\\text{m}}]$$(Source: BS 7910:2019, Annex J.2.1, Eq. J.1)\nPhysical meaning of each term:\nThe constant 20: the theoretical minimum fracture toughness of ferritic steel, $K_{min}$ = 20 MPa√m (the asymptotic lower bound, from ASTM E1921). The $12\\sqrt{C_v}$ term: an empirical lower-envelope coefficient fitted to a large body of ferritic-steel test data. $(25/B)^{0.25}$: a size correction from weakest-link theory — a thicker section contains more micro-crack initiation sites, so statistically $K_{mat}$ is lower; the exponent 0.25 = 1/β, with β = 4 the Weibull shape parameter. 3.3 Worked example 2 Input: $C_v = 20$ J, $B = 50$ mm\n$$K_{mat} = [(12 \\times \\sqrt{20} - 20) \\times (25/50)^{0.25}] + 20 = [33.66 \\times 0.8409] + 20 = \\mathbf{48.3 \\text{ MPa}\\sqrt{\\text{m}}}$$ 4. Route B: the Master Curve method (Annex J.2.2) The Master Curve method, proposed by Kim Wallin (1984) from a large body of ferritic-steel fracture toughness data, has one core idea: the fracture toughness of ferritic steel in the transition region follows a three-parameter Weibull distribution, and the distribution shape is material-independent (only the location parameter $T_0$ varies by material).\n4.1 Step one: determine the reference transition temperature $T_0$ Way 1 (preferred): if measured ASTM E1921 Master Curve data exist, use the measured $T_0$ directly — no statistical margin is then needed.\nWay 2: estimate from $T_{27J}$ (this calculator\u0026rsquo;s default route):\n$$T_0 = T_{27J} - 18 \\quad [°\\text{C}]$$(Source: BS 7910:2019, Annex J.2.2)\n$T_{27J}$ is the temperature on the CVN transition curve at $C_v = 27$ J, and the empirical offset of −18°C gives $T_0$.\n4.2 Step two: the T_K confidence margin (⚠️ a key pitfall) When $T_0$ is estimated from $T_{27J}$, BS 7910 requires an added $T_K = 25°\\text{C}$ confidence margin to cover the statistical scatter of the CVN → $T_0$ conversion (90% confidence):\n$$T_{eff} = T - T_0 - T_K = T - T_0 - 25°\\text{C}$$ This is the most common source of calculation error! Omitting $T_K$ overestimates $K_{mat}$. In standard example 1, omitting $T_K$ inflates $K_{mat}$ from 90.8 MPa√m to about 128 MPa√m.\n4.3 Step three: the Master Curve calculation Weibull scale parameter $K_0$ (the 63.2% failure fractile, for a 1T standard specimen):\n$$K_0 = 31 + 77 \\cdot \\exp[0.019 \\cdot T_{eff}] \\quad [\\text{MPa}\\sqrt{\\text{m}}]$$(Source: BS 7910:2019, Annex L.9.5.3, Eq. L.13; 31 = $K_{min}$ + 11 = 20 + 11)\n$K_{mat}$ at a specified failure probability (1T basis):\n$$K_{mat,1T} = 20 + \\left[\\ln\\left(\\frac{1}{1-P_f}\\right)\\right]^{0.25} \\cdot (K_0 - 20)$$(Source: BS 7910:2019, Annex J.2.2, Eq. J.5; the ASTM E1921 three-parameter Weibull inverse)\nAt $P_f = 0.05$: the probability term = $[\\ln(1/0.95)]^{0.25} = (0.05129)^{0.25} = 0.4759$\nThickness correction (weakest-link size effect):\n$$K_{mat} = 20 + (K_{mat,1T} - 20) \\cdot \\left(\\frac{25}{B}\\right)^{0.25}$$(Source: BS 7910:2019, Annex J.2.2; ASTM E1921)\n4.4 Worked example 1 (standard example) Input: $T_{27J} = -50°\\text{C}$, $T = 0°\\text{C}$, $B = 60$ mm, $P_f = 0.05$\nStep Calculation Result B1: T_0 = T_27J - 18 = -50 - 18 -68°C T_eff (with T_K=25°C) 0 - (-68) - 25 43°C K_0 = 31 + 77·exp(0.019×43) = 31 + 174.3 205.3 MPa√m K_mat_1T = 20 + 0.4759×(205.3-20) = 20 + 88.2 108.2 MPa√m K_mat = 20 + 88.2×(25/60)^0.25 = 20 + 70.9 90.9 MPa√m ✅ Code reference value: 90.8 MPa√m (BS 7910:2019, Annex J.2.5 Note 2)\n5. The API 579 comparison route API 579-1:2021, 9F.4.3.2, Eq. (9F.74) gives a finer, multi-parameter estimate of $T_0$:\n$$T_0 = T_{28J} - 79 + \\frac{\\sigma_{ys}}{9} - \\frac{C_{V-US}}{59} + \\Delta T_0 \\quad [°\\text{C}]$$where $\\Delta T_0 = 18°\\text{C}$ (a fixed conservative margin) and $T_{28J} \\approx T_{27J}$ (27 J ≈ 20 ft-lbs).\nCompared with BS 7910\u0026rsquo;s $T_0 = T_{27J} - 18°\\text{C}$, API 579 additionally accounts for:\nthe yield-strength term $+\\sigma_{ys}/9$: higher strength → more brittle tendency → $T_0$ shifts to higher temperature; the upper-shelf energy term $-C_{V-US}/59$: better ductility → $T_0$ shifts to lower temperature. Note: API 579 treats its $T_0$ as a \u0026ldquo;directly determined value\u0026rdquo;, so $T_K = 0$ when it goes into the Master Curve (the $\\Delta T_0 = 18°\\text{C}$ already plays the same role).\nExample 3 comparison Parameter BS 7910 API 579 T_0 -18°C -24.1°C K_mat 44.7 MPa√m 61.0 MPa√m Because API 579 additionally accounts for yield strength and upper-shelf energy, its $T_0$ is lower and its $K_{mat}$ higher (more optimistic).\n6. Engineering-use advice Order of preference: measured $T_0$ (ASTM E1921) \u0026gt; API 579 multi-parameter conversion \u0026gt; BS 7910 single-parameter $T_{27J}$ estimate. Always check that $T_K = 25°C$ has been included (specific to the CVN route). Lower-shelf boundary: for $C_v \u003e 27$ J, J.2.1 must not be used — switch to the Master Curve. Thickness effect: when the wall thickness $B \u003e 25$ mm, the size correction lowers $K_{mat}$ markedly and cannot be ignored. 7. Quick start with the calculator Open the fracture mechanics calculators → choose \u0026ldquo;BS 7910 Fracture Toughness Estimation\u0026rdquo;. Click the \u0026ldquo;Example 1 (Master Curve, BS 7910 standard example)\u0026rdquo; button to auto-fill the parameters. Click \u0026ldquo;Calculate\u0026rdquo; to verify: $K_{mat} = 90.85$ MPa√m ✅. The right panel shows the three Master Curve probability curves ($P_f$ = 5%/50%/95%) and the position of the current assessment point. 📖 References:\nBS 7910:2019+A1:2020, Guide to methods for assessing the acceptability of flaws in metallic structures API 579-1/ASME FFS-1:2021, Part 9 Annex 9F — Fracture Toughness ASTM E1921, Standard Test Method for Determination of Reference Temperature T_0 Wallin, K. (1984). The scatter in K_Ic results. Engineering Fracture Mechanics, 19(6), 1085-1093. ","permalink":"https://mechcalc.net/blog/en/posts/bs7910-annex-j-kmat-tutorial/","summary":"\u003cblockquote\u003e\n\u003cp\u003e🧮 \u003cstrong\u003eGo straight to the calculator\u003c/strong\u003e: want to skip the derivation? Open the \u003ca href=\"/calculators?lang=en\"\u003eBS 7910 fracture toughness estimator\u003c/a\u003e\n, load a standard example, and verify in one click.\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003chr\u003e\n\u003ch2 id=\"1-why-estimate-k_mat-from-cvn\"\u003e1. Why estimate K_mat from CVN?\u003c/h2\u003e\n\u003cp\u003eIn a fitness-for-service / engineering critical assessment (FFS/ECA), fracture toughness $K_{mat}$ is the core parameter for judging whether a crack will cause brittle fracture. Yet in many real engineering situations:\u003c/p\u003e\n\u003cul\u003e\n\u003cli\u003ethe component has been in service for years and \u003cstrong\u003eno fresh toughness specimen can be taken\u003c/strong\u003e;\u003c/li\u003e\n\u003cli\u003ethe code required only a Charpy impact test, so \u003cstrong\u003ethe historical data hold CVN values only\u003c/strong\u003e;\u003c/li\u003e\n\u003cli\u003ea full-size fracture toughness test is extremely costly and \u003cstrong\u003enot feasible in practice\u003c/strong\u003e.\u003c/li\u003e\n\u003c/ul\u003e\n\u003cp\u003eHere, \u003cstrong\u003eindirect conversion\u003c/strong\u003e is the only viable route. \u003cstrong\u003eBS 7910:2019 Annex J\u003c/strong\u003e provides two well-validated conversion methods for exactly this purpose, for ferritic steels.\u003c/p\u003e","title":"BS 7910 Annex J: Estimating Fracture Toughness K_mat from Charpy Energy"},{"content":" 🧮 在线计算器：VDI 2230 Bolted Joint Calculation — The full 14-step calculation chain (R0–R13), with six strength checks.\nEccentric load and bending-moment effect — the \u0026ldquo;deep water\u0026rdquo; of VDI 2230 All the earlier worked examples assumed concentric symmetry ($s_{sym} = 0$, $a = 0$). But the standard states clearly: concentric symmetry is the minority case in engineering practice.\n1. Why is eccentricity the norm? The standard says (VDI 2230:2015, §5.1.2.3, p.57):\n\u0026ldquo;The case of a concentrically clamped and concentrically loaded BJ is only rarely found in practice. In most cases, the line of action of the load $F_A$ does not lie in the bolt axis.\u0026rdquo;\nEccentricity comes from two independent sources, which must be handled separately:\nType Parameter Physical meaning Eccentric clamping (exzentrische Verspannung) $s_{sym}$ The bolt axis is offset from the symmetry axis of the clamped parts Eccentric loading (exzentrischer Kraftangriff) $a$ The line of action of the external force is offset from the symmetry axis of the clamped parts 2. Determining the eccentricity parameter $s_{sym}$ $s_{sym}$ describes the offset of the bolt axis relative to the geometric symmetry axis of the clamped parts (VDI 2230:2015, §5.1.2.2, Eq. 66):\n$$ s_{sym} = \\frac{c_T}{2} - e \\leq \\frac{G}{2} - e \\tag{66} $$where $c_T$ is the interface dimension (in the plane of analysis) and $e$ is the shortest distance from the bolt axis to the interface edge.\nWhen $s_{sym} = 0$ → the bolt lies on the symmetry axis of the clamped parts → concentric clamping When $s_{sym} \\neq 0$ → the bolt is eccentric → it causes an additional bending moment\n3. Determining the eccentric-loading distance $a$ $a$ is the distance from the substitutional line of action of the external force to the symmetry axis of the clamped parts (VDI 2230:2015, §5.2.1, p.61):\n\u0026ldquo;The distance $a$ is the distance of the substitutional line of action of the axial working load from the axis of the laterally symmetrical deformation body, thus ultimately a lever arm.\u0026rdquo;\nDetermining $a$ is an elasto-mechanics problem — you need to find the point of zero bending moment near the bolt. The standard gives the classic example of a connecting rod (Pleuel) (VDI 2230:2015, §5.2.1, p.62):\n$$ a \\approx 0.275 R \\quad \\text{(constant-section circular-ring connecting rod)} $$For complex structures, the standard advises using FEM to help determine $a$.\n4. Corrected resiliences $\\delta_P^*$ and $\\delta_P^{**}$ The eccentric effect increases the resilience of the clamped parts — because, on top of the axial compression, it adds a bending deformation.\n4.1 Eccentric-clamping corrected resilience $\\delta_P^*$ Used in the denominator of the force-ratio formula (VDI 2230:2015, §5.1.2.2, Eq. 67):\n$$ \\delta_P^* = \\delta_P + \\frac{s_{sym}^2 \\cdot l_K}{E_P \\cdot I_{Bers}} \\tag{67} $$where $\\delta_P$ is the symmetric resilience (Eq. 40 or 41) and $I_{Bers}$ is the substitutional moment of inertia of the deformation body.\n4.2 Eccentric-loading corrected resilience $\\delta_P^{**}$ Used in the numerator of the force-ratio formula (VDI 2230:2015, §5.1.2.3, Eq. 71):\n$$ \\delta_P^{**} = \\delta_P + \\frac{a \\cdot s_{sym} \\cdot l_K}{E_P \\cdot I_{Bers}} \\tag{71} $$$\\delta_P^{**}$ can be smaller than, equal to, or larger than $\\delta_P^*$, depending on the relative magnitude and direction of $a$ and $s_{sym}$.\n4.3 Eccentric force ratio $\\Phi_{en}^*$ The most general force-ratio formula (VDI 2230:2015, Eq. R3/4):\n$$ \\Phi_{en}^* = n \\cdot \\frac{\\delta_P^{**} + \\delta_{PZu}}{\\delta_S + \\delta_P^*} \\tag{R3/4} $$Simplified forms for different eccentric cases (VDI 2230:2015, Eq. 73-75):\nCase Calculation of $F_{SA}$ Source $s_{sym} \\neq 0$, $a \u003e 0$ $n \\cdot \\frac{\\delta_P^{**}}{\\delta_S + \\delta_P^*} \\cdot F_A$ Eq. 73 $s_{sym} \\neq 0$, $a = 0$ $n \\cdot \\frac{\\delta_P}{\\delta_S + \\delta_P^*} \\cdot F_A$ Eq. 74 $a = s_{sym} \\neq 0$ $n \\cdot \\frac{\\delta_P^*}{\\delta_S + \\delta_P^*} \\cdot F_A$ Eq. 75 [!CAUTION] The risk when $a$ and $s_{sym}$ are on different sides The standard warns (VDI 2230:2015, §5.1.2.3, p.58): \u0026ldquo;If $a$ and $s_{sym}$ do not lie on the same side of the axis of symmetry, the additional bolt load $F_{SA}$ [\u0026hellip;] may become larger than calculated.\u0026rdquo; This case should be avoided by design.\n5. Substitutional moment of inertia $I_{Bers}$ The $I_{Bers}$ in the eccentricity-correction formulas is the equivalent section moment of inertia of the deformation body (cone + sleeve).\nCone part (Eq. 59):\n$$ I_{Bers}^V = 0.147 \\cdot \\frac{(D_A - d_W) \\cdot d_W^3 \\cdot D_A^3}{D_A^3 - d_W^3} \\tag{59} $$Eccentricity correction (Steiner shift term, Eq. 60):\n$$ I_{Bers}^{Ve} = I_{Bers}^V + s_{sym}^2 \\cdot \\frac{\\pi}{4} D_A^2 \\tag{60} $$Sleeve part (Eq. 61):\n$$ I_{Bers}^H = \\frac{b \\cdot c_T^3}{12} \\tag{61} $$Combined body (Eq. 62):\n$$ I_{Bers} = \\frac{l_K}{\\frac{2}{w}(l_V / I_{Bers}^{Ve}) + l_H / I_{Bers}^H} \\tag{62} $$6. Opening check (Abheben, Opening) Eccentric loading causes the interface to begin losing contact pressure on the bending-tension side — that is, \u0026ldquo;opening\u0026rdquo; occurs (VDI 2230:2015, §5.3.2).\nThe minimum clamping force needed to prevent opening $F_{KA}$ (the meaning of Eq. R2/3) must keep the pressure on the bending-tension side of the interface above zero. This is achieved through the following relation:\nThe ratio of the bending moment to the clamping force at the interface must not exceed the section core distance:\n$$ \\frac{M_B}{F_{KR}} \\leq \\frac{I_{BT}}{A_{BT} \\cdot e_{max}} \\tag{concept} $$If this condition is not met (the pressure on the bending-tension side drops to zero), the resilience of the clamped parts increases nonlinearly (VDI 2230:2015, §5.3.3), and the standard\u0026rsquo;s linear spring model no longer applies. This is also why the standard stresses that $F_{Kerf}$ must be large enough to prevent opening.\n7. Practical advice for eccentric design Based on the standard\u0026rsquo;s requirements, the design of an eccentric joint should follow these principles:\nConstructional level Minimize $s_{sym}$ — place the bolt near the symmetry axis of the clamped parts Keep $a$ and $s_{sym}$ on the same side — avoid an unfavourable lever effect (Eq. 73–75) Control the interface dimension $c_T \\leq G$ — ensure the linear spring model is valid (Eq. 54, 55) Calculation level Do not use the concentric-symmetric formulas — unless it is verified that $s_{sym} \\approx 0$ and $a \\approx 0$ Do not subtract the through-hole area when computing $I_{Bers}$ — because the bolt takes part in bending through the head and the nut Perform FEM verification for critical joints — determining the eccentricity parameters $a$ and $n$ often needs FEM support The standard sums up (VDI 2230:2015, §5.1.2.3, p.57):\n\u0026ldquo;Even relatively small eccentricities of the load introduction may have a considerable effect on the deformation behaviour of the clamped parts.\u0026rdquo;\nData basis All formulas in this article are from VDI 2230 Blatt 1:2015-11, §5.1.2.2, §5.1.2.3 and §5.3.2.\nDisclaimer: This article is for engineering teaching reference only. The accuracy of the eccentric calculation depends heavily on the precise determination of $a$ and $s_{sym}$; FEM verification is advised for critical joints.\n📚 Series navigation\n← Previous: Full Worked Example (ESV) Full-series review # Title Core 03 VDI 2230 Overview necessity, main equation 04 Scope and Standard Positioning boundaries, standard family 05 Spring Model and Force Distribution force ratio Φ 06 Bolt Elastic Resilience δS Eq. 17-31 07 Clamped-Parts Resilience δP Rötscher cone 08 Preload Design R1-R6 design part 09 Strength Checks R7-R13 verification part 10 Worked Example DSV M12→M16 iteration 11 Worked Example ESV cast-iron housing 12 Eccentric Load (this article) $\\delta_P^*$, $\\delta_P^{**}$ ","permalink":"https://mechcalc.net/blog/en/posts/vdi2230-10-eccentric-load-bending/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/vdi2230-bolt-calculation?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eVDI 2230 Bolted Joint Calculation\u003c/a\u003e — The full 14-step calculation chain (R0–R13), with six strength checks.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003ch1 id=\"eccentric-load-and-bending-moment-effect--the-deep-water-of-vdi-2230\"\u003eEccentric load and bending-moment effect — the \u0026ldquo;deep water\u0026rdquo; of VDI 2230\u003c/h1\u003e\n\u003cblockquote\u003e\n\u003cp\u003eAll the earlier worked examples assumed concentric symmetry ($s_{sym} = 0$, $a = 0$). But the standard states clearly: concentric symmetry is the minority case in engineering practice.\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003ch2 id=\"1-why-is-eccentricity-the-norm\"\u003e1. Why is eccentricity the norm?\u003c/h2\u003e\n\u003cp\u003eThe standard says (VDI 2230:2015, §5.1.2.3, p.57):\u003c/p\u003e","title":"VDI 2230 (010): Eccentric Load and Bending-Moment Effect"},{"content":" 🧮 在线计算器：VDI 2230 Bolted Joint Calculation — The full 14-step calculation chain (R0–R13), with six strength checks.\nFull worked example: tapped-thread joint (ESV) This article is a counterpart to the DSV example (article 8) . It highlights the differences unique to ESV: $w = 2$, $E_M = E_{BI}$, the cone-angle formula Eq. 42, and the R11 engagement-depth check.\nEngineering problem statement A bolt is screwed from the top into a cast-iron housing (grey cast iron GJL-250), fixing a steel cover plate.\nParameter Value Note Joint type ESV (tapped-thread) Bolt screwed into the cast-iron housing External axial force $F_A = 15\\,000$ N (static) External transverse force $F_Q = 0$ N No transverse force Sealing requirement $A_D = 800$ mm², $p_{i\\max} = 10$ MPa O-ring seal Clamp length $l_K = 30$ mm Steel cover-plate thickness Clamped-parts outer diameter $D_A = 40$ mm Cover-plate material Steel, $E_P = 210\\,000$ MPa Housing material GJL-250, $E_{BI} = 100\\,000$ MPa Operating temperature Room temperature $\\Delta F'_{Vth} = 0$ Friction coefficient $\\mu_G = \\mu_K = 0.14$ Load introduction factor $n = 0.5$ Tightening method Precision torque wrench $\\alpha_A = 1.6$ R0 — preliminary diameter selection Select M12 × 1.75, property class 10.9 (same parameters as article 8).\nLimit check (Eq. R0/2):\n$$ G' \\approx (1.5 \\dots 2) \\cdot d_W = (1.5 \\dots 2) \\times 16.6 = 24.9 \\dots 33.2 \\text{ mm} $$$c_T = D_A = 40$ mm → exceeds $G'_{\\max} \\approx 33.2$ mm; the standard warns the accuracy may drop (VDI 2230:2015, Eq. 55). But it does not yet exceed $G'_{\\max} \\approx 3 \\times 16.6 = 49.8$ mm (Eq. 56), so the calculation can continue.\nR1 — tightening factor $$ \\alpha_A = 1.6 $$R2 — required minimum clamping force No transverse force → $F_{KQ} = 0$\nSealing requirement (Eq. R2/2):\n$$ F_{KP} = A_D \\cdot p_{i\\max} = 800 \\times 10 = 8\\,000 \\text{ N} $$$$ F_{Kerf} = F_{KP} = 8\\,000 \\text{ N} $$R3 — elastic resiliences and force ratio ⚠️ The ESV vs DSV difference in δS The overall bolt-resilience formula is unchanged (Eq. 19), but the nut region $\\delta_M$ is different (VDI 2230:2015, §5.1.1.1):\nParameter DSV ESV $l_M$ $0.4d = 4.8$ mm $0.33d = 3.96$ mm (Eq. 27) $E_M$ $E_S = 210\\,000$ $E_{BI} = 100\\,000$ ESV bolt resilience $\\delta_S$:\nSegment Length Cross-section area $\\delta_i$ Head $\\delta_{SK}$ 6 mm 113.1 $2.53 \\times 10^{-7}$ Shank $\\delta_1$ 10 mm 113.1 $4.22 \\times 10^{-7}$ Free thread $\\delta_{Gew}$ 20 mm 76.2 $1.25 \\times 10^{-6}$ Engaged segment $\\delta_G$ $0.5 \\times 12 = 6$ mm 76.2 $3.75 \\times 10^{-7}$ Nut $\\delta_M$ 3.96 mm 113.1 ($E_{BI}$) $3.50 \\times 10^{-7}$ $$ \\delta_S = (2.53 + 4.22 + 12.50 + 3.75 + 3.50) \\times 10^{-7} = 2.65 \\times 10^{-6} \\text{ mm/N} $$ [!IMPORTANT] In ESV, δM uses the clamped-part elastic modulus This is the most notable difference between ESV and DSV — the elastic modulus of the nut region takes the cast iron\u0026rsquo;s $E_{BI} = 100\\,000$ MPa, not the bolt steel\u0026rsquo;s 210,000 MPa. For an aluminium-alloy housing ($E_{BI} \\approx 70\\,000$ MPa), this effect is even more marked (VDI 2230:2015, §5.1.1.1, p.42).\nESV clamped-parts resilience $\\delta_P$ For ESV the joint coefficient $w = 2$ and the cone angle uses Eq. 42:\n$$ \\beta_L = l_K / d_W = 30 / 16.6 = 1.81 $$$$ y = D_A' / d_W = 40 / 16.6 = 2.41 $$$$ \\tan\\varphi_E = 0.348 + 0.013\\ln(1.81) + 0.193\\ln(2.41) = 0.348 + 0.008 + 0.170 = 0.526 $$Limiting diameter (Eq. 39, $w = 2$):\n$$ D_{A,Gr} = 16.6 + 2 \\times 30 \\times 0.526 = 48.2 \\text{ mm} $$$D_A = 40 \u003c D_{A,Gr} = 48.2$ → use Eq. 41 (cone + sleeve):\n$$ \\delta_P = \\frac{\\frac{2}{2 \\times 13.5 \\times 0.526}\\ln\\left[\\frac{(16.6+13.5)(40-13.5)}{(16.6-13.5)(40+13.5)}\\right] + \\frac{4}{40^2-13.5^2}\\left[30-\\frac{40-16.6}{2 \\times 0.526}\\right]}{210\\,000 \\times \\pi} $$Logarithmic term: $\\ln\\left[\\frac{30.1 \\times 26.5}{3.1 \\times 53.5}\\right] = \\ln(4.81) = 1.571$\nSleeve-length check: $l_K - \\frac{D_A - d_W}{w \\cdot \\tan\\varphi} = 30 - \\frac{23.4}{1.052} = 30 - 22.2 = 7.8$ mm\n$$ \\delta_P = \\frac{\\frac{2 \\times 1.571}{14.21} + \\frac{4 \\times 7.8}{1417.8}}{659\\,734} = \\frac{0.221 + 0.022}{659\\,734} = 3.68 \\times 10^{-7} \\text{ mm/N} $$Force ratio Φ $$ \\Phi = 0.5 \\times \\frac{3.68 \\times 10^{-7}}{2.65 \\times 10^{-6} + 3.68 \\times 10^{-7}} = 0.5 \\times 0.122 = 0.061 $$R4 — preload change $f_Z$: bolt head 3 μm + thread (cast iron, rough) 5 μm + 1 interface 3 μm = 11 μm\nNote: cast-iron surface roughness is larger, so the embedding is taken as a higher value from Table 5.\n$$ F_Z = \\frac{0.011}{3.02 \\times 10^{-6}} = 3\\,642 \\text{ N} $$R5 — minimum assembly preload $$ F_{M\\min} = 8\\,000 + (1-0.061) \\times 15\\,000 + 3\\,642 = 8\\,000 + 14\\,085 + 3\\,642 = 25\\,727 \\text{ N} $$R6 — maximum assembly preload $$ F_{M\\max} = 1.6 \\times 25\\,727 = 41\\,163 \\text{ N} $$R7 — assembly stress check $F_{MTab}$(M12-10.9, $\\mu = 0.14$) ≈ 57 000 N (Table A1)\n$$ F_{Mzul} \\approx 57\\,000 \\geq 41\\,163 \\quad → \\quad ✅ $$M12-10.9 is enough! (In contrast to the DSV example — for ESV the transverse force is zero and the sealing clamping force is smaller, so M12 passes.)\nR8–R12 check summary Step Check item Computed value Allowable value Safety factor Result R8 Service stress $F_{S\\max} = 57\\,000 + 0.061 \\times 15\\,000 = 57\\,915$ N $R_{p0.2} \\times A_S = 79\\,242$ N $S_F = 1.37$ ✅ R9 Fatigue $\\sigma_a = 0$ (static load) — — ✅ R10 Bearing pressure $p = 57\\,000 / A_{p\\min}$ $p_G$(GJL-250) ≈ 460 MPa needs verification ✅ R11 Engagement depth see below R12 Slip $F_Q = 0$ → not needed — — — R11 — engagement-depth check (ESV-specific) This is a check step unique to ESV (VDI 2230:2015, §5.5.5, Eq. R11/1).\nFor GJL-250 (grey cast iron, $R_m \\approx 250$ MPa), the standard Bild 36 gives:\n$$ m_{eff}/d \\approx 1.8 \\quad \\text{(steel bolt 10.9 into grey cast iron)} $$$$ m_{eff\\min} = 1.8 \\times 12 = 21.6 \\text{ mm} $$Design recommendation: an engagement depth of at least 22 mm, to make sure the thread is not stripped.\n[!IMPORTANT] ESV engagement depth varies greatly with the housing material\nSteel on steel: $m_{eff}/d \\approx 0.8 \\sim 1.0$ Steel on cast iron: $m_{eff}/d \\approx 1.5 \\sim 2.0$ Steel on aluminium alloy: $m_{eff}/d \\approx 2.0 \\sim 2.5$ The softer the housing material, the greater the required engagement depth (VDI 2230:2015, §5.5.5, Bild 36).\nR13 — tightening torque $$ M_A = 57\\,000 \\times [0.16 \\times 1.75 + 0.58 \\times 10.863 \\times 0.14 + \\frac{14.9}{2} \\times 0.14] $$$$ = 57\\,000 \\times [0.28 + 0.882 + 1.043] \\times 10^{-3} = 57\\,000 \\times 2.205 \\times 10^{-3} $$$$ M_A \\approx 126 \\text{ N·m} $$ DSV vs ESV comparison summary Item Article 8 DSV (M16-10.9) This article ESV (M12-10.9) Joint coefficient $w$ 1 2 Cone-angle formula Eq. 43 Eq. 42 $E_M$ (nut region) $E_S = 210\\,000$ $E_{BI} = 100\\,000$ $l_M$ (nut substitutional length) $0.4d$ $0.33d$ R11 engagement depth ❌ not applicable ✅ must be checked Governing design factor Transverse force $F_{KQ}$ Sealing $F_{KP}$ Data basis and accuracy statement All formulas in this article are from VDI 2230 Blatt 1:2015-11. The thread parameters are from DIN 13-1. The GJL-250 mechanical properties are from DIN EN 1561.\nDisclaimer: This article is for engineering teaching reference only.\n📚 Series navigation\n← Previous: Full Worked Example (DSV) ｜ Next: Eccentric Load and Bending-Moment Effect → ","permalink":"https://mechcalc.net/blog/en/posts/vdi2230-09-worked-example-esv/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/vdi2230-bolt-calculation?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eVDI 2230 Bolted Joint Calculation\u003c/a\u003e — The full 14-step calculation chain (R0–R13), with six strength checks.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003ch1 id=\"full-worked-example-tapped-thread-joint-esv\"\u003eFull worked example: tapped-thread joint (ESV)\u003c/h1\u003e\n\u003cblockquote\u003e\n\u003cp\u003eThis article is a counterpart to the \u003ca href=\"/blog/en/posts/vdi2230-08-worked-example-dsv/\"\u003eDSV example (article 8)\u003c/a\u003e\n. It highlights the differences unique to ESV: $w = 2$, $E_M = E_{BI}$, the cone-angle formula Eq. 42, and the R11 engagement-depth check.\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003ch2 id=\"engineering-problem-statement\"\u003eEngineering problem statement\u003c/h2\u003e\n\u003cp\u003eA bolt is screwed from the top into a cast-iron housing (grey cast iron GJL-250), fixing a steel cover plate.\u003c/p\u003e","title":"VDI 2230 (009): Full Worked Example (ESV)"},{"content":" 🧮 在线计算器：VDI 2230 Bolted Joint Calculation — The full 14-step calculation chain (R0–R13), with six strength checks.\nFull worked example: through-bolt joint (DSV) This article ties all the theory of the previous 7 articles into one complete calculation chain. Using a concrete engineering case, we go through every step of R0–R13.\nEngineering problem statement A steel flange joint using a single bolt to fix a cover plate. The design requirements are:\nParameter Value Note Joint type DSV (through-bolt) The bolt passes through two plates, fastened with a nut External axial force $F_A = 20\\,000$ N (static) Working load along the bolt axis External transverse force $F_Q = 5\\,000$ N Perpendicular to the bolt axis Clamp length $l_K = 40$ mm Total thickness of the two clamped parts Clamped-parts outer diameter $D_A = 44$ mm Equivalent outer diameter at the interface Clamped-parts material Steel, $E_P = 210\\,000$ MPa Operating temperature Room temperature (no thermal effect) $\\Delta F'_{Vth} = 0$ Interface friction coefficient $\\mu_T = 0.12$ Steel-on-steel interface Thread friction coefficient $\\mu_G = 0.12$ Bearing-surface friction coefficient $\\mu_K = 0.12$ Number of interfaces $q_F = 1$ (single shear) Load introduction factor $n = 0.5$ Force introduced near the interface Tightening method Precision torque wrench (μ known) R0 — preliminary diameter selection Based on experience or a quick pre-selection (see the pre-selection article ), preliminarily select:\nBolt: M12 × 1.75, property class 10.9 Thread geometry (DIN 13-1): $d = 12$ mm, $d_2 = 10.863$ mm, $d_3 = 9.853$ mm, $P = 1.75$ mm Cross-section areas: $A_S = 84.3$ mm², $A_{d_3} = 76.2$ mm², $A_N = 113.1$ mm² Bolt head: hexagon head DIN EN ISO 4017, $d_W = 16.6$ mm, $k = 7.5$ mm Through-hole diameter (DIN EN 20273, medium): $d_h = 13.5$ mm Material properties (DIN EN ISO 898-1): $R_{p0.2\\min} = 940$ MPa, $R_m = 1040$ MPa Limit check (Eq. R0/1):\n$$ G = h_{\\min} + d_W = 40 + 16.6 = 56.6 \\text{ mm} $$$c_T = D_A = 44 \\text{ mm} \u003c G = 56.6$ mm → ✅ satisfied\nR1 — tightening factor Precision torque wrench (friction coefficient known), from Table A8 (VDI 2230:2015):\n$$ \\alpha_A = 1.6 $$ R2 — required minimum clamping force This case has a transverse force to transmit by friction (Eq. R2/1):\n$$ F_{KQ} = \\frac{F_{Q\\max}}{q_F \\cdot \\mu_{T\\min}} = \\frac{5\\,000}{1 \\times 0.12} = 41\\,667 \\text{ N} $$No sealing requirement, no bending moment → $F_{KP} = 0$, $F_{KA} = 0$\n$$ F_{Kerf} = F_{KQ} = 41\\,667 \\text{ N} \\tag{R2/4} $$ R3 — elastic resiliences and force ratio Bolt resilience $\\delta_S$ Assume the bolt consists of: head + 20 mm plain shank + 20 mm free thread + engaged segment + nut region.\nSegment Length Cross-section area $\\delta_i$ (mm/N) Head $\\delta_{SK}$ $0.5 \\times 12 = 6$ mm $A_N = 113.1$ $2.53 \\times 10^{-7}$ Shank $\\delta_1$ 20 mm $A_N = 113.1$ $8.43 \\times 10^{-7}$ Free thread $\\delta_{Gew}$ 20 mm $A_{d_3} = 76.2$ $1.25 \\times 10^{-6}$ Engaged segment $\\delta_G$ $0.5 \\times 12 = 6$ mm $A_{d_3} = 76.2$ $3.75 \\times 10^{-7}$ Nut $\\delta_M$ $0.4 \\times 12 = 4.8$ mm $A_N = 113.1$ $2.02 \\times 10^{-7}$ $$ \\delta_S = (2.53 + 8.43 + 12.50 + 3.75 + 2.02) \\times 10^{-7} = 2.93 \\times 10^{-6} \\text{ mm/N} $$Clamped-parts resilience $\\delta_P$ First compute the cone angle (Eq. 43, 44, 45):\n$$ \\beta_L = l_K / d_W = 40 / 16.6 = 2.41 $$$$ y = D_A' / d_W = 44 / 16.6 = 2.65 $$$$ \\tan\\varphi_D = 0.362 + 0.032 \\ln(2.41/2) + 0.153 \\ln(2.65) = 0.362 + 0.006 + 0.149 = 0.517 $$Limiting diameter (Eq. 39, $w = 1$ for DSV):\n$$ D_{A,Gr} = 16.6 + 1 \\times 40 \\times 0.517 = 37.3 \\text{ mm} $$$D_A = 44 \u003e D_{A,Gr} = 37.3$ → use Eq. 40 (pure cone):\n$$ \\delta_P = \\frac{2 \\ln\\left[\\frac{(16.6+13.5)(16.6+1 \\times 40 \\times 0.517-13.5)}{(16.6-13.5)(16.6+1 \\times 40 \\times 0.517+13.5)}\\right]}{1 \\times 210\\,000 \\times \\pi \\times 13.5 \\times 0.517} $$$$ = \\frac{2 \\ln\\left[\\frac{30.1 \\times 23.8}{3.1 \\times 50.8}\\right]}{5\\,770\\,000} = \\frac{2 \\ln(4.55)}{5\\,770\\,000} = \\frac{2 \\times 1.515}{5\\,770\\,000} = 5.25 \\times 10^{-7} \\text{ mm/N} $$Force ratio Φ Concentric symmetric loading, taking $\\delta_{PZu} = 0$ (simplification) (Eq. R3/3):\n$$ \\Phi_n = n \\cdot \\frac{\\delta_P}{\\delta_S + \\delta_P} = 0.5 \\times \\frac{5.25 \\times 10^{-7}}{2.93 \\times 10^{-6} + 5.25 \\times 10^{-7}} = 0.5 \\times \\frac{0.525}{3.455} = 0.076 $$Force ratio $\\Phi = 0.076$ — the bolt takes only 7.6% of the external force!\nAdditional bolt force and plate unloading force:\n$$ F_{SA} = 0.076 \\times 20\\,000 = 1\\,520 \\text{ N} $$$$ F_{PA} = (1 - 0.076) \\times 20\\,000 = 18\\,480 \\text{ N} $$ R4 — preload change Embedding loss $f_Z$ (Table 5): bolt head 3 μm + nut 3 μm + 1 interface 3 μm = 9 μm\n$$ F_Z = \\frac{f_Z}{\\delta_S + \\delta_P} = \\frac{0.009}{3.455 \\times 10^{-6}} = 2\\,604 \\text{ N} \\tag{R4/1} $$Thermal expansion Room-temperature case → $\\Delta F'_{Vth} = 0$\nR5 — minimum assembly preload $$ F_{M\\min} = 41\\,667 + (1 - 0.076) \\times 20\\,000 + 2\\,604 + 0 = 62\\,751 \\text{ N} \\tag{R5/1} $$ R6 — maximum assembly preload $$ F_{M\\max} = 1.6 \\times 62\\,751 = 100\\,402 \\text{ N} \\tag{R6/1} $$ R7 — assembly stress check Allowable assembly preload (Eq. R7/2, $\\nu = 0.9$, $d_0 = d_S \\approx 10.36$ mm):\n$$ F_{Mzul} = 84.3 \\times \\frac{0.9 \\times 940}{\\sqrt{1 + 3\\left[\\frac{3}{2} \\times \\frac{10.863}{10.36}\\left(\\frac{1.75}{\\pi \\times 10.863} + 1.155 \\times 0.12\\right)\\right]^2}} $$$$ = 84.3 \\times \\frac{846}{\\sqrt{1 + 3 \\times [1.574 \\times (0.0513 + 0.1386)]^2}} $$$$ = 84.3 \\times \\frac{846}{\\sqrt{1 + 3 \\times 0.0898}} = 84.3 \\times \\frac{846}{\\sqrt{1.269}} = 84.3 \\times 751 = 63\\,309 \\text{ N} $$ In Table A1 ($\\mu_G = \\mu_K = 0.12$, $\\nu = 0.9$), the $F_{MTab}$ for M12-10.9 is $\\approx 63\\,000$ N, consistent with the calculation.\nCheck (Eq. R7/3):\n$$ F_{Mzul} = 63\\,309 \\text{ N} \\quad \\text{vs} \\quad F_{M\\max} = 100\\,402 \\text{ N} $$$$ 63\\,309 \u003c 100\\,402 \\quad → \\quad ❌ \\text{ not satisfied!} $$ Iteration: upgrade to M16 M12 is not enough; upgrade to M16 × 2.0, property class 10.9:\nParameter M16 × 2.0 value Source $d_2$ 14.701 mm DIN 13-1 $d_3$ 13.546 mm DIN 13-1 $A_S$ 157 mm² DIN 13-1 $A_{d_3}$ 144.1 mm² $\\pi/4 \\times 13.546^2$ $A_N$ 201.1 mm² $\\pi/4 \\times 16^2$ $d_W$ 21.9 mm DIN EN ISO 4017 $d_h$ 17.5 mm DIN EN 20273 $F_{MTab}$ ≈ 119 kN Table A1, $\\mu = 0.12$ Recompute from R2 (with $F_{Kerf}$ unchanged):\nδS (M16):\nSegment Length Cross-section area $\\delta_i$ Head $0.5 \\times 16 = 8$ mm $A_N = 201.1$ $1.89 \\times 10^{-7}$ Shank 16 mm $A_N = 201.1$ $3.79 \\times 10^{-7}$ Free thread 24 mm $A_{d_3} = 144.1$ $7.93 \\times 10^{-7}$ Engaged segment $0.5 \\times 16 = 8$ mm $A_{d_3} = 144.1$ $2.64 \\times 10^{-7}$ Nut $0.4 \\times 16 = 6.4$ mm $A_N = 201.1$ $1.52 \\times 10^{-7}$ Total $1.78 \\times 10^{-6}$ δP (M16):\n$\\beta_L = 40/21.9 = 1.83$, $y = 44/21.9 = 2.01$\n$\\tan\\varphi_D = 0.362 + 0.032\\ln(0.91) + 0.153\\ln(2.01) = 0.362 - 0.003 + 0.107 = 0.466$\n$D_{A,Gr} = 21.9 + 40 \\times 0.466 = 40.5$ mm\n$D_A = 44 \u003e 40.5$ → Eq. 40:\n$$ \\delta_P = \\frac{2\\ln\\left[\\frac{(21.9+17.5)(21.9+40 \\times 0.466-17.5)}{(21.9-17.5)(21.9+40 \\times 0.466+17.5)}\\right]}{210\\,000 \\times \\pi \\times 17.5 \\times 0.466} = 3.72 \\times 10^{-7} \\text{ mm/N} $$Φ (M16):\n$$ \\Phi = 0.5 \\times \\frac{3.72 \\times 10^{-7}}{1.78 \\times 10^{-6} + 3.72 \\times 10^{-7}} = 0.5 \\times 0.173 = 0.086 $$R4 (M16): $F_Z = 0.009 / (2.15 \\times 10^{-6}) = 4\\,186$ N\nR5: $F_{M\\min} = 41\\,667 + 0.914 \\times 20\\,000 + 4\\,186 = 64\\,133$ N\nR6: $F_{M\\max} = 1.6 \\times 64\\,133 = 102\\,613$ N\nR7: $F_{Mzul}(M16) \\approx 119\\,000$ N (Table A1)\n$$ 119\\,000 \\geq 102\\,613 \\quad → \\quad ✅ \\text{ satisfied!} $$ R8–R12 check summary (M16-10.9) Step Check item Computed value Allowable value Safety factor Result R8 Service stress $F_{S\\max} = 119\\,000 + 0.086 \\times 20\\,000 = 120\\,720$ N $R_{p0.2} \\times A_S = 147\\,580$ N $S_F = 1.22$ ✅ R9 Fatigue $\\sigma_a = 0$ (static load) — — ✅ R10 Bearing pressure $p = 119\\,000 / A_{p\\min}$ $p_G$ needs table ✅ R11 Engagement depth Not applicable for DSV (through-bolt) — — — R12 Slip $F_{KR\\min} = 119\\,000/1.6 - 0.914 \\times 20\\,000 - 4\\,186 = 52\\,064$ N $F_{KQerf} = 41\\,667$ N $S_G = 1.25$ ✅ R13 — tightening torque (Eq. R13/1, M16, $\\mu_G = \\mu_K = 0.12$, $D_{Km} \\approx 19.6$ mm):\n$$ M_A = 119\\,000 \\times [0.16 \\times 2.0 + 0.58 \\times 14.701 \\times 0.12 + \\frac{19.6}{2} \\times 0.12] $$$$ = 119\\,000 \\times [0.32 + 1.023 + 1.176] = 119\\,000 \\times 2.519 \\times 10^{-3} \\text{ m} $$$$ M_A \\approx 300 \\text{ N·m} $$ Result summary Item Value Final selection M16 × 2.0 — 10.9 Minimum assembly preload $F_{M\\min}$ 64 133 N Maximum assembly preload $F_{M\\max}$ 102 613 N Allowable assembly preload $F_{Mzul}$ ≈ 119 000 N Force ratio $\\Phi$ 0.086 Safety factor $S_F$ (service) 1.22 Safety factor $S_G$ (slip) 1.25 Tightening torque $M_A$ ≈ 300 N·m [!IMPORTANT] The lesson from M12 to M16 On the surface, the 20 kN external force in this example is not large, but the clamping force needed for transverse-force transfer $F_{KQ} = 41.7$ kN is the governing factor of the design. Although M12 has enough tensile strength, its $F_{Mzul}$ is not enough to cover $F_{M\\max}$. This is exactly the value of the systematic VDI 2230 method — it reveals the weak link that intuition easily overlooks.\nData basis and accuracy statement All formulas in this article are from VDI 2230 Blatt 1:2015-11. The thread parameters are from DIN 13-1, the bolt-head geometry from DIN EN ISO 4017, and the through-hole diameter from DIN EN 20273.\nDisclaimer: This article is for engineering teaching reference only. Some simplifying assumptions are used in the calculation (concentric symmetric, $\\delta_{PZu} = 0$). In real engineering use, all parameters should be verified against the specific conditions.\n📚 Series navigation\n← Previous: Strength Checks R7–R13 ｜ Next: Full Worked Example (ESV) → ","permalink":"https://mechcalc.net/blog/en/posts/vdi2230-08-worked-example-dsv/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/vdi2230-bolt-calculation?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eVDI 2230 Bolted Joint Calculation\u003c/a\u003e — The full 14-step calculation chain (R0–R13), with six strength checks.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003ch1 id=\"full-worked-example-through-bolt-joint-dsv\"\u003eFull worked example: through-bolt joint (DSV)\u003c/h1\u003e\n\u003cblockquote\u003e\n\u003cp\u003eThis article ties all the theory of the previous 7 articles into one complete calculation chain. Using a concrete engineering case, we go through every step of R0–R13.\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003ch2 id=\"engineering-problem-statement\"\u003eEngineering problem statement\u003c/h2\u003e\n\u003cp\u003eA steel flange joint using a single bolt to fix a cover plate. The design requirements are:\u003c/p\u003e","title":"VDI 2230 (008): Full Worked Example (DSV)"},{"content":" 🧮 在线计算器：VDI 2230 Bolted Joint Calculation — The full 14-step calculation chain (R0–R13), with six strength checks.\nSix strength checks and the tightening torque: R7–R13 R1–R6 answered \u0026ldquo;how much preload to apply\u0026rdquo;; R7–R13 answer \u0026ldquo;can the bolt take it\u0026rdquo;.\n1. R7 — assembly stress check (§5.5.1) This is the first gate: the bolt does not exceed yield strength during assembly.\nThe standard allows the use of a fraction of the yield strength (usually $\\nu = 0.9$, i.e. 90%); the allowable assembly comparison stress is (VDI 2230:2015, §5.5.1, Eq. R7/1):\n$$ \\sigma_{red,Mzul} = \\nu \\cdot R_{p0.2\\min} \\tag{R7/1} $$The allowable assembly preload $F_{Mzul}$ accounts for the combined stress of tension and torsion that the bolt carries at the same time during tightening (VDI 2230:2015, §5.5.1, Eq. R7/2):\n$$ F_{Mzul} = A_0 \\cdot \\frac{\\nu \\cdot R_{p0.2\\min}}{\\sqrt{1 + 3 \\left[\\frac{3}{2} \\frac{d_2}{d_0}\\left(\\frac{P}{\\pi \\cdot d_2} + 1.155 \\mu_{G\\min}\\right)\\right]^2}} \\tag{R7/2} $$Check condition (Eq. R7/3):\n$$ F_{Mzul} \\geq F_{M\\max} \\tag{R7/3} $$If it is not met, the standard requires explicitly (VDI 2230:2015, §5.5.1, p.23):\n\u0026ldquo;If the requirement is not met, a larger bolt nominal diameter is to be selected and the calculation repeated from R2.\u0026rdquo;\n2. R8 — service stress check (§5.5.2) The total bolt force under service contains the preload and the additional bolt force (VDI 2230:2015, §5.5.2, Eq. R8/1):\n$$ F_{S\\max} = F_{Mzul} + \\Phi_{en}^{*} \\cdot F_{A\\max} - \\Delta F_{Vth} \\tag{R8/1} $$ [!IMPORTANT] Sign convention for the thermal effect Note: here, if $\\Delta F_{Vth} \u003e 0$ (the thermal effect reduces the preload), set $\\Delta F_{Vth} = 0$ — this is the conservative direction, making sure the check is based on the maximum possible bolt force (VDI 2230:2015, R8, p.24).\nThe maximum tensile stress and torsional stress (Eq. R8/2, R8/3):\n$$ \\sigma_{z\\max} = F_{S\\max} / A_0 \\tag{R8/2} $$$$ \\tau_{\\max} = M_G / W_P \\tag{R8/3} $$The standard recommends a reduction factor $k_\\tau = 0.5$ to account for the decay of torsional stress during operation — because embedding and vibration partly release the torque introduced at assembly (VDI 2230:2015, §5.5.2, Eq. R8/4):\n$$ \\sigma_{red,B} = \\sqrt{\\sigma_{z\\max}^2 + 3(k_\\tau \\cdot \\tau_{\\max})^2} \\tag{R8/4} $$Check condition and safety factor (Eq. R8/5):\n$$ S_F = \\frac{R_{p0.2\\min}}{\\sigma_{red,B}} \\geq 1.0 \\tag{R8/5-2} $$3. R9 — fatigue stress check (§5.5.3) If the joint carries a dynamic load (the external force alternates between $F_{A\\max}$ and $F_{A\\min}$), fatigue must be checked.\nThe stress amplitude (VDI 2230:2015, §5.5.3, Eq. R9/1):\n$$ \\sigma_a = \\frac{F_{SAo} - F_{SAu}}{2 A_S} \\tag{R9/1} $$where $F_{SAo}$ and $F_{SAu}$ are the maximum and minimum additional bolt forces.\nThe standard gives empirical formulas for the bolt fatigue limit ($N_D \\geq 2 \\times 10^6$):\nRolled after heat treatment (SV) (VDI 2230:2015, Eq. R9/5-1):\n$$ \\sigma_{ASV} = 0.85 \\cdot (150/d + 45) \\tag{R9/5-1} $$Rolled before heat treatment (SG) (Eq. R9/5-2):\n$$ \\sigma_{ASG} = (2 - F_{Sm}/F_{0.2\\min}) \\cdot \\sigma_{ASV} \\tag{R9/5-2} $$Check condition (Eq. R9/3, R9/4):\n$$ \\sigma_a \\leq \\sigma_{AS} \\tag{R9/3} $$$$ S_D = \\frac{\\sigma_{AS}}{\\sigma_a} \\geq 1.0 \\quad (\\text{recommended} \\geq 1.2) \\tag{R9/4} $$4. R10 — bearing pressure check (§5.5.4) The bearing pressure under the bolt head / nut must not exceed the limiting surface pressure $p_G$ of the clamped-part material, otherwise it causes creep and preload loss.\nThe standard explains (VDI 2230:2015, §5.5.4, p.26):\n\u0026ldquo;Surface pressures which cause creep (time-dependent plastic flowing) in conjunction with a loss of preload should not become effective.\u0026rdquo;\nAssembly state (Eq. R10/1):\n$$ p_{M\\max} = F_{Mzul} / A_{p\\min} \\leq p_G \\tag{R10/1} $$Service state (Eq. R10/2):\n$$ p_{B\\max} = (F_{V\\max} + F_{SA\\max} - \\Delta F_{Vth}) / A_{p\\min} \\leq p_G \\tag{R10/2} $$Safety factor (Eq. R10/4):\n$$ S_p = p_G / p_{M/B\\max} \\geq 1.0 \\tag{R10/4} $$5. R11 — minimum engagement depth (§5.5.5) For a tapped-thread joint (ESV), the thread engagement depth must be sufficient to avoid stripping the internal thread (VDI 2230:2015, §5.5.5, Eq. R11/1):\n$$ F_{mS} \\leq \\min(F_{mGM},\\; F_{MGS}) \\tag{R11/1} $$That is, the maximum breaking force of the bolt must be less than the stripping force of the internal thread or the bolt thread.\nIn Bild 36 the standard gives a chart of the minimum relative engagement depth $m_{eff}/d$ for standard threads M4–M39; typically:\nSteel on steel: $m_{eff} \\approx 0.8d \\sim 1.0d$ Steel on cast iron / aluminium: $m_{eff} \\approx 1.5d \\sim 2.5d$ 6. R12 — slip safety (§5.5.6) The transverse force must be transmitted through the friction at the interface. The minimum residual clamping force (VDI 2230:2015, §5.5.6, Eq. R12/1):\n$$ F_{KR\\min} = \\frac{F_{Mzul}}{\\alpha_A} - (1 - \\Phi_{en}^{*}) F_{A\\max} - F_Z - \\Delta F_{Vth} \\tag{R12/1} $$Slip safety factor (Eq. R12/4):\n$$ S_G = \\frac{F_{KR\\min}}{F_{KQerf}} \u003e 1.0 \\tag{R12/4} $$Recommended values from the standard (VDI 2230:2015, §5.5.6, p.28):\nStatic load: $S_G \\geq 1.2$ Alternating transverse force: $S_G \\geq 1.8$ If $S_G$ is not met, the standard also provides a shear check as a backup — checking whether the bolt shank would be sheared off at the interface (Eq. R12/5–R12/7).\n7. R13 — tightening torque (§5.4.3) The final step turns the design result into installation parameters the workshop can carry out (VDI 2230:2015, §5.4.3, Eq. R13/1):\n$$ M_A = F_{Mzul} \\left[0.16 \\cdot P + 0.58 \\cdot d_2 \\cdot \\mu_{G\\min} + \\frac{D_{Km}}{2} \\cdot \\mu_{K\\min}\\right] \\tag{R13/1} $$The three terms represent, respectively:\nTerm Physical meaning $0.16 \\cdot P$ helix-angle effect (pitch contribution) $0.58 \\cdot d_2 \\cdot \\mu_{G\\min}$ thread-flank friction torque $\\frac{D_{Km}}{2} \\cdot \\mu_{K\\min}$ bolt-head / nut bearing-surface friction torque [!NOTE] The make-up of the torque In a typical case, about 50% of the total tightening torque is spent on the bearing-surface friction, about 40% on the thread friction, and only about 10% is turned into effective preload. This is why the friction coefficient is so critical to preload design.\n8. The logical hierarchy of the six checks R7 assembly stress check → bolt does not exceed yield at assembly (static, one-off) R8 service stress check → does not exceed yield in service (static, continuous) R9 fatigue stress check → no fatigue fracture under alternating load (dynamic, long-term) R10 bearing pressure check → no crushing under the head / nut (material, creep) R11 engagement depth check → internal thread not stripped (geometry, ESV only) R12 slip safety check → interface does not slip (friction, transverse force) R13 tightening torque → turned into workshop parameters (output) If R7 is not met → select a larger bolt and recompute from R2 If R8–R12 are not met → adjust bolt size / property class / tightening method / friction coefficient / external force\nData basis and accuracy statement All formulas in this article are from VDI 2230 Blatt 1:2015-11, §4.2 calculation summary and §5.5. The recommended safety-factor values are from the respective sections of the standard.\nDisclaimer: This article is for engineering teaching reference only.\n📚 Series navigation\n← Previous: Preload Design R1–R6 ｜ Next: Full Worked Example (DSV) → ","permalink":"https://mechcalc.net/blog/en/posts/vdi2230-07-strength-check-r7-r13/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/vdi2230-bolt-calculation?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eVDI 2230 Bolted Joint Calculation\u003c/a\u003e — The full 14-step calculation chain (R0–R13), with six strength checks.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003ch1 id=\"six-strength-checks-and-the-tightening-torque-r7r13\"\u003eSix strength checks and the tightening torque: R7–R13\u003c/h1\u003e\n\u003cblockquote\u003e\n\u003cp\u003eR1–R6 answered \u0026ldquo;how much preload to apply\u0026rdquo;; R7–R13 answer \u0026ldquo;can the bolt take it\u0026rdquo;.\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003ch2 id=\"1-r7--assembly-stress-check-551\"\u003e1. R7 — assembly stress check (§5.5.1)\u003c/h2\u003e\n\u003cp\u003eThis is the first gate: \u003cstrong\u003ethe bolt does not exceed yield strength during assembly\u003c/strong\u003e.\u003c/p\u003e\n\u003cp\u003eThe standard allows the use of a fraction of the yield strength (usually $\\nu = 0.9$, i.e. 90%); the allowable assembly comparison stress is (VDI 2230:2015, §5.5.1, Eq. R7/1):\u003c/p\u003e","title":"VDI 2230 (007): The Six Strength Checks R7–R13"},{"content":" 🧮 在线计算器：VDI 2230 Bolted Joint Calculation — The full 14-step calculation chain (R0–R13), with six strength checks.\nPreload design: from functional requirement to assembly preload (R1–R6) The 14-step VDI 2230 calculation chain splits into a design part (R0–R6) and a verification part (R7–R13). This article explains the core steps of the design part.\n1. R0 — preliminary diameter selection The task of R0 is to preliminarily fix the nominal diameter $d$ based on experience or a simplified method (such as the Kübler equation), and to check the interface limiting dimension (VDI 2230:2015, §4.2, Eq. R0/1, R0/2):\n$$ G = h_{\\min} + d_W \\tag{R0/1} $$If the interface dimension $c_T$ exceeds $G$, the standard\u0026rsquo;s cone-resilience formula produces a notable error and FEM verification should be considered.\n2. R1 — tightening factor αA The tightening factor $\\alpha_A$ describes the ratio of the maximum to the minimum preload that the same tightening method can produce under the same conditions (VDI 2230:2015, §5.4.3, Eq. R1/1):\n$$ \\alpha_A = \\frac{F_{M\\max}}{F_{M\\min}} \\tag{R1/1} $$The standard gives reference values for different tightening methods in Table A8:\nTightening method $\\alpha_A$ Source Manual torque wrench 2.5–4.0 Table A8 Precision torque wrench (μ known) 1.4–1.6 Table A8 Yield-point-controlled tightening 1.0 Table A8 Angle-controlled tightening 1.0 Table A8 Physical meaning: the larger $\\alpha_A$, the greater the uncertainty of the preload. The design must check strength against the maximum preload $F_{M\\max}$ and function against the minimum preload $F_{M\\min}$ — both ends must be safe.\n3. R2 — required minimum clamping force FKerf $F_{Kerf}$ is the starting point of the whole design calculation — it answers \u0026ldquo;at least how much residual clamping force does the interface need to meet the functional requirement\u0026rdquo;. The standard gives three functional sources (VDI 2230:2015, §4.2, Eq. R2/1–R2/4):\n3.1 Transmitting transverse force and/or torque $$ F_{KQ} = \\frac{F_{Q\\max}}{q_F \\cdot \\mu_{T\\min}} + \\frac{M_{Y\\max}}{q_M \\cdot r_a \\cdot \\mu_{T\\min}} \\tag{R2/1} $$where $q_F$ and $q_M$ are the numbers of interfaces taking part in slipping (single shear $q = 1$, double shear $q = 2$), and $\\mu_{T\\min}$ is the minimum interface friction coefficient.\n3.2 Sealing function $$ F_{KP} = A_D \\cdot p_{i\\max} \\tag{R2/2} $$where $A_D$ is the sealing area and $p_{i\\max}$ is the maximum medium pressure.\n3.3 Preventing opening $$ F_{KA} \\tag{R2/3} $$The specific value of $F_{KA}$ depends on the eccentric geometry and the bending-moment effect, see §5.3.2.\n3.4 Total requirement $$ F_{Kerf} \\geq \\max\\left(F_{KQ};\\; F_{KP} + F_{KA}\\right) \\tag{R2/4} $$ [!NOTE] Choosing the design starting point Many people get stuck at R2 — because they do not know what value $F_{Kerf}$ should take. The standard\u0026rsquo;s logic is: start from the function. If your joint must both transmit transverse force and seal, take the larger of the two. If none of the three items above applies (pure axial tension with no sealing), then $F_{Kerf} = 0$.\n4. R3 — force ratio Φ (external-force distribution) This step was discussed in detail in the spring model article and the bolt-resilience article . A recap of the key formulas:\nConcentric symmetric (VDI 2230:2015, Eq. R3/3):\n$$ \\Phi_n = n \\cdot \\frac{\\delta_P + \\delta_{PZu}}{\\delta_S + \\delta_P} \\tag{R3/3} $$Eccentric (VDI 2230:2015, Eq. R3/4):\n$$ \\Phi_{en}^{*} = n \\cdot \\frac{\\delta_P^{**} + \\delta_{PZu}}{\\delta_S + \\delta_P^{*}} \\tag{R3/4} $$The output of this step also includes the additional bolt force and the plate unloading force (Eq. R3/1, R3/2):\n$$ F_{SA} = \\Phi \\cdot F_A \\tag{R3/1} $$$$ F_{PA} = (1 - \\Phi) \\cdot F_A \\tag{R3/2} $$5. R4 — preload change (the \u0026ldquo;preload thieves\u0026rdquo;) 5.1 Embedding loss $$ F_Z = \\frac{f_Z}{\\delta_S + \\delta_P} \\tag{R4/1} $$The standard explains (VDI 2230:2015, §5.4.2, R4):\n\u0026ldquo;The guide values for the amounts of embedding $f_Z$ in the case of bolts, nuts and compact clamped parts made of steel can be taken from Table 5.\u0026rdquo;\nTypical values from Table 5: under the bolt head ≈ 3 μm, nut / internal thread ≈ 3 μm, each internal interface ≈ 3 μm.\n5.2 Thermal expansion difference $$ \\Delta F'_{Vth} = \\frac{l_K \\cdot (\\alpha_S \\cdot \\Delta T_S - \\alpha_P \\cdot \\Delta T_P)}{\\delta_S \\frac{E_{SRT}}{E_{ST}} + \\delta_P \\frac{E_{PRT}}{E_{PT}}} \\tag{R4/2} $$Symbols: $\\alpha_S, \\alpha_P$ are the thermal expansion coefficients, $\\Delta T$ is the temperature rise, and the subscripts RT/T denote the elastic-modulus correction at room / operating temperature.\n[!IMPORTANT] A key point to note The standard states clearly in R5: if $\\Delta F'_{Vth} \u003c 0$ (i.e. the thermal effect increases the preload), then when computing $F_{M\\min}$ set $\\Delta F'_{Vth} = 0$. This is a conservative treatment — do not rely on the thermal effect to keep the clamping force (VDI 2230:2015, R5, p.25).\n6. R5 — minimum assembly preload $$ F_{M\\min} = F_{Kerf} + (1 - \\Phi_{en}^{*}) \\cdot F_{A\\max} + F_Z + \\Delta F'_{Vth} \\tag{R5/1} $$This is the version of the VDI 2230 main equation (Eq. 16) with the $\\alpha_A$ factor removed. Its logic is (VDI 2230:2015, §5.4.3, R5):\n\u0026ldquo;The required minimum assembly preload is obtained while taking into account preload changes and assuming the greatest possible relief of the joint.\u0026rdquo;\nRead it right to left: the preload must be large enough that, after subtracting the external-force unloading $(1-\\Phi)F_A$, the embedding loss $F_Z$ and the thermal change $\\Delta F'_{Vth}$, a clamping force of $F_{Kerf}$ still remains.\n7. R6 — maximum assembly preload $$ F_{M\\max} = \\alpha_A \\cdot F_{M\\min} \\tag{R6/1} $$Source (VDI 2230:2015, §5.4.3, R6):\n\u0026ldquo;Taking into account (R1/1), the possible maximum assembly preload is calculated.\u0026rdquo;\n$F_{M\\max}$ is the maximum preload that can occur at assembly (because $\\alpha_A \u003e 1$); the later strength checks (R7–R8) are all based on this worst case.\n8. The design logic chain of R0–R6 R0 → preliminary d R2 → FKerf (functional requirement) R3 → Φ (resilience ratio → force distribution) needs δS(R3→§5.1.1), δP(R3→§5.1.2) R4 → FZ, ΔF\u0026#39;Vth (preload loss) needs δS, δP, fZ(Table 5), thermal parameters R5 → FMmin = FKerf + (1-Φ)FA + FZ + ΔF\u0026#39;Vth R1 → αA (tightening scatter) R6 → FMmax = αA · FMmin ↓ enter the checks R7–R13 The key insight: R1–R6 is a from-requirement-to-preload derivation chain. Its core idea is \u0026ldquo;work backward from the functional bottom line of the interface\u0026rdquo; — the exact opposite of the traditional \u0026ldquo;select the bolt first, then check\u0026rdquo; approach.\nData basis and accuracy statement All formulas in this article are from VDI 2230 Blatt 1:2015-11, §4.2 calculation summary and §5.4.\nDisclaimer: This article is for engineering teaching reference only.\n📚 Series navigation\n← Previous: Clamped-Parts Resilience δP ｜ Next: Strength Checks R7–R13 → ","permalink":"https://mechcalc.net/blog/en/posts/vdi2230-06-preload-design-r1-r6/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/vdi2230-bolt-calculation?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eVDI 2230 Bolted Joint Calculation\u003c/a\u003e — The full 14-step calculation chain (R0–R13), with six strength checks.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003ch1 id=\"preload-design-from-functional-requirement-to-assembly-preload-r1r6\"\u003ePreload design: from functional requirement to assembly preload (R1–R6)\u003c/h1\u003e\n\u003cblockquote\u003e\n\u003cp\u003eThe 14-step VDI 2230 calculation chain splits into a \u003cstrong\u003edesign part\u003c/strong\u003e (R0–R6) and a \u003cstrong\u003everification part\u003c/strong\u003e (R7–R13). This article explains the core steps of the design part.\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003ch2 id=\"1-r0--preliminary-diameter-selection\"\u003e1. R0 — preliminary diameter selection\u003c/h2\u003e\n\u003cp\u003eThe task of R0 is to \u003cstrong\u003epreliminarily fix the nominal diameter $d$\u003c/strong\u003e based on experience or a simplified method (such as the Kübler equation), and to check the interface limiting dimension (VDI 2230:2015, §4.2, Eq. R0/1, R0/2):\u003c/p\u003e","title":"VDI 2230 (006): Preload Design R1–R6"},{"content":" 🧮 在线计算器：VDI 2230 Bolted Joint Calculation — The full 14-step calculation chain (R0–R13), with six strength checks.\nClamped-parts resilience δP and the Rötscher cone model The previous article settled how \u0026ldquo;soft\u0026rdquo; the bolt is ($\\delta_S$); this one answers how \u0026ldquo;soft\u0026rdquo; the clamped parts are ($\\delta_P$) — the other half of the force ratio $\\Phi$ calculation.\n1. Why is δP harder to compute than δS? The bolt resilience $\\delta_S$ can be broken simply into series cylindrical segments that add up. But the clamped parts are a completely different case — the standard states (VDI 2230:2015, §5.1.2, p.45):\n\u0026ldquo;The calculation of the elastic resilience $\\delta_P$ of the parts preloaded by the bolt [\u0026hellip;] proves to be difficult on account of the three-dimensional stress and deformation state which forms when preload is applied.\u0026rdquo;\nThe preload spreads outward from the bearing surface under the bolt head; the compression zone widens gradually from the bearing surface toward the interface, in a shape close to a paraboloid of revolution (Rotationsparaboloid) (VDI 2230:2015, §5.1.2, p.45, [7; 9; 10]).\nVDI 2230\u0026rsquo;s solution is: use an equivalent deformation cone (Ersatzverformungskegel) to replace the real paraboloid compression zone, so that it has the same elastic resilience — this is the so-called Rötscher cone model.\n2. Core formula: computing δP The general integral form of the clamped-parts resilience is (VDI 2230:2015, §5.1.2, Eq. 38):\n$$ \\delta_P = \\int_{y=0}^{y=l_K} \\frac{\\mathrm{d}y}{E(y) \\cdot A(y)} \\tag{38} $$Integrating over the cone, the standard gives a closed-form analytical solution.\n2.1 Case one: $D_A \\geq D_{A,Gr}$ (cone does not reach the outer wall) The cone can fully develop, unconstrained by the outer wall (VDI 2230:2015, §5.1.2.1, Eq. 40):\n$$ \\delta_P = \\frac{2 \\ln \\left[ \\frac{(d_W + d_h)(d_W + w \\cdot l_K \\cdot \\tan\\varphi - d_h)}{(d_W - d_h)(d_W + w \\cdot l_K \\cdot \\tan\\varphi + d_h)} \\right]}{w \\cdot E_P \\cdot \\pi \\cdot d_h \\cdot \\tan\\varphi} \\tag{40} $$2.2 Case two: $d_W \u003c D_A \u003c D_{A,Gr}$ (cone + sleeve combination) After the cone develops to the outer wall, the remaining part becomes a hollow sleeve of constant wall thickness (VDI 2230:2015, §5.1.2.1, Eq. 41):\n$$ \\delta_P = \\frac{\\frac{2}{w \\cdot d_h \\cdot \\tan\\varphi} \\ln\\left[\\frac{(d_W + d_h)(D_A - d_h)}{(d_W - d_h)(D_A + d_h)}\\right] + \\frac{4}{D_A^2 - d_h^2}\\left[l_K - \\frac{D_A - d_W}{w \\cdot \\tan\\varphi}\\right]}{E_P \\cdot \\pi} \\tag{41} $$2.3 The decision boundary: limiting diameter $D_{A,Gr}$ The standard uses a limiting diameter to decide which formula to use (VDI 2230:2015, §5.1.2, Eq. 39):\n$$ D_{A,Gr} = d_W + w \\cdot l_K \\cdot \\tan\\varphi \\tag{39} $$where the joint coefficient is:\nJoint type $w$ Source ESV (tapped-thread) 2 §5.1.2 DSV (through-bolt) 1 §5.1.2 $D_A \\geq D_{A,Gr}$: use Eq. 40 (pure cone) $d_W \u003c D_A \u003c D_{A,Gr}$: use Eq. 41 (cone + sleeve) $d_W \\geq D_A$: sleeve only (hollow cylinder) 3. Cone angle φ — not a constant! This is the most subtle part of the VDI 2230 resilience model. The cone angle $\\varphi$ is not a fixed value but a variable determined by the geometry of the clamped parts (VDI 2230:2015, §5.1.2.1, Eq. 42-45):\nTapped-thread joint ESV / TTJ:\n$$ \\tan\\varphi_E = 0.348 + 0.013 \\ln\\beta_L + 0.193 \\ln y \\tag{42} $$Through-bolt joint DSV / TBJ:\n$$ \\tan\\varphi_D = 0.362 + 0.032 \\ln(\\beta_L / 2) + 0.153 \\ln y \\tag{43} $$where:\n$$ \\beta_L = l_K / d_W \\tag{44} $$$$ y = D_A' / d_W \\tag{45} $$The standard notes that, as an approximation, $\\tan\\varphi = 0.6$ can be taken within the following dimension-ratio ranges (VDI 2230:2015, §5.1.2.1, p.51):\nESV: $\\beta_L = 4 \\dots 6$, $y = 2.5 \\dots 4$ DSV: $\\beta_L = 0.5 \\dots 4$, $y = 4 \\dots 6$ \u0026ldquo;In this case, the maximum error when calculating the plate resilience is about 5%.\u0026rdquo; (VDI 2230:2015, §5.1.2.1, p.51)\n4. Handling multi-layer plates When the clamped parts are made of multiple layers of different materials (different $E_P$), the standard requires the cone and sleeve to be further split into sub-segments matching each layer (VDI 2230:2015, §5.1.2.1, Eq. 52-53):\nRecursion of the bearing diameter of each cone sub-segment (Eq. 52):\n$$ d_{W,i} = d_W + 2 \\cdot \\tan\\varphi \\cdot \\sum_{i=1}^{j} l_{i-1} \\tag{52} $$The total resilience is the sum of the sub-segments (Eq. 53):\n$$ \\delta_P = \\sum_{i=1}^{j} \\delta_{Pi}^V + \\sum_{i=j+1}^{m} \\delta_{Pi}^H \\tag{53} $$5. Summary of DSV vs ESV differences in the δP calculation Parameter DSV (through-bolt) ESV (tapped-thread) Source Joint coefficient $w$ 1 2 §5.1.2, Eq. 39 Cone-angle formula Eq. 43 Eq. 42 §5.1.2.1 Number of deformation cones 2 (one at head + one at nut) 1 (one at head) Bild 8, 9 Nut region $E_M$ $E_S$ (bolt material) $E_{BI}$ (clamped-part material) §5.1.1.1 The ESV $w = 2$ means only one full cone extends from the bolt head to the interface, while the DSV $w = 1$ means the head and the nut each contribute one cone, meeting in the middle of the clamp length.\n6. Numerical sensitivity note The logarithmic term in Eq. 40 has $(d_W - d_h)$ in the denominator. When $d_W \\approx d_h$ (the bearing-surface outer diameter is close to the through-hole diameter), this difference approaches zero and can cause numerical instability. In engineering practice this means: when the bearing surface is too small, the clamped-parts resilience rises sharply — which is exactly the expected physical behaviour (the compression zone degenerates into an extremely thin ring).\nData basis and accuracy statement All formulas in this article are from VDI 2230 Blatt 1:2015-11, §5.1.2. The accuracy of the cone-angle regression formulas (Eq. 42/43) is about ±5% (§5.1.2.1).\nDisclaimer: This article is for engineering teaching reference only. The final responsibility for engineering safety verification rests with the user.\n📚 Series navigation\n← Previous: Bolt Elastic Resilience δS ｜ Next: Force Ratio Φ and Preload Design R1–R6 → ","permalink":"https://mechcalc.net/blog/en/posts/vdi2230-05-clamped-parts-resilience-cone/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/vdi2230-bolt-calculation?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eVDI 2230 Bolted Joint Calculation\u003c/a\u003e — The full 14-step calculation chain (R0–R13), with six strength checks.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003ch1 id=\"clamped-parts-resilience-δp-and-the-rötscher-cone-model\"\u003eClamped-parts resilience δP and the Rötscher cone model\u003c/h1\u003e\n\u003cblockquote\u003e\n\u003cp\u003eThe previous article settled how \u0026ldquo;soft\u0026rdquo; the bolt is ($\\delta_S$); this one answers how \u0026ldquo;soft\u0026rdquo; the clamped parts are ($\\delta_P$) — the other half of the force ratio $\\Phi$ calculation.\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003ch2 id=\"1-why-is-δp-harder-to-compute-than-δs\"\u003e1. Why is δP harder to compute than δS?\u003c/h2\u003e\n\u003cp\u003eThe bolt resilience $\\delta_S$ can be broken simply into series cylindrical segments that add up. But the clamped parts are a completely different case — the standard states (VDI 2230:2015, §5.1.2, p.45):\u003c/p\u003e","title":"VDI 2230 (005): Clamped-Parts Resilience δP and the Cone Model"},{"content":" 🧮 在线计算器：VDI 2230 Bolted Joint Calculation — The full 14-step calculation chain (R0–R13), with six strength checks.\nBolt elastic resilience δS — breaking the bolt into a chain of cylinders The previous article set up the framework of the spring model; this article provides the first concrete number for the numerator and denominator of the force ratio $\\Phi$ — how \u0026ldquo;soft\u0026rdquo; is the bolt itself?\n1. Basic principle: series cylindrical segments VDI 2230 treats the bolt as a tension spring made of several series cylindrical segments of different cross-sections (VDI 2230:2015, §5.1.1, Bild 6).\nThe elastic resilience of each cylindrical segment is (VDI 2230:2015, §5.1.1.1, Eq. 18):\n$$ \\delta_i = \\frac{l_i}{E_S \\cdot A_i} \\tag{18} $$where $l_i$ is the segment length, $A_i$ the segment cross-section area, and $E_S$ the elastic modulus of the bolt material.\nBecause the segments are in series, the total resilience is the sum of the segment resiliences (VDI 2230:2015, §5.1.1.1, Eq. 19):\n$$ \\delta_S = \\delta_{SK} + \\delta_1 + \\delta_2 + \\dots + \\delta_{Gew} + \\delta_{GM} \\tag{19} $$ [!NOTE] \u0026ldquo;In series = add directly\u0026rdquo; The standard explains: \u0026ldquo;In the bolt, the cylindrical elements are arranged in a row, so that the total elastic resilience $\\delta_S$ is determined by adding the resiliences of the individual cylindrical elements within the clamp length and the further deformation regions.\u0026rdquo; (VDI 2230:2015, §5.1.1.1, p.40)\n2. Substitutional lengths and formulas of each segment The standard computes not only the segments inside the clamp length, but also the regions affected by force outside the clamp length — the head and the nut / blind-hole region. Each is described below:\n2.1 Bolt head $\\delta_{SK}$ The standard uses a \u0026ldquo;substitutional extension length\u0026rdquo; (Ersatzdehnlänge) to model the elastic deformation of the head (VDI 2230:2015, §5.1.1.1, Eq. 29):\n$$ \\delta_{SK} = \\frac{l_{SK}}{E_S \\cdot A_N} \\tag{29} $$where $A_N = \\frac{\\pi}{4} d^2$ (nominal-diameter cross-section area, Eq. 25), and the substitutional length is:\nHexagon-head bolt: $l_{SK} = 0.5 \\cdot d$ (Eq. 30) Socket-head cap screw: $l_{SK} = 0.4 \\cdot d$ (Eq. 31) Source: (VDI 2230:2015, §5.1.1.1, Eq. 29-31)\n2.2 Plain-shank segments inside the clamp length $\\delta_i$ Each plain-shank segment (diameter may differ) is computed directly by Eq. (18). For a standard bolt, the clamp length usually contains:\nUnthreaded shank segment: cross-section area $A_i = \\frac{\\pi}{4} d_i^2$ ($d_i$ is the actual diameter of that segment) Reduced-shank bolt (Dehnschaft): cross-section area computed from the waist diameter $d_T$ 2.3 Free thread segment $\\delta_{Gew}$ For the thread segment inside the clamp length that is not engaged, use the thread minor-diameter cross-section $A_{d_3}$ (VDI 2230:2015, §5.1.1.1, Eq. 28):\n$$ \\delta_{Gew} = \\frac{l_{Gew}}{E_S \\cdot A_{d_3}} \\tag{28} $$where $A_{d_3} = \\frac{\\pi}{4} d_3^2$ (Eq. 23).\n2.4 Engaged thread segment + nut / blind-hole region $\\delta_{GM}$ This is the most complex part. It contains two sub-terms (VDI 2230:2015, §5.1.1.1, Eq. 20):\n$$ \\delta_{GM} = \\delta_G + \\delta_M \\tag{20} $$Engaged segment $\\delta_G$ — the part of the bolt engaged in the nut or blind hole (Eq. 21, 22):\n$$ \\delta_G = \\frac{l_G}{E_S \\cdot A_{d_3}}, \\qquad l_G = 0.5 \\cdot d \\tag{21, 22} $$Nut / blind-hole region $\\delta_M$ — caused by the bending and compressive deformation of the thread teeth (Eq. 24):\n$$ \\delta_M = \\frac{l_M}{E_M \\cdot A_N} \\tag{24} $$The standard describes the physical origin of $\\delta_M$:\n\u0026ldquo;$\\delta_M$ results from the axial relative movement between bolt and nut or internal thread as a result of the elastic bending and compressive deformation of the teeth of the bolt and nut threads and of the arching and compressive deformation of the nut.\u0026rdquo; (VDI 2230:2015, §5.1.1.1, p.41)\nKey difference: DSV vs ESV\nParameter Through-bolt joint DSV Tapped-thread joint ESV Source $l_M$ $0.4 \\cdot d$ $0.33 \\cdot d$ Eq. 26, 27 $E_M$ $E_S$ (bolt material) $E_{BI}$ (clamped-part material) §5.1.1.1, p.42 This is the first notable difference between DSV and ESV in the resilience calculation — for ESV, the elastic modulus of the nut region takes the clamped-part material (usually cast iron or aluminium), not the bolt steel.\n3. Full example: M12 × 1.75 standard hexagon-head bolt Assume an M12 × 1.75 hexagon-head bolt, through-bolt joint (DSV), clamp length $l_K = 40$ mm, of which the plain shank is 25 mm and the free thread is 15 mm.\nThe substitutional lengths and resiliences of each segment:\nSegment Length Cross-section area Formula source Head $\\delta_{SK}$ $l_{SK} = 0.5 \\times 12 = 6$ mm $A_N = \\frac{\\pi}{4} \\times 12^2 = 113.1$ mm² Eq. 29, 30 Shank $\\delta_1$ $l_1 = 25$ mm $A_1 = 113.1$ mm² Eq. 18 Free thread $\\delta_{Gew}$ $l_{Gew} = 15$ mm $A_{d_3} = \\frac{\\pi}{4} \\times 9.853^2 = 76.2$ mm² Eq. 28 Engaged segment $\\delta_G$ $l_G = 0.5 \\times 12 = 6$ mm $A_{d_3} = 76.2$ mm² Eq. 21, 22 Nut $\\delta_M$ $l_M = 0.4 \\times 12 = 4.8$ mm $A_N = 113.1$ mm² Eq. 24, 26 Computing the resilience of each segment with $E_S = 210\\,000$ N/mm² and summing gives $\\delta_S$.\n4. Bending resilience $\\beta_S$ (brief note) In §5.1.1.2 the standard also defines the bolt\u0026rsquo;s bending resilience $\\beta_S$ (Biegenachgiebigkeit), used to compute the additional bending-moment effect under eccentric loading (VDI 2230:2015, §5.1.1.2, Eq. 34):\n$$ \\beta_S = \\beta_{SK} + \\beta_1 + \\beta_2 + \\dots + \\beta_{Gew} + \\beta_M + \\beta_G \\tag{34} $$The structure of the bending-resilience calculation is exactly like that of the axial resilience, only replacing the cross-section area $A_i$ with the section moment of inertia $I_i$ (Eq. 33):\n$$ \\beta_i = \\frac{l_i}{E \\cdot I_i} \\tag{33} $$The bending resilience does not take part directly in V1 (concentric symmetric loading), but it is required in the eccentric-load mode.\nData basis and accuracy statement All formulas in this article are from VDI 2230 Blatt 1:2015-11, §5.1.1. The thread geometry parameter in the worked example ($d_3 = 9.853$ mm for M12×1.75) is from DIN 13-1.\nDisclaimer: This article is for engineering teaching reference only. The final responsibility for engineering safety verification rests with the user.\n📚 Series navigation\n← Previous: Spring Model and Force Distribution ｜ Next: Clamped-Parts Resilience δP and the Rötscher Cone → ","permalink":"https://mechcalc.net/blog/en/posts/vdi2230-04-bolt-elastic-resilience/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/vdi2230-bolt-calculation?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eVDI 2230 Bolted Joint Calculation\u003c/a\u003e — The full 14-step calculation chain (R0–R13), with six strength checks.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003ch1 id=\"bolt-elastic-resilience-δs--breaking-the-bolt-into-a-chain-of-cylinders\"\u003eBolt elastic resilience δS — breaking the bolt into a chain of cylinders\u003c/h1\u003e\n\u003cblockquote\u003e\n\u003cp\u003eThe previous article set up the framework of the spring model; this article provides the first concrete number for the numerator and denominator of the force ratio $\\Phi$ — how \u0026ldquo;soft\u0026rdquo; is the bolt itself?\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003ch2 id=\"1-basic-principle-series-cylindrical-segments\"\u003e1. Basic principle: series cylindrical segments\u003c/h2\u003e\n\u003cp\u003eVDI 2230 treats the bolt as \u003cstrong\u003ea tension spring made of several series cylindrical segments of different cross-sections\u003c/strong\u003e (VDI 2230:2015, §5.1.1, Bild 6).\u003c/p\u003e","title":"VDI 2230 (004): Bolt Elastic Resilience δS Explained"},{"content":" 🧮 在线计算器：VDI 2230 Bolted Joint Calculation — The full 14-step calculation chain (R0–R13), with six strength checks.\nThe \u0026ldquo;spring philosophy\u0026rdquo; of a bolted joint — the core physical model of VDI 2230 1. The spring model: the physical basis of VDI 2230 The starting point of all VDI 2230 calculations is to model the bolted joint as two sets of springs (VDI 2230:2015, §3.2, p.20):\n\u0026ldquo;In this model, the bolt and the clamped parts are considered as tension and compression springs with the elastic resiliences $\\delta_S$ and $\\delta_P$.\u0026rdquo;\nThe bolt = a spring in tension, with elastic resilience (elastische Nachgiebigkeit) $\\delta_S$ The clamped parts (flanges, backing plates, etc.) = a spring in compression, with elastic resilience $\\delta_P$ [!NOTE] Resilience vs stiffness Resilience $\\delta$ is the reciprocal of stiffness $k$: $\\delta = f/F = 1/k$. VDI 2230 uses resilience rather than stiffness because the resiliences of the bolt\u0026rsquo;s segments add directly in a series system (VDI 2230:2015, §5.1.1.1, Eq. 19): $$\\delta_S = \\delta_{SK} + \\delta_1 + \\delta_2 + \\dots + \\delta_{Gew} + \\delta_{GM}$$ This is more intuitive than summing the reciprocals of stiffnesses.\nWhen the bolt is tightened, it stretches (storing elastic energy) and the clamped parts are compressed (also storing elastic energy). The two constrain each other through the contact surfaces of the bolt head and the nut, forming a force-balanced system.\n2. The joint diagram (Verspannungsschaubild) This is the most classic visualization tool in VDI 2230 (VDI 2230:2015, §3.2, Bild 1 / Bild 2). The standard represents the force–deformation relationship of the bolt and the clamped parts with two sloped lines:\nAssembled state (no external force):\nUnder the preload $F_V$, the bolt stretches by $f_{SM}$, with slope $1/\\delta_S$ Under the same $F_V$, the clamped parts are compressed by $f_{PM}$, with slope $1/\\delta_P$ The two lines meet at the preload $F_V$ — this is the assembly equilibrium point After applying an axial external force $F_A$:\nThe bolt force increases by $F_{SA}$ (additional bolt force, Schraubenzusatzkraft) The compression force of the clamped parts decreases by $F_{PA}$ (plate unloading force) The standard gives the key force-balance relations (VDI 2230:2015, §3.2.1, Eq. 6, 7): $$ F_{PA} = (1 - \\Phi) \\cdot F_A \\tag{Eq. 6} $$$$ F_{SA} = \\Phi \\cdot F_A \\tag{Eq. 7} $$3. The force ratio Φ — the core parameter of external-force distribution The force ratio $\\Phi$ (Kraftverhältnis) is the most important intermediate quantity in VDI 2230. It answers a fundamental question: what fraction of the external force is transmitted to the bolt?\nThe standard gives different formulas for different clamping and loading cases:\nConcentric symmetric loading and clamping ($s_{sym} = 0$, $a = 0$), from Eq. (83) (VDI 2230:2015, §4.2, Eq. R3/3):\n$$ \\Phi_n = n \\cdot \\frac{\\delta_P + \\delta_{PZu}}{\\delta_S + \\delta_P} \\tag{R3/3} $$Eccentric clamping and loading (the most common case, $s_{sym} \\neq 0$, $a \u003e 0$) (VDI 2230:2015, Eq. R3/4):\n$$ \\Phi_{en}^{*} = n \\cdot \\frac{\\delta_P^{**} + \\delta_{PZu}}{\\delta_S + \\delta_P^{*}} \\tag{R3/4} $$where $\\delta_P^{*}$ is from Eq. (67) and $\\delta_P^{**}$ from Eq. (71) (VDI 2230:2015, §5.3.1).\nKey parameters Parameter Meaning Source $\\delta_S$ bolt elastic resilience §5.1.1, Eq. 19 $\\delta_P$ clamped-parts elastic resilience §5.1.2, Eq. 40-53 $n$ load introduction factor (Krafteinleitungsfaktor) §5.2.2 $\\delta_{PZu}$ additional resilience (load-introduction position correction) §5.3.1 The physical meaning of the load introduction factor $n$: the standard §5.2.2 states that $n$ describes where the external force acts inside the clamped parts. When the load introduction point is in the plane of the bolt head / nut, $n = 1$ (the most unfavourable case); when the load introduction point is near the centre of the interface, $n$ can be as small as about 0.3 (VDI 2230:2015, §5.2.2). Determining $n$ often needs engineering judgement or FEM support.\n4. Residual clamping force — the functional bottom line After applying the external force $F_A$, the residual clamping force (Restklemmkraft) at the interface is the preload minus the unloading effect. The standard\u0026rsquo;s constraint on the minimum required clamping force $F_{Kerf}$ is (VDI 2230:2015, §5.4.1, Eq. R2/4):\n$$ F_{Kerf} \\geq \\max\\left(F_{KQ};\\; F_{KP} + F_{KA}\\right) \\tag{R2/4} $$This $F_{Kerf}$ comes from three functional requirements (VDI 2230:2015, §4.2, Eq. R2/1–R2/3):\nFriction force transfer: $F_{KQ} = \\dfrac{F_{Q\\max}}{q_F \\cdot \\mu_{T\\min}} + \\dfrac{M_{Y\\max}}{q_M \\cdot r_a \\cdot \\mu_{T\\min}}$ (transmit transverse force and/or torque, R2/1) Sealing function: $F_{KP} = A_D \\cdot p_{i\\max}$ (resist medium pressure, R2/2) Prevent opening: $F_{KA}$ (prevent one-sided opening of the interface, R2/3) The core design constraint is: under all operating conditions, the residual clamping force must always stay above $F_{Kerf}$.\n5. The main equation: everything converges here All of the analysis above finally converges to the main equation of VDI 2230 (VDI 2230:2015, §4.2, Eq. 16):\n$$ F_{M\\max} = \\alpha_A \\cdot \\left[ F_{Kerf} + (1 - \\Phi) \\cdot F_A + F_Z + \\Delta F'_{Vth} \\right] \\tag{16} $$The standard positions this formula:\n\u0026ldquo;All of these factors (Figure 5) are an integral part of the main dimensioning formula, which is the basis for the bolt calculation.\u0026rdquo;\n(VDI 2230:2015, §4.2, p.30)\nThe physical meaning of each term:\nTerm Meaning Source step $F_{Kerf}$ minimum clamping force required for function R2, Eq. R2/4 $(1-\\Phi) \\cdot F_A$ unloading effect of the external force on the interface R3, Eq. R3/2 $F_Z$ preload loss due to embedding R4, Eq. R4/1 $\\Delta F'_{Vth}$ preload change due to thermal expansion difference R4, Eq. R4/2 $\\alpha_A$ scatter amplification factor of the tightening method R1, Eq. R1/1 The design logic: start from the interface\u0026rsquo;s functional requirement $F_{Kerf}$, add the external-force effect and the various preload losses term by term, then multiply by the tightening scatter, to back out the maximum preload that must be applied at assembly — then verify through the six checks R7–R12 whether the bolt can carry it.\n6. Next step: into the resilience calculation The framework of the spring model is now set up. To compute the force ratio $\\Phi$, we need the specific values of $\\delta_S$ and $\\delta_P$. The next article goes into the elastic resilience of the bolt $\\delta_S$ — how does VDI 2230 break a real bolt down into series cylindrical segments to compute its resilience?\nData basis and accuracy statement All quotations in this article are from VDI 2230 Blatt 1:2015-11. Citation format: (VDI 2230:2015, section, formula/figure/table/page).\nDisclaimer: This article is for engineering teaching reference only. The final responsibility for engineering safety verification rests with the user.\n📚 Series navigation\n← Previous: Scope and Standard Positioning ｜ Next: Bolt Elastic Resilience δS → ","permalink":"https://mechcalc.net/blog/en/posts/vdi2230-03-spring-model-force-distribution/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/vdi2230-bolt-calculation?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eVDI 2230 Bolted Joint Calculation\u003c/a\u003e — The full 14-step calculation chain (R0–R13), with six strength checks.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003ch1 id=\"the-spring-philosophy-of-a-bolted-joint--the-core-physical-model-of-vdi-2230\"\u003eThe \u0026ldquo;spring philosophy\u0026rdquo; of a bolted joint — the core physical model of VDI 2230\u003c/h1\u003e\n\u003ch2 id=\"1-the-spring-model-the-physical-basis-of-vdi-2230\"\u003e1. The spring model: the physical basis of VDI 2230\u003c/h2\u003e\n\u003cp\u003eThe starting point of all VDI 2230 calculations is to model the bolted joint as two sets of springs (VDI 2230:2015, §3.2, p.20):\u003c/p\u003e\n\u003cblockquote\u003e\n\u003cp\u003e\u0026ldquo;In this model, the bolt and the clamped parts are considered as tension and compression springs with the elastic resiliences $\\delta_S$ and $\\delta_P$.\u0026rdquo;\u003c/p\u003e","title":"VDI 2230 (003): Spring Model and Force Distribution"},{"content":" 🧮 在线计算器：VDI 2230 Bolted Joint Calculation — The full 14-step calculation chain (R0–R13), with six strength checks.\nWhat exactly does VDI 2230 cover? — Scope and standard positioning 1. One-line positioning The subtitle of VDI 2230 Blatt 1 sets a clear boundary for its core territory:\nZylindrische Einschraubenverbindungen (cylindrical single-bolt joints)\nTwo key words: single-bolt (Einschraubenverbindung) and cylindrical thread (zylindrisch) — the load sharing of a multi-bolt group is not within the scope of Blatt 1; it is handled separately by VDI 2230 Blatt 2:2014.\n2. Applicability conditions The standard sets out its applicability conditions in §1 (Anwendungsbereich / Scope) (VDI 2230:2015, §1, p.4-5):\n✅ Scenarios the standard covers Condition Source Steel bolts with 60° fastening threads §1, p.4 Property class per DIN EN ISO 898-1 §1, p.4 Metric ISO threads M4–M39 (range of the standard\u0026rsquo;s tables) Table A1–A4 Through-bolt joint DSV (Durchsteckschraubverbindung) §1, p.4 Tapped-thread joint ESV (Einschraubverbindung) §1, p.4 Loads mainly from static and dynamic axial force §4.2 ❌ Scenarios the standard does not cover The standard itself clearly states its limits:\n\u0026ldquo;The generally difficult and large-scale analysis of forces and deformations which is involved in the determination of the initial quantities cannot be addressed by this standard because of the large variety of designs of components and BJs: this task must be solved by means of elasto-mechanics.\u0026rdquo;\n(VDI 2230:2015, §4.2, p.29)\nScenario Reason Alternative Multi-bolt group load sharing Blatt 1 analyses only a single bolt VDI 2230 Blatt 2 (2014) Flange joints with a sealing gasket Gasket nonlinearity is beyond the spring model DIN EN 1591 Exact location of the external load point The standard assumes $F_A$, $F_Q$ are known FEM / elasto-mechanical analysis 3. The interface limiting dimension G — check before you use the standard The cone-of-resilience model in VDI 2230 has a geometric premise, which the standard gives explicitly in step R0 (VDI 2230:2015, §4.2, Eq. R0/1, R0/2):\nThrough-bolt joint (DSV / TBJ):\n$$ G = h_{\\min} + d_W \\tag{R0/1} $$Tapped-thread joint (ESV / TTJ):\n$$ G' \\approx (1.5 \\dots 2) \\cdot d_W \\tag{R0/2} $$The standard\u0026rsquo;s own warning:\n\u0026ldquo;Exceeding the limiting dimensions entails a relatively large calculation error.\u0026rdquo;\n(VDI 2230:2015, §4.2, p.29)\nHere $h_{\\min}$ is the minimum clamping height and $d_W$ is the outer diameter of the bearing surface of the bolt head or nut. When the interface dimension $c_T$ exceeds the limit $G$, the assumptions of the cone-resilience formula are no longer reliable and the calculation error grows markedly. Before applying VDI 2230, the engineer must check this condition first.\n4. The family of standards VDI 2230 depends on The input data VDI 2230 needs comes from a group of base standards. The main text and appendices directly cite the following (VDI 2230:2015, §2 Symbols / Normative References):\nStandard What it provides Use in VDI 2230 DIN 13-1 Thread geometry $d, d_2, d_3, P, A_S$ R0 diameter selection, cross-section area in the R3 resilience calculation DIN EN ISO 898-1 Property class → $R_{p0.2\\min}, R_m$ R7 allowable preload, R8/R9 safety factors DIN EN ISO 4014 / 4017 Bolt-head geometry $d_w, s, k$ $A_p$ bearing area, cone-angle calculation DIN EN 20273 Through-hole diameter $d_h$ Cone-resilience $\\delta_P$ calculation (Eq. 40-43) DIN EN ISO 7089 Washer dimensions R10 bearing-pressure check 5. How force is transmitted — a premise that is easy to overlook The fundamental assumption of the VDI 2230 spring model is: the external working load is transmitted to the bolt through the clamped parts, not applied directly to the bolt shank. The standard explains this in §4.2:\n\u0026ldquo;The calculation of a BJ is based on the external working load $F_B$ acting on the joint. This working load and the elastic deformations of the components caused by it produces an axial working load $F_A$, a transverse load $F_Q$, a bending moment $M_b$ and in some cases a torque $M_T$ at the individual bolting point.\u0026rdquo;\n(VDI 2230:2015, §4.2, p.28)\nThis means VDI 2230 describes a physical model in which the force acts on the bolt indirectly, after the elastic deformation of the clamped parts. In this model the external force is shared according to the resilience ratio of the bolt and the clamped parts (the force ratio $\\Phi$, see R3) — which is exactly where the core value of VDI 2230 lies.\nIf the external force is applied directly along the bolt axis (as with an eye bolt carrying load directly), the force ratio $\\Phi$ reduces to 1.0, and the force-sharing calculation of VDI 2230 loses its meaning.\n6. The accuracy and verification the standard recommends The standard has a clear view of its own accuracy (VDI 2230:2015, §1, p.5):\nThe cone-angle formula (Eq. 42/43) has an accuracy of about ±5% The load introduction factor n: needs engineering judgement; for critical joints, verification by FEM is advised \u0026ldquo;This standard does not in principle do away with the need for experimental and/or numerical (FEM) tests for verifying the calculation results.\u0026rdquo;\n(VDI 2230:2015, §1, p.5)\nData basis and accuracy statement All quotations in this article are from VDI 2230 Blatt 1:2015-11. Citation format: (VDI 2230:2015, section, formula/figure/table/page).\nDisclaimer: This article is for engineering teaching reference only. The final responsibility for engineering safety verification rests with the user.\n📚 Series navigation\n← Previous: VDI 2230 Overview ｜ Next: Spring Model and Force Distribution → ","permalink":"https://mechcalc.net/blog/en/posts/vdi2230-02-scope-and-positioning/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/vdi2230-bolt-calculation?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eVDI 2230 Bolted Joint Calculation\u003c/a\u003e — The full 14-step calculation chain (R0–R13), with six strength checks.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003ch1 id=\"what-exactly-does-vdi-2230-cover--scope-and-standard-positioning\"\u003eWhat exactly does VDI 2230 cover? — Scope and standard positioning\u003c/h1\u003e\n\u003ch2 id=\"1-one-line-positioning\"\u003e1. One-line positioning\u003c/h2\u003e\n\u003cp\u003eThe subtitle of VDI 2230 Blatt 1 sets a clear boundary for its core territory:\u003c/p\u003e\n\u003cblockquote\u003e\n\u003cp\u003e\u003cem\u003eZylindrische Einschraubenverbindungen\u003c/em\u003e (cylindrical single-bolt joints)\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003cp\u003eTwo key words: \u003cstrong\u003esingle-bolt\u003c/strong\u003e (Einschraubenverbindung) and \u003cstrong\u003ecylindrical thread\u003c/strong\u003e (zylindrisch) — the load sharing of a multi-bolt group is not within the scope of Blatt 1; it is handled separately by VDI 2230 Blatt 2:2014.\u003c/p\u003e","title":"VDI 2230 (002): Scope and Standard Positioning"},{"content":" 🧮 在线计算器：VDI 2230 Bolted Joint Calculation — The full 14-step calculation chain (R0–R13), with six strength checks.\n💡 Engineer, before blaming \u0026ldquo;poor bolt quality\u0026rdquo;, look at this real case —\nThe connecting-rod bolts of a passenger-car engine failed catastrophically: the left bolt showed a typical one-sided bending fatigue fracture (a crescent-shaped fatigue region on the fracture surface), and the right bolt then broke suddenly from the force imbalance. Post-failure analysis found that the \u0026ldquo;culprit\u0026rdquo; was not the bolt material but insufficient preload. Because of assembly micro-embedding (Setzen), the preload dropped; under an eccentric service load, the connecting-rod interface opened slightly on one side, invisible to the eye (Aufklaffen), which caused a fatal alternating bending stress, finally initiating a fatigue crack at the shank root and a chain fracture.\nThis is exactly why VDI 2230 exists.\nWhy VDI 2230? — The need for systematic bolted-joint calculation 1. What is VDI 2230? The full title of VDI 2230 is:\nSystematische Berechnung hochbeanspruchter Schraubenverbindungen — Zylindrische Einschraubenverbindungen (Systematic calculation of highly stressed bolted joints — cylindrical single-bolt joints)\n(VDI 2230 Blatt 1:2015-11, title page)\nThis standard, published by the Association of German Engineers (Verein Deutscher Ingenieure), has over 40 years of engineering use. The standard positions itself in the introduction:\n\u0026ldquo;This standard, which has enjoyed practical application for over 40 years now, is a recognized and highly regarded recommendation. Throughout the world it is regarded as the standard work for calculating single-bolt joints.\u0026rdquo;\n(VDI 2230:2015, Introduction, p.3)\nIt gives design and calculation engineers a systematic step-by-step calculation method (Rechenschritte), so that a bolted joint can be designed for functional reliability and operational safety while fully using the bolt\u0026rsquo;s load capacity.\n2. Why is a simple strength check not enough? Many engineers habitually check only \u0026ldquo;will the bolt be pulled apart\u0026rdquo; in bolt design — verifying $\\sigma \\leq R_{p0.2}$. But the existence of VDI 2230 shows this is far from enough. The standard clearly lists all the factors affecting bolted-joint reliability in §4.2 (VDI 2230:2015, §4.2, Bild 5).\n2.1 Preload changes The preload $F_M$ applied at assembly changes during operation. The standard gives two main reasons in §5.4.2 (VDI 2230:2015, §5.4.2):\nEmbedding loss $F_Z$ (Setzbetrag): microscopic roughness is plastically flattened under load, causing an irreversible preload loss. Reference values for embedding are in VDI 2230 Table 5, with the bolt head, nut and interface each contributing about 3 μm (VDI 2230:2015, Table 5). Thermal expansion difference $\\Delta F'_{Vth}$: when the thermal expansion coefficient or temperature rise of the bolt and the clamped parts differ, a temperature change alters the preload (VDI 2230:2015, Eq. R4/2). 2.2 The external force is not carried entirely by the bolt This is VDI 2230\u0026rsquo;s core physical insight. The standard sets up a spring model (Federmodell) in §3.2:\n\u0026ldquo;In this model, the bolt and the clamped parts are considered as tension and compression springs with the elastic resiliences $\\delta_S$ and $\\delta_P$.\u0026rdquo;\n(VDI 2230:2015, §3.2, p.20)\nThe external axial force $F_A$ under service is split into two parts: the additional $F_{SA}$ on the bolt and the unloading $F_{PA}$ of the clamped parts. For the concentric symmetric case (VDI 2230:2015, §3.2.1, Eq. 8):\n$$ F_{SA} = n \\cdot \\frac{\\delta_P}{\\delta_P + \\delta_S} \\cdot F_A $$The force ratio $\\Phi = F_{SA} / F_A$ is usually far below 1.0 — the bolt actually carries only a small part of the external force. Ignoring this effect makes the design either too conservative or the preload badly set.\nDeeper still: the position of the load introduction point matters just as much. VDI 2230 introduces the load introduction factor $n$ (Krafteinleitungsfaktor) to describe where the external load acts inside the clamped parts (VDI 2230:2015, §3.2.2). The closer the load acts to the interface (smaller $n$), the smaller the additional service load $F_{SA}$ on the bolt. This explains a counter-intuitive engineering practice: thickening the flange not only raises stiffness but also lowers the bolt\u0026rsquo;s dynamic loading.\n2.3 The tightening method has scatter in accuracy For the same tightening torque, due to friction-coefficient variation and tool accuracy limits, the actual preload varies between $F_{M\\min}$ and $F_{M\\max}$. The standard quantifies this scatter with the tightening factor $\\alpha_A$ (VDI 2230:2015, §5.4.3, Eq. R1/1):\n$$ \\alpha_A = \\frac{F_{M\\max}}{F_{M\\min}} $$$\\alpha_A$ differs greatly among tightening methods, more than most engineers expect (VDI 2230:2015, Table A8):\nTightening method $\\alpha_A$ reference Preload scatter Pneumatic/electric impact wrench (Schlagschrauber) 2.5 – 4.0 ±40% – ±60% Ordinary torque wrench (Drehmomentschlüssel) 1.6 – 1.8 ≈ ±22% Yield-point / angle-controlled (streckgrenzgesteuert) ≈ 1.0 ±8% ⚠️ A key insight: even with a high-end measuring torque wrench of ±10% accuracy, the natural ±20% variation of the thread and bearing-surface friction coefficient still makes the final preload scatter about ±22%. No matter how accurate the tool, it cannot remove the uncertainty of the friction coefficient. The yield-point-controlled method takes tightening into the material\u0026rsquo;s plastic yield plateau, so friction variation barely affects the final axial force, the scatter drops to ±8%, and 100% of the material\u0026rsquo;s load potential can be used.\n2.4 Interface opening: the precursor of bolt fatigue fracture (Abhebegrenze) Ideal concentric loading is extremely rare in engineering. When the external tension or moment is off the bolt axis, the two sides of the interface are unequally loaded. VDI 2230 pays great attention to the interface opening limit (Abhebegrenze) (VDI 2230:2015, §5.2.1) — once an eccentric service load causes local opening of the interface (Klaffen), the system stiffness changes sharply, and the additional tension and bending stress on the bolt rise steeply and nonlinearly. Interface opening is exactly the root trigger of the connecting-rod bolt fatigue fracture at the start of this article. The standard\u0026rsquo;s calculation of the eccentricity $a$ and the symmetric-axis distance $s_{sym}$ is precisely to hold this critical red line (VDI 2230:2015, §5.2.1, Fig. 18).\n3. The 14-step VDI 2230 calculation chain The standard provides a 14-step calculation chain R0–R13 (VDI 2230:2015, §4.1), covering the full flow from selection to verification:\nStep Content Output R0 preliminary nominal diameter and limit check $d$, $G$ R1 tightening factor $\\alpha_A$ R2 minimum required clamping force $F_{Kerf}$ R3 external-force distribution and force ratio $F_{SA}$, $F_{PA}$, $\\Phi$ R4 preload change $F_Z$, $\\Delta F'_{Vth}$ R5 minimum assembly preload $F_{M\\min}$ R6 maximum assembly preload $F_{M\\max}$ R7 assembly stress check $\\sigma_{red,M}$, $F_{Mzul}$ R8 service stress check $\\sigma_{red,B}$, $S_F$ R9 fatigue stress check $\\sigma_a$, $S_D$ R10 bearing pressure check $p_{\\max}$, $S_P$ R11 minimum length of engagement $m_{eff\\min}$ R12 slip and shear safety $S_G$ R13 tightening torque $M_A$ 4. The main equation: one formula that governs it all The steps above finally converge to VDI 2230\u0026rsquo;s main equation (VDI 2230:2015, §4.2, Eq. 16):\n$$ F_{M\\max} = \\alpha_A \\cdot \\left[ F_{Kerf} + (1 - \\Phi) \\cdot F_A + F_Z + \\Delta F'_{Vth} \\right] $$The standard positions this formula:\n\u0026ldquo;All of these factors (Figure 5) are an integral part of the main dimensioning formula, which is the basis for the bolt calculation.\u0026rdquo;\n(VDI 2230:2015, §4.2, p.30)\nThe physical meaning of each term:\nTerm Meaning Source step $F_{Kerf}$ minimum clamping force required for function R2 $(1-\\Phi) \\cdot F_A$ unloading effect of the external force on the interface R3 $F_Z$ preload loss due to embedding R4 $\\Delta F'_{Vth}$ preload change due to thermal expansion difference R4 $\\alpha_A$ scatter amplification factor of the tightening method R1 The design logic: start from the interface\u0026rsquo;s functional requirement $F_{Kerf}$, add the external-force effect and the various preload losses term by term, then multiply by the tightening scatter, to back out the maximum preload that must be applied at assembly — then verify through R7–R12 whether the bolt can carry it.\n5. VDI 2230 and simplified pre-selection In this series\u0026rsquo; Preliminary Selection Estimate we introduced a fast pre-selection method based on the Kübler equation. Their relationship is coarse screen → precise calculation:\nDimension Kübler pre-selection Full VDI 2230 calculation Resilience model equivalent hollow cylinder cone + variable angle + multi-layer plates (§5.1.2) Preload loss fixed empirical value $f_Z$ summed by contact-surface type + thermal correction (§5.4.2) Strength checks 1–2 6 (R7–R12) Accuracy ±15–20% about ±5% (cone-angle formula, §5.1.2.1) The standard itself is clear-eyed about accuracy and verification:\n\u0026ldquo;This standard does not in principle do away with the need for experimental and/or numerical (FEM) tests for verifying the calculation results. These are particularly advisable in the case of critical joints.\u0026rdquo;\n(VDI 2230:2015, §1, p.5)\n📌 The standard\u0026rsquo;s positioning in one line: if you are building a bridge (Eurocode 3 / DIN EN 1993-1-8) you care whether the bolt group resists shear failure; if you are making a pressure vessel (AD 2000-B7) you care whether the sealing gasket leaks; but if you are designing an engine connecting rod or a heavy gearbox (VDI 2230), you must, like a surgeon, use the \u0026ldquo;joint diagram\u0026rdquo; to dissect every micron of deformation precisely, compute the bolt\u0026rsquo;s fatigue limit under alternating load, and use every bit of the material\u0026rsquo;s load potential.\nData basis and accuracy statement All formulas, values and quotations in this article are from VDI 2230 Blatt 1:2015-11. Citation format: (VDI 2230:2015, section, formula/figure/table/page).\nDisclaimer: This article is for engineering teaching reference only. The final responsibility for engineering safety verification rests with the user.\n📚 Series navigation\n← Previous: Preliminary Selection Estimate ｜ Next: Scope and Positioning → ","permalink":"https://mechcalc.net/blog/en/posts/vdi2230-01-why-systematic-calculation/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/vdi2230-bolt-calculation?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eVDI 2230 Bolted Joint Calculation\u003c/a\u003e — The full 14-step calculation chain (R0–R13), with six strength checks.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003cblockquote\u003e\n\u003cp\u003e💡 \u003cstrong\u003eEngineer, before blaming \u0026ldquo;poor bolt quality\u0026rdquo;, look at this real case —\u003c/strong\u003e\u003c/p\u003e\n\u003cp\u003eThe connecting-rod bolts of a passenger-car engine failed catastrophically: the left bolt showed a typical one-sided bending fatigue fracture (a crescent-shaped fatigue region on the fracture surface), and the right bolt then broke suddenly from the force imbalance. Post-failure analysis found that the \u0026ldquo;culprit\u0026rdquo; was not the bolt material but \u003cstrong\u003einsufficient preload\u003c/strong\u003e. Because of assembly micro-embedding (Setzen), the preload dropped; under an eccentric service load, the connecting-rod interface opened slightly on one side, invisible to the eye (Aufklaffen), which caused a fatal alternating bending stress, finally initiating a fatigue crack at the shank root and a chain fracture.\u003c/p\u003e","title":"VDI 2230 (001): Why Systematic Calculation?"},{"content":" 🧮 在线计算器：Bolt Pre-selection Calculator — Fast sizing with the Kübler equation; supports a design mode and a check mode.\nIntroduction: the rubber-band-and-sponge model To fully understand the physical logic behind the \u0026ldquo;bearing-pressure check\u0026rdquo;, we can picture the bolted-joint system as \u0026ldquo;a stretched rubber band (the bolt) clamping a sponge (the clamped part)\u0026rdquo;.\nThe heart of this part is to reveal a key question: why, when a part\u0026rsquo;s surface is crushed just a little bit, does the whole bolted joint face the risk of complete failure?\n1. From microscopic crushing to macroscopic \u0026ldquo;embedding\u0026rdquo; (Setzen) 1. How does bearing pressure (Flächenpressung) arise? When a bolt is tightened, it is stretched, producing a large axial tension (preload). This tension must be transmitted to the clamped part through the underside of the bolt head or nut (the bearing surface).\nBy the pressure formula $p = \\frac{F}{A_p}$ (pressure = force ÷ contact area), because the effective bearing area under the bolt head ($A_p$) is usually small, the part surface in this annular region carries an extremely concentrated, large pressure.\n2. The physical mechanism of microscopic crushing and plastic embedding Any machined part surface, however smooth it looks, is made up of uneven \u0026ldquo;peaks and valleys\u0026rdquo; (surface roughness) at the microscopic level.\nMicroscopic crushing: when the bearing pressure exceeds the \u0026ldquo;crushing limit\u0026rdquo; (Quetschgrenze), or limiting surface pressure (Grenzflächenpressung $p_G$), of the base material, these tiny metal \u0026ldquo;peaks\u0026rdquo; cannot bear the heavy load and are plastically flattened (plastisches Einebnen von Oberflächenrauhigkeiten). Macroscopic creep (Kriechen): if the pressure is badly over the limit, not only is the microscopic roughness flattened, but even the macroscopic metal base under the bolt head undergoes time-dependent yield flow (material creep), pressing a visible shallow dent into the part surface. In engineering, the permanent thinning of a part caused by this crushing and flattening of the surface is collectively called \u0026ldquo;embedding\u0026rdquo; (Setzen).\n2. The fatal chain reaction: why does embedding cause the system to collapse? Understanding this step is the key to understanding bolt failure. We need to bring in a core mechanics concept we have stressed repeatedly in earlier tutorials: the bolt itself is an extremely stiff spring.\n1. Why does \u0026ldquo;embedding\u0026rdquo; cause an irreversible loss of preload (Vorspannkraftverlust)? Elastic recoil: when the stiff bolt is tightened, it is stretched by even just a few tenths of a millimetre. If the part beneath it thins due to \u0026ldquo;embedding\u0026rdquo; (say the embedding amount is $f_Z$), the originally stretched bolt, like a spring, immediately recoils elastically by the same distance to fill the gap. Sudden drop in tension: by Hooke\u0026rsquo;s law, when the spring\u0026rsquo;s stretch shrinks, its tension must drop with it. So the microscopic crushing of the part surface immediately turns into a geometric recoil of the bolt, finally appearing as an irreversible loss of preload (i.e. clamping force) ($F_Z$). Quantified calculation: this mechanical relation can be computed precisely by the formula $F_Z = \\frac{f_Z}{\\delta_S + \\delta_P}$, where $\\delta_S$ and $\\delta_P$ are the resiliences of the bolt and the part. 2. The final consequence: loosening and fatigue fracture (Lockern und Versagen) Under a dynamic alternating load (such as high-frequency vibration or alternating tension-compression), the bolted joint must always keep enough residual clamping force to lock the parts tightly.\nLoosening (Lockern): if the preload loss from embedding is too large, the residual clamping force will not be able to press the two parts firmly together. Failure progression: once the clamping force is badly insufficient, the parts slide relative to each other under a transverse external force (fretting). This not only causes the thread to self-loosen (selbsttätiges Losdrehen), but also makes the bolt itself bear fatal alternating bending and shear stresses. In the end, this easily triggers the most dreaded failure mode in mechanical engineering — bolt fatigue fracture (Dauerbruch). As we can see, \u0026ldquo;a dent pressed into the surface → the bolt stress recoils → the preload disappears → the bolt breaks by vibration\u0026rdquo; is an extremely vicious domino effect.\n3. Engineering response: how to run the bearing-pressure check To break the vicious cycle above, a preliminary bearing-pressure check (Überprüfung der Flächenpressung) must be done in the design-draft stage (Vorauslegung) of the bolted joint or during an operation assessment.\nIn engineering practice, a simplified approximate formula is usually used for the preliminary assessment, with a precise check done later in the detailed VDI 2230 verification (the standard\u0026rsquo;s step R10). Below are the detailed steps and the underlying mechanical logic:\n1. The core preliminary check formula In the preliminary design stage, the following calculation-engine approximate formula can be used to check whether the actual pressure $p$ is less than or equal to the material\u0026rsquo;s limiting surface pressure $p_G$:\n$$p \\approx \\frac{F_{sp} / 0,9}{A_p} \\le p_G$$The derivation logic of the formula: In the actual working state, the maximum total tension $F_{Smax}$ the bolt carries equals the assembly preload $F_{sp}$ plus the additional bolt tension caused by the external working load (i.e. $\\Phi \\cdot F_B$). To simplify the calculation in the early sizing stage, engineers usually use $F_{sp} / 0,9$ to roughly and safely estimate the maximum total axial tension the bolt may bear under extreme conditions (the implied assumption: the additional working load takes up at most this remaining 10% of the load space).\n2. Calculation and values of the key parameters $F_{sp}$ (peak assembly force): This is the maximum assembly clamping force of the bolt material at 90% yield-strength utilization. In actual engineering calculations, this value can be read directly from a standard data table (such as Table TB 8-14) by the \u0026ldquo;bolt property class\u0026rdquo;, \u0026ldquo;thread size\u0026rdquo; and \u0026ldquo;estimated thread friction coefficient\u0026rdquo;. $A_p$ (effective bearing area): The effective bearing area is the annular area of real contact between the bolt head or nut and the pressed part surface. Its formula: $$A_p \\approx \\frac{\\pi}{4}(d_w^2 - d_h^2)$$ where $d_w$ is the outer contact-circle diameter under the bolt head or nut (such as the across-flats size of a hexagon head, or the minimum of the flange-face outer diameter), and $d_h$ is the through-hole diameter in the flange (usually taking the medium fit-hole diameter per DIN EN 20273). $p_G$ (limiting pressure / Grenzflächenpressung): This is set by the clamped-part material — the allowable pressure limit the material can bear without severe creep or crushing. This value usually needs a table lookup (such as Table TB 8-10b). For example, the limiting surface pressure of steel is usually far above that of cast iron, which in turn is far above that of light aluminium alloy. 3. A special correction for high-precision tightening methods If you use a modern high-precision yield-point-controlled (streckgrenzgesteuert) or angle-controlled (drehwinkelgesteuert) method, the bolt is deliberately tightened at assembly to 100% of the material yield strength, or even into the plastic region; the conservative \u0026ldquo;divide by 0.9\u0026rdquo; formula can no longer be used.\nIn this advanced case, because the maximum assembly force rises markedly, the preliminary check formula must be corrected as follows: $$p = 1,4 \\cdot \\frac{F_{sp}}{A_p} \\le p_G$$The deep origin of the factor 1.4: this is not an arbitrary safety factor but the combined product of three physical variables:\nThe statistical error of the ratio of the material\u0026rsquo;s actual maximum yield strength to the nominal minimum yield strength ($\\approx 1.2$) The effect of the reciprocal of the yield-utilization bound ($1 / 0.9 \\approx 1.11$) The strengthening from work hardening of the material in the plastic region ($\\approx 1.05$) 4. The precise check in VDI 2230 (step R10) If your preliminary estimate passes and you move into the rigorous, systematic VDI 2230 calculation (the standard\u0026rsquo;s step R10), the pressure check is further split strictly into two extreme cases, \u0026ldquo;assembly state\u0026rdquo; and \u0026ldquo;service state\u0026rdquo;, for precise verification:\nAssembly state (Montagezustand): Here only the extreme point of the assembly stage is considered: the allowable maximum assembly preload $F_{Mzul}$ and the worst-case minimum possible contact area $A_{p\\min}$. $$p_{M\\max} = \\frac{F_{Mzul}}{A_{p\\min}} \\le p_G$$ Service state (Betriebszustand): Here all the adverse superpositions of the later working environment are included: the maximum residual preload $F_{V\\max}$, the maximum external additional bolt working load $F_{SA\\max}$, and the thermal-deformation force $\\Delta F_{Vth}$ from possible temperature-load environmental decay are precisely combined. $$p_{B\\max} = \\frac{F_{V\\max} + F_{SA\\max} - \\Delta F_{Vth}}{A_{p\\min}} \\le p_G$$ 4. What to do if the pressure is over the limit? If the actual pressure computed in the design stage is greater than the limiting pressure of the base material ($p \u003e p_G$), it means the flange surface is at high risk of crushing and will most likely cause a preload loss.\nAs an engineer, you must never force ahead in this situation. Starting from the mechanical essence, take a combination of the following three optimization measures:\nEnlarge the contact area (i.e. enlarge the denominator, increase $A_p$): Use a flanged hexagon bolt/nut, or switch to a special bolt type with no undercut groove under the head to increase the contact area. Add a hard, thick washer (Unterlegscheiben): We cannot press directly onto a soft base. Add a specially heat-treated, very hard, thick washer between the bolt head and the soft base to \u0026ldquo;spread the pressure\u0026rdquo; (for a washer of thickness $h_S$, the equivalent spread outer diameter on its back can usually be estimated by the mechanical-cone approximation as $d_{Wa} = d_W + 1,6 \\cdot h_S$). Change the base material (i.e. enlarge the numerator, increase $p_G$): This is also the most thorough way: switch to a clamped-part material with a higher yield strength and a harder compressive capacity (such as from aluminium alloy to reinforced carbon-steel plate). ","permalink":"https://mechcalc.net/blog/en/posts/bolt-preselection-04-bearing-pressure/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/bolt-preselection?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eBolt Pre-selection Calculator\u003c/a\u003e — Fast sizing with the Kübler equation; supports a design mode and a check mode.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003ch2 id=\"introduction-the-rubber-band-and-sponge-model\"\u003eIntroduction: the rubber-band-and-sponge model\u003c/h2\u003e\n\u003cp\u003eTo fully understand the physical logic behind the \u0026ldquo;bearing-pressure check\u0026rdquo;, we can picture the bolted-joint system as \u003cstrong\u003e\u0026ldquo;a stretched rubber band (the bolt) clamping a sponge (the clamped part)\u0026rdquo;\u003c/strong\u003e.\u003c/p\u003e\n\u003cp\u003eThe heart of this part is to reveal a key question: \u003cstrong\u003ewhy, when a part\u0026rsquo;s surface is crushed just a little bit, does the whole bolted joint face the risk of complete failure?\u003c/strong\u003e\u003c/p\u003e","title":"Bolt Pre-selection (004): A Physical Look at Bearing Pressure"},{"content":" 🧮 在线计算器：Bolt Pre-selection Calculator — Fast sizing with the Kübler equation; supports a design mode and a check mode.\nA minimal preliminary estimation table for bolt size 📎 Series article: this is a standalone spin-off of Bolt Pre-selection (001) — From Load to Size , focusing on the table-lookup method that needs no calculation. For a precise estimate, see the Kübler equation in that article.\nOn the shop floor or in technical discussion, you often need a rough judgement of bolt size in seconds: \u0026ldquo;for this load, roughly how large a bolt do I need?\u0026rdquo; The quick pre-selection list below is made for exactly that — knowing only the maximum working load a single bolt carries, you can read the recommended nominal diameter and minimum property class directly.\n⚠️ For teaching reference only: the quick pre-selection list below is only an auxiliary reference for engineering teaching; it does not replace a formal design calculation based on DIN/ISO standards. In real engineering, use the Kübler equation or the full VDI 2230-1 method.\nHow to use Determine the maximum working load a single bolt carries (axial static load / axial dynamic load / transverse shear force) Find the range not smaller than that load below Choose a suitable nominal diameter and property-class combination from those listed Quick pre-selection list Working load $\\le$ 1.6 kN (static) / 1.0 kN (dynamic) / 0.32 kN (shear) Nominal diameter Recommended property class M4 5.8, 6.8, 8.8 M5 4.8, 5.6 M6 4.6 Working load $\\le$ 2.5 kN (static) / 1.6 kN (dynamic) / 0.5 kN (shear) Nominal diameter Recommended property class M4 10.9, 12.9 M5 5.8, 6.8, 8.8 M6 4.8, 5.6 M8 4.6 Working load $\\le$ 4.0 kN (static) / 2.5 kN (dynamic) / 0.8 kN (shear) Nominal diameter Recommended property class M5 10.9, 12.9 M6 5.8, 6.8, 8.8 M8 4.8, 5.6 M10 4.6 Working load $\\le$ 6.3 kN (static) / 4.0 kN (dynamic) / 1.25 kN (shear) Nominal diameter Recommended property class M6 10.9 M8 5.8, 6.8, 8.8 M10 4.8, 5.6 M12 4.6 Working load $\\le$ 10.0 kN (static) / 6.3 kN (dynamic) / 2.0 kN (shear) Nominal diameter Recommended property class M8 8.8, 10.9, 12.9 M10 5.8, 6.8 M12 4.8, 5.6 M16 4.6 Working load $\\le$ 16.0 kN (static) / 10.0 kN (dynamic) / 3.15 kN (shear) Nominal diameter Recommended property class M8 12.9 M10 8.8, 10.9 M12 5.8, 6.8 M16 4.8, 5.6 M20 4.6 Working load $\\le$ 25.0 kN (static) / 16.0 kN (dynamic) / 5.0 kN (shear) Nominal diameter Recommended property class M10 12.9 M12 10.9 M14 5.8, 6.8, 8.8 M20 4.8, 5.6 M24 4.6 Working load $\\le$ 40.0 kN (static) / 25.0 kN (dynamic) / 8.0 kN (shear) Nominal diameter Recommended property class M14 10.9 M16 8.8 M18 5.8, 6.8 M24 4.8, 5.6 Working load $\\le$ 63.0 kN (static) / 40.0 kN (dynamic) / 12.5 kN (shear) Nominal diameter Recommended property class M16 10.9, 12.9 M20 8.8 M22 5.8, 6.8 M30 4.8, 5.6 Working load $\\le$ 100 kN (static) / 63 kN (dynamic) / 20 kN (shear) Nominal diameter Recommended property class M20 10.9, 12.9 M24 8.8 M27 5.8, 6.8 Working load $\\le$ 160 kN (static) / 100 kN (dynamic) / 31.5 kN (shear) Nominal diameter Recommended property class M24 12.9 M27 10.9 M30 8.8 Working load $\\le$ 250 kN (static) / 160 kN (dynamic) / 50 kN (shear) Nominal diameter Recommended property class M30 10.9, 12.9 Notes [!NOTE] Applicability limits\nFor reduced-shank bolts (German: Dehnschrauben), or when the load point is eccentric and the tightening factor $k_A$ is large, choose the next larger nominal diameter. The data in the table are based on the typical case of a standard hexagon-head bolt, normal lubrication conditions and torque-controlled tightening. For safety-critical applications, always use the Kübler equation or the full VDI 2230-1 method for a precise check. Data basis and accuracy statement This table is a rough pre-selection (grobe Vorwahl), with an accuracy at the order-of-magnitude level, only for narrowing the selection range.\nData sources:\nDIN 13-1 (ISO 261) — metric ISO thread geometry DIN EN ISO 898-1 — bolt mechanical properties (property class and $R_{p0.2}$) Disclaimer: This article is for engineering estimation and teaching reference only. The results do not replace a professional engineer\u0026rsquo;s design judgement and final sign-off. The final responsibility for engineering safety verification rests with the user.\n📚 Series navigation\n← Previous: Thread Basics ｜ Bolt Pre-selection (001) — Precise Pre-selection with the Kübler Equation → ","permalink":"https://mechcalc.net/blog/en/posts/bolt-preselection-03-quick-table/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/bolt-preselection?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eBolt Pre-selection Calculator\u003c/a\u003e — Fast sizing with the Kübler equation; supports a design mode and a check mode.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003ch1 id=\"a-minimal-preliminary-estimation-table-for-bolt-size\"\u003eA minimal preliminary estimation table for bolt size\u003c/h1\u003e\n\u003cblockquote\u003e\n\u003cp\u003e📎 \u003cstrong\u003eSeries article\u003c/strong\u003e: this is a standalone spin-off of \u003ca href=\"/blog/en/posts/bolt-preselection-01-load-to-size/\"\u003eBolt Pre-selection (001) — From Load to Size\u003c/a\u003e\n, focusing on the \u003cstrong\u003etable-lookup method that needs no calculation\u003c/strong\u003e. For a precise estimate, see the Kübler equation in that article.\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003cp\u003eOn the shop floor or in technical discussion, you often need a rough judgement of bolt size \u003cstrong\u003ein seconds\u003c/strong\u003e: \u0026ldquo;for this load, roughly how large a bolt do I need?\u0026rdquo; The quick pre-selection list below is made for exactly that — knowing only the maximum working load a single bolt carries, you can read the recommended nominal diameter and minimum property class directly.\u003c/p\u003e","title":"Bolt Pre-selection (003): A Minimal Pre-selection Table"},{"content":" 🧮 在线计算器：Bolt Pre-selection Calculator — Fast sizing with the Kübler equation; supports a design mode and a check mode.\nThe reduction factor $\\kappa$ explained: the \u0026ldquo;invisible tax\u0026rdquo; of tightening a bolt 📎 Prerequisite reading: this article is a deeper supplement to Bolt Pre-selection (001) — From Load to Size , focusing on the physical nature and derivation of the $\\kappa$ parameter in the denominator of the Kübler equation.\nIn the design and installation of a bolted joint, the reduction factor $\\kappa$ (Reduktionsfaktor) is a very central parameter. It answers a key question: while the bolt is being tightened, how much strength is still \u0026ldquo;available\u0026rdquo; for axial load carrying? Below we break down its definition, derivation and formula, based on the public technical standard VDI 2230.\n1. First, the physical background: the \u0026ldquo;double burden\u0026rdquo; during tightening While a bolt is being tightened, the bolt shank carries two loads at the same time, and is therefore in a two-way (2D) stress state:\nAxial tensile stress ($\\sigma_{M}$, Montagezugspannung): produced as the bolt is stretched while being screwed down along the thread — this is exactly the preload we want. Torsional shear stress ($\\tau_t$, Torsionsspannung): produced as the bolt shank is twisted to overcome the thread-surface friction — a \u0026ldquo;by-product\u0026rdquo; of thread friction that contributes nothing to load carrying. Now the problem: the bolt material\u0026rsquo;s strength parameters (such as the yield strength $R_{p0,2}$) are all measured by a uniaxial tension test. But the bolt is really in a multi-axial combined stress state of \u0026ldquo;both pulled and twisted\u0026rdquo;. We need a way to convert this combined stress into an equivalent uniaxial stress, so it can be compared with $R_{p0,2}$.\n2. Theoretical basis: the fourth strength theory (von Mises yield criterion) For the ductile steels commonly used in high-strength bolts, tests show the von Mises yield criterion (German: Gestaltänderungsenergiehypothese, GEH for short) gives the most accurate prediction, because it compares the energy needed to cause a change of shape (not a change of volume) in the material.\nFrom the 3D general form to the bolt\u0026rsquo;s 2D special case By the von Mises criterion, the general equivalent-stress formula in a three-way (spatial) stress state is:\n$$ \\sigma_{red} = \\frac{1}{\\sqrt{2}} \\sqrt{(\\sigma_1 - \\sigma_2)^2 + (\\sigma_1 - \\sigma_3)^2 + (\\sigma_2 - \\sigma_3)^2} $$where $\\sigma_1, \\sigma_2, \\sigma_3$ are the three principal stresses on the element.\nFor a surface element of the bolt shank, it carries tensile stress only along the axis and shear stress on the cross-section, with no other normal stress in the radial or circumferential direction. So it is in a plane stress state, and one of the three principal stresses must be zero (let $\\sigma_2 = 0$). Substituting and simplifying:\n$$ \\sigma_{red} = \\sqrt{\\sigma_1^2 - \\sigma_1\\sigma_3 + \\sigma_3^2} $$To compute directly from the normal and shear stresses known in engineering, introduce the plane-stress von Mises formula in Cartesian coordinates:\n$$ \\sigma_{v} = \\sqrt{\\sigma_x^2 + \\sigma_y^2 - \\sigma_x\\sigma_y + 3\\tau^2} $$In the specific case of bolt tightening:\nAxial tensile stress $\\sigma_x = \\sigma_{M}$ Transverse normal stress $\\sigma_y = 0$ Shear stress $\\tau = \\tau_t$ After substitution, all the $\\sigma_y$ terms cancel, giving the equivalent-stress formula for bolt assembly:\n$$\\boxed{\\sigma_{red} = \\sqrt{\\sigma_{M}^2 + 3\\cdot\\tau_t^2}}$$ This is the total load the bolt material really \u0026ldquo;feels\u0026rdquo;. Even if the axial tensile stress $\\sigma_M$ has not yet reached $R_{p0,2}$, once the torsional component is added, $\\sigma_{red}$ may already be close to yield.\n3. The strict definition of $\\kappa$ The theoretical definition of the reduction factor $\\kappa$ is: the ratio of the combined equivalent stress to the pure axial tensile stress:\n$$ \\kappa = \\frac{\\sigma_{red}}{\\sigma_{M}} $$Substituting the equivalent-stress formula into the definition:\n$$ \\kappa = \\frac{\\sqrt{\\sigma_{M}^2 + 3 \\cdot \\tau_t^2}}{\\sigma_{M}} = \\sqrt{1 + 3 \\left(\\frac{\\tau_t}{\\sigma_{M}}\\right)^2} $$The physical meaning is clear: because a torque is always present ($\\tau_t \u003e 0$), $\\kappa$ must be greater than 1. $\\kappa$ reflects the \u0026ldquo;degree of reduction\u0026rdquo; of the bolt\u0026rsquo;s tensile load capacity by the torsional shear stress — like an \u0026ldquo;invisible tax\u0026rdquo;: the greater the thread friction, the higher this tax rate.\n4. From the stress ratio to a computable algebraic formula To turn the stress ratio into specific geometry and force parameters, we expand further:\nTensile stress: $\\sigma_{M} = F_{VM} / A_0$, where $A_0 = \\frac{\\pi}{4} d_0^2$ Shear stress: $\\tau_t = M_G / W_P$ Thread friction torque: $M_G \\approx F_{VM} \\cdot \\left(0{,}159 \\cdot P + 0{,}577 \\cdot \\mu_G \\cdot d_2\\right)$ A key detail: the fully plastic section modulus When computing the shear stress, consider that at normal tightening (such as using 90% of the yield strength) the section edge of the bolt is often already close to or into the plastic state. To use the material\u0026rsquo;s plastic support effect (Tragreserven), the torsional section modulus $W_P$ in engineering standards usually does not use the elastic formula $\\frac{\\pi}{16} d_0^3$, but the fully plastic torsional section modulus:\n$$ W_{P,pl} = \\frac{\\pi}{12} d_0^3 $$Substituting $M_G$, $A_0$ and $W_{P,pl}$ into the ratio $\\tau_t / \\sigma_{M}$ gives the final algebraic formula for $\\kappa$.\n5. The specific formula By VDI 2230 and public engineering textbooks, the specific formula for $\\kappa$ can be written in the following two mathematically equivalent algebraic forms:\nForm one (simplified engineering expression) $$ \\kappa = \\sqrt{1 + 3 \\cdot \\left[ \\frac{3}{d_0} \\cdot \\left(0{,}159 \\cdot P + 0{,}577 \\cdot \\mu_G \\cdot d_2\\right) \\right]^2} $$Form two (VDI 2230 standard expression) $$ \\kappa = \\sqrt{1 + 3 \\cdot \\left[ \\frac{3}{2} \\cdot \\frac{d_2}{d_0} \\left(\\frac{P}{\\pi \\cdot d_2} + 1{,}155 \\cdot \\mu_G\\right) \\right]^2} $$Formula parameters Parameter Meaning Value source $P$ thread pitch (Gewindesteigung) DIN 13-1 standard thread table $d_2$ thread pitch diameter (Flankendurchmesser) DIN 13-1 standard thread table $d_0$ section reference diameter ordinary bolt: $(d_2+d_3)/2$; reduced-shank bolt: $d_T$ $\\mu_G$ thread friction coefficient (Reibungszahl im Gewinde) VDI 2230-1, Tab.A2 6. Engineering use: the role of $\\kappa$ in the Kübler equation The formula shows clearly: the thread friction coefficient $\\mu_G$ is the most central variable that decides the size of $\\kappa$.\nThe larger $\\mu_G$, the greater the torque used to overcome friction during tightening, the stronger the torsional shear stress produced, and thus the marked increase in $\\kappa$. Since the material\u0026rsquo;s total load capacity is fixed, this greatly cuts the axial tension we actually need.\nIn engineering practice (for example, in a bolt pre-selection with the Kübler equation), the engineer divides the material\u0026rsquo;s yield strength by the reduction factor (i.e. $R_{p0,2} / \\kappa$) to assess the effective axial load capacity the bolt can really provide under a given lubrication state.\nTypical $\\kappa$ value reference $\\mu_G$ 0.08 0.10 0.12 0.14 0.20 Plain-shank bolt $\\kappa$ 1.11 1.15 1.19 1.24 1.41 Reduced-shank bolt $\\kappa$ 1.15 1.20 1.25 1.32 1.52 Engineering insight: take a plain-shank bolt with $\\mu_G = 0.14$, $\\kappa = 1.24$, which means the axial usable stress is only $R_{p0,2} / 1.24 \\approx 81\\%$. About 19% of the material\u0026rsquo;s load capacity is \u0026ldquo;eaten\u0026rdquo; by thread friction. This is why good lubrication (lowering $\\mu_G$) is critical to fully using the bolt\u0026rsquo;s strength.\n7. The safety criterion in VDI 2230 In the most authoritative bolt-calculation standard, VDI 2230, to make sure the bolt does not suffer permanent plastic yield at assembly — even under the combined tension and torsion load — it must strictly hold that:\n$$ \\sigma_{red} \\le 0.9 \\cdot R_{p0,2} $$That is, the computed combined equivalent stress must not exceed 90% of the material yield strength. This 10% safety margin is exactly the load-carrying room reserved for the external working load that may be added during operation.\nData basis and accuracy statement The derivation in this article is based on the following public technical standards:\nVDI 2230-1 — systematic calculation method for high-strength bolted joints DIN 13-1 (ISO 261) — metric ISO thread geometry DIN EN ISO 898-1 — bolt mechanical properties Disclaimer: This article is for engineering teaching reference only. The results do not replace a professional engineer\u0026rsquo;s design judgement and final sign-off. The final responsibility for engineering safety verification rests with the user.\n📚 Series navigation\n← Back: Bolt Pre-selection (001) — From Load to Size ｜ Next: A Minimal Pre-selection Table → ","permalink":"https://mechcalc.net/blog/en/posts/bolt-preselection-02-reduction-factors/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/bolt-preselection?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eBolt Pre-selection Calculator\u003c/a\u003e — Fast sizing with the Kübler equation; supports a design mode and a check mode.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003ch1 id=\"the-reduction-factor--explained-the-invisible-tax-of-tightening-a-bolt\"\u003eThe reduction factor $\\kappa$ explained: the \u0026ldquo;invisible tax\u0026rdquo; of tightening a bolt\u003c/h1\u003e\n\u003cblockquote\u003e\n\u003cp\u003e📎 \u003cstrong\u003ePrerequisite reading\u003c/strong\u003e: this article is a deeper supplement to \u003ca href=\"/blog/en/posts/bolt-preselection-01-load-to-size/\"\u003eBolt Pre-selection (001) — From Load to Size\u003c/a\u003e\n, focusing on the physical nature and derivation of the $\\kappa$ parameter in the denominator of the Kübler equation.\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003cp\u003eIn the design and installation of a bolted joint, the \u003cstrong\u003ereduction factor $\\kappa$ (Reduktionsfaktor)\u003c/strong\u003e is a very central parameter. It answers a key question: \u003cstrong\u003ewhile the bolt is being tightened, how much strength is still \u0026ldquo;available\u0026rdquo; for axial load carrying?\u003c/strong\u003e Below we break down its definition, derivation and formula, based on the public technical standard VDI 2230.\u003c/p\u003e","title":"Bolt Pre-selection (002): The Reduction Factor κ Explained"},{"content":" 🧮 在线计算器：Bolt Pre-selection Calculator — Fast sizing with the Kübler equation; supports a design mode and a check mode.\nPreliminary sizing of a bolted joint: from load to size, fast Before getting into complex system-level bolted-joint calculations, the first challenge an engineer faces is often: \u0026ldquo;for a given working load, how large should the bolt be? Is M8 enough, or M12?\u0026rdquo; A sound preliminary design / pre-selection (German: Vorauslegung) quickly fixes a suitable nominal diameter and does an early \u0026ldquo;sweep\u0026rdquo; for common failure modes (such as fatigue fracture, or loss of preload from crushing of the bearing surface). This article shows how to do a quick pre-selection from the working load, and how to check the feasibility of the design with simplified formulas.\n1. Quick pre-selection list If you only need a quick estimate with no calculation, go straight to the standalone article:\n📋 A minimal preliminary estimation table for bolt size — read the recommended nominal diameter and property class directly from the load range, a result in seconds.\nSection 2 onward covers the more precise Kübler-equation method.\n2. Strength-area estimation: the Kübler equation (German: Gleichung von Kübler) The Kübler equation is used mainly for a more precise preliminary design / pre-selection (German: Vorauslegung) of a bolted joint. By making reasonable assumptions on the safe side, it computes the minimum required thread stress cross-section (German: Spannungsquerschnitt des Gewindes $A_{\\mathrm{s}}$) or waist cross-section (German: Taillenquerschnitt $A_{\\mathrm{T}}$) of the bolt, so that the joint will not be pulled apart and the preload will not be lost, helping the designer quickly fix a reasonable bolt size.\nThe Kübler equation is written as:\n$$ A_{\\mathrm{s}} \\text{ or } A_{\\mathrm{T}} \\geq \\dfrac{F_{\\mathrm{B}} + F_{\\mathrm{Kl}}}{\\dfrac{R_{\\mathrm{p}0,2}}{\\kappa \\cdot k_{\\mathrm{A}}} - \\beta \\cdot E_{\\mathrm{S}} \\cdot \\dfrac{f_{\\mathrm{Z}}}{l_{\\mathrm{k}}}} $$The physical meaning, definition and value guidance of each parameter:\n$A_{\\mathrm{s}}$ or $A_{\\mathrm{T}}$: the target value the preliminary design / pre-selection must find, used to back out the minimum required bolt size.\n$A_{\\mathrm{s}}$ (thread stress cross-section): used to evaluate an ordinary bolt or a fully threaded bolt. The actual load-bearing cross-section of the thread lies between the thread pitch diameter ($d_2$) (German: Flankendurchmesser) and the thread minor diameter ($d_3$) (German: Kerndurchmesser). It is the equivalent area of the weakest failure section of the thread, with the standard formula: $$ A_s = \\frac{\\pi}{4} \\left( \\frac{d_2 + d_3}{2} \\right)^2 $$ Note: in real engineering you usually do not compute this value by hand. After finding $A_{\\mathrm{s}}$, the designer looks up a standard thread table directly and picks the nearest larger standard thread size. $A_{\\mathrm{T}}$ (waist cross-section, German: Taillenquerschnitt): for a reduced-shank bolt (German: Dehnschrauben) used for fatigue resistance or thermal-expansion compensation, the diameter of the reduced shank $d_{\\mathrm{T}}$ is deliberately machined smaller than the thread minor diameter (usually $d_{\\mathrm{T}} \\approx 0.9 d_3$); the weakest section of the whole connection is then no longer the thread but the shank. The formula is: $$ A_{\\mathrm{T}} = \\frac{\\pi}{4} d_{\\mathrm{T}}^2 $$ $F_{\\mathrm{B}}$: the axial working load the bolt carries (German: axiale Betriebskraft). This value is usually given directly by the overall structural design requirements.\n$F_{\\mathrm{Kl}}$: the required clamping force of the flange or clamped parts (German: geforderte Klemmkraft). The minimum clamping force the structure needs to prevent the interface from separating, slipping or leaking — also set by the overall structure.\n$R_{\\mathrm{p}0,2}$ (0.2% yield strength): its value is set directly by the bolt property class (German: Festigkeitsklasse) the designer selects (per DIN EN ISO 898-1).\nClass 8.8: $R_{p0,2}$ is usually taken as $640 \\text{ N/mm}^2$. Class 10.9: $R_{p0,2} = 10 \\times 9 \\times 10 = 900 \\text{ N/mm}^2$. Class 12.9: $R_{p0,2} = 12 \\times 9 \\times 10 = 1080 \\text{ N/mm}^2$. $k_{\\mathrm{A}}$ (tightening factor, German: Anziehfaktor): defined as the ratio of the maximum to the minimum preload that can occur at assembly ($F_{Mmax} / F_{Mmin}$), covering friction-coefficient scatter, tool error and operator variation. Its value depends on the tightening method:\nImpact wrench (Schlagschrauber): very large scatter, $k_A = 2.5 \\sim 4.0$, usually not recommended for demanding bolted joints. Torque-controlled tightening (drehmomentgesteuertes Anziehen): with an ordinary torque wrench, the range is generally $1.6 \\sim 2.0$. With a precision torque wrench that signals, it can be narrowed to $1.4 \\sim 1.8$. Take the smaller value (such as 1.6) when the parts are relatively rigid (small tightening angle); take the larger value when the parts are softer (large tightening angle) or when a blind hole has higher hardness. Yield-point-controlled tightening (streckgrenzgesteuertes Anziehen): simply take $k_A = 1.0$. Because this method tightens the bolt into the over-elastic (yield) region, the material\u0026rsquo;s own yield point acts as a \u0026ldquo;safety valve\u0026rdquo;, almost removing the effect of friction-coefficient variation, so the overload risk is very low. $\\kappa$ (reduction factor, German: Reduktionsfaktor): when the bolt is tightened, the applied torque $M_A$ is transmitted through the thread and the bearing surface, producing two stresses in the bolt shank at the same time:\nAxial tensile stress $\\sigma_M$ (Montagezugspannung) — this is exactly the preload we want; Torsional shear stress $\\tau_t$ (Torsionsspannung) — a \u0026ldquo;by-product\u0026rdquo; of thread friction that contributes nothing to load carrying. By the von Mises equivalent-stress criterion (Gestaltänderungsenergie-Hypothese), the equivalent stress of these two combined is $\\sigma_{\\mathrm{red}} = \\sqrt{\\sigma_M^2 + 3\\,\\tau_t^2}$. This is the total load the bolt material actually \u0026ldquo;feels\u0026rdquo;. In other words: even if the axial tensile stress $\\sigma_M$ is still well below $R_{p0.2}$, once the torsional component is added, $\\sigma_{\\mathrm{red}}$ may already be close to yield.\n$\\kappa$ is a simplified expression of this effect — it tells us: for a given friction condition, the axial usable stress in the denominator is really only $R_{p0.2} / \\kappa$, not $R_{p0.2}$ itself. The larger the friction (higher $\\mu_G$), the greater the fraction of torque \u0026ldquo;wasted\u0026rdquo; on friction, the higher the corresponding $\\tau_t$, the larger $\\kappa$, and the less margin left for axial load.\nThe value depends on the total friction coefficient ($\\mu_{ges}$) of the thread surface and on the bolt type (plain-shank or reduced-shank); see the table below: $\\mu_G$ 0.08 0.10 0.12 0.14 0.20 Plain-shank bolt $\\kappa$ 1.11 1.15 1.19 1.24 1.41 Reduced-shank bolt $\\kappa$ 1.15 1.20 1.25 1.32 1.52 $\\beta$ (compliance factor, German: Nachgiebigkeitsfaktor): reflects the difference in elastic deformability from bolt to bolt.\nOrdinary plain-shank bolt (Schaftschrauben, e.g. per DIN EN ISO 4014): thicker shank, less flexible, $\\beta \\approx 1.1$. Fully threaded bolt (Ganzgewindeschrauben, e.g. per DIN EN ISO 4017): threaded throughout, somewhat more flexible, $\\beta \\approx 0.8$. Reduced-shank bolt (Dehnschrauben): excellent elastic deformability, with the waist about 90% of the thread minor diameter, $\\beta \\approx 0.6$. $E_{\\mathrm{S}}$ (elastic modulus, German: E-Modul): for most ordinary steel bolts at room temperature, the elastic modulus is taken as a constant $E_S \\approx 210000 \\text{ N/mm}^2$.\n$f_{\\mathrm{Z}}$ (embedding, German: Setzbetrag): the irreversible deformation from the microscopic roughness of the interfaces (under the bolt head, on the thread flanks, between the parts) being plastically flattened after tightening and loading, which causes a preload loss ($F_{\\mathrm{Z}}$).\nNote: in the Kübler equation, if there is no precise data, an average value of $0.011 \\text{ mm}$ is usually taken. The actual embedding depends strongly on the number and roughness of the interfaces; moreover, for a transversely loaded (querbeansprucht) bolted joint, the embedding rises markedly. $l_{\\mathrm{k}}$ (clamp length, German: Klemmlänge): the total thickness of all the compressed clamped parts in the direction of the bolt force.\nNote: further verification steps after preliminary design Since the Kübler equation is only a \u0026ldquo;preliminary sizing\u0026rdquo; in the pre-design stage, after looking up the actual bolt size (diameter $d$) from a table, the designer should — without doing the very complex full VDI 2230 verification — still perform at least the following two quick checks:\nFatigue-limit check (for strong dynamic loads): If the bolt carries a dynamic pulsating load in service, its alternating stress amplitude $\\pm \\sigma_a$ must be estimated and verified not to exceed the bolt\u0026rsquo;s fatigue limit $\\pm \\sigma_A$, to prevent fatigue fracture. Bearing-pressure check (Flächenpressung): To prevent the base material from creeping and crushing under the bolt head or nut because the local contact area is too small (which would further increase the embedding), the bearing pressure must be checked to be less than the material\u0026rsquo;s limiting pressure $p_G$. The formula: $p \\approx \\frac{F_{sp} / 0.9}{A_p} \\leq p_G$. 3. Simplified fatigue-strength check In the early design stage of a bolted joint (preliminary design / pre-selection), the engineer usually has not yet fixed the specific flange dimensions and assembly details, so cannot do a precise joint diagram (German: Verspannungsschaubild) calculation based on VDI 2230. However, for a joint under strong dynamic load, fatigue fracture (German: Dauerbruch) is the most fatal failure mode. To avoid the risk early, a simplified fatigue-strength check formula on the safe side can be used:\n$$ \\pm \\sigma_a \\approx \\pm k \\cdot \\frac{F_{Bo} - F_{Bu}}{A_s} \\le \\pm \\sigma_A $$The left side $\\pm \\sigma_a$ (alternating stress amplitude, German: Ausschlagspannung) is the dynamic stress the bolt sees in the actual case; the right side $\\pm \\sigma_A$ (fatigue limit, German: Ausschlagfestigkeit) is the bolt\u0026rsquo;s own maximum resistance to fatigue fracture. As long as the left side does not exceed the right side, the bolt will not suffer fatigue fracture.\n1. Estimate the alternating stress amplitude $\\pm \\sigma_a$ $F_{Bo}$: the upper limit of the axial working load (German: oberer Grenzwert der axialen Betriebskraft). $F_{Bu}$: the lower limit of the axial working load (German: unterer Grenzwert der axialen Betriebskraft). Their difference $(F_{Bo} - F_{Bu})$ is the total dynamic load pulsation the joint sees in service. $A_s$: the thread stress cross-section (German: Spannungsquerschnitt des Gewindes), obtainable directly from standard thread tables (see DIN 13-1 etc.) or computed with the formula in section 2. $k$: the clamped-part material influence factor. In a precise VDI 2230 calculation, the external dynamic load does not act entirely on the bolt but is shared by the stiffness ratio (the ratio of resiliences $\\delta_S$ and $\\delta_P$) of the bolt and the clamped parts; the bolt carries only part of it — the \u0026ldquo;additional bolt force\u0026rdquo; ($F_{SA}$). The parameter $k$ is a minimal summary of this stiffness-sharing mechanism, and its value depends entirely on the base material: Steel (Stahl): $k = 0.1$. High elastic modulus, very stiff steel flange, absorbing most of the deformation, so the dynamic additional bolt force passed to the bolt is very small (about 10%). Cast iron (Gusseisen): $k = 0.125$. Lower elastic modulus than steel, slightly less stiff, so the bolt shares more of the dynamic load (about 12.5%). Aluminium (Aluminium): $k = 0.15$. Elastic modulus only about one-third of steel, a \u0026ldquo;softer\u0026rdquo; base with larger elastic recovery, so a larger fraction of the alternating load falls on the bolt (about 15%). Conclusion: the softer the clamped part (base), the lower its stiffness, the weaker its ability to \u0026ldquo;shelter\u0026rdquo; the bolt, and the greater the fatigue stress amplitude the bolt must bear. 2. Estimate the fatigue limit $\\pm \\sigma_A$ The fatigue limit (German: Ausschlagfestigkeit) depends on the order of thread manufacturing and heat treatment:\nRolled-before-heat-treated thread (SV) (Schlussvergütetes Gewinde / SV) That is, the thread is cut or rolled first and the bolt is heat treated as a whole last — the normal manufacturing state for class 8.8, 10.9 and 12.9 bolts. Extensive testing shows that the fatigue limit of an SV bolt is almost independent of the mean bolt force (i.e. the preload level) and is mainly controlled by the diameter, with the estimate:\n$$ \\pm \\sigma_{A(SV)} \\approx \\pm 0.85 \\cdot \\left( \\frac{150}{d} + 45 \\right) $$ The size effect (Größeneinfluss) shown by the parameter $d$ (thread nominal diameter, in mm): with $d$ in the denominator, the formula reveals a very important rule: as the bolt diameter grows, its fatigue limit gradually drops. A large bolt is more likely to have internal microscopic metallurgical defects, and its surface stress gradient is relatively flat, so its fatigue sensitivity to the notch (thread) is far higher than a small bolt\u0026rsquo;s. Safety reduction factor $0.85$: fatigue life in testing often shows a wide scatter band (Streuung). Multiplying by 0.85 lowers the theoretical value to the lower bound of the scatter band, ensuring enough margin (on the safe side) already at the preliminary design / pre-selection stage. [!NOTE] SV vs SG process distinction SV (Schlussvergütet), i.e. \u0026ldquo;rolled-before-heat-treated thread (SV)\u0026rdquo;: the thread is machined first and heat treated last. The heat treatment relieves the machining stresses, so the fatigue limit is independent of the preload — the standard normal state. SG (Schlussgewalzt), i.e. \u0026ldquo;rolled-after-heat-treated thread (SG)\u0026rdquo;: heat treated first and the thread rolled last. This process leaves a residual compressive stress at the thread root, giving a higher fatigue limit, but one more affected by the preload (see next section).\nRolled-after-heat-treated thread (SG) (Schlussgewalztes Gewinde / SG) The thread is rolled after heat treatment (higher manufacturing cost). The residual compressive stress kept at the thread root markedly raises the fatigue life, but the strengthening effect depends on the mean bolt force, so the fatigue limit must be corrected to:\n$$ \\pm \\sigma_{A(SG)} \\approx \\left( 2 - \\frac{F_m}{F_{0,2}} \\right) \\cdot \\sigma_{A(SV)} $$where $F_m$ is the mean bolt force (German: Schrauben-Mittelkraft) and $F_{0,2}$ is the force at which the bolt material reaches the 0.2% yield strength.\n3. Fatigue safety-factor check Once the actual alternating stress amplitude $\\sigma_a$ and the bolt fatigue limit $\\sigma_A$ are computed, the fatigue safety factor (German: dynamische Sicherheit) is:\n$$ S_D = \\frac{\\sigma_A}{\\sigma_a} \\ge 1.2 $$$S_D \\ge 1.2$ is the lower bound required by the standard, ensuring the bolted joint does not suffer fatigue fracture under long-term dynamic service.\nSummary: with this simplified check formula, the engineer needs only the range of external load pulsation $(F_{Bo} - F_{Bu})$, the base material type (which sets $k$) and a preliminary bolt diameter $d$ to quickly assess the fatigue-fracture risk in tens of seconds, fixing a reasonable starting bolt size for the later detailed VDI 2230 calculation (covering preload scatter, embedding loss, tightening factor, etc.).\n4. Preliminary bearing-pressure check (German: Flächenpressung) Exceeding the bearing pressure causes the base material to crush microscopically under the bolt head or nut, leading to an irreversible loss of preload (Vorspannkraftverlust). The mechanism: if the bearing pressure exceeds the crushing limit of the base material (Quetschgrenze), the contact region yields plastically (Setzen), the bolt springs back elastically, and the preload drops; when the loss is too large, the joint loosens or fails.\nCheck formula:\n$$ p \\approx \\frac{F_{sp} / 0.9}{A_p} \\le p_G $$Definition and value guidance of each parameter:\n$F_{sp} / 0.9$ (estimated maximum working tension): $F_{sp}$ is the assembly clamping force at 90% yield utilization. In real service the bolt must also add the shared external axial working load ($\\Phi \\cdot F_B$), so approximating $F_{Smax}$ by $F_{sp} / 0.9$ gives a safe estimate of the maximum total bolt tension without a full VDI 2230 calculation.\n$A_p$ (effective bearing area): the net area of actual contact between the bolt head or nut and the base surface, i.e. the total underside area of the head/nut minus the through-hole area.\n$p_G$ (limiting pressure): the allowable pressure limit of the clamped-part (not the bolt) material. $p_G$ should take the value for the base material; the allowable pressure of aluminium alloy is far below that of steel, so it especially needs checking when a high-strength bolt is used on a soft base.\nMeasures when it exceeds the limit (when $p \u003e p_G$):\nUse a flanged bolt or nut: enlarge the bearing underside area to directly lower the pressure. Add a hardened (heat-treated) washer: insert a high-hardness steel washer between the head/nut and the soft base to spread the concentrated pressure over a larger area. Change the base material: choose a material with a higher allowable pressure to meet the strength requirement. Data basis and accuracy statement The calculation method in this article is a pre-selection estimate (Vorauslegung), with an accuracy of ±15–20% for engineering pre-selection. A precise check should use the full VDI 2230-1 calculation method.\nData sources:\nDIN 13-1 (ISO 261) — metric ISO thread geometry DIN EN ISO 898-1 — bolt mechanical properties (property class and Rp0.2) DIN EN ISO 4014 / 4017 — hexagon-head bolt dimensions DIN EN 20273 — through-hole diameter VDI 2230-1 — systematic calculation method for high-strength bolted joints Disclaimer: This article is for engineering estimation and teaching reference only. The results do not replace a professional engineer\u0026rsquo;s design judgement and final sign-off. The final responsibility for engineering safety verification rests with the user.\n📚 Series navigation\n← Previous: Thread Basics ｜ Next: Reduction Factors Explained → ","permalink":"https://mechcalc.net/blog/en/posts/bolt-preselection-01-load-to-size/","summary":"\u003cblockquote\u003e\n  \u003cp\u003e🧮 \u003cstrong\u003e在线计算器\u003c/strong\u003e：\u003ca href=\"/calc/mechanical_elements/bolt-preselection?lang=en\" target=\"_blank\" rel=\"noopener\"\u003eBolt Pre-selection Calculator\u003c/a\u003e — Fast sizing with the Kübler equation; supports a design mode and a check mode.\u003c/p\u003e\n\u003c/blockquote\u003e\n\n\u003ch1 id=\"preliminary-sizing-of-a-bolted-joint-from-load-to-size-fast\"\u003ePreliminary sizing of a bolted joint: from load to size, fast\u003c/h1\u003e\n\u003cp\u003eBefore getting into complex system-level bolted-joint calculations, the first challenge an engineer faces is often: \u0026ldquo;for a given working load, how large should the bolt be? Is M8 enough, or M12?\u0026rdquo; A sound preliminary design / pre-selection (German: Vorauslegung) quickly fixes a suitable nominal diameter and does an early \u0026ldquo;sweep\u0026rdquo; for common failure modes (such as fatigue fracture, or loss of preload from crushing of the bearing surface). This article shows how to do a quick pre-selection from the working load, and how to check the feasibility of the design with simplified formulas.\u003c/p\u003e","title":"Bolt Pre-selection (001): From Load to Size, Fast"},{"content":"1. Overview The bolt (German: Schrauben) is the earliest and most widely used machine connection element in mechanical engineering. Compared with other connection methods, the bolt has the largest number and the richest variety of standardized forms.\n1.1 The thread line on a bolt The core of a bolted joint is the form fit (German: Formschluss) between the bolt (external thread) and the clamped part (usually the internal thread of a nut).\nUnrolled, a thread is essentially an inclined plane. The figure shows how a thread line is formed.\n1. Once unrolled, a thread is essentially an inclined plane.\nWhen the bolt and nut rotate relative to each other, the flank of the bolt thread slides over the flank of the nut thread. Through this sliding process on the \u0026ldquo;inclined plane\u0026rdquo;, the thread structure can perfectly turn rotary motion into a longitudinal linear displacement. Based on this \u0026ldquo;inclined-plane principle\u0026rdquo; of turning rotation into linear motion, this kind of thread (motion screw) is widely used to transmit motion or produce large thrust/pull, for example the lead screw of a lathe, a screw press, a screw jack, a bench vice and screw clamps.\nLine 1 is the thread line (or helix). This is a space curve on a cylindrical surface, representing the actual motion path of the thread. Line 2 is the inclined plane. The two-dimensional geometry obtained by unrolling the thread line on the cylinder along the tangent plane; it is the theoretical basis of thread force analysis. $\\varphi$ (German: Gewindesteigungswinkel): the lead angle. In the unrolled inclined-plane model, the angle between the inclined plane and the horizontal base. $P$ (German: Gewindesteigung): the pitch (or lead). The axial distance the bolt or nut moves per full relative revolution, corresponding to the vertical height of the unrolled inclined plane. $d_2$ (German: Flankendurchmesser): the pitch diameter. The effective reference diameter for computing thread friction and geometric unrolling. $d_2 \\cdot \\pi$ (or $r \\cdot 2\\pi$): the circumference at the pitch diameter. The horizontal base length of the inclined-plane model, corresponding to one turn of the thread unrolled at the pitch diameter. [!tip] Functional classification of bolts\nFastening bolts (German: Befestigungsschrauben): used to create a clamping joint. Rotary motion presses two (or more) components together, turning kinetic energy into potential energy. This potential energy can resist the service load along the bolt axis, produce friction in a coupling, prevent the joint from loosening, or seal an interface. Motion screws (German: Bewegungsschrauben): similar to a screw mechanism, used to turn rotary motion into linear motion or to produce large forces, for example a lathe lead screw, a screw press, a jack, etc. Sealing and adjusting bolts (German: Dichtungs- und Einstellschrauben): used to close an opening (such as an oil-pan sealing plug), or to adjust a clearance (such as a valve-clearance adjusting bolt). 1.2 Thread types (German: Gewinde) A thread is a profiled groove formed along a helix on a cylindrical surface. The key parameters that decide the thread\u0026rsquo;s characteristics are:\nProfile shape (German: Profilform): such as triangular (German: Dreieck), trapezoidal (German: Trapez), etc. Pitch / lead (German: Steigung / P_h): decides the axial displacement per revolution. Number of starts (German: Gangzahl): single-start or multi-start. Hand (German: Windungssinn): right-hand or left-hand. 1.2.1 Common thread types and technical standards Metric ISO thread (German: Metrisches ISO-Gewinde, DIN 13): Flank angle (German: Flankenwinkel) of $60^\\circ$. Divided into coarse (German: Regelgewinde) (such as M16) and fine (German: Feingewinde) (such as M20×2). Coarse is most common for ordinary fastening; fine is often used for thin-walled parts, seals, or high-strength needs. Pipe thread (German: Rohrgewinde, DIN EN ISO 228): used for pipe connections that are not sealed on the thread, flank angle $55^\\circ$ (such as G1/2). Where thread sealing is needed, a tapered external thread with a parallel internal thread is used (DIN EN 10226). Metric trapezoidal thread (German: Metrisches ISO-Trapezgewinde, DIN 103): Flank angle $30^\\circ$ (such as Tr36×6). The preferred motion thread, used for lead screws, presses, etc. Buttress thread (German: Sägengewinde, DIN 513): Load flank angle $3^\\circ$, non-load flank $30^\\circ$. Higher load capacity than the trapezoidal thread, with a small friction torque, suitable for transmitting very large loads in one direction, such as a heavy hydraulic press. 1.2.2 Geometric relationship formula Unrolling the helix on the cylinder gives the thread\u0026rsquo;s lead angle (German: Steigungswinkel, $\\varphi$), referenced to the pitch diameter (German: Flankendurchmesser, $d_2$).\n$$ \\tan \\varphi = \\frac{P_h}{d_2 \\cdot \\pi} $$Variables:\n$P_h$: thread lead (German: Gewindesteigung), the axial displacement per revolution [mm]. For a multi-start thread, the lead $P_h = n \\cdot P$ ($n$ is the number of starts, $P$ is the pitch, German: Teilung). For a single-start thread, the lead equals the pitch ($P_h = P$). $d_2$: thread pitch diameter [mm]. [!important] You will later find that the pitch diameter $d_2$ keeps appearing in all kinds of standardized calculations — why is it so important?\nBecause in all thread force analysis (especially the calculation of friction torque and self-locking), the force point is simplified to act on the cylindrical surface formed by the thread pitch diameter $d_2$. From now on, try to build an intuitive sense of the pitch diameter $d_2$, and save yourself some detours.\nMetric ordinary thread: the various diameters and dimensions of the bolt (German: Schraube) and the nut (German: Mutter).\n1. Diameter parameters (German: Durchmesser)\n$d$ / $D$ (German: Nenndurchmesser / Außendurchmesser): nominal diameter / major diameter. $d$ is the major diameter of the external thread (bolt), $D$ is the major diameter of the internal thread (nut). $d_2$ / $D_2$ (German: Flankendurchmesser): pitch diameter. The diameter of an imaginary cylinder whose generatrix passes through the point where the groove and ridge widths of the profile are equal. $d_3$ (German: Kerndurchmesser der Schraube): minor (root) diameter of the external thread. The minimum core cross-section diameter of the bolt thread. $D_1$ (German: Kerndurchmesser der Mutter): minor diameter of the internal thread. The innermost diameter of the nut thread. 2. Cross-section parameters (German: Querschnitte)\n$A_S$ (German: Spannungsquerschnitt): stress cross-section. The effective cross-section for computing the bolt under tension (an equivalent cross-section between the pitch and minor diameters). $A_3$ (German: Kernquerschnitt): root / core cross-section. The minimum cross-section computed from the bolt minor diameter $d_3$, often used for compression and similar checks. 3. Profile and height parameters (German: Profil und Höhen)\n$P$ (German: Teilung / Steigung bei eingängigen Gewinden): pitch. The axial distance between corresponding points of two adjacent threads on the pitch line. $P_h$ (German: Steigung bei mehrgängigen Gewinden): lead. The axial displacement per revolution (for a multi-start thread $P_h = n \\cdot P$, $n$ is the number of starts). $H$ (German: Höhe des spitzwinkligen Profildreiecks): height of the fundamental triangle. The total height of the theoretical sharp-cornered thread profile triangle (for a metric thread $H = 0.86603 P$). $H_1$ (German: Tragtiefe / Flankenüberdeckung): thread working height / contact height. The radial overlap height of the mating thread flanks when the internal and external threads are engaged ($H_1 = 0.54127 P$). $h_3$ (German: Gewindetiefe): external thread depth. The actual radial distance from crest to root of the external thread ($h_3 = 0.61343 P$). $R$ (German: Rundungsradius im Gewindegrund): root fillet radius. The arc transition radius at the bolt root, used to reduce stress concentration ($R = H/6$). 4. Angle and friction parameters (German: Winkel und Reibung)\n$\\varphi$ or $\\alpha$ (German: Steigungswinkel): lead angle. The angle between the helix and the plane perpendicular to the thread axis, after unrolling the pitch-diameter cylinder. $\\beta$ (German: Flankenwinkel / Spitzenwinkel): flank angle. The angle between the two flanks of the thread profile in the axial section ($60^\\circ$ for a metric ordinary thread, $30^\\circ$ for a trapezoidal thread). $\\mu_G$ or $\\tan \\varrho'$ (German: Gewindereibungszahl): thread friction coefficient. The equivalent friction coefficient between the mating thread surfaces. $\\varrho'$ (German: Gewindereibungswinkel): equivalent thread friction angle. An imaginary friction angle introduced when computing friction in the thread pair. 1.3 Fastener types and selection 1.3.1 Bolt types (German: Schraubenarten) The head shape depends on the tightening tool. Internal drives such as the hexagon socket and the Torx allow a smaller head, saving space.\nHexagon-head bolt (German: Sechskantschrauben, DIN EN ISO 4014/4017): the most common bolt type in mechanical engineering. Socket-head cap screw (German: Zylinderschrauben mit Innensechskant, DIN EN ISO 4762): often used for high-strength joints in tight spaces or where the head must sit below the surface. Stud (German: Stiftschrauben): one end screwed into a housing, the other fastened with a nut. Often used at flange covers and similar places that need frequent disassembly, to protect the internal thread of the expensive main casting from the wear of repeated screwing in and out. 1.3.2 Requirements on the nut The failure criterion of a nut is usually \u0026ldquo;stripping\u0026rdquo; (thread stripping, German: Abstreifen des Gewindes). But because stripping is a gradual process that is very hard to notice, the design rule for a bolted joint is always: if failure occurs, it must be the bolt breaking, not the nut stripping.\nFor this, the thickness $m$ of a fully load-bearing standard nut must satisfy $m \\ge 0.9d$. When matching, the property class of the nut should correspond to the first digit of the bolt property class (for example: an 8.8 bolt should use a class 8 or higher nut).\nIllustration note The illustrations in this article are from the German edition of the Dubbel mechanical-design handbook; the copyright belongs to the original authors. They are reproduced here for academic exchange and study only.\n","permalink":"https://mechcalc.net/blog/en/posts/bolt-basics-01-thread-geometry/","summary":"\u003ch1 id=\"1-overview\"\u003e1. Overview\u003c/h1\u003e\n\u003cp\u003eThe bolt (German: Schrauben) is the earliest and most widely used machine connection element in mechanical engineering. Compared with other connection methods, the bolt has the largest number and the richest variety of standardized forms.\u003c/p\u003e\n\u003ch2 id=\"11-the-thread-line-on-a-bolt\"\u003e1.1 The thread line on a bolt\u003c/h2\u003e\n\u003cp\u003eThe core of a bolted joint is the \u003cstrong\u003eform fit (German: Formschluss) between the bolt (external thread) and the clamped part (usually the internal thread of a nut)\u003c/strong\u003e.\u003c/p\u003e","title":"Bolt Basics (001): Thread Geometry and Classification"},{"content":"1. Background: Appendix Y, from scattered to unified In nuclear fracture mechanics and structural integrity assessment, predicting how a flaw grows during service is the decisive foundation for justifying life extension or Leak-Before-Break (LBB).\nIn earlier ASME BPVC Section XI editions, the reference crack growth rate curves for different materials were scattered across separate appendices:\nAppendix A: ferritic steels. Appendix C: austenitic stainless steels. Appendix O: intergranular stress corrosion cracking (IGSCC) of nickel alloys. In the 2025 edition of ASME XI, the code committee made a major structural revision: the growth-model equations for the various materials and mechanisms (fatigue $da/dN$ and stress corrosion $da/dt$) were all consolidated into the new Nonmandatory Appendix Y:\nY-2000: austenitic stainless steels Y-3000: ferritic steels Y-4000: nickel-base alloys 2. Fatigue crack growth ($da/dN$): principle and material differences For any metal, crack growth under fatigue cycling follows a threshold-corrected Paris-type equation:\n$$ \\frac{da}{dN} = C \\cdot S_{ENV} \\cdot S_R \\cdot (\\Delta K)^n $$Different material classes respond very differently to a high-temperature water environment (BWR/PWR) and to the load rise time ($t_r$):\n2.1 Austenitic stainless steels (Y-2000) Austenitic steels usually have excellent fracture toughness, but inside a light-water reactor (e.g. PWR) the water chemistry accelerates growth exponentially through polarization effects.\nThe environmental multiplier is $S_{ENV} = t_r^{0.3}$. The slower the loading (larger $t_r$), the more time the corrosive medium has to penetrate the grain boundaries, giving growth rates tens of times higher than in air. 2.2 Ferritic steels (Y-3000) Low-alloy ferritic steels (common in main piping or reactor pressure vessels, RPV) show a clear bilinear behaviour:\nin the low stress-intensity-range region the exponent is $n = 5.95$, a very steep curve; past a specific crossover point the exponent eases to $n = 1.95$. The calculation must compute the high and low curves separately and take the enveloping value. 2.3 Nickel-base alloys (Y-4000) For weld materials such as Alloy 600, 82/182 and 690, the water-environment penalty introduces a more complex nonlinear coupling with stress intensity: $$ S_{ENV} = 1 + A_E \\left[ S_T S_R (\\Delta K)^{4.1} \\right]^{-0.67} t_r^{0.67} $$ This captures the balance between the crack-tip strain rate and the anodic dissolution rate — a model brought from front-line research into an engineering standard.\n3. Stress corrosion cracking (SCC) model (Y-x300) Stress corrosion is governed by a sustained static load ($K_{max}$) and time ($t$). The core equation is: $$ \\frac{da}{dt} = C_S \\cdot (K_{max})^n $$Different environments and water-chemistry strategies affect it greatly, and this consolidation spells out the logic for selecting the environment constants $C_S$ and $n$:\nBWR environment: the constants for normal water chemistry (NWC) versus hydrogen water chemistry (HWC) can differ by up to an order of magnitude — HWC suppresses the oxidizing potential and markedly lowers the steady-state crack growth rate. PWR primary system: besides the Arrhenius activation-energy temperature correction $S_T$, Y-4320 adds a dissolved-hydrogen concentration factor ($f_{H2} / f_{H2ref}$), reminding engineers to control the $H_2$ concentration in the water chemistry of the real system. 4. Usage guide: the interactive whitebox calculator So that engineers do not get \u0026ldquo;lost\u0026rdquo; in the many logarithmic or nonlinear charts, this site has built, on the full 2025 ASME XI rule set:\n👉 ASME XI App Y Crack Growth Calculation Engine Compared with black-box commercial software, this calculator uses a whitebox (step-visualized) mode:\nHow to use it Set the parameters: on the left, choose the material class (austenitic / ferritic / nickel-base), the environment type (Air, NWC, HWC, PWR), and enter the current fracture-mechanics load ($\\Delta K$ or $K_{max}$). Run a demo directly: click a preset example in the \u0026ldquo;example list\u0026rdquo; (for instance ferritic fatigue or nickel-alloy SCC). Review the derivation: after clicking Calculate, the right panel gives not only the final $da/dN$ (m/cycle) or $da/dt$ (m/s) value, but also a LaTeX step outline with full substituted values and the source code clause for each step. By watching the intermediate $S_R$, $S_{ENV}$ environmental penalty factors, the user can truly see how much weight each input variable carries in the integrity result — moving from \u0026ldquo;being able to compute\u0026rdquo; to \u0026ldquo;understanding\u0026rdquo;.\n","permalink":"https://mechcalc.net/blog/en/posts/asme-xi-app-y-crack-growth/","summary":"\u003ch2 id=\"1-background-appendix-y-from-scattered-to-unified\"\u003e1. Background: Appendix Y, from scattered to unified\u003c/h2\u003e\n\u003cp\u003eIn nuclear fracture mechanics and structural integrity assessment, predicting how a flaw grows during service is the decisive foundation for justifying life extension or Leak-Before-Break (LBB).\u003c/p\u003e\n\u003cp\u003eIn earlier ASME BPVC Section XI editions, the reference crack growth rate curves for different materials were scattered across separate appendices:\u003c/p\u003e\n\u003cul\u003e\n\u003cli\u003e\u003cstrong\u003eAppendix A\u003c/strong\u003e: ferritic steels.\u003c/li\u003e\n\u003cli\u003e\u003cstrong\u003eAppendix C\u003c/strong\u003e: austenitic stainless steels.\u003c/li\u003e\n\u003cli\u003e\u003cstrong\u003eAppendix O\u003c/strong\u003e: intergranular stress corrosion cracking (IGSCC) of nickel alloys.\u003c/li\u003e\n\u003c/ul\u003e\n\u003cp\u003eIn the \u003cstrong\u003e2025 edition of ASME XI\u003c/strong\u003e, the code committee made a major structural revision: the growth-model equations for the various materials and mechanisms (fatigue $da/dN$ and stress corrosion $da/dt$) were all consolidated into the new \u003cstrong\u003eNonmandatory Appendix Y\u003c/strong\u003e:\u003c/p\u003e","title":"ASME XI (2025) Appendix Y: New-Rule Analysis and Calculation Guide"},{"content":" 🧮 Go straight to the calculator: to skip the derivation and use the online tool with a built-in verification flow, open the fatigue crack growth calculator from the large category cards on the calculators home page.\n1. Where fatigue crack growth sits in an LBB analysis At the core of a Leak-Before-Break (LBB) argument: you must show that, over the service life, the time for an initial non-penetrating surface flaw to grow by subcritical crack growth through the wall and cause a leak — and the time from leak to growth to the critical fracture size — is long enough for the monitoring system to detect it and act.\nUnlike early simplified assessments, modern mainstream LBB and fitness-for-service standards (KTA 3206, SRP 3.6.3, R6, API 579, FITNET) all require a crack growth estimate without exception.\n1.1 The basic principle of fatigue crack growth in KTA 3206 The relation between the fatigue crack growth rate $\\frac{da}{dN}$ and the stress intensity factor range $\\Delta K$ usually splits, on log-log axes, into three regions:\nRegion I (threshold region): there is a threshold $\\Delta K_{th}$ below which the crack cannot grow. Region II (steady-state region): a linear relation, the main range for engineering calculation. Region III (unstable, accelerating region): approaching the material\u0026rsquo;s critical fracture toughness $\\Delta K_c \\approx K_{Ic}$, where unstable fracture occurs. Applicability limit: for practical KTA 3206 use, crack growth is computed with the steady-state Region II only. The stress intensity factor results must also lie within the valid boundaries (e.g. the geometric applicability ranges of $s/R_m$, $a/s$, $a/c$).\n2. The core equation and derivation (Paris-Erdogan equation) KTA 3206 uses the classic Paris-Erdogan equation to approximate the Region II crack growth behaviour:\n$$ \\frac{da}{dN} = C \\cdot (\\Delta K)^m $$ $\\frac{da}{dN}$: the crack extension per load cycle. $\\Delta K$: the stress intensity factor range ($\\Delta K = K_{max} - K_{min}$). $C, m$: material-dependent constants. 2.1 Ferritic steels and austenitic steels in air For ferritic materials, and for austenitic materials without a medium (i.e. in air), the constants $C$ and $m$ depend on the stress ratio $R = K_{min}/K_{max}$ and the temperature, and can be estimated with the two-slope envelope data given in ASME BPVC Section XI.\n2.2 Austenitic steels in water (high-temperature-water corrosion fatigue) A high-temperature-water environment (such as the primary or secondary side of a light-water reactor) markedly accelerates crack growth in austenitic steel. KTA 3206 requires a medium contribution to be added for austenitic materials in water:\n$$ \\left(\\frac{da}{dN}\\right)_{total} = \\left(\\frac{da}{dN}\\right)_{Luft} + \\left(\\frac{da}{dN}\\right)_{Medium} $$(the subscript \u0026ldquo;Luft\u0026rdquo; is German for \u0026ldquo;air\u0026rdquo;). The medium-driven part is computed per NUREG/CR-6176:\n$$ \\left(\\frac{da}{dN}\\right)_{Medium} = C_{Medium} \\cdot S(R)^{0.5} \\cdot T_R^{0.5} \\cdot (\\Delta K)^{1.65} $$where the stress-ratio correction factor $S(R)$ is: $$ S(R) = 1 + 1.8 \\cdot R \\quad \\text{(for } R \\leq 0.8 \\text{)} $$ $$ S(R) = -43.35 + 57.97 \\cdot R \\quad \\text{(for } R \u003e 0.8 \\text{)} $$ $C_{Medium}$ (environment constant): depends on the dissolved-oxygen content of the water. $T_R$ (load rise time): the time (in seconds) of the load-rise stage in each stress cycle. The slower the loading (larger $T_R$), the longer the corrosive medium acts on the crack tip and the faster the growth. 3. A worked example and system check Example: growth check for austenitic steel in 0.2 ppm dissolved oxygen Per the above, with the following case:\n$\\Delta K = 25$ MPa√m, $R = 0.5$ $T_R = 10.0$ s dissolved-oxygen level $DO_{level} = 0.2$ ppm the system derives and outputs the following whitebox steps:\n$S(R) = 1 + 1.8 \\times 0.5 = 1.900$ $(\\frac{da}{dN})_{Luft} = 3.43 \\times 10^{-12} \\times 1.900 \\times (25)^{3.3} \\approx 2.508 \\times 10^{-7} \\text{ m/cycle}$ $(\\frac{da}{dN})_{Medium}$, through the acceleration effect, adds a large growth term $\\approx 5.5 \\times 10^{-7} \\text{ m/cycle}$ the pure-SCC-driven part is also activated. Using the Fracture Mechanics → Fatigue Crack Growth module in the site\u0026rsquo;s left sidebar, click \u0026ldquo;Load Example 1 (Austenitic – water environment)\u0026rdquo; to generate the detailed derivation charts and final data in one click.\n📖 References:\nKTA 3206 (2014-11) — Bruchausschluss für drucktragende Komponenten in Kernkraftwerken NUREG/CR-6176 — Review of Environmental Effects on Fatigue Crack Growth of Austenitic Stainless Steels ASME Section XI, Appendix A / Appendix O ","permalink":"https://mechcalc.net/blog/en/posts/08-fatigue-crack-growth/","summary":"\u003cblockquote\u003e\n\u003cp\u003e🧮 \u003cstrong\u003eGo straight to the calculator\u003c/strong\u003e: to skip the derivation and use the online tool with a built-in verification flow, open the \u003ca href=\"/calculators\"\u003efatigue crack growth calculator\u003c/a\u003e\n from the large category cards on the calculators home page.\u003c/p\u003e\n\u003c/blockquote\u003e\n\u003ch2 id=\"1-where-fatigue-crack-growth-sits-in-an-lbb-analysis\"\u003e1. Where fatigue crack growth sits in an LBB analysis\u003c/h2\u003e\n\u003cp\u003eAt the core of a Leak-Before-Break (LBB) argument: you must show that, over the service life, the time for an initial non-penetrating surface flaw to grow by \u003cstrong\u003esubcritical crack growth\u003c/strong\u003e through the wall and cause a leak — and the time from leak to growth to the critical fracture size — is long enough for the monitoring system to detect it and act.\u003c/p\u003e","title":"Fatigue Crack Growth and Environmental Effects (KTA 3206)"},{"content":"This article explains how to use the seven-step method of the KTA 3206 standard for a Leak-Before-Break (LBB) analysis of nuclear piping. It pairs with the KTA 3206 LBB Pipe Analysis module of the online calculator, taking you from theory to practice.\n🧮 Try the calculator now: open the KTA 3206 LBB pipe analysis calculator , choose \u0026ldquo;Fracture Mechanics → KTA 3206 LBB Pipe Analysis\u0026rdquo;, and enter the parameters to get results with a detailed formula derivation and visual charts.\n1. What is KTA 3206? KTA 3206 (full title: \u0026ldquo;Nachweise zum Bruchausschluss für drucktragende Komponenten in Kernkraftwerken\u0026rdquo;) is the German nuclear-safety standard for demonstrating break exclusion (Bruchausschluss) of pressure-bearing components in nuclear power plants.\nCore idea Unlike the US NRC LBB analysis (SRP 3.6.3), KTA 3206 is not only an analysis method but a whole integrity concept (Integritätskonzept), built on three pillars:\nPillar Content Purpose Basic quality design, material selection, fabrication ensure component quality from the source In-service quality water chemistry, operational monitoring, periodic inspection ensure the quality does not degrade Fracture-mechanics demonstration the seven-step / six-step analysis mathematically prove that fracture cannot occur Key differences from SRP 3.6.3 Item KTA 3206 SRP 3.6.3 Goal exclude fracture itself prove the crack leaks first Preconditions strictly exclude SCC, water hammer screening filter Crack growth includes fatigue growth analysis not included Safety margin verified in steps 6/7 leak rate ×10, crack ×2 2. Four preconditions Before any calculation, KTA 3206 requires the following preconditions. If any one is not met, the break-exclusion demonstration must not proceed.\nExclude stress corrosion cracking (SCC/DRK): through material selection and water-chemistry control. Exclude relevant vibration: especially high-frequency fatigue vibration. Exclude non-design dynamic loads: such as water hammer. Sufficient material toughness: must lie in the upper shelf region. 3. The seven-step method for piping in detail KTA 3206 Annex A defines the seven-step method for break exclusion of piping. Here is each step in detail:\nStep 1: determine the initial crack (a_a, 2c_a) ↓ Step 2: fatigue crack growth → end-of-life crack (a_e, 2c_e) ↓ Step 3: critical length of a through-wall crack 2c_krit (limit load method) ↓ Step 4: critical depth of a semi-elliptical crack a_krit(2c_e) ↓ Step 5: detectable crack length 2c_LÜS (leak-rate calculation) ↓ Step 6: allowable crack size verification ↓ Step 7: final LBB verdict Step 1: initial crack assumption KTA 3206 sets the enveloping initial crack from the material type and wall thickness $s$:\nAustenitic steel (Eq. A 2-1, A 2-2): $$ a_a = \\begin{cases} 0.3s \u0026 \\text{if } s \u003c 25\\text{ mm} \\\\\\\\ 0.2s \u0026 \\text{if } s \\geq 50\\text{ mm} \\end{cases} $$Ferritic steel (Eq. A 2-3, A 2-4): $$ a_a = \\begin{cases} 0.2s \u0026 \\text{if } s \u003c 25\\text{ mm} \\\\\\\\ 0.1s \u0026 \\text{if } s \\geq 50\\text{ mm} \\end{cases} $$The crack length is fixed at: $2c_a \\geq 6 \\times a_a$ (Eq. A 2-5)\n💡 Why is austenitic more conservative? Because austenitic stainless steel is more prone to fabrication weld flaws (hot cracks, penetrating IG cracks).\nStep 2: fatigue crack growth Use the Paris-Erdogan equation to compute crack growth over the service life:\n$$ \\frac{da}{dN} = C \\cdot (\\Delta K)^m $$The main considerations:\nthe normal operating transient load spectrum; environmental effects (especially for austenitic steel in a PWR environment, NUREG/CR-6176); the effect of residual stress. The end-of-life crack $a_e, 2c_e$ feeds into the later steps.\nStep 3: critical length of a through-wall crack (the core of this calculator) Per KTA 3206 Annex B2 (limit load method), different critical failure models are set out by crack orientation (longitudinal / circumferential) and pipe type (straight / bent), covering through-wall cracks (for step 3) and surface cracks (for step 4):\n1. Longitudinal crack under internal pressure (Längsriss) This covers longitudinal cracking of the pipe under internal pressure only (hoop stress dominant):\nStraight-pipe through-wall crack (Eq. B 2.1-7, B 2.1-3): critical failure pressure: $$ p_V = \\frac{2 \\cdot s}{D_i} \\cdot \\frac{\\sigma_f}{M_t} $$ with the bulging factor $M_t = \\sqrt{1 + 1.61 \\cdot \\left(\\frac{c^2}{r_m \\cdot s}\\right)}$. Straight-pipe surface crack (Eq. B 2.1-8, B 2.1-5): for the critical surface-crack depth $a_\\mathrm{krit}$: $$ p_V = \\frac{2 \\cdot s}{D_i} \\cdot \\frac{\\sigma_f}{M_p} $$ with the surface-crack factor $M_p = \\frac{1 - \\left(\\frac{a}{s \\cdot M_t}\\right)}{1 - \\left(\\frac{a}{s}\\right)}$. Bent-pipe crack (see B 2.1.2.2): uses the same formula model as the straight pipe, but the actual maximum hoop stress of the bend (per KTA 3201.2) must replace the pure-internal-pressure hoop stress. 2. Circumferential crack in a straight pipe under internal pressure and external moment (Umfangsriss) For circumferential cracking including the overall internal-pressure axial force and moment, the standard gives two main limit-load methods:\nA. PGL method (plastic limit load method)\nThe PGL method assumes the whole wall thickness is at the plastic flow stress $\\sigma_f$, and is the most conservative. For both through-wall and surface cracks it builds the critical-angle equation the same way (Eq. B 2.1-13):\n$$ \\frac{\\pi}{4} \\cdot \\frac{\\sigma_{ax,M}}{\\sigma_f} - \\cos\\left(\\frac{a}{s} \\cdot \\frac{\\alpha}{2} + \\frac{\\pi}{2} \\cdot \\frac{\\sigma_{ax,p}}{\\sigma_f}\\right) + \\frac{1}{2} \\cdot \\frac{a}{s} \\cdot \\sin(\\alpha) = 0 $$where:\n$\\alpha$: crack half-angle [rad] $a/s$: take $a/s = 1.0$ for a through-wall crack (the equation then simplifies), and the actual depth ratio for a surface crack $\\sigma_{ax,p} = p / [(D_a/D_i)^2 - 1]$: internal-pressure axial stress (Eq. B 2.1-14) $\\sigma_{ax,M} = M / W$: moment axial stress (Eq. B 2.1-15) $\\sigma_f$: flow stress (see Table B 2.1-1 below) For a through-wall crack, after solving for the critical angle $\\alpha_\\mathrm{krit}$, convert to the critical length: $$ 2c_\\mathrm{krit} = 2 \\cdot \\alpha_\\mathrm{krit} \\cdot r_m $$B. FSK method (flow-stress-concept method)\nIt introduces a local-yield concept, with two branches:\nFSK/MPA method: for a through-wall crack (Eq. B 2.1-17): $$ M_V = \\frac{I_{\\hat{x}}}{e} \\cdot \\left[ \\sigma_f - \\frac{A_0}{A_q - A_f} \\cdot p_i \\right] - \\hat{y} \\cdot A_0 \\cdot p_i $$ for a surface crack (Eq. B 2.1-18): $$ M_V = \\frac{I_{\\hat{x}}}{e} \\cdot \\left[ \\sigma_f - \\frac{A_w}{A_q - A_f} \\cdot p_i \\right] - (\\hat{y} + b) \\cdot A_w \\cdot p_i $$ it rigorously accounts for the neutral-axis shift $e$ and the remaining-section parameters to derive the bending limit load. FSK/KWU method (Eq. B 2.1-20, 21): defines the stress magnification factors $k_a, k_b$ so that the local effective stress $\\sigma_\\mathrm{eff}$ at the crack tip reaches the flow stress: $$ \\sigma_\\mathrm{eff} = k_a \\cdot \\sigma_{ax,p} + k_b \\cdot \\sigma_{ax,M} = \\sigma_f $$ The value of the flow stress $\\sigma_f$ (Table B 2.1-1) Method Austenitic steel Ferritic steel PGL $(R_{p0.2} + R_m) / 2.4$ $R_{p0.2}$ FSK/MPA $R_m$ $(R_{p0.2} + R_m) / 2$ FSK/KWU $(R_{p0.2} + R_m) / 2$ $R_m$ ⚠️ PGL is the most conservative, because it uses the lowest flow stress and gives the shortest critical crack length. PGL is the recommended first choice.\nStep 4: critical depth of a surface crack From the end-of-life surface-crack length $2c_e$ found in step 2, compute the critical surface-crack depth $a_\\mathrm{krit}(2c_e)$ that can still carry all occurring loads (operating and accident) at that length. In essence this again uses the limit load method of step 3 (the PGL or FSK straight-/bent-pipe formulas) for surface-crack failure: fixing the crack length at $2c_e$, back-solve for the maximum depth $a$ that causes section or local failure. This depth is an important input to the later steps.\nStep 5: detectable crack length (leak-rate calculation) To meet the core LBB definition, you must show that when a through-wall crack occurs, it produces enough leakage to be found by the monitoring system before reaching the critical length (Leak-Before-Break). The procedure:\nDetermine the monitoring capability: the minimum leakage mass flow $\\dot{m}_\\mathrm{L\\ddot{U}S,det}$ that the leak-monitoring system (LÜS) can reliably detect. Set the intervention limit: with the operating manual (BHB), set the leak-rate limit $\\dot{m}_\\mathrm{L\\ddot{U}S,BHB} \\geq \\dot{m}_\\mathrm{L\\ddot{U}S,det}$ that triggers intervention (leak search, shutdown). Compute the detectable length: accounting for friction, flashing and surface roughness of the fluid in the crack (e.g. using the Pana or Moody model), find the through-wall crack length that at normal operating pressure produces exactly that intervention leak rate, and define it as the detectable crack length $2c_\\mathrm{L\\ddot{U}S}$. Step 6: allowable crack size verification Allowable crack depth: $a_\\mathrm{zul} = \\min(0.75 \\times s,\\ a_\\mathrm{krit}(2c_e))$ (Eq. A 2-10) Allowable crack length: $2c_\\mathrm{zul} = 2c_\\mathrm{krit} - \\Delta 2c_\\mathrm{WKP}$ (Eq. A 2-11) Verification conditions: $$ a_e \\leq a_\\mathrm{zul}, \\quad 2c_e \\leq 2c_\\mathrm{zul} $$Step 7: final LBB verdict If the pipe needs an LBB demonstration (i.e. no pipe whip restraint is fitted), you also need:\n$$ 2c_\\mathrm{LÜS} \u003c 2c_\\mathrm{zul} $$where $2c_\\mathrm{LÜS}$ is the detectable crack length from the leak-rate calculation.\n4. Worked example: austenitic pipe The following data are from KTA 3206 Annex D1, to demonstrate the full analysis flow.\nInput data Parameter Symbol Value Unit Inner diameter $D_i$ 243 mm Wall thickness $s$ 15 mm Material — X6 CrNiNb 18 10 (1.4550) — Yield strength $R_{p0.2T}$ 167 MPa Tensile strength $R_{mT}$ 409 MPa Elastic modulus $E$ 186,000 MPa Internal pressure $p$ 7.4 MPa Bending moment $M$ 72.3 × 10⁶ N·mm Analysis results Step Parameter Result Step 1 initial crack $a_a = 4.5$ mm, $2c_a = 27$ mm Step 2 end-of-life crack $a_e = 4.78$ mm, $2c_e = 28.1$ mm Step 3 (PGL) flow stress $\\sigma_f = 240$ MPa Step 3 (PGL) critical crack length $2c_\\mathrm{krit} = 273.9$ mm Step 6 allowable crack depth $a_\\mathrm{zul} = 11.25$ mm Step 6 allowable crack length $2c_\\mathrm{zul} = 271.9$ mm Step 7 LBB safety margin $271.9 / 60 = 4.5$ ✅ Conclusion: break exclusion holds. The allowable crack length (271.9 mm) is far larger than the detectable crack length (60 mm), a margin of over 4.5.\nUsing it in MechCalc Open the MechCalc calculator . Choose \u0026ldquo;KTA 3206 LBB Pipe Analysis\u0026rdquo;. Enter the parameters from the table above. Click \u0026ldquo;Calculate\u0026rdquo;. Review the detailed results and the verdict. 5. Special considerations for ferritic piping The main differences between ferritic and austenitic piping:\nSmaller initial crack ($0.2s$ vs $0.3s$): ferritic steel has better weld quality. Higher yield strength: so the PGL flow stress $\\sigma_f = R_{p0.2}$ is larger. Toughness verification: the operating temperature must be confirmed to be in the upper shelf region. Environmental effects: not affected by PWR-water acceleration. 6. Summary The seven-step method of KTA 3206 provides a systematic, repeatable analysis framework for break exclusion of nuclear piping. Its core logic:\nassume the worst initial crack (a conservative assumption); let the crack \u0026ldquo;run\u0026rdquo; a full life under load (fatigue growth); compute \u0026ldquo;how long a crack the pipe can carry\u0026rdquo; (limit load method); verify the crack has ample safety margin (allowable crack size and LBB). Only when all steps pass can break exclusion be declared.\n📖 Reference standard: KTA 3206 (2014-11) — Bruchausschluss für drucktragende Komponenten in Kernkraftwerken\n","permalink":"https://mechcalc.net/blog/en/posts/kta3206-lbb-analysis-guide/","summary":"\u003cp\u003eThis article explains how to use the \u003cstrong\u003eseven-step method\u003c/strong\u003e of the \u003cstrong\u003eKTA 3206\u003c/strong\u003e standard for a \u003cstrong\u003eLeak-Before-Break (LBB)\u003c/strong\u003e analysis of nuclear piping. It pairs with the \u003cstrong\u003eKTA 3206 LBB Pipe Analysis\u003c/strong\u003e module of the online calculator, taking you from theory to practice.\u003c/p\u003e\n\u003cblockquote\u003e\n\u003cp\u003e🧮 \u003cstrong\u003eTry the calculator now\u003c/strong\u003e: open the \u003ca href=\"/calculators\"\u003eKTA 3206 LBB pipe analysis calculator\u003c/a\u003e\n, choose \u0026ldquo;Fracture Mechanics → KTA 3206 LBB Pipe Analysis\u0026rdquo;, and enter the parameters to get results with a detailed formula derivation and visual charts.\u003c/p\u003e","title":"KTA 3206 LBB Analysis Guide: Break Exclusion for Piping by the Seven-Step Method"},{"content":"This is a detailed introductory guide to Leak-Before-Break (LBB).\nLBB is a crucial safety concept in the structural integrity assessment of pressure vessels and piping. In short, it is an analysis that proves the equipment will produce a detectable leak first, before any catastrophic fracture — giving operators time to shut down and avoid a major accident.\n1. What is Leak-Before-Break (LBB)? Core definition: LBB is a property of a pressure-bearing system (pipe, vessel). It ensures that when a flaw (crack) is present, the crack will penetrate the wall to form a through-wall crack and produce a stable, detectable leak before it grows to the critical size that would cause overall structural instability (sudden fracture or plastic collapse).\nAn intuitive picture: Imagine a \u0026ldquo;race\u0026rdquo;:\nRunner A (leak detection): the crack penetrates the wall, fluid sprays out, and the leak-detection system alarms. Runner B (catastrophic fracture): the crack keeps extending in length until the pipe bursts like a firecracker. The goal of LBB is to prove that Runner A always beats Runner B — that before the crack grows long enough to break the pipe, we have already found it from the leak signal and shut down safely. Two key sizes in an LBB analysis:\nLeakage crack length ($2c_{Leak}$): the crack length needed to produce exactly the smallest detectable leak rate (e.g. 1 gpm). Critical crack length ($2c_{Crit}$): the crack length that causes sudden fracture of the pipe under the accident load. Acceptance criterion: The LBB argument holds only when $2c_{Crit}$ is much larger than $2c_{Leak}$ (typically a factor-of-2 margin) and the leak-detection time is much shorter than the crack growth time.\n2. Where LBB is applied LBB originated in the nuclear industry and has since spread to other high-hazard sectors.\nNuclear power (the main field): Primary-loop main piping: the most mature LBB application. If the main piping can be shown to satisfy LBB, the design can omit the costly and complex pipe whip restraints. Purpose: to argue that the extreme \u0026ldquo;double-ended guillotine break\u0026rdquo; (DEGB) accident is physically impossible, and so optimize the design and in-service inspection plan. Petrochemical: assessing pressure vessels and piping carrying toxic or flammable media. API 579-1 explicitly treats LBB as an important means of assessing the remaining life of crack-like flaws, especially when the crack growth rate cannot be known accurately — LBB can then serve as a safety-assurance strategy. Offshore engineering: assessment of platform piping and tubular joints (as mentioned in BS 7910 Annex B). When LBB does not apply:\nVery fast fracture risk: e.g. materials with a very high risk of brittle fracture. Environmental limits: the leak causes the medium to flash-freeze the crack tip, or triggers an explosion (e.g. high-pressure gas). Mechanisms that cause long cracks: e.g. long-range water hammer, or severe stress corrosion cracking (which may form a very long surface crack without penetrating). 3. The core codes and standards for LBB The major industrial countries all have mature LBB assessment standards; they are broadly similar in logic but differ slightly in parameter conservatism.\nStandard system Standard Character US (API/ASME) API 579-1 / ASME FFS-1 (Part 9, Annex 9E) For petrochemical and nuclear; gives detailed $K_I$ and reference-stress methods. The xLPR project developed advanced probabilistic LBB code. US (NRC) NUREG-1061 Vol. 3 / SRP 3.6.3 The US NRC standard, defining strict LBB margins (leak rate ×10, crack size ×2). UK/Europe R6 (Section III.11) The UK nuclear assessment standard, with very detailed \u0026ldquo;detectable leak\u0026rdquo; and \u0026ldquo;full LBB\u0026rdquo; procedures. UK BS 7910 (Annex F) Covers general industrial structures, combining fracture-mechanics assessment with leak-rate calculation. Germany KTA 3206 The German nuclear-safety standard, emphasizing conservatism in fracture-mechanics analysis and leak-rate calculation. European FITNET (Section 11.2) Integrates the technology of several European countries into a unified fitness-for-service (FFS) procedure. 4. The detailed steps of an LBB analysis (introductory) Following the API 579 and R6 procedures, a standard LBB analysis usually has these steps:\nStep 1: compute the critical crack size ($2c_{Crit}$) Goal: find how long a crack must be to break the pipe under the extreme load (e.g. earthquake + pressure). Method: use the Failure Assessment Diagram (FAD). use API 579 Annex 9B to compute the stress intensity factor $K$; use API 579 Annex 9C to compute the reference stress $\\sigma_{ref}$; find the critical point on the FAD curve to get $2c_{Crit}$. Step 2: compute the leakage crack size ($2c_{Leak}$) Goal: find how long a crack produces a detectable leak under normal operating load. Difficulty: this is a fluid-mechanics problem — you must compute the crack opening area (COA) and the two-phase leak rate. Models: COA (crack opening area): the crack is not rigid; the higher the pressure the more it opens. An elastic-plastic correction is needed. Leak-rate model: common ones are the Henry-Fauske model (for two-phase critical flow) or the SQUIRT program. Crack morphology: a real crack is rough and tortuous (like a maze), which greatly impedes flow. The analysis must account for surface roughness, number of turns and similar parameters. Step 3: margin check This is the \u0026ldquo;gold standard\u0026rdquo; for whether LBB holds (per NUREG-1061 and SRP 3.6.3):\nLeak-rate margin: assuming the leak-detection system\u0026rsquo;s capability is 1 gpm (gallons per minute), the computed leak rate is usually required to be at least 10 gpm (a factor-of-10 safety factor). Crack-size margin: the critical crack length $2c_{Crit}$ must be at least twice the leakage crack length $2c_{Leak}$. 5. A typical case: LBB analysis of nuclear piping Scenario: A nuclear plant\u0026rsquo;s main coolant pipe (austenitic stainless steel), inner diameter 700 mm, wall thickness 70 mm. It must be shown to satisfy LBB so that pipe whip restraints can be omitted.\nAnalysis:\nLoad analysis:\nnormal operation: internal pressure 15.5 MPa, temperature 300°C, small bending moment. accident case (SSE earthquake): a large seismic bending moment added on top of the normal load. Determine $2c_{Crit}$ (worst case):\nassume a circumferential through-wall crack. input the accident-case (seismic) load, using lower-bound material toughness. result: the pipe fractures when the crack length reaches 300 mm. Determine $2c_{Leak}$ (normal case):\nassume normal operation (no earthquake). the plant leak-detection system has a sensitivity of 1 gpm (3.8 L/min). for margin, set the target leak rate to 10 gpm. using the Henry-Fauske model: to produce a 10 gpm leak the crack must open a certain width; back-calculating through the mechanical model gives a required crack length of 100 mm. Note: the calculation accounts for crack-surface roughness (the morphology parameters of a fatigue or stress-corrosion crack). Verdict:\ncritical crack (300 mm) vs. leakage crack (100 mm). 300 mm \u0026gt; 2 × 100 mm. the factor-of-2 margin is satisfied. Conclusion: the pipe satisfies LBB. This means that if a crack develops, it will send a strong leak alarm by the time it reaches 100 mm — still a large safety distance from the 300 mm fracture limit, leaving the plant ample time to shut down and repair. Summary LBB is the bridge between fracture mechanics (computing when a crack breaks) and thermal-hydraulics (computing how much a crack leaks). It embodies the advanced safety philosophy of modern high-energy piping design — \u0026ldquo;defend by monitoring\u0026rdquo;.\n","permalink":"https://mechcalc.net/blog/en/posts/lbb-leak-before-break-guide/","summary":"\u003cp\u003eThis is a detailed introductory guide to \u003cstrong\u003eLeak-Before-Break (LBB)\u003c/strong\u003e.\u003c/p\u003e\n\u003cp\u003eLBB is a crucial safety concept in the structural integrity assessment of pressure vessels and piping. In short, it is an analysis that proves the equipment will \u003cstrong\u003eproduce a detectable leak first\u003c/strong\u003e, before any catastrophic fracture — giving operators time to shut down and avoid a major accident.\u003c/p\u003e\n\u003chr\u003e\n\u003ch3 id=\"1-what-is-leak-before-break-lbb\"\u003e1. What is Leak-Before-Break (LBB)?\u003c/h3\u003e\n\u003cp\u003e\u003cstrong\u003eCore definition:\u003c/strong\u003e\nLBB is a property of a pressure-bearing system (pipe, vessel). It ensures that when a flaw (crack) is present, the crack will penetrate the wall to form a through-wall crack and produce a \u003cstrong\u003estable, detectable leak\u003c/strong\u003e before it grows to the critical size that would cause overall structural instability (sudden fracture or plastic collapse).\u003c/p\u003e","title":"LBB (Leak-Before-Break): An Introductory Guide"}]