🧮 Online calculator: VDI 2230-1 Single-bolt Joint Calculation — The full 14-step calculation chain (R0–R13), with six strength checks. Use the inputs in the section “Re-run it in the calculator” to check this example step by step.
Full worked example: through-bolt joint (DSV)
This article ties the theory of the previous 7 articles into one calculation chain. It works a through-bolt joint (DSV) example of our own through R0–R13 of VDI 2230, step by step; it matches the VDI 2230-1 single-bolt joint calculator on this site. All inputs are listed in the section “Re-run it in the calculator” at the end, so you can check the numbers yourself.
Engineering problem statement
A steel flange joint using a single bolt to fix a cover plate. The design requirements are:
| Parameter | 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 (static) | Perpendicular to the bolt axis, carried by friction at the interface |
| Clamp length | $l_K = 40$ mm | Two clamped plates, 20 mm each |
| Clamped-parts outer diameter | $D_A = 44$ mm | Equivalent outer diameter at the interface |
| Clamped-parts material | S355 JO, $E_P = 205\,000$ MPa | $E_P$ and limiting surface pressure $p_G = 760$ MPa, both from VDI 2230:2015, Tabelle A9 |
| Operating temperature | Room temperature (no thermal effect) | $\Delta F'_{Vth} = 0$ |
| Coefficient of friction at the interface | $\mu_T = 0.12$ | Steel-on-steel interface |
| Coefficient of friction in the thread | $\mu_G = 0.12$ | |
| Coefficient of friction under the head and nut | $\mu_K = 0.12$ | Under the bolt head and the nut |
| Number of interfaces | $q_F = 1$ (single shear) | |
| Load introduction factor | $n = 0.5$ | Force introduced near the interface |
| Tightening method | Torque wrench, target torque found by tests | $\alpha_A$ see R1 |
R0 — preliminary diameter selection
Based on experience or a quick pre-selection (see the pre-selection article ), preliminarily select:
- Bolt: 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, bearing diameter $d_W = 17.0$ mm (the default value for a hexagon head in our calculator)
- 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
Limiting size $G$: this joint is concentrically clamped and concentrically loaded. The limit $G = h_{\min} + d_W$ of (R0/1) is a validity limit for eccentrically clamped or eccentrically loaded joints, so it does not need to be checked here (VDI 2230:2015, R0). Note also that $h_{\min}$ in this formula is the thickness of the thinner of the two plates (20 mm here), not the clamp length $l_K$ (VDI 2230:2015, §5.1.2.2, Gl.(54)).
[!note] What the calculator’s design mode would do The design mode of our calculator estimates the diameter from Tabelle A7 and gives M14 for this case. But M14 also fails R7: $F_{Mzul} = 86\,869$ N is smaller than $F_{M\max} = 101\,767$ N, and the safety margin against slipping $S_G$ is only 0.78. So below we go straight from M12 to M16; M14 does not need to be tried.
R1 — tightening factor
Torque-controlled tightening with a torque wrench, target torque found by tests: Tabelle A8 gives $\alpha_A = 1.4 \sim 1.6$ for this class (VDI 2230:2015, Tabelle A8). We take the upper value to stay on the safe side:
$$ \alpha_A = 1.6 $$R2 — required minimum clamp load
This case has a transverse force to transmit by friction (Eq. R2/1):
$$ F_{KQ} = \frac{F_{Q\max}}{q_F \cdot \mu_{T\min}} = \frac{5\,000}{1 \times 0.12} = 41\,667 \text{ N} $$No sealing requirement, no bending moment → $F_{KP} = 0$, $F_{KA} = 0$
$$ F_{Kerf} = F_{KQ} = 41\,667 \text{ N} \tag{R2/4} $$R3 — elastic resiliences and load factor
Bolt resilience
The bolt consists of: head + 20 mm plain shank + 20 mm free thread + engaged thread + nut region.
| Segment | 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 | $A_N = 113.1$ | $8.42 \times 10^{-7}$ |
| Free thread $\delta_{Gew}$ | 20 mm | $A_{d_3} = 76.2$ | $1.249 \times 10^{-6}$ |
| Engaged thread $\delta_G$ | $0.5 \times 12 = 6$ mm | $A_{d_3} = 76.2$ | $3.75 \times 10^{-7}$ |
| Nut $\delta_M$ | $0.4 \times 12 = 4.8$ mm | $A_N = 113.1$ | $2.02 \times 10^{-7}$ |
Clamped-parts resilience
First compute the cone angle (Eq. 43, 44, 45). No separate outer diameter of the base body is given, so $D_A' = D_A$:
$$ \beta_L = l_K / d_W = 40 / 17.0 = 2.353 $$$$ y = D_A' / d_W = 44 / 17.0 = 2.588 $$$$ \tan\varphi_D = 0.362 + 0.032 \ln(2.353/2) + 0.153 \ln(2.588) = 0.362 + 0.0052 + 0.1455 = 0.5127 $$Limiting diameter (Eq. 39, $w = 1$ for DSV):
$$ D_{A,Gr} = 17.0 + 1 \times 40 \times 0.5127 = 37.5 \text{ mm} $$$D_A = 44 \text{ mm} \geq D_{A,Gr} = 37.5$ mm, so there is no deformation sleeve; use Eq. 40 (pure cone):
$$ \delta_P = \frac{2 \ln\left[\frac{(17.0+13.5)(17.0+1 \times 40 \times 0.5127-13.5)}{(17.0-13.5)(17.0+1 \times 40 \times 0.5127+13.5)}\right]}{1 \times 205\,000 \times \pi \times 13.5 \times 0.5127} $$$$ = \frac{2 \ln\left[\frac{30.5 \times 24.01}{3.5 \times 51.01}\right]}{4\,458\,000} = \frac{2 \ln(4.102)}{4\,458\,000} = \frac{2 \times 1.411}{4\,458\,000} = 6.33 \times 10^{-7} \text{ mm/N} $$Load factor Φ
Concentric clamping and loading, so (R3/3) applies. For DSV $w=1$, and by Gl.(81) $\delta_{PZu}=(w-1)\delta_M=0$, so (R3/3) becomes:
$$ \Phi_n = n \cdot \frac{\delta_P}{\delta_S + \delta_P} = 0.5 \times \frac{6.33 \times 10^{-7}}{2.92 \times 10^{-6} + 6.33 \times 10^{-7}} = 0.5 \times \frac{6.33}{35.54} = 0.0891 $$Load factor $\Phi = 0.0891$ — the bolt takes only 8.9% of the external force!
Additional bolt load and plate unloading force:
$$ F_{SA} = 0.0891 \times 20\,000 = 1\,782 \text{ N} $$$$ F_{PA} = (1 - 0.0891) \times 20\,000 = 18\,218 \text{ N} $$R4 — preload change
Preload loss due to embedding
Take the amount of embedding (Setzen) $f_Z$ from Tabelle 5. The interface roughness here is $R_z < 10\ \mu$m; the transverse force is carried by friction at the interface, so we use the shear-load row. A DSV has two bearing surfaces (bolt head and nut) and 1 inner interface:
$f_Z$ = thread 3 μm + head bearing surface 3 μm + nut bearing surface 3 μm + 1 inner interface 2 μm = 11 μm (VDI 2230:2015, §5.4.2.1, Tabelle 5)
$$ F_Z = \frac{f_Z}{\delta_S + \delta_P} = \frac{0.011}{3.554 \times 10^{-6}} = 3\,095 \text{ N} \tag{R4/1} $$Thermal expansion
Room-temperature case → $\Delta F'_{Vth} = 0$
R5 — minimum assembly preload
$$ F_{M\min} = 41\,667 + (1 - 0.0891) \times 20\,000 + 3\,095 + 0 = 62\,980 \text{ N} \tag{R5/1} $$R6 — maximum assembly preload
$$ F_{M\max} = 1.6 \times 62\,980 = 100\,768 \text{ N} \tag{R6/1} $$R7 — assembly stress check
Permissible assembly preload (Eq. R7/2, $\nu = 0.9$, $d_0 = d_S = (d_2 + d_3)/2 = 10.358$ mm, $A_0 = \frac{\pi}{4} d_0^2 = 84.26$ mm²):
$$ F_{Mzul} = 84.26 \times \frac{0.9 \times 940}{\sqrt{1 + 3\left[\frac{3}{2} \times \frac{10.863}{10.358}\left(\frac{1.75}{\pi \times 10.863} + 1.155 \times 0.12\right)\right]^2}} $$$$ = 84.26 \times \frac{846}{\sqrt{1 + 3 \times [1.573 \times (0.0513 + 0.1386)]^2}} $$$$ = 84.26 \times \frac{846}{\sqrt{1 + 3 \times 0.0892}} = 84.26 \times \frac{846}{\sqrt{1.268}} = 84.26 \times 751.4 = 63\,315 \text{ N} $$The table values $F_{MTab}$ in Tabelle A1–A4 are calculated with this same formula at 90 % use of the yield point (rounding error no more than ±2.5 %), so for $\nu = 0.9$ you can take $F_{Mzul}$ straight from the table (VDI 2230:2015, §5.5.1).
Check (Eq. R7/3):
$$ F_{Mzul} = 63\,315 \text{ N} \quad \text{vs} \quad F_{M\max} = 100\,768 \text{ N} $$$$ 63\,315 < 100\,768 \;\Rightarrow\; F_{Mzul} < F_{M\max} $$❌ Not satisfied: even at the permissible assembly preload, M12 cannot cover the upper end of the tightening scatter.
Iteration: upgrade to M16
M12 is not enough; upgrade to M16 × 2.0, property class 10.9:
| Parameter | 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 | $\pi/4 \times 16^2$ |
| $d_W$ | 22.5 mm | Default hexagon-head value in our calculator |
| $d_h$ | 17.5 mm | DIN EN 20273 |
| $F_{Mzul}$ | 118 928 N | Same R7/2 formula, $\mu_G = 0.12$ |
Recompute from R2 (with $F_{Kerf}$ unchanged):
δS (M16):
| Segment | 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 thread | $0.5 \times 16 = 8$ 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 | $1.777 \times 10^{-6}$ |
δP (M16):
$\beta_L = 40/22.5 = 1.778$, $y = 44/22.5 = 1.956$
$\tan\varphi_D = 0.362 + 0.032\ln(0.889) + 0.153\ln(1.956) = 0.362 - 0.0038 + 0.1026 = 0.4608$
$D_{A,Gr} = 22.5 + 40 \times 0.4608 = 40.9$ mm
$D_A = 44 \text{ mm} \geq 40.9$ mm → Eq. 40:
$$ \delta_P = \frac{2\ln\left[\frac{(22.5+17.5)(22.5+40 \times 0.4608-17.5)}{(22.5-17.5)(22.5+40 \times 0.4608+17.5)}\right]}{205\,000 \times \pi \times 17.5 \times 0.4608} = \frac{2 \ln(3.208)}{5\,194\,000} = 4.49 \times 10^{-7} \text{ mm/N} $$Φ (M16):
$$ \Phi = 0.5 \times \frac{4.49 \times 10^{-7}}{1.777 \times 10^{-6} + 4.49 \times 10^{-7}} = 0.5 \times 0.202 = 0.1008 $$R4 (M16): $F_Z = 0.011 / (2.226 \times 10^{-6}) = 4\,941$ N
R5: $F_{M\min} = 41\,667 + (1 - 0.1008) \times 20\,000 + 4\,941 = 41\,667 + 17\,984 + 4\,941 = 64\,591$ N
R6: $F_{M\max} = 1.6 \times 64\,591 = 103\,346$ N
R7: $F_{Mzul}(\text{M16}) = 118\,928$ N
$$ 118\,928 \geq 103\,346 \;\Rightarrow\; F_{Mzul} \geq F_{M\max} $$✅ Satisfied.
R8–R12 check summary (M16-10.9)
| Step | Check item | Computed value | Permissible value | Safety factor | Result |
|---|---|---|---|---|---|
| R8 | Service stress | $\sigma_{red,B} = 811.1$ N/mm² | $R_{p0.2} = 940$ N/mm² | $S_F = 1.16$ | ✅ |
| R9 | Fatigue | Static load, $\sigma_a = 0$ | — | — | not needed |
| R10 | Surface pressure | No washer: $p_{B\max} = 770$ N/mm² | $p_G = 760$ N/mm² | $S_{pB} = 0.99$ | ❌ |
| R10 | Surface pressure (with washer) | $p_{M\max} = 345$ N/mm² | $p_G = 760$ N/mm² | $S_P = 2.20$ | ✅ |
| R11 | Length of engagement | Not applicable for DSV (standard nut) | — | — | — |
| R12 | Slip | $F_{KR\min} = 51\,405$ N | $F_{KQerf} = 41\,667$ N | $S_G = 1.23$ (required ≥ 1.2) | ✅ |
R8 — service stress
After torque-controlled tightening, the torsional stress from tightening stays in the bolt, so it must be included in the comparative stress by (R8/4). In service the torsional stress is reduced with $k_\tau = 0.5$ (VDI 2230:2015, R8):
$$ F_{S\max} = F_{Mzul} + \Phi F_A = 118\,928 + 0.1008 \times 20\,000 = 120\,945 \text{ N} \tag{R8/1} $$$$ \sigma_{z\max} = \frac{F_{S\max}}{A_0} = \frac{120\,945}{156.7} = 772.0 \text{ N/mm}^2, \qquad \tau_{\max} = \frac{M_G}{W_P} = \frac{159\,017}{553.2} = 287.5 \text{ N/mm}^2 $$$$ \sigma_{red,B} = \sqrt{\sigma_{z\max}^2 + 3(k_\tau \tau_{\max})^2} = \sqrt{772.0^2 + 3 \times (0.5 \times 287.5)^2} = 811.1 \text{ N/mm}^2 \tag{R8/4} $$$$ S_F = \frac{R_{p0.2\min}}{\sigma_{red,B}} = \frac{940}{811.1} = 1.16 \geq 1.0 $$The tension-only check (R8/6), $R_{p0.2} A_0 \geq F_{S\max}$, is only for the case where the torsional stress is fully gone or the bolt is tightened without torsion. Used wrongly here, it would make $S_F$ about 5 % too high.
R10 — surface pressure
The bearing area is taken as a flat ring, $A_{p\min} = \frac{\pi}{4}(d_{Wa}^2 - D_{Ki}^2)$ (Gl.(193)). No hole chamfer is given, so the inner diameter $D_{Ki}$ is taken as the through-hole diameter $d_h = 17.5$ mm; this makes the area a little too large, so the result is on the unsafe side.
Without a washer ($d_{Wa} = d_W = 22.5$ mm, $A_{p\min} = 157.1$ mm²):
- Assembly state: $p_{M\max} = F_{Mzul}/A_{p\min} = 118\,928/157.1 = 757$ N/mm², $S_P = 760/757 = 1.00$ — right at the limit;
- Working state: $p_{B\max} = (F_{V\max} + F_{SA})/A_{p\min} = (118\,928 + 2\,017)/157.1 = 770$ N/mm², $S_{pB} = 0.99$ — fails ($F_{V\max}$ is taken as $F_{Mzul}$, on the safe side).
Above the limiting surface pressure, the clamped part creeps under the head and the preload keeps dropping. The fix is a hardened washer, $h_S = 3$ mm thick, under the head and under the nut. For the pressure calculation its outer diameter is (VDI 2230:2015, §5.5.4, Gl.(194)):
$$ d_{Wa} = d_W + 1.6\,h_S = 22.5 + 1.6 \times 3 = 27.3 \text{ mm} $$$A_{p\min} = \frac{\pi}{4}(27.3^2 - 17.5^2) = 344.8$ mm², $p_{M\max} = 345$ N/mm², $S_P = 2.20$; $p_{B\max} = 351$ N/mm², $S_{pB} = 2.17$. ✅ Passed. The washer itself must also be strong and thick enough (VDI 2230:2015, §5.5.4).
R12 — slip resistance
The residual clamp load is taken for the worst case: the lowest assembly preload $F_{Mzul}/\alpha_A$, minus the plate unloading force and the preload loss due to embedding:
$$ F_{KR\min} = \frac{F_{Mzul}}{\alpha_A} - (1-\Phi)F_A - F_Z = \frac{118\,928}{1.6} - (1 - 0.1008) \times 20\,000 - 4\,941 = 51\,405 \text{ N} $$$$ S_G = \frac{F_{KR\min}}{F_{KQerf}} = \frac{51\,405}{41\,667} = 1.23 \geq 1.2 $$For static load, the usual required value of $S_G$ is 1.2 (VDI 2230:2015, R12). This joint just passes. The other two R12 checks, bolt shear $S_A = 20.2$ and bolt bearing pressure $S_L = 9.18$, have large margins and do not govern.
R13 — tightening torque
(Eq. R13/1, M16, $\mu_G = \mu_K = 0.12$). Friction diameter of the bearing surface $D_{Km} = (d_W + D_{Ki})/2 = (22.5 + 17.5)/2 = 20.0$ mm:
$$ M_A = 118\,928 \times \left[0.16 \times 2.0 + 0.58 \times 14.701 \times 0.12 + \frac{20.0}{2} \times 0.12\right] $$$$ = 118\,928 \times [0.320 + 1.023 + 1.200] = 118\,928 \times 2.543 = 302\,458 \text{ N·mm} $$$$ M_A \approx 302 \text{ N·m} $$Result summary
| Item | Value |
|---|---|
| Final selection | M16 × 2.0 — 10.9, with a 3 mm hardened washer under the head and under the nut |
| Minimum assembly preload $F_{M\min}$ | 64 591 N |
| Maximum assembly preload $F_{M\max}$ | 103 346 N |
| Permissible assembly preload $F_{Mzul}$ | 118 928 N |
| Load factor $\Phi$ | 0.1008 |
| Safety margin against exceeding the yield point $S_F$ (service stress) | 1.16 |
| Safety factor $S_P$ (surface pressure, with washer) | 2.20 |
| Safety factor $S_G$ (slip, required ≥ 1.2) | 1.23 |
| Tightening torque $M_A$ | ≈ 302 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 clamp load 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 $F_{M\max}$. With M16 the bolt itself is fine, but the surface pressure is just over the limit, and a washer is needed. This is exactly the value of the systematic VDI 2230 method — it reveals the weak link that intuition easily overlooks.
Re-run it in the calculator
Open the VDI 2230-1 single-bolt joint calculator , fill in the inputs below, keep all other inputs at their defaults, and click Run Calculation. This is our own example, so the table gives every input.
| Input (label in the calculator) | Value | Note |
|---|---|---|
| Calculation Mode | Verification Mode (specified) | Design mode gives M14, see the note under R0 |
| Joint Type | TBJ — Through bolt/DSV | |
| Bolt size | M16 | Enter M12 first to see R7 fail |
| Head Type | Hexagon head | $d_W$ uses the default 22.5 mm (17.0 mm for M12) |
| Clamped layers | Two layers, 20 mm each | For each layer click the material button and pick S355 JO from Table A9 (fills in $E_P = 205\,000$ and $p_G = 760$) |
| Equivalent OD at the interface $D_A$ | 44 mm | |
| Through-hole Diameter $d_h$ | 17.5 mm | 13.5 for M12 |
| Unthreaded shank length in the clamp $l_1$ | 16 mm | 20 for M12 |
| Washer thickness under the head $h_S$ | 3 mm | Leave empty to see the R10 result without a washer |
| Head-side bearing-face inner diameter $D_{Ki}$ | empty | The calculator uses $d_h$ and says so in the steps |
| Axial Load $F_A$ / Transverse Load $F_Q$ | 20 000 N / 5 000 N | |
| Load Introduction Factor $n$ | 0.5 | |
| Strength Class | 10.9 | $R_{p0,2} = 940$ and $R_m = 1040$ are the defaults |
| Thread / Head / Interface Friction | 0.12 / 0.12 / 0.12 | Leave the friction class unselected |
| Number of force-transmitting inner interfaces $q_F$ | 1 | |
| Tightening Method / Tightening Factor $\alpha_A$ | Torque controlled / 1.6 | |
| Utilisation $\nu$ | 0.9 | |
| Embedding $f_Z$ | 0.011 mm | The 11 μm embedding amount from R4 |
| Include Thermal Effects / Fatigue Check | No / No | Static load at room temperature |
How the results map to this article (M16 with washer):
| Calculator output | Calculator value | Where in this article |
|---|---|---|
| Bolt resilience $\delta_S$ / clamped-part resilience $\delta_P$ | $1.777 \times 10^{-6}$ / $4.489 \times 10^{-7}$ mm/N | Iteration M16 |
| Load factor $\Phi$ | 0.1008 | Iteration M16 |
| Preload loss due to embedding $F_Z$ | 4 941 N | R4 |
| $F_{M\min}$ / $F_{M\max}$ / $F_{Mzul}$ | 64 591 / 103 346 / 118 928 N | R5–R7 |
| Maximum bolt load $F_{S\max}$ / comparative stress $\sigma_{red,B}$ | 120 945 N / 811.1 N/mm² | R8 |
| Assembly-state / working-state pressure | 344.9 / 350.7 N/mm² | R10 |
| Minimum residual clamp load $F_{KR\min}$ | 51 405 N | R12 |
| Tightening torque $M_A$ | 302 458 N·mm | R13 |
| $S_F$ / $S_P$ / $S_{pB}$ / $S_G$ / $S_A$ / $S_L$ | 1.16 / 2.20 / 2.17 / 1.23 / 20.20 / 9.18 | Summary |
The calculator also gives one more value: the concentric opening limit $F_{Aab} = 77\,169$ N (Gl.(87)), far above $F_A = 20\,000$ N, so the interface does not open.
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 through-hole diameter from DIN EN 20273, and the elastic modulus and limiting surface pressure of the clamped parts from VDI 2230 Tabelle A9. The bearing diameter $d_W$ of the hexagon head is the default value in our calculator.
Disclaimer: This article is for engineering teaching reference only. The calculation assumes concentric clamping and concentric loading. In real engineering use, all parameters should be verified against the specific conditions.
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