🧮 在线计算器：VDI 2230 Bolted Joint Calculation — The full 14-step calculation chain (R0–R13), with six strength checks.
Full 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.
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 | 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:
- 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 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):
$$ G = h_{\min} + d_W = 40 + 16.6 = 56.6 \text{ mm} $$$c_T = D_A = 44 \text{ mm} < G = 56.6$ mm → ✅ satisfied
R1 — tightening factor
Precision torque wrench (friction coefficient known), from Table A8 (VDI 2230:2015):
$$ \alpha_A = 1.6 $$R2 — required minimum clamping force
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 force ratio
Bolt resilience $\delta_S$
Assume the bolt consists of: head + 20 mm plain shank + 20 mm free thread + engaged segment + 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 | $A_N = 113.1$ | $8.43 \times 10^{-7}$ |
| 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 | $A_N = 113.1$ | $2.02 \times 10^{-7}$ |
Clamped-parts resilience $\delta_P$
First compute the cone angle (Eq. 43, 44, 45):
$$ \beta_L = l_K / d_W = 40 / 16.6 = 2.41 $$$$ y = D_A' / d_W = 44 / 16.6 = 2.65 $$$$ \tan\varphi_D = 0.362 + 0.032 \ln(2.41/2) + 0.153 \ln(2.65) = 0.362 + 0.006 + 0.149 = 0.517 $$Limiting diameter (Eq. 39, $w = 1$ for DSV):
$$ D_{A,Gr} = 16.6 + 1 \times 40 \times 0.517 = 37.3 \text{ mm} $$$D_A = 44 > D_{A,Gr} = 37.3$ → use Eq. 40 (pure cone):
$$ \delta_P = \frac{2 \ln\left[\frac{(16.6+13.5)(16.6+1 \times 40 \times 0.517-13.5)}{(16.6-13.5)(16.6+1 \times 40 \times 0.517+13.5)}\right]}{1 \times 210\,000 \times \pi \times 13.5 \times 0.517} $$$$ = \frac{2 \ln\left[\frac{30.1 \times 23.8}{3.1 \times 50.8}\right]}{5\,770\,000} = \frac{2 \ln(4.55)}{5\,770\,000} = \frac{2 \times 1.515}{5\,770\,000} = 5.25 \times 10^{-7} \text{ mm/N} $$Force ratio Φ
Concentric symmetric loading, taking $\delta_{PZu} = 0$ (simplification) (Eq. R3/3):
$$ \Phi_n = n \cdot \frac{\delta_P}{\delta_S + \delta_P} = 0.5 \times \frac{5.25 \times 10^{-7}}{2.93 \times 10^{-6} + 5.25 \times 10^{-7}} = 0.5 \times \frac{0.525}{3.455} = 0.076 $$Force ratio $\Phi = 0.076$ — the bolt takes only 7.6% of the external force!
Additional bolt force and plate unloading force:
$$ F_{SA} = 0.076 \times 20\,000 = 1\,520 \text{ N} $$$$ F_{PA} = (1 - 0.076) \times 20\,000 = 18\,480 \text{ N} $$R4 — preload change
Embedding loss
$f_Z$ (Table 5): bolt head 3 μm + nut 3 μm + 1 interface 3 μm = 9 μm
$$ F_Z = \frac{f_Z}{\delta_S + \delta_P} = \frac{0.009}{3.455 \times 10^{-6}} = 2\,604 \text{ N} \tag{R4/1} $$Thermal expansion
Room-temperature case → $\Delta F'_{Vth} = 0$
R5 — 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):
$$ F_{Mzul} = 84.3 \times \frac{0.9 \times 940}{\sqrt{1 + 3\left[\frac{3}{2} \times \frac{10.863}{10.36}\left(\frac{1.75}{\pi \times 10.863} + 1.155 \times 0.12\right)\right]^2}} $$$$ = 84.3 \times \frac{846}{\sqrt{1 + 3 \times [1.574 \times (0.0513 + 0.1386)]^2}} $$$$ = 84.3 \times \frac{846}{\sqrt{1 + 3 \times 0.0898}} = 84.3 \times \frac{846}{\sqrt{1.269}} = 84.3 \times 751 = 63\,309 \text{ N} $$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.
Check (Eq. R7/3):
$$ F_{Mzul} = 63\,309 \text{ N} \quad \text{vs} \quad F_{M\max} = 100\,402 \text{ N} $$$$ 63\,309 < 100\,402 \quad → \quad ❌ \text{ not satisfied!} $$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 | $\pi/4 \times 16^2$ |
| $d_W$ | 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):
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 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):
$\beta_L = 40/21.9 = 1.83$, $y = 44/21.9 = 2.01$
$\tan\varphi_D = 0.362 + 0.032\ln(0.91) + 0.153\ln(2.01) = 0.362 - 0.003 + 0.107 = 0.466$
$D_{A,Gr} = 21.9 + 40 \times 0.466 = 40.5$ mm
$D_A = 44 > 40.5$ → Eq. 40:
$$ \delta_P = \frac{2\ln\left[\frac{(21.9+17.5)(21.9+40 \times 0.466-17.5)}{(21.9-17.5)(21.9+40 \times 0.466+17.5)}\right]}{210\,000 \times \pi \times 17.5 \times 0.466} = 3.72 \times 10^{-7} \text{ mm/N} $$Φ (M16):
$$ \Phi = 0.5 \times \frac{3.72 \times 10^{-7}}{1.78 \times 10^{-6} + 3.72 \times 10^{-7}} = 0.5 \times 0.173 = 0.086 $$R4 (M16): $F_Z = 0.009 / (2.15 \times 10^{-6}) = 4\,186$ N
R5: $F_{M\min} = 41\,667 + 0.914 \times 20\,000 + 4\,186 = 64\,133$ N
R6: $F_{M\max} = 1.6 \times 64\,133 = 102\,613$ N
R7: $F_{Mzul}(M16) \approx 119\,000$ N (Table A1)
$$ 119\,000 \geq 102\,613 \quad → \quad ✅ \text{ satisfied!} $$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):
$$ M_A = 119\,000 \times [0.16 \times 2.0 + 0.58 \times 14.701 \times 0.12 + \frac{19.6}{2} \times 0.12] $$$$ = 119\,000 \times [0.32 + 1.023 + 1.176] = 119\,000 \times 2.519 \times 10^{-3} \text{ m} $$$$ M_A \approx 300 \text{ N·m} $$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.
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 bolt-head geometry from DIN EN ISO 4017, and the through-hole diameter from DIN EN 20273.
Disclaimer: 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.
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