🧮 在线计算器：VDI 2230 Bolted Joint Calculation — The full 14-step calculation chain (R0–R13), with six strength checks.
Eccentric load and bending-moment effect — the “deep water” 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.
1. Why is eccentricity the norm?
The standard says (VDI 2230:2015, §5.1.2.3, p.57):
“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.”
Eccentricity comes from two independent sources, which must be handled separately:
| Type | 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):
$$ s_{sym} = \frac{c_T}{2} - e \leq \frac{G}{2} - e \tag{66} $$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.
When $s_{sym} = 0$ → 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
3. 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):
“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.”
Determining $a$ 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):
$$ a \approx 0.275 R \quad \text{(constant-section circular-ring connecting rod)} $$For complex structures, the standard advises using FEM to help determine $a$.
4. 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.
4.1 Eccentric-clamping corrected resilience $\delta_P^*$
Used in the denominator of the force-ratio formula (VDI 2230:2015, §5.1.2.2, Eq. 67):
$$ \delta_P^* = \delta_P + \frac{s_{sym}^2 \cdot l_K}{E_P \cdot I_{Bers}} \tag{67} $$where $\delta_P$ is the symmetric resilience (Eq. 40 or 41) and $I_{Bers}$ is the substitutional moment of inertia of the deformation body.
4.2 Eccentric-loading corrected resilience $\delta_P^{**}$
Used in the numerator of the force-ratio formula (VDI 2230:2015, §5.1.2.3, Eq. 71):
$$ \delta_P^{**} = \delta_P + \frac{a \cdot s_{sym} \cdot l_K}{E_P \cdot I_{Bers}} \tag{71} $$$\delta_P^{**}$ can be smaller than, equal to, or larger than $\delta_P^*$, depending on the relative magnitude and direction of $a$ and $s_{sym}$.
4.3 Eccentric force ratio $\Phi_{en}^*$
The most general force-ratio formula (VDI 2230:2015, Eq. R3/4):
$$ \Phi_{en}^* = n \cdot \frac{\delta_P^{**} + \delta_{PZu}}{\delta_S + \delta_P^*} \tag{R3/4} $$Simplified forms for different eccentric cases (VDI 2230:2015, Eq. 73-75):
| Case | Calculation of $F_{SA}$ | Source |
|---|---|---|
| $s_{sym} \neq 0$, $a > 0$ | $n \cdot \frac{\delta_P^{**}}{\delta_S + \delta_P^*} \cdot F_A$ | 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): “If $a$ and $s_{sym}$ do not lie on the same side of the axis of symmetry, the additional bolt load $F_{SA}$ […] may become larger than calculated.” This case should be avoided by design.
5. 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).
Cone part (Eq. 59):
$$ I_{Bers}^V = 0.147 \cdot \frac{(D_A - d_W) \cdot d_W^3 \cdot D_A^3}{D_A^3 - d_W^3} \tag{59} $$Eccentricity correction (Steiner shift term, Eq. 60):
$$ I_{Bers}^{Ve} = I_{Bers}^V + s_{sym}^2 \cdot \frac{\pi}{4} D_A^2 \tag{60} $$Sleeve part (Eq. 61):
$$ I_{Bers}^H = \frac{b \cdot c_T^3}{12} \tag{61} $$Combined body (Eq. 62):
$$ I_{Bers} = \frac{l_K}{\frac{2}{w}(l_V / I_{Bers}^{Ve}) + l_H / I_{Bers}^H} \tag{62} $$6. Opening check (Abheben, Opening)
Eccentric loading causes the interface to begin losing contact pressure on the bending-tension side — that is, “opening” occurs (VDI 2230:2015, §5.3.2).
The 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:
The ratio of the bending moment to the clamping force at the interface must not exceed the section core distance:
$$ \frac{M_B}{F_{KR}} \leq \frac{I_{BT}}{A_{BT} \cdot e_{max}} \tag{concept} $$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’s linear spring model no longer applies. This is also why the standard stresses that $F_{Kerf}$ must be large enough to prevent opening.
7. Practical advice for eccentric design
Based on the standard’s requirements, the design of an eccentric joint should follow these principles:
Constructional 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):
“Even relatively small eccentricities of the load introduction may have a considerable effect on the deformation behaviour of the clamped parts.”
Data 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.
Disclaimer: 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.
📚 Series navigation
← 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) | $\delta_P^*$, $\delta_P^{**}$ |