Bolted Circular Flange

Bolted Circular Flange

In engineering, it is very common to find a circular flange fixed with bolts subjected to a bending moment. This is the case for pipe joints or plates supporting a bearing for rotating movements.

If there is a moment centered with the flange, since the joint bolts are at different distances from the axis of application of said moment, each will support a different tensile force. Therefore, it is necessary to determine these loads to properly dimension the bolted joint.

To begin with, the bolts furthest from the moment axis will support a maximum load (Fmax), which we will simply call F, while the bolts crossing the moment axis will have zero tensile load.

Bolted circular flange 2 suavizada
Load distribution

The load on the other bolts between these two positions depends linearly on their distance to the moment axis, if we look at the flange profile, but their contribution to the moment depends circularly on the position they occupy on the flange circumference.

Bolted circular flange 2b
Load distribution scheme
Thus, the distance of each bolt to the moment axis can be calculated as:

  \normalsize d_i = R sin (\alpha_i)

Where R is the radius of the bolt circle and α is the angle with respect to the moment axis.

  \Large \frac {F}{R} = \frac {F_i}{d_i}

Where Fi is the force on each bolt, with the other variables already defined. Solving for Fi, we have:
And the force of each bolt can be calculated by similar triangles in the flange profile as:

  \normalsize M = \sum_{i=0}^n F sin (\alpha_i) \cdot R  sin (\alpha_i) = F  R  \sum_{i=0}^n sin^2 (\alpha_i) = \Large \frac {F R n} {2}

So the maximum force on the bolt furthest from the moment axis will be:

  \normalsize F_i = F sin (\alpha_i)

The total moment on the flange (M) will be the summation of the force of each bolt multiplied by its distance to the moment axis, from i=0 to n.

  F = \Large  \frac {2  M} {n  R}

And the forces in each bolt will be the maximum force multiplied by the sine of the angle (alpha) it occupies on the circumference.

  \normalsize F_i = F sin (\alpha_i) = \Large  \frac {2  M} {n  R} \normalsize sin (\alpha_i)  

It should be noted that the moment on the flange affects the bolts in tension only on one half of it, since the other half, subjected to compression, tries to join both plates of said flange, so the bolts in this zone do not support load due to the moment.

Beyond Static Formulas: Real-world Application

The theoretical formula F = \Large \frac {2 M} {n R} provides a solid basis for static pre-sizing. However, in critical heavy machinery, relying solely on static calculations is a risk.

Case Study: Fatigue in Slewing Rings (Tecma Drive)
We tackled a complex challenge for Tecma Drive S.L.: validating fatigue life of fasteners in large slewing bearings (a massive circular flange application).

Using the Soderberg Criterion and custom M-Ra Curves (Moment vs. Axial Load), we analyzed grades 8.8, 10.9, and 12.9 to ensure safety under cyclic loads, preventing catastrophic failure in rotating equipment.

grafica-fatiga-corona-tornilleria-momento-carga-axial-compresion-traccion
M-Ra Fatigue Analysis developed for Tecma Drive

Mathematical Development

\normalsize M = \sum_{i=0}^n F sin (\alpha_i) \cdot R  sin (\alpha_i) = FR \sum_{i=0}^n sin^2 (\alpha_i) = 
(1) =  FR \int_0^{2\pi} sin^2 (\alpha) \large \frac {n}{2\pi} \normalsize d\alpha = (2) = \frac {FRn}{2\pi} \int_0^{2\pi} \frac{1}{2}(1-cos (2\alpha)) \normalsize d\alpha =
= \frac {1}{2} \frac {FRn}{2\pi}\{ [\alpha ]_{0}^{\rm 2\pi} - \cancel{[\frac{1}{2} sin(2\alpha)]_{0}^{\rm 2\pi}}\} =\large \frac {FRn}{2}
\normalsize For  i = 0 \longrightarrow \alpha = 0 
\normalsize For  i = n \longrightarrow \alpha = 2\pi 
\alpha = \large \frac{2\pi}{n} \normalsize i
  i = \large \frac{n}{2\pi} \normalsize \alpha  ;         di = \large \frac{n}{2\pi} \normalsize d\alpha
\normalsize sin^2 (\alpha) + cos^2 (\alpha) = 1 
\normalsize sin^2 (\alpha) = 1 - cos^2 (\alpha)   (A)
\normalsize cos (2\alpha) = cos^2 (\alpha) - sin^2 (\alpha)
\normalsize  cos^2 (\alpha) = cos (2\alpha) - sin^2 (\alpha)   (B)
Substituting B into A:
\normalsize sin^2 (\alpha) = \frac{1}{2} (1 - cos (2\alpha))   (2) 
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