# What does SFC on a bolt mean

## Strength calculation of a bolt and pin connection

If you want to determine the strength of a bolt or pin connection through a stress calculation, this process is complex. This has the following causes:

- the
**preload**as a result of the impact and the superimposed deformation of the pin and component under the load - the appearance
**plastic deformations**without affecting the transmission behavior - the
**Deformation-dependent adjusting contact and notch stresses**for bolt-tab connections with play **uneven distribution of the load**on the individual pins in the case of several rows of pins, according to the deformation behavior of the components

In order to still achieve a reliable result, assumptions and simplifications are made.

**Assumptions:**

- Neglect of deformation
- linear stress distribution present

### Simplifications regarding the cause of failure shear

In the next figure you can see a shaft-hub connection that is ensured by a bolt.

The mean shear stress is defined by:

**mean shear stress:**$ \ tau = \ frac {F} {A} = \ frac {4 \, \ cdot \, F} {\ pi \, \ cdot \, d ^ 2} $

For transverse pins in shaft-hub connections, the associated peripheral force $ F_u $ is calculated at the interface. The circumferential force is defined by:

**Circumferential force:**$ F_u = \ frac {2 \, \ cdot \, T} {D} $ with $ T $ = torque

The circumferential force $ F_u $ is divided accordingly into $ 2 \ cdot \ frac {F_u} {2} $.

For this reason, the shear stress equation is obtained as follows:

**Shear stress:**$ \ tau = \ frac {F_u} {2 \, \ cdot \, A} = \ frac {T} {A \, \ cdot \, D} = \ frac {4 \, T} {\ pi \, \ cdot \, d ^ 2 \, \ cdot \, D} $

The following applies to the permissible shear stress $ \ tau_ {perm} $:

**permissible shear stress:**$ \ tau_ {zul} = \ frac {\ tau_F} {\ nu} \, \, \, $ with $ \, \, \, \ nu = 2 $ to $ 4 $

$ \ nu $ is the required security.

**Double fit**before, both parts must be able to withstand the same load, otherwise there will be an asymmetry.

### Simplifications regarding the cause of failure shear and bending

In the next illustration you can see a hammered bolt that is loaded by a force $ F $.

Both shear stress and bending stress occur.

The **Shear stress** results as above from $ \ tau_a = \ frac {F} {A} $.

The additionally occurring bending stresses due to the force $ F $ are new. The bending stress is formally described by:

**Bending stress:**$ \ sigma_b = \ frac {M_b} {W_b} = \ frac {F \, \ cdot \, l_F \, \ cdot \, 32} {\ pi \, \ cdot \, d ^ 3} $

- $ l_F $ = length of the lever arm
- $ M_b $ = bending moment
- $ W_b $ = bending resistance moment

According to the shape energy change hypothesis (GEH), the following applies to the equivalent stress:

**Equivalent stress:**$ \ sigma_v = \ sqrt {\ sigma_b ^ 2 + 3 \ cdot \ tau ^ 2} \, \, \, $ with $ \, \, \, \ nu = 2 $ to $ 4 $

If there is only a very short restraint, as in the next figure, the bending is neglected in pin connections.

The effective support is estimated for the simplified calculation of the bending moment. In our illustration, the maximum bending stress is $ \ sigma_ {max} $ at point A of the clevis shown:

**maximum bending stress:**$ \ sigma_ {max} = \ frac {M_b} {W_b} = \ frac {1} {2} \ cdot F \ cdot (\ frac {a} {2} + \ frac {b} {4} \ cdot \ frac {1} {W_b}) $

At the interface B, however, the bending stress $ \ sigma_ {bB} $ is defined by:

**Bending stress:**$ \ sigma_ {bB} = \ frac {M_b} {W_b} = \ frac {1} {2} \ cdot F \ cdot \ frac {a} {2} \ cdot \ frac {1} {W_b} $

### Simplifications regarding the cause of failure surface pressure

In this load variant, surface pressure results from the superposition of shear force and bending.

In the illustration you will again see a pin with a spring attached to its head, at the end of which a force $ F $ acts.

The shear force is calculated from:

**Shear force:**$ P_d = \ frac {F} {s \, \ cdot \, d} $

The bend $ P_b $ is obtained by rearranging the equation of moments. The equation of moments around the point (yellow) has the form:

**Moment equation:**$ P_b \ cdot \ frac {s} {2} \ cdot \ frac {2} {3} \ cdot d \ cdot \ frac {s} {2} = F \ cdot (l_F + \ frac {s} {2} ) $

If one solves the moment equation for $ P_b $, one obtains:

**Bend:**$ P_b = 6 \ cdot F \ cdot \ frac {l_F + \ frac {s} {2}} {s ^ 2 \, \ cdot \, d} $

With the equations for the shear force and the bending, we can finally set up the equation for the maximum surface pressure, which results from both:

**maximum surface pressure:**$ P_ {max} = P_d + P_b = \ frac {F} {s \, \ cdot \, d} (1 + 6 \ frac {l_F + \ frac {s} {2}} {s}) $

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