# Redistributing Shear Stresses from Null Elements

### Technical Article

SHAPE-THIN allows you to calculate section properties and stresses of any cross‑sections. If a flange or a web is weakened by bolt holes, you can consider this by using null elements. The stresses are subsequently recalculated with the reduced cross‑section values. In this case, it is necessary to pay special attention to shear stresses. By default, these are set to zero in the area of the null elements. When recalculating shear stresses with the reduced cross‑section values and without further adaptation, it turns out that the integral of the shear stresses is no longer equal to the applied shear force. The following example shows in detail how to calculate the shear stress.

#### Example Calculation

Given an element of length l = 200 mm and thickness t = 8 mm. The shear force is applied with 120 kN. This results in the following diagrams for static moment, shear force and shear stress. The second moment of area is I _{y} = 533 cm ^{4} .

Figure 01 - Result Diagrams of Gross Cross-Section

The shear force is the shear stress multiplied by the length and the thickness of the respective element. The integral is calculated as follows:

$\mathrm V\;=\;\mathrm t\;\cdot\;\int\frac{\mathrm Q\;\cdot\;\left(\mathrm t\;\cdot\;\mathrm z\;\cdot\;\left({\displaystyle\frac{\mathrm l}2\;-\;\frac{\mathrm z}2}\right)\right)}{{\mathrm I}_\mathrm y\;\cdot\;\mathrm t}\;\mathrm{dz}$

where z is the value of the z-coordinate

By adding three forces resulting from the division of the elements, you obtain the shear force of 120 kN.

In the next step, the middle element with a length of 20 mm is converted into a null element. This corresponds to the hole mentioned above. The second moment of area is I _{y} = 469 cm ^{4} . The shear stresses of the null element must now be redistributed to the remaining elements. For this purpose, a correction factor k is determined, which describes the ratio of the shear force to the effective shear force components.

$\begin{array}{l}\mathrm k\;=\;\frac{\mathrm{Querkraft}}{\mathrm{Summe}\;\mathrm{der}\;\mathrm{noch}\;\mathrm{wirksamen}\;\mathrm{Schubkraftanteile}\;\mathrm{am}\;\mathrm{Brutto}-\mathrm{Querschnitt}}\;=\\=\;\frac{120}{101,1\;+\;7,3}\;=\;1,11\end{array}$

Then, the shear force is multiplied by this factor:

Q = 120 ∙ 1.11 = 133.2 kN

With this modified shear force, the shear stresses on the weakened cross-section are now calculated. The following diagrams result for the static moment, the shear force and the shear stress.

Figure 02 - Result Diagrams of Weakened Cross-Section

If you add the shear forces, you get back the applied shear force of 120 kN. The components from the null element have been completely redistributed.

#### Reference

#### Keywords

Shear stress Shear force Null element

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