# Shear Stress in Beams

To calculate the shear stress in the beam at point $$P$$, given that the beam is subject to a resultant internal vertical shear force of $$V = 3 kN$$.

The second moment of area of the cross-section of the beam about the neutral axis:

$$I = \frac{bd^3}{12} = \frac{1}{12}(100mm)(125mm)^3 = 16.28E6 mm^4$$

A horizontal line is drawn through point $$P$$ and the partial area $$A'$$ is shown shaded.

$$Q = y_c A'$$

Where $$y_c$$ is the distance between the partial area's neutral axis and the neutral axis of the beam, $$A'$$ is the area of the partial.

$$Q = y_c A' = [12.5mm + \frac{1}{2} (50mm)] (50mm)(100mm) = 18.75E4 mm^3$$

Applying the shear formula:

$$\tau = \frac{VQ}{It} = \frac{(2kN)(18.75E4mm^3)}{(16.28E6mm^4)(100mm)} = 0.346MPa$$

The maximum shear stress in the beam can be calculated in the same way with the partial area as exactly half of the beam,

$$Q = y_c A' = \frac{62.5mm}{2} (100mm)(62.5mm) = 19.53E4 mm^3$$

$$\tau = \frac{VQ}{It} = \frac{(2kN)(19.53E4mm^3)}{(16.28E6mm^4)(100mm)} = 0.360 MPa$$