# 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,

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

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