Question

The maximum shear stress in a rectangular cross section ______________per cent greater than the average shear stress on the cross section.

• A. 100
• B. 20
• C. 30
• D. 50

Answer 1

Answer:

D. 50

The average shear stress in a rectangular cross-section is given by $\tau =\frac{F(A\overline y)}{Ib}$ where F = shear force, $A \overline y$ = moment of the area taken into consideration, I = moment of inertia, b = width of the section. For maximum shear stress, y = 0 Therefore,$\tau_{max} =1.5 \tau{avg}$ . Substituting values, $\tau =\frac {6F}{b h ^ 3} \times (\frac{h ^ 2}{4} + 0) = \frac {6F}{b h ^ 3} \times\frac {h ^ 2}{4} = \frac{3F}{2(bh)} = \frac {3F}{2A}$ . Hence, the maximum shear stress in a rectangular cross-section is 50% greater than the average shear stress on the cross-section.