Question

For a circular cross section beam is subjected to a shearing force F, the maximum shear stress induced will be (where d = diameter)

• A. $$\frac {F}{\pi d^2}$$
• B. $$\frac {16F}{3\pi d^2}$$
• C. $$\frac {2F}{\pi d^2}$$
• D. $$\frac {F}{4 d^2}$$

Answer 1

Answer:

$$\frac {16F}{3\pi d^2}$$

For a circular cross section beam subjected to a shearing force F, the maximum shear stress induced is $\frac{16F} {3 \pi d ^ 2}$ The average shear stress is $\frac{F} {\frac{\pi d ^ 2}{ 4}}$ Using the relation between maximum and average shear stress for a circular section, we get the maximum shear stress as 4/3 times the average shear stress. Hence, substituting the value of average shear stress, we get the maximum shear stress as$\frac {16F}{3πd^2}$.