Question

A rod of cross-sectional area '$A$' is subjected to equal and opposite forces '$F$' at its end, as shown in figure. Consider a plane through the rod making angle $\theta$ with cross-sectional plane, if $\theta ={30}^{\circ }$, then

• A. tensile stress at this plane is $\frac{3F}{4A}$.
• B. shear stress at this plane is $\frac{\sqrt{3}}{4}\frac{F}{A}$.
• C. maximum shear stress in the entire range of $\theta$ is $\frac{F}{2A}$.
• D. tensile stress at this plane is $\frac{\sqrt{3}}{2}\frac{F}{A}$.

Answer 1

Answer: (A), (B), (C)

maximum shear stress in the entire range of $\theta$ is $\frac{F}{2A}$.


$A^{{}^{\prime }}\cos \theta =A$

$A^{{}^{\prime }}=\frac{A}{\cos \theta }$

${\sigma }_{Tensile}=\frac{Fcos\theta }{\frac{A}{cos\theta }}=\frac{Fco{s}^2\theta }{A}=\frac{3F}{4A}$

${\sigma }_{Shear}=\frac{F\sin \theta }{\frac{A}{\cos \theta }}=\frac{F}{A}\sin \theta \cos \theta =\frac{Fsin2\theta }{2A}=\frac{\sqrt{3}F}{4A}$

Maximum shear stress occures at $\theta ={45}^{\circ }$ and it is equal to $\frac{F}{2A}$.