Question

A bar is subjected to equal and opposite forces as shown in the figure. PQRS is a plane making angle $\theta$ with the cross-section of the bar. If the area of cross-section be ‘A’, then what is the tensile stress on PQRS

• A. $\frac{F}{A}$
  
• B. $\frac{Fcos\theta }{A}$
  
• C. $\frac{Fco{s}^2\theta }{A}$
  
• D. $\frac{F}{Acos\theta }$

Answer 1

The correct option is C $\frac{Fco{s}^2\theta }{A}$


As tensile stress = $\frac{Normalforce}{Area}=\frac{F_N}{A_N}$
and here $A_N=\left(\frac{A}{cos\theta }\right),F_N=Normalforce=Fcos\theta$
So, Tensile stress $=\frac{Fcos\theta }{\frac{A}{cos\theta }}=\frac{Fco{s}^2\theta }{A}$