A bar of cross section `A` is subjected to equal and opposite tensile force at its ends. Consider a plane section of the bar whose normal makes an angle `theta` with the axis of the bar.

a. What is the tensile stress on the plane?
b. What is the shearing stress on the this plane?
c. For what value of `theta` is the tensile stress maximum?
d. For what value of `theta` is the shearing stress maximum?

Explanation

a. The resolved part of `F` along the normal is the tensile force on this plane and the resolved part parallel to the plane is the shearing force on this plane.

Therefore, tensile stress `=("Force")/("Area")=(Fcostheta)/(Asectheta)=F/Acos^(2)theta`
`(:'` Area of section `= Asectheta)`
b. Shearing stress `=("Force")/("Area")=(Fsintheta)/(Asectheta)=F/(2A)sin2theta`
c. Obviously, tensile stress on the plane is maximum when `cos^(2)theta` is maximum that is `costheta=1` or `theta=0^@`.
d.Obviously shearing stress is maximum when is `sin2theta` is maximum that is
`sin2theta=1` or `2theta=90^@` or `theta=45^@`