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A bar of cross-section A is subjected to equal and opposite tensile forces at its ends. Consider a plane section of the bar whose normal makes an angle `theta` with the axis of the bar .
(i) What is the tensile stress on this plane?
(ii) What is the shearing stress on this plane?
(iii) For what value of `theta` is the tensile stress maximum
(iv) For what value of `theta` is the shearing stress maximum?

Text Solution

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(i) 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 the plane.
`therefore` Area of plane section `=Asectheta`
Tensile stress `=("Force")/("Area")=(F costheta)/(Asectheta)" " =(F)/(A)cos^(2)theta`
(ii) Shearing stress applied on the plane
So , `F=Fsintheta`
Shearing stress `=("Force")/("Area")=(Fsintheta)/(Asectheta)=(F)/(A)sin thetacostheta`
`=(F)/(2A)sin2theta " " [becausesin2theta=2sinthetacostheta]`
(iii) Tensile stress will be maximum when `cos^(2)theta` is maximum i.e., `costheta=1 or theta=0^(@)` (iv) Shearing stress will be maximum when `sin2theta` is maximum i.e . `sin2theta=1` or `2theta=90^(@)ortheta=45^(@)` .
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