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Question
Consider a long steel bar under a tensile stress due to forces F acting at the edges along the length of the bar (Figure). Consider a plane making an angle θ with the length. What are the tensile and shearing stresses on this plane
- For what angle is the tensile stress a maximum?
- For what angle is the shearing stress a maximum?
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Solution
Consider the adjacent diagram,
Let the cross-sectional area of the bar be A. COnsider the equilibrium of the plane aa'. A force F must be acting on this plane making an angle `pi/2 - theta` with the normal ON. Resolving F into components, along the plane (FP) and normal to the plane.
`F_p = F cos θ`
`F_N = F sin θ`
Let the area of the face aa' be A, then
`A/A^' = sin θ`
∴ `A^' = A/sin θ`
The tensile stress = `"Normal force"/"Area" = (F sin θ)/A^'`
= `(F sin θ)/(A/sin θ)`
= `F/A sin^2 θ`
Shearing stress = `"Parallel force"/"Area"`
= `(F sin θ)/(A/sin θ)`
= `F/A sin θ. cos θ`
= `F/(2A) (2 sin θ. cos θ)`
= `F/(2A) sin 2θ`
a. For tensile stress to be maximum, `sin^2theta` = 1
⇒ `sin θ` = 1
⇒ θ = `pi/2`
b. For shearing stress to be maximum,
sin 2θ = 1
⇒ 2θ = `pi/2`
⇒ θ = `pi/4`
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