Consider a long steel bar under a tensile-stress due to forces F- acting at the edges along the length of the bar fig- Consider a plane making an angle - with the length- What are the tensile and shearing stresses on this plane-i At what angle is the tensile stress is maximum-ii At what angle is the shearing stress is maximum-

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Shear Modulus

Consider a lo...

Question

Consider a long steel bar under a tensile-stress due to forces $\to F$ acting at the edges along the length of the bar (fig.). Consider a plane making an angle $θ$ with the length. What are the tensile and shearing stresses on this plane?
(i) At what angle is the tensile stress is maximum?
(ii) At what angle is the shearing stress is maximum?

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Solution

According Problem $F$ is applied along horizontal so resolving into two component one is parallel to inclined one and other is perpendicular to inclined me
$F$ parallel $= F_{11} = F cos θ$
$F$ perpendicular $= F_{⊥} = F sin θ$
$F_{⊥}$ Producer tensile stren , $F_{11}$ producer shear stren
Area of focus $a a^{1}$ then $sin θ = \frac{A}{A^{1}} \Rightarrow A^{1} = A / sin θ$
Tensile stren on the plane $a a^{1}$
$a a^{1} = \frac{F_{⊥}}{A^{1}} \Rightarrow \frac{F sin θ}{A / sin θ} = \frac{F}{A} {sin}^{2} θ$
Shearing stren on the plane $a a^{1}$
Shearing stren $= \frac{P a r a l l e l f o r c e}{a r e a}$
$= \frac{f_{11}}{A^{1}} = \frac{F cos θ}{A / sin θ} = \frac{F sin θ cos θ}{A} = \frac{F (2 sin θ cos θ)}{2 A}$
$= \frac{F sin 2 θ}{2 A}$
(a) For tensile stren to be maximum
${sin}^{2} θ = 1$ $sin θ = 1$ $θ = \frac{π}{2}$ or $90^{o}$
(b) For shearing stren to be maximum
$sin 2 θ = 1$ $2 θ = \frac{π}{2}$ $θ = \frac{π}{4}$ or $45^{o}$

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