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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 $\theta$ with the length. What are the tensile and shearing stresses on this plane?
(a) For what angle is the tensile stress a maximum?
(b) For what angle is the shearing stress a maximum?
(a) For what angle is the tensile stress a maximum?
(b) For what angle is the shearing stress a maximum?
Solution:
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Verified Answer
Consider the given diagram.
Let the cross-sectional area of the bar be $A$. Consider the equilibrium of the plane $a a^{\prime}$. A force $F$ must be acting on this plane making an angle $\frac{\pi}{2}-\theta$ with the normal $O N$. Resolving $F$ into components, along the plane $(F P)$ and normal to the plane.
$$
\begin{aligned}
&F_P=F \cos \theta \\
&F_N=F \sin \theta
\end{aligned}
$$
Let the area of cross section plane cut along the face $a a^{\prime}$ be $A^{\prime}$, then $\therefore \frac{A}{A^{\prime}}=\sin \theta$ $\therefore A^{\prime}=\frac{A}{\sin \theta}$ The tensile stress $=\frac{\text { Normal force to surface }\left(F_P\right)}{\text { Area of } \operatorname{surface}}=\frac{F \sin \theta}{A^{\prime}}$ $=\frac{F \sin \theta}{A / \sin \theta}=\frac{F}{A} \sin ^2 \theta$
Shearing stress
$$
\begin{aligned}
&=\frac{\text { [Parallel force along the plane } \left.\left(F_P\right)\right]}{\text { Area of plane }} \\
&=\frac{F \cos \theta}{A / \sin \theta}=\frac{F}{A} \sin \theta \cdot \cos \theta \\
&=\frac{F}{2 A}(2 \sin \theta \cdot \cos \theta)=\frac{F}{2 A} \sin 2 \theta
\end{aligned}
$$
(a) For maximum tensile stress, $\sin ^2 \theta=1$
$$
\begin{aligned}
&\Rightarrow \sin \theta=1 \Rightarrow \sin \theta=\sin 90^{\circ} \\
&\Rightarrow \theta=\frac{\pi}{2}
\end{aligned}
$$
(b) For maximum shearing stress
$$
=\sin 2 \theta=1 \Rightarrow 2 \theta=\frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4}
$$
Let the cross-sectional area of the bar be $A$. Consider the equilibrium of the plane $a a^{\prime}$. A force $F$ must be acting on this plane making an angle $\frac{\pi}{2}-\theta$ with the normal $O N$. Resolving $F$ into components, along the plane $(F P)$ and normal to the plane.
$$
\begin{aligned}
&F_P=F \cos \theta \\
&F_N=F \sin \theta
\end{aligned}
$$
Let the area of cross section plane cut along the face $a a^{\prime}$ be $A^{\prime}$, then $\therefore \frac{A}{A^{\prime}}=\sin \theta$ $\therefore A^{\prime}=\frac{A}{\sin \theta}$ The tensile stress $=\frac{\text { Normal force to surface }\left(F_P\right)}{\text { Area of } \operatorname{surface}}=\frac{F \sin \theta}{A^{\prime}}$ $=\frac{F \sin \theta}{A / \sin \theta}=\frac{F}{A} \sin ^2 \theta$
Shearing stress
$$
\begin{aligned}
&=\frac{\text { [Parallel force along the plane } \left.\left(F_P\right)\right]}{\text { Area of plane }} \\
&=\frac{F \cos \theta}{A / \sin \theta}=\frac{F}{A} \sin \theta \cdot \cos \theta \\
&=\frac{F}{2 A}(2 \sin \theta \cdot \cos \theta)=\frac{F}{2 A} \sin 2 \theta
\end{aligned}
$$
(a) For maximum tensile stress, $\sin ^2 \theta=1$
$$
\begin{aligned}
&\Rightarrow \sin \theta=1 \Rightarrow \sin \theta=\sin 90^{\circ} \\
&\Rightarrow \theta=\frac{\pi}{2}
\end{aligned}
$$
(b) For maximum shearing stress
$$
=\sin 2 \theta=1 \Rightarrow 2 \theta=\frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4}
$$
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