A bar of cross-sectional area A is subjected | vedclass

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(A) Let the cross-sectional area of the bar be $A$. The force $F$ acts perpendicular to the cross-section of area $A$.Consider a plane $BB'$ inclined

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$	an\, 	heta$

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(A) Let the cross-sectional area of the bar be $A$. The force $F$ acts perpendicular to the cross-section of area $A$.Consider a plane $BB'$ inclined at an angle $	heta$ with the length of the bar. The normal to this plane makes an angle $	heta$ with the direction of the force $F$.The area of the inclined plane $BB'$ is $A' = A / \sin 	heta$.The force $F$ can be resolved into two components relative to the plane $BB'$:$1$. Normal force component: $F_N = F \sin 	heta$ (acting perpendicular to the plane $BB'$).$2$. Tangential (shearing) force component: $F_S = F \cos 	heta$ (acting along the plane $BB'$).The tensile stress (normal stress) on plane $BB'$ is $\sigma = F_N / A' = (F \sin 	heta) / (A / \sin 	heta) = (F / A) \sin^2 	heta$.The shearing stress on plane $BB'$ is $	au = F_S / A' = (F \cos 	heta) / (A / \sin 	heta) = (F / A) \sin 	heta \cos 	heta$.The ratio of tensile stress to shearing stress is:$\frac{\sigma}{	au} = \frac{(F/A) \sin^2 	heta}{(F/A) \sin 	heta \cos 	heta} = \frac{\sin 	heta}{\cos 	heta} = 	an 	heta$.

$A$ bar of cross-sectional area $A$ is subjected to two equal and opposite tensile forces at its ends as shown in the figure. Consider a plane $BB'$ making an angle $\theta$ with the length. The ratio of tensile stress to the shearing stress on the plane $BB'$ is

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