Consider a long steel bar under a tensile stress due to forces F acting at the edges along the length of the bar. Consider a plane making an angle \theta with the length. What are the tensile and shearing stresses on this planet?

a) for what angle is the tensile stress a maximum?

b) for what angle is the shearing stress a maximum?

Answers (1)

a) Tensile stress = normal forces to the surface of plane F/ area

A = area of cross-section which is perpendicular to the bar

A’ = the area of plane cut along aa’ of the cross section

\sin \theta = \frac{A}{A^{'}}

 

The force perpendicular component along A or AA^{'}=\sin \theta

Tensile stress = \left ( \frac{F\sin \theta}{A} \right ) \times \sin \theta

For max value, sin θ should be equal to 1. Hence, \theta =\frac{\pi}{2}

b)  Shearing stress = forces along the plane F/ area = F\frac{\cos \theta}{A^{'}}=\left ( \frac{F}{2A} \right )\times 2 \sin \theta \cos \theta = \left ( \frac{F}{2A} \right )\sin 2\theta

for shearing stress to be maximum, \sin 2 \theta =1

which gives us \theta =\frac{\pi}{4}

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