Ncert Exercises & Solutions

Q.24 Consider a long steel bar under tensile stress due to force 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?

(a) For what angle is the tensile stress a maximum?

(b) For what angle is the shearing stress a maximum?

Hint: For stress, consider the force perpendicular to the plane.

Step 1: Find the tensile and shearing stresses.

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

\frac{π}{2} - θ

with the normal ON. Resolving F into components along the plane (FP) and normal to the plane.

F_{p} = F cosθ

F_{N} = F sinθ

Let the area of the face aa' be A', then,

\frac{A}{A'} = \sin θ A' = \frac{A}{\sin θ}

The tensile stress

= \frac{Normal force}{Area} = \frac{F sinθ}{A'}

= \frac{F sinθ}{A / sinθ} = \frac{F}{A} \sin^{2} θ

Shearing stress =

\frac{Parallel force}{Area}

= \frac{F cosθ}{A / Sinθ} = \frac{F}{A} sinθ . cosθ

= \frac{F}{2 A} (2 sinθ . cosθ) = \frac{F}{2 A} \sin 2 θ

Step 2: Find the maximum tensile stress.

(a) For tensile stress to be maximum,

\sin^{2} θ = 1

\Rightarrow sinθ = 1

\Rightarrow θ = \frac{π}{2}

Step 3: Find the minimum shearing stress.

(b) For shearing stress to be maximum,

\sin 2 θ = 1 \Rightarrow 2 θ = \frac{π}{2} \Rightarrow θ = \frac{π}{4}

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