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NCERT Exemplar · Q24

Q.Consider a long steel bar of uniform cross-sectional area AA (measured perpendicular to its length) placed under a tensile stress by two equal and opposite forces F\mathbf{F} applied at its two ends, each directed outward along the length (axis) of the bar. Consider an internal plane that cuts across the bar making an angle θ\theta with the length (axis) of the bar. Find the tensile (normal) and shearing (tangential) stresses acting on this plane, and hence determine:

(a) the angle for which the tensile stress is a maximum;
(b) the angle for which the shearing stress is a maximum.
Jammu Kashmir JkboseLong· 5mImportance★★★★★est
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Cutting the bar along a plane tilted at angle θ\theta to its axis, the axial pulling force resolves into one part perpendicular to the plane (giving normal/tensile stress) and one part along the plane (giving shear stress). Working these out gives a tensile stress FAsin⁡2θ\frac{F}{A}\sin^{2}\theta, maximum when the plane is perpendicular to the axis (θ=90∘\theta=90^\circ), and a shear stress F2Asin⁡2θ\frac{F}{2A}\sin 2\theta, maximum on a 45∘45^\circ plane.

Concept

Stress on a plane = (force component on that plane) ÷\div (area of that plane). The bar's normal cross-section (perpendicular to the axis) has area AA. A plane tilted so that it makes an angle θ\theta with the axis is larger.

Area of the oblique plane

If a plane makes angle θ\theta with the length of the bar, its area is

Aθ=Asin⁡θ.A_\theta=\frac{A}{\sin\theta}.

(Check: at θ=90∘\theta=90^\circ the plane is the normal cross-section, Aθ=AA_\theta=A; as θ→0\theta\to0 the plane becomes nearly parallel to the axis and its area grows without bound.)

Resolving the force

The force FF acts along the axis. Relative to the oblique plane:

  • component normal to the plane: Fn=Fsin⁡θF_n=F\sin\theta,
  • component tangential to (along) the plane: Ft=Fcos⁡θF_t=F\cos\theta.

Stresses

Tensile (normal) stress:

σn=FnAθ=Fsin⁡θA/sin⁡θ=FAsin⁡2θ.\sigma_n=\frac{F_n}{A_\theta}=\frac{F\sin\theta}{A/\sin\theta}=\frac{F}{A}\sin^{2}\theta.

Shearing (tangential) stress:

σt=FtAθ=Fcos⁡θA/sin⁡θ=FAsin⁡θcos⁡θ=F2Asin⁡2θ.\sigma_t=\frac{F_t}{A_\theta}=\frac{F\cos\theta}{A/\sin\theta}=\frac{F}{A}\sin\theta\cos\theta=\frac{F}{2A}\sin 2\theta.

(a) Maximum tensile stress …

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