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Principal Stress and Principal Planes MCQs – Part 1

15 questions. Tap an option to check it instantly, then read the solution.

Practice SSC JE and other civil engineering exam questions on principal stresses and principal planes: stresses on an inclined plane under uniaxial and biaxial loading, pure shear, maximum shear stress and its plane, Mohr's circle, the ellipse of stress, and the theories of failure (Rankine, St. Venant, Guest–Tresca, Haigh, von Mises), plus a few related questions the book places in this chapter. Tap an option to check it, then open the full explanation for each question.

Question 1Medium
Two simply supported beams carry central concentrated loads. Beam A: length 5 m, rigidity EI, load 2 kN. Beam B: length 10 m, rigidity 2EI, load 1 kN. Compare the slopes at the ends.
Full explanation

Answer: A. Slope of beam A = slope of beam B

θ = WL²/16EI. A: 2 × 25/16EI = 25/8EI. B: 1 × 100/(16 × 2EI) = 25/8EI. Equal. (The commission's key is reported as (d).)

Question 2Easy
Which of the following is an INCORRECT statement?
Full explanation

Answer: B. A slab is primarily designed for torsion

A slab is a flexural member designed mainly for bending (and checked for shear and deflection), not torsion.

Question 3Easy
The property of a material or structure indicating the extent to which it can deform beyond the yield limit before failure or fracture is called:
Full explanation

Answer: B. ductility

Ductility is the capacity for large plastic deformation beyond yield before fracture.

Question 4Easy
Which of the following statements is correct with respect to the modulus of rigidity?
Full explanation

Answer: A. It is the ratio of shear stress to shear strain.

G = shear stress / shear strain (within the elastic limit).

Question 5Medium
A state of pure shear stress is produced by:
Full explanation

Answer: A. tension in one direction and equal compression in the perpendicular direction

σ₁ = +σ, σ₂ = −σ: on the 45° planes σn = 0 and τ = σ — pure shear.

Question 6Easy
Which of the following statements is true?
Full explanation

Answer: A. Shear stress on principal planes is zero

Principal planes are defined as the planes on which the shear stress is zero; the normal stress there is a maximum or minimum.

Question 7Medium
If the normal stresses due to longitudinal and transverse loads on a bar are σ₁ and σ₂ respectively, the normal component of stress on a plane inclined at θ to the longitudinal load is:
Full explanation

Answer: B. σ₁ sin²θ + σ₂ cos²θ

A plane at θ to σ₁ has its normal at (90° − θ) to σ₁, so σn = σ₁ sin²θ + σ₂ cos²θ.

Question 8Easy
The angle between a principal plane and the plane of maximum shear is:
Full explanation

Answer: B. 135°

Planes of maximum shear are at 45° (equivalently 135°) to the principal planes. 45° is not offered, so 135° is the answer.

Question 9Hard
Under an axial force, the tangential component of stress on an inclined plane whose normal makes an angle θ with the direction of the force may be obtained by multiplying the normal component of stress on that plane by:
Full explanation

Answer: C. tan θ

With θ measured so that σn = σ cos²θ and τ = σ sinθ cosθ, τ = σn × tanθ. (Book key: sin²θ.)

Question 10Medium
The ratio of the tangential and normal components of stress on an inclined plane through θ to the direction of the force is:
Full explanation

Answer: C. tan θ

σn = σ cos²θ, τ = σ sinθ cosθ ⇒ τ/σn = tan θ (θ taken as the angle between the force and the plane's normal, as the book does).

Question 11Medium
If the normal stresses due to longitudinal and transverse loads on a bar are σ₁ and σ₂ respectively, the tangential component of stress on a plane inclined at θ to the longitudinal load is:
Full explanation

Answer: C. (σ₁ − σ₂) sin 2θ / 2

τ = ((σ₁ − σ₂)/2) sin 2θ.

Question 12Easy
When a member is subjected to an axial tensile load, the greatest normal stress is equal to:
Full explanation

Answer: C. twice the maximum shear stress

Uniaxial σ: τmax = σ/2 on the 45° planes, so σmax = 2τmax.

Question 13Hard
At a point in a strained material, two mutually perpendicular tensile stresses of 2000 kg/cm² and 1000 kg/cm² act. The intensity of shear (tangential) stress on a plane inclined at 15° to the axis of the minor stress is:
Full explanation

Answer: B. 250 kg/cm²

The plane's normal is at 15° to the major stress: τ = ((2000 − 1000)/2) sin 30° = 250 kg/cm². (Its normal stress is 1933 kg/cm², not offered.)

Question 14Medium
A material may fail if (σ₀ = yield stress in simple tension):
Full explanation

Answer: D. all options are correct

(a) is Rankine's criterion, (b) St. Venant's, (c) Guest–Tresca's. Each is a recognised failure criterion, so all are correct.

Question 15Easy
The normal and tangential components of stress on a plane inclined at θ to the direction of the force are equal when θ is:
Full explanation

Answer: A. 45°

σ cos²θ = σ sinθ cosθ ⇒ tanθ = 1 ⇒ θ = 45°.