Home/MCQ Practice/Strength of Materials MCQs/Principal Stress and Principal Planes

Structures·MCQ practice

Principal Stress and Principal Planes MCQs – Part 2

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

Question 16Medium
The locus of the end point of the resultant of the normal and tangential components of stress on an inclined plane is:
Full explanation

Answer: C. an ellipse

The resultant stress vector on planes of every orientation traces the ellipse of stress, with semi-axes σ₁ and σ₂.

Question 17Medium
Mohr's strength theory is based on which of the following facts?
Full explanation

Answer: D. All options are correct

Mohr's theory assumes shear failure on a slip plane, governed by σ and τ on that plane, using the largest Mohr circle (σ₁, σ₃) — so σ₂ has no effect. All three are its bases.

Question 18Medium
In a Mohr's circle drawn on the σ–τ plane (σ = normal stress, τ = shear stress), the vertical diameter represents:
Full explanation

Answer: A. maximum shear stress

The ends of the vertical diameter are the points of maximum shear (±τmax = ±R); its half-length is τmax.

Question 19Easy
The angle between a principal plane and the plane of maximum shear is:
Full explanation

Answer: A. 45°

τ = (σ/2) sin 2θ is maximum at 2θ = 90°, i.e. θ = 45°.

Question 20Easy
A square element carries a tensile stress σ in the horizontal direction and a compressive stress σ in the vertical direction (normal stresses only). The radius of its Mohr's circle is:
Full explanation

Answer: A. σ

R = (σ₁ − σ₂)/2 = (σ − (−σ))/2 = σ.

Question 21Easy
The total strain energy theory for the failure of a material at the elastic limit is known as:
Full explanation

Answer: D. Haigh's theory

Total strain energy theory = Haigh's (Beltrami–Haigh) theory.

Question 22Easy
The maximum shear stress theory for the failure of a material at the elastic limit is known as:
Full explanation

Answer: A. Guest's or Tresca's theory

Maximum shear stress theory = Guest's or Tresca's theory: failure when τmax = σy/2.

Question 23Medium
According to the maximum strain energy theory, the yield locus is:
Full explanation

Answer: D. an ellipse

σ₁² + σ₂² − 2μσ₁σ₂ = σy² is an ellipse in the σ₁–σ₂ plane.

Question 24Easy
In the case of biaxial stress, the maximum value of shear stress is given by:
Full explanation

Answer: D. none of these

τmax = (σ₁ − σ₂)/2 — half the difference, which is not listed.

Question 25Easy
The maximum strain theory for the failure of a material at the elastic limit is known as:
Full explanation

Answer: B. St. Venant's theory

Maximum principal strain theory = St. Venant's theory.

Question 26Medium
A block of brittle material of cross-section 4 m² carries an axial tensile load of 20 tonnes. The maximum shear stress in the block is:
Full explanation

Answer: D. none of these

A = 40 000 cm², W = 20 000 kg ⇒ σ = 0.5 kg/cm² ⇒ τmax = σ/2 = 0.25 kg/cm². Not listed.

Question 27Easy
The maximum principal stress theory for the failure of a material at the elastic limit is known as:
Full explanation

Answer: C. Rankine's theory

Maximum principal stress theory = Rankine's theory: failure when σ₁ reaches σy.

Question 28Easy
The shear strain energy theory for the failure of a material at the elastic limit is due to:
Full explanation

Answer: D. von Mises

Shear strain energy (distortion energy) theory = von Mises (von Mises–Hencky) theory.

Question 29Easy
Modulus of rigidity is expressed as:
Full explanation

Answer: C. shear stress / shear strain

G = shear stress / shear strain.

Question 30Easy
In a tensile test, when the material is stressed beyond the elastic limit, the tensile strain ______ compared with the stress.
Full explanation

Answer: D. increases more quickly

Beyond the elastic limit, Hooke's law no longer holds: strain grows much faster than stress (plastic deformation).