Physics

Solid Mechanics

537 Questions

Solid mechanics questions evaluate the understanding of stress, strain, and material deformation under various loads. Topics include analyzing beams, cantilevers, and structural steel designs using specific industrial standards. These principles are strictly examined in engineering and civil services preliminary tests.

Stress and strainBeam analysisPrestressed concreteMaterial strengthStructural design

Solid Mechanics Questions

Multiple choice

Orthogonal turning is performed on a cylindrical work piece with shear strength of 250 MPa. The following conditions are used cutting velocity is 180 m/min, feed is 0.20 mm/rev, depth of cut is 3 mm, chip thickness ratio = 0.5. The orthogonal rake angle is 70. Apply merchant's theory for analysis.

The shear plane angle (in degrees) and the shear force respectively are

  1. 52; 320 N

  2. 52; 400 N

  3. 28; 400 N

  4. 28; 320 N
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Multiple choice

In the deep drawing of cups, blank show a tendency to wrinkle up around the periphery (flange). The most likely cause and remedy of the phenomenon are, respectively,

  1. buckling due to circumferential compression; Increase blank holder pressure

  2. high blank holder pressure and high friction; Reduce blank holder pressure and apply Iubricant

  3. high temperature causing increase in circumferential length; Apply coolant to blank

  4. buckling due to circumferential compression; decrease blank holder pressure
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Seamless cylinders and tubes can be made by hot drawing or cupping The thickness of the cup is reduced and its length increased by drawing it through a series of dies having reduced clearance between the die and the punch. Due to reduction in its thickness, blanks shows a tendency to wrinkle up around the periphery because of buckling due to circumferential compreession an due to this compression blank holder pressure increases

Multiple choice

A pin-ended column of length L, modulus of elasticity E and second moment of the cross-sectional area I is loaded centrically by a compressive load P. The critical buckling load $|(P_{cr})|$ is given by
  1. $P_{cr}=\frac{EI}{x^2L^2}$
  2. $P_{cr}=\frac{x^2EI}{3L^2}$
  3. $P_{cr}=\frac{\pi EI}{L^2}$
  4. $P_{cr}=\frac{x^2EI}{L^2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\text{Here both ends are hinged,}\\ So,\hspace{2.6cm}C=1\\ \text{Substitute in equation(i), we get} \\ \hspace{3cm}P_{cr}=\frac{\pi^2EI}{L^2}$

Multiple choice

According to Von-Mises' distortion energy theory, the distortion energy under three dimensional stress state is represented by
  1. $\frac{1}{2E}[\sigma_1^2+\sigma_2^2+\sigma_2^2-2V(\sigma_1\sigma_2+\sigma_3\sigma_2+\sigma_1\sigma_3)]$
  2. $\frac{1+v}{2E}[\sigma_1^2+\sigma_2^2+\sigma_2^2-(\sigma_1\sigma_2+\sigma_3\sigma_2+\sigma_1\sigma_3)]$
  3. $\frac{1+v}{3E}[\sigma_1^2+\sigma_2^2+\sigma_2^2-(\sigma_1\sigma_2+\sigma_3\sigma_2+\sigma_1\sigma_3)]$
  4. $\frac{1}{3E}[\sigma_1^2+\sigma_2^2+\sigma_2^2-V(\sigma_1\sigma_2+\sigma_3\sigma_2+\sigma_1\sigma_3)]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Multiple choice

The ultimate tensile strength of a material is 400 MPa and the elongation up to maximum load is 35%. If the material obeys power law of hardening, then the true stress-true strain relation (stress in MPa) in the plastic deformation range is
  1. $\sigma=540\in ^{0.30}$
  2. $\sigma=775\in ^{0.30}$
  3. $\sigma=540\in ^{0.35}$
  4. $\sigma=775\in^{0.35}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\hspace{5cm}\sigma=K\epsilon^n\hspace{9cm}...(i)\\ \hspace{5cm}K=\frac{\sigma}{\epsilon^n}=\frac{540}{(0.30)^{0.30}}=7774.92\simeq775\\ So,\text{From equation (i)}\hspace{0.7cm} \sigma=775\epsilon ^{0.3} $