Physics

Solid Mechanics

510 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

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} $

Multiple choice

Two identical cantilever beams are supported as shown, with their free ends in contact through a rigid roller. After the load P is applied, the free ends will have

  1. equal deflections but not equal slopes

  2. equal slopes but not equal deflections

  3. equal slopes as well as equal deflections

  4. neither equal slopes nor equal deflections
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From the figure, we can say that load P applies a force on upper cantilever and the reaction force also applied on upper cantilever by the rigid roller. Due to this, deflections occur in both the cantilever, which are equal in amount. But because of different forces applied by the P and rigid roller, the slopes are unequal.