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
  1. Shear reinforcement should be designed for 175 kN for beam P and the section for beam Q should be revised.

  2. Nominal shear reinforcement is required for beam P and the shear reinforcement should be designed for 120 kN for beam Q.

  3. Shear reinforcement should be designed for 175 kN for beam P and the shear reinforcement should be designed for 525 kN for beam Q.

  4. The sections for both beams P and Q need to be revised.

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 

Multiple choice
  1. (i) RS should be provided under the concentrated load only. (ii) PQ should be placed in the tension side of the flange.

  2. (i) RS helps to prevent local buckling of the web. (ii) PQ should be placed in the compression side of the flange.

  3. (i) RS should be provided at supports. (ii) PQ should be placed along the neutral axis.

  4. (i) RS should be provided away from points of action of concentrated loads. (ii) PQ should be provided on the compression side of the flange.

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 

Multiple choice
  1. Maximum of $\Big(\frac{\pi}{4}D^2\sigma_b\Big) and (\pi DL \sigma_{st})$
  2. Maximum of $\Big(\frac{\pi}{4}D^2\sigma_{st}\Big) and (\pi DL \sigma_b)$
  3. Minimum of $\Big(\frac{\pi}{4}D^2\sigma_{st}\Big) and (\pi DL \sigma_b)$
  4. Minimum of $\Big(\frac{\pi}{4}D^2\sigma_b\Big) and (\pi DL \sigma_{st})$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 

Multiple choice
  1. $\delta_1 = \frac{2}{5} \Big(\frac{2P}{k}\Big) and \delta_2 \frac{4}{5}\Big(\frac{2P}{k}\Big)$
  2. $\delta_1 = \frac{2}{5} \Big(\frac{P}{k}\Big) and \delta_2 \frac{4}{5}\Big(\frac{P}{k}\Big)$
  3. $\delta_1 = \frac{2}{5} \Big(\frac{P}{\sqrt{2}k}\Big) and \delta_2 \frac{4}{5}\Big(\frac{P}{\sqrt{2}k}\Big)$
  4. $\delta_1 = \frac{2}{5} \Big(\frac{\sqrt{2}P}{k}\Big) and \delta_2 \frac{4}{5}\Big(\frac{\sqrt{2}P}{k}\Big)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 

Multiple choice
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that brittle materials do not show elongation when external load is applied on them and they get altered by certain amount of load and thus, get cracked from unexpected points. The amount of ultimate load and point of fracture cannot be predetermined and elongation is also very, very less. These properties are best described by option (4).

Multiple choice
  1. M20 for both

  2. M40 and M30

  3. M15 and M20

  4. M30 and M40

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 As per code minimum grade concrete used for post-tension is M-30 and pre-tension is M-40. Reason for this is that in pre-tension losses are more than post-tension and a high strength concrete required for pre-stress so that it can safely bear the high local stress.

Multiple choice
  1. $\frac{16TL}{\pi d^4 G}$
  2. $\frac{32TL}{\pi d^4 G}$
  3. $\frac{64TL}{\pi d^4 G}$
  4. $\frac{128TL}{\pi d^4 G}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The angle of twist for a circular shaft is given by the formula theta = TL / (GJ), where J is the polar moment of inertia. For a solid circular shaft, J = (pi * d^4) / 32. Substituting J gives theta = TL / (G * pi * d^4 / 32) = 32TL / (pi * d^4 * G).