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

A three-span continuous beam has a internal hinge at B. Section B is at the mindspan of AC. Section R is at the midspan of CG. The 20 kN load is applied at section B whereas 10 kN loads are applied at sections D and F as shown in the figure. Span GH is subjected to uniformly distributed load of magnitude 5 kN/m. For the loading shown, shear force immediate to the right of section E is 9.84 kN upwards and the sagging moment at section E is 10.31 kN-m.

The magnitude of the shear force immediate to the left and immediate to the right of section B are, respectively

  1. 0 and 20 kN

  2. 10 kN and 10 kN

  3. 20 kN and 0

  4. 9.84 kN and 10.16 kN
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

At the limit state of collapse, an R.C. beam is subjected to flexural moment 200 kN-m, shear force 20 kN and torque 9 kN-m. The beam is 300 mm wide and has a gross depth of 425 mm, with an effective cover of 25 mm. The equivalent nominal shear stress $(\tau_{ve})$as calculated by using the design code turns out to be lesser than the design shear strength $(\tau_C)$of the concrete.

The equivalent flexural moment (Mel) for designing the longitudinal tension steel is

  1. 187 kN-m

  2. 200 kN-m

  3. 29 kN-m

  4. 213 kN-m
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equivalent moment Me = M + T(1 + D/b) / 1.7. Plugging in values: 200 + 9 * (1 + 425/300) / 1.7 = 200 + 9 * (2.416) / 1.7 = 200 + 12.79 = 212.79 kN-m, rounded to 213 kN-m.

Multiple choice

At the limit state of collapse, an R.C. beam is subjected to flexural moment 200 kN-m, shear force 20 kN and torque 9 kN-m. The beam is 300 mm wide and has a gross depth of 425 mm, with an effective cover of 25 mm. The equivalent nominal shear stress $(\tau_{ve})$as calculated by using the design code turns out to be lesser than the design shear strength $(\tau_C)$of the concrete.

The equivalent shear force (Vc) is

  1. 20 kN

  2. 54 kN

  3. 56 kN

  4. 68 kN
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equivalent shear force Ve = V + 1.6 * T / b. Ve = 20 + 1.6 * 9000 / 300 = 20 + 48 = 68 kN.

Multiple choice

Muller Breslau principle in structural analysis is used for

  1. drawing influence line diagram for any force function

  2. writing virtual work equation

  3. super position of load effects

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 Muller Breslau used for drawing influence line diagram for any force function by means of deflected diagram.

Multiple choice

Top ring beam of an Intze tank carries a hoop tension of 120 kN. The beam cross section is 250 mm wide and 400 mm deep and it is reinforced with 4 bars of 20 mm diameter of Fe 415 grade. Modular ratio of the concrete is 10. The tensile stress in N/mm2 in the concrete is

  1. 1.02

  2. 1.07

  3. 1.20

  4. 1.32
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The hoop tension is resisted by the concrete and the steel reinforcement. Using the modular ratio m=10, the equivalent area of concrete is A_eq = A_c + (m-1)A_s. The tensile stress in concrete is sigma_c = T / A_eq.