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

A simply supported prestressed concrete beam is 6 m long and 300 mm wide. Its gross depth is 600 mm. It is prestressed by horizontal cable tendons at a uniform eccentricity of 100 mm. The prestressing tensile force in the cable tendons is 1000 kN. Neglect the self weight of beam.

The maximum normal compressive stress in the beam at transfer is

  1. Zero

  2. 5.55 N/mm2

  3. 11.11 N/mm2

  4. 15.68 N/mm2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The normal stress at any section is given by (P/A) + (Pe/Z) or (P/A) - (Pe/Z). For a rectangular beam, A = b*d and Z = (b*d^2)/6. Calculating these values with P=1000 kN, e=100 mm, b=300 mm, and d=600 mm yields the maximum compressive stress.

Multiple choice

A circular solid shaft of span L = 5 m is fixed at one end and free at the other end. A twisting moment T = 100 kN-m is applied at the free end. The torsional rigidity Gj is 50000 kN-m2/red. Following statements are made for this shaft. (I) The maximum rotation is 0.01 rad. (II) The torsional strain energy is 1 kN-m. With reference to the above statements, which of the following applies?

  1. Both statements are true.

  2. Statement I is true but II is false.

  3. Statement II is true but I is false.

  4. Both the statements are false.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Rotation theta = TL/GJ = (100 * 5) / 50000 = 0.01 rad, so Statement I is true. Torsional strain energy U = (T^2 * L) / (2 * GJ) = (100^2 * 5) / (2 * 50000) = 0.5 kN-m, so Statement II is false.

Multiple choice

A moment M of magnitude 50 kN-m is transmitted to a column flange through a bracket by using four 20 mm diameter rivets as shown in the figure. The shear force induced in the rivet A is

  1. 250 kN

  2. 175.8 kN

  3. 125 kN

  4. 88.4 kN
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The moment M induces a shear force F = M*r / sum(r^2) in each rivet, where r is the distance from the centroid. With four rivets, the shear force is calculated by considering the moment arm and the geometry of the rivet group.

Multiple choice

The plane frame below is analysed by neglecting axial deformations. The following statements are made with respect to the analysis (I) Column AB carries axial force only. (II) Vertical deflection at the center of beam BC is 1 mm. With reference to the above statements, which of the following applies? El = 81380 kN-m2

  1. Both the statements are true.

  2. Statement I is true but II is false.

  3. Statement II is true but I is false.

  4. Both the statements are false.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a frame with symmetric loading and geometry, the column axial forces can be determined by statics. The deflection of the beam can be found using standard beam deflection formulas or moment distribution methods.

Multiple choice

An R.C. short column with 300 mm x 300 mm square cross-section is made of M20 grade concrete and has 4 members, 20 mm diameter longitudinal bars of Fe 415 steel. It is under the action of a concentric axial compressive load. Ignoring the reduction in the area of concrete due to steel bars, the ultimate axial load carrying capacity of the column is

  1. 1659 kN

  2. 1548 kN

  3. 1198 kN

  4. 1069 kN
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The ultimate load capacity of a short column is given by 0.4*fck*Ac + 0.67*fy*Asc. Using the provided dimensions and material strengths, the calculation yields approximately 1069 kN.

Multiple choice

A strut in a steel truss is composed of two equal angles ISA 150 mm x 150 mm of thickness 100 mm connected back-to-back to the same side of a gusset plate. The cross sectional area of each angle is 2921 mm2 and moment of inertia (Ixx = Iyy) is 6335000 mm4. The distance of the centroid of the angle from its surface (Cx = Cy) is 40.8 mm. The minimum radius of gyration of the strut is

  1. 93.2 mm

  2. 62.7 mm

  3. 46.6 mm

  4. 29.8 mm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The radius of gyration is sqrt(I/A). For a built-up section, calculate the combined moment of inertia about the axis of interest using the parallel axis theorem and divide by the total area.

Multiple choice

A propped cantilever of span L is carrying a vertical concentrated load acting at midspan. The plastic moment of the section is Mp. The magnitude of the collapse load is
  1. $\frac{8M_P}{L}$
  2. $\frac{6M_P}{L}$
  3. $\frac{4M_P}{L}$
  4. $\frac{2M_P}{L}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a propped cantilever with a concentrated load at midspan, the collapse mechanism involves plastic hinges at the support and under the load. Equating internal work to external work gives the collapse load.

Multiple choice

Two equal angles ISA 100 mm x 100 mm of thickness 10 mm are placed back-to-back and connected to the either side of a gusset plate through a single row of 16 mm diameter rivets in double shear. The effective areas of the connected and unconnected legs of each of these angles are 775 mm2 and 950 mm2, respectively. If these angles are NOT tackriveted, the net effective area of this pair of angles is

  1. 3650 mm2

  2. 3450 mm2

  3. 3076 mm2

  4. 2899 mm2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The net effective area is calculated by subtracting the area of the rivet holes from the gross area of the connected leg and adding the net area of the unconnected leg.

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 vertical reaction at support H is

  1. 15 kN upwards

  2. 9.84 kN upwards

  3. 15 kN downwards

  4. 9.84 kN downwards
Reveal answer Fill a bubble to check yourself
B Correct answer
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.