McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Civil Engineering .
| 51. |
Which type of load is applied in tensile testing? |
| A. | Axial load |
| B. | Shear load |
| C. | Transverse load |
| D. | Longitudinal load |
| Answer» C. Transverse load | |
| 52. |
Which law is also called as the elasticity law? |
| A. | Bernoulli’s law |
| B. | Stress law |
| C. | Hooke’s law |
| D. | Poisson’s law |
| Answer» C. Hooke’s law | |
| 53. |
The materials which have the same elastic properties in all directions are called __________ |
| A. | Isotropic |
| B. | Brittle |
| C. | Homogeneous |
| D. | Hard |
| Answer» A. Isotropic | |
| 54. |
The calculation of the moment of the body due to the loadings involve a quantity called ____________ |
| A. | Moment |
| B. | Inertia |
| C. | Moment of Inertia |
| D. | Rotation |
| Answer» C. Moment of Inertia | |
| 55. |
Moment of Inertia is the integration of the square of the distance of the centroid and the del area along the whole area of the structure. |
| A. | True |
| B. | False |
| C. | none |
| D. | all |
| Answer» A. True | |
| 56. |
What is parallel axis theorem and to whom it is applied? |
| A. | Theorem used to add the two mutually perpendicular moment of inertias for areas |
| B. | Theorem used to add the two mutually perpendicular moment of inertias for volumes |
| C. | Theorem used to add the two mutually perpendicular moment of inertias |
| D. | Theorem used to add the two mutually perpendicular moment of inertias for vectors |
| Answer» A. Theorem used to add the two mutually perpendicular moment of inertias for areas | |
| 57. |
The parallel axis theorem gives the moment of inertia ______________ to the surface of considerance. |
| A. | Linear |
| B. | Non-Linear |
| C. | Perpendicular |
| D. | Parallel |
| Answer» C. Perpendicular | |
| 58. |
In the calculation of the radius of gyration, we use intensity of loadings. So whenever the distributed loading acts perpendicular to an area its intensity varies __________ |
| A. | Linearly |
| B. | Non-Linearly |
| C. | Parabolically |
| D. | Cubically |
| Answer» A. Linearly | |
| 59. |
Elongation of a bar of uniform cross section of length „L‟, due to its own weight „W‟ is given by |
| A. | 2WL/E |
| B. | WL/E |
| C. | WL/2E |
| D. | WL/3E |
| Answer» C. WL/2E | |
| 60. |
steel bar 10 mm x 10 mm cross section is subjected to an axial tensile load of 20kN. If the length of bar is 1 m and E = 200 GPa, then elongation of the bar is: |
| A. | 1 mm |
| B. | 0.5 mm |
| C. | 0.75 mm |
| D. | 1.5 mm |
| Answer» A. 1 mm | |
| 61. |
The modulus of rigidity and poisson‟s ratio of a material are 80 GPa and 0.3 respectively. Its young‟s modulus will be |
| A. | 160 GPa |
| B. | 208 GPa |
| C. | 120 GPa |
| D. | 104 GPa |
| Answer» D. 104 GPa | |
| 62. |
If the value of poisson‟s ratio is zero |
| A. | the lateral strain is high |
| B. | the material is perfectly elastic |
| C. | there is no linear strain in the material |
| D. | none of the above |
| Answer» C. there is no linear strain in the material | |
| 63. |
The ratio between direct stress and volumetric strain is: |
| A. | Bulk modulus |
| B. | Poisson’s ratio |
| C. | Factor of safety |
| D. | Modulus of rigidity |
| Answer» A. Bulk modulus | |
| 64. |
Young‟s modulus of a material which gives 2 kN/mm2 stress at 0.01 strain is |
| A. | 20kN/mm2 |
| B. | 0.02kN/mm2 |
| C. | 200 kN/mm2 |
| D. | 2000kN/mm2 |
| Answer» C. 200 kN/mm2 | |
| 65. |
The Young‟s modulus of elasticity of a material is 2.5 times its modulus of rigidity. The Poisson‟s ratio for the material will be |
| A. | 0.25 |
| B. | 0.33 |
| C. | 0.50 |
| D. | 0.75 |
| Answer» A. 0.25 | |
| 66. |
Consider a 250mmx15mmx10mm steel bar which is free to expand is heated from 150C to 400C. what will be developed? |
| A. | Compressive stress |
| B. | Tensile stress |
| C. | Shear stress |
| D. | No stress |
| Answer» D. No stress | |
| 67. |
The safe stress for a hollow steel column which carries an axial load of 2100 kN is 125 MN/m2. if the external diameter of the column is 30cm, what will be the internal diameter? |
| A. | 25 cm |
| B. | 26.19cm |
| C. | 30.14 cm |
| D. | 27.9 cm |
| Answer» B. 26.19cm | |
| 68. |
The percentage reduction in area of a cast iron specimen during tensile test would be of the order of |
| A. | more than 50% |
| B. | 25—50% |
| C. | 10—25% |
| D. | negligible. |
| Answer» D. negligible. | |
| 69. |
In a tensile test, near the elastic limit zone, the |
| A. | tensile strain increases more quickly |
| B. | tensile strain decreases more quickly |
| C. | tensile strain increases in proportion to the stress |
| D. | tensile strain decreases in proportion to the stress |
| Answer» A. tensile strain increases more quickly | |
| 70. |
The stress necessary to initiate yielding is |
| A. | considerably greater than that necessary to continue it |
| B. | considerably lesser than that necessary to continue it |
| C. | greater than that necessary to stop it |
| D. | lesser than that necessary to stop it |
| Answer» A. considerably greater than that necessary to continue it | |
| 71. |
Rupture stress is |
| A. | breaking stress |
| B. | maximum load/original cross-sectional area |
| C. | load at breaking point/A |
| D. | load at breaking point/neck area |
| Answer» D. load at breaking point/neck area | |
| 72. |
stress at which extension of material takes place more quickly as compared to increase in load is called |
| A. | elastic point of the material |
| B. | plastic point of the material |
| C. | breaking point of the material |
| D. | yielding point of the material |
| Answer» D. yielding point of the material | |
| 73. |
The energy absorbed in a body, when it is strained within the elastic limits, is known as |
| A. | strain energy |
| B. | resilience |
| C. | proof resilience |
| D. | modulus of resilience |
| Answer» A. strain energy | |
| 74. |
Resilience of a material is considered when it is subjected to |
| A. | frequent heat treatment |
| B. | fatigue |
| C. | creep |
| D. | shock loading |
| Answer» D. shock loading | |
| 75. |
The maximum strain energy that can be stored in a body is known as |
| A. | impact energy |
| B. | resilience |
| C. | proof resilience |
| D. | modulus of resilience |
| Answer» C. proof resilience | |
| 76. |
The total strain energy stored in a body is termed as |
| A. | resilience |
| B. | proof resilience |
| C. | modulus of resilience |
| D. | toughness |
| Answer» A. resilience | |
| 77. |
Proof resilience per material is known as |
| A. | resilience |
| B. | proof resilience |
| C. | modulus of resilience |
| D. | toughness |
| Answer» C. modulus of resilience | |
| 78. |
The stress induced in a body due to suddenly applied load compared to when it is applied gradually is |
| A. | same |
| B. | half |
| C. | two times |
| D. | four times |
| Answer» C. two times | |
| 79. |
strain energy stored in a body due to suddenly applied load compared to when it is applied gradually is |
| A. | same |
| B. | twice |
| C. | four times |
| D. | eight times |
| Answer» C. four times | |
| 80. |
During a tensile test on a specimen of 1 cm cross-section, maximum load observed was 8 tonnes and area of cross-section at neck was 0.5 cm2. Ultimate tensile strength of specimen is |
| A. | 4 tonnes/cm2 |
| B. | 8 tonnes/cm2 |
| C. | 16 tonnes/cm2 |
| D. | 22 tonnes/cm2 |
| Answer» B. 8 tonnes/cm2 | |
| 81. |
Tensile strength of a material is obtained by dividing the maximum load during the test by the |
| A. | area at the time of fracture |
| B. | original cross-sectional area |
| C. | average of (a) and (b) |
| D. | minimum area after fracture |
| Answer» B. original cross-sectional area | |
| 82. |
An axial pull of 50 KN is suddenly applied to a steel bar 2 m long and 1000 mm2 in cross-section. If modulus of elasticity is 200 GPa, find strain energy stored in the bar |
| A. | 10,000 N.mm |
| B. | 20,000 N.mm |
| C. | 25,000 N.mm |
| D. | 50,000 N.mm |
| Answer» D. 50,000 N.mm | |
| 83. |
A simply supported beam 6 m long and of effective depth 50 cm, carries a uniformly distributed load 2400 kg/m including its self weight. If the lever arm factor is 0.85 and permissible tensile stress of steel is 1400 kg/cm2, the area of steel required, is |
| A. | 14 cm2 |
| B. | 15 cm2 |
| C. | 16 cm2 |
| D. | 17 cm2 |
| Answer» C. 16 cm2 | |
| 84. |
A 10 m long mild steel rail section is fixed at 300 K temperature. If temperature increases by 60 K, find stress in rail section if ends are not yielded. Coefficient of thermal expansion is 12×10-6/K. |
| A. | 72 N/mm2 |
| B. | 144 N/mm2 |
| C. | 120 N/mm2 |
| D. | 240 N/mm2 |
| Answer» B. 144 N/mm2 | |
| 85. |
The ultimate shear stress of a mild steel plate of 10 mm thickness is 350 N/mm2. Calculate the diameter of the hole that can be punched to it without exceeding a compressive stress of 700 N/mm2. |
| A. | 10 mm |
| B. | 20 mm |
| C. | 7 mm |
| D. | 35 mm |
| Answer» B. 20 mm | |
| 86. |
A bar 2 m long and 20 mm diameter is subjected to an axial pull of 125.6 KN. Due to this load, length increases by 4 mm and diameter reduce by 0.012 mm. Find Poison‟s ratio. |
| A. | 0.2 |
| B. | 0.25 |
| C. | 0.3 |
| D. | 0.35 |
| Answer» C. 0.3 | |
| 87. |
A composite section of R.C.C. column 300mm×300mm in section having 20mm diameter 4 bars, one at each corner. Strength of concrete is 5 N/mm2 and modular ratio Es/Ec=9. Calculate load taken by column. |
| A. | 150 KN |
| B. | 200 KN |
| C. | 400 KN |
| D. | 500 KN |
| Answer» D. 500 KN | |
| 88. |
The moment of inertia of a triangular section of base 3 unit and height 2 unit, about an axis passing through its base is . |
| A. | 6 |
| B. | 9 |
| C. | 8 |
| D. | 2 |
| Answer» D. 2 | |
| 89. |
Moment of inertia of a square of side 1 unit about an axis through its center of gravity, is . |
| A. | 1 |
| B. | 1/12 |
| C. | 1/3 |
| D. | 1/4 |
| Answer» B. 1/12 | |
| 90. |
The axis about which moment of area is taken is known as . |
| A. | Axis of area |
| B. | Axis of moment |
| C. | Axis of reference |
| D. | Axis of rotation |
| Answer» C. Axis of reference | |
| 91. |
What is the formula of theorem of parallel axis? |
| A. | Iab = Ig + ah |
| B. | Iab = ah2 + Ig |
| C. | Iab = Ig – ah2 |
| D. | Izz = Iyy + Ixx |
| Answer» B. Iab = ah2 + Ig | |
| 92. |
Moment of inertia of a circular section of 2 cm diameter, about an axis through its centre of gravity, is . |
| A. | π/64 |
| B. | π/4 |
| C. | π/16 |
| D. | π/2 |
| Answer» B. π/4 | |
| 93. |
What is the unit of section modulus? |
| A. | mm |
| B. | mm2 |
| C. | mm3 |
| D. | mm4 |
| Answer» C. mm3 | |
| 94. |
What is the formula of theorem of perpendicular axis? |
| A. | Izz = Ixx – Iyy |
| B. | Izz = Ixx + Ah2 |
| C. | Izz – Ixx = Iyy |
| D. | None of the above |
| Answer» C. Izz – Ixx = Iyy | |
| 95. |
What is the unit of moment of inertia? |
| A. | mm |
| B. | mm2 |
| C. | mm3 |
| D. | mm4 |
| Answer» D. mm4 | |
| 96. |
What is the unit of Radius of gyration? |
| A. | mm |
| B. | mm2 |
| C. | mm3 |
| D. | mm4 |
| Answer» A. mm | |
| 97. |
What is the formula of radius of gyration? |
| A. | k2 = I/A |
| B. | k2 = I2/A |
| C. | k2 = I2/A2 |
| D. | k2 = (I/A)1/2 |
| Answer» A. k2 = I/A | |
| 98. |
What will be the radius of gyration of a circular plate of diameter 10cm? |
| A. | 1.5cm |
| B. | 2.0cm |
| C. | 2.5cm |
| D. | 3.0cm |
| Answer» C. 2.5cm | |
| 99. |
Moment of inertia of any section about an axis passing through its C.G is |
| A. | Maximum |
| B. | Minimum |
| C. | Depends upon the dimensions of the section |
| D. | Depends upon the shape of the section |
| Answer» B. Minimum | |
| 100. |
The moment of inertia of a triangular section of base „b‟ and height „h‟ about an axis passing through its base is ……. times the moment of inertia about an axis passing through its C.G. and parallel to the base |
| A. | 9 |
| B. | 4 |
| C. | 2 |
| D. | 3 |
| Answer» D. 3 | |
Done Studing? Take A Test.
Great job completing your study session! Now it's time to put your knowledge to the test. Challenge yourself, see how much you've learned, and identify areas for improvement. Don’t worry, this is all part of the journey to mastery. Ready for the next step? Take a quiz to solidify what you've just studied.