McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Civil Engineering .
| 51. |
In flat slab design, When drop panels are used, the thickness of drop panel for determination of area of reinforcement shall be |
| A. | equal to thickness of drop |
| B. | equal to thickness of slab plus one quarter the distance between edge of drop and edge of capital |
| C. | lesser of (a) and (b) |
| D. | greater of (a) and (b) |
| Answer» C. lesser of (a) and (b) | |
| 52. |
In flat slab design, let τv = shear stress at critical section and τc = permissible shear stress in concrete , then no shear reinforcement is required |
| A. | if τv < τc |
| B. | if τc < τv < 1.5 τc |
| C. | if τv > τc |
| D. | if τv > 1.5τc |
| Answer» A. if τv < τc | |
| 53. |
In flat slab design, let τv = shear stress at critical section and τc = permissible shear stress in concrete , then shear reinforcement shall be provided |
| A. | if τv < τc |
| B. | if τc < τv < 1.5 τc |
| C. | if τv > τc |
| D. | if τv > 1.5τc |
| Answer» B. if τc < τv < 1.5 τc | |
| 54. |
In flat slab design, let τv = shear stress at critical section and τc = permissible shear stress in concrete , then flat slab is redesigned |
| A. | if τv < τc |
| B. | if τc < τv < 1.5 τc |
| C. | if τv > τc |
| D. | if τv > 1.5τc |
| Answer» D. if τv > 1.5τc | |
| 55. |
In flat slab design, the moment at the support of column strip is |
| A. | 0 |
| B. | positive |
| C. | negative |
| D. | may be positive or negative |
| Answer» C. negative | |
| 56. |
In limit state method of design of flat slab, τc = permissible shear stress in concrete |
| A. | τc = 0.25 √fck |
| B. | τc = 0.16 √fck |
| C. | τc = 0.45 √fck |
| D. | τc = 0.70 √fck |
| Answer» A. τc = 0.25 √fck | |
| 57. |
In working method of design of flat slab, τc = permissible shear stress in concrete |
| A. | τc = 0.25 √fck |
| B. | τc = 0.16 √fck |
| C. | τc = 0.45 √fck |
| D. | τc = 0.70 √fck |
| Answer» B. τc = 0.16 √fck | |
| 58. |
In direct design method of flat slab, total design moment Mo is 945 kNm then negative design moment in middle strip is |
| A. | 368.55 knm |
| B. | 245.70 knm |
| C. | 198.45 knm |
| D. | 132.30 knm |
| Answer» B. 245.70 knm | |
| 59. |
In direct design method of flat slab, total design moment Mo is 945 kNm then negative design moment in column strip is |
| A. | 368.55 knm |
| B. | 245.70 knm |
| C. | 198.45 knm |
| D. | 132.30 knm |
| Answer» A. 368.55 knm | |
| 60. |
In direct design method of flat slab, total design moment Mo is 945 kNm then positive design moment in middle strip is |
| A. | 330.75 knm |
| B. | 614.25 knm |
| C. | 198.45 knm |
| D. | 132.30 knm |
| Answer» D. 132.30 knm | |
| 61. |
In direct design method of flat slab, total design moment Mo is 945 kNm then positive design moment in column strip is |
| A. | 330.75 knm |
| B. | 614.25 knm |
| C. | 198.45 knm |
| D. | 132.30 knm |
| Answer» C. 198.45 knm | |
| 62. |
In direct design method of flat slab, total design moment Mo is 945 kNm then positive design moment is |
| A. | 330.75 knm |
| B. | 614.25 knm |
| C. | 236.25 knm |
| D. | 708.75 knm |
| Answer» A. 330.75 knm | |
| 63. |
In direct design method of flat slab, total design moment Mo is 945 kNm then negative design moment is |
| A. | 330.75 knm |
| B. | 614.25 knm |
| C. | 236.25 knm |
| D. | 708.75 knm |
| Answer» B. 614.25 knm | |
| 64. |
The pressure exerted by the retained material on the retaining wall is called |
| A. | active earth pressure |
| B. | earth pressure |
| C. | passive earth pressure |
| D. | both (a) and (b) |
| Answer» D. both (a) and (b) | |
| 65. |
A retaining wall which resist the earth pressure due to backfill by its dead weight is called |
| A. | cantilever retaining wall |
| B. | gravity wall |
| C. | counterfort retaining wall |
| D. | buttress retaining wall |
| Answer» A. cantilever retaining wall | |
| 66. |
Cantilever RC retaining wall proves to be economical for height |
| A. | 5m to 7m |
| B. | 8m to 10m |
| C. | 11 m to 15m |
| D. | more than 15m |
| Answer» A. 5m to 7m | |
| 67. |
Let H= height of retaining wall, ϒ=unit weight of backfill and ka = coefficient of active earth pressure, kp = coefficient of passive earth pressure, then the intensity of active earth pressure per unit area of wall at any depth ‘h’ below top of the wall is given by |
| A. | pa = ka ϒ h |
| B. | pa = kp ϒ h |
| C. | pa = ka ϒ h2 /2 |
| D. | pa = ka ϒ h3 /6 |
| Answer» A. pa = ka ϒ h | |
| 68. |
Let H= height of retaining wall, ϒ=unit weight of backfill and ka = coefficient of active earth pressure, kp = coefficient of passive earth pressure, then total pressure at any height ‘h’ below top of the wall is given by |
| A. | pa = ka ϒ h |
| B. | pa = kp ϒ h |
| C. | pa = ka ϒ h2 /2 |
| D. | pa = ka ϒ h3 /6 |
| Answer» C. pa = ka ϒ h2 /2 | |
| 69. |
Let H= height of retaining wall, ϒ=unit weight of backfill and ka = coefficient of active earth pressure, kp = coefficient of passive earth pressure, then bending moment at any height ‘h’ below top of the wall is given by |
| A. | pa = ka ϒ h |
| B. | pa = kp ϒ h |
| C. | pa = ka ϒ h2 /2 |
| D. | pa = ka ϒ h3 /6 |
| Answer» D. pa = ka ϒ h3 /6 | |
| 70. |
Coefficient of active earth pressure ka |
| A. | ka = 1-sinϕ / 1+sinϕ |
| B. | ka = 1-sin2ϕ / 1+sin2ϕ |
| C. | ka = 1+sinϕ / 1-sinϕ |
| D. | ka = 1+sin2ϕ / 1-sin2ϕ |
| Answer» A. ka = 1-sinϕ / 1+sinϕ | |
| 71. |
Coefficient of passive earth pressure kp |
| A. | kp = 1-sinϕ / 1+sinϕ |
| B. | kp = 1-sin2ϕ / 1+sin2ϕ |
| C. | kp = 1+sinϕ / 1-sinϕ |
| D. | kp = 1+sin2ϕ / 1-sin2ϕ |
| Answer» C. kp = 1+sinϕ / 1-sinϕ | |
| 72. |
The relation between ka = coefficient of active earth pressure and kp = coefficient of passive earth pressure is |
| A. | kp =3 x ka |
| B. | ka =3 x kp |
| C. | kp =9 x ka |
| D. | ka =9 x kp |
| Answer» C. kp =9 x ka | |
| 73. |
The vertical stem of cantilever retaining wall is subjected to |
| A. | varying earth pressure developing tensile stresses on earth side |
| B. | varying earth pressure developing tensile stresses on opposite side of earth side |
| C. | varying large upward soil pressure |
| D. | downward force due to self-weight of slab |
| Answer» A. varying earth pressure developing tensile stresses on earth side | |
| 74. |
The heel slab of cantilever retaining wall is subjected to
|
| A. | 1 ,2 and 3 |
| B. | only 2 and 3 |
| C. | only 1 and 3 |
| D. | 2, 3 and 4 |
| Answer» D. 2, 3 and 4 | |
| 75. |
The toe slab of cantilever retaining wall is subjected to
|
| A. | 1 ,2 and 3 |
| B. | only 2 and 3 |
| C. | only 1 and 3 |
| D. | 2, 3 and 4 |
| Answer» C. only 1 and 3 | |
| 76. |
To stabilize a concrete cantilever retaining wall against sliding, the ratio of sliding force to resisting force should be |
| A. | ≥ 1.55 |
| B. | ≤ 1.55 |
| C. | ≥ 1.0 |
| D. | ≤ 0.645 |
| Answer» D. ≤ 0.645 | |
| 77. |
To stabilize a concrete cantilever retaining wall against sliding, the ratio of resisting force to sliding force should be |
| A. | ≥ 1.55 |
| B. | ≤ 1.55 |
| C. | ≥ 1.0 |
| D. | ≤ 0.645 |
| Answer» A. ≥ 1.55 | |
| 78. |
In retaining wall to prevent the sliding of wall sometimes |
| A. | shear key is provided |
| B. | bending key is provided |
| C. | ankle key is provided |
| D. | bearings are provided |
| Answer» A. shear key is provided | |
| 79. |
If the angle of repose is 31º the coefficient of active earth pressure is |
| A. | 0.29 |
| B. | 0.32 |
| C. | 0.3 |
| D. | 0.22 |
| Answer» B. 0.32 | |
| 80. |
The temperature and shrinkage reinforcement provided in retaining wall for mild steel |
| A. | 0.12% of gross sectional area |
| B. | 0.15% of gross sectional area |
| C. | 0.51% of gross sectional area |
| D. | 0.21% of gross sectional area |
| Answer» B. 0.15% of gross sectional area | |
| 81. |
The temperature and shrinkage reinforcement provided in retaining wall for HYSD reinforcement is |
| A. | 0.12% of gross sectional area |
| B. | 0.15% of gross sectional area |
| C. | 0.51% of gross sectional area |
| D. | 0.21% of gross sectional area |
| Answer» A. 0.12% of gross sectional area | |
| 82. |
For stability of retaining wall against retaining wall the factor of safety against overturning |
| A. | should not less than 1.55 |
| B. | should not more than 1.55 |
| C. | should not less than 1.00 |
| D. | 1 |
| Answer» A. should not less than 1.55 | |
| 83. |
If embankment is sloping at an angle of 18º to the horizontal, the coefficient of active earth pressure is |
| A. | 0.3 |
| B. | 0.36 |
| C. | 3.6 |
| D. | 3 |
| Answer» B. 0.36 | |
| 84. |
If angle of repose is 30º then Coefficient of active earth pressure ka |
| A. | 3 |
| B. | 9 |
| C. | 1/3 |
| D. | 1/9 |
| Answer» C. 1/3 | |
| 85. |
If angle of repose is 30º then Coefficient of passive earth pressure kp |
| A. | 3 |
| B. | 9 |
| C. | 1/3 |
| D. | 1/9 |
| Answer» A. 3 | |
| 86. |
The maximum permissible eccentricity of a retaining wall of width B to avoid failure in tension is |
| A. | b/2 |
| B. | b/3 |
| C. | b/6 |
| D. | b/12 |
| Answer» C. b/6 | |
| 87. |
Let height of retaining wall is 5.1m, ϒ=unit weight of backfill is 18kN/m3 and ka = coefficient of active earth pressure is 0.32, then total pressure at height 5.1m below top of the wall is given by |
| A. | 74.90 kn |
| B. | 79.40 kn |
| C. | 94.70 kn |
| D. | 97.40 kn |
| Answer» A. 74.90 kn | |
| 88. |
Let height of retaining wall is 5.1m, ϒ=unit weight of backfill is 18kN/m3 and ka = coefficient of active earth pressure is 0.32, then bending moment at height 5.1m below top of the wall is given by |
| A. | 123.74 knm |
| B. | 137.24 knm |
| C. | 127.34 knm |
| D. | 124.73 knm |
| Answer» C. 127.34 knm | |
| 89. |
In axially prestressed concrete members, the steel is under |
| A. | compression |
| B. | tension |
| C. | torsion |
| D. | shear |
| Answer» B. tension | |
| 90. |
In axially prestressed members, the concrete is under |
| A. | tension |
| B. | compression |
| C. | torsion |
| D. | shear |
| Answer» B. compression | |
| 91. |
Prestressing is possible by using |
| A. | mild steel |
| B. | high-strength deformed bars |
| C. | high-tensile steel |
| D. | all of the above |
| Answer» C. high-tensile steel | |
| 92. |
Prestressing steel has an ultimate tensile strength nearly |
| A. | twice that of hysd bars |
| B. | thrice that of mild steel reinforcements |
| C. | four times that of hysd bars |
| D. | six times that of hysd bars |
| Answer» C. four times that of hysd bars | |
| 93. |
Prestressing is economical for members of |
| A. | long span |
| B. | medium span |
| C. | short span |
| D. | all of the above |
| Answer» A. long span | |
| 94. |
Linear prestressing is adopted in |
| A. | circular tanks |
| B. | pipes |
| C. | beams |
| D. | both a and b |
| Answer» C. beams | |
| 95. |
Circular prestressing is advantageous in |
| A. | beams |
| B. | columns |
| C. | pipes and tanks |
| D. | both a and b |
| Answer» C. pipes and tanks | |
| 96. |
Prestressing wires in electric poles are |
| A. | concentric |
| B. | eccentric |
| C. | parabolic |
| D. | biaxial |
| Answer» A. concentric | |
| 97. |
In the construction of large circular water tanks, it is economical to adopt |
| A. | reinforced concrete |
| B. | prestressed concrete |
| C. | steel |
| D. | none of the above |
| Answer» B. prestressed concrete | |
| 98. |
In cable-stayed bridges, the cables supporting the deck of the bridge are under |
| A. | compression |
| B. | torsion |
| C. | shear |
| D. | tension |
| Answer» D. tension | |
| 99. |
The grade of concrete for prestressed members should be in the range of |
| A. | m-20 to m-30 |
| B. | m-80 to m-100 |
| C. | m-30 to m-60 |
| D. | m-60 to m-80 |
| Answer» C. m-30 to m-60 | |
| 100. |
High-strength mixes should have a water/cement ratio of |
| A. | 0.6 to 0.8 |
| B. | 0.3 to 0.4 |
| C. | 0.2 to 0.3 |
| D. | 0.4 to 0.6 |
| Answer» B. 0.3 to 0.4 | |
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