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This section includes 124 Mcqs, each offering curated multiple-choice questions to sharpen your Civil Engineering knowledge and support exam preparation. Choose a topic below to get started.
51. |
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» D. six times that of hysd bars | |
52. |
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» D. ka =9 x kp | |
53. |
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» D. kp = 1+sin2ϕ / 1-sin2ϕ | |
54. |
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» C. 0.3 | |
55. |
After the tensioning operation, the space between the tendons and the ducts are |
A. | layered |
B. | grouted |
C. | cemented |
D. | drilled |
Answer» C. cemented | |
56. |
The bond of prestressing wires in Hoyer’s system can be formed by |
A. | helical crimping |
B. | tangential crimping |
C. | circular crimping |
D. | diode crimping |
Answer» B. tangential crimping | |
57. |
A flat slab can be design by |
A. | direct design method |
B. | equivalent frame method |
C. | both (a) and (b) |
D. | eulers method |
Answer» D. eulers method | |
58. |
The panels shall be rectangular, and the ratio of the longer span to the shorter span within a panel shall |
A. | not be less than 3.0 |
B. | not be greater than 2.0 |
C. | not be greater than 3.0 |
D. | not be less than 2.0 |
Answer» C. not be greater than 3.0 | |
59. |
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» D. may be positive or negative | |
60. |
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» C. if τv > τc | |
61. |
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» D. m-60 to m-80 | |
62. |
Prestressing is possible by using |
A. | mild steel |
B. | high-strength deformed bars |
C. | high-tensile steel |
D. | all of the above |
Answer» D. all of the above | |
63. |
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» B. if τc < τv < 1.5 τc | |
64. |
In direct design method of flat slab design, At an exterior support, the column strip shall be designed to resist the |
A. | total negative moment in the panel at that support. |
B. | total positive moment in the panel at that support. |
C. | 75 % of total negative moment in the panel at that support. |
D. | 75 % of total positive moment in the panel at that support. |
Answer» B. total positive moment in the panel at that support. | |
65. |
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» E. | |
66. |
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» B. 245.70 knm | |
67. |
Hoyer’s system of pre tensioning is generally adopted for |
A. | small scale members |
B. | large scale members |
C. | middle span members |
D. | end members |
Answer» C. middle span members | |
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» D. pa = ka ϒ h3 /6 | |
69. |
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» B. gravity wall | |
70. |
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» E. | |
71. |
Tendon splices are required in |
A. | hoyer’s long line method of pretensioning |
B. | short span prestressed members |
C. | long span continuous members |
D. | all of the above |
Answer» D. all of the above | |
72. |
The tendon splice preferred in the case of long span continuous prestressed members is |
A. | wrapped splice |
B. | clamp splice |
C. | torpedo splice |
D. | none of the above |
Answer» D. none of the above | |
73. |
In prestressed members requiring very large forces, the tendons preferred are |
A. | wires |
B. | bars |
C. | strands |
D. | none of the above |
Answer» D. none of the above | |
74. |
In post-tensioning system |
A. | wires are fist tensioned followed by concreting |
B. | tensioning of wires and concreting is simultaneously done |
C. | the wires are tensioned against hardened concrete |
D. | none of the above |
Answer» D. none of the above | |
75. |
In long line method or pretensioning, |
A. | only one member can be cast |
B. | two members can be cast |
C. | several members can be cast |
D. | none of the above |
Answer» D. none of the above | |
76. |
The creep coefficient depends upon the |
A. | span of the structural member |
B. | notional size of the member |
C. | depth of the member |
D. | none of the above |
Answer» C. depth of the member | |
77. |
Target mean strength of concrete depends upon |
A. | water/cement ratio |
B. | type of aggregate |
C. | characteristic strength of concrete |
D. | all of the above |
Answer» D. all of the above | |
78. |
Shrinkage of concrete in a structural member is due to |
A. | dead load on the member |
B. | live load |
C. | loss of moisture and drying of concrete |
D. | both a and b |
Answer» D. both a and b | |
79. |
Creep of concrete in a structural member is due to |
A. | elastic strain |
B. | elasto-plastic strain |
C. | inelastic strain |
D. | plastic strain |
Answer» D. plastic strain | |
80. |
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» C. 0.2 to 0.3 | |
81. |
Let height of retaining wall is 5.1m, Ï’=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» D. 124.73 knm | |
82. |
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» C. 0.3 | |
83. |
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» C. 0.51% of gross sectional area | |
84. |
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» B. ≤ 1.55 | |
85. |
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» B. bending key is provided | |
86. |
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» B. 8m to 10m | |
87. |
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» E. | |
88. |
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» D. 132.30 knm | |
89. |
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» C. if Ï„v > Ï„c | |
90. |
In direct design method of flat slab design, At an interior support, the column strip shall be designed to resist |
A. | 75 percent of the total positive moment in the panel at that support |
B. | 25 percent of the total negative moment in the panel at that support |
C. | 75 percent of the total negative moment in the panel at that support |
D. | 65 percent of the total negative moment in the panel at that support |
Answer» D. 65 percent of the total negative moment in the panel at that support | |
91. |
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» C. 198.45 knm | |
92. |
In flat slab design, in an interior span total design moment Mo shall be distributed in proportion |
A. | 25 % negative design moment & 75 % positive design moment |
B. | 75 % negative design moment & 255 % positive design moment |
C. | 35 % negative design moment & 65 % positive design moment |
D. | 65 % negative design moment & 35 % positive design moment |
Answer» E. | |
93. |
According to IS 456: 2000, a flat slab can be design by direct design method if there are continuous span in each direction |
A. | minimum 3 |
B. | maximum 3 |
C. | minimum 4 |
D. | no limitation on spans |
Answer» B. maximum 3 | |
94. |
The bond of prestressing wires in Hoyer’s system can be formed by |
A. | helical crimping |
B. | tangential crimping |
C. | circular crimping |
D. | diode crimping |
Answer» B. tangential crimping | |
95. |
Which is one of the systems used for pretensioning? |
A. | magnel-balton system |
B. | freyssinet system |
C. | gifford-udall system |
D. | hoyer’s long line method |
Answer» E. | |
96. |
The permissible stress in prestressing steel should not exceed |
A. | 80% |
B. | 60% |
C. | 50% |
D. | 70% |
Answer» B. 60% | |
97. |
When the concrete attains sufficient strength, which elements are released? |
A. | jacks |
B. | casting bed |
C. | tendons |
D. | beams |
Answer» B. casting bed | |
98. |
The method of prestressing the concrete after it attains its strength is known as |
A. | pre tensioning |
B. | post tensioning |
C. | chemical prestressing |
D. | axial prestressing |
Answer» C. chemical prestressing | |
99. |
In pre-tensioning system, after curing and hardening of concrete the reinforcement is set |
A. | free |
B. | fixed |
C. | locked |
D. | jacked |
Answer» C. locked | |
100. |
The compression in concrete and tension in steel are developed by? |
A. | joint cements |
B. | expansion cements |
C. | water cement ratio |
D. | hardened cements |
Answer» C. water cement ratio | |