McqMate
| 1. |
The phenomena of development of internal tensile stresses in a concrete member by means of tensioning devices are called as |
| A. | pre-tensioning |
| B. | post-tensioning |
| C. | prestressing of concrete |
| D. | thermoelectric prestressing |
| Answer» C. prestressing of concrete | |
| 2. |
In reinforced concrete members the prestress is commonly introduced by |
| A. | tensioning the steel reinforcement |
| B. | tendons |
| C. | shortening of concrete |
| D. | rings |
| Answer» A. tensioning the steel reinforcement | |
| 3. |
Which of the following basic concept is involved in the analysis of prestressed concrete members? |
| A. | combined and bending stresses |
| B. | principle stresses |
| C. | shear stresses |
| D. | overhead stresses |
| Answer» A. combined and bending stresses | |
| 4. |
The prestressing of concrete member is carried out to reduce |
| A. | compressive stresses |
| B. | tensile stresses |
| C. | bending stresses |
| D. | shear force |
| Answer» A. compressive stresses | |
| 5. |
The earliest examples of wooden barrel construction by force-fitting of metal bands and shrink-fitting of metal tiers of wooden wheels indicate the art of |
| A. | prestressing |
| B. | tensioning |
| C. | stress |
| D. | straining |
| Answer» A. prestressing | |
| 6. |
The concept is used in many branches of civil engineering and building construction? |
| A. | reinforced concrete |
| B. | prestressed concrete |
| C. | steel concrete |
| D. | lump sum concrete |
| Answer» B. prestressed concrete | |
| 7. |
The attempt to introduce permanently acting forces in concrete to resist the elastic forces is |
| A. | prestressing |
| B. | loading |
| C. | pre-straining |
| D. | bending |
| Answer» A. prestressing | |
| 8. |
In reinforced concrete members, the prestress commonly introduced is |
| A. | tensioning steel reinforcement |
| B. | tensioning wood reinforcement |
| C. | tensioning rings |
| D. | tensioning plates |
| Answer» A. tensioning steel reinforcement | |
| 9. |
Development of early cracks in reinforced concrete is due to |
| A. | strains of steel |
| B. | stresses of steel |
| C. | ultimate load |
| D. | bending of steel |
| Answer» A. strains of steel | |
| 10. |
The significant observations which resulted from the pioneering research on prestressed concrete were |
| A. | high strength steel and losses of prestress |
| B. | high strength tendon and losses of creep |
| C. | high strength bars and losses of strain |
| D. | high strength rings and losses of shrinkage |
| Answer» A. high strength steel and losses of prestress | |
| 11. |
The necessity of high strength concrete in prestressed concrete is due to |
| A. | shear and bonding |
| B. | loading and unloading |
| C. | cracking |
| D. | bending |
| Answer» A. shear and bonding | |
| 12. |
In the zone of anchorages the material preferred to minimize costs is |
| A. | high strength steel |
| B. | high strength bars |
| C. | high strength tendons |
| D. | high strength concrete |
| Answer» D. high strength concrete | |
| 13. |
The length of the prestressing tendon between the end of the member and the point where the steel attains its stress is called |
| A. | anchorage |
| B. | de bonding |
| C. | cracking load |
| D. | transmission length |
| Answer» D. transmission length | |
| 14. |
In cab cable, the curved portion of the tendon and anchors lie in |
| A. | compression and tension zone |
| B. | cracking zone |
| C. | tension and compression zone |
| D. | loading zone |
| Answer» C. tension and compression zone | |
| 15. |
The load at which the prestressed member develops its first crack is called as |
| A. | transfer load |
| B. | creep load |
| C. | bending load |
| D. | cracking load |
| Answer» D. cracking load | |
| 16. |
In circular prestressing members, the tendons are supplied in form of |
| A. | cables |
| B. | bars |
| C. | wires |
| D. | rings |
| Answer» D. rings | |
| 17. |
In case of continuous prestressed concrete members to gain continuity, splicing is done by |
| A. | reinforcement |
| B. | steel |
| C. | concrete |
| D. | tendons |
| Answer» D. tendons | |
| 18. |
The phenomena of drying process of contraction concrete refer to |
| A. | moisture loss |
| B. | shrinkage of concrete |
| C. | drying process |
| D. | weight loss |
| Answer» B. shrinkage of concrete | |
| 19. |
The ratio between the creep strain and elastic strain of concrete is defined as |
| A. | creep ratio |
| B. | creep elasticity |
| C. | creep coefficient |
| D. | creep factor |
| Answer» C. creep coefficient | |
| 20. |
The phenomena of reduction of stress in steel at a constant strain are known as |
| A. | reduction of stress |
| B. | relaxation of stress |
| C. | de bonding |
| D. | proof stress |
| Answer» C. de bonding | |
| 21. |
A device which helps the tendons to transmit prestress to the member and maintain it for the design period is? |
| A. | cab cable |
| B. | anchorage |
| C. | tendon |
| D. | transfer |
| Answer» C. tendon | |
| 22. |
Which of the following type of prestress applied to concrete in which tensile stresses to a limited degree are permitted is known as |
| A. | moderate prestressing |
| B. | partial prestressing |
| C. | full prestressing |
| D. | axial prestressing |
| Answer» B. partial prestressing | |
| 23. |
Prevention of bond between the steel and concrete is known as |
| A. | bond prestressed concrete |
| B. | axial prestressing |
| C. | de bonding |
| D. | proof stress |
| Answer» C. de bonding | |
| 24. |
Which one of the following is the basic assumption involved in designing of prestressed concrete members? |
| A. | plane member remains plane before and after bending |
| B. | variation of stresses in tensile reinforcement |
| C. | development of principle stresses |
| D. | hooke’s law is not valid for prestressing |
| Answer» A. plane member remains plane before and after bending | |
| 25. |
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» B. expansion cements | |
| 26. |
In pre-tensioning system, after curing and hardening of concrete the reinforcement is set |
| A. | free |
| B. | fixed |
| C. | locked |
| D. | jacked |
| Answer» B. fixed | |
| 27. |
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» B. post tensioning | |
| 28. |
The ultimate strength of high tensile steel is |
| A. | 1100 |
| B. | 2100 |
| C. | 1500 |
| D. | 1250 |
| Answer» B. 2100 | |
| 29. |
The high tensile steel is obtained by increasing content of |
| A. | carbon content in steel |
| B. | aluminium content in steel |
| C. | manganese content in steel |
| D. | sulphur content in steel |
| Answer» A. carbon content in steel | |
| 30. |
The permissible stress in prestressing steel should not exceed |
| A. | 80% |
| B. | 60% |
| C. | 50% |
| D. | 70% |
| Answer» A. 80% | |
| 31. |
When the concrete attains sufficient strength, which elements are released? |
| A. | jacks |
| B. | casting bed |
| C. | tendons |
| D. | beams |
| Answer» A. jacks | |
| 32. |
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» D. hoyer’s long line method | |
| 33. |
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» B. large scale members | |
| 34. |
The transfer of prestress of concrete is achieved by |
| A. | plates |
| B. | rings |
| C. | steel bars |
| D. | jacks |
| Answer» D. jacks | |
| 35. |
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» A. helical crimping | |
| 36. |
The Hoyer’s method of prestressing is done by |
| A. | pulling out of wires |
| B. | pushing wires |
| C. | heating of wires |
| D. | stressing of wires |
| Answer» A. pulling out of wires | |
| 37. |
Hoyer’s system of pretensioning can be done for beams. |
| A. | 2 |
| B. | more than 2 |
| C. | less than 2 |
| D. | 3 |
| Answer» B. more than 2 | |
| 38. |
In post tensioning, the concrete units are cast by |
| A. | ducts |
| B. | jacks |
| C. | anchorages |
| D. | wedges |
| Answer» A. ducts | |
| 39. |
After the tensioning operation, the space between the tendons and the ducts are |
| A. | layered |
| B. | grouted |
| C. | cemented |
| D. | drilled |
| Answer» B. grouted | |
| 40. |
A slab without beam is called as |
| A. | bubble deck slab |
| B. | grid slab |
| C. | flat slab |
| D. | both (a) and (c) |
| Answer» C. flat slab | |
| 41. |
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» A. minimum 3 | |
| 42. |
A flat slab can be design by |
| A. | direct design method |
| B. | equivalent frame method |
| C. | both (a) and (b) |
| D. | eulers method |
| Answer» C. both (a) and (b) | |
| 43. |
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» B. not be greater than 2.0 | |
| 44. |
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» D. 65 % negative design moment & 35 % positive design moment | |
| 45. |
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» C. 75 percent of the total negative moment in the panel at that support | |
| 46. |
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» A. total negative moment in the panel at that support. | |
| 47. |
In flat slab design, The drops when provided shall be rectangular in plan, and have a length in each direction |
| A. | not less than three fourth of the panel length in that direction |
| B. | not less than one fourth of the panel length in that direction |
| C. | not less than one half of the panel length in that direction |
| D. | not less than one third of the panel length in that direction |
| Answer» D. not less than one third of the panel length in that direction | |
| 48. |
In flat slab design, the column strip shall be designed to resist |
| A. | 60 percent of the total positive moment in the panel |
| B. | 40 percent of the total positive moment in the panel |
| C. | 35 percent of the total negative moment in the panel |
| D. | 65 percent of the total negative moment in the panel |
| Answer» A. 60 percent of the total positive moment in the panel | |
| 49. |
In flat slab design, the middle strip shall be designed to resist |
| A. | 60 percent of the total positive moment in the panel |
| B. | 40 percent of the total positive moment in the panel |
| C. | 35 percent of the total negative moment in the panel |
| D. | 65 percent of the total negative moment in the panel |
| Answer» B. 40 percent of the total positive moment in the panel | |
| 50. |
In design of flat slab, The critical section for shear shall be at a distance |
| A. | effective depth /2 from the periphery of the column/capital/drop panel |
| B. | effective depth from the periphery of the column/capital/drop panel |
| C. | face of the column/capital/drop panel |
| D. | none of the above |
| Answer» A. effective depth /2 from the periphery of the column/capital/drop panel | |
| 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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