McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Civil Engineering .
| 51. |
What will be the rotation of beam BC at point C? |
| A. | 0.2 |
| B. | -0.2 |
| C. | 0 |
| D. | can’t say |
| Answer» C. 0 | |
| Explanation: there won’t be any rotation at point c as it is a fixed support. | |
| 52. |
What will be the extra condition, which we will get if we conserve moment near joint B? |
| A. | mba + mca = 0 |
| B. | mba + mcb = 0 |
| C. | mba + mbc = 0 |
| D. | mab + mbc = 0 |
| Answer» C. mba + mbc = 0 | |
| Explanation: just cut a small part near joint b and conserve moment around that joint. | |
| 53. |
Total how many equations will be generated? |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 4 |
| Answer» D. 4 | |
| Explanation: two separate fixed beams are | |
| 54. |
What will be the value of mAB, after solving these equations? |
| A. | 3.09 |
| B. | 1.54 |
| C. | 12.86 |
| D. | -3.09 |
| Answer» B. 1.54 | |
| Explanation: just substitute the value of rotation at point b in 1st equation. | |
| 55. |
What will be the value of mBC, after solving these equations? |
| A. | 3.09 |
| B. | 1.54 |
| C. | 12.86 |
| D. | -3.09 |
| Answer» A. 3.09 | |
| Explanation: just substitute the value of rotation at point b in 3rd equation. | |
| 56. |
How many compatibility equations should be written if we have n no. of redundant reactions? |
| A. | n – 1 |
| B. | n |
| C. | n + 1 |
| D. | n + 2 |
| Answer» B. n | |
| Explanation: no. of redundant reactions and compatibility equations are equal. | |
| 57. |
Flexibility matrix is always:- |
| A. | symmetric |
| B. | non-symmetric |
| C. | anti-symmetric |
| D. | depends upon loads applied |
| Answer» A. symmetric | |
| Explanation: flexibility matrixes are always symmetric as a consequence of betti’s law. | |
| 58. |
Which of the following primary structure is best for computational purposes? |
| A. | symmetric |
| B. | non-symmetric |
| C. | anti-symmetric |
| D. | depends upon loads applied |
| Answer» A. symmetric | |
| Explanation: it is easier to compute solutions for flexibility coefficient matrix in that case. | |
| 59. |
For computational purposes, deflected primary structure ans actual structure should be |
| A. | as different as possible |
| B. | as similar as possible |
| C. | it doesn’t matter |
| D. | in between |
| Answer» B. as similar as possible | |
| Explanation: this leads to small corrections induced by redundants. | |
| 60. |
In general, any structure can be classified as a symmetric one :- |
| A. | when its structure is symmetric |
| B. | when its loading is symmetric |
| C. | when its supports are symmetric |
| D. | when it develops symmetric internal loading and deflections |
| Answer» D. when it develops symmetric internal loading and deflections | |
| Explanation: it is deemed as symmetric when when it develops symmetric internal loading and deflections about central axis. | |
| 61. |
Normally, which of the following things may/may not be symmetric to develop symmetricity? |
| A. | material |
| B. | geometry |
| C. | loading |
| D. | dki |
| Answer» D. dki | |
| Explanation: composition, geometry and loading are generally require to be symmetric. | |
| 62. |
The degree of freedom for a pin jointed plane frame is given by 3j – m – r. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: the degree of freedom for a pin jointed plane frame is given by 2j–r, where j is the number of joint and r is the number of reactions. | |
| 63. |
In case of pin jointed plane frame, rotational displacement of the nodes are not considered. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: in case of pin jointed plane frame, rotational displacement of the nodes are not considered because member are not subjected to bending. members are subjected to axial forces only. | |
| 64. |
Identify the correct statements of the followings. |
| A. | two degrees of freedom are available at each joint of pin jointed plane frame |
| B. | six degrees of freedom are available at each joint of pin jointed space frame |
| C. | four degrees of freedom are available at each joint of rigid jointed plane frame |
| D. | three degrees of freedom are available at each joint of pin jointed space frame |
| Answer» A. two degrees of freedom are available at each joint of pin jointed plane frame | |
| Explanation: each joint of pin jointed plane frame, allows horizontal and vertical sway of the joint. therefore, degree of freedom is two at a joint of pin jointed plane frame. | |
| 65. |
Calculate the kinematic indeterminacy of the following pin jointed plane frame. |
| A. | 2 |
| B. | 6 |
| C. | 12 |
| D. | 18 |
| Answer» C. 12 | |
| Explanation: the truss is supported by hinged support at both ends. hinged support in a pin jointed plane frame does not offers any degree of freedom as rotation is not considered. for the given truss, it consists of six pin joint offering two degree of freedom each. therefore, the degree of freedom is 12. | |
| 66. |
Calculate the kinematic indeterminacy of the following pin jointed plane frame. |
| A. | 12 |
| B. | 13 |
| C. | 14 |
| D. | 15 |
| Answer» B. 13 | |
| Explanation: the truss is supported by hinged support at one end and roller support at other end. hinged support in a pin jointed plane frame does not offers any degree of freedom as rotation is not considered. but roller support offers horizontal movement and hence degree of freedom is 1. for the given truss, it consists of six pin joint offering two degree of freedom each. therefore, the degree of freedom is 12 + 1 = 13. | |
| 67. |
Calculate the kinematic indeterminacy of the following pin jointed plane frame. |
| A. | 8 |
| B. | 9 |
| C. | 10 |
| D. | 12 |
| Answer» B. 9 | |
| Explanation: the truss is supported by hinged support at one end and roller support at other end. hinged support in a pin jointed plane frame does not offers any degree of freedom as rotation is not considered. but roller support offers horizontal movement and hence degree of freedom is 1. for the given truss, it consists of four pin joint offering two degree of freedom each. therefore, the degree of freedom is 8 + 1 = 9. | |
| 68. |
Calculate the kinematic indeterminacy of the following pin jointed plane frame. |
| A. | 8 |
| B. | 10 |
| C. | 12 |
| D. | 14 |
| Answer» A. 8 | |
| Explanation: the truss is supported by hinged support at both ends. hinged support in a pin jointed plane frame does not offers any degree of freedom as rotation is not | |
| 69. |
Calculate the kinematic indeterminacy of the following pin jointed plane frame. |
| A. | 4 |
| B. | 5 |
| C. | 6 |
| D. | 8 |
| Answer» A. 4 | |
| Explanation: the truss is supported by hinged support at both ends. hinged support in a pin jointed plane frame does not offers any degree of freedom as rotation is not considered. for the given truss, it consists of two pin joint offering two degree of freedom each. therefore, the degree of freedom is 4. | |
| 70. |
Calculate the kinematic indeterminacy of the following pin jointed plane frame. |
| A. | 4 |
| B. | 6 |
| C. | 8 |
| D. | 10 |
| Answer» C. 8 | |
| Explanation: the truss is supported by hinged support at both ends. hinged support in a pin jointed plane frame does not offers any degree of freedom as rotation is not considered. for the given truss, it consists of four pin joint offering two degree of freedom each. therefore, the degree of freedom is 8. | |
| 71. |
Total how many equations will be generated? |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 4 |
| Answer» C. 3 | |
| Explanation: beam ab will give 2 equations, but ultimately beam bc will give only one equation. | |
| 72. |
What will be the value of mAB, after solving these equations? |
| A. | 108 |
| B. | 72 |
| C. | -72 |
| D. | -108 |
| Answer» D. -108 | |
| Explanation: just substitute the value of rotation at point b in 1st equation. | |
| 73. |
What will be the value of mBC, after solving these equations? |
| A. | 108 |
| B. | 72 |
| C. | -72 |
| D. | -108 |
| Answer» C. -72 | |
| Explanation: just substitute the value of rotation at point b in 3rd equation. | |
| 74. |
What will be the value of mCB, after solving these equations? |
| A. | 3.09 |
| B. | 1.54 |
| C. | 12.86 |
| D. | 0 |
| Answer» D. 0 | |
| Explanation: it is already assumed that its | |
| 75. |
What will be the value of mBA, after solving these equations? |
| A. | 108 |
| B. | 72 |
| C. | -72 |
| D. | -108 |
| Answer» B. 72 | |
| Explanation: it will be inverse of mbc as shown above. | |
| 76. |
What will be the support reaction at point B? |
| A. | 25.5 |
| B. | 22.5 |
| C. | 15 |
| D. | 37.5 |
| Answer» D. 37.5 | |
| Explanation: find shear at point b in both beam ab and bc and then cut a small part near support b and conserve force in vertical | |
| 77. |
There will be one point of discontinuity in the bending moment diagram of this question. State whether the above statement is true or false. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: this statement is false as no external moment is applied in between the beam. | |
| 78. |
How many separate parts will be required for this question? |
| A. | 0 |
| B. | 1 |
| C. | 2 |
| D. | 3 |
| Answer» D. 3 | |
| Explanation: since, 4 supports are there we will divide it into 2 separate parts to solve. | |
| 79. |
Distribution factor is the ratio in which force sharing capacity of various members meeting at a rigid joint. |
| A. | true |
| B. | false |
| C. | answer: b |
| D. | explanation: distribution factor is the ratio in which moment sharing capacity of various members meeting at a rigid joint. |
| Answer» D. explanation: distribution factor is the ratio in which moment sharing capacity of various members meeting at a rigid joint. | |
| Explanation: no force is acting on beam bc. | |
| 80. |
What will be the FEM at point A in beam AB? |
| A. | 453.1 |
| B. | -453.1 |
| C. | 72 |
| D. | -72 |
| Answer» D. -72 | |
| Explanation: formula for moment for unit load applied at mid point is wl2.12 and direction will be anti clockwise. | |
| 81. |
What will be the FEM at point B in beam AB? |
| A. | 453.1 |
| B. | -453.1 |
| C. | 72 |
| D. | -72 |
| Answer» C. 72 | |
| Explanation: formula for moment for unit load applied at mid point is wl^2.12 and direction will be clockwise. | |
| 82. |
What will be the rotation of beam CD at point D? |
| A. | 0.2 |
| B. | -0.2 |
| C. | 0 |
| D. | can’t say |
| Answer» C. 0 | |
| Explanation: there won’t be any rotation at point c as it is a fixed support. | |
| 83. |
After using all the joint conditions, how many unknowns are still left? |
| A. | 0 |
| B. | 1 |
| C. | 2 |
| D. | 3 |
| Answer» C. 2 | |
| Explanation: rotation at point b and at point c in either of beam is not known (they are both equal). | |
| 84. |
What will be one of the extra condition, which we will get if we conserve moment near joint B? |
| A. | mba + mca = 0 |
| B. | mba + mcb = 0 |
| C. | mba + mbc = 0 |
| D. | mab + mbc = 0 |
| Answer» C. mba + mbc = 0 | |
| Explanation: just cut a small part near joint b and conserve moment around that joint. | |
| 85. |
Total how many equations will be generated? |
| A. | 3 |
| B. | 4 |
| C. | 5 |
| D. | 6 |
| Answer» D. 6 | |
| Explanation: three separate fixed beams are | |
| 86. |
If in a pin-jointed plane frame (m + r) > 2j, then the frame is |
| A. | Stable and statically determinate |
| B. | Stable and statically indeterminate |
| C. | Unstable |
| D. | None of the above |
| Answer» B. Stable and statically indeterminate | |
| 87. |
Principle of superposition is applicable when |
| A. | Deflections are linear functions of applied forces |
| B. | Material obeys Hooke's law |
| C. | The action of applied forces will be affected by small deformations of the structure |
| D. | None of the above |
| Answer» A. Deflections are linear functions of applied forces | |
| 88. |
The Castigliano's second theorem can be used to compute deflections |
| A. | In statically determinate structures only |
| B. | For any type of structure |
| C. | At the point under the load only |
| D. | For beams and frames only |
| Answer» B. For any type of structure | |
| 89. |
When a uniformly distributed load, longer than the span of the girder, moves from left to right, then the maximum bending moment at mid section of span occurs when the uniformly distributed load occupies |
| A. | Less than the left half span |
| B. | Whole of left half span |
| C. | More than the left half span |
| D. | Whole span |
| Answer» D. Whole span | |
| 90. |
Which of the following methods of structural analysis is a force method? |
| A. | Slope deflection method |
| B. | Column analogy method |
| C. | Moment distribution method |
| D. | None of the above |
| Answer» B. Column analogy method | |
| 91. |
Which of the following is not the displacement method? |
| A. | Equilibrium method |
| B. | Column analogy method |
| C. | Moment distribution method |
| D. | Kani's method |
| Answer» B. Column analogy method | |
| 92. |
For a two-hinged arch, if one of the supports settles down vertically, then the horizontal thrust |
| A. | Is increased |
| B. | Is decreased |
| C. | Remains unchanged |
| D. | Becomes zero |
| Answer» C. Remains unchanged | |
| 93. |
The deflection at any point of a perfect frame can be obtained by applying a unit load at the joint in |
| A. | Vertical direction |
| B. | Horizontal direction |
| C. | Inclined direction |
| D. | The direction in which the deflection is required |
| Answer» D. The direction in which the deflection is required | |
| 94. |
The principle of virtual work can be applied to elastic system by considering the virtual work of |
| A. | Internal forces only |
| B. | External forces only |
| C. | Internal as well as external forces |
| D. | None of the above |
| Answer» C. Internal as well as external forces | |
| 95. |
If in a rigid-jointed space frame, (6m + r) < 6j, then the frame is |
| A. | Unstable |
| B. | Stable and statically determinate |
| C. | Stable and statically indeterminate |
| D. | None of the above |
| Answer» A. Unstable | |
| 96. |
The three moments equation is applicable only when |
| A. | The beam is prismatic |
| B. | There is no settlement of supports |
| C. | There is no discontinuity such as hinges within the span |
| D. | The spans are equal |
| Answer» C. There is no discontinuity such as hinges within the span | |
| 97. |
Which of the following methods of structural analysis is a displacement method? |
| A. | Moment distribution method |
| B. | Column analogy method |
| C. | Three moment equation |
| D. | None of the above |
| Answer» A. Moment distribution method | |
| 98. |
The fixed support in a real beam becomes in the conjugate beam a |
| A. | Roller support |
| B. | Hinged support |
| C. | Fixed support |
| D. | Free end |
| Answer» D. Free end | |
| 99. |
For a symmetrical two hinged parabolic arch, if one of the supports settles horizontally, then the horizontal thrust |
| A. | Is increased |
| B. | Is decreased |
| C. | Remains unchanged |
| D. | Becomes zero |
| Answer» B. Is decreased | |
| 100. |
In the displacement method of structural analysis, the basic unknowns are |
| A. | Displacements |
| B. | Force |
| C. | Displacements and forces |
| D. | None of the above |
| Answer» A. Displacements | |
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