McqMate
| 1. |
All reinforced concrete buildings are most of times:- |
| A. | statically determinate |
| B. | statically indeterminate |
| C. | mixture |
| D. | unstable |
| Answer» B. statically indeterminate | |
| Explanation: this is because columns and beams are continuous in these cases over joints and supports. | |
| 2. |
in most cases, for a given loading maximum stress and deflection of an indeterminate structure are than that of a determinate one. |
| A. | larger |
| B. | smaller |
| C. | larger for small load |
| D. | smaller for larger load |
| Answer» B. smaller | |
| Explanation: indeterminate structure deflects lesser than a determinate one. | |
| 3. |
Which structure will perform better during earthquake? |
| A. | statically determinate |
| B. | statically indeterminate |
| C. | both |
| D. | depends upon magnitude of earthquake |
| Answer» B. statically indeterminate | |
| Explanation: indeterminate structure has a tendency to redistribute its load to its redundant supports in case of overloading. | |
| 4. |
There are two beams of equal length L and a load P is acting on centre of both beams. One of them is simply supported at both ends while the other one is fixed at both ends. Deflection of centre of simply supported beam will be times that of defection of centre of fixed beam. |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 4 |
| Answer» D. 4 | |
| Explanation: maximum moment developed in simply supported beam will be twice that of fixed supported and hence, we can find deflections. | |
| 5. |
Which type of structure would cost less in terms of materials? |
| A. | statically determinate |
| B. | statically indeterminate |
| C. | both will cost equally |
| D. | depends upon loading |
| Answer» B. statically indeterminate | |
| Explanation: statically indeterminate would cost less as they can support a loading with thinner members and increased stability. | |
| 6. |
Which type of structure would cost less in terms of supports? |
| A. | statically determinate |
| B. | statically indeterminate |
| C. | both will cost equally |
| D. | depends upon loading |
| Answer» A. statically determinate | |
| Explanation: supports and joints of indeterminate structures are costly compared to that of a determinate one. | |
| 7. |
Differential settlement is problematic to which type of structure? |
| A. | statically determinate |
| B. | statically indeterminate |
| C. | both |
| D. | neither |
| Answer» B. statically indeterminate | |
| Explanation: it causes development of internal stress in statically indeterminate structures. | |
| 8. |
Fabrication errors don’t cause additional stresses in statically indeterminate structures. State whether the above statement is true or false. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: fabrication errors do cause generation of additional stresses in statically indeterminate structures. | |
| 9. |
If in the above problem in Q5, if load P is excessively increased in simply supported beam, then where would a hinge like point form? |
| A. | at one of the ends |
| B. | at both ends |
| C. | at centre |
| D. | at centre as well as both ends |
| Answer» B. at both ends | |
| Explanation: on application of excessive load, a hinge/pin like point forms at the centre of beam. | |
| 10. |
What will be the value of mCB, after solving these equations? |
| A. | 3.09 |
| B. | 1.54 |
| C. | 12.86 |
| D. | -3.09 |
| Answer» D. -3.09 | |
| Explanation: just substitute the value of rotation at point b in 4th equation. | |
| 11. |
What will be the value of mBA, after solving these equations? |
| A. | 3.09 |
| B. | 1.54 |
| C. | 12.86 |
| D. | -3.09 |
| Answer» C. 12.86 | |
| Explanation: it will be inverse of mbc as shown above. | |
| 12. |
What will be the support reaction at point B? |
| A. | 4.95 |
| B. | -5.95 |
| C. | 3.65 |
| D. | can’t say |
| Answer» A. 4.95 | |
| 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 direction. | |
| 13. |
How many separate parts will be required for this question? |
| A. | 0 |
| B. | 1 |
| C. | 2 |
| D. | 3 |
| Answer» C. 2 | |
| Explanation: since, 3 supports are there we will divide it into 2 separate parts to solve. | |
| 14. |
What will be the deflection of beam AB? |
| A. | 0 |
| B. | 1/ei |
| C. | 2/ei |
| D. | can’t say |
| Answer» A. 0 | |
| Explanation: as, point a is a fixed support, it won’t allow any deflection to take place. | |
| 15. |
What will be the deflection of beam BC? |
| A. | 0 |
| B. | 1/ei |
| C. | 2/ei |
| D. | can’t say |
| Answer» A. 0 | |
| Explanation: as, point b is a fixed support, it won’t allow any deflection to take place. | |
| 16. |
What will be the rotation of beam BC at point C? |
| A. | 0.2 |
| B. | -0.2 |
| C. | 0 |
| D. | it will depend upon the rotation of beam bc at point b |
| Answer» D. it will depend upon the rotation of beam bc at point b | |
| Explanation: according to equations, it will depend upon the rotation of beam bc at point c. | |
| 17. |
After using all the joint conditions, how many unknowns are still left? |
| A. | 0 |
| B. | 1 |
| C. | 2 |
| D. | 3 |
| Answer» B. 1 | |
| Explanation: rotation at point b in either of beam is not known (they are both equal). | |
| 18. |
While using slope deflection method, in which direction is moment taken as positive? |
| A. | clockwise |
| B. | anti-clockwise |
| C. | depends upon case |
| D. | depends upon loading |
| Answer» A. clockwise | |
| Explanation: clockwise moments are always taken as positive while theses equations. | |
| 19. |
While drawing BMD after using these equations, in which direction is moment taken as positive? |
| A. | clockwise |
| B. | anti-clockwise |
| C. | depends upon case |
| D. | depends upon loading |
| Answer» C. depends upon case | |
| Explanation: while drawing bmd after using these equations we use our usual notations for sign of moments. | |
| 20. |
If support B settles by 1mm downward, what is direction of rotation at point A? |
| A. | +ve |
| B. | -ve |
| C. | can’t say |
| D. | depends upon loading at point a |
| Answer» A. +ve | |
| Explanation: in this case, ab rotates clockwise about point a, so rotation is +ve. | |
| 21. |
If support A settles by 1mm downward, what is direction of rotation at point A? |
| A. | +ve |
| B. | -ve |
| C. | can’t say |
| D. | depends upon loading at point a |
| Answer» B. -ve | |
| Explanation: in this case, ab rotates anti clockwise about point a, so rotation is +ve. | |
| 22. |
If support A settles by 1mm downward, what is direction of rotation at point B? |
| A. | +ve |
| B. | -ve |
| C. | can’t say |
| D. | depends upon loading at point a |
| Answer» B. -ve | |
| Explanation: in this case, ab rotates anti clockwise about point b, so rotation is +ve. | |
| 23. |
If a beam has 4 external supports, then how many parts would it be divided in case of using slope deflection equations? |
| A. | 0 |
| B. | 1 |
| C. | 2 |
| D. | 3 |
| Answer» D. 3 | |
| Explanation: we divide each part between two supports, so there will be 3 parts as 4 supports are there. | |
| 24. |
How many sde (slope deflection equations) are possible if 4 supports are there? |
| A. | 0 |
| B. | 3 |
| C. | 4 |
| D. | 6 |
| Answer» D. 6 | |
| Explanation: each part will give 2 unique equations, so there will be a total of 6 equations. | |
| 25. |
Initially, how many total unknowns will be there in 6 equations? |
| A. | 3 |
| B. | 6 |
| C. | 9 |
| D. | 12 |
| Answer» C. 9 | |
| Explanation: total 9 unknowns will be there, 2 rotations and one deflection in each part. | |
| 26. |
After writing sde, what is the second step? |
| A. | use stress-strain relationship |
| B. | use equilibrium equations |
| C. | use compatibility equations |
| D. | can’t say |
| Answer» B. use equilibrium equations | |
| Explanation: after writing sde, we use equilibrium equations so as to lessen the no. of unknowns. | |
| 27. |
The maximum negative bending moment in fixed beam carrying udl occurs at |
| A. | mid span |
| B. | 1/3 of the span |
| C. | supports |
| D. | half of the span |
| Answer» C. supports | |
| Explanation: in case of fixed beam subjected to gravity loads maximum hogging or negative bending moment develops at the supports. at centre, the maximum bending moment is reduced. | |
| 28. |
A fixed beam of the uniform section is carrying a point load at the centre, if the moment of inertia of the middle half portion is reduced to half its previous value, then the fixed end moments will |
| A. | increase |
| B. | remains constant |
| C. | decrease |
| D. | change their direction |
| Answer» A. increase | |
| Explanation: the flexural rigidity value is reduced in middle half portion of the second case fixed end moments which have developed in a beam section will be increases. | |
| 29. |
In propped cantilevers, the prop reaction is 3/8 wl. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: in propped cantilever beam net deflection at fixed end is zero therefore rl3/3ei = wl4/8ei | |
| 30. |
A propped cantilever beam carrying total load “W” distributed evenly over its entire length calculate the vertical force required in the prop. |
| A. | 3/4 w |
| B. | w |
| C. | 5/8 w |
| D. | 3/8 w |
| Answer» D. 3/8 w | |
| Explanation: therefore total load on beam = w = wl | |
| 31. |
A mouthpiece is a short length of a pipe which is not more than times its diameter. |
| A. | 3-4 |
| B. | 5-6 |
| C. | 1 -2 |
| D. | 2-3 |
| Answer» D. 2-3 | |
| Explanation: a mouth piece is defined as a short length of a pipe which is not more than two or three times its diameter, fitted to an orifice of same diameter provided especially in a tank containing liquid. | |
| 32. |
The section which has a minimum cross sectional are in a flow is known as |
| A. | vena contracta |
| B. | thyrocade |
| C. | submergent |
| D. | upstream edge |
| Answer» A. vena contracta | |
| Explanation: the section of the jet, at which the flow in a liquid has a minimum cross | |
| 33. |
Which of the following is not a hydraulic coefficient? |
| A. | coefficient of contraction |
| B. | coefficient of discharge |
| C. | coefficient of viscosity |
| D. | coefficient of velocity |
| Answer» C. coefficient of viscosity | |
| Explanation: coefficient of viscosity can be defined as the shear stress required producing unit rate of angular deformation. it is also called as dynamic viscosity. | |
| 34. |
Bell mouthed orifices can be categorised in according to |
| A. | size |
| B. | shape |
| C. | shape of upstream |
| D. | nature of discharge |
| Answer» C. shape of upstream | |
| Explanation: the orifices are classified on the basis of their size, shape, shape of upstream edge and discharge conditions. | |
| 35. |
The Cv taken for sharp edged orifice generally is |
| A. | 0.97 |
| B. | 0.98 |
| C. | 0.95 |
| D. | 0.99 |
| Answer» B. 0.98 | |
| Explanation: the cv taken for sharp edged orifice generally is 0.98. | |
| 36. |
The relation between hydraulic coefficients is Cd = Cc × Cv. |
| A. | false |
| B. | true |
| Answer» B. true | |
| Explanation: cd = qa / qth | |
| 37. |
Deflection Equations method was developed by:- |
| A. | mohr |
| B. | bernoulli |
| C. | maxwell |
| D. | mohr and manderla |
| Answer» D. mohr and manderla | |
| Explanation: this method was developed by mohr and manderla. | |
| 38. |
What is the Degree of freedom of this beam? |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 4 |
| Answer» C. 3 | |
| Explanation: rotation at both ends and relative displacement in y direction due to settlement of one of the ends. | |
| 39. |
If point A goes down and point B goes up, then this will be:- |
| A. | positive rotation |
| B. | negative rotation |
| C. | can’t say |
| D. | depends upon magnitude |
| Answer» B. negative rotation | |
| Explanation: it will cause span’s cord angle to rotate anti-clockwise. | |
| 40. |
What will be MBA in this case? |
| A. | ei θa/l |
| B. | 2ei θa/l |
| C. | 3ei θa/l |
| D. | 4ei θa/l |
| Answer» B. 2ei θa/l | |
| Explanation: use conjugate beam method and since displacement at both ends is zero, equate both end moments of conjugate beam to zero. | |
| 41. |
What will be MAB in this case? |
| A. | ei θa/l |
| B. | 2ei θa/l |
| C. | 3ei θa/l |
| D. | 4ei θa/l |
| Answer» D. 4ei θa/l | |
| Explanation: use conjugate beam method and since displacement at both ends is zero, equate both end moments of conjugate beam to zero. | |
| 42. |
What will be MBA in this case? |
| A. | ei θb/l |
| B. | 2ei θb/l |
| C. | 3ei θb/l |
| D. | 4ei θb/l |
| Answer» D. 4ei θb/l | |
| Explanation: use conjugate beam method and since displacement at both ends is zero, equate both end moments of conjugate beam to zero. | |
| 43. |
What will be the difference in moment acting at both the ends? |
| A. | ei Δ/l2 |
| B. | 2ei Δ/l2 |
| C. | 3ei Δ/l2 |
| D. | 0 |
| Answer» D. 0 | |
| Explanation: both ends are not rotating. | |
| 44. |
What will be the difference in shear force acting at both the ends? |
| A. | ei Δ/l2 |
| B. | 2ei Δ/l2 |
| C. | 3ei Δ/l2 |
| D. | 0 |
| Answer» D. 0 | |
| Explanation: both ends are not rotating. | |
| 45. |
Moment Distribution Method is applicable to the determinate and indeterminate structure. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: moment distribution method developed by hardy cross to analyze the indeterminate structures like beams and rigid jointed frame with internal hinges also. | |
| 46. |
Carryover Moment is defined as |
| A. | the moment applied at one end to cause unit slope at the support |
| B. | the additional moment applied at one end to completely resist the rotation caused due to external loading |
| C. | the moment developed or induced at one end due to a moment at another end |
| D. | the moment applied at one end to cause unit slope at another end |
| Answer» C. the moment developed or induced at one end due to a moment at another end | |
| Explanation: carryover moment is defined as the moment developed or induced at one end due to a moment at another end. it is useful in calculating stiffness factor and moment distribution for a particular joint. | |
| 47. |
Carryover Moment at end B due to moment M applied at end A for the given non-prismatic beam is |
| A. | 0 |
| B. | m |
| C. | greater than m/2 |
| D. | lesser than m/2 |
| Answer» C. greater than m/2 | |
| Explanation: moment applied at end a would be combinedly resisted by both of the support i.e. a and b. however, for the given non-prismatic member the cross section at the support is more and hence offers more resistance to the external bending moment. | |
| 48. |
3 SUPPORT SETTLEMENT - SYMMETRIC FRAMES WITH SYMMETRIC AND SKEW-SYMMETRIC LOADINGS. |
| A. | 0 |
| B. | 10 |
| C. | 20 |
| D. | 30 |
| Answer» A. 0 | |
| Explanation: since, there is no external load acting on beam ab, there won’t be any fem at point a. | |
| 49. |
What will be the FEM at point B in beam AB? |
| A. | 0 |
| B. | 10 |
| C. | 20 |
| D. | 30 |
| Answer» A. 0 | |
| Explanation: since, there is no external load acting on beam ab, there won’t be any fem at point b in beam ab. | |
| 50. |
What will be the rotation of beam AB at point A? |
| A. | 0.2 |
| B. | -0.2 |
| C. | 0 |
| D. | can’t say |
| Answer» C. 0 | |
| Explanation: there won’t be any rotation at point a as it is a fixed support. | |
| 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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