# 130+ Structural Analysis 1 Solved MCQs

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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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Question and answers in Structural Analysis 1, Structural Analysis 1 multiple choice questions and answers, Structural Analysis 1 Important MCQs, Solved MCQs for Structural Analysis 1, Structural Analysis 1 MCQs with answers PDF download