Category: Structural Analysis Objective Questions

250+ TOP MCQs on Live Load Reduction and Answers

Structural Analysis Multiple Choice Questions on “Live Load Reduction”.

1. If influence area contributing to a particular load increases, then what happens to the possibility of having full design load on each square meter of area?
a) Increases
b) Decreases
c) Doesn’t affect
d) Depends on case
Answer: b
Clarification: As area increases, likelihood of having max. Load decreases due to known forces.

2. Building codes usually permit reduction in live load when small areas are concerned.
State whether this statement is true or false
a) True
b) False
Answer: b
Clarification: Possibility of having maximum live load decreases as area increases. So, reduction in live load is permitted in cases of large area, not small area.

3. Reduction factors given in Section 4.8 of ASCE 7-02 & in Section 1607.9.1 of IBC-2003 are:-
a) Different
b) Same
c) Vary in some cases
d) There is no reduction factor in Section 4.8 of ASCE 7-02
Answer: b
Clarification: Both have same reduction factor which is L = Lo(0.25 + (15/((KLL*AT)0.5))).

4. What is the minimum limit of influence area (in sq. ft) till which live loads can be reduced?
a) 100
b) 200
c) 300
d) 400
Answer: d
Clarification: From the equation of reduction factor, it can be shown that the lower limit is 400sq.ft.

5. In the case of a structural member supporting more than one floor with load exceeding 100psf., what is the maximum permitted reduction %?
a) 10
b) 15
c) 20
d) 25
Answer: c
Clarification: In buildings supporting more than one floor, studies have shown that rarely it is loaded (a floor) with more than 80% of its design load.

6. Loads used to obtain column design forces and to obtain beam design forces are:-
a) Necessarily same
b) Necessarily different
c) Most likely same
d) Most likely different
Answer: d
Clarification: This happens because the reduction factor are most likely to be different.

7. When column supports the top floor of a building, then live load reduction is:-
a) Permitted
b) Not permitted
c) Depends upon type of beam
d) Depends upon amount of load
Answer: b
Clarification: When column support a floor, it should be thought of as supporting a single floor only.

8. Live load element factor KLL remains constant for all types of structure.
State whether this statement is true or false.
a) True
b) False
Answer: b
Clarification: Live load element factor KLL depends upon type of structure in question.

250+ TOP MCQs on Sections Method Numericals and Answers

Structural Analysis Multiple Choice Questions on “Sections Method Numericals”.

1. Method of the section for determinate truss analysis is time effective.
a) True
b) False
Answer: a
Clarification: Method of the section does not require the calculation of forces in the member prior to it. To compute force in any member, we cut the section and establish equilibrium conditions and hence it is time effective method.

2. To find the forces in the vertical member, preferable equilibrium equations are __________
a) ∑H = 0 & ∑M =0
b) ∑H = 0 & ∑V = 0
c) ∑H = 0, ∑V = 0 & ∑M = 0
d) ∑V = 0 & ∑M = 0
Answer: d
Clarification: Vertical unknowns at a particular selection should be preferably solved for ∑V = 0 first and later for ∑M = 0. Vertical Member does not have possesses any horizontal component of the forces hence ∑H = 0 is irrelevant to the context.

3. What is the force in member CG?

a) 0
b) W
c) W/2
d) W/√2
Answer: d
Clarification: ∑V = 0 & ∑M = 0
W/2 = FCG SIN 45°
FCG = W/√2.

4. What is the force in member CD?

a) 0
b) W (Compressive)
c) W (Tensile)
d) W/2 (Compressive)
Answer: b
Clarification: ∑MG = 0
FCD X L + (W/2) X 2L = 0

FCD = W (Compressive).

5. What is the force in member FG?

a) 0
b) W/2 (Tension)
c) W/2 (Compression)
d) W (Compressive)
Answer: b
Clarification: ∑MC = 0
FFG X L + (W/2) X L = 0
FFG = W/2.

6. What is the force in member EH?

a) 0
b) W ( Compressive)
c) W ( Tension )
d) W/2 (Compressive)
Answer: a
Clarification: ∑MB = 0
FHE X L = 0
FHE = 0.

7. Find the Force in member DG.

a) W (Tensile)
b) W (Compressive)
c) 0
d) W/2 (Compressive)
Answer: b
Clarification: ∑V = 0
W + FDG = 0
FDG = W (Compressive).

8. Method of the section can always be used to calculate the force in any members.
a) True
b) False
Answer: b
Clarification: Method of the section has its own limitation. It cannot be used to compute the force of the member attached to a joint where already forces in more than one member is unknown.

9. Methods of Section can be used to evaluate which of the following truss?
a)
b)
c)
d)
Answer: a
Clarification: Truss A is externally and internally determinate and thus it can be evaluated by Method of Section. But Truss B and C are internally indeterminate, Truss D is externally indeterminate and Hence truss B, C, and D cannot be evaluated by Method of Section.

250+ TOP MCQs on Qualitative Influence Lines and Answers

Structural Analysis Multiple Choice Questions on “Qualitative Influence Lines”.

While writing influence line equations, left most point is always considered as origin and following sign convention is followed.

1. If we require to construct ILD of vertical support at a pin joint, then according to Muller-Breslau principle, by which type of support should it be replaced?
a) Roller guide
b) Pin roller
c) Fixed support
d) Hinge
Answer: a
Clarification: We need to remove the force for which we need to construct ILD and roller guide would remove the vertical reaction.

Following figure is for Q2-Q5.
A is a pin support, B is a hinge, C and D are roller type support.
AB = BC = 1m, CD = 2m

2. For ILD of shear at a point just left to C, what will be the equation for it on BC part of beam?
a) X
b) -X
c) 2X
d) 1
Answer: d
Clarification: On applying Muller-Breslau principle, we will see that the part right to the point can’t move as point C is right next to it. So, ILD will have to remain parallel to x axis.

3. If we draw ILD of shear at appoint just right to point C, then what will be its slope at BC part of beam?
a) 0.5
b) -.5
c) 1
d) -1
Answer: b
Clarification: Slope of ILD in part CD of beam will be -0.5 as initially it will be at 1 and finally at 0. Now, due to Muller-Breslau principle, both lines will be parallel.

4. If we draw ILD for shear at a point E (lying in between points C & D), then at how many points would his curve attain its peak?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: At E, value will be +0.5 and since slope of two lines will be parallel, value of ILD at B too will be 0.5.

5. What will be the lowest point of ILD curve for moment at a point just left to point C?
a) -1
b) -2
c) -3
d) -4
Answer: b
Clarification: For moment, we rotate it by 1 rad. So, min. point will be -2 at hinge B as after that the curve will change its slope.

Point A is fixed support, B and D are hinges and C and E are pin roller supports.
AB = BC = CD = DE=1m

6. What will be the equation for ILD of vertical reaction at point A for AB part of beam?
a) 1X
b) 2X
c) 1
d) 2
Answer: c
Clarification: Since, point A is a fixed support, ILD will be parallel to x axis.

7. What will be the maximum point of ILD of vertical reaction at point C?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: It will be 1 at point C and will be 2 at point D after which it will change its slope as D is a hinge.

8. What will be the area under the ILD curve if we make it for the vertical reaction at a point just left to point C?
a) 1
b) 1.5
c) 2
d) 2.5
Answer: b
Clarification: It will be a triangle with base 2m and height 1 for CE and triangle with base 1 and height 1 for BC. It will be 0 for AB as A is a fixed support.

9. What will be the area under the ILD curve if we make it for the vertical reaction at a point just right to point C?
a) 1
b) 1.5
c) 2
d) 4
Answer: b
Clarification: It will be a triangle with 1*1 dimension for CD and a triangle with base 1 and height 1 for DE. It will be 0 for rest of the beam.

250+ TOP MCQs on Method of Virtual Work: Trusses and Castigliano’s Theorem and Answers

Structural Analysis Multiple Choice Questions on “Method of Virtual Work: Trusses and Castigliano’s Theorem”.

N = normal force
n = Internal virtual normal force
Δ = Displacement of joints by real loads
L = length of a member
A = cross-sectional area of member
E = modulus of elasticity of a member

1. Internal deformation caused by real loads will be in a linear elastic member:-
a) 1⁄4 NL/AE
b) 1⁄3 NL/AE
c) 1⁄2 NL/AE
d) NL/AE
Answer: d
Clarification: Since it is a linear elastic material, we can use various relationships.

2. What will be the value of Δ in a member:-
a) Σ 1⁄4 nNL/AE
b) Σ 1⁄3 nNL/AE
c) Σ 1⁄2 nNL/AE
d) Σ nNL/AE
Answer: d
Clarification: Just substituting the earlier equation in the main equation, we can get it.

3. What is change in length of member if temperature increases by ΔT and expansion coefficient is ά?
a) 1⁄4 ά ΔTL
b) 1⁄3 ά ΔTL
c) 1⁄2 ά ΔTL
d) ά ΔTL
Answer: d
Clarification: Change in length is directly proportional to change in temperature and expansion coefficient with 1 as proportionality coefficient.

4. What will be the value of Δ in a member:-
a) Σ 1⁄4 nά ΔTL
b) Σ 1⁄3 nά ΔTL
c) Σ 1⁄2 nά ΔTL
d) Σ nά ΔTL
Answer: d
Clarification: Just substituting the earlier equation in the main equation, we can get it.

5. What is the unit of virtual unit load?
a) N
b) Lb
c) kip
d) Anything
Answer: d
Clarification: Its unit can be anything as it will cancel with that of n.

Δ = displacement caused when force is increased by a small amount.

6. This theorem is applicable when temperature is varying. State whether the above sentence is true or false.
a) True
b) False
Answer: b
Clarification: It is applicable only when temperature is not changing.

7. In which of the following cases, is this theorem applicable?
a) Yielding support, non-linear elastic material
b) Non-yielding support, linear elastic material
c) Yielding support, linear elastic material
d) Non-yielding support, non-linear elastic material
Answer: b
Clarification: It is applicable in cases of non-yielding support and non-linear elastic material.

8. If any of the external forces acting increases, then internal energy would:-
a) Decrease
b) Increase
c) Not change
d) Become -ve
Answer: b
Clarification: Due to increase in force, external work done would increase which would cause an increase in strain energy.

9. What will be Δ if change in force is DP and du is change in internal energy?
a) 1⁄4 du/dp
b) 1⁄3 du/dp
c) 1⁄2 du/dp
d) du/dp
Answer: d
Clarification: On equating internal energy after changing order of application of forces.

10. This theorem is applicable when non-conservative forces are applied.
State whether the above statement is true or file.
a) True
b) False
Answer: b
Clarification: It is only applicable when forces are of conservative type.

250+ TOP MCQs on Loading Conditions for Allowable Stress Design and Answers

Structural Analysis Multiple Choice Questions on “Loading Conditions for Allowable Stress Design”.

1. With the allowable stress method, which conditions are computed?
a) Most severe loading conditions and inelastic stresses
b) Most severe loading conditions and elastic stresses
c) Feeble loading conditions and inelastic stresses
d) Feeble loading conditions and elastic stresses
Answer: b
Clarification: This method computes most severe loading conditions and elastic stresses but appreciably below the ultimate stresses.

2. Allowable stress method considers which type of loadings?
a) Simultaneous
b) Non- simultaneous
c) Mixture of both
d) Only dead loads
Answer: a
Clarification: It considers simultaneous loading possibilities to determine most severe loading conditions.

3. Live load and roof live loads are represented by the same symbol.
State whether the above written statement is true or false.
a) True
b) False
Answer: a
Clarification: L represents live load, while Lr is used to represent roof live load.

4. Which of the following simultaneous loading conditions are not necessary to be considered for the most severe situations:-
a) D + F
b) D + H + F + L + T
c) D + H + F
d) 0.6D + W + H
Answer: c
Clarification: Necessary conditions are mentioned in Section 2 of ASCE 7-02. These can be matched from there.

5. With which loads are impact effects considered?
a) Rain
b) Snow
c) Live
d) Fluids
Answer: c
Clarification: Impact effects occur when live loads are quickly applied. So, they are considered with them only.

6. Which of the following load do not vary appreciably with time?
a) Snow
b) Dead
c) Rain
d) Wind
Answer: b
Clarification: There is never a permanent presence of snow on a structure, nor is that of rain and wind.

7. If a full dead load is not acting during an earthquake or a severe wind storm, then chances of overturning of building will:-
a) Decrease
b) Increase
c) Remain same
d) Depend upon case
Answer: b
Clarification: Earthquake and wind load acts in lateral direction, enhancing overturning. While, dead load acts in the vertical downward direction resisting overturning.

8. When two or more loads are acting on a structure in addition to dead load, then ASCE permits the loads other than dead loads to be multiplied by a factor of (provided the result is not less than that produced by dead load and the load causing greatest effect)
a) 0.60
b) 0.65
c) 0.70
d) 0.75
Answer: d
Clarification: In these conditions, most likely loads other than dead one do not achieve their maximum values simultaneously. This assumption is validated by load surveys.

9. In the above question, using factor of 0.75 is:-
a) Compulsory
b) Compulsory in some case
c) Depends upon the engineer
d) Depends upon loads
Answer: c
Clarification: Factor of 0.75 is only helping in listing the minimum conditions to be considered.

250+ TOP MCQs on Space Trusses and Answers

Structural Analysis Interview Questions and Answers for freshers on “Space Trusses”.

1. What is the simplest element of a space truss
a) triangle
b) tetrahedron
c) octahedron
d) pyramid
Answer: b
Clarification: Simplest element of a space truss is built on a basic triangle.

2. How many additional members are required to make a simple space truss from a basic tetrahedral element?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: Three additional members forming 1 extra joint are needed to form multi-connected tetrahedrons aka simple space truss.

3. How many equations are solved per joints while solving space trusses?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: 3 equations are solved per joint. Forces are conserved in all the three directions.
B= no. of bars of the truss
R= total no. of external support reaction
J= total no. of joints.

4. If B=6, R=6 and J= 4, then the truss is:-
a) statically determinate
b) statically indeterminate
c) stable
d) unstable
Answer: a
Clarification: B + R = 12 = 3*J. So, truss is statically determinate. Without further insight, we can’t predict stability.

5. If, in the above question R= 7 then the truss are:-
a) statically determinate
b) statically indeterminate
c) stable
d) unstable
Answer: b
Clarification: B + R = 13 > 3*J. So, truss is statically indeterminate. Without further insight, we can’t predict stability.

6. If, in the above question R= 5 then the truss is:-
a) statically determinate
b) statically indeterminate and stable
c) stable
d) unstable
Answer: d
Clarification: B + R = 11 < 3*J. So, the truss will be unstable.

7. The external stability of the space truss requires that the support reactions keep the truss in force equilibrium about all the axes.
State whether the above statement is true or false.
a) true
b) false
Answer: b
Clarification: They require moment equilibrium about all the axes as well.

8. What is the degree of freedom of space roller joint?
a) 0
b) 1
c) 2
d) 3
Answer: b
Clarification: It can move along a slot in the base plane.

9. According to assumptions, which type of joints are used in space truss?
a) pin joint
b) ball and socket joint
c) fixed joint
d) roller joint
Answer: b
Clarification: To treat every member of a space truss as axial- force members, every joint is assumed to be of ball & socket type.

10. How many reaction forces are involved in a short link type of joint?
a) 1
b) 2
c) 3
d) 4
Answer: a
Clarification: One force is unknown which acts in line with the link.

11. How many reaction forces are involved in a roller type of joint?
a) 1
b) 2
c) 3
d) 4
Answer: a
Clarification: One unknown force acts perpendicular to the plane of roller.

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