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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250+ TOP MCQs on Influence Lines for Floor Girders and Maximum Influence at a Point due to a Series of Concentrated Loads and Answers

Structural Analysis test on “Influence Lines for Floor Girders and Maximum Influence at a Point due to a Series of Concentrated Loads”.

1. Floor loads are transmitted from slabs to which part in a floor girder system?
a) Floor beams
b) Side girders
c) Supporting columns
d) Nowhere
Answer: a
Clarification: Loads get transmitted to floor beams after slabs.

2. Which part transmits floor loads to side girders in a floor girder system?
a) Floor beams
b) Slabs
c) Supporting columns
d) Nowhere
Answer: a
Clarification: Floor beams transmit loads to side girder.

3. Supporting columns are last element of load transfer in a floor girder system.
State whether the above system is true or false.
a) True
b) False
Answer: a
Clarification: Support columns receives floor load at last of the transfer chain.

4. What is the main load carrying member in this system?
a) Floor beams
b) Side girders
c) Supporting columns
d) Nowhere
Answer: b
Clarification: Girders are the main load carrying member and its study is important for the analysis of floor load transmission.

5. Load transfer to girders happen at only some points. What are they called?
a) Girder points
b) Panel points
c) Column points
d) Side points
Answer: b
Clarification: They are called panel points and members between these points are called panels.

6. Generally, which parts’ ILD is given the most preference?
a) Floor beams
b) Side girders
c) Supporting columns
d) Nowhere
Answer: b
Clarification: Side girders are the main load carrying member of the system, so their ILD is given the most preference.

7. Assuming that we need to determine maximum shear at a point C of a beam, which of the following sentences will always be true for that condition?
a) At least one load at end point
b) At least two load at both the end points
c) One of the loads should be at a point in proximity to point C
d) None of the loads should be at a point in proximity to point C
Answer: c
Clarification: For maximum shear at point C one of the loads should be at a point in proximity (just left or just right) to point C.

8. For a cantilevered beam, where would absolute maximum shear occur?
a) At midpoint
b) Depends upon position of load applied
c) Just next to free end
d) Just next to fixed support
Answer: d
Clarification: We can use method of section to prove the above result while applying load at any arbitrary point.

9. For a simply supported beam, where would absolute maximum shear occur?
a) At midpoint
b) Depends upon position of load applied
c) Just next to one of ends
d) At both of the ends
Answer: c
Clarification: Maximum shear would occur depending upon position load, but it will occur at one of the ends.

10. For a cantilevered beam, where would absolute maximum moment occur?
a) At midpoint
b) Depends upon position of load applied
c) Just next to free end
d) Just next to fixed support
Answer: d
Clarification: It would occur at same point as max. Shear, but loads should be applied at far other end of beam so as to maximize it.

11. For a simply supported beam, where would absolute maximum moment occur?
a) At midpoint
b) Depends upon position of load applied
c) Just next to one of ends
d) At both of the ends
Answer: b
Clarification: One will have to analyze properly to find that point in this case.

12. If there are three point loads acting simultaneously on a simply supported beam, where would absolute max. Moment occur?
a) Beneath smallest force
b) Beneath med. force
c) Beneath largest force
d) Can‘t say
Answer: d
Clarification: One can’t say as it would require analysis, although generally it comes under the greatest force.

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

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

Δ = displacement caused when force is increased by a small amount
P = external force applied
N = internal force in the member force applied
L = length of member
A = cross-sectional area of member
E = Modulus of elasticity
Same symbol is used for partial and total differentiation and they are pretty obvious.

1. What will be Δ in case of straight members using theorem?
a) 14 ΣN(dN/dP)L/AE
b) 13 ΣN(dN/dP)L/AE
c) 12 ΣN(dN/dP)L/AE
d) ΣN(dN/dP)L/AE
Answer: d
Clarification: On substituting value of internal energy in earlier theorem, we can get this.

2. P is treated here as:-
a) constant
b) variable
c) it doesn’t matter
d) depends upon load
Answer: b
Clarification: P is treated as variable and N is expressed in its term for partial differentiation.

3. Force P is applied in the direction of Δ
State whether the above statement is true or false.
a) true
b) false
Answer: a
Clarification: P is applied in above said direction. That is how we have been calculating the work done till now.

4. N is caused by:-
a) constant forces
b) variable forces
c) both
d) neither
Answer: c
Clarification: It is caused by both the constant external force and variable P.
A beam has been subjected to gradually applied load P1 and P2 causing deflection Δ1 and Δ2.
Gradual increase of dp1 causes subsequent deflection of dΔ1 and dΔ2.

5. What will be the external work performed during application of load?
a) 12 (p1 Δ1 + p2 Δ2)
b) 12 (p2 Δ1 + p1 Δ2)
c) p1 Δ1 + p2 Δ2
d) p2 Δ1 + p1 Δ2
Answer: a
Clarification: Since loads are gradually applied, work done will be average load times deflection. We can also find by integration.

6. What will be the work done during additional application of dp1?
a) p1 dΔ1 + p2 dΔ2 + dp1d Δ1
b) p1 dΔ1 + p2 dΔ2 + 12 dp1d Δ1
c) p1 dΔ1 + 12 p2 dΔ2 + dp1d Δ1
d) 12 p1 dΔ1 + p2 dΔ2 + dp1d Δ1
Answer: b
Clarification: At this time p1 and p2 are already applied, only dp1 is gradually applied.

7. Additional work done due to application of dp1 is p1 dΔ1 + p2 dΔ2.
Sate whether the above statement is true or false.
a) true
b) false
Answer: a
Clarification: It is true as the third term can be ignored as it is very small.

8. What will be the work done if all three forces are place at once on the beam?
a) (p1 + dp1)(Δ1 + dΔ1) + (p2)( Δ2 + dΔ2)
b) (p1 + dp1)(Δ1 + dΔ1) + 12 (p2)( Δ2 + dΔ2)
c) 12 (p1 + dp1)(Δ1 + dΔ1) + (p2)( Δ2 + dΔ2)
d) 12 (p1 + dp1)(Δ1 + dΔ1) + 12 (p2)( Δ2 + dΔ2)
Answer: d
Clarification: Now, since all the loads are gradually applied, all will have a factor of half.

9. What will be change in work done in both case on initial application of load?
a) p1dΔ1 + dp1 Δ1 + p2dΔ2
b) 12 p1dΔ1 + dp1 Δ1 + p2dΔ2
c) 12 p1dΔ1 + 12 dp1 Δ1 + p2dΔ2
d) 12 p1dΔ1 + 12 dp1 Δ1 + 12 p2dΔ2
Answer: d
Clarification: We will get this by just subtracting two works done. This will be termed as dw.

10. Which of the following is equal to Δ1?
a) dw/dp2
b) dw/p1
c) dw/p2
d) dw/dp1
Answer: d
Clarification: Just substitute value of p2d Δ2 in dw using one of the earlier equation.

X is taken along the axis of beam
1 = external virtual unit load acting on the beam with direction same as that of Δ.
m = internal virtual moment in beam.
Δ = external displacement of the point caused by the real loads.
M = internal moment caused by the real loads.
E = modulus of elasticity .
I = moment of inertia of cross-sectional area.

11. Which of the following term is integrated to calculate Δ.
a) mM/EI
b) M/mEI
c) E/mMI
d) I/EMm
Answer: a
Clarification: To calculate Δ we equate work done on both side which will mean m multiplied by angular displacement which is M/EI.

12. If L is the length of beam, then what are the upper and lower limits of the above integration?
a) –L, L
b) –L, 0
c) 0, L
d) ½ L, L
Answer: c
Clarification: Integration is done all over the beam, as it will give the work done.

13. Generally, in doing such integrations in which of the following’s term is m expressed?
a) M
b) E
c) I
d) x
Answer: d
Clarification: Since we have to integrate wrt x, we express m in terms of x.

14. Which of the following term does 1.Δ represents?
a) work done by actual forces
b) virtual strain energy stored in beam
c) real strain energy stored in beam
d) total work done by actual and virtual forces
Answer: b
Clarification: Term shown above basically reprents virual load multiplied by displacement.

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250+ TOP MCQs on Loading Conditions for Strength Design and Answers

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

1. By this method, the load factors are:-
a) Smaller than 0.5
b) 0.5 < 0.75
c) 0.75 < 1
d) Larger than 1
Answer: d
Clarification: Resulting factored load are used to have a design to support the ultimate loads. So, factors are > 1.

2. Purpose of load factors is to account for:-
a) Lapse in designing
b) Lapse in constructing
c) Lapse in funding
d) Lapse in predicting magnitudes of dead or live load
Answer: d
Clarification: It accounts for uncertainties related to estimation of magnitude of dead or live loads.

3. Load factors used for live load are _____ than that used for dead loads.
a) Smaller
b) Larger
c) Depends upon case
d) Depends upon loading
Answer: b
Clarification: Magnitudes of dead loads can be predicted more accurately than magnitudes of live loads.

4. The load factor to be used with strength design depends upon:-
a) Amount of load
b) Is constant
c) Depends upon engineer
d) Depends upon type of structure
Answer: d
Clarification: Load factors are determined statistically and type of structures is considered.

5. Which of the following load combinations is recommended for building structures?
a) D + F
b) 1.4D + F
c) D + 1.4F
d) 1.4(D + F)
Answer: d
Clarification: These are mentioned inn ASCE 7-02 and can be verified from there.

6. When larger dead loads tend to reduce the effects of other loads, what is the recommended factor that is used with the dead load?
a) 0.9
b) 0.8
c) 0.7
d) 0.6
Answer: a
Clarification: 0.9D + 1.6W + 1.6H & 0.9D + 1E + 1.6H are the examples.

7. Load factors do not vary in relation to the seriousness of failure.
Is the above statement true or false?
a) True
b) False
Answer: a
Clarification: Load factors were developed on the assumption that designers would consider the seriousness of all the possible failure in specifying loads. Load factors are minimum values.

8. In the equation U-0.9D + 1.6W + 1.6H, what is the load factor for H if the structural action of H counteracts that due to W or E?
a) 0.1
b) 0.05
c) 0.005
d) 0
Answer: d
Clarification: H should be included in design resistance here as lateral earth pressure opposes action of other forces.

9. In case of wind and seismic loads, how many times do we have to apply applicable equations?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: Wind and seismic loads can have two values depending upon direction of those forces, and it is possible for the sign of them to be different.

250+ TOP MCQs on Kinematic Indeterminacy of Beams & Frames and Answers

Structural Analysis Multiple Choice Questions on “Kinematic Indeterminacy of Trusses”.

1. The degree of freedom for the vertical guided roller is ________
a) 0
b) 1
c) 2
d) 3
Answer: b
Clarification: Vertical guided roller resists horizontal/axial force and bending moment. However, it cannot resist vertical forces as it slides down along the plane. Hence it has 1 degree of freedom.

2. What is kinematic indeterminacy for the given figure considering axial deformation?

a) 0
b) 2
c) 4
d) 6
Answer: c
Clarification: The given beam is supported by roller support at both of its ends. Roller support, if considering axial deformation has two degrees of freedom i.e. Rotation and Horizontal sway. Therefore, the degree of freedom of the beam is 4.

3. What is kinematic indeterminacy for the given figure without considering axial deformation?

a) 0
b) 2
c) 4
d) 6
Answer: b
Clarification: The given beam is supported by roller support at both of its ends. Roller support, if not considering axial deformation has one degree of freedom i.e. Rotation. Therefore, the degree of freedom of the beam is 2.

4. What is kinematic indeterminacy for the given figure considering axial deformation?

a) 0
b) 4
c) 6
d) 10
Answer: a
Clarification: The given beam is supported by fixed support at both of it ends. Fixed support, if not considering axial deformation does not any degree of freedom. Therefore, the degree of freedom of the beam is 0.

5. What is kinematic indeterminacy for the given figure?

a) 0
b) 1
c) 2
d) 3
Answer: b
Clarification: The given beam is supported by fixed supports at both of its ends and intermediary roller support. Fixed does not provide any degree of freedom. Whereas, roller support provides both rotation and horizontal sway. But horizontal sway is already restricted by the ends fixed support. Thus, the only degree of freedom is the rotation about roller support.

6. The degree of freedom of a joint for the rigid jointed joint plane frame is _____
a) 0
b) 2
c) 3
d) 6
Answer: c
Clarification: Number of degree of freedom of a joint for the rigid jointed plane frame is 3 i.e. Horizontal sway, Vertical sway, and Rotation.

7. What is kinematic indeterminacy for the given figure without considering axial deformation?

a) 0
b) 2
c) 4
d) 6
Answer: d
Clarification: The given set of frame consists of fixed end supports with two intermediate rigid joints. Each rigid joint allows horizontal sway, vertical sway, and rotation. Therefore, two rigid joints will have six degrees of freedom.

8. What is kinematic indeterminacy for the given figure without considering axial deformation?

a) 0
b) 2
c) 3
d) 6
Answer: c
Clarification: The given set of frame consists of fixed end supports with two intermediate rigid joints. Each rigid joint allows horizontal sway, vertical sway, and rotation. But since the axial deformation is to be neglected. Therefore, vertical sway of the joints and either horizontal sway of a joint is neglected. Hence, degree of freedom is three.

9. The degree of freedom for a rigid jointed plane frame without axial deformation is given by 3j – m – r.
a) True
b) False
Answer: b
Clarification: The degree of freedom for a rigid jointed plane frame without axial deformation is given by 3j – r, where j is the number of joint and r is the number of reactions.

10. The degree of freedom of the given typical joint is ______

a) 1
b) 2
c) 3
d) 4
Answer: d
Clarification: Horizontal hinged joint provides four degrees of freedom. Four freedom at the horizontal hinged joint are horizontal sways of both connected members, vertical deflection of the joint and rotation of the joint.

250+ TOP MCQs on Influence Line Diagram – Numericals and Answers

Structural Analysis Multiple Choice Questions on “Influence Line Diagram – Numericals”.

1. Concept of ILD is only applicable to the static indeterminate structure.
a) True
b) False
Answer: b
Clarification: Concept of ILD for calculating shear force and bending moment is applicable to both statically determinate and indeterminate structures.

2. ILD for bending moment diagram of the cantilever beam is _______

a)
b)
c)
d)
Answer: d
Clarification: Maximum bending moment for cantilever beam occurs when the load is placed at free end, therefore ordinate of ILD of bending moment diagram will be maximum at a free end. When the load is at support itself, then the bending moment produced is zero, hence ordinate of ILD is zero.

3. ILD for the shear force at the support of the cantilever beam is __________

a)
b)
c)
d)
Answer: b
Clarification: For drawing ILD of shear force at the support we must neglect the support at the given point and then lift the member providing it unit displacement upward and let the rest of the beam follow the suit.

4. ILD for the shear force at section C for the given simply supported beam is _________

a)
b)
c)
d)
Answer: a
Clarification: For drawing ILD of shear force at section, any section other than the support, Assume roller at that section and apply downward push on the left arm of the section and upward push to the right section of the arm. Hence assuming roller at C, Downward and Upward push to the left and right of section respectively.

5. ILD for the bending moment at section C for the given simply supported beam is ________

a)
b)
c)
d)
Answer: b
Clarification: To draw ILD for Bending Moment at section E, we must assume internal hinge at the section and apply anticlockwise rotation to the left arm and clockwise rotation to the right of the arm. Thus, assuming an internal hinge at section C and applying rotation to the arms gives us figure B as ILD to bending moment at section C.

6. ILD for Shear force at section E is __________

a)
b)
c)
d)
Answer: a
Clarification: For drawing ILD of shear force at section, any section other than the support, Assume roller at that section and apply downward push on the left arm of the section and upward push to the right section of the arm. Hence assuming roller at E, Downward and Upward push to the left and right of section respectively.


7. ILD for Shear force at support A is _______

a)
b)
c)
d)
View Answer

Answer: c
Clarification: For drawing ILD of shear force at the support we must neglect the support at the given point and then lift the member providing it unit displacement upward and let the rest of the beam follow the suit.


8. ILD for Shear force at support B is _______

a)
b)
c)
d)
View Answer

Answer: d
Clarification: For drawing ILD of shear force at the support we must neglect the support at the given point and then lift the member providing it unit displacement upward and let the rest of the beam follow the suit.

9. ILD for Shear force at the section just right to the point A is _______

a)
b)
c)
d)
Answer: b
Clarification: For drawing ILD of shear force at section, any section other than the support, Assume roller at that section and apply downward push on the left arm of the section and upward push to the right section of the arm. But since the ILD is to be drawn for just to the right to A, we have to divide the section to the just right of A.

10. ILD for Bending Moment at section E is ________

a)
b)
c)
d)
Answer: b
Clarification: To draw ILD for Bending Moment at section E, we must assume internal hinge at the section and apply anticlockwise rotation to the left arm and clockwise rotation to the right of the arm. Thus, assuming an internal hinge at section E and applying rotation to the arms gives us figure B as ILD to bending moment at section E.


11. ILD for the member DI for the given truss if the unit load rolls along beam AB is ________

a)
b)
c)
d)
View Answer

Answer: a
Clarification: Member DI is a zero force member if the load is supposed to be rolled along bottom beam AB. Therefore, ILD for member DI is zero throughout.


12. ILD for the member CH for the given truss if the unit load rolls along AB is _________

a)
b)
c)
d)
View Answer

Answer: b
Clarification: If unit load is to roll along bottom beam AB, Force produces in CH will be zero if it is at supports or anywhere between joint I to B. However force in member CH will be maximum when load is placed at joint H. Since the truss as a whole structure will sag under the action of load, the member CH will be in tension.


13. ILD for the member CI for the given truss if the unit load rolls along AB is ________

a)
b)
c)
d)
View Answer

Answer: d
Clarification: If a unit load is to roll along bottom beam AB, Force produces in CI will be zero if it is at supports. However, the force in the inclined member CI will change its nature as the load passes through joint H to Joint I. Since the truss as a whole structure will sag under the action of load, the joint C will be under compression and I will be in tension.

14. To draw qualitative ILD of indeterminate structure, which of the following concept is used.
a) Unit Load Method
b) Castigilano’s First energy theorem
c) Mullers Breslou’s Principle
d) Kani’s Method
Answer: c
Clarification: Mullers Breslou’s Principle is very useful in performing a qualitative analysis of ILD for indeterminate structure. Unit Load Method, Castigliano’s First Theorem and Kani’s Method are among various methods to determine displacement and rotation in a structure.

15. ILD for a fixed beam is determined by Mullers Brelou’s Principle.
a) True
b) False
Answer: a
Clarification: Mullers Breslou’s Principle is very useful in performing a qualitative analysis of ILD for indeterminate structure. The fixed beam is an indeterminate structure and thus, ILD for a fixed beam is determined by Mullers Brelou’s Principle.