Category: Structural Analysis Objective Questions

250+ TOP MCQs on Classification of Coplanar Trusses and Answers

Structural Analysis Multiple Choice Questions on “Classification of Coplanar Trusses”.

1. Simple trusses consist entirely of triangle. True or false?
a) True
b) False
Answer: a
Clarification: It can consists of any other shaped intermediate parts, as long as it is stable.

2. Trusses and frames are different as:-
a) Trusses can bend, frames can’t
b) Both can bend
c) None of them can bend
d) Trusses can’t but frame can
Answer: d
Clarification: Trusses can’t bend but, frames can. Due to this, frames can have bending moments.

3. What is the major difference between truss and beam?
a) Beam can’t transmit load in vertical direction while truss can
b) Truss can’t transmit load in vertical direction while beam can
c) Beam can’t transmit load in axial direction while truss can
d) Truss can’t transmit load in axial direction while beam can
Answer: b
Clarification: Due to its geometry, all bending loads get converted to compressive or tensile load in trusses.

4. A truss formed by joining two or more simple trusses is called:-
a) Simple
b) Compound
c) Complex
d) None of the mentioned
Answer: b
Clarification: Truss formed by joining two or more simple trusses is called compound trusses.

5. If a truss has two more members surpassing each other, then it is:-
a) Simple
b) Compound
c) Complex
d) None of the mentioned
Answer: c
Clarification: Trusses falling into neither simple nor compound are categorized into complex trusses.
B= no. of bars of the truss
R= total no. of external support reaction
J= total no. of joints.

6. If B = 5, R = 3 and J = 4, then the truss is:-
a) Statically determinate
b) Statically indeterminate
c) Stable
d) Unstable
Answer: a
Clarification: B + R = 8 = 2*J. So, truss is statically determinate. Without further insight, we can’t predict stability.

7. If, in the above question R= 4 then the truss are:-
a) Statically determinate
b) Statically indeterminate
c) Stable
d) Unstable
Answer: b
Clarification: B + R = 9 > 2*J. So, truss is statically indeterminate. Without further insight, we can’t predict stability.

8. If, in the above question R = 3 then the truss is:-
a) Statically determinate
b) Statically indeterminate and stable
c) Stable
d) Unstable
Answer: d
Clarification: B + R = 7 < 2*J. So, the truss will be unstable.

9. A simple truss will be always internally stable.
State whether the above sentence is true or false.
a) True
b) False
Answer: a
Clarification: A simple truss starts with a triangle which is internally stable. Then we add rigid elements to it, so there is no scope of relative movement/deformation.

10. If a truss is internally unstable, then we should use it when it is:-
a) Statically determinate
b) Statically indeterminate
c) Statically determinate or statically indeterminate
d) We must never use it
Answer: d
Clarification: If a truss is internally unstable then its use must always be avoided.

250+ TOP MCQs on Cable Subjected to a Uniform Distributed Load and Answers

Structural Analysis Multiple Choice Questions on “Cable Subjected to a Uniform Distributed Load”.

1. Cable is a tension member.
a) True
b) False
Answer: a
Clarification: Cable if a flexible member and hence it cannot resist bending moment. Cable cannot bear any compressive loading but it transfers the tensile loading through it. Thus, cable is a tensile member.

2. The shape of the cable is a funicular polygon.
a) True
b) False
Answer: a
Clarification: Cable when loaded takes the shape of bending moment diagram that would be formed if the same set of the load is applied on to the simply supported beam of same length and properties. Thus, the shape of the cable is a funicular polygon.

3. The shape of the cable, when loaded with uniformly distributed load throughout the span is _____
a) Linear always
b) Parabolic always
c) Trapezoidal always
d) Linear or parabolic depending upon the intensity of loading
Answer: b
Clarification: Cable when loaded takes the shape of bending moment diagram that would be formed if the same set of the load is applied on to the simply supported beam of same length and properties. Thus, the shape of cable loaded with uniformly distributed load throughout is parabolic.

4. The horizontal thrust produced at supports of cable when loaded with uniformly throughout the span is ____
a) (frac{WL^2}{32H} )
b) (frac{WL^2}{16H} )
c) (frac{WL^2}{8H} )
d) (frac{WL^2}{2H} )
Answer: c
Clarification:

∑H = 0
HA = HB = 0
∑M = 0
H = (frac{WL^2}{8H}. )

5. Minimum tension in the cable when loaded uniformly throughout the span is ____
a) (frac{WL^2}{32H} )
b) (frac{WL^2}{16H} )
c) (frac{WL^2}{8H} )
d) (frac{WL^2}{2H} )
Answer: c
Clarification:

∑H = 0
HA = HB = 0
∑M = 0
H = (frac{WL^2}{8H} )
Minimum tension in the cable equals to the horizontal thrust produced in the cable. Therefore, minimum tension in the cable is (frac{WL^2}{8H}. )

6. Bending Moment at point D for the shown figure is ____

a) 0
b) (frac{WL^2}{8} )
c) (frac{WL^2}{2} )
d) (frac{WH^2}{2} )
Answer: a
Clarification: Bending Moment at any point in the cable loaded with uniformly distributed load is zero. Therefore, Bending Moment at point D is zero.

7. Maximum tension in the cable when loaded uniformly throughout the span is ____
a) (frac{WL}{2}(sqrt{1+frac{L^2}{2H^2}}) )
b) (frac{WL}{2}(sqrt{1+frac{L^2}{4H^2}}) )
c) (frac{WL}{2}(sqrt{1+frac{L^2}{8H^2}}) )
d) (frac{WL}{2}(sqrt{1+frac{L^2}{16H^2}}) )
Answer: d
Clarification: Maximum tension in the cable is the square root of the sum of the squares of maximum vertical and maximum horizontal loads.
Maximum Vertical Load – V = (frac{WL}{2} )
Maximum Horizontal Load – H = (frac{WL^2}{8H} )
Therefore, Maximum Tension is (frac{WL}{2}(sqrt{1+frac{L^2}{16H^2}}) )

8. Equilibrium equations used to analyse cable is _____
a) ∑H = 0 only
b) ∑H = 0 and ∑V = 0 only
c) ∑H = 0, ∑V = 0 and ∑M = 0
d) ∑V = 0 only
Answer: c
Clarification: To analyse the frame completely, we are required with all the three of equilibrium equations i.e. ∑H = 0, ∑V = 0 and ∑M = 0.

250+ TOP MCQs on Conjugate-Beam Method and Answers

Structural Analysis Multiple Choice Questions on “Conjugate-Beam Method”.

1. Conjugate-Beam method was developed by:-
a) Hooke
b) Otto mohr
c) Charles E.greene
d) H.muller-Breslau
Answer: d
Clarification: Moment area method was developed by mohr and Greene.

2. In this method, shear compare with:-
a) Slope
b) Moment
c) Displacement
d) External load
Answer: a
Clarification: They have similar equations. In one we find area under load, while in other we find area under M/EI.

3 In this method, moment compare with:-
a) Slope
b) Shear
c) Displacement
d) External load
Answer: c
Clarification: They have similar equations as they both are doubly integrated.

4. If L is length of conjugate beam and l is length of real beam then:
a) L > l
b) L < l
c) L = l
d) Can’t say
Answer: c
Clarification: Length remains same while converting a beam into its conjugate.

5. While converting a beam into its conjugate one, end supports remain same. This statement is:-
a) Always true
b) Always false
c) Can’t say
d) Depends upon type of load
Answer: c
Clarification: To account for slope and displacement of real beam, support changes of conjugate beam but not every time.

6. Pin joint is replaced by __________ in conjugate beam.
a) Roller
b) Pin
c) Fixed support
d) Link
Answer: b
Clarification: Since pin has non-zero slope but zero displacement, conjugate beam has got to have zero moment.

7. Roller is replaced by fixed joint in conjugate beam.
State whether the above statement is true or false.
a) True
b) False
Answer: b
Clarification: It is replaced by a roller only. Logic of pin applies here also.

8. Fixed joint is replaced by ___________ in conjugate beam.
a) Roller
b) Pin
c) Free end
d) Link
Answer: c
Clarification: Since fixed support has non-zero slope and non-zero displacement, conjugate beam has got to be free.

9. Free end is replaced by __________ in conjugate beam.
a) Roller
b) Pin
c) Fixed support
d) Link
Answer: c
Clarification: Since a free end has non-zero slope and non- zero displacement, conjugate beam has got to have non-zero moment and non-zero slope.

10. Internal pin is replaced by __________ in conjugate beam.
a) Roller
b) Pin
c) Fixed support
d) Hinge
Answer: d
Clarification: Since int. pin has non-zero slope but zero displacement, conjugate beam has got to have zero moment.

11. Hinge is replaced by ___________ in conjugate beam.
a) Roller
b) Hinge
c) Fixed support
d) Link
Answer: a
Clarification: Since hinge has non-zero slope and non- zero displacement, conjugate beam has got to have non-zero moment and non-zero shear.

12. In conjugate beam, load is always away from beam.
State whether the above statement is true or false.
a) True
b) False
Answer: a
Clarification: If M/EI is +ve then it acts upward, otherwise downward.

250+ TOP MCQs on Method of Joints and Answers

Structural Analysis Multiple Choice Questions on “Method of Joints”.

1. How many equilibrium equations do we need to solve generally on each joint of a truss?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: Summation of forces in x and y direction should be equated to 0. Since there is no bending moments in trusses, we don’t need to solve the third equation.

2. If a member of a truss is in compression, then what will be the direction of force that it will apply to the joints?
a) Outward
b) Inward
c) Depends on case
d) No force will be there
Answer: a
Clarification: Member will apply outward force. Joint will in turn apply inward force resulting in compression of the member.

3. If a member of a truss is in tension, then what will be the direction of force that it will apply to the joints?
a) Outward
b) Inward
c) Depends on case
d) No force will be there
Answer: b
Clarification: Member will apply inward force. Joint will in turn apply outward force resulting in compression of the member.

4. What should be ideally the first step to approach to a problem using method of joints?
a) Draw fbd of each joint
b) Draw fbd of overall truss
c) Identify zero force members
d) Determine external reaction forces
Answer: c
Clarification: Identifying zero force members should always be the first step to approach any truss problem as it eliminate a lot of variables and is fairly easy.

5. What should be the angle (in degrees) in the given system (part of a bigger system) if both of the members have to be a zero force member?

a) 22.5
b) 45
c) 67.5
d) 90
Answer: d
Clarification: 90o would mean that without any external force, each one would carry no force to satisfy equations of equilibrium.


In the above figure, force is applied at joint c and its magnitude is 10N with downward direction. This question is used for Q6-Q9.

6. Which of the following are 0 force members?
a) FG, HI, HJ
b) HI, HJ, AE
c) HI, HJ, HE
d) HI, HJ, FH
Answer: a
Clarification: FH, HE and AE are non-zero force member as there are directly transmitting load from the external support. So, by option elimination we can say that the answer is (a).

7. What will the magnitude of force (in N) transmitted by FI?
a) 0
b) 1
c) 2
d) 3
Answer: a
Clarification: GF is a zero force member as stated in earlier question. Now, in joint F, BF and FH are in a line. This means that the only remaining member FI which is not in line will transmit zero force.

8. What will the magnitude of force (in N) transmitted by IC?
a) 0
b) 1
c) 2
d) 3
Answer: a
Clarification: IH is a zero member force as is FI. So, IC too will be zero force members.

9. What is total no. of zero force members in the above given system?
a) 7
b) 8
c) 9
d) 10
Answer: c
Clarification: Following are the zero force member based on the logics explained above: – GF, HI, HJ, ED, FI, IC, CH, JE and JC.

250+ TOP MCQs on Arches and Answers

Structural Analysis Multiple Choice Questions on “Arches”.

1. An arch is a beam except for ____
a) It does not resist inclined load
b) It does not resist transverse forces
c) It does not allow rotation at any point
d) It does not allow horizontal movement
Answer: d
Clarification: An arch is a curved member in which horizontal displacements are prevented at the supports/springings/abutments.

2. An arch is more economical than a beam for a shorter span length.
a) True
b) False
Answer: a
Clarification: Bending Moment for an arch is given by the bending moment produced in simply supported for same loading minus bending moment produced due to horizontal thrust. Since the bending moment produced is lower for the same loading, it is more economical than the beam.

3. Two hinged arches is a determinate structure.
a) True
b) False
Answer: b
Clarification: Two hinged arches is an indeterminate structure. We can calculate vertical reactions by using ∑M = 0 and ∑V = 0 but the horizontal reaction cannot be computed by any of equilibrium equations. Thus, two hinged arches is an indeterminate structure.

4. Calculate the horizontal thrust for the two hinged parabolic arch loaded uniformly throughout with distributed load.

a) (frac{WL^2}{32H} )
b) (frac{WL^2}{16H} )
c) (frac{WL^2}{8H} )
d) (frac{WL^2}{2H} )
Answer: c
Clarification: ∑H = 0
H = (frac{∫M.y dy}{∫y^2 dy})
Where, y=(frac{4 H x ( L-x )}{L^2})
Hence, H = (frac{WL^2}{8H} )

5. Calculate the horizontal thrust for the two hinged parabolic arch loaded uniformly for the left half span of the arch with distributed load.

a) (frac{WL^2}{32H} )
b) (frac{WL^2}{16H} )
c) (frac{WL^2}{8H} )
d) (frac{WL^2}{2H} )
Answer: b
Clarification: ∑H = 0
H = (frac{∫M.y dy}{∫y^2 dy})
Where, y=(frac{4 H x ( L-x )}{L^2})
Hence, H = (frac{WL^2}{16H} )

6. Calculate the horizontal thrust for the two hinged semicircular arch loaded uniformly throughout with distributed load.

a) (frac{W}{pi})
b) (frac{W}{pi}) sin2∞
c) (frac{4RW}{3pi})
d) (frac{W}{2pi})
Answer: c
Clarification: ∑H = 0
H = (frac{∫M.y dy}{∫y^2 dy})
Y = (sqrt{R^2- x^2} – sqrt{R^2-(frac{L^2}{2})} )
Hence, H = (frac{4RW}{3pi}.)

7. Calculate the horizontal thrust for the two hinged semicircular arch loaded with point load at its crown.

a) (frac{W}{pi})
b) (frac{W}{pi}) 2∞
c) (frac{4RW}{3pi})
d) (frac{W}{2pi})
Answer: a
Clarification: ∑H = 0
H = (frac{∫M.y dy}{∫y^2 dy})
Y = (sqrt{R^2- x^2} – sqrt{R^2-(frac{L^2}{2})} )
Hence, H = (frac{W}{pi})

8. Calculate the horizontal thrust for the two hinged parabolic arch loaded with point load at its crown.

a) (frac{W}{pi})
b) (frac{W}{pi})sin2∞
c) (frac{4RW}{3pi})
d) (frac{25WL}{128H})
Answer: d
Clarification: ∑H = 0
H = (frac{∫M.y dy}{∫y^2 dy})
Where, y=(frac{4 H x (L-x)}{L^2})
Hence, H = (frac{25WL}{128H}.)

9. Calculate the horizontal thrust for the two hinged semicircular arch loaded with point load at inclination of α with horizontal axis on the left span.

a) (frac{W}{pi})
b) (frac{W}{pi})sin2∞
c) (frac{4RW}{3pi})
d) (frac{25WL}{128H})
Answer: b
Clarification: ∑H = 0
H = (frac{∫M.y dy}{∫y^2 dy})
Y = (sqrt{R^2- x^2} – sqrt{R^2-(frac{L}{2^2})} )
Hence, H = (frac{W}{pi})sin2∞.

10. Identify the incorrect statement according to the hinged arches.
a) Three hinged arch is a statically determinate structure
b) To analyze three hinged arch, equlibrium equations are sufficient
c) For three hinged parabolic arch subjected to u.d.l over the entire span, the bending moment is constant throughout the span
d) For two hinged parabolic arch subjected to u.d.l over the entire span, the bending moment is zero throughout the span
Answer: c
Clarification: For three hinged parabolic arch subjected to u.d.l over the entire span, the bending moment and radial shear at any section is zero throughout the span.

250+ TOP MCQs on Conjugate-Beam Method – 2 and Answers

Structural Analysis MCQs on “Conjugate-Beam Method – 2”.

B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.

structural-analysis-questions-answers-mcqs-q1

1. What will be the reaction force at support C?
a) 20
b) 40
c) 80
d) 120
Answer: c
Clarification: Balance moment about point B.

2. What will be the shape of SFD in this case?
a) Linear
b) Parabolic
c) Linear with discontinuity
d) Arbitrary curve
Answer: c
Clarification: Since loads are not uniform, SFD will be linear and support at point C and B will lead to discontinuity.

3. What is the shape of BMD for this diagram?
a) Rectangular
b) Triangular
c) Parabolic
d) Arbitrary curve
Answer: b
Clarification: It will be 0 till point B, and then will increase till C and then again decrease till D (every time linearly), thus making it triangular.

4. What will be the peak value of SFD?
a) 20
b) 40
c) 60
d) 80
Answer: c
Clarification: It will decrease to 20 at point B and then increase by 80 at point C and then will remain constant. So, peak value will be 80-20 i.e. 60KN.

5. Where would peak value of BMD lie?
a) A
b) B
c) C
d) D
Answer: c
Clarification: It will be 0 till point B, and then will increase till C and then again decrease till D (every time linearly), thus making it triangular.

6. Which type of joint would replace point A in its conjugate beam?
a) roller
b) pin
c) hinge
d) fixed
Answer: d
Clarification: Since point A is a free end, a fixed joint would replace it in the conjugate beam.

7. How many fixed joint will be there in conjugate beam?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: Since there are 2 fixed ends in initial beam, there will be two fixed joints.

8. There won’t be any hinge in the conjugate beam.
State whether the above statement is true or false.
a) true
b) false
Answer: b
Clarification: Initial beam comprises of one internal pin and one internal roller, so conjugate beam would contain two hinges.

9. What will be the shear developed at hinge B in conjugate beam?
a) 30/EI
b) 40/EI
c) 60/EI
d) 80/EI
Answer: a
Clarification: Balance moment about point C in the beam BC.

10. What will be the shear developed at hinge C in conjugate beam?
a) 30/EI
b) 40/EI
c) 60/EI
d) 80/EI
Answer: c
Clarification: Balance moment about point B in the beam BC.

11. What will be the modulus of slope at point A?
a) 90/EI
b) 40/EI
c) 30/EI
d) 20/EI
Answer: c
Clarification: Just balance shear forces in the conjugate beam AB.

12. What will be the modulus of slope at point D?
a) 90/EI
b) 40/EI
c) 30/EI
d) 20/EI
Answer: a
Clarification: Just balance shear forces in the conjugate beam CD after balancing moment.

13. What will be the modulus of deflection of point A?
a) 20/EI
b) 30/EI
c) 80/EI
d) 90/EI
Answer: b
Clarification: Balance moment about point B in beam AB.

14. What will be the modulus of deflection of point D?
a) 20/EI
b) 30/EI
c) 80/EI
d) 90/EI
Answer: c
Clarification: Balance moment about point C in beam CD.

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