Category: Structural Analysis Objective Questions

Structural Analysis Multiple Choice Questions on “Influence Lines”.

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

1. In BMD and SFD :-
a) Points remain fixed, position of load changes
b) Points change, position of loads remain fixed
c) Both of them changes
d) Neither of them changes
Answer: b
Clarification: In BMD and SFD, we analyze the structure by fixing loads initially.

2. in influence line diagrams (ILD) :-
a) Points remain fixed, position of load changes
b) Points change, position of loads remain fixed
c) Both of them changes
d) Neither of them changes
Answer: a
Clarification: In ILD we analyze effect of a force or moment on a fixed point by constantly varying point of application of load/moment.

3. For drawing ILD, what value of test load is assumed?
a) 1 unit
b) Arbitrary
c) Depends upon structure
d) 0
Answer: a
Clarification: 1 unit load is assumed as calculations are easy then. Actual loads are multiplies with the results obtained to calculate further.

4. ILD of statically determinate beams consists of curve as well as straight lines.
State whether the above statement is true or false.
a) True
b) False
Answer: b
Clarification: ILD of statically determinate beams consist of only straight lines.

Following figure is for questions 5 and 5.

AB is of length 1m.
5. What will be the equation for vertical reaction at point A (RAY)?
a) RAY = 1-X
b) RAY = 2-X
c) RAY = 3-X
d) RAY = 4-X
Answer: a
Clarification: Place unit load at any variable distance X and calculate RAY by conserving moment about point B.

6. What will be the shape of ILD curve for vertical reaction at point A (RAY)?
a) Triangular
b) Circular
c) Rectangular
d) Trapezoidal
Answer: a
Clarification: On plotting the above given equation, shape of curve will come out to be triangular.

Following figure is for questions 7 and 8.

AB = BC = 1m

7. What will be the equation for vertical reaction at point B (RBY)?
a) RBY = 4X
b) RBY = 2X
c) RBY = 3X
d) RBY = X
Answer: d
Clarification: Take load at any variable distance X and conserve load about point A.

8. Where will be the maximum point of ILD lie?
a) A
b) B
c) C
d) Between A and B
Answer: c
Clarification: On drawing curve based on the above equation, we will find that maximum point will lie at C.

9. Maximum point of ILD always lies at the point at which load is applied.
State whether the above statement is true or false.
a) True
b) False
Answer: b
Clarification: In the above example above sentence is proved wrong.

Structural Analysis Multiple Choice Questions on “External Work and Strain Energy”.

U_e= work done by all external forces
U_i = internal work or strain energy
Δ = final elongation of bar
Θ = final angular deflection

1. U_i is not developed when:-
a) Structure elongates
b) Structure bends
c) Structure deforms
d) External force is zero
Answer: d
Clarification: U_i is developed when structure deforms, which is happening in other three options.

2. What will be the value of Ue if material is linear elastic? Axial force is increased from 0 to P gradually.
a) ¹⁄₄ P Δ
b) ¹⁄₃P Δ
c) ¹⁄₂P Δ
d) P Δ
Answer: c
Clarification: Due to linear elasticity, we can substitute force in terms of P and Δ and then integrate wrt x to get the final answer.

3. What will be the work done force P if another load external load F’ causes deflection Δ’ in the above question?
a) ¹⁄₄P Δ’
b) ¹⁄₃P Δ’
c) ¹⁄₂P Δ’
d) P Δ’
Answer: d
Clarification: Here, P will remain constant. So, it will be a simple integration from 0 to Δ’.

4. What will be the work done by F’?
a) ¹⁄₄F’ Δ’
b) ¹⁄₃F’ Δ’
c) ¹⁄₂F’ Δ’
d) F’ Δ’
Answer: c
Clarification: Here, P will remain constant. So, it will be a simple integration from 0 to Δ’.

5. What will be the value of U_e if material is linear elastic? Moment is increased from 0 to m gradually.
a) ¹⁄₄ M θ
b) ¹⁄₃ M θ
c) ¹⁄₂ M θ
d) M θ
Answer: c
Clarification: Due to linear elasticity, we can substitute moment in terms of M and θ and then integrate wrt x to get the final answer. Mdθ is done for moment to calculate work done.

6. What will be the work done force M if another load external load M’ causes deflection θ’ in the above question?
a) ¹⁄₄ M θ’
b) ¹⁄₃ M θ’
c) ¹⁄₂ M θ’
d) M θ’
Answer: d
Clarification: Here, P will remain constant. So, it will be a simple integration from 0 to Δ’.

7. What will be the work done by M’ in above question?
a) ¹⁄₄ M θ’
b) ¹⁄₃ M θ’
c) ¹⁄₂ M θ’
d) M θ’
Answer: c
Clarification: Due to linear elasticity, we can substitute moment in terms of M and θ and then integrate wrt x to get the final answer. Mdθ is done for moment to calculate work done.

8. If an axial force N is applied gradually to a bar which is linear elastic and has a constant cross sectional area A and length L, what will be Δ?
a) ¹⁄₄ NL/AE
b) ¹⁄₃ NL/AE
c) ¹⁄₂ NL/AE
d) NL/AE
Answer: d
Clarification: Hooke’s law will be valid here as material is linear elastic.

9. In the above question, what will be the value of U_i ?
a) ¹⁄₄ N²L/AE
b) ¹⁄₃ N²L/AE
c) ¹⁄₂ N²L/AE
d) N²L/AE
Answer: c
Clarification: Once deformation is known, we can calculate the work done using earlier equations and then U_e = U_i.

10. What will be the value of dU_i in terms of E and I?
a) ¹⁄₄ M²dx/EI
b) ¹⁄₃ M²dx/EI
c) ¹⁄₂ M²dx/EI
d) M²dx/EI
Answer: c
Clarification: Relation between dθ and M/EI is known. So, we can use that to get results.

Structural Analysis Multiple Choice Questions on “Tributary Areas”.

1. The tributary is an unloaded area. This statement is:-
a) True
b) False
Answer: b
Clarification: The tributary area is the loaded area of a particular structure directly contributing to the load applied.

2. The tributary area lines between two columns/beams are at how much distance from one of the beam?
a) 0.2 of the total in between length
b) 0.4 of the total in between length
c) 0.5 of the total in between length
d) Depends on the amount of load carried
Answer: c
Clarification: Tributary area is bounded by lines lying halfway from next beam/column, irrespective of load carried.

3. For an interior girder, what is the shape of tributary area?
a) Rectangular
b) Triangular
C) Circular
d) Depends upon case
Answer: d
Clarification: If the girder surrounding the girder in question is similar to it, then it will be a triangle, otherwise it might be trapezium in some cases.

4. What is the maximum no. of structures to which an area can be a tributary are to?
a) 1
b) 2
c) 3
d) 4
Answer: a
Clarification: A particular area can’t be a tributary area to more than one member of a structure at one time.

5. Which out of the following statements is true?
a) Tributary area > influence area
b) Tributary area <= influence area
c) Tributary area >= influence area
d) No relation can be defined
Answer: b
Clarification: An influence area is always a tributary area. So, tributary area <= influence area.

6. If in an interior beam, adjacent structures are exactly similar then the tributary area is:-
a) Trapezium
b) Acute triangle
c) Obtuse triangle
d) Right angled triangle
Answer: d
Clarification: Since, the structures are similar; angles of triangle will be 45 and 90 degrees.

7. One area can serve as an influence area to more than one member of a structure at the same time.
State whether the above statement is true or false.
a) True
b) False
Answer: a
Clarification: Unlike tributary areas influences can serve to more than one as they can affect more than one forces on loaded.

8. 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.

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

1. How many equilibrium equations are used in method of sections?
a) 2
b) 4
c) 3
d) 5
Answer: c
Clarification: Moments too can be conserved along with forces in both directions. So, total no. of equations are three.

2. In trusses, a member in the state of tension is subjected to:-
a) push
b) pull
c) lateral force
d) either pull or push
Answer: b
Clarification: Pull is for tension, while push is for compression.

3. In method of sections, what is the maximum no. of unknown members through which the imaginary section can pass?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: Since we have three equilibrium equations, so we can have maximum 3 unknown forces/members through which imaginary section can pass.

4. Method of substitute members is use for which type of trusses?
a) complex
b) compound
c) simple
d) simple and compound
Answer: a
Clarification: Method of substitute members is used to solve problems involving complex trusses.

5. First step to solve complex truss using Method of substitute members is to convert it into unstable simple truss.
State whether the above statement is true or false.
a) true
b) false
Answer: b
Clarification: First step is to convert it to stable simple truss.
Shear force is represented by V
Bending moment is represented by M
Distance along the truss is represented by X
W is the uniform load applied.

6. On differentiating V wrt X we will get:-
a) W
b) -W
c) M
d) None of the mentioned
Answer: b
Clarification: On applying equilibrium equation, V – W(x)Δx – V – ΔV = 0.

7. On differentiating M wrt X we will get:-
a) W
b) -W
c) V
d) None of the mentioned
Answer: c
Clarification: On applying equilibrium equation, M + VΔx – M – ΔM = 0.

8. 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.

Structural Analysis Multiple Choice Questions on “Influence Lines for Beams”.

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

1. If on ILD analysis peak force comes out to be 2 KN, then what will be the peak force if loading is 2KN?
a) 1 KN
b) 2 KN
c) 3 KN
d) 4 KN
Answer: d
Clarification: Peak force will be load multiplied by earlier peak load i.e. 2*2.

Following figure is for Q2-Q7.
AC= 1m, CB =3 m
C is just an arbitrary point. A is pin support and B is a roller type support.

2. What will be the equation of ILD of shear force at point C for CB part?
a) 0.75 – 0.375X
b) 0.75 – 0.475X
c) 0.85 – 0.375X
d) 0.75 – 0.1375X
Answer: a
Clarification: Just assume force at any point between BC and conserve moment about point B.

3. What will be the equation of ILD of shear force at point C for AC part?
a) .25X – 1.25
b) .25X – 2.25
c) .25X – .25
d) .25X + .25
Answer: c
Clarification: Just assume force at any point between AC and conserve moment about point A.

4. If we have to apply a concentrated load in the above shown beam, such that shear at C becomes max. , where should we apply that load?
a) At A
b) At B
c) At C
d) Midway between A and C
Answer: c
Clarification: If we draw ILD according to the above given equations, we will see that peak of ILD comes at point C.

5. If a concentrated load of 50KN is applied at point C, then what will be the shear developed at point C?
a) 17.5 KN
b) 27.5 KN
c) 37.5 KN
d) 47.5 KN
Answer: c
Clarification: Position of ILD at point C is 0.75 (peak). So, shear developed will be 0.75 multiplied by 50KN.

6. What will be the shear developed at point C if a uniform load of 10KN/m is applied between point B and C?
a) 10.25 KN
b) 11.25 KN
c) 12.25 KN
d) 13.25 KN
Answer: b
Clarification: In case of uniform load, area of ILD curve multiplied by uniform load gives the shear.

7. If both, a load of 50KN at point C and a uniform load of 10KN/m between CB acts, then what will be the shear generated at point C?
a) 48.75
b) 50.75
c) 46.75
d) 52.75
Answer: a
Clarification: Net shear generated will be the sum of individually generated shear which has been already calculated earlier.

Following figure is for Q8-Q10.
AB= 2m, BC= 3m, CD= 3m
B is pin support, D is roller and C is just an arbitrary point.

8. What will be the ILD equation for ILD of shear at point B?
a) 1.33 – 0.116625X
b) 2.33 – 0.16625X
c) 3.33 – 0.16625X
d) 1.33 – 0.16625X
Answer: d
Clarification: Apply unit load at any point at a distance X and conserve moment about point D.

9. What will be the ILD equation for ILD of shear at point C for AB part of beam?
a) -0.33 + 0.165X
b) -0.33 + 0.265X
c) -0.43 + 0.165X
d) -0.33 + 0.365X
Answer: a
Clarification: Apply unit load between point A and B and conserve moment about point B.

10. What will be the ILD equation for ILD of shear at point D?
a) -.43 + 0.16625X
b) -.33 + 0.16625X
c) -.53 + 0.16625X
d) -.33 + 0.216625X
Answer: b
Clarification: Apply load at any point and conserve moment about point B.