[MCQ’s]Structural Analysis

Module-1

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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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. Which of the following are 0 force members?
a) FG, HI, HJ
b) HI, HJ, AE
c) HI, HJ, HE
d) HI, HJ, FH
Explanation: 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).

6. What will the magnitude of force (in N) transmitted by FI?
a) 0
b) 1
c) 2
d) 3
Explanation: 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.

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

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

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

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

11. 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
Explanation: Since we have three equilibrium equations, so we can have maximum 3 unknown forces/members through which imaginary section can pass.

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

13. 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
Explanation: 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.

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

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

16. 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
Explanation: Member will apply outward force. Joint will in turn apply inward force resulting in compression of the member.

17. Top most part of an arch is called ________
a) Sofit
b) Crown
c) Center
d) Abutment
Explanation: Top most part is called crown. Other options are name of other part.

18. Which of the following is true in case of stone brick?
a) They are weak in compression and tension
b) They are good in compression and tension
c) They are weak in compression and good in tension
d) They are good in compression but weak in tension
Explanation: Stone brick is very good in compression but weak in tension.

19. Shape of three-hinged arch is always :-
a) Hyperbolic
b) Circular
c) Parabolic
d) Can be any arbitrary curve
Explanation: Three hinged arch is always made in parabolic shape.

20. Internal bending moment generated in a three hinged arch is always:-
a) 0
b) Infinite
c) Varies
d) Non zero value but remains constant
Explanation: Due to its geometry, bending moment always comes out to be zero in 3 hinged arches.

21. Internal shear force generated in a three hinged arch is always:-
a) 0
b) Infinite
c) Varies
d) Non zero value but remains constant
Explanation: Due to its geometry, shear force always comes out to be zero in 3 hinged arches.

22. What is the degree of indeterminacy of a fixed arch?
a) 1
b) 2
c) 3
d) 4
Explanation: It is indeterminate to 3 degrees.

23. What is the degree of indeterminacy of a two hinged arch?
a) 1
b) 2
c) 3
d) 4
Explanation: It is indeterminate to 2 degrees. It consists of two pin supports at both of its ends.

24. In real life, bending stresses are zero in a three hinge arch.
State whether the above statement is true or false.
a) True
b) False
Explanation: In real life, loads are moving which results in generation of bending stresses.

25. 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
Explanation: An arch is a curved member in which horizontal displacements are prevented at the supports/springings/abutments.

26. An arch is more economical than a beam for a shorter span length.
a) True
b) False
Explanation: 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.

27. Two hinged arches is a determinate structure.
a) True
b) False
Explanation: 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.

28. 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
Explanation: 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.

Module-2

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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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

8. Where will be the maximum point of ILD lie?
a) A
b) B
c) C
d) Between A and B
Explanation: 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
Explanation: In the above example above sentence is proved wrong.

While writing influence line equations, left most point is always considered as origin and following sign convention is followed. 10. 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
Explanation: 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. 11. 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
Explanation: Just assume force at any point between BC and conserve moment about point B.

12. 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
Explanation: Just assume force at any point between AC and conserve moment about point A.

13. 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
Explanation: If we draw ILD according to the above given equations, we will see that peak of ILD comes at point C.

14. 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
Explanation: Position of ILD at point C is 0.75 (peak). So, shear developed will be 0.75 multiplied by 50KN.

16. 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
Explanation: In case of uniform load, area of ILD curve multiplied by uniform load gives the shear.

17. 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
Explanation: 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. 18. 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
Explanation: Apply unit load at any point at a distance X and conserve moment about point D.

19. 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
Explanation: Apply unit load between point A and B and conserve moment about point B.

20. 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
Explanation: Apply load at any point and conserve moment about point B.

Module-3

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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: It is only applicable when forces are of conservative type.

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.

11. What will be Δ in case of straight members using theorem?
a) 1⁄4 ΣN(dN/dP)L/AE
b) 1⁄3 ΣN(dN/dP)L/AE
c) 1⁄2 ΣN(dN/dP)L/AE
d) ΣN(dN/dP)L/AE
Explanation: On substituting value of internal energy in earlier theorem, we can get this.

12. P is treated here as:-
a) constant
b) variable
c) it doesn’t matter
Explanation: P is treated as variable and N is expressed in its term for partial differentiation.

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

14. N is caused by:-
a) constant forces
b) variable forces
c) both
d) neither
Explanation: 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.

15. What will be the external work performed during application of load?
a) 1⁄2 (p1 Δ1 + p2 Δ2)
b) 1⁄2 (p2 Δ1 + p1 Δ2)
c) p1 Δ1 + p2 Δ2
d) p2 Δ1 + p1 Δ2
Explanation: Since loads are gradually applied, work done will be average load times deflection. We can also find by integration.

16. What will be the work done during additional application of dp1?
a) p1 dΔ1 + p2 dΔ2 + dp1d Δ1
b) p1 dΔ1 + p2 dΔ2 + 1⁄2 dp1d Δ1
c) p1 dΔ1 + 1⁄2 p2 dΔ2 + dp1d Δ1
d) 1⁄2 p1 dΔ1 + p2 dΔ2 + dp1d Δ1
Explanation: At this time p1 and p2 are already applied, only dp1 is gradually applied.

17. 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
Explanation: It is true as the third term can be ignored as it is very small.

18. 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) + 1⁄2 (p2)( Δ2 + dΔ2)
c) 1⁄2 (p1 + dp1)(Δ1 + dΔ1) + (p2)( Δ2 + dΔ2)
d) 1⁄2 (p1 + dp1)(Δ1 + dΔ1) + 1⁄2 (p2)( Δ2 + dΔ2)
Explanation: Now, since all the loads are gradually applied, all will have a factor of half.

19. What will be change in work done in both case on initial application of load?
a) p1dΔ1 + dp1 Δ1 + p2dΔ2
b) 1⁄2 p1dΔ1 + dp1 Δ1 + p2dΔ2
c) 1⁄2 p1dΔ1 + 1⁄2 dp1 Δ1 + p2dΔ2
d) 1⁄2 p1dΔ1 + 1⁄2 dp1 Δ1 + 1⁄2 p2dΔ2
Explanation: We will get this by just subtracting two works done. This will be termed as dw.

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

21. Most of the real world structures are statically determinate.
State whether the above statement is true or false.
a) True
b) False
Explanation: Most of the real world structures are statically indeterminate owing to added supports or member or general form.

22. All reinforced concrete buildings are most of times:-
a) Statically determinate
b) Statically indeterminate
c) Mixture
d) Unstable
Explanation: This is because columns and beams are continuous in these cases over joints and supports.

23. 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
Explanation: Indeterminate structure deflects lesser than a determinate one.

24. Which structure will perform better during earthquake?
a) Statically determinate
b) Statically indeterminate
c) Both
d) Depends upon magnitude of earthquake
Explanation: Indeterminate structure has a tendency to redistribute its load to its redundant supports in case of overloading.

25. 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
Explanation: Maximum moment developed in simply supported beam will be twice that of fixed supported and hence, we can find deflections.

26. Which type of structure would cost less in terms of materials?
a) Statically determinate
b) Statically indeterminate
c) Both will cost equally
Explanation: Statically indeterminate would cost less as they can support a loading with thinner members and increased stability.

27. Which type of structure would cost less in terms of supports?
a) Statically determinate
b) Statically indeterminate
c) Both will cost equally
Explanation: Supports and joints of indeterminate structures are costly compared to that of a determinate one.

28. Differential settlement is problematic to which type of structure?
a) Statically determinate
b) Statically indeterminate
c) Both
d) Neither
Explanation: It causes development of internal stress in statically indeterminate structures.

29. Fabrication errors don’t cause additional stresses in statically indeterminate structures.
State whether the above statement is true or false.
a) True
b) False
Explanation: Fabrication errors do cause generation of additional stresses in statically indeterminate structures.

30. 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
Explanation: On application of excessive load, a hinge/pin like point forms at the centre of beam.

31. The above structure is statically indeterminate.
State whether the above sentence is true or false.
a) True
b) False
Explanation: 4 unknown forces and 3 reactions are there. So, it is statically indeterminate.

32. This structure is made redundant by temporarily removing how many support reactions?
a) 1
b) 2
c) 3
d) 4
Explanation: As the degree of indeterminacy is one, one of the support reactions is chosen to be redundant.

33. Which of the following forces can’t be chosen to be redundant?
a) Vertical support at point A
b) P
c) Vertical support at B
d) Moment at A
Explanation: Load P can’t be chosen as this is the load causing deflection. Rest all support forces can be chosen to be redundant.

34. Δ/ bb refers to displacement cause by By.
What does first b in Δ/ bb stands for?
a) Where unknown reaction acts
b) Point whose deflection is considered
c) Can be anything
Explanation: First b refers to the point where load is specified, second B refers to the point where load is applied.

35. What is the compatibility equation for the above mentioned condition?
a) Δ/ bb – Δb = 0
b) Δ/ bb + Δb = 0
c) Δ/ bb – Δb > 0
d) Δ/ bb – Δb < 0
Explanation: Both the displacement will be opposite in direction and equal in magnitude.

36. How much load acts and in which direction at point B for the displacement to be fbb?
a) Unit, opposite to By
b) Unit, same as by
c) Small, opposite to By
d) Large, same as By
Explanation: Displacement caused by unit load acting in the direction of By is termed as fbb.

37. Which one of the following describes fbb?
a) Δ/ bb / By
b) By/ Δ / bb
c) Δ/ bb – By
d) Δ/ bb + By
Explanation: Δ/ bb is actual displacement caused by By, and fbb is caused by unit load. So, its fbb times By Δ/ bb/By.

38. According to maxwell’s theorem:-
a) fab = faa
b) fbb = faa
c) fba = fab
d) fba = fbb
Explanation: The displacement of a point B on a structure due to a unit load acting at point A is equal to the displacement of point A when the unit load is acting at point B.

39. Maxwell’s theorem doesn’t apply when external moments are placed on the beam.
State whether the above statement is true or false.
a) true
b) false
Explanation: It is also applicable when external moments are acting on the structure instead of forces.

40. Maxwell’s reciprocal theorem is applicable for elastic materials only.
a) True
b) False
Explanation: Maxwell’s reciprocal theorem is applicable for elastic materials that too which follows Hooke’s law.

41. Maxwell’s reciprocal theorem is applicable for only prismatic members.
a) True
b) False
Explanation: Maxwell’s reciprocal theorem is applicable for both prismatic as well as non-prismatic members as long as they are made of elastic materials and obeys Hooke’s Law.

42. Identify the incorrect statement from the following regarding applicability of Maxwell’s reciprocal theorem.
a) It is applicable for elastic member’s only
b) The temperature must remain constant throughout
c) Supports of the member should be unyielding
d) It is applicable for prismatic member only
Explanation: Maxwell’s reciprocal theorem is applicable for both prismatic as well as non-prismatic members as long as they are made of elastic materials and obeys Hooke’s Law.

43. Law of reciprocal deflections was given by ______
a) E. Betti
b) James Clerk Maxwell
c) Alberto Castigliano
d) Clayperon
Explanation: Law of reciprocal deflections is also known as Maxwell’s Reciprocal Deflection Theorem and was offered by James Clerk Maxwell in 1864.

44. Maxwell’s reciprocal theorem is applicable for elastic materials only.
a) True
b) False
Explanation: Maxwell’s reciprocal theorem is applicable for elastic materials that too which follows Hooke’s law.

45. Maxwell’s reciprocal theorem is applicable for only prismatic members.
a) True
b) False
Explanation: Maxwell’s reciprocal theorem is applicable for both prismatic as well as non-prismatic members as long as they are made of elastic materials and obeys Hooke’s Law.

46. Identify the incorrect statement from the following regarding applicability of Maxwell’s reciprocal theorem.
a) It is applicable for elastic member’s only
b) The temperature must remain constant throughout
c) Supports of the member should be unyielding
d) It is applicable for prismatic member only
Explanation: Maxwell’s reciprocal theorem is applicable for both prismatic as well as non-prismatic members as long as they are made of elastic materials and obeys Hooke’s Law.

47. Betti’s Theorem is based on ______
a) Balancing of external and internal forces
c) Balancing of external and internal moments
Explanation: Betti’s Theorem is derived by balancing the work done produced by external and internal loadings.

48. Law of reciprocal deflections was given by ______
a) E. Betti
b) James Clerk Maxwell
c) Alberto Castigliano
d) Clayperon
Explanation: Law of reciprocal deflections is also known as Maxwell’s Reciprocal Deflection Theorem and was offered by James Clerk Maxwell in 1864.

Module-4

State whether the above statement is true or false
a) true
b) false

2. 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
Explanation: No. of redundant reactions and compatibility equations are equal.

3. Flexibility matrix is always:-
a) symmetric
b) non-symmetric
c) anti-symmetric
Explanation: Flexibility matrixes are always symmetric as a consequence of Betti’s law.

4. Numerical accuracy of the solution increases if flexibility coefficients with larger values are located:-
a) near main diagonal
b) near edges
c) in between
d) near side middles
Explanation: Numerical accuracy increases if larger coefficients are located near the main diagonal of matrix.

5. Which of the following primary structure is best for computational purposes?
a) symmetric
b) non-symmetric
c) anti-symmetric
Explanation: It is easier to compute solutions for the flexibility coefficient matrix in that case.

6. 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
Explanation: This leads to small corrections induced by redundant.

7. Indeterminate analysis of asymmetrical structure is difficult from that of a non-symmetric one.
State whether the above statement is true or false.
a) true
b) false
Explanation: Symmetricity makes the indeterminate analysis easier.

8. In general, any structure can be classified as a symmetric one :-
a) when its structure is symmetric
c) when its supports are symmetric
Explanation: It is deemed as symmetric when when it develops symmetric internal loading and deflections about central axis.

9. Normally, which of the following things may/may not be symmetric to develop symmetricity?
a) material
b) geometry
d) dki

10. Indeterminate analysis of an anti-symmetrically loaded structure is difficult from that of non-symmetric one.
State whether the above statement is true or false.
a) true
b) false
Explanation: Solving one side would give modulus of deflections of other side as well.

Module-5

1. For stable structures, one of the important properties of flexibility and
stiffness matrices are that the elements on the main diagonal
(i) Of a stiffness matrix must be positive
(ii) Of a stiffness matrix must be negative
(iii) Of a flexibility matrix must be positive
(iv) Of a flexibility matrix must be negative
a. (i) and (iii)
b. (ii) and (iii)
c. (i) and (iv)
d. (ii) and (iv)

2. Study the following statements.
i) The displacement method is more useful when degree of kinematic
indeterminacy is greater than the degree of static indeterminacy.
ii) The displacement method is more useful when degree of kinematic
indeterminacy is less than the degree of static indeterminacy.
iii) The force method is more useful when degree of static indeterminacy
is greater than the degree of kinematic indeterminacy.
iv) The force method is more useful when degree of static indeterminacy
is less than the degree of kinematic indeterminacy.
a. (i) and (iii)
b. (ii) and (iii)
c. (i) and (iv)
d. (ii) and (iv)

3. Muller Breslau’s principle for obtaining influence lines is applicable to
i) trusses
ii) statically determinate beams and frames
iii) statically indeterminate structures, the material of which is elastic and
follows Hooke’s law
iv) any statically indeterminate structure
a. (i), (ii) and (iii)
b. (i), (ii) and (iv)
c. (i) and (ii)
d. only (i)

4. In the slope deflection equations, the deformations are considered to be
caused by
i) bending moment
ii) shear force
iii) axial force
a. only (i)
b. (i)and(ii)
c. (ii) and (iii)
d. (i), (ii) and (iii)

5. Select the correct statement
a. Flexibility matrix is a square symmetrical matrix
b. Stiffness matrix is a square symmetrical matrix
c. both (a) and (b)
d. none of the above

6. Stiffness matrix is a
a) Symmetric matrix.
b) Skew symmetric matrix.
d) Augmented matrix.
a) Symmetric matrix.
Stiffness matrix is a symmetric matrix.

7. The local x-axis of a member is always parallel to the ____ of the member.
a) Longitudinal axis.
b) Vertical axis.
c) Transverse axis.
d) Horizontal axis.
a) Longitudinal axis.
The local x-axis of a member is always parallel to the longitudinal axis of the member.

8. The symmetry of stiffness matrix proves
a) Castigliano’s theorem.
b) Maxwell reciprocal theorem.
c) Betti’s law.
d) Hardy cross theorem.
c) Betti’s law.

9. Size of the stiffness matrix _____ if the number of force component increase.
a) Increase.
b) Decrease.
c) Remain unchanged.
d) Proportionally increases.
d) Proportionally increases.
Size of the stiffness matrix proportionally increases if the number of force component increase.

10. To describe material non-linearity, __ should be non-linear.
a) Constitutive relationship.
b) Strain-displacement relationship.
c) Shape function.
d) Element type.
a) Constitutive relationship.

11. FEA is a/ an
a) Analytical technique.
b) Computational technique.
c) Differential technique.
d) None of these.
Computational technique.
FEA(Finite Element Analysis) is a computational technique.

12. For a state of stable equilibrium, the potential energy should be
a) Maximum.
b) Minimum.
c) Zero.
d) Cannot be specified.
a) Maximum.
For a state of stable equilibrium, the potential energy should be maximum.

13. The field variables are the
a) Independent variable.
b) Neither dependent nor independent variable.
c) Dependent variables.
d) None of these.
c) Dependent variables.
The field variables are the dependent variables.

Module-6

1. Moment Distribution Method is applicable to the determinate and indeterminate structure.
a) True
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.

2. 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
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.

3. Moment Distribution Method does not consider axial and shear effects for the displacement calculations for the given structure.
a) True
b) False
Explanation: The moment distribution method developed by Hardy Cross in 1930 is useful for analysis of indeterminate beams and frames. The method considers the flexural effect and ignores any effect caused due to shear and axial loadings.

4. Carryover Moment at end B due to moment M applied at end A for the given beam is _______ a) 0
b) +M
c) -M
d) +M2
Explanation: For a moment applied M at free end with hinged support at another end, as the hinged support cannot carry or resist any bending moment thus carry over moment is zero.

5. 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
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. Thus support B resists moment greater than M/2.

b = no. of members in a structure
r = no. of unknown reactions in member
j = no. of joints in the structure
c = no. of extra equations in the structure

6. Which of the following is the correct equation for DSI of a truss?
a) b + r – 2J
b) b + r + 2J
c) b – r – 2J
d) b – r – 2J
Explanation: b and r gives no. of unknown forces and 2j is the no. of equations available.

7. Which of the following is a correct equation for the DSI of a frame?
a) 3b + r + 3j
b) 3b + r – 3j
c) 3b – r – 3j
d) depends upon structure
Explanation: It depends upon whether that frame is planar or space.

8. Which of the following is a correct equation for the DSI of a planar frame?
a) 3b + r + 3j-c
b) 3b + r – 3j-c
c) 3b – r – 3j-c
d) depends upon structure
Explanation: Each member will give 3 unknowns and each joint will 3 equations.

9. Tree method is used to find DSI.
State whether the above statement is true or false.
a) true
b) false
Explanation: Tree method is used to find DSI in planar frame structures.

10. How many degrees of freedom are counted for a fixed support while calculating DKI?
a) 0
b) 1
c) 2
d) 3
Explanation: In fixed support, there is no possibility of deflection, so no degree of freedom is counted.

11. How many degrees of freedom are counted for a pin support while calculating DKI?
a) 0
b) 1
c) 2
d) 3
Explanation: In fixed support, there is one possibility of rotation, so no degree of freedom is counted.

12. How many degrees of freedom are counted for a roller support while calculating DKI?
a) 0
b) 1
c) 2
d) 3
Explanation: In roller support, there are two possibilities of deflection, so two degree of freedom is counted.

13. How many degrees of freedom are counted for a fixed support while calculating DKI?
a) 0
b) 1
c) 2
d) 3
Explanation: In fixed support, there is no possibility of deflection, so no degree of freedom is counted.

14. How many degrees of freedom are counted for an internal pin support while calculating DKI?
a) 0
b) 1
c) 2
d) 3
Explanation: In internal pin support, there is only one possibility of deflection, so no degree of freedom is counted.

15. When a structural member of the uniform section is subjected to a moment at one end only, then the moment required so as to rotate that end to produce a unit slope, is called _____
a) Resistance of member
b) Stiffness of member
c) Capacity of member
d) Potential of member
Explanation: When a structural member of the uniform section is subjected to a moment at one end only, then the moment required so as to rotate that end to produce a unit slope, is called stiffness of the member. Stiffness is basically a measure of the ability of the member to resist rotational deformation.

16. Distribution factor is the ratio in which force sharing capacity of various members meeting at a rigid joint.
a) True
b) False
Explanation: Distribution factor is the ratio in which moment sharing capacity of various members meeting at a rigid joint.

17. Moment Distribution Method is applicable to the determinate and indeterminate structure.
a) True
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.

18. 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