Methods of Structural Analysis MCQ Quiz - Objective Question with Answer for Methods of Structural Analysis - Download Free PDF

Last updated on Jun 13, 2025

Latest Methods of Structural Analysis MCQ Objective Questions

Methods of Structural Analysis Question 1:

The fixed support in a real beam becomes in the conjugate beam as

  1. Free support
  2. Hinged support
  3. Roller support
  4. Fixed support

Answer (Detailed Solution Below)

Option 1 : Free support

Methods of Structural Analysis Question 1 Detailed Solution

Explanation:

  • In the conjugate beam method, the bending moment diagram of the real beam is treated as the load on the conjugate beam.

  • The support conditions in the conjugate beam change according to the boundary conditions of the real beam:

    • A fixed support in the real beam becomes a hinged support in the conjugate beam.

    • This is because a fixed end in the real beam (zero displacement and rotation) corresponds to a point in the conjugate beam where the slope is zero but rotation is allowed, like a hinge.

  • This transformation helps analyze beam deflections and rotations using simpler static analysis of the conjugate beam.

 Additional InformationTable depicting real beam and its corresponding conjugate beam: 

Screenshot 2025-06-04 182138

Methods of Structural Analysis Question 2:

The rotational stiffness coefficient K11 for the frame having two members of equal \(EI/L\) is given by

  1. \(7 EI/L\)
  2. \(9 EI/L\)
  3. \(8 EI/L\)
  4. \(6 EI/L\)

Answer (Detailed Solution Below)

Option 1 : \(7 EI/L\)

Methods of Structural Analysis Question 2 Detailed Solution

Explanation:

KCB = 4EI/L (Far end is fixed)

KCA = 3EI/L (Far end is pinned)

K11 = KCB KCA 

      = 4EI/L + 3EI/L  = 7EI/L

F1 A.M Madhu 04.05.20 D 10

Methods of Structural Analysis Question 3:

If one end of a prismatic beam AB with fixed ends is given a transverse displacement (\(\Delta\)) without rotation, then the moment induced at A or B due to the displacement is

  1. \(12EI \Delta / L^2\)
  2. \(EI \Delta / L^3\)
  3. \(12EI \Delta / L^3\)
  4. \(6EI \Delta / L^2\)

Answer (Detailed Solution Below)

Option 4 : \(6EI \Delta / L^2\)

Methods of Structural Analysis Question 3 Detailed Solution

Explanation:

This is a case of pure transverse displacement of a fixed-fixed prismatic beam with no rotation — typically referred to as "sway" in structural analysis.

  • Beam AB is fixed at both ends.

  • One end is displaced transversely by Δ \Delta" id="MathJax-Element-27-Frame" role="presentation" style="position: relative;" tabindex="0">Δ \Delta (without rotation).

  • This introduces equal and opposite end moments to maintain equilibrium.

For such a case, the induced moment at each end (A or B) is:\(6EI \Delta / L^2\)

Additional Information

  • Transverse displacement refers to the movement of a structural member perpendicular to its longitudinal axis.
  • For a horizontal beam, transverse displacement means vertical movement.

  • For a vertical column, transverse displacement means horizontal movement (also called sway).

  • A beam fixed at both ends is displaced sideways (laterally) at one end without rotation.

  • This displacement is parallel to the direction of the applied shear or lateral force, not along the length of the beam.

  • To maintain fixity (zero rotation), internal moments develop at both supports.

Methods of Structural Analysis Question 4:

For a linear elastic structural element

  1. Stiffness is directly proportional to the flexibility
  2. Stiffness may not be correlated with flexibility
  3. Stiffness is equal to flexibility
  4. Stiffness is inversely proportional to the flexibility

Answer (Detailed Solution Below)

Option 4 : Stiffness is inversely proportional to the flexibility

Methods of Structural Analysis Question 4 Detailed Solution

Explanation:

  • Stiffness (K) measures a structure’s resistance to deformation under load.

  • Flexibility (F) measures how much a structure deforms under a given load.

  • For a linear elastic element, these two are reciprocals of each other.

  • Therefore, if stiffness increases, flexibility decreases, and vice versa.

Additional Information

  • This inverse relationship is fundamental in structural analysis and design.

  • Stiffness depends on material properties (like Young’s modulus) and geometry.

  • Flexibility matrices and stiffness matrices are key concepts in methods like the Flexibility Method and Stiffness Method of structural analysis.

Methods of Structural Analysis Question 5:

A moment 'K' required to rotate near end of a prismatic beam through a unit angle without translation, the far end being freely supported is given by:

  1. \(K = \dfrac{3EI}{L}\)
  2. \(K = \dfrac{4EI}{L}\)
  3. \(K = \dfrac{EI}{L}\)
  4. \(K = \dfrac{L}{EI}\)
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : \(K = \dfrac{3EI}{L}\)

Methods of Structural Analysis Question 5 Detailed Solution

Explanation:

Stiffness (S): It is defined as the force per unit deflection or moment required per unit rotation.

 

Case 1:

For the far end being Hinged Supported or freely supported:

RRB JE CE 23 10Q Free FT Slot 7 Hindi - Final 16

If a unit rotation is to be caused at an end A for the far end being hinged support. Moment of 3EI/L is to be applied at end and hence stiffness for the member is said to be 3EI/L

Case 2:

For the far end being Fixed Supported:

RRB JE CE 23 10Q Free FT Slot 7 Hindi - Final 15

If a unit rotation is to be caused at an end A for the far end being fixed supported. Moment of 4EI/L is to be applied at end and hence stiffness for the member is said to be 4EI/L

Case 3: when the far end is free

Beam offers no resistance to the moment.

S = 0

Top Methods of Structural Analysis MCQ Objective Questions

Slope = area of BMD/EI, is the relation given by:

  1. Mohr's first theorem
  2. Mohr's second therorem
  3. Castigliano's therorem
  4. Macaulay's theorem

Answer (Detailed Solution Below)

Option 1 : Mohr's first theorem

Methods of Structural Analysis Question 6 Detailed Solution

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Mohr’s Theorem I:

The angle between the two tangents drawn on the elastic line is equal to the area of the Bending Moment Diagram between those two points divided by flexural rigidity.

RRB JE CE 39 17Q Structure Chapter Test 2 (1)(Hindi) - Final images Q17

\(\theta = \frac{{\left[ {Area\;of\;bending\;moment\;diagram} \right]}}{{EI}}\)

Mohr’s Theorem II:

The deviation of a point away from the tangent drawn from the other point is given by the moment of area of bending moment diagram about the first point divided by flexural rigidity.

\(\delta = BB' = \frac{{\left[ {Area\;of\;bending\;moment\;diagram} \right] \times \bar x}}{{EI}}\)

The ratio of the deflections of the free end of a cantilever due to an isolated load at 1/3rd and 2/3rd of the span is

  1. 1/7
  2. 2/7
  3. 3/7
  4. 2/5

Answer (Detailed Solution Below)

Option 2 : 2/7

Methods of Structural Analysis Question 7 Detailed Solution

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Ist condition:

For a cantilever beam subjected to load W at distance of L/3 from free end, the deflection is given by:

F1 N.M Madhu 25.02.20 D4

\({\rm{y}}{{\rm{c}}_1} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{2{\rm{L}}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{2{\rm{L}}}}{3}} \right)^2} \times \frac{{\rm{L}}}{3} = {\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}^3}\)

IInd condition:

For a cantilever beam subjected to load W at distance of 2L/3 from free end:

F1 N.M Madhu 25.02.20 D5

\({\rm{y}}{{\rm{c}}_2} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{\rm{L}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{\rm{L}}}{3}} \right)^2} \times \frac{{2{\rm{L}}}}{3} = \frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}\)

\({\rm{Ratio\;}}\left( {\rm{r}} \right) = \frac{{{\rm{y}}{{\rm{c}}_1}}}{{{\rm{y}}{{\rm{c}}_2}}} = \frac{{{{\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}}^3}}}{{\frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}}} = \frac{{28}}{8} = \frac{7}{2}\)

\(\therefore \frac{{{\rm{y}}{{\rm{c}}_2}}}{{{\rm{y}}{{\rm{c}}_1}}} = \frac{2}{7}\)

In column analogy method, the area of an analogous column for a fixed beam of span L and flexural rigidity EI is taken as

  1. L/EI
  2. L/2EI
  3. L/4EI
  4. L/8EI

Answer (Detailed Solution Below)

Option 1 : L/EI

Methods of Structural Analysis Question 8 Detailed Solution

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Explanation:

  • The column analogous method is the force method used to analyze the indeterminate structure.
  • In the method of column analogous, the actual structure is considered under the action of applied loads and the redundant acting simultaneously.The load on the top of the analogous column is usually the B.M.D. due to applied loads on simple spans and therefore the reaction to this applied load is the B.M.D. due to redundant on simple spans.
  • It is based on the analogy between moments at ends and pressure at the edges of the short column. If both ends of the beam are fixed then the end moments are called fixed end moments.
  • An analogous column will be a short column with cross-section dimensions L and 1/EI.

F1 A.M Madhu 09.04.20 D7

so, area of analogous column = L/EI

The moment distribution method is best suited for

  1. in determinate pin jointed truss
  2. rigid frames
  3. space frames
  4. trussed beam

Answer (Detailed Solution Below)

Option 2 : rigid frames

Methods of Structural Analysis Question 9 Detailed Solution

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Concept:

  • The moment distribution method is best suited for the rigid 2D frame as for rigid 2D frame because only one kind of moment (Mz) is acting at every joint of the structure the moment developed at the joint is distributed to all connected members at that particular joint.
  • The moment distribution method only takes into account the moment effect not the axial force effect.
  • As trusses are designed to carry axial force, so it is not desired to do moment distribution for pin-jointed truss or trussed beam.
  • For space frame, there are three kinds of the moment (Mx, My & Mz) acting at every joint. So, it is very difficult and inconvenient to do moment distribution for spaced frame.

A fixed beam of uniform section is carrying a point load at its mid span. If the moment of inertia of the middle half-length is now reduced to half of its precious value then the fixed end moments will -

  1. Increase
  2. Decrease
  3. Remain constant
  4. Change their direction

Answer (Detailed Solution Below)

Option 1 : Increase

Methods of Structural Analysis Question 10 Detailed Solution

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Explanation:

As we know that 

Fixed end moments correspond to reactions at support such that the slope at supports will be zero 

As conjugate beam method 

The slope at supports is equal to the reaction at supports of the conjugate beam.

So we can predict that the reaction ay support of the conjugate beam is inversely proportional to the moment of inertia (MOI)

Greater MOI  of the beam, lesser will be support reaction of conjugate beam hence lesser will be the slope.

So slope decreased with an increase in MOI the moment required to make slope zero will be lesser means the fixed end moment will increase. 

Which of the following is not the displacement method?

  1. Slope deflection method
  2. Column analogy method
  3. Moment distribution method
  4. Kani’s method

Answer (Detailed Solution Below)

Option 2 : Column analogy method

Methods of Structural Analysis Question 11 Detailed Solution

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Concept:

When in addition to equilibrium equations, compatibility equations are used to evaluate the unknown reactions and internal forces in any structure, such analysis is called indeterminate analysis, and such structure is called indeterminate structure.

We have two distinct methods of analysis for such an indeterminate structure:

a. Force method of analysis

b. Displacement method of analysis

Difference between force method and Displacement method:

Force Method / Flexibility method

Displacement Method / Stiffness method

Used when DS < Dk

Used when Ds > Dk

Forces are redundant or unknowns

Displacements are redundant or unknowns

Equilibrium equations and compatibility equations are required for the solution.

Equilibrium equations and compatibility equations are required for the solution

Starts with equilibrium equations and Then forces are found using compatibility equations.

Starts with compatibility equations and then displacements are found using equilibrium equations.

No. of unknowns = D(Degree of static indeterminacy)

No. of unknowns = D(Degree of kinematic indeterminacy)

Not suitable for programming as no iterations required.

Suitable for programming

Examples:

1. Method of consistent deformation

2. The theorem of least work

3. Column analogy method

4.Flexibility matrix method

Examples:

1.Slope deflection method

2.Moment distribution method

3. Kani’s method

4.Stiffness matrix method

A fixed beam AB is subjected to a triangular load varying from zero at end A to W per unit length at end B. What is the ratio of fixed end moment at A to B?

  1. 1/3
  2. 1
  3. 2/3
  4. 3/2

Answer (Detailed Solution Below)

Option 3 : 2/3

Methods of Structural Analysis Question 12 Detailed Solution

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Concept:-

The given figure shows the fixed end moments under given loading:-

F1 Killi 01-02-21 Savita D1

 

The fixed end moment at end A

\(M_A=-{wL^2\over 30}\)

\(M_B={wL^2\over 20}\)

\(\frac{{{M_A}}}{{{M_B}}}\; = {{wL^2\over 30}\over{wL^2\over 20}} \;\frac{2}{3}\)

Hence, the ratio of the moment will be 2/3.

If W is the load on a circular slab of radius R, the maximum circumferential moment at the centre of the slab is

A. \(\frac{{W{R^2}}}{{16}}\)

B. \(\frac{{2W{R^2}}}{{16}}\)

C. \(\frac{{3W{R^2}}}{{16}}\)

  1. A Only
  2. B Only
  3. C Only
  4. Zero

Answer (Detailed Solution Below)

Option 3 : C Only

Methods of Structural Analysis Question 13 Detailed Solution

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\({M_\theta }\; = \frac{W}{{16}} \times \left( {{R^2}\left( {3 + \mu } \right) - {r^2}\left( {1 + 3\mu } \right)} \right)\)

Where Mθ = Circumferential moment

R = radius of slab

μ = Poisson’s ratio

r = any section at a distance r from the centre of the slab

W = load on circular slab

For the maximum circumferential moment at centre

r = 0 and μ = 0

therefore, \({M_\theta } = \frac{W}{{16}} \times 3 \times {R^2} = \frac{{3W{R^2}}}{{16}}\)

Fixed end moment of a propped cantilever due to a concentrated load P at a distance ‘a’ form fixed ends as shown in figure is given by

F3 Savita Engineering 10-8-22 D1

  1. \(\rm \frac{{Pab\left( {L + b} \right)}}{{2{L^2}}}\)
  2. \(\rm \frac{{Pab\left( {L + b} \right)}}{{{L^2}}}\)
  3. \(\rm \frac{{ - Pab\left( {L + b} \right)}}{{2{L^2}}}\)
  4. \(\rm \frac{{ - Pab\left( {L + a} \right)}}{{{L^2}}}\)

Answer (Detailed Solution Below)

Option 3 : \(\rm \frac{{ - Pab\left( {L + b} \right)}}{{2{L^2}}}\)

Methods of Structural Analysis Question 14 Detailed Solution

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Explanation:

F3 Savita Engineering 10-8-22 D2

 

F3 Savita Engineering 10-8-22 D3

fixed end moments

Fixed end

\(M_{FAB}= \frac{{Pab(L + b)}}{{2{L^2}}}\)

So \(M_{FAB} = \frac{-{Pab(L + b)}}{{2{L^2}}}\)

Negative sign as anticlockwise rotation is taken

A beam of span 5 m, fixed at A and B, carries a point load of 50 kN at 2 m from 'A'. The fixed end moments at the supports 'A' and 'B', respectively, are;

  1. 24 kNm clockwise and 36 kNm clockwise
  2. 24 kNm anticlockwise and 36 kNm anticlockwise
  3. 36 kNm clockwise and 24 kNm anticlockwise
  4. 36 kNm anticlockwise and 24 kNm clockwise

Answer (Detailed Solution Below)

Option 4 : 36 kNm anticlockwise and 24 kNm clockwise

Methods of Structural Analysis Question 15 Detailed Solution

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Concept:

The FEM’s when a beam is subjected to point load and clear loading such as uDL is given below:

F2 Abhishek M 11.2.21 Pallavi D5

 

F2 Abhishek M 11.2.21 Pallavi D6

Calculation:

Given:

F2 Abhishek M 11.2.21 Pallavi D7

a = 2m

b = 3m

L = a + b = 5m

P = 50 kN

\(M_{AB}^F = - \frac{{P{b^2}a}}{{{L^2}}} = - \frac{{50\; \times\; {{\left( 3 \right)}^2}\; \times \;2}}{{{{\left( 5 \right)}^2}}} = - 36\;kNm\;\left( \uparrow \right)\)

(-ve sign means that \(M_{AB}^F\) is in Anticlockwises

\(M_{AB}^F = + \frac{{P{a^2}b}}{{{L^2}}} = \frac{{50\; \times \;{{\left( 2 \right)}^2}\;\times\; 3}}{{{{\left( 5 \right)}^2}}} = + 24\;kN\;M\left( \downarrow \right)\)

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