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

Explanation:

• In an internally indeterminate truss, the forces depend on the overall geometry and load, not just on the cross-sectional area of individual members.

• Changing the area of a redundant member affects its stiffness, but the force in that member is determined by equilibrium and compatibility conditions.

• Increasing the cross-sectional area does not directly change the internal force; it changes the member's ability to carry that force safely (strength and stiffness), not the force itself.

• Therefore, doubling the area does not affect the force in that member.

 Additional Information

• A truss is said to be statically determinate if its internal forces can be found using only equilibrium equations.

• Redundancy refers to the presence of extra members or supports beyond what is needed for static determinacy.

• A truss with redundant members is called internally indeterminate, meaning forces cannot be found by equilibrium alone; compatibility and deformation analysis are also required.

• Redundant members provide additional load paths, improving structural reliability and stiffness.

• However, redundancy complicates analysis and design because the structure’s response depends on material properties and deformation compatibility.

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: 

Explanation:

K_CB = 4EI/L (Far end is fixed)

KCA = 3EI/L (Far end is pinned)

K₁₁ = KCB + KCA 

      = 4EI/L + 3EI/L  = 7EI/L

Explanation:

Truss:

• They efficiently transfer loads through axial forces in members (tension or compression).
• They reduce self-weight compared to solid beams or girders.

• They provide greater stiffness and strength with less material—thus more economical.

• Commonly used in industrial buildings, railway bridges, airplane hangars, etc.

Additional Information

• Beam: Good for short spans; becomes too deep and heavy for long spans, increasing material cost.

• Plate Girder: Used for medium to long spans, but heavier and more expensive than a truss.

• Arch: Structurally efficient but requires horizontal thrust resistance and complex supports, making it less economical in general.

Explanation:

In a two-hinged parabolic arch, temperature rise causes the arch to expand horizontally. Since the supports are hinged and cannot move, this thermal expansion leads to:

• Increased horizontal thrust to maintain the equilibrium.

• This is due to the development of internal compressive forces along the arch axis.

Additional Information

• Two-hinged arches are statically indeterminate.

• When temperature increases, the crown tries to rise due to thermal expansion.

• But the hinges resist this deformation → hence internal force builds up → horizontal thrust increases.

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.

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.

Calculation:

ΣV = 0, V_A + V_B = 150 kN

ΣH = 0, H_A = H_B = H

At support B, ΣM_B = 0

V_A × 20 - 150 × 16 = 0

V_A = 120 kN

V_B = 150 - 120 = 30 kN

At central hinge, ΣM_C = 0

V_A × 10 - H × 4 - 150 × 6 = 0

H × 4 = 120 × 10 - 150 × 6

H = 75 kN

So, the vertical reaction at A and the horizontal thrust are 120 kN and 75kN respectively.

Concept:

A two-dimensional structure in general is classified as a statically indeterminate structure if it cannot be analyzed by conditions of static equilibrium.

The conditions of equilibrium for 2D structures are:

1. The Sum of vertical forces is zero (∑Fy = 0).
2. The Sum of horizontal forces is zero (∑Fx = 0).

• The Sum of moments of all the forces about any point in the plane is zero (∑Mz = 0).

​
Simply supported beam:

Number of unknowns = 3

Degree of static indeterminacy = 3 - 3 = 0. Hence it is statically determinate.

Cantilever beam:

Number of unknowns = 3

Degree of static indeterminacy = 3 - 3 = 0. Hence it is statically determinate.

Three hinged arches:

Number of unknown = 4

Degree of static indeterminacy = 4 - 3 -1 = 0. (Additional equation due to internal hinge ∵ B.M = 0)

Hence it is statically determinate.

Two hinged arches:

Number of unknown = 4

Degree of static indeterminacy = 4 - 3 = 1.

Hence it is statically indeterminate.

Concept:

Kinematic Indeterminacy:

It is the total number of possible degrees of freedom of all the joints.

Dk = 3J - r + h (For beam & portal frame)

Dk = 2J - r + h (For truss structure)

Where,

Dk = Kinematic Indeterminacy,

r = No. of unknown reactions

h = No. of plastic hinges

J = No. of joints

Calculation:

Given;

J = 2

r = 1 + 3 = 4 (1 vertical reaction at roller support, and 1 vertical, 1 horizontal and 1 moment reaction at fixed support)

h = 0

D_k = 3 × 2 - 4 = 2

D_k = 2

Concept:

For a truss, Degree of static indeterminacy = m + r - 2j

Where,

m = number of members, r = number of reactions, and j = number of Joints

Calculation:

In the given truss,

Number of members(m) = 11, 

Number of Joints(j) = 6,

number of reactions(r) = 4

Degree of static indeterminacy = m + r - 2j

= 11 + 4 - (2 × 6)

= 15 - 12

= 3.

Hence, In the figure, the static degree of indeterminacy is 3.

For the external stability of structures following conditions should be satisfied:

1) All reactions should not be parallel

2) All reactions should not be concurrent

3) The reaction should be nontrivial

4) There should be a minimum number of externally independent support reactions

5) For stability in 3D structures, all reactions should be non-coplanar, non-concurrent and non-parallel

∴ If all the reactions acting on a planar system are concurrent in nature, then the system is unstable.

Concept:

Influence line diagram:

An influence line for a given function, such as a reaction, axial force, shear force, or bending moment, is a graph that shows the variation of that function at any given point on a structure due to the application of a unit load at any point on the structure. ILD can be drawn for statically determinate as well as indeterminate structures.

Advantages of drawing ILD are as follows:

i) To determine the value of a quantity (shear force, bending moment, deflection, etc.) for a given system of loads on the span of the structure.

ii) To determine the position of a live load for the quantity to have the maximum value and hence to compute the maximum value of the quantity.

Explanation:

Rolling load = 40 kN

Maximum positive shear force:

When rolling load will be acting at point A then the shear force will be maximum.

Maximum shear force = Magnitude of load × ordinate of ILD under the load

= 40 × 1 = 40 kN (+)

Maximum negative shear force:

When rolling load will be acting at point B then the shear force will be maximum.

Maximum shear force = Magnitude of load × ordinate of ILD under the load

= 40 × 1 = 40 kN (-).

Explanation

Given, 3 joints and 4 members so it signifies it a frame

For a given frame:

We know in a frame the relation between members and joints is given by 

m = 2j - 3 

Where m = members , j = joints

Given, m = 4, j = 3 

Let's check the relation

m = 2 × 3 - 3 = 3, so we get m = 3

But we have 4 members i.e 1 in excess

∴ the answer is redundant.

I^st condition:

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

II^nd condition:

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

Explanation:

Given:

L = 30 m

h = 5 m

∑F_H = 0 

H_A = H_B = H

∑F_V = 0 

R_A + R_B = 40 kN     ---(1)

∑M_B = 0 

R_A × 30 - 40 × 20 = 0

R_A = 26.67 kN