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

Explanation:

• The cantilever method is a simplified approximate method used to analyze lateral loads (wind, earthquake) in multi-storey buildings.

• The method assumes that the building behaves like a vertical cantilever beam, with the base fixed and the top free.

• The columns act like longitudinal fibers of the cantilever — resisting bending.

• The method assumes that the beams do not resist axial force — only flexure.

• The axial force in each column is assumed proportional to its horizontal distance from the centroid of all columns in the storey.

• This models the "fiber" action of the cantilever analogy.

Additional Information

• The Portal Method is an approximate, simplified analysis method used to analyze lateral forces (like wind or earthquake) in low to medium-rise framed buildings.

• It is typically used for rigid frames in portal-shaped structures or for frames with a series of columns and beams.

• The structure is assumed to behave like a series of interconnected portals.

• For preliminary analysis and design of multistorey buildings subjected to lateral loads.

• Suitable for buildings with 3 to 5 storeys.

• Common in situations where full structural analysis (moment distribution or matrix methods) is too time-consuming.

Explanation:

• A space frame (or space truss) is a three-dimensional structural system where members are arranged in a triangular configuration to efficiently carry loads in all directions.

• The simplest form of a space truss uses six members and four joints, forming a tetrahedron.

• A tetrahedron is a solid with four triangular faces, four nodes (joints), and six edges (members), making it the most stable 3D shape.

 Additional Information

• A truss is a structural framework made up of straight members connected at joints (nodes), designed to support loads primarily through axial forces—either tension or compression.

• Trusses are commonly used in bridges, roofs, towers, and industrial buildings, where they provide high strength with minimal material by efficiently distributing loads.

• The joints in a truss are typically assumed to be pinned (hinged), meaning the members only carry axial forces and do not resist bending or shear.

• Trusses are classified based on their geometry:

  • Plane trusses lie in a two-dimensional plane (e.g., roof trusses).

  • Space trusses are three-dimensional and can resist forces in all directions (e.g., space frames, tetrahedral units).

Explanation:

m > 2j − 3

• This is the condition for a redundant (statically indeterminate) plane truss.
• When the number of members exceeds what is required for static determinacy, the structure becomes statically indeterminate.
• In such cases, additional compatibility equations are required to solve internal forces.

 Additional Informationm < 2j − 3

• This condition indicates a mechanism or unstable structure, where the number of members is insufficient to maintain structural integrity.
• Such frames will not be able to carry loads effectively without deformation.

• • Condition for a perfect (just-rigid) frame:

  $m=2j-3$

• • Redundant (over-rigid) frame:

  $$
  m > 2j - 3

• • Deficient (unstable) frame:

  $m<2j-3$

Explanation:

The Principle of Superposition applies to linear elastic systems and states that:

When a structure is subjected to multiple forces, the total deformation is the algebraic sum of the deformations caused by each individual force acting alone.

Mathematical Representation:

If a bar is subjected to forces F1​,F2​,F3​,… on sections of length L1​,L2​,L3​,… and cross-sectional areas A1​,A2​,A3​,…, and if the material has modulus of elasticity $E$, then:

Here, each term represents the deformation in an individual segment, and the total deformation is their algebraic (not arithmetic or geometric) sum.

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.

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.

$\theta =\frac{\left[Areaofbendingmomentdiagram\right]}{EI}$

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.

$\delta =B{B}^{\prime }=\frac{\left[Areaofbendingmomentdiagram\right]\times \overline{x}}{EI}$

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:

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:

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:

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.

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

$R_A={\displaystyle \frac{80}{3}}=26.67kN$

R_A = 26.67 kN

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:

$\mathrm{y}{\mathrm{c}}_1=\frac{\mathrm{W}}{3\mathrm{E}\mathrm{I}}\times {\left(\frac{2\mathrm{L}}{3}\right)}^3+\frac{\mathrm{W}}{2\mathrm{E}\mathrm{I}}{\left(\frac{2\mathrm{L}}{3}\right)}^2\times \frac{\mathrm{L}}{3}={\frac{28\mathrm{W}\mathrm{L}}{162\mathrm{E}\mathrm{I}}}^3$

II^nd condition:

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

$\mathrm{y}{\mathrm{c}}_2=\frac{\mathrm{W}}{3\mathrm{E}\mathrm{I}}\times {\left(\frac{\mathrm{L}}{3}\right)}^3+\frac{\mathrm{W}}{2\mathrm{E}\mathrm{I}}{\left(\frac{\mathrm{L}}{3}\right)}^2\times \frac{2\mathrm{L}}{3}=\frac{8\mathrm{W}{\mathrm{L}}^3}{162\mathrm{E}\mathrm{I}}$

$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\left(\mathrm{r}\right)=\frac{\mathrm{y}{\mathrm{c}}_1}{\mathrm{y}{\mathrm{c}}_2}=\frac{{\frac{28\mathrm{W}\mathrm{L}}{162\mathrm{E}\mathrm{I}}}^3}{\frac{8\mathrm{W}{\mathrm{L}}^3}{162\mathrm{E}\mathrm{I}}}=\frac{28}{8}=\frac{7}{2}$

$\therefore \frac{\mathrm{y}{\mathrm{c}}_2}{\mathrm{y}{\mathrm{c}}_1}=\frac{2}{7}$