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

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:

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:

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

 Concept: 

The principle of virtual work for deformable bodies says that the external virtual work applied to a structure must equal the internal virtual work that is caused within the structure,  Wv,e = Wv,i

External Virtual Work = Internal Virtual Work

Virtual External Forces×  Real external deflections = Virtual Internal forces × Real Internal Deformations

Explanation:

• In the Moment Distribution Method, the Distribution Factor (DF) for a member is a measure of how the joint's moment is distributed among the various connected members.
• It helps in determining the proportion of the joint's moment that will be distributed to each member.

The formula to calculate the Distribution Factor (DF) for a member is:

 \(DF = \frac{K_i}{\sum K} \)

Where:

• Ki K_i" id="MathJax-Element-28-Frame" role="presentation" style="position: relative;" tabindex="0">KiKi K_i" id="MathJax-Element-1-Frame" role="presentation" style="position: relative;" tabindex="0">Ki$K_i$ $K_i$ $K_i$  is the stiffness of the considered member at the joint.

• ∑K \sum K" id="MathJax-Element-29-Frame" role="presentation" style="position: relative;" tabindex="0">∑K∑K \sum K" id="MathJax-Element-2-Frame" role="presentation" style="position: relative;" tabindex="0">∑K$\sum K$ $\sum K$ $\sum K$  is the total stiffness of all the members meeting at the joint.

 Additional Information

1. The Moment Distribution Method is an approximate iterative technique used to analyze indeterminate structures, especially continuous beams and frames, that cannot be solved easily using simple static equilibrium equations.
2. It was developed by Dr. Hardy Cross in the early 20th century.
3. The method works by redistributing the bending moments in a structure in such a way that equilibrium is gradually achieved through an iterative process.

Key Concepts in the Moment Distribution Method:

1. Fixed-End Moment (FEM):
   The moment at the ends of a beam or frame when both ends are assumed to be fixed, considering external loads applied to the structure. It is an initial moment, calculated without considering the flexibility of the joints.

2. Distribution Factor (DF):
   The factor used to distribute the moment at a joint to the connected members, based on the stiffness of the members. It represents the proportion of the joint's moment each member will take.

3. Carry-Over Factor:
   After distributing the moment, the unbalanced moment at the joint is "carried over" to the connected members in the next iteration. The carry-over factor is typically 0.5 0.5" id="MathJax-Element-30-Frame" role="presentation" style="position: relative;" tabindex="0">0.50.5 0.5" id="MathJax-Element-3-Frame" role="presentation" style="position: relative;" tabindex="0">0.5$0.5$ $0.5$ $0.5$ for beams with simple supports.

4. Convergence:
   The iterative process continues until the moment values at the joints do not change significantly between iterations. This indicates that the structure is in equilibrium, and the final moment distribution is achieved.

Explanation:

• The Slope-Deflection Method is more efficient than the Force Method because it reduces the number of equations and unknowns.
• It directly works with the displacement (rotation and deflection) at the supports, which simplifies the solution process when compared to the Force Method, where you need to deal with unknown forces and compatibility equations.

 Additional InformationSummary of Slope-Deflection Method 

• Used for analyzing indeterminate beams and frames.

• Based on the relationship between end moments of members and the rotations and displacements at their ends.

• Primary unknowns are joint rotations and displacements, not internal forces.

• Uses slope-deflection equations to express member end moments in terms of rotations, deflections, and fixed-end moments.

• Moment equilibrium is applied at joints to solve for unknown rotations.

• Involves fewer equations than the Force Method, making it more efficient for analyzing complex frames.

• Best suited for structures with multiple spans, frames with or without lateral sway, and beams under various loading conditions.

• Can become tedious for very large or highly complex structures, where matrix methods are preferred.

• Commonly introduced in structural analysis as a foundational method before moving to advanced techniques.

Explanation:

• Clapeyron's Theorem of Three Moments is a method used in structural analysis to calculate the bending moments at the supports of a continuous beam.
• It is a fundamental tool in analyzing statically indeterminate structures, specifically for continuous beams with multiple spans.

 Additional InformationClapeyron's Theorem of Three Moments

• Purpose:

  • It is a method used to calculate the bending moments at the supports of a continuous beam.

  • Helps in analyzing statically indeterminate structures.

• Conditions:

  • Applicable to continuous beams (i.e., beams with more than two supports).

  • Requires knowledge of the beam's span lengths, external loads, and reactions.

• Theorem Statement:

  • The theorem relates the bending moments at three consecutive supports of a continuous beam and provides an equation for each of the internal moments.

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.

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

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.

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

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.

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:

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

II^nd condition:

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

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

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

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

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

\(R_A = \dfrac{80}{3} = 26.67 \ kN\)

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 (-).