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All questions of Structural Analysis for Civil Engineering (CE) Exam

If a system has more equations of equilibrium than no. of forces, then the system is:-
  • a)
    Improperly constrained
  • b)
    Partially constrained
  • c)
    Stable
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Neha Mukherjee answered
Let Ds = static Indeterminacy
Ds = R - r
R → no. of unknowns
r → No. of equilibrium equations available
If r > R
Ds < 0, system is partially constraintIf
Ds > 0, then system is over stiff.
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Consider the following statements:
The principle of superposition is not applicable when
1. The material does not obey Hooke’s law
2. The effect of temperature changes are taken into consideration
3. The structure is being analysed for the effect of support settlement
Which of these statement (s) is/are correct?
  • a)
    1 only
  • b)
    1 and 2
  • c)
    2 and 3
  • d)
    1, 2 and 3
Correct answer is option 'D'. Can you explain this answer?

Jaideep Malik answered
According to the principle of superposition, for a linearly elastic structure, the load effects caused by two or more loadings are the sum of the load effects caused by each loading separately. Note that the principle is limited to:
• Linear material behaviour only;
• Structures undergoing small deformations only (linear geometry).
It is not applicable when:
1. The material does not obey Hooke’s law.
2. The effect of temperature changes are taken into consideration.
3. The structure is being analysed for the effect of support settlement.

A two hinge arch of span 40 m carries a point load of 62.8 KN at its crown. The Horizontal thrust in the arch is (in KN)
  • a)
    20
  • b)
    40
  • c)
    31
  • d)
    62.8
Correct answer is option 'A'. Can you explain this answer?

Sparsh Unni answered
Given data:
Span of the arch (L) = 40 m
Point load at the crown (P) = 62.8 kN

Calculation:
To find the horizontal thrust in the arch, we can use the formula:
Horizontal thrust (H) = P(L/2) / L

Substitute the values:
H = 62.8 * (40/2) / 40
H = 62.8 * 20 / 40
H = 31.4 kN
Therefore, the horizontal thrust in the arch is 31.4 kN, which is closest to option 'A' (20 kN).

Clapeyron’s theorem is associated with the analysis of
  • a)
    simply supported beams
  • b)
    fixed beams
  • c)
    continuous beams
  • d)
    cantilever beams
Correct answer is option 'C'. Can you explain this answer?

Rajat Patel answered
  • Beams that have more than one span are defined as continuous beams. Continuous beams are very common in the bridge and building structures.
  • When a beam is continuous over many supports and the moment of inertia of different spans is different, the force method of analysis becomes quite cumbersome.
  • However, the force method of analysis could be further simplified for this particular case (continuous beam) by choosing the unknown bending moments at the supports as unknowns.
  • One compatibility equation is written at each intermediate support of a continuous beam in terms of the loads on the adjacent span and bending moment at left, center (the support where the compatibility equation is written) and rigid supports.
  • Two consecutive spans of the continuous beam are considered at one time. Since the compatibility equation is written in terms of three moments, it is known as the equation of three moments.
  • In this manner, each span is treated individually as a simply supported beam with external loads and two end support moments.
  • For each intermediate support, one compatibility equation is written in terms of three moments. Thus, we get as many equations as there are unknowns. Each equation will have only three unknowns.
  • Clapeyron first proposed this method in 1857 and it is widely known as the Clapeyron theorem.

The unit load method used in structural analysis is
  • a)
    applicable only to statically indeterminate structures
  • b)
    another name for stiffness method
  • c)
    an extension of Maxwell’s reciprocal theorem
  • d)
    derived from Castigliano’s theorem
Correct answer is option 'D'. Can you explain this answer?

Rajat Patel answered
The unit load method is extensively used in the calculation of deflection of beams, frames and trusses. Theoretically this method can be used to calculate deflections in statically determinate and indeterminate structures. However it is extensively used in evaluation of deflections of statically determinate structures only as the method requires a priori knowledge of internal stress resultants. The unit load method used in structural analysis is derived from Castigliano’s theorem.

Which of the following material is not used in making trusses?
  • a)
    Wooden struts
  • b)
    Metal bars
  • c)
    Channel
  • d)
    Concrete
Correct answer is option 'D'. Can you explain this answer?

Ashwin Gupta answered
Concrete is a brittle material and is unable to take tension. In Truss both tension and compression force is available, that’s why concrete is not used in making truss. While metal bars and channels are made of material which have both tensile as well as compressive strength.

Force in the member BC of the truss shown in the given figure is
  • a)
    5 t, tensile
  • b)
    zero
  • c)
    2.88 t, compressive
  • d)
    5 t, compressive
Correct answer is option 'D'. Can you explain this answer?

Prem Prakash answered
If we take joint B and applied joint method as 
submission of horizontal is equal to zero
5t+Fbcsin30-Fabsin30=0
Fbc-Fab = 10 equation 1
submission of vertical force equal to zero 
Fabcos30+Fbccos30=0
Fab=-Fbc equation 2
if we solve equation 1 and 2 we get Fbc = 5t

The force in CD of the truss shown in Fig, is
  • a)
    3t compression
  • b)
    3t tension
  • c)
    zero
  • d)
    1.5t compression
Correct answer is option 'C'. Can you explain this answer?

Telecom Tuners answered
In a truss, the method of joints can be used to find the forces in the truss members. At joint C, since there are only two members meeting (CD and CE) and assuming there are no external loads or supports at C, the sum of horizontal forces must equal zero. Member CD is horizontal, so the force in member CD must be zero to satisfy equilibrium (since there's no other horizontal member to balance a force in CD). This is why the force in member CD is zero.

The Muller-Breslau principle in structure analysis is used for
  • a)
    Drawing influence line diagram for any force function
  • b)
    Superimposition of load effects
  • c)
    Writing virtual work equation
  • d)
    None of the above
Correct answer is option 'A'. Can you explain this answer?

Rohan Singh answered
Muller Breslau Principle: The Muller-Breslau principle states that the ILD for any stress function in a structure is represented by its deflected shape obtained by removing the restrained offered by that stress function and introducing a directly related generalised unit displacement in the direction of that stress function.

What is the degree of static indeterminacy of the structure shown in figure below?
  • a)
    3
  • b)
    4
  • c)
    1
  • d)
    2
Correct answer is option 'A'. Can you explain this answer?

Crack Gate answered
The degree of static indeterminacy (DSI) of a structure is the number of extra reactions or internal forces that are not required to maintain equilibrium. It can be calculated using the formula DSI=r−s, where r is the number of reaction components (including internal hinges) and s is the number of equilibrium equations (which is 3 for planar structures). The DSI indicates how many additional equations would be needed to solve for the reactions using statics alone. A structure is statically indeterminate if DSI>0.

When on end of a fixed beam deflects by δ, then the bending moment at deflected end is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Engineers Adda answered
The bending moment at the deflected end of a fixed beam due to a deflection δ is a theoretical concept that is derived from the beam's stiffness (which depends on the modulus of elasticity E and moment of inertia I) and the geometry of the beam (length L).

What are the bending moments at ends A and B of uniform fixed beam AB as shown in figure when two concentrated loads acts at 1/3 spans?
  • a)
    2/9 wL
  • b)
    4/9 wL
  • c)
    6/9 wL
  • d)
    8/9 wL
Correct answer is option 'A'. Can you explain this answer?

Gate Funda answered
we have a fixed beam AB with two concentrated loads acting at one-third of the span from each support.
To find the bending moment at ends A and B, we can use the principle of superposition, which is applicable to linear systems like this one. The total moment at any section is the algebraic sum of the moments caused by each load acting separately.
The fixed-end moments (FEM) for a point load W at a distance 'a' from the support on a beam of length 'L' is given by: 
For the given case, 'a' is L/3 and 'W' is the load applied.

Chapter doubts & questions for Structural Analysis - Mock Test Series of SSC JE Civil Engineering 2025 2025 is part of Civil Engineering (CE) exam preparation. The chapters have been prepared according to the Civil Engineering (CE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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