Test: Theory of Structures - SSC JE MCQ

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.

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).

For equilibrium of any body.

(i) ∑F_H = 0

(ii) ∑F_V = 0

(iii) ∑M_Z = 0

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.

For the two-hinge arch with a span of 40 m and a point load of 62.8 kN at its crown, the horizontal thrust can be determined by recognizing that the arch is symmetric, and the point load is at the crown, causing equal reactions at the supports. The horizontal thrust in a two-hinged arch under a central point load is equal to the vertical reaction times the horizontal distance from the reaction to the load divided by the rise of the arch. If the arch is a perfect semi-circle, the horizontal thrust would be equal to the vertical reaction. a semi-circular arch or similar conditions where the horizontal thrust equals the vertical reaction.

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.

Let D_s = static Indeterminacy

D_s = R - r

R → no. of unknowns

r → No. of equilibrium equations available

If r > R

D_s < 0, system is partially constraintIf

D_s > 0, then system is over stiff.

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.

Beam are the horizontal members which can transmit the load perpendicular to their longitudinal direction and not in axial direction while truss can transmit the load in axial direction only.

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.

D_s = m – 2j + r_e

m = 23

j = 11

∴ D_s = 23 – 22 + 5 = 6

Three-hinged arch is a determinate structure and rise in temperature do not produce any stress.

Force method is useful when D_s < D_k

Displacement method is useful when D_k < D_s­

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.

D_k = 3j – r_e + r_r

= 3 x 4 – 5 + 1 = 8

If members are considered axially rigid

D_k = 8 – 3 = 5

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.

• 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.

Due to sway, the deflection of point B will be equal to that of point C.