← MES Jr. Grade

Past Questions — 2016

MES Jr. Grade • 2016 • Civil Engineering Paper-I

A beam of length L is fixed at one end and simply supported at the other end. If EI is the flexural rigidity, the stiffness factor of the beam is

A) EIL
B) 2EI/L
C) 3EI/L
D) 4EI/L
For a beam fixed at one end and simply supported at the other, the end stiffness factor is the same as the fixed-end stiffness when the far end is partially restrained. The standard stiffness factor for a fixed-pinned beam is 4EI/L. By comparison, a cantilever fixed-free beam has stiffness 3EI/L, and a beam fixed at both ends has higher rotational restraint. Therefore, the correct answer is 4EI/L.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

The slenderness ratio of a steel member carrying compressive loads resulting from dead and imposed loads shall not exceed

A) 160
B) 180
C) 200
D) 220
For steel members carrying compressive loads due to dead and imposed loads, the permissible slenderness ratio is limited to 160 as per standard steel design provisions. This limit is used to prevent excessive buckling and ensure adequate stability under compression. Higher values like 180, 200, and 220 are not the prescribed limit for such compression members.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

“If three coplanar forces acting at a point on a body keep it in equilibrium then each force is proportional to the sine of the angle between the other two forces”. The above statement is known as

A) Triangle law of forces
B) Lami’s theorem
C) Varignon’s theorem
D) Polygon law of forces
The statement says that when three coplanar concurrent forces are in equilibrium, each force is proportional to the sine of the angle between the other two forces. This is exactly **Lami’s theorem**: F1/sin(α) = F2/sin(β) = F3/sin(γ) where each angle is the angle between the other two forces. Why others are incorrect: - **Triangle law of forces**: gives the resultant of two forces by the third side of a triangle, not this sine relationship. - **Varignon’s theorem**: deals with moments of forces, not equilibrium of three concurrent forces. - **Polygon law of forces**: states that if several forces acting at a point are in equilibrium, they can be represented by the sides of a closed polygon. So the correct answer is **Lami’s theorem**.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

Which of the following is the reason of unsymmetrical bending?

A) The section is symmetrical but the load line is inclined to both principal axes.
B) The section is unsymmetrical and the load line is along centroidal axis.
C) neither (a) nor (b)
D) both (a) and (b)
Unsymmetrical bending occurs when the bending load does not act in a principal plane or when the section itself is unsymmetrical so that bending is not about a single principal axis. In option (a), even though the section is symmetrical, the load line is inclined to both principal axes, causing bending about more than one axis. In option (b), an unsymmetrical section under a load along the centroidal axis can also produce bending about non-principal axes. Therefore, both situations can lead to unsymmetrical bending.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

The ratio of linear compressive stress and compressive strain is termed as

A) Modulus of rigidity
B) Modulus of elasticity
C) Poisson’s ratio
D) Bulk Modulus of elasticity
Stress is force per unit area, and strain is the corresponding deformation per unit original dimension. The ratio of linear compressive stress to compressive strain gives Young’s modulus, also called the modulus of elasticity. Modulus of rigidity relates shear stress to shear strain, Poisson’s ratio is the ratio of lateral strain to longitudinal strain, and bulk modulus relates hydrostatic stress to volumetric strain.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

Generally influence lines are drawn for a/an X load moving on the span. Find the correct replacement for X.

A) zero
B) unit
C) eccentric
D) uniformly distributed
Influence lines are generally drawn for a **unit load** moving over the span. The purpose is to determine how a response quantity such as reaction, shear force, or bending moment at a point varies as the load moves. Using a unit load makes the influence ordinate directly equal to the effect at that point. Hence, the correct replacement for X is **unit**. The other options are incorrect because influence lines are not typically drawn for zero load, eccentric load, or uniformly distributed load in the general definition.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

Equilibrium conditions for coplanar concurrent forces are

A) ΣFx=0
B) ΣFy=0
C) both (a) and (b)
D) none of these
For a system of coplanar concurrent forces to be in equilibrium, the resultant force must be zero in both the x-direction and the y-direction. Hence, the equilibrium conditions are ΣFx = 0 and ΣFy = 0. Option (a) alone is incomplete, option (b) alone is incomplete, so the correct choice is both (a) and (b).
MES Jr. Grade • 2016 • Civil Engineering Paper-I

A limit state design should satisfy the serviceability requirements such as deflection and

A) flexure
B) shear
C) torsion
D) cracking
In limit state design, serviceability requirements deal with the satisfactory performance of a structure under normal use. Typical serviceability checks include deflection, cracking, and vibration. Among the given options, cracking is the correct companion requirement to deflection. Flexure, shear, and torsion are modes of structural action or strength limit states, not serviceability requirements.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

Forces whose line of action pass through a common point are called

A) non-parallel forces
B) non-concurrent forces
C) coplanar forces
D) concurrent forces
Forces whose lines of action pass through a common point are called concurrent forces. They may be in the same plane or different planes, but the key condition is that all the lines intersect at one point. The other options are incorrect: non-parallel forces only means the forces are not parallel; non-concurrent forces do not meet at one point; coplanar forces only means the forces lie in the same plane.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

Coal, petroleum, air and water are used in the production of this type of plastics. Identify the type.

A) Acrylic
B) Vinyl
C) Poly Vinyl Acetate (PVA)
D) Poly Vinyl Chloride (PVC)
Acrylic plastics are synthetic plastics produced from petrochemical raw materials, commonly described in exam contexts as being made from coal, petroleum, air, and water. Among the options, acrylic is the correct type. Vinyl, PVA, and PVC are other plastic types, but this specific raw-material description is classically associated with acrylic plastics.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

The permissible stresses in steel reinforcement (in N/mm²) in Mild Steel Bars Conforming to Grade I of IS 432 (Part I) in Tension upto and including 20 mm diameter is

A) 130
B) 140
C) 190
D) 230
For Mild Steel Bars conforming to Grade I of IS 432 (Part I), the permissible tensile stress for bars up to and including 20 mm diameter is 130 N/mm². The other values correspond to different categories or are not the standard permissible stress for this specific grade and diameter range.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

For practical purposes, Portland cement may be considered as composed of four principal compounds namely Lime, Silica, Iron Oxide and

A) Magnesia
B) Soda
C) Alumina
D) Sulphur Oxide
Portland cement is practically considered to be composed of four principal compounds: lime, silica, alumina, and iron oxide. Hence the missing component is alumina. Magnesia, soda, and sulphur oxide are not counted among the four principal compounds in this standard classification.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

Sometimes the light entering from the windows in the walls of a building is inadequate. In this case, some more windows are provided on the flat roofs to admit more light into the room. This type of window is known as

A) Skylight
B) Dormer window
C) Clerestorey window
D) Lantem light
A window provided on a flat roof to admit additional light into a room is called a skylight. Dormer windows project from a sloping roof, clerestory windows are placed in the upper part of walls, and lantern lights are raised structures on roofs used for lighting and ventilation.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

This is the most common type of floor covering in Indian buildings and it is also known as Indian Patent Stone flooring. Identify the type.

A) Concrete floor covering
B) Stone floor covering
C) Tile floor covering
D) Mosaic floor covering
Indian Patent Stone flooring is another name for mosaic flooring. It is a common floor finish in Indian buildings, made by laying chips of marble, stone, or other aggregates in cement mortar and polishing the surface. Concrete flooring, stone flooring, and tile flooring are different types and are not known by this name.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

A type of roof, commonly adopted, sloping in two directions is known as

A) Shed roof
B) Gabble roof
C) Hiproof
D) Mansard roof
A roof sloping in two directions and meeting at a ridge is a gable roof (spelled here as 'Gabble roof'). It is one of the most commonly adopted roof types. Shed roof slopes in only one direction, hip roof slopes in four directions, and mansard roof is a double-sloped roof on each side.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

In the case of sections near supports of continuous beams, which of the following statements is true?

A) compression at top fibre
B) tension at top fibre
C) compression at neutral axis
D) tension at neutral axis
In a continuous beam, the regions near supports usually develop negative bending moment (hogging moment). Under hogging moment, the top fibres are in tension and the bottom fibres are in compression. Therefore, near the supports of continuous beams, the correct statement is tension at top fibre. The other options are incorrect because compression at the top fibre is associated with sagging moment regions, and stress is not stated correctly in terms of the neutral axis.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

A rigid body in translation

A) can move along a straight or curved path
B) can only move in a straight line
C) cannot move on a circular path
D) must undergo curvilinear motion
In pure translation, a rigid body moves without rotation, so every line in the body remains parallel to its original position. The path of motion may be straight or curved. Hence, a rigid body in translation can move along a straight or curved path. Option (b) is too restrictive, (c) is incorrect because a translated body may follow a circular path, and (d) is wrong because curvilinear motion is not compulsory.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

If b = width and d = depth of section, Moment of inertia of a rectangle about its XX — axis is given by

A) db²/12
B) bd³/12
C) bd³/6
D) bd³/6
For a rectangular section, the moment of inertia about the centroidal XX-axis (axis parallel to the width) is I = bd^3/12, where b is the width and d is the depth. The depth is cubed because the resistance to bending depends on the dimension perpendicular to the axis. Option (a) is incorrect because it gives b² instead of d³, and options (c) and (d) are incorrect because the denominator should be 12, not 6.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

Bending moment at supports in case of simply supported beams is always

A) zero
B) unity
C) less than unity
D) more than unity
In a simply supported beam, the supports are idealized as a pin and a roller, which allow rotation and do not resist bending. Therefore, the bending moment at both supports is always zero. Hence, option (a) is correct. The other options are incorrect because bending moment at supports of a simply supported beam is neither unity, nor generally less than or more than unity.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

If Poisson’s ratio = 1/m, E = Modulus of Elasticity and C = Modulus of Rigidity, then their relation is given by

A) E=C(1+ 1/m)
B) E=2C(1+ 1/m)
C) E=C(1+2/m)
D) E=2C(1+2/m)
For an isotropic elastic material, the relation between Young’s modulus (E), modulus of rigidity (C or G), and Poisson’s ratio (ν) is: E = 2G(1 + ν) Here, Poisson’s ratio ν = 1/m and modulus of rigidity is given as C. So, E = 2C(1 + 1/m) Hence, option (b) is correct. The other options do not match the standard elastic relation.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

The sum of rotation factors at a point in Kani’s Method of design is replacement for X.

A) 0
B) -1/2
C) 1/2
D) 1
In Kani’s method, the rotation factor is defined as the fraction of the unbalanced moment carried over to the adjoining member. At a joint, the algebraic sum of the rotation factors of all members meeting there is 1. Hence the sum of rotation factors at a point is replaced by X = 1. Therefore, option (d) is correct.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

A beam of length L, fixed at both ends, carries a uniformly distributed load of w per unit length throughout the span. The bending moment at the ends is

A) wL²/8
B) wL²/12
C) wL²/16
D) wL²/24
For a beam fixed at both ends and carrying a uniformly distributed load over the full span, the standard fixed-end moment at each support is wL²/12 (hogging). This is a well-known result from strength of materials. Therefore, the bending moment at the ends is wL²/12. The other options correspond to different loading/support conditions and are not correct here.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

The force polygon for coplanar concurrent forces

A) must close
B) must not close
C) may or may not close
D) any of these
For coplanar concurrent forces, the force polygon is a graphical representation of the vector sum of the forces. If the forces are in equilibrium, their resultant is zero, so the polygon closes. If the forces are not in equilibrium, the resultant is not zero, so the polygon does not close. Therefore, the force polygon may or may not close depending on whether the system is in equilibrium.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

While planning buildings, the minimum distance of a septic tank from any source of drinking water must be

A) 22 metres
B) 20 metres
C) 18 metres
D) 16 metres
The recommended minimum separation between a septic tank and any source of drinking water is 20 metres to prevent contamination of the water supply. Among the given options, 20 metres is the standard safe distance. The other values (22 m, 18 m, 16 m) are not the commonly prescribed minimum in sanitary engineering practice.
MES Jr. Grade • 2016 • Civil Engineering Paper-I

As per IS 456 : 2000, for spans upto 10 m, the vertical deflections limits may generally be assumed to be satisfied provided that the span to effective depth ratio in case of simply supported beams is not greater than

A) 7
B) 15
C) 20
D) 26
As per IS 456:2000, the basic span-to-effective depth ratio for deflection control in a simply supported beam is 20. However, for spans up to 10 m, this value may generally be increased by 30%, giving 20 × 1.3 = 26. Hence, the limit is 26. The other options do not match the code provision.
Page 1
← Prev Next →