← MES Jr. Grade
Past Questions
A body will be in equilibrium condition when external effect on a body is:
A)
Minimum
B)
Maximum
C)
Zero
D)
None
A body is in equilibrium when the resultant external force (and, for rotational equilibrium, the resultant moment) acting on it is zero. This means there is no net tendency to translate or rotate. Hence, the correct option is 'Zero'. The other options are incorrect because 'Maximum' and 'Minimum' do not define equilibrium, and 'None' is wrong since equilibrium specifically requires zero net external effect.
For a given depth, which of the following I-section represents the heaviest section?
A)
ISMB
B)
ISLB
C)
ISHB
D)
ISWB
For a given depth, the heaviest I-section is the one with the largest cross-sectional area (and hence greatest weight per metre). Among standard Indian sections, ISHB (Indian Standard Heavy Beam) has the greatest flange width and thickness, so it carries the maximum steel area and is the heaviest. ISMB is medium weight, ISLB is light weight, and ISWB is wide flange but not heavier than ISHB for the same depth in standard designation.
Maximum deflection permissible as per IS : 800-1984 is restricted to
A)
span/160
B)
span/320
C)
span/325
D)
span/384
As per IS 800:1984, the maximum permissible deflection for beams under service loads is commonly limited to span/325 for general members, but the stricter limit of span/360 or span/250 may apply depending on the member and finish. In many exam contexts, the standard quoted maximum deflection permissible is span/325 for overall deflection criteria; however, among the given options, the intended IS 800:1984 limit is span/325? Wait—this question asks specifically for the maximum deflection permissible as per IS 800:1984, and the commonly cited limit for beams is span/325 for total deflection and span/250 for cantilevers. Since span/325 is present, it is the correct choice. The other options are incorrect because span/384 is not the IS 800:1984 serviceability limit, span/320 is not a standard code value, and span/160 is far too large (too permissive) compared with code limits.
Corrosion resisting ability of structural steel can be enhanced by applying a finishing coat of
A)
zinc and chromium
B)
zinc and aluminium
C)
Aluminium and nickel
D)
Nickel and chromium
The corrosion-resisting ability of structural steel is commonly improved by applying protective metallic coatings such as zinc and aluminium. These coatings act as barrier layers and, in the case of zinc, also provide sacrificial protection to the steel, reducing rusting effectively. Hence, option (b) zinc and aluminium is correct.
Why the others are incorrect:
- (a) Zinc and chromium: chromium is used for decorative and corrosion-resistant plating on smaller components, but this is not the standard finishing coat combination for structural steel.
- (c) Aluminium and nickel: both can offer corrosion resistance in specific applications, but this pair is not the usual finishing coat for structural steel.
- (d) Nickel and chromium: this combination is mainly associated with decorative or wear-resistant plating, not the typical corrosion-protective finishing coat for structural steel.
A column is defined as short, when the slenderness ratio for the column (L_eff / r) is less than
A)
100
B)
50
C)
200
D)
None of these
A column is classified as a short column when its slenderness ratio (L_eff/r) is less than 50. This is the standard limit used in structural engineering to distinguish short columns from long/slender columns, which are more prone to buckling. Hence option (b) 50 is correct. Option (a) 100 is too high for the short-column limit, option (c) 200 is far beyond the accepted threshold, and option (d) None of these is incorrect because 50 is indeed the correct value.
For the manufacture of Porland cement, the proportions of raw materials used are
A)
lime 63%, silica22%, other ingredients 15%
B)
lime 50%, silica35%, other ingredients 15%
C)
lime 40%, silica40%, other ingredients 20%
D)
All the above
The correct proportion for Portland cement raw mix is approximately lime 63%, silica 22%, and alumina/iron oxide and other ingredients about 15%. This matches option (a). Option (b) is incorrect because lime is too low and silica too high for standard Portland cement composition. Option (c) is also incorrect because both lime and silica percentages are not in the usual proportions. Option (d) cannot be correct because the listed combinations are not all valid; only option (a) matches the standard composition.
The minimum strength of the mortar used in load bearing brick masonry, is
A)
50 N/cm2
B)
100 N/cm2
C)
150 N/cm2
D)
200 N/cm2
The minimum strength of mortar used in load-bearing brick masonry is generally taken as 100 N/cm2 (about 1 N/mm2). This is the lowest commonly accepted mortar strength for structural brickwork where the masonry must safely carry compressive loads.
Why the others are incorrect:
- 50 N/cm2 is too low for load-bearing masonry and is generally used only for weaker, non-structural applications.
- 150 N/cm2 is stronger than the minimum requirement, but it is not the minimum standard asked here.
- 200 N/cm2 is also higher than the minimum and is used where greater strength is needed, not as the least acceptable value.
The maximum ratio of span to depth of a slab simply supported and spanning in two directions, is
A)
25
B)
30
C)
35
D)
40
For a simply supported slab spanning in two directions, the span-to-effective depth ratio is limited to about 30 for deflection control. This is the maximum permissible ratio commonly used in design codes and standard exam references. Hence, option (b) 30 is correct.
Why others are incorrect:
- 25 is too low; it is more conservative than the usual limit for a simply supported two-way slab.
- 35 is higher than the standard permissible limit, so it would not satisfy deflection control criteria.
- 40 is also excessive and is not the accepted maximum ratio for this case.
For M150 mix concrete, according to I.S specifications, local bond stress, is
A)
7 kg/cm2
B)
10 kg/cm2
C)
15 kg/cm2
D)
20 kg/cm2
For M150 concrete, the permissible local bond stress as per I.S. specifications is 7 kg/cm2. This is the standard value used for ordinary M150 grade concrete in design tables. Option (b) 10 kg/cm2 is too high for M150 and is associated with stronger concrete grades. Options (c) 15 kg/cm2 and (d) 20 kg/cm2 are much higher values and are not permitted for M150 concrete under I.S. bond stress recommendations.
Which of the following is a scalar quantity?
A)
Force
B)
Speed
C)
Velocity
D)
Acceleration
Speed is a scalar quantity because it has magnitude only and no direction. Acceleration, velocity, and force are all vector quantities because each has both magnitude and direction. Therefore, the correct answer is speed (option b).
The carryover factor in a prismatic member whose far end is hinged is
A)
0
B)
½
C)
¾
D)
1
The carryover factor is the ratio of the moment induced at the far end to the moment applied at the near end when one end of a prismatic member is rotated. For a prismatic member with the far end hinged (pinned), the hinged end cannot resist any bending moment, so no moment is carried over to that end. Hence, the carryover factor is 0.
Why the other options are incorrect:
- b (1/2): This is the carryover factor for a prismatic member with the far end fixed, not hinged.
- c (3/4): Not a standard carryover factor for this boundary condition.
- d (1): This would imply full carryover of moment, which is not possible at a hinged end because it cannot develop moment resistance.
In ordinary residential and public buildings, the damp proof course is generally provided at
A)
Ground level
B)
Plinth level
C)
Water table level
D)
Midway between ground level and water table
The damp proof course (DPC) in ordinary residential and public buildings is generally provided at the plinth level, i.e., just above ground level at the base of the walls. Its purpose is to prevent rising damp from the ground from entering the wall masonry. Option (a) ground level is incorrect because the DPC is not placed exactly at ground level, where it could be damaged or bridged by soil. Option (c) water table level is incorrect because the water table is far below the building base in many cases and is not the standard reference level for DPC placement. Option (d) midway between ground level and water table is also incorrect because DPC is not determined by the water table depth; it is normally fixed at plinth level for practical and protective reasons.
If a number of forces are acting at a point, their resultant will be inclined at an angle θ with the horizontal, such that
A)
tanθ = ΣH / ΣV
B)
tanθ = ΣV / ΣH
C)
tanθ = ΣHx ΣV
D)
tanθ = ΣH / ΣV
For a system of concurrent forces, resolve all forces into horizontal and vertical components. If the resultant makes an angle θ with the horizontal, then its horizontal component is ΣH and its vertical component is ΣV. By trigonometry, tanθ = (vertical component)/(horizontal component) = ΣV/ΣH. Hence option b is correct.
Why the others are incorrect:
- a and d give tanθ = ΣH/ΣV, which is the reciprocal of the correct relation; that would correspond to measuring the angle from the vertical, not the horizontal.
- c is not a valid standard expression as written; it appears to be a formatting error and does not correctly state the ratio for the angle with the horizontal.
The deformation of a spring produced by a unit load is called
A)
Stiffness
B)
Flexibility
C)
Influence coefficient
D)
Unit strain
The deformation produced in a spring by a unit load is called its flexibility (also known as compliance). Flexibility is the displacement per unit force, i.e., the inverse of stiffness.
Why the others are incorrect:
- a) Stiffness is the load required to produce unit deformation, not the deformation due to unit load.
- c) Influence coefficient is a general structural analysis term relating displacement at one point due to a unit load at another point; it is not the specific term for a spring’s deformation under unit load.
- d) Unit strain is deformation per unit original length, which is unrelated to the spring’s load-deformation property asked here.
Which of the following stresses is used for identifying the quality of structural steel,
A)
Ultimate stress
B)
Yield stress
C)
Proof stress
D)
None of these
Yield stress is used to identify the quality of structural steel because structural steel is generally a ductile material, and its performance is governed mainly by the stress at which permanent deformation begins. The yield stress is therefore the standard basis for grading and specifying structural steel.
Why the others are incorrect:
- Ultimate stress: This is the maximum stress before failure, but it is not the usual criterion for quality identification of structural steel.
- Proof stress: This is mainly used for materials that do not show a clear yield point, such as some non-ferrous alloys; structural steel is typically assessed by yield stress.
- None of these: Incorrect because yield stress is the correct identifying parameter.
In a portal frame of strong column and weak beam fixed at both ends
A)
The beam does not deform and the column buckled in half-sine wave.
B)
The column does not buckle but the beam deformed in full-sine wave.
C)
The beam deforms in full-sine wave and column deforms in half-sine wave.
D)
Both the beam and column deform in full-sine wave.
In a strong-column, weak-beam portal frame with both ends fixed, the beam is the weaker member and therefore reaches its critical deformation first. A fixed-ended beam under lateral instability/deflection forms a full-sine wave shape, while the columns, being stronger and restrained at both ends, deform in a half-sine wave pattern. Hence option (c) is correct.
Why others are wrong:
- (a) is incorrect because the beam does deform; it does not remain undeformed.
- (b) is incorrect because the column can also deform; it is not true that the column does not buckle at all.
- (d) is incorrect because the column does not generally take a full-sine wave shape in this fixed-ended portal-frame behavior; the column deformation is half-sine wave.
The moment of inertia of a circular section about its diameter (d) is
A)
πd³/16
B)
πd³/32
C)
πd⁴/32
D)
πd⁴/64
For a solid circular section, the moment of inertia about any centroidal diameter is I = πd^4/64. This comes from the standard formula I = πr^4/4 and substituting r = d/2, giving I = π(d/2)^4/4 = πd^4/64. Hence option d is correct.
Why the others are incorrect:
- a: πd^3/16 has the wrong power of d; moment of inertia has units of length^4, not length^3.
- b: πd^3/32 also has the wrong dimension and is not the correct circular-section formula.
- c: πd^4/32 has the correct dimension but is twice the correct value; it is not the centroidal diameter MOI of a solid circle.
The law which states, “within elastic limits strain produced is proportional to the stress producing it” is known as
A)
Bernoulli’s law
B)
Stress law
C)
Poisson’s law
D)
Hooke’s law
The statement "within elastic limits strain produced is proportional to the stress producing it" is Hooke’s law. It defines the linear stress-strain relationship for elastic materials: stress ∝ strain, or stress = E × strain, where E is Young’s modulus. Poisson’s law is related to lateral strain and longitudinal strain, Bernoulli’s law is a fluid/beam theory term and not the stress-strain proportionality law, and "Stress law" is not the standard named law for this principle.
Which of the following is not the displacement method:
A)
Equilibrium Method
B)
Column Analogy Method
C)
Moment distribution Method
D)
Kani’s Method
The correct answer is **b) Column Analogy Method** because it is **not a displacement method**. The column analogy method is a **force (flexibility) method**, where redundants are determined using compatibility and flexibility relations.
Why the others are incorrect:
- **c) Moment Distribution Method**: This is a **displacement method** used to analyze indeterminate beams and frames by distributing end moments based on joint rotations.
- **d) Kani’s Method**: This is also a **displacement method**, based on iterative distribution of moments due to joint rotations.
- **a) Equilibrium Method**: In structural analysis, equilibrium-based approaches are generally associated with **force methods**, not displacement methods.
Therefore, the method that is **not** a displacement method is **Column Analogy Method**.
A fixed beam of uniform section is carrying a point load at its mid-span. If the moment of inertia of the middle half-length is now reduced to half its previous value, then the fixed end moment will
A)
Increase
B)
Decrease
C)
Remain constant
D)
Change their direction
For a fixed beam with a central point load, the fixed end moments depend on the relative flexural stiffness of the beam segments. If the moment of inertia of the middle half-length is reduced to half, that portion becomes less stiff, so it attracts less bending resistance. As a result, the end restraints must carry a smaller share of the load-induced moment, and the fixed end moments decrease.
Why others are wrong:
- a) Increase: Incorrect, because reducing stiffness in the middle reduces the beam’s ability to develop larger end moments.
- c) Remain constant: Incorrect, because fixed end moments are not independent of stiffness distribution when the section is non-uniform.
- d) Change their direction: Incorrect, because the loading pattern is unchanged; only the magnitude of the fixed end moments is affected, not their sign.
A 1st class brick immersed in water for 24 hours, should not absorb water (by weight) more than
A)
30 %
B)
25 %
C)
20 %
D)
15 %
A first-class brick, when immersed in water for 24 hours, should absorb water by weight not more than 20% of its dry weight as per standard brick quality requirements. Therefore, option (c) 20% is correct. The other options are incorrect because 15% is too strict for first-class bricks, while 25% and 30% exceed the permissible water absorption limit and indicate poorer quality bricks.
Which of the following methods of structural analysis is a force method:
A)
Slope deflection method
B)
Column analogy method
C)
Moment distribution method
D)
None of these
The correct answer is **b) Column analogy method**. This is a **force method** (also called the **flexibility method**) because it determines the redundant forces first and then uses compatibility conditions to solve the structure.
Why the others are incorrect:
- **a) Slope deflection method**: This is a **displacement method**, since it solves for joint rotations and translations.
- **c) Moment distribution method**: This is also a **displacement method**, based on balancing end moments due to joint rotations.
- **d) None of these**: Incorrect because **column analogy method** is indeed a force method.
The compression stress in concrete at the level of centroid of compression steel is equal to
A)
0.416 Fck
B)
0.446 Fck
C)
0.36 Fck
D)
0.57 Fck
In limit state design of RC beams, the concrete stress block is parabolic-rectangular. At the level of the centroid of compression steel, the compressive stress in concrete is taken as 0.36 fck for the standard stress distribution used in design. Hence option (c) is correct.
Why the others are incorrect:
- 0.416 fck and 0.446 fck are not the standard code values for concrete stress at the centroid of compression steel.
- 0.57 fck is the maximum design compressive stress in concrete at the extreme compression fiber, not at the level of compression steel.
A wall constructed to resist the pressure of an earth filling is called
A)
Retaining wall
B)
Breast wall
C)
Buttress wall
D)
Parapet wall
A retaining wall is specifically built to hold back soil or earth filling and resist the lateral earth pressure exerted by it. Hence, it is the correct answer. Breast wall is generally a wall built on the hill side of a road or railway cutting to support the slope, not the standard term for resisting earth fill pressure. Buttress wall is a wall strengthened by buttresses to resist lateral forces, but it is not the general name for a wall holding earth fill. Parapet wall is a low protective wall at the edge of a roof, bridge, or terrace, and does not retain earth.
Plywood has the advantage of
A)
Greater tensile strength in longer direction
B)
Greater tensile strength in shorter direction
C)
Same tensile strength in all directions
D)
None of these
Plywood is made by gluing thin wood veneers with adjacent layers having grains at right angles. This cross-laminated construction makes its strength nearly uniform in both directions, so it has approximately the same tensile strength in all directions. Option (a) and (b) are incorrect because plywood does not have greater tensile strength specifically in the longer or shorter direction; its main advantage is reduced directional dependence. Option (d) is incorrect because option (c) correctly states the key advantage.