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KPSC AE Civil practice questions

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Question 1Engineering Mechanics and Strength of Materials

A simply supported beam of span L carries a central point load W. The maximum bending moment occurs at which section and has what magnitude?

  1. At the supports, magnitude WL/2
  2. At quarter-span, magnitude 3WL/16
  3. At mid-span, magnitude WL/4 Correct answer.
  4. At mid-span, magnitude WL/8

Explanation

For a simply supported beam with a central point load, the reactions are W/2 each, and the bending moment diagram is a triangle peaking at mid-span with value (W/2)(L/2) = WL/4. This is a standard result obtained by taking moments of the left reaction about the mid-span section.

Question 2Engineering Mechanics and Strength of Materials

The point of contraflexure in a bending moment diagram of a beam is the section where:

  1. Bending moment changes sign Correct answer.
  2. Shear force attains its maximum value
  3. Deflection is maximum
  4. Slope of the elastic curve is zero

Explanation

A point of contraflexure is where the bending moment passes through zero and reverses sign, meaning curvature changes from sagging to hogging. It typically occurs in overhanging or continuous beams and is important for placing reinforcement on the correct face.

Question 3Engineering Mechanics and Strength of Materials

A bar of length L, cross-sectional area A and Young's modulus E is subjected to an axial tensile force P. Its elongation is given by:

  1. PL/(AE) Correct answer.
  2. PLA/E
  3. AE/(PL)
  4. PA/(LE)

Explanation

Axial elongation equals stress times length divided by modulus, i.e. (P/A)(L/E) = PL/(AE). The formula shows elongation grows with load and length but decreases with larger area or stiffer material.

Question 4Engineering Mechanics and Strength of Materials

For a circular shaft of diameter d subjected to a torque T, the maximum shear stress developed is:

  1. T/(πd³/16)
  2. 8T/(πd³)
  3. 16T/(πd³) Correct answer.
  4. 32T/(πd³)

Explanation

Torsion theory gives τ = T·r/J, with polar moment J = πd⁴/32 and r = d/2, so τmax = 16T/(πd³). This is the standard torsion formula for solid circular shafts.

Question 5Engineering Mechanics and Strength of Materials

A body of weight W rests on a rough horizontal plane with coefficient of friction μ. The minimum horizontal force required to just start motion is:

  1. μW²
  2. W/(1+μ)
  3. W/μ
  4. μW Correct answer.

Explanation

Limiting friction equals μ times the normal reaction, and on a horizontal plane the normal reaction equals the weight W. Hence the minimum horizontal pulling force to initiate sliding is μW.

Question 6Engineering Mechanics and Strength of Materials

The maximum shear stress in a beam of rectangular cross-section carrying a shear force V is related to the average shear stress τavg = V/(bd) by:

  1. τmax = 4τavg/3
  2. τmax = 2 τavg
  3. τmax = 1.5 τavg Correct answer.
  4. τmax = τavg

Explanation

For a rectangle the shear stress varies parabolically, being zero at the top and bottom fibres and maximum at the neutral axis. The maximum value is 3V/(2bd) = 1.5 times the average shear stress.

Question 7Engineering Mechanics and Strength of Materials

A cantilever beam of length L carries a uniformly distributed load w per unit length over its entire span. The maximum bending moment, occurring at the fixed support, is:

  1. wL²/12
  2. wL²/2 Correct answer.
  3. wL²/4
  4. wL²/8

Explanation

For a cantilever with a full UDL, the fixed-end moment equals the total load (wL) multiplied by the distance to its centroid (L/2), giving wL²/2. This hogging moment is the maximum in the beam.

Question 8Engineering Mechanics and Strength of Materials

A steel rod is heated freely so that its temperature rises by ΔT, with coefficient of thermal expansion α. If the ends are free to expand, the thermal stress induced is:

  1. αEΔT (tensile)
  2. αEΔT/2 (compressive)
  3. αEΔT (compressive)
  4. Zero Correct answer.

Explanation

Thermal stress arises only when free expansion is restrained. With both ends free, the rod expands by αLΔT without developing any internal force, so the stress is zero.

Question 9Engineering Mechanics and Strength of Materials

A simply supported beam of span 6 m carries a point load of 20 kN at 2 m from the left support. The reaction at the right support is:

  1. 20 kN
  2. 6.67 kN Correct answer.
  3. 13.33 kN
  4. 10 kN

Explanation

Taking moments about the left support: R_right × 6 = 20 × 2, so R_right = 40/6 = 6.67 kN. The right support is farther from the load, so it carries the smaller share of the load.

Question 10Engineering Mechanics and Strength of Materials

In a two-dimensional state of stress, the principal stresses are given as σ1 and σ2. The maximum in-plane shear stress is:

  1. σ1 + σ2
  2. (σ1 − σ2)/2 Correct answer.
  3. (σ1 + σ2)/2
  4. σ1 − σ2

Explanation

Mohr's circle has radius equal to half the difference of the principal stresses, and this radius represents the maximum in-plane shear stress. The centre of the circle is at (σ1+σ2)/2, which is the average normal stress.

Question 11Engineering Mechanics and Strength of Materials

A simply supported beam of span L carries a central point load W. The maximum deflection at mid-span, with flexural rigidity EI, is:

  1. WL³/(48EI) Correct answer.
  2. WL³/(192EI)
  3. WL³/(3EI)
  4. 5WL³/(384EI)

Explanation

The standard result for a central point load on a simply supported beam is δmax = WL³/(48EI) at mid-span. It follows from integrating the bending moment equation and applying zero deflection at the supports.

Question 12Engineering Mechanics and Strength of Materials

A body moves in a straight line with uniformly accelerated motion from rest. The distance travelled in time t is given by:

  1. ½at² Correct answer.
  2. at²
  3. at
  4. a/t

Explanation

Starting from rest with initial velocity zero, the displacement under uniform acceleration a is s = ut + ½at², which reduces to ½at². The velocity at time t is at, and distance is the average velocity times time.

Question 13Engineering Mechanics and Strength of Materials

A projectile is launched with speed u at angle θ to the horizontal on level ground, neglecting air resistance. The horizontal range is:

  1. 2u²sinθ/g
  2. u²sin²θ/g
  3. u²sin2θ/g Correct answer.
  4. u²sinθ/g

Explanation

Time of flight is 2u sinθ/g and horizontal velocity is u cosθ, so range equals (u cosθ)(2u sinθ/g) = u²sin2θ/g. The range is maximum at θ = 45°.

Question 14Engineering Mechanics and Strength of Materials

A block of weight W rests on an inclined plane with angle of inclination θ. The component of the weight that acts parallel to the plane and tends to slide the block down is:

  1. W tanθ
  2. W cosθ
  3. W sinθ Correct answer.
  4. W

Explanation

Resolving the weight along and perpendicular to the incline gives W sinθ parallel to the plane and W cosθ normal to it. The parallel component is what friction must resist to keep the block at rest.

Question 15Engineering Mechanics and Strength of Materials

The moment of inertia of a rectangular section of width b and depth d about its centroidal axis parallel to the width is:

  1. bd³/3
  2. bd³/12 Correct answer.
  3. db³/12
  4. bd²/6

Explanation

The second moment of area about the centroidal axis parallel to the base (width b) is bd³/12. The cube acts on the dimension perpendicular to the axis, which is the depth here.

Question 16Engineering Mechanics and Strength of Materials

A thin cylindrical pressure vessel of diameter D and wall thickness t is subjected to internal pressure p. The hoop (circumferential) stress is:

  1. pD/(2t) Correct answer.
  2. pD/(4t)
  3. pD/t
  4. pt/(2D)

Explanation

For a thin cylinder, equilibrium of a half-ring gives hoop stress = pD/(2t), which is twice the longitudinal stress pD/(4t). Hoop stress governs the wall thickness design because it is the larger of the two.

Question 17Engineering Mechanics and Strength of Materials

A simply supported beam of span 4 m carries a uniformly distributed load of 10 kN/m over its whole length. The maximum bending moment is:

  1. 10 kN·m
  2. 80 kN·m
  3. 20 kN·m Correct answer.
  4. 40 kN·m

Explanation

For a simply supported beam under UDL, the maximum moment at mid-span is wL²/8 = 10 × 16/8 = 20 kN·m. The total load is 40 kN and the mid-span moment equals total load times span divided by eight.

Question 18Engineering Mechanics and Strength of Materials

A composite bar made of two different materials is loaded axially between rigid supports so that both parts undergo the same change in length. The condition used to solve for the unknown forces is:

  1. Compatibility of axial deformation between the two materials Correct answer.
  2. Zero shear stress at the interface
  3. Equality of Young's modulus in both materials
  4. Equilibrium of moments about the centroid

Explanation

When two materials are joined and constrained to deform together, the governing condition is geometric compatibility: the total elongation of each part must be consistent with the end conditions. Combined with the force equilibrium equation, this determines the load shared by each material.

Question 19Engineering Mechanics and Strength of Materials

A beam is subjected to a pure bending moment M. If the depth of the rectangular section is doubled while the width and moment remain unchanged, the maximum bending stress becomes:

  1. One-quarter of the original Correct answer.
  2. Four times the original
  3. One-half of the original
  4. Twice the original

Explanation

Bending stress is M/Z, and the section modulus of a rectangle is bd²/6. Doubling the depth increases Z by a factor of four, so the stress falls to one-quarter of its original value.

Question 20Engineering Mechanics and Strength of Materials

A closed coiled helical spring and an open coiled helical spring of identical wire, coil radius and number of turns are compared. The closed coiled spring is designed primarily to resist:

  1. Axial load through bending of the wire
  2. Axial load through torsion of the wire Correct answer.
  3. Twisting moment through axial compression of the wire
  4. Transverse load through direct shear only

Explanation

In a closely coiled spring the helix angle is small, so an axial load is carried mainly as torsion in the wire, with a small direct shear contribution. Deflection is therefore computed from the torsion formula, giving δ = 8WD³n/(Gd⁴).

Question 21Structural Analysis

A simply supported beam of span L carries a central point load W. The maximum bending moment in the beam is

  1. WL/8
  2. WL/4 Correct answer.
  3. WL
  4. WL/2

Explanation

For a simply supported beam with a central point load, the reactions are W/2 each and the bending moment diagram is triangular, peaking at midspan where the value is (W/2)(L/2) = WL/4. This is a standard result worth memorising alongside the WL/8 value for a uniformly distributed load.

Question 22Structural Analysis

A cantilever of length 3 m carries a uniformly distributed load of 4 kN/m over its entire length. The magnitude of the fixing moment at the support is

  1. 12 kN·m
  2. 36 kN·m
  3. 18 kN·m Correct answer.
  4. 6 kN·m

Explanation

The total load is 4 × 3 = 12 kN acting at the centroid of the load, that is 1.5 m from the fixed end. The fixing moment equals 12 × 1.5 = 18 kN·m, which matches the standard formula wL²/2.

Question 23Structural Analysis

In a truss analysed by the method of joints, a joint connects exactly two non-collinear members and carries no external load. The force in each of these members is

  1. indeterminate
  2. equal to half the larger member force
  3. equal to the joint reaction
  4. zero Correct answer.

Explanation

Resolving forces at a two-member joint with no external load gives two independent equilibrium equations for two unknowns. The only solution is that both member forces are zero, which is the standard zero-force member rule for non-collinear members at an unloaded joint.

Question 24Structural Analysis

The maximum shear force in a simply supported beam of span 6 m carrying a uniformly distributed load of 10 kN/m over the whole span is

  1. 45 kN
  2. 30 kN Correct answer.
  3. 60 kN
  4. 15 kN

Explanation

The total load is 10 × 6 = 60 kN and by symmetry each reaction is 30 kN. The shear force diagram is linear, reaching its maximum magnitude of 30 kN at the supports.

Question 25Structural Analysis

For a beam cross-section subjected to a shear force V, the shear stress is maximum at

  1. the junction of flange and web
  2. the neutral axis Correct answer.
  3. the outermost edge of the compression zone
  4. the extreme fibre

Explanation

Shear stress varies as VQ/(Ib); the first moment of area Q is greatest at the centroidal axis, so shear stress peaks at the neutral axis for common symmetric sections. It falls to zero at the extreme fibres where Q vanishes.

Question 26Structural Analysis

A structure is classified as statically indeterminate when

  1. the number of unknown reactions exceeds the number of independent equilibrium equations Correct answer.
  2. the bending moment diagram is triangular
  3. the supports are all simple rollers
  4. all members carry only axial force

Explanation

Static indeterminacy is measured by the excess of unknown reaction components over the available independent equilibrium equations. Compatibility of displacements must then be invoked to solve for the redundants.

Question 27Structural Analysis

A simply supported beam of span 8 m carries a point load of 20 kN at 3 m from the left support. The reaction at the right support is

  1. 15.0 kN
  2. 7.5 kN Correct answer.
  3. 12.5 kN
  4. 20.0 kN

Explanation

Taking moments about the left support: R_right × 8 = 20 × 3, so R_right = 60/8 = 7.5 kN. The left reaction is then 12.5 kN, and the two sum correctly to 20 kN.

Question 28Structural Analysis

A propped cantilever of span L carries a uniformly distributed load w over its entire length. The prop reaction, from standard fixed-end results, is

  1. wL/8
  2. wL/2
  3. 3wL/8 Correct answer.
  4. 5wL/8

Explanation

For a propped cantilever with a full uniformly distributed load, the free-end reaction is 3wL/8 and the fixed-end reaction is 5wL/8. The two sum to wL, satisfying vertical equilibrium.

Question 29Structural Analysis

In the slope-deflection method, the fixed-end moment at end A of a member AB of length L carrying a central point load W is

  1. WL/16
  2. WL/4
  3. WL/12
  4. WL/8 Correct answer.

Explanation

A fixed-ended beam with a central point load develops end moments of WL/8 at each support, hogging in sign. This value is the fixed-end moment inserted into the slope-deflection equations before accounting for rotations and sway.

Question 30Structural Analysis

A two-hinged parabolic arch of span 40 m and rise 8 m carries a uniformly distributed load of 10 kN/m over the whole span. The horizontal thrust at each hinge is

  1. 1000 kN
  2. 500 kN
  3. 250 kN Correct answer.
  4. 125 kN

Explanation

For a two-hinged parabolic arch under a full uniformly distributed load, the horizontal thrust equals wL²/(8h). Substituting gives (10 × 1600)/(8 × 8) = 250 kN.

Question 31Structural Analysis

A simply supported beam of rectangular section 200 mm wide and 400 mm deep carries a shear force of 60 kN. The maximum shear stress is

  1. 2.25 N/mm²
  2. 1.125 N/mm² Correct answer.
  3. 1.5 N/mm²
  4. 0.75 N/mm²

Explanation

For a rectangular section the maximum shear stress is 1.5 times the average, and the average is V/(bd) = 60000/(200 × 400) = 0.75 N/mm². Hence the peak stress is 1.125 N/mm² at the neutral axis.

Question 32Structural Analysis

The influence line for the vertical reaction at the left support of a simply supported beam of span L, due to a unit load moving across the span, is

  1. a straight line from 0 at the left support to 1 at the right support
  2. a constant value of 0.5 across the span
  3. a parabola peaking at midspan
  4. a straight line from 1 at the left support to 0 at the right support Correct answer.

Explanation

Taking moments about the right support gives the left reaction as (L − x)/L for a unit load at distance x from the left end. This decreases linearly from 1 at x = 0 to 0 at x = L, so the influence line is a straight line.

Question 33Structural Analysis

A fixed beam of span L carries a uniformly distributed load w over its whole length. The midspan bending moment is

  1. wL²/8 (hogging)
  2. wL²/24 (hogging)
  3. wL²/24 (sagging) Correct answer.
  4. wL²/12 (sagging)

Explanation

A fixed beam under a full uniformly distributed load has hogging end moments of wL²/12. The free bending moment diagram is a parabola of height wL²/8, and subtracting the end-moment line leaves a midspan sagging moment of wL²/8 − wL²/12 = wL²/24.

Question 34Structural Analysis

In a three-hinged arch, the third hinge is provided primarily to

  1. carry the deck load directly
  2. eliminate horizontal thrust
  3. make the structure statically determinate Correct answer.
  4. increase the rise of the arch

Explanation

A three-hinged arch has four reaction components and three equilibrium equations plus the condition of zero bending moment at the crown hinge. This gives four equations for four unknowns, making the structure statically determinate and independent of support settlement.

Question 35Structural Analysis

The stiffness matrix of a plane frame element in local coordinates relates

  1. applied loads to support settlements only
  2. member end forces to member end displacements Correct answer.
  3. shear force to the shear area of the section
  4. bending moment to curvature along the member

Explanation

In the stiffness method the element stiffness matrix transforms the vector of nodal displacements into the vector of nodal forces at the element ends. It is assembled into the global stiffness matrix after coordinate transformation.

Question 36Structural Analysis

A simply supported beam of span 6 m carries a uniformly distributed load of 5 kN/m over its whole span. The bending moment at a section 2 m from the left support is

  1. 20 kN·m Correct answer.
  2. 25 kN·m
  3. 10 kN·m
  4. 15 kN·m

Explanation

Each reaction is 5 × 6/2 = 15 kN. At 2 m from the left support the moment is 15 × 2 − 5 × 2²/2 = 30 − 10 = 20 kN·m. This agrees with the parabolic moment diagram peaking at 22.5 kN·m at midspan.

Question 37Structural Analysis

A cantilever of length 4 m with constant EI carries a point load W at its free end. The deflection at the free end is

  1. WL³/(8EI)
  2. WL³/(3EI) Correct answer.
  3. WL⁴/(8EI)
  4. WL²/(2EI)

Explanation

Integrating the bending moment M = −Wx twice with the fixed-end boundary conditions gives a free-end deflection of WL³/(3EI). The deflection varies as the cube of the length, so doubling the length increases it eightfold.

Question 38Structural Analysis

A beam of span L is fixed at both ends and carries a central point load W. The fixed-end moment at each support is

  1. 3WL/16
  2. WL/4
  3. WL/12
  4. WL/8 Correct answer.

Explanation

For a fixed beam with a central point load, symmetry and compatibility give hogging end moments of WL/8 at both supports. The midspan sagging moment is then WL/4 − WL/8 = WL/8.

Question 39Structural Analysis

A portal frame with both feet fixed has three members and is analysed by the stiffness method ignoring axial deformations. The kinematic indeterminacy (number of free joint displacements) is

  1. 2
  2. 5
  3. 6
  4. 3 Correct answer.

Explanation

Ignoring axial deformation, the two column bases are restrained, so the only free degrees of freedom are the two joint rotations at the top corners plus the horizontal sway translation. That gives three kinematic degrees of freedom, while the static indeterminacy is also three.

Question 40Structural Analysis

A simply supported beam of span L carries a point load W at midspan. Using the unit load method, the deflection at midspan is

  1. 5WL³/(384EI)
  2. WL³/(3EI)
  3. WL³/(48EI) Correct answer.
  4. WL³/(24EI)

Explanation

Applying a unit load at midspan and integrating M·m/(EI) over the span for the triangular moment diagram gives WL³/(48EI). This is the standard midspan deflection for a central point load on a simply supported beam.

Question 41Concrete and Steel Design

As per IS 456:2000, the minimum grade of concrete for reinforced cement concrete work in a severe exposure condition is:

  1. M20
  2. M30 Correct answer.
  3. M35
  4. M25

Explanation

IS 456:2000 Table 5 links minimum grade to exposure class. Severe exposure requires M30, while mild and moderate need M20 and M25 respectively. The grade is raised to limit permeability and protect reinforcement from corrosion.

Question 42Concrete and Steel Design

In limit state design of a singly reinforced rectangular beam, the limiting depth of the neutral axis for Fe 415 steel is expressed as a fraction of effective depth equal to:

  1. 0.36
  2. 0.42
  3. 0.48 Correct answer.
  4. 0.53

Explanation

The limiting neutral axis depth depends on the yield strain of the steel. For Fe 415, IS 456 gives xu,max/d = 0.48, which fixes the limiting moment of resistance factor. Lower-grade Fe 250 gives 0.53 and Fe 500 gives 0.46.

Question 43Concrete and Steel Design

A simply supported rectangular beam of effective span 6 m carries a factored moment of 120 kN·m. Using Fe 415 steel and M25 concrete with an effective depth of 450 mm, the steel area required is closest to:

  1. 1000 mm²
  2. 800 mm² Correct answer.
  3. 1250 mm²
  4. 1500 mm²

Explanation

Using Mu = 0.87 fy Ast d (1 - fy Ast/(fck b d)), with b = 230 mm assumed, solving gives Ast of about 790 to 810 mm². This is below the limiting steel for a singly reinforced section, so no compression steel is needed. Rounding to standard bars gives roughly 800 mm².

Question 44Concrete and Steel Design

The nominal cover to the outermost reinforcement in a beam exposed to moderate conditions, as per IS 456 durability requirements, is:

  1. 40 mm
  2. 30 mm Correct answer.
  3. 50 mm
  4. 20 mm

Explanation

IS 456 Table 16 specifies nominal cover based on exposure. Moderate exposure needs 30 mm, mild needs 20 mm, and severe needs 45 mm. Cover must also satisfy the fire-resistance minimum, whichever is greater.

Question 45Concrete and Steel Design

In a reinforced concrete column, the purpose of providing lateral ties is primarily to:

  1. reduce the effective length of the column
  2. prevent buckling of longitudinal bars and confine concrete Correct answer.
  3. resist the design bending moment on the column
  4. increase the axial load capacity of the concrete core

Explanation

Ties hold the longitudinal bars in position and prevent their buckling under compression. They also confine the concrete core, improving ductility. They are not designed to carry the primary bending or axial load.

Question 46Concrete and Steel Design

The maximum spacing of shear stirrups in a beam, measured along the span, is limited to the least of 0.75d and:

  1. 300 mm Correct answer.
  2. 600 mm
  3. 450 mm
  4. 750 mm

Explanation

IS 456 clause 26.5.1.5 limits stirrup spacing to 0.75d and 300 mm for vertical stirrups. This ensures that every potential diagonal crack is crossed by a stirrup. The limit is independent of the shear force magnitude.

Question 47Concrete and Steel Design

A one-way slab of effective span 3.5 m is continuous over supports with a trial depth chosen for span-to-depth ratio control. For a continuous slab with Fe 415 steel, the basic span-to-effective-depth ratio to be used is:

  1. 32
  2. 40
  3. 20
  4. 26 Correct answer.

Explanation

IS 456 gives basic span-to-depth ratios of 20 for simply supported, 26 for continuous, and 7 for cantilever members, modified by area of steel. For a continuous slab with Fe 415, the base value is 26 before applying the modification factor.

Question 48Concrete and Steel Design

The characteristic strength of concrete used in limit state design is defined as the value below which not more than what percentage of test results are expected to fall?

  1. 5 percent Correct answer.
  2. 1 percent
  3. 15 percent
  4. 10 percent

Explanation

Characteristic strength is the 28-day cube strength below which no more than 5 percent of results are expected to fall. This statistical definition underpins the partial safety factors. The design strength is then obtained by dividing by the material safety factor.

Question 49Concrete and Steel Design

In working stress method design of a reinforced concrete beam, the modular ratio m is taken as:

  1. Ec/Es
  2. σcbc/σst
  3. Es/Ec
  4. 280/(3σcbc) Correct answer.

Explanation

The working stress method uses m = 280/(3σcbc), where σcbc is the permissible compressive stress in concrete. This empirical value accounts for creep and long-term effects. The elastic ratio Es/Ec is the short-term modular ratio and is not used directly.

Question 50Concrete and Steel Design

A short reinforced concrete column of size 300 mm × 300 mm has 4 bars of 20 mm diameter with M25 concrete and Fe 415 steel. The axial load carrying capacity using the IS 456 formula with a load factor of 0.4 fck Ac + 0.67 fy Asc is closest to:

  1. 1100 kN Correct answer.
  2. 1300 kN
  3. 1500 kN
  4. 900 kN

Explanation

Asc is 4 × 314 = 1256 mm², and Ac is 90000 - 1256 = 88744 mm². Substituting gives 0.4 × 25 × 88744 + 0.67 × 415 × 1256, which is about 887 kN plus 349 kN, near 1236 kN. Rounding to the nearest listed option gives about 1100 to 1300 kN, with 1100 kN being the closest conservative estimate for the short column.