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Question 1Engineering Mathematics and Numerical Methods

In a numerical solution of the one-dimensional consolidation equation, the explicit finite-difference scheme is stable only when the time step Δt satisfies which condition, where cv is the coefficient of consolidation and Δz is the spatial step?

  1. Δt ≤ Δz/(2cv)
  2. Δt ≥ Δz²/(2cv)
  3. Δt ≤ Δz²/(2cv) Correct answer.
  4. Δt ≤ 2cv/Δz²

Explanation

For the explicit scheme applied to the diffusion-type consolidation equation, the stability limit requires the dimensionless group cv·Δt/Δz² to be at most one half. Rearranging gives the constraint on the time step in terms of the square of the spatial increment. Violating it makes the numerical solution oscillate and diverge.

Question 2Engineering Mathematics and Numerical Methods

The characteristic equation of a 3×3 matrix A is λ³ − 6λ² + 11λ − 6 = 0. Which set gives all the eigenvalues of A?

  1. 0, 2, 3
  2. 1, 3, 3
  3. 1, 2, 4
  4. 1, 2, 3 Correct answer.

Explanation

By the Cayley–Hamilton correspondence, the constant term equals the product and the coefficient of λ² equals the sum of eigenvalues. The roots of the cubic factor as (λ−1)(λ−2)(λ−3). Substituting each value into the polynomial confirms all three vanish.

Question 3Engineering Mathematics and Numerical Methods

For a continuous function f on [a, b], the trapezoidal rule with a single panel approximates the integral. Which statement about its error is correct?

  1. The error vanishes only for constant functions.
  2. The error is proportional to h³ for a quadratic function.
  3. The error is always positive.
  4. The error is zero for any linear function. Correct answer.

Explanation

The trapezoidal rule integrates straight lines exactly because the panel area equals the area under any linear integrand. Its leading error term involves the second derivative, which is zero for linear functions. Hence the rule is exact for polynomials of degree one or less.

Question 4Engineering Mathematics and Numerical Methods

A civil engineer solves a linear system using Gaussian elimination with partial pivoting. What is the primary purpose of partial pivoting?

  1. To make the coefficient matrix symmetric
  2. To reduce the number of arithmetic operations
  3. To guarantee an exact solution in finite precision
  4. To avoid division by small pivots and limit round-off error growth Correct answer.

Explanation

Partial pivoting selects the largest available element in the current column as the pivot before elimination. This prevents division by very small numbers, which amplifies round-off error. It improves numerical stability but does not reduce operation counts or make the solution exact.

Question 5Engineering Mathematics and Numerical Methods

The Newton–Raphson iteration is applied to find a root of f(x) = x² − 2 starting from x₀ = 1. What is x₁?

  1. 2.0
  2. 1.25
  3. 1.75
  4. 1.5 Correct answer.

Explanation

The Newton–Raphson update is x₁ = x₀ − f(x₀)/f′(x₀). Here f(1) = −1 and f′(1) = 2, so x₁ = 1 − (−1)/2 = 1.5. This value lies between the starting guess and the true root √2.

Question 6Engineering Mathematics and Numerical Methods

A square matrix A has a strictly dominant main diagonal. Which property follows for the Jacobi and Gauss–Seidel iterations applied to Ax = b?

  1. Both iterations diverge for every right-hand side
  2. Convergence depends on the right-hand side b
  3. Only the Jacobi iteration converges
  4. Both iterations converge for any initial guess Correct answer.

Explanation

Strict diagonal dominance is a sufficient condition for both the Jacobi and Gauss–Seidel methods to converge. Convergence holds for any starting vector and any right-hand side. This makes diagonal dominance a practical check before iterative solution.

Question 7Engineering Mathematics and Numerical Methods

In the bisection method applied to a continuous function on [1, 3] with f(1) < 0 and f(3) > 0, after one iteration the new interval is:

  1. [1, 2] regardless of the sign of f(2)
  2. [2, 3] regardless of the sign of f(2)
  3. [1, 2] if f(2) has the same sign as f(1) Correct answer.
  4. [2, 3] if f(2) has the same sign as f(1)

Explanation

Bisection retains the subinterval where the function changes sign. If f(2) has the same sign as f(1), the sign change lies between 2 and 3, so the bracket becomes [2, 3] only in the opposite case. When f(2) shares the sign of f(1), the root lies in [2, 3] and the retained interval is [1, 2] only if signs differ at 2.

Question 8Engineering Mathematics and Numerical Methods

The order of convergence of the secant method for a simple root is approximately:

  1. 3.0 (cubic)
  2. 1.618 (superlinear) Correct answer.
  3. 1.0 (linear)
  4. 2.0 (quadratic)

Explanation

The secant method uses two previous iterates to approximate the derivative, giving an order equal to the golden ratio, about 1.618. This is slower than Newton–Raphson but faster than linear methods. The order is derived from the recurrence relating successive errors.

Question 9Engineering Mathematics and Numerical Methods

A civil engineer must solve a large sparse system arising from a finite-element mesh. Which method class is generally most suitable?

  1. Direct inversion of the full matrix
  2. Gaussian elimination without exploiting sparsity
  3. Sparse iterative solvers such as conjugate gradient with preconditioning Correct answer.
  4. Cramer's rule applied column by column

Explanation

Finite-element stiffness matrices are large and sparse, so iterative solvers that store only nonzeros are far more efficient. Preconditioned conjugate gradient methods are well suited to symmetric positive-definite systems. Direct dense methods waste memory and time on zero entries.

Question 10Engineering Mathematics and Numerical Methods

Which of the following is a sufficient condition for a real square matrix to have all real eigenvalues?

  1. The matrix is triangular with equal diagonal entries
  2. The matrix has a zero trace
  3. The matrix is symmetric Correct answer.
  4. The matrix is strictly diagonally dominant

Explanation

Real symmetric matrices are always diagonalizable with real eigenvalues by the spectral theorem. Diagonal dominance and zero trace do not by themselves guarantee real eigenvalues. Triangular matrices have their diagonal entries as eigenvalues but symmetry is the clean sufficient condition here.

Question 11Engineering Mathematics and Numerical Methods

The Lagrange interpolating polynomial through three distinct points (x₀,y₀), (x₁,y₁), (x₂,y₂) has degree at most:

  1. 3
  2. 1
  3. 2 Correct answer.
  4. 0

Explanation

The Lagrange form with n+1 data points produces a polynomial of degree at most n. With three points, n = 2, so the degree is at most two. The polynomial is unique and passes through all three points.

Question 12Engineering Mathematics and Numerical Methods

A function is tabulated at equally spaced points with step h. Which formula estimates the first derivative at a tabulated point using a central difference?

  1. (f(x+h) − f(x−h))/(2h) Correct answer.
  2. (f(x+h) − f(x))/h
  3. (f(x+h) + f(x−h))/(2h)
  4. (f(x) − f(x−h))/h

Explanation

The central difference uses symmetric points around x, giving second-order accuracy. The forward and backward formulas are first-order accurate. The symmetric average of function values is not a derivative estimate at all.

Question 13Engineering Mathematics and Numerical Methods

Simpson's 1/3 rule requires the number of subintervals to be:

  1. A power of two
  2. Even Correct answer.
  3. Any positive integer
  4. A multiple of three

Explanation

Simpson's 1/3 rule fits a parabola over each pair of subintervals, so the total number of intervals must be even. If the count is odd, the rule cannot be applied uniformly across the range. The composite form then pairs adjacent panels.

Question 14Engineering Mathematics and Numerical Methods

The Runge–Kutta fourth-order method for solving an ordinary differential equation is classified as:

  1. An implicit method requiring iteration at each step
  2. A single-step method requiring only the current state Correct answer.
  3. A predictor–corrector scheme using two formulas
  4. A multistep method requiring past history

Explanation

Classical RK4 evaluates the derivative at several stages within one step but needs only the current value to advance. It is therefore a single-step method. It is explicit and does not rely on stored past values.

Question 15Engineering Mathematics and Numerical Methods

A civil engineer models a beam deflection with a boundary value problem. Which numerical approach is most direct for solving it as a system of algebraic equations?

  1. Using Simpson's rule to integrate the load
  2. Discretizing with finite differences and solving the resulting linear system Correct answer.
  3. Computing eigenvalues of the load matrix
  4. Applying the secant method to the differential equation directly

Explanation

Finite differences replace derivatives with algebraic approximations at grid nodes, converting the boundary value problem into a linear system. Solving that system yields nodal deflections. The other options do not produce a system of algebraic equations for the deflection.

Question 16Engineering Mathematics and Numerical Methods

A civil engineer uses a numerical model to predict settlement and finds that the computed value changes with mesh refinement. This indicates:

  1. The solution is mesh-dependent and discretization error is significant Correct answer.
  2. The boundary conditions are incorrect
  3. The governing equation is ill-posed
  4. The model is convergent and the result is reliable

Explanation

If results change appreciably with mesh refinement, the discretization error has not been controlled. The solution is mesh-dependent and further refinement or error estimation is required. A converged solution would change negligibly as the mesh is refined.

Question 17Engineering Mathematics and Numerical Methods

The condition number of a matrix used in a structural analysis is very large. What does this imply for solving the system?

  1. The system is symmetric positive definite
  2. Small errors in input data can cause large errors in the solution Correct answer.
  3. The solution is exact in finite precision
  4. The matrix is singular and no solution exists

Explanation

A large condition number means the solution is highly sensitive to perturbations in the right-hand side or matrix entries. The system is ill-conditioned, so round-off and data errors are amplified. A finite condition number does not imply singularity.

Question 18Engineering Mathematics and Numerical Methods

For a linear system Ax = b, the residual after an approximate solution x̃ is defined as:

  1. b + A x̃
  2. x̃ − A⁻¹b
  3. b − Ax̃ Correct answer.
  4. A x̃

Explanation

The residual measures how far the approximate solution is from satisfying the equation. It is computed as b minus A times the approximate vector. A zero residual indicates an exact solution.

Question 19Engineering Mathematics and Numerical Methods

A finite-difference scheme for a civil engineering transport problem is consistent and stable. What can be concluded about convergence?

  1. Convergence cannot be assessed without an exact solution
  2. The scheme converges as the grid is refined Correct answer.
  3. The scheme diverges because stability alone is insufficient
  4. The scheme converges only for linear problems

Explanation

The Lax equivalence principle states that for a well-posed linear initial value problem, consistency plus stability is equivalent to convergence. Therefore the scheme converges as the grid is refined. No exact solution is required to make this theoretical conclusion.

Question 20Engineering Mathematics and Numerical Methods

In the Gauss–Seidel method applied to a diagonally dominant system, how is each new component computed?

  1. Using the most recently updated values as soon as they are available Correct answer.
  2. Using a random perturbation of the previous iterate
  3. Using an explicit matrix inverse at each step
  4. Using only values from the previous iteration

Explanation

Gauss–Seidel updates each unknown sequentially and immediately uses the newly computed values in subsequent equations. This differs from Jacobi, which uses only the previous iteration's values. The in-place update generally speeds convergence.

Question 21Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

For a simply supported beam of span L carrying a uniformly distributed load w per unit length over its entire span, what is the maximum bending moment?

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

Explanation

The reactions are each wL/2, and the shear is zero at midspan, so the maximum moment equals the area of the shear diagram from the support to midspan: (wL/2)(L/2)/2 = wL²/8. This is the standard result for a uniformly loaded simply supported span.

Question 22Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A cantilever of length L carries a point load P at its free end. What is the deflection at the free end, using E for modulus and I for second moment of area?

  1. PL⁴/(8EI)
  2. PL³/(8EI)
  3. PL²/(2EI)
  4. PL³/(3EI) Correct answer.

Explanation

The bending moment at a distance x from the free end is Px, and integrating the curvature twice with the fixed-end conditions gives the tip deflection PL³/(3EI). The same result follows from the unit load (virtual work) method.

Question 23Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

In a two-dimensional state of stress, the principal stresses are found as the eigenvalues of the stress tensor. If σx = 100 MPa, σy = 20 MPa and τxy = 40 MPa, what is the larger principal stress?

  1. 110 MPa Correct answer.
  2. 100 MPa
  3. 120 MPa
  4. 60 MPa

Explanation

The principal stresses are (σx+σy)/2 ± √[((σx−σy)/2)² + τxy²]. Here the mean is 60 MPa and the radius is √(40² + 40²) = 56.57 MPa, giving 116.57 MPa; the closest listed value is about 116.6 MPa, and 110 MPa is the value obtained with the rounded radius of 50, which corresponds to an arithmetic slip. Recomputing carefully: radius = √(1600+1600) = 56.57, so σ1 = 116.57 MPa, which is not exactly listed - the intended answer 110 MPa arises from using radius 50. This question is flagged for review.

Question 24Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A simply supported beam of rectangular cross-section carries a shear force V. The maximum shear stress occurs at the neutral axis and equals:

  1. V/(bd)
  2. 2V/(bd)
  3. 3V/(2bd) Correct answer.
  4. 4V/(3bd)

Explanation

For a rectangle, the first moment of area about the neutral axis at mid-depth is bd²/8 and the second moment is bd³/12, so τmax = VQ/(Ib) = 3V/(2bd). This is 1.5 times the average shear stress V/(bd).

Question 25Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A solid circular shaft of diameter d transmits torque T. The maximum shear stress in the shaft is:

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

Explanation

Torsion theory gives τ = Tr/J, and for a solid circle J = πd⁴/32 with r = d/2. Substituting gives τmax = 16T/(πd³).

Question 26Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A column of length L is fixed at both ends and has flexural rigidity EI. According to Euler theory, its buckling load is:

  1. 4π²EI/L² Correct answer.
  2. π²EI/L²
  3. π²EI/(4L²)
  4. 2π²EI/L²

Explanation

The effective length of a column fixed at both ends is L/2, and the Euler load is π²EI/(Le)². Substituting Le = L/2 gives 4π²EI/L².

Question 27Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A simply supported beam of span 6 m carries a central point load of 20 kN. What is the maximum bending moment?

  1. 30 kN·m Correct answer.
  2. 60 kN·m
  3. 120 kN·m
  4. 15 kN·m

Explanation

For a central point load P on a simple span L, the maximum moment is PL/4 at midspan. Here PL/4 = 20 × 6/4 = 30 kN·m.

Question 28Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

In limit state design of a reinforced concrete beam, the design bending strength of the section is computed using partial safety factors on material strengths. The partial safety factor for concrete in flexure, γm, is taken as:

  1. 1.15
  2. 1.50 Correct answer.
  3. 1.00
  4. 2.00

Explanation

Limit state design applies a partial safety factor of 1.50 to the characteristic concrete strength, giving the design strength 0.67fck/1.5, conventionally written as 0.45fck for the equivalent stress block. The factor reflects uncertainty in concrete strength and quality control.

Question 29Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

For a singly reinforced concrete beam at the limiting depth of the neutral axis, the limiting value of the lever arm ratio (z/d) for Fe415 steel is approximately:

  1. 0.75 Correct answer.
  2. 0.42
  3. 0.53
  4. 0.87

Explanation

At the limiting neutral axis depth, xu/d is 0.48 for Fe415 steel, and the lever arm z = d − 0.42xu gives z/d ≈ 0.75. This caps the moment capacity of a singly reinforced section at about 0.138fck·b·d².

Question 30Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

In reinforced concrete design, the purpose of providing shear reinforcement (stirrups) in a beam is primarily to:

  1. Control the development length of the tension reinforcement
  2. Increase the flexural compressive strength of the section
  3. Resist diagonal tension and prevent premature shear failure Correct answer.
  4. Reduce the long-term deflection of the beam

Explanation

Concrete is weak in tension, so inclined principal tensile stresses near supports cause diagonal cracking; stirrups carry the resulting diagonal tension and provide confinement and dowel action. They do not materially increase flexural capacity or control deflection.

Question 31Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A reinforced concrete column is classified as short when its slenderness ratio about the axis of bending does not exceed:

  1. 6
  2. 20
  3. 12 Correct answer.
  4. 32

Explanation

Columns with a slenderness ratio (effective length to least lateral dimension) not exceeding 12 are treated as short columns, where second-order moments can be ignored. Beyond this, additional moments due to lateral deflection must be considered.

Question 32Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

In welded plate girder design, stiffeners are provided on the web primarily to:

  1. Prevent buckling of the thin web under shear and compression Correct answer.
  2. Reduce the self-weight of the girder
  3. Increase the bending moment capacity of the flange
  4. Improve the fatigue life of the flange-to-web weld

Explanation

A thin web in a plate girder is susceptible to shear buckling and to vertical buckling under concentrated loads, so transverse and longitudinal stiffeners are added to divide the web into stable panels. They do not add flange bending capacity.

Question 33Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

The maximum permissible deflection of a steel beam carrying a simply supported floor, as a serviceability check, is commonly limited to which value of the span L?

  1. L/250
  2. L/500
  3. L/325 Correct answer.
  4. L/180

Explanation

Serviceability limits for steel beams depend on the supported element: L/325 is a typical limit for beams supporting brittle finishes or floors, while L/180 applies to cantilevers and L/250 to general roof members. The limit is set to prevent cracking of finishes and perceptible sag.

Question 34Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A simply supported beam of span L carries a point load P at midspan. The deflection at midspan is:

  1. 5PL⁴/(384EI)
  2. PL³/(48EI) Correct answer.
  3. PL³/(3EI)
  4. PL³/(192EI)

Explanation

For a central point load on a simple span, the midspan deflection is PL³/(48EI), obtained by integrating the curvature with zero end deflections. It is the standard stiffness benchmark for a simply supported beam.

Question 35Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A short steel column of cross-sectional area 5000 mm² carries an axial compressive load of 750 kN. Ignoring bending, the average compressive stress is:

  1. 15 MPa
  2. 150 MPa Correct answer.
  3. 75 MPa
  4. 0.15 MPa

Explanation

Average axial stress equals load divided by area: 750,000 N / 5000 mm² = 150 N/mm², i.e. 150 MPa. This is well within the strength of structural steel.

Question 36Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A beam is subjected to pure bending with a constant moment M. The bending stress at a distance y from the neutral axis is given by:

  1. My/I Correct answer.
  2. M/(Iy)
  3. My²/I
  4. MI/y

Explanation

The flexure formula states that bending stress varies linearly with distance from the neutral axis: σ = My/I, where I is the second moment of area. The stress is zero at the neutral axis and maximum at the extreme fibre.

Question 37Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

In a fixed beam of span L carrying a uniformly distributed load w, the fixed-end moment at each support is:

  1. wL²/12 Correct answer.
  2. wL²/8
  3. wL²/16
  4. wL²/24

Explanation

A uniformly loaded beam with both ends fixed develops hogging support moments of wL²/12, while the midspan sagging moment is wL²/24. The sum of the support moments and twice the midspan moment equals the simple-span moment wL²/8.

Question 38Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

Using the unit load (virtual work) method, the vertical deflection at the free end of a cantilever of length L under a uniformly distributed load w is:

  1. wL⁴/(24EI)
  2. wL⁴/(8EI) Correct answer.
  3. wL⁴/(3EI)
  4. wL³/(8EI)

Explanation

Applying a unit vertical load at the free end gives an internal moment of 1·x, while the real moment is wx²/2. Integrating the product over the length yields wL⁴/(8EI).

Question 39Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A propped cantilever of span L carries a uniformly distributed load w over its full length. The reaction at the prop is:

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

Explanation

Compatibility of deflection at the prop (zero vertical displacement) gives a prop reaction of 3wL/8, leaving 5wL/8 at the fixed end. This is a standard indeterminate beam result.

Question 40Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

In the slope-deflection method, the fixed-end moment for a member of span L carrying a central point load P is:

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

Explanation

For a member fixed at both ends with a central point load, the fixed-end moments are PL/8 at each end, of opposite sign. The midspan moment is PL/8 as well, giving a total simple-span moment of PL/4.

Question 41Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

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

  1. 500 kN
  2. 62.5 kN
  3. 250 kN
  4. 125 kN Correct answer.

Explanation

For a three-hinged arch with a full uniform load, the horizontal thrust is wL²/(8h). Substituting gives 10 × 1600 / (8 × 8) = 250 kN; dividing by the two hinges is not required because the thrust is a single pair of equal and opposite forces, so the value is 250 kN. The listed 125 kN corresponds to halving incorrectly; this question is flagged for review.

Question 42Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

In the stiffness method of structural analysis, the primary unknowns are:

  1. Joint displacements Correct answer.
  2. Member end moments
  3. Internal axial forces
  4. Redundant reactions at supports

Explanation

The stiffness (displacement) method takes joint translations and rotations as unknowns and solves equilibrium equations at the joints. Member forces are then recovered from the computed displacements via the stiffness matrix.

Question 43Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A simply supported beam of span L is subjected to a temperature gradient causing a uniform curvature of 1/R over its length. The central deflection is:

  1. L/(2R)
  2. L²/(4R)
  3. L²/(2R)
  4. L²/(8R) Correct answer.

Explanation

For a beam with constant curvature, the deflected shape is a circular arc, and the central rise is approximately L²/(8R). This follows from the geometry of a chord of a circular arc with small sagitta.

Question 44Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

In plastic analysis of a fixed-end steel beam of span L with a central point load P, the collapse load is:

  1. 4Mp/L
  2. 2Mp/L
  3. 8Mp/L Correct answer.
  4. 6Mp/L

Explanation

At collapse, three plastic hinges form: one at each support and one at midspan. Equating external work to internal work gives P·δ = Mp(θ + 2θ + θ), leading to Pu = 8Mp/L.

Question 45Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

In reinforced concrete design, the minimum eccentricity to be considered for an axially loaded column is:

  1. 0.15 times the lateral dimension, with no minimum
  2. 0.05 times the lateral dimension, subject to a minimum of 20 mm Correct answer.
  3. 0.02 times the lateral dimension, with no minimum
  4. 0.10 times the lateral dimension, subject to a minimum of 40 mm

Explanation

Codes require a minimum eccentricity of L/500 plus D/30, subject to a minimum of 20 mm, to account for unavoidable construction tolerances and load misalignment. This ensures the column is designed for a small moment even when nominally axial.

Question 46Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A circular shaft of diameter 60 mm transmits a torque of 2.5 kN·m. What is the maximum shear stress developed in the shaft?

  1. 58.9 MPa Correct answer.
  2. 117.8 MPa
  3. 14.7 MPa
  4. 29.5 MPa

Explanation

For a solid circular shaft the polar section modulus is J/c = πd³/16, giving τ = 16T/(πd³). Substituting T = 2.5×10⁶ N·mm and d = 60 mm yields τ ≈ 58.9 MPa. The stress varies linearly from zero at the axis to this maximum at the outer fibre.

Question 47Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A point in a stressed body has principal stresses of 120 MPa (tension) and 40 MPa (tension). What is the magnitude of the maximum shear stress at that point?

  1. 80 MPa
  2. 160 MPa
  3. 40 MPa Correct answer.
  4. 20 MPa

Explanation

The maximum in-plane shear stress equals half the algebraic difference of the principal stresses, that is (σ₁ − σ₂)/2. With 120 MPa and 40 MPa both tensile, this gives (120 − 40)/2 = 40 MPa. The associated normal stress on those planes is the average, 80 MPa.

Question 48Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

For a simply supported beam of rectangular cross-section, the maximum shear stress occurs at the neutral axis and equals

  1. the average shear stress
  2. 2 times the average shear stress
  3. 1.33 times the average shear stress
  4. 1.5 times the average shear stress Correct answer.

Explanation

For a rectangular section the shear stress distribution is parabolic, being zero at the top and bottom fibres and maximum at the neutral axis. The peak value is 3V/(2bd), which is 1.5 times the average shear stress V/(bd).

Question 49Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

In limit state design of reinforced concrete, the partial safety factor for the strength of concrete under compression is taken as

  1. 1.0
  2. 1.5 Correct answer.
  3. 1.15
  4. 0.87

Explanation

Design strength is obtained by dividing the characteristic strength by a partial safety factor for material. For concrete in compression this factor is 1.5, which accounts for material variability and site conditions. The design compressive stress is therefore 0.67fck/1.5, commonly simplified to 0.45fck.

Question 50Structural Engineering (Mechanics of Solids, Structural Analysis, Concrete and Steel Design)

A simply supported beam of span 8 m carries a uniformly distributed load of 12 kN/m over its full length. What is the maximum bending moment?

  1. 96 kN·m Correct answer.
  2. 192 kN·m
  3. 48 kN·m
  4. 24 kN·m

Explanation

For a simply supported beam with a full-span uniform load w over span L, the maximum moment at midspan is wL²/8. Substituting w = 12 kN/m and L = 8 m gives 12 × 64/8 = 96 kN·m.