Question 1

Section properties – polar moment:
For a solid circular shaft of diameter D, what is the polar moment of inertia J about its own axis?

Accepted Answer

Introduction / Context: The polar moment of inertia J quantifies a shaft’s resistance to torsional deformation. For circular shafts, J is directly used in torsion equations to compute shear stress and angle of twist, which are critical in power transmission and drive-train design. Given Data / Assumptions: Circular cross-section, solid, diameter D. Homogeneous, isotropic material. Axis of interest is the shaft’s centroidal axis. Concept / Approach: For circular sections, J equals the sum of the second moments of area about two orthogonal centroidal axes: J = I_x + I_y. For a solid circle, I_x = I_y = (pi / 64) * D^4. Therefore, J = 2 * (pi / 64) * D^4 = (pi / 32) * D^4. Step-by-Step Solution: Recall: I_x(solid circle) = (pi / 64) * D^4.By symmetry: I_y = I_x.Compute J: J = I_x + I_y = 2 * (pi / 64) * D^4 = (pi / 32) * D^4.Use in torsion: tau_max = T * R / J, theta = T * L / (G * J). Verification / Alternative check: Dimensionally, J has units of length^4, matching D^4. Values appear in design tables for quick reference and align with standard formulae. Why Other Options Are Wrong: (pi / 64) * D^4: this is I_x or I_y for a solid circle, not J. (pi / 16) * D^4 and (pi / 8) * D^4: overestimate J by factors of 2 and 4, respectively. Common Pitfalls: Confusing polar moment J with area moment I. Remember: torsion uses J, bending uses I. Final Answer: J = (pi / 32) * D^4

Question 2

Pure bending – stress distribution:
In a beam subjected to pure bending, the intensity of normal stress in any fibre is __________ the distance of that fibre from the neutral axis.

Accepted Answer

Introduction / Context: Understanding how bending stress varies across a beam’s depth is central to safe and economical design. Under pure bending (constant bending moment, zero shear), the normal stress varies linearly with the distance from the neutral axis, reaching zero at the neutral axis and maximum at the extreme fibres. Given Data / Assumptions: Euler–Bernoulli beam theory applies. Plane sections remain plane and perpendicular to the neutral axis after bending. Material behavior is linear elastic and homogeneous. Concept / Approach: The flexure formula provides the relationship between bending stress and the fibre distance y from the neutral axis: sigma = M * y / I, where M is the bending moment and I is the second moment of area about the neutral axis. This shows a direct proportionality to y. Step-by-Step Solution: Start from curvature–strain relation: epsilon = y / rho.Hooke’s law: sigma = E * epsilon = E * y / rho.From moment–curvature: M / I = E / rho → E / rho = M / I.Combine: sigma = (M / I) * y → sigma ∝ y. Verification / Alternative check: The linear stress distribution is consistent with zero stress at the neutral axis and equal magnitudes of tension and compression at symmetric extreme fibres for symmetric sections under pure bending. Why Other Options Are Wrong: Equal to / less than / more than: these are not general relationships; the correct dependence is proportional, not a fixed inequality. Common Pitfalls: Forgetting sign convention: fibres on one side are in tension (positive), and on the other in compression (negative). Magnitudes still scale linearly with |y|. Final Answer: Directly proportional to

Question 3

Elastic constants – E and K:
Select the correct relationship between Young's modulus (E) and bulk modulus (K) for an isotropic material in terms of Poisson's ratio v.

Accepted Answer

Introduction / Context: In isotropic linear elasticity, the elastic constants E (Young’s modulus), G (shear modulus), K (bulk modulus), and v (Poisson’s ratio) are interrelated. Knowing one pair allows computation of the others, which is especially useful when test data report different constants than those needed for analysis. Given Data / Assumptions: Material is homogeneous and isotropic. Small strains and linear elastic behavior. Standard definitions of elastic constants apply. Concept / Approach: The classical relationships among elastic constants are: E = 2 * G * (1 + v) and K = E / (3 * (1 - 2v)). Rearranging the second yields E = 3 * K * (1 - 2v). These formulas ensure consistency between volumetric and deviatoric responses in isotropic media. Step-by-Step Solution: Start from bulk modulus definition in isotropy: K = E / (3 * (1 - 2v)).Rearrange to solve for E: E = 3 * K * (1 - 2v).Optionally cross-check with E, G, v relation: E = 2 * G * (1 + v) and K = (2 * G * (1 + v)) / (3 * (1 - 2v)).Thus, the correct relation between E and K is E = 3 * K * (1 - 2v). Verification / Alternative check: Dimensional consistency is satisfied. For typical metals with v ≈ 0.3, the factor (1 - 2v) ≈ 0.4, giving E ≈ 1.2K, which aligns with typical ranges. Why Other Options Are Wrong: E = 2 * K * (1 + v): confuses K with G; the coefficient 2 belongs with G. E = K / (1 - 2v): missing factor 3; overestimates E. E = 9 * K * v / (1 + v): not a standard relation between E and K. Common Pitfalls: Using these isotropic relations for anisotropic materials (e.g., composites) where they do not hold. Final Answer: E = 3 * K * (1 - 2v)

Question 4

Bending sign convention:
Across the neutral axis of a beam under bending, the fibres on one side are in tension and on the other side are in compression. Do you agree?

Accepted Answer

Introduction / Context: When beams bend under transverse loads, internal normal stresses develop that vary from compression on one side to tension on the other, with a zero-stress surface known as the neutral axis in between. Understanding this distribution is fundamental to sizing members and checking serviceability. Given Data / Assumptions: Euler–Bernoulli beam assumptions (plane sections remain plane). Linear elastic, homogeneous material. Conventional sagging moment (concave up) places top in compression and bottom in tension for prismatic, horizontally oriented beams. Concept / Approach: From the flexure formula sigma = M * y / I, stress is proportional to the distance y from the neutral axis. The sign depends on the curvature and the side of the axis. Hence, one side experiences positive (tension) and the opposite negative (compression) stress, separated by the neutral axis where sigma = 0. Step-by-Step Solution: Identify the bending moment sign (sagging or hogging).Apply sigma = M * y / I with sign convention for y above or below the neutral axis.Conclude that stresses have opposite signs across the neutral axis.Note that the neutral axis passes through the centroid for homogeneous sections under pure bending. Verification / Alternative check: Deflection shape visualization: fibres on the concave side shorten (compression), while those on the convex side lengthen (tension), confirming the opposite nature of stress. Why Other Options Are Wrong: Disagree / conditional options: The statement holds broadly for beams in bending regardless of support type, provided material is homogeneous and within elastic range. Common Pitfalls: Confusing sign conventions or misidentifying the neutral axis location in non-homogeneous or composite sections. Final Answer: Agree

Question 5

Torsion capacity comparison:
Two shafts of the same material are compared. Shaft A is solid with diameter 100 mm. Shaft B is hollow with outer diameter 100 mm and inner diameter 50 mm. The torque that shaft B can transmit (at the same allowable she…

Accepted Answer

Introduction / Context: Designers often prefer hollow shafts to save weight while maintaining torsional strength. Torque capacity at a given allowable shear stress is proportional to the polar moment of inertia J of the section. Comparing J for solid and hollow circular sections quantifies the efficiency of material distribution. Given Data / Assumptions: Same material and allowable shear stress for both shafts. Solid shaft A: diameter D = 100 mm. Hollow shaft B: outer diameter D_o = 100 mm, inner diameter D_i = 50 mm. Length and boundary conditions are comparable so that torque capacity scales with J. Concept / Approach: For a circular shaft, maximum shear stress tau_max = T * R / J. At a fixed tau_allow, torque capacity T_allow ∝ J / R. For shafts with the same outer radius R, the ratio of allowable torques equals the ratio of J values. Here, both have the same outer radius, so T ratio simplifies to J ratio. Step-by-Step Solution: Solid shaft: J_s = (pi / 32) * D^4.Hollow shaft: J_h = (pi / 32) * (D_o^4 − D_i^4).With D_o = 100 and D_i = 50: J_h / J_s = (100^4 − 50^4) / 100^4.Compute: 50^4 = (1/16) * 100^4 → J_h / J_s = 1 − 1/16 = 15/16. Verification / Alternative check: Numeric check confirms 0.9375 fraction. The modest reduction in capacity versus large weight saving highlights the advantage of hollow shafts. Why Other Options Are Wrong: 1/6, 1/8, 1/4: greatly underestimate hollow shaft capacity; they do not reflect the fourth-power diameter dependence. Common Pitfalls: Comparing torque per unit weight without accounting for equal outer diameter; or forgetting to use the fourth power in J calculations. Final Answer: 15/16

Question 6

Neutral axis behavior:
At the neutral axis of a beam under bending, which statement best describes the state of the fibres?

Accepted Answer

Introduction / Context: The neutral axis (NA) is a central concept in beam theory. It is the locus of points within the cross-section that experience zero normal stress during bending. Correctly identifying conditions at the NA helps in understanding strain distribution and in locating the centroidal axis for homogeneous sections. Given Data / Assumptions: Linear elastic bending assumptions (plane sections remain plane). Homogeneous material, no residual stresses. Pure bending or general bending with NA defined by zero normal stress. Concept / Approach: Normal strain varies linearly through the depth: epsilon = y / rho. At y = 0 (the NA), epsilon = 0. By Hooke’s law, sigma = E * epsilon, the normal stress is also zero at the NA. Thus, fibres along the NA neither elongate nor shorten; they undergo no normal strain. Step-by-Step Solution: Take curvature 1 / rho from bending.Relate strain to distance from NA: epsilon = y / rho.At NA: y = 0 → epsilon = 0 → sigma = 0.Therefore, the layers at the NA neither tensile nor compressive strain—no normal strain develops there. Verification / Alternative check: Experimental strain gauge readings across a beam depth show zero reading at the NA, confirming theory. Why Other Options Are Wrong: Maximum/minimum bending stress occurs at extreme fibres, not at the NA. Compression at NA is incorrect; stress changes sign across the NA and is zero at it. Common Pitfalls: Assuming the NA always passes through the geometric centroid; for composite or non-homogeneous sections, the NA may shift relative to the centroid. Final Answer: The layers do not undergo any strain

Question 7

Spring terminology:
The load (force) required to produce a unit deflection of a spring is termed as what?

Accepted Answer

Introduction / Context: Springs are characterized by how much they deflect under load. The direct proportionality constant between load and deflection in the elastic range is the spring stiffness, a critical parameter for designing suspensions, valves, and vibration isolators. Given Data / Assumptions: Small deflections in the elastic range (Hooke’s law holds). Deflection measured along the spring’s working direction. Material and geometry fixed. Concept / Approach: For linear springs, F = k * x, where F is load, x is deflection, and k is spring stiffness (spring rate). Therefore, the load required for unit deflection is simply k. Other listed terms are stiffness-related but apply to beams or torsional members, not directly to overall spring response in translation. Step-by-Step Solution: Define spring law: F = k * x.Set x = 1 (unit deflection): F_unit = k.Hence, the required term is spring stiffness.For helical compression springs, k = G * d^4 / (8 * D^3 * N) (for reference), where d = wire diameter, D = mean coil diameter, N = active coils, and G = shear modulus. Verification / Alternative check: Dimensionally, k has units of force per unit displacement (e.g., N/mm), matching “load per unit deflection.” Why Other Options Are Wrong: Flexural rigidity (E * I) applies to beam bending stiffness, not directly to a spring’s axial rate. Torsional rigidity (G * J) is a torsional stiffness measure of a shaft or rod. Young’s modulus is a material constant, not the overall element stiffness. Common Pitfalls: Confusing material modulus with system stiffness; stiffness depends on geometry and boundary conditions in addition to material properties. Final Answer: Spring stiffness

Question 8

Beam diagrams – UDL case:
For a simply supported beam carrying a uniformly distributed load of intensity w per unit length over the whole span, the bending moment diagram has which shape?

Accepted Answer

Introduction / Context: Shear force and bending moment diagrams are essential tools for beam design. Recognizing the characteristic shapes for common loading scenarios allows quick checks and aids in sizing members without re-deriving equations each time. Given Data / Assumptions: Simply supported beam. Uniformly distributed load (UDL) of intensity w acting over the full span. Static, linear elastic behavior. Concept / Approach: Relationships between load w(x), shear force V(x), and bending moment M(x) are: dV/dx = −w(x) and dM/dx = V(x). For constant w, shear varies linearly with x, and bending moment varies quadratically, resulting in a parabolic bending moment diagram. Step-by-Step Solution: Load: w(x) = w (constant).Integrate once: V(x) = R_A − w * x → linear variation from support A to B.Integrate again: M(x) = R_A * x − (w * x^2) / 2 → quadratic in x.Hence, the bending moment diagram is a parabola with maximum at midspan M_max = w * l^2 / 8. Verification / Alternative check: Symmetry dictates V is zero at midspan for symmetric UDL; substituting x = l/2 into M(x) yields M_max = w * l^2 / 8, consistent with standard formulas and confirming the parabolic shape. Why Other Options Are Wrong: Horizontal/vertical line: would imply zero or undefined change; inconsistent with quadratic bending moment under constant load. Inclined straight line: corresponds to constant shear (e.g., point load case), not UDL. Common Pitfalls: Confusing the shapes: for a point load at midspan, the bending moment diagram is triangular (piecewise linear), while for UDL it is parabolic. Final Answer: A parabolic curve

Question 9

Fundamentals of stress and strain: Which statement below is incorrect? Select the single wrong definition among linear strain, Poisson’s ratio, volumetric strain, and bulk modulus.

Accepted Answer

Introduction / Context: Engineering mechanics relies on precise terminology. Confusing basic definitions of strain measures and elastic moduli leads to wrong designs. This question checks conceptual clarity on linear strain, Poisson’s ratio, volumetric strain, and bulk modulus. Given Data / Assumptions: Small deformations (engineering strain) are considered. Isotropic, homogeneous materials in the elastic range. Standard sign conventions (tension positive). Concept / Approach: Recall the standard definitions:Linear (longitudinal) strain ε = ΔL / L.Poisson’s ratio ν = − (lateral strain / longitudinal strain); the magnitude equals lateral/longitudinal.Volumetric strain ε_v = ΔV / V.Bulk modulus K = hydrostatic (volumetric) stress / volumetric strain.Young’s modulus E, not K, is the ratio of linear stress to linear strain in uniaxial loading. Step-by-Step Solution: Check option (a): matches ε = ΔL / L → correct.Check option (b): definition aligns with ν when longitudinal strain is used → correct in magnitude.Check option (c): ε_v definition is standard → correct.Check option (d): this is the definition of E, not K → wrong. Verification / Alternative check: Elasticity relations: σ = E * ε (uniaxial); p = K * ε_v (hydrostatic). These confirm the mismatch in option (d). Why Other Options Are Wrong: They are not wrong; (a), (b), and (c) are standard textbook definitions.“None of these is wrong” is invalid because (d) is incorrect. Common Pitfalls: Mixing up E and K; treating Poisson’s ratio without noting that lateral strain is of opposite sign; confusing volumetric and linear strains. Final Answer: Bulk modulus is the ratio of linear stress to linear strain.

Question 10

Composite bar of copper and steel heated to 40°C: What is the nature of stress induced in the copper bar (assuming the two are rigidly bonded and free expansion is restrained)?

Accepted Answer

Introduction / Context: Temperature changes in composite members cause self-equilibrating thermal stresses when free expansion is restrained. Understanding the sign of thermal stress in each material is essential for safe composite design. Given Data / Assumptions: Copper bar is rigidly bonded to a steel bar (composite system). Uniform heating to 40°C above the reference temperature. Free expansion is prevented by compatibility (bonding) and overall restraint. Typical properties: α_cu > α_steel, and E_cu, E_steel finite and positive. Concept / Approach: Free thermal strain for each material is epsilon_free = α * ΔT. Because α_cu > α_steel, copper wants to expand more than steel. Bonding enforces equal total strain, so copper is forced to expand less than it would freely → copper experiences compressive mechanical strain; steel is forced to expand more than freely → tensile mechanical strain. Step-by-Step Solution: Compute free strains: ε_cu_free = α_cu * ΔT, ε_st_free = α_st * ΔT.Compatibility: ε_cu_total = ε_st_total.Mechanical strains: ε_mech = ε_total − ε_free.Since ε_cu_free > ε_total, copper’s ε_mech is negative → compression; steel’s ε_mech is positive → tension. Verification / Alternative check: Force equilibrium requires internal compressive force in copper to balance tensile force in steel. Analytical solution yields sigma_cu = − (E_cu * E_st / (E_cu A_cu + E_st A_st)) * (α_cu − α_st) * ΔT * A_st in compression (sign negative). Why Other Options Are Wrong: Tensile/Zero/Shear: contradict thermal compatibility and equilibrium for α_cu > α_steel.“Depends only on cross-section” is false; coefficients of expansion are decisive. Common Pitfalls: Assuming no stress appears under uniform heating; ignoring restraint due to bonding; mixing sign convention for tension/compression. Final Answer: Compressive

Question 11

Short column criterion: When is a column classified as “short”? Choose the standard slenderness-based definition.

Accepted Answer

Introduction / Context: Columns are classified as short or long based on their tendency to fail by crushing or by buckling. Slenderness ratio governs the dominant failure mode and the analysis method (material strength vs. stability). Given Data / Assumptions: Slenderness ratio = L / k_min, where L is effective length and k_min is least radius of gyration. Typical steel design heuristics: short if L/k_min ≲ 80 (some codes use different thresholds). Elastic behaviour up to failure criterion. Concept / Approach: Low slenderness implies stocky columns with minimal buckling tendency; failure is governed by material crushing (compressive stress limit). High slenderness increases susceptibility to elastic instability (Euler buckling), requiring stability checks. Step-by-Step Solution: Compute slenderness: λ = L / k_min.If λ is small (≈ 80 or below for steel), classify as short.Use compressive strength-based design (including end conditions, imperfections as per code). Verification / Alternative check: Compare with Euler critical load formula P_cr = π^2 E I / L_e^2; small L/k increases P_cr sharply, making buckling less critical, consistent with “short column.” Why Other Options Are Wrong: Absolute length or area alone (a, b) do not decide buckling behaviour.“More than 80” (d) corresponds to a long/slender column.Load magnitude (e) is not a classification criterion. Common Pitfalls: Ignoring effective length (end restraints); using area instead of radius of gyration; applying the same slenderness limit to all materials without checking the relevant code. Final Answer: When the slenderness ratio (L / k_min) is less than about 80

Question 12

Thick-walled cylinder under internal pressure p: The circumferential (tangential) stress is tensile while the radial stress is compressive across the thickness. Is this statement correct?

Accepted Answer

Introduction / Context: Pressurized cylinders (gun barrels, pressure vessels, pipes) exhibit non-uniform stress fields. Distinguishing tangential (hoop) and radial stresses is crucial in thick-cylinder design using Lame’s equations. Given Data / Assumptions: Internal pressure p acts; external pressure may be zero unless stated. Thick cylinder theory (Lame) applies; elastic, isotropic material. Ends closed; plane strain in axial direction if long. Concept / Approach: In a thick tube:sigma_theta (hoop) is tensile and highest near the inner radius.sigma_r (radial) is compressive, equal to −p at the inner surface and reduces in magnitude toward the outer surface (often to zero if external pressure is zero). Step-by-Step Solution: Lame’s forms: sigma_r = A − B/r^2, sigma_theta = A + B/r^2.At r = r_i: sigma_r = −p (compressive), sigma_theta is tensile and maximum.At r = r_o: if p_ext = 0, sigma_r = 0, while sigma_theta remains tensile but smaller than at r_i. Verification / Alternative check: Plotting sigma_r across thickness shows negative values increasing toward zero; sigma_theta always positive for internal pressure only, confirming the statement. Why Other Options Are Wrong: “Incorrect” contradicts thick-cylinder solutions.Dependence on external pressure alone is false; sign pattern holds for p_ext = 0.Truth limited to a surface is incorrect; the sign trend holds throughout thickness. Common Pitfalls: Using thin-cylinder formulas; confusing sign conventions; assuming radial stress is tensile at the outer surface without external pressure. Final Answer: Correct

Question 13

Welded joints: Shear strength of a double parallel fillet weld (two sides) of length l and effective throat s, using allowable shear stress τ, is equal to?

Accepted Answer

Introduction / Context: Designing fillet welds requires computing load capacity from effective throat area and allowable shear stress. A double parallel fillet weld places fillets on two opposite sides of a joint, doubling the effective area. Given Data / Assumptions: Two parallel fillet welds of length l each. Effective throat size = s (for a right-angled fillet, s ≈ 0.707 * leg size). Allowable shear stress in weld metal = τ. Concept / Approach: Shear strength = allowable shear stress * effective shear area. For parallel fillets on two sides, total effective throat area equals 2 * s * l (neglecting end returns). Therefore capacity equals P_allow = 2 * s * l * τ. Step-by-Step Solution: Compute area for one fillet: A_1 = s * l.Two fillets in parallel: A_total = 2 * s * l.Strength: P_allow = A_total * τ = 2 * s * l * τ. Verification / Alternative check: Compare with code formulas where effective throat = 0.707 * leg; substituting s = 0.707 a gives the same scaling factor of 2 for two weld lines. Why Other Options Are Wrong: 0.5 s l τ and s l τ underestimate by ignoring the second fillet or using an incorrect factor.4 s l τ overestimates; there are only two parallel fillets, not four.Option “2.s.l.τ” is the same as the correct expression but duplicated; we select the standard formatting in option (c). Common Pitfalls: Confusing leg size with throat size; forgetting that two-sided fillets double the area; neglecting reduction factors for intermittent welds or eccentric loading (not applicable here). Final Answer: 2 s l τ

Question 14

Mohr’s circle basics: The radius of Mohr’s circle equals which combination of the two principal stresses?

Accepted Answer

Introduction / Context: Mohr’s circle graphically represents plane stress transformation. Knowing the center and radius allows immediate computation of principal stresses and maximum shear stress. Given Data / Assumptions: Principal stresses σ1 and σ2 are real values (plane stress state). Standard plotting: σ-axis horizontal, τ-axis vertical. Positive shear convention consistent with Mohr’s construction. Concept / Approach: The circle’s center is at C = (σ1 + σ2)/2 on the σ-axis, and the radius is R = (σ1 − σ2)/2. The maximum in-plane shear equals this radius. Step-by-Step Solution: Plot points A(σ1, 0) and B(σ2, 0) on the σ-axis.Center C is the midpoint of AB → (σ1 + σ2)/2.Radius R is half the length of AB → (σ1 − σ2)/2. Verification / Alternative check: General stress components (σx, σy, τxy) reduce to principal form by rotation; analytical solutions give σ1,2 = (σx + σy)/2 ± √[ ((σx − σy)/2)^2 + τxy^2 ], where the radical equals the radius R. Why Other Options Are Wrong: Sum or half-sum describe the center, not the radius.Full difference is twice the radius.Product has no direct geometric meaning in Mohr’s circle radius. Common Pitfalls: Mixing center and radius formulas; sign mistakes in τ that reflect points across the σ-axis. Final Answer: Half the difference of the two principal stresses

Question 15

Thick cylinder under internal pressure only: How does radial stress vary across the wall thickness?

Accepted Answer

Introduction / Context: Design of thick-walled pressure components requires correct understanding of radial and hoop stress distributions. Errors in the sign or boundary conditions can seriously compromise safety. Given Data / Assumptions: Internal pressure p acts at the bore; external pressure is zero unless stated. Elastic, isotropic material; Lame’s solution applies. Static loading; no thermal gradients. Concept / Approach: From Lame’s equations: sigma_r = A − B/r^2. Boundary conditions for internal-only pressure are sigma_r(r_i) = −p (compressive) and sigma_r(r_o) = 0. Thus the largest magnitude of radial stress occurs at the inner radius and decreases to zero at the outer radius. Step-by-Step Solution: Apply inner boundary: sigma_r(r_i) = −p (maximum compressive).Apply outer boundary: sigma_r(r_o) = 0 (free surface).Therefore, radial stress varies from −p at r_i to 0 at r_o; it is not uniform. Verification / Alternative check: Plotting sigma_r versus r yields a monotonic curve increasing from −p to 0. Hoop stress sigma_theta remains tensile and larger than |sigma_r| near the inner radius. Why Other Options Are Wrong: Options (a) and (c) reverse the boundary conditions.Option (b) is incomplete; it omits that sigma_r at the outer surface equals zero when external pressure is absent.Uniform distribution is a thin-wall assumption for hoop stress, not for radial stress in thick walls. Common Pitfalls: Assuming thin-wall results; forgetting that radial stress at a free outer surface is zero when no external pressure acts. Final Answer: Maximum at the inner surface and zero at the outer surface (when no external pressure)

Question 16

Pure torsion of circular shaft: Under a twisting moment, every cross-section of the shaft is primarily subjected to which type of stress?

Accepted Answer

Introduction / Context: Shafts transmit torque in machines (motors, gearboxes, rotors). The dominant stress under pure torsion for circular shafts is shear, which defines the torsional strength and angle of twist. Given Data / Assumptions: Solid or hollow circular shaft. Pure torque (no bending, no axial load). Linear elastic, isotropic material; Saint-Venant torsion holds. Concept / Approach: In circular shafts, warping is negligible and cross-sections remain plane. The shear stress varies linearly with radius: tau(r) = T * r / J where T is torque and J is polar second moment of area. Normal stresses (tension/compression) arise only if bending or axial loads are present. Step-by-Step Solution: Identify loading: twisting moment T only.Use torsion formula: tau_max at outer radius r_o is tau_max = T * r_o / J.Angle of twist: theta = T * L / (G * J), confirming shear-dominated behaviour (G is shear modulus). Verification / Alternative check: Mohr’s circle for pure shear shows principal normal stresses ±tau at 45°, but the applied stress state on the cross-section is shear. Why Other Options Are Wrong: Tension/Compression/Bending are not produced by pure torque in a circular shaft.Hydrostatic stress requires equal normal stresses in all directions, unrelated to torsion. Common Pitfalls: Confusing principal normal stresses at 45° with applied shear on the cross-section; applying noncircular-shaft concepts where warping matters. Final Answer: Shear stress

Question 17

Combined stress state: A body has direct tensile stress σ = 300 MPa (one plane) and simple shear τ = 200 MPa. What is the maximum in-plane shear stress?

Accepted Answer

Introduction / Context: When both normal and shear stresses act, the maximum shear stress must be computed using stress transformation or Mohr’s circle. This governs yielding under Tresca criterion and influences fatigue in multiaxial states. Given Data / Assumptions: Plane stress with σx = 300 MPa, σy = 0 MPa, τxy = 200 MPa. Linear elasticity; sign conventions standard. We seek the maximum in-plane shear stress. Concept / Approach: Maximum in-plane shear is the radius of Mohr’s circle: tau_max = √[ ((σx − σy)/2)^2 + τxy^2 ]. This reduces to the pure-shear value σ/2 only when τxy = 0. Step-by-Step Solution: Compute half-difference: (σx − σy)/2 = (300 − 0)/2 = 150 MPa.Apply formula: tau_max = √(150^2 + 200^2) MPa.Evaluate: 150^2 = 22500; 200^2 = 40000; sum = 62500.tau_max = √62500 = 250 MPa. Verification / Alternative check: Construct Mohr’s circle with points (300, 200) and (0, −200); the circle radius equals 250 MPa, confirming the calculation. Why Other Options Are Wrong: −100 MPa: shear magnitude cannot be negative; sign denotes sense only.300 or 400 MPa: overestimates; not aligned with the geometric radius.200 MPa: equals applied shear but ignores accompanying normal stress. Common Pitfalls: Using σ/2 blindly; neglecting the influence of τxy; arithmetic slips when squaring and adding. Final Answer: 250 MPa

Question 18

Simply supported beam of span l under uniformly distributed load w (per unit length): The location of maximum deflection and its value—does the maximum occur at midspan?

Accepted Answer

Introduction / Context: Serviceability design limits deflection to protect finishes and function. For common loading cases, knowing where the maximum deflection occurs avoids unnecessary full integrations. Given Data / Assumptions: Straight, prismatic beam; simply supported ends. Uniformly distributed load w along the whole span. Elastic analysis; constant E and I. Concept / Approach: The classical solution for a simply supported beam with full-span UDL gives maximum deflection at the midspan (x = l/2) with value:delta_max = 5 w l^4 / (384 E I). Symmetry of loading and supports ensures the extremum is at the center. Step-by-Step Solution: Write bending moment: M(x) = (w/2)(l x − x^2).Curvature: d^2y/dx^2 = M(x)/(E I).Integrate twice, apply boundary conditions y(0) = 0, y(l) = 0.Evaluate deflection at x = l/2 → delta_max = 5 w l^4 / (384 E I). Verification / Alternative check: Use standard beam tables or Castigliano’s theorem; both return the same midspan maximum and value. Why Other Options Are Wrong: “False” contradicts the symmetric solution.Linear varying w changes the position; but here w is uniform.End fixity is a different support condition; not applicable.Cantilevers have different deflection profiles; the maximum is at the free end, not midspan. Common Pitfalls: Mixing formulas for point loads and UDLs; forgetting the 5/384 factor; confusing simply supported with fixed-ended results. Final Answer: True

Question 19

Cantilever beam with linearly varying load

A cantilever beam of length l carries a triangular distributed load that varies from zero at the free end to w per unit length at the fixed end. Identify the correct nature of the bending moment diagram …

Accepted Answer

Introduction / Context: This question checks understanding of how distributed load shapes translate into shear force and bending moment diagrams in strength of materials. Recognizing whether the bending moment diagram (BMD) is linear, parabolic, or cubic is critical for sizing beams and predicting maximum stresses and deflections. Given Data / Assumptions: Cantilever beam of length l, fixed at one end and free at the other. Transverse load varies linearly along the span: zero at the free end, w per unit length at the fixed end. Small-deflection beam theory; sign conventions standard. Concept / Approach: For any beam: load intensity w(x) integrates to shear V(x), and shear integrates to bending moment M(x). If w(x) is uniform (constant), V is linear and M is parabolic. If w(x) is linear (triangular), then V is quadratic in x and M is cubic in x. Step-by-Step Solution: Let w(x) = k x (with x measured from the free end toward the fixed end).Shear force: V(x) = ∫ w(x) dx = ∫ k x dx = (k/2) x^2 + C → quadratic.Bending moment: M(x) = ∫ V(x) dx = ∫ (k/2) x^2 dx = (k/6) x^3 + C′ → cubic.Therefore the BMD is a cubic curve (not a parabola). Verification / Alternative check: Dimensional progression rule: increase the polynomial order by one upon integration. Triangular load (order 1) → shear (order 2) → moment (order 3). This quick rule confirms the cubic nature. Why Other Options Are Wrong: Agree: incorrect because it calls the BMD parabolic. The BMD is cubic. The options about dependence on elastic modulus or shear shape are distractors; the curve family depends directly on the load distribution and basic equilibrium relations. Common Pitfalls: Confusing the UDL case (parabolic M) with triangular loading (cubic M); mixing the coordinate origin; forgetting that shear is the integral of load and moment is the integral of shear. Final Answer: Disagree

Question 20

Hollow circular shaft – torque capacity expression

For a hollow circular shaft of outer diameter D and inner diameter d subjected to torsion, which expression gives the transmitted torque T in terms of maximum shear stress τ_max?

Accepted Answer

Introduction / Context: Design of shafts under torsion requires the correct relationship among torque, section geometry, and shear stress. For hollow shafts, capacity depends on the polar section modulus rather than simple area. Given Data / Assumptions: Hollow circular shaft with outer diameter D and inner diameter d. Maximum shear stress in the material is τ_max. Elastic torsion theory applies (linear elastic, Saint-Venant assumptions). Concept / Approach: The torsion formula is T / J = τ / R = G θ / L. The torque–stress relation uses the polar second moment J and radius to the outer fiber R = D/2. The polar section modulus Z_p = J / R directly gives T = τ_max * Z_p. Step-by-Step Solution: Polar second moment: J = (π/32) * (D^4 - d^4).Outer radius: R = D / 2.Polar section modulus: Z_p = J / R = [(π/32) * (D^4 - d^4)] / (D/2) = (π/16) * (D^4 - d^4) / D.Torque capacity at τ_max: T = τ_max * Z_p = (π/16) * τ_max * (D^4 - d^4) / D. Verification / Alternative check: For a solid shaft (d = 0), the formula reduces to T = (π/16) * τ_max * D^3, which is the standard solid-shaft result (since J = π D^4 / 32 and Z_p = J/R = π D^3 / 16). Why Other Options Are Wrong: (a) omits division by R and thus gives J scaled by τ rather than T; (c) uses area terms D^2 - d^2 which are not appropriate for torsion; (d) has an incorrect constant; (e) is dimensionally inconsistent. Common Pitfalls: Confusing polar second moment with area; forgetting to divide by the outer radius; mixing diameter and radius. Final Answer: T = (π/16) * τ_max * (D^4 - d^4) / D

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