Question 1

Effect of section modulus on bending stress

If the section modulus Z of a beam is increased while bending moment M remains the same, how will the maximum bending stress change?

Accepted Answer

Introduction / Context: Beam design frequently leverages the relationship σ = M / Z. Increasing section modulus by changing cross-section shape or size is a primary strategy to cut bending stress for a given moment. Given Data / Assumptions: Bending moment M is fixed by loading and span. Section modulus Z is altered via geometry (e.g., larger depth, flanges). Linear elastic bending; plane sections remain plane. Concept / Approach: Maximum bending stress in elastic range is σ_max = M / Z. This is an inverse proportionality: higher Z means lower σ for the same M. Step-by-Step Solution: Start with σ = M / Z.Increase Z → denominator increases.Therefore σ decreases proportionally.No sign change occurs; stress nature (tension/compression) is set by loading, not Z. Verification / Alternative check: Example: if Z doubles, σ halves. This is why I-sections (large Z for given area) are efficient for bending resistance. Why Other Options Are Wrong: Not change: contradicts σ = M / Z. Increase: opposite of the formula. Become zero or reverse sign: impossible without changing M or load sense. Common Pitfalls: Confusing section modulus Z with second moment I; Z = I / c, so increasing depth c while adjusting I is key to stress reduction. Final Answer: decrease

Question 2

Helical spring deflection ratio – effect of mean coil diameter

Two closely coiled helical springs A and B are made of the same material, have the same number of active turns, and the same wire diameter. Under the same axial load W, the mean coil …

Accepted Answer

Introduction / Context: Spring sizing depends strongly on geometry. For closely coiled helical springs, deflection varies with the cube of the mean coil diameter, making diameter a dominant factor in compliance. Given Data / Assumptions: Springs A and B: same material (same shear modulus G), same number of active turns n, and same wire diameter (common symbol often d_w). Both carry the same axial load W. Mean coil diameters: D_A = 2 D_B. Concept / Approach: The deflection of a closely coiled helical spring under axial load is δ = 8 W D^3 n / (G d_w^4). With all factors except D identical, δ ∝ D^3. Step-by-Step Solution: δ_A / δ_B = (D_A^3) / (D_B^3).Given D_A = 2 D_B → δ_A / δ_B = (2^3) = 8.Therefore δ_B / δ_A = 1 / 8. Verification / Alternative check: Dimensional check: the formula shows cubic dependence on D, consistent with torsional deformation of coils. Why Other Options Are Wrong: 1/4, 2, 4, 8 result from linear or squared assumptions about diameter influence; the correct relationship is cubic. Common Pitfalls: Confusing wire diameter with mean coil diameter; using outer instead of mean diameter; ignoring inactive end coils (not affecting the ratio here since n is the same). Final Answer: 1/8

Question 3

Axial deformation and linear strain of a rectangular bar

A rectangular bar of length l, breadth b, and thickness t is subjected to an axial tensile pull P. What is the linear (engineering) strain ε along the length in terms of P, b, t, and modulu…

Accepted Answer

Introduction / Context: Linear strain connects applied load to material stiffness and cross-section. This relationship underpins elongation predictions for bars, tie rods, and machine members subjected to direct tension. Given Data / Assumptions: Prismatic bar with constant rectangular cross-section A = b * t. Axial tensile load P; homogeneous, isotropic material. Linear elastic behavior (Hooke's law) and small strains. Concept / Approach: Normal stress σ = P / A. Engineering strain ε = σ / E in the linear elastic range. Substituting A gives ε = (P / (b * t)) / E = P / (b * t * E). Step-by-Step Solution: Area: A = b * t.Stress: σ = P / A = P / (b * t).Hooke’s law: ε = σ / E = [P / (b * t)] / E.Therefore: ε = P / (b * t * E). Verification / Alternative check: Elongation Δl = (P * l) / (A * E). Dividing by original length l yields ε = Δl / l = P / (A * E) = P / (b * t * E), matching the result. Why Other Options Are Wrong: (b) inverts the relationship; (c) mixes incompatible terms; (d) ignores area and is dimensionally incorrect; (e) gives displacement-like units rather than dimensionless strain. Common Pitfalls: Forgetting to use cross-sectional area; confusing strain with elongation; mixing true and engineering strain definitions. Final Answer: ε = P / (b * t * E)

Question 4

Thermal stress in a restrained bar – dependence on area

In a prismatic bar fully restrained against expansion and subjected to a uniform temperature rise ΔT, the induced thermal stress will

Accepted Answer

Introduction / Context: Thermal stresses arise when free thermal expansion or contraction is prevented. Designers must know whether bar size alters the induced stress under full restraint to select materials and allowances properly. Given Data / Assumptions: Uniform temperature change ΔT. Bar is perfectly restrained axially (no net change in length). Linear elastic material with Young’s modulus E and thermal expansion coefficient α. Concept / Approach: Free thermal strain would be ε_free = α ΔT. Full restraint enforces ε_total = 0, so mechanical strain must cancel thermal strain. Hence ε_mech = −α ΔT, and the induced stress is σ = E * ε_mech = −E α ΔT, independent of area. Step-by-Step Solution: Free strain: ε_free = α ΔT.Restraint condition: ε_total = ε_mech + ε_free = 0.Therefore ε_mech = −α ΔT.Stress from Hooke’s law: σ = E * ε_mech = −E α ΔT (sign indicates compression for heating). Verification / Alternative check: Although the reaction force depends on area (F = σ * A), the stress level σ does not. Two bars of different areas but same material and ΔT develop the same thermal stress under identical perfect restraint. Why Other Options Are Wrong: (a) confuses force with stress; (c) moment of inertia matters for bending, not pure axial thermal stress; (d) zero stress occurs only when expansion is unrestrained; (e) length influences free expansion, but not the stress under perfect restraint. Common Pitfalls: Mixing up stress and force; assuming partial restraint behaves the same as full restraint; neglecting temperature gradients which can induce bending. Final Answer: does not depend on the cross-sectional area of the bar

Question 5

Definition of neutral axis in beam bending

The line formed by the intersection of the neutral layer with any normal cross-section of a beam is called the neutral axis. State whether this statement is correct.

Accepted Answer

Introduction / Context: Understanding the neutral axis helps visualize stress distributions in bending. At the neutral axis, fiber strain and bending stress are zero, transitioning from compression to tension across the section. Given Data / Assumptions: Beam under pure or combined bending (elastic range). Plane sections remain plane; material is homogeneous and isotropic. Concept / Approach: The neutral layer (or surface) is the locus of points within the beam that experience zero longitudinal strain during bending. Intersecting this surface with a cross-section produces a line: the neutral axis. Step-by-Step Solution: Strain distribution in bending is linear: ε = (y / R).At y = 0 (neutral layer), ε = 0 and hence σ = E ε = 0.Intersection with a cross-section gives the neutral axis, lying typically through the centroid for homogeneous beams under pure bending. Verification / Alternative check: For symmetric homogeneous sections, the neutral axis passes through the centroidal axis. For non-symmetric sections, it still exists but its position relative to geometric axes depends on stiffness distribution. Why Other Options Are Wrong: Limiting the definition to cantilevers, prismatic, or symmetrical sections is unnecessary; the concept applies broadly in elastic bending. Common Pitfalls: Confusing neutral axis with zero shear line; assuming neutral axis always aligns with a geometric symmetry axis in composite or nonhomogeneous beams. Final Answer: True

Question 6

Principal normal stress under uniaxial tension

A body is subjected only to a direct tensile stress σ (no other stresses). The maximum normal stress in the body equals:

Accepted Answer

Introduction / Context: Principal stresses represent extreme normal stresses at a point. For basic stress states, being able to identify principal values without constructing Mohr’s circle is a key skill in mechanics of materials. Given Data / Assumptions: Only one nonzero stress component: σ_x = σ (tension); σ_y = 0; τ_xy = 0. Continuum is homogeneous; small deformations. Concept / Approach: Principal stresses for a 2D state satisfy τ_n = 0 on principal planes. With σ_x = σ, σ_y = 0, τ_xy = 0, the principal stresses are simply σ_1 = σ and σ_2 = 0. Therefore, the maximum normal stress equals σ. Step-by-Step Solution: General principal stress formula: σ_{1,2} = (σ_x + σ_y)/2 ± sqrt[((σ_x − σ_y)/2)^2 + τ_xy^2].Substitute σ_x = σ, σ_y = 0, τ_xy = 0 → σ_{1,2} = σ/2 ± sqrt[(σ/2)^2] = σ and 0.Hence maximum normal stress = σ. Verification / Alternative check: Mohr’s circle reduces to a circle with center at σ/2 on the σ-axis and radius σ/2. The rightmost point corresponds to σ. Why Other Options Are Wrong: σ/2 confuses center value with principal maximum; 2σ is impossible without other stresses; zero is only the minor principal stress; dependence on Poisson’s ratio is irrelevant to principal stress magnitude here. Common Pitfalls: Mixing true and engineering stress; assuming lateral contraction affects principal stress magnitude; forgetting that shear is zero on principal planes. Final Answer: the direct tensile stress σ

Question 7

Unriveted (solid) plate strength per pitch length in riveted joint design

For a plate of thickness t, pitch p, and allowable tensile stress σ_t, what is the strength of the unriveted (solid) plate per pitch length?

Accepted Answer

Introduction / Context: In riveted joint design, capacities per pitch length are compared for several failure modes. The strength of the unriveted (solid) plate is the baseline tensile capacity if no rivet hole weakened the section. Given Data / Assumptions: Plate thickness t and pitch p between rivets. Allowable tensile stress in the plate σ_t. Linear elastic, uniform stress distribution assumed for nominal capacity comparison. Concept / Approach: The unriveted (solid) plate per pitch carries tension across the full pitch area A_solid = p * t. Its tensile capacity equals allowable stress times this area. Step-by-Step Solution: Solid area per pitch: A_solid = p * t.Allowable tensile capacity: F = σ_t * A_solid.Therefore, solid plate strength per pitch = p * t * σ_t. Verification / Alternative check: Compare with net section capacity (p − d) * t * σ_t used for plate tearing around the rivet; the solid capacity is larger because no hole reduces the area. Why Other Options Are Wrong: (a) and (d) relate to bearing or shear modes; (c) gives net section (tearing) strength, not the unriveted plate; (e) uses circular area, irrelevant for plate tension over pitch width. Common Pitfalls: Confusing solid, net, and bearing capacities; mixing symbols σ_t (tension), σ_c (bearing/crushing), and τ (shear). Final Answer: p * t * σ_t

Question 8

Columns under compressive loading: What do we call a column that primarily fails by direct compressive stress (crushing) rather than by elastic buckling?

Accepted Answer

Introduction / Context: Columns may fail either by direct compressive crushing or by elastic instability (buckling). Correctly identifying the governing failure mode is essential for sizing members and choosing the right formula (crushing stress versus Euler or Rankine theory). Given Data / Assumptions: Prismatic column under axial compression. Low slenderness ratio (small effective length to radius of gyration). Material behaves elastically up to crushing strength. Concept / Approach: Failure mode depends primarily on slenderness ratio. Short, stocky members (small slenderness) develop high compressive stresses before they can deflect laterally; thus they fail by direct compression (crushing). Long, slender members fail by buckling at stresses well below crushing strength. Intermediate columns require combined formulas (e.g., Rankine–Gordon). Step-by-Step Solution: Classify by slenderness ratio: lambda = Le / r.If lambda is small (stocky), buckling load is very high and crushing governs.Therefore, a column failing by direct stress is termed a short column. Verification / Alternative check: Compare Euler load P_cr ∝ 1 / (Le^2) with material crushing load P_c = A * sigma_c. For small Le, P_cr ≫ P_c, so crushing occurs first, confirming the classification as a short column. Why Other Options Are Wrong: Long column: dominated by buckling, not direct crushing. Medium column: mixed behavior; not necessarily direct-stress failure. Weak column: vague term; weakness is not the defining criterion. Common Pitfalls: Using Euler's formula for stocky members; ignoring end conditions that change effective length; forgetting to check material crushing stress first for low slenderness ratios. Final Answer: short column

Question 9

Thick cylinder under internal pressure p: What is the maximum radial stress at the inner surface of a thick cylindrical shell (sign convention: tensile positive)?

Accepted Answer

Introduction / Context: Lame's theory for thick-walled cylinders provides radial and hoop stresses that vary across the thickness. Knowing boundary values, especially at the bore where stresses are critical, is fundamental for pressure vessel design and safety assessment. Given Data / Assumptions: Thick-walled, cylindrical geometry; internal pressure p only. Elastic, isotropic material; plane stress in the cylinder wall. Sign convention: tensile positive, compressive negative. Concept / Approach: Lame's solution gives radial stress sigma_r and hoop stress sigma_theta as functions of radius. Boundary condition at the inner radius r_i is sigma_r(r_i) = -p (compressive), and at the outer radius r_o (if no external pressure) sigma_r(r_o) = 0. Thus the maximum magnitude of radial stress occurs at the inner surface and equals the internal pressure but compressive in sign. Step-by-Step Solution: Apply boundary: sigma_r(r_i) = -p.Across thickness, sigma_r increases from -p at r_i to 0 at r_o.Therefore, maximum radial stress (by magnitude) at the inner surface is -p (compressive). Verification / Alternative check: Physical reasoning: internal pressure pushes the wall inward; on the inner face the fluid contact produces compressive normal stress equal to p on the solid surface, matching sigma_r = -p. Why Other Options Are Wrong: Zero or +p contradict boundary conditions. 2p or p/2 are not consistent with Lame's boundary values for radial stress. Common Pitfalls: Confusing radial stress with hoop stress (which is tensile and typically larger in magnitude); mixing sign conventions; assuming thin cylinder relations where radial stress is approximated as negligible across thickness. Final Answer: -p (compressive)

Question 10

Weld design (double fillet, lap/tee joint): Using the conventional leg size s and weld length l, the tensile strength (capacity) of a double-fillet welded joint is approximated as:

Accepted Answer

Introduction / Context: Design of fillet welds often uses simplified strength expressions in terms of leg size s and weld length l for quick sizing. For symmetric double-fillet joints (e.g., lap or tee with fillets on both sides), the effective area is roughly doubled compared to a single fillet, leading to a doubled tensile capacity under direct load sharing. Given Data / Assumptions: Double fillet of leg size s on both sides, length l. Allowable tensile stress for weld metal = σt. Conventional simplified design (leg-size basis), ignoring detailed throat coefficient in this question's formulation. Concept / Approach: In basic textbook treatments aligned with these options, the tensile capacity is taken proportional to weld effective area. For a single fillet, capacity is approximately s * l * σt; for a double fillet, area doubles, so capacity ≈ 2 * s * l * σt. (More rigorous practice uses the throat thickness t_throat = 0.707 s, giving 2 * 0.707 s * l * σ_allow, but such detail is not reflected in the given options.) Step-by-Step Solution: Single fillet capacity (simplified) ≈ s * l * σt.Double fillet → two such areas in parallel → 2 * s * l * σt.Therefore, the tensile strength of the double-fillet joint is 2 s * l * σt. Verification / Alternative check: Throat-based check: using t_throat = 0.707 s gives capacity ≈ 2 * 0.707 s * l * σ_allow = 1.414 s l σ_allow, close to but slightly less than 2 s l σt when σ_allow definitions differ. The problem's answer set aligns with the simplified leg-size expression. Why Other Options Are Wrong: 0.5 s l σt and s l σt correspond to smaller capacities (single fillet or half value). 0.707 s l σt is a single-fillet throat approximation, not double fillet. 2.0 s l σt duplicates option C; both indicate the same correct value. Common Pitfalls: Confusing leg size with throat thickness; mixing allowable weld metal stress with base metal allowable; forgetting that two fillets share the load. Final Answer: 2 s * l * σt

Question 11

Thin cylindrical pressure vessel: For a cylinder of diameter d and thickness t under internal pressure p, what is the longitudinal (axial) membrane stress in the shell?

Accepted Answer

Introduction / Context: Thin-walled pressure vessels develop two principal membrane stresses: circumferential (hoop) and longitudinal (axial). Correct formulas are vital for sizing thickness and checking allowable stress compliance in boilers, tanks, and pipelines. Given Data / Assumptions: Thin cylinder, t ≪ d. Internal pressure p; ends closed. Uniform membrane stresses; neglect radial stress. Concept / Approach: Equilibrium on a free-body of half the vessel gives hoop stress sigma_h = p d / (2 t). Equilibrium on the end cap gives longitudinal stress sigma_L = p d / (4 t). The longitudinal stress is half the hoop stress in a thin cylinder with closed ends. Step-by-Step Solution: Hoop: sigma_h = p d / (2 t).End equilibrium: p * (π d^2 / 4) balanced by 2 * sigma_L * t * (d / 2) * π → sigma_L = p d / (4 t).Hence the longitudinal stress is p d / (4 t). Verification / Alternative check: Check ratio: sigma_h / sigma_L = 2, consistent with thin cylinder theory. Why Other Options Are Wrong: p d / t and 2 p d / t are excessive (do not match equilibrium). p d / (2 t) is the hoop, not longitudinal, stress. p d / (6 t) has no basis in thin-cylinder derivation. Common Pitfalls: Using hoop formula for both directions; forgetting that end closures create axial stress; applying thick-cylinder relations to thin shells. Final Answer: p d / (4 t)

Question 12

Plane stress with in-plane shear: Given normal stresses σx and σy on perpendicular planes and shear τxy, what is the maximum normal (principal) stress?

Accepted Answer

Introduction / Context: Transformations of plane stress determine principal stresses and directions. The maximum normal stress controls yielding under maximum normal stress criteria and is widely used to assess safety margins in flat plates and machine elements. Given Data / Assumptions: Plane stress state: σx, σy, τxy known. Homogeneous, isotropic, linear elastic material. Standard sign conventions. Concept / Approach: From Mohr's circle or transformation equations, principal stresses are σ1,2 = (σx + σy)/2 ± sqrt( ((σx − σy)/2)^2 + τxy^2 ) The maximum normal stress is σ1, obtained using the plus sign. Step-by-Step Solution: Compute average normal stress: (σx + σy)/2.Compute radius R = sqrt( ((σx − σy)/2)^2 + τxy^2 ).Maximum principal stress: σ1 = (σx + σy)/2 + R. Verification / Alternative check: Special cases: if τxy = 0, σ1 = max(σx, σy); if σx = σy, σ1 = σx + |τxy|, matching Mohr's circle for pure shear superposed on hydrostatic stress. Why Other Options Are Wrong: (σx + σy)/2 is the center value, not the maximum principal stress. |σx − σy|/2 is valid only when τxy = 0 and then equals the radius, not σ1 directly. sqrt( σx^2 + σy^2 + τxy^2 ) is not a principal-stress formula. (σx − σy)/2 + τxy is not invariant and lacks proper vector combination. Common Pitfalls: Dropping the 1/2 inside the square root; mixing up radius and diameter on Mohr's circle; sign mistakes for τxy. Final Answer: (σx + σy) / 2 + sqrt( ((σx − σy) / 2)^2 + τxy^2 )

Question 13

Earth pressure on a retaining wall: The total horizontal thrust per metre length (active condition) on a wall of height h retaining cohesionless soil (unit weight w, angle of repose φ) is

Accepted Answer

Introduction / Context: Retaining structures must resist lateral earth pressures. The Rankine active earth pressure theory provides a convenient closed-form for cohesionless backfill, frequently used in preliminary design and checks. Given Data / Assumptions: Backfill is dry, cohesionless, with unit weight w and angle of internal friction φ (equal to angle of repose). Vertical, smooth wall; horizontal backfill surface (Rankine conditions). Active state of earth pressure (wall yields sufficiently). Concept / Approach: Rankine active pressure coefficient K_a = tan^2(45° − φ/2) = (1 − sin φ)/(1 + sin φ). Resultant active thrust P_a per metre length is P_a = (1/2) * w * h^2 * K_a acting at h/3 above the base. Substituting K_a gives the expression in terms of φ. Step-by-Step Solution: Compute K_a = tan^2(45° − φ/2).Total thrust: P_a = (w * h^2 / 2) * K_a.Hence P_a = (w * h^2 / 2) * tan^2(45° − φ/2). Verification / Alternative check: Using identity K_a = (1 − sin φ)/(1 + sin φ), P_a = (w h^2 / 2) * (1 − sin φ)/(1 + sin φ), consistent with Rankine theory. Why Other Options Are Wrong: tan^2(45° + φ/2) corresponds to K_p (passive), not active. w h^2 tan φ lacks the correct coefficient and dimensions. (w h^2 / 3) times the ratio mixes location of resultant with magnitude. (w h^2 / 2) ignores soil friction angle entirely. Common Pitfalls: Confusing K_a and K_p signs; forgetting that the resultant acts at h/3; applying Coulomb theory coefficients to Rankine conditions without accounting for wall friction or backface inclination. Final Answer: (w * h^2 / 2) * tan^2(45° − φ/2)

Question 14

Axial extension of a circular bar tapering uniformly from diameter d1 to diameter d2 over length L under axial pull P (E = Young's modulus): Which expression gives the total extension?

Accepted Answer

Introduction / Context: Members with varying cross-section require integration of strain along the length to compute elongation under axial load. A classic result is the extension of a circular bar that tapers linearly from diameter d1 to d2. Given Data / Assumptions: Bar length L; diameters d1 (one end) to d2 (other end), linearly varying. Axial tensile load P; Young's modulus E constant. Small deformations; axial stress uniform over each section. Concept / Approach: For an element at distance x with local diameter d(x), area A(x) = π [d(x)]^2 / 4. Strain at x is P / (E * A(x)). Total extension is the integral of strain over length L. With linear taper, d(x) varies linearly → integration yields a simple closed form. Step-by-Step Solution: Let d(x) = d1 + ( (d2 − d1) / L ) * x.A(x) = π [d(x)]^2 / 4.Incremental extension: du = (P / (E * A(x))) dx.Integrate u = ∫(0→L) (4 P / (π E [d(x)]^2)) dx.For linear d(x), the integral evaluates to u = (4 P L) / (π E d1 d2). Verification / Alternative check: Check limits: if d1 = d2 = d (uniform bar) → u = (4 P L)/(π E d^2) = (P L)/(A E), since A = π d^2/4. This confirms correctness. Why Other Options Are Wrong: Using average area (A_avg) is an approximation; exact integral gives 1/(d1 d2) dependence, not function of (d1 + d2)^2. Forms with d1^2 + d2^2 do not reduce properly for d1 = d2. Factor 2 in the numerator (option E) underestimates extension by half. Common Pitfalls: Arithmetic slips when integrating 1/[d(x)]^2; forgetting to verify the uniform-bar limit; using arithmetic average of areas instead of the correct harmonic-type relation. Final Answer: (4 * P * L) / (π * E * d1 * d2)

Question 15

Beam under tent-shaped load: A simply supported beam carries a load that varies gradually from zero at both supports to w per metre at midspan (symmetric). Is the statement “the bending moment at the centre is w l / 4” correct?

Accepted Answer

Introduction / Context: Different load distributions produce different shear and bending moment diagrams. Dimensional consistency and known reference cases help validate formulas. The given statement resembles the classic point-load case at midspan rather than a distributed tent-shaped load. Given Data / Assumptions: Simply supported beam of span l. Load intensity is zero at both supports and increases linearly to w per metre at midspan, then symmetrically decreases. w denotes load intensity (force per unit length). Concept / Approach: For a central point load W on a simply supported beam, M_max at midspan equals W * l / 4 (units of forcelength). For a uniformly distributed load q, M_max = q * l^2 / 8 (forcelength). For a tent-shaped distributed load peaking at midspan, the correct midspan moment involves w * l^2 multiplied by a coefficient distinct from 1/8 and certainly not w * l / 4, which is dimensionally inconsistent if w is per metre. Step-by-Step Solution: Check dimensions: bending moment units should be force * length.Given w is force/length, w * l / 4 has units of force (incorrect). Correct form must include l^2.Therefore, the stated expression cannot be correct for a distributed load described. Verification / Alternative check: Detailed integration of the symmetric triangular intensity gives M_center proportional to w * l^2 with a coefficient less than that for UDL. Regardless, w * l / 4 is invalid dimensionally and numerically. Why Other Options Are Wrong: “Yes” contradicts both dimensional analysis and known beam results. “Only for very small spans” or “Depends on end fixity” are irrelevant; the form remains wrong. “Only if w is total load” would redefine symbols, but the stem clearly defines w as intensity. Common Pitfalls: Memorizing W l / 4 (point load) and misapplying it to distributed loads; ignoring unit consistency; mixing up W (total load) with w (load per unit length). Final Answer: No

Question 16

Buckling versus crushing: For long (slender) columns, how does the Euler buckling load compare with the material's crushing (compressive) load?

Accepted Answer

Introduction / Context: Long columns fail by elastic instability at stresses far below the compressive strength of the material. Recognizing this distinction prevents unsafe designs that rely solely on compressive strength without checking Euler stability limits. Given Data / Assumptions: Slender column, length and end conditions such that Euler theory applies. Elastic behavior up to buckling. Crushing load refers to A * σ_c (material strength), far higher than elastic buckling load for slender members. Concept / Approach: Euler buckling load is P_cr = (π^2 * E * I) / (Le^2). As slenderness increases (Le large), P_cr drops rapidly. For sufficiently slender members, P_cr is much less than A * σ_c; thus buckling precedes material crushing. Step-by-Step Solution: Express P_cr and P_crushing = A * σ_c.Note P_cr ∝ 1 / (Le^2) → small for large Le.Hence for long columns, P_cr < P_crushing. Verification / Alternative check: Numerical example: For a steel strut with large slenderness, Euler critical stress may be a fraction (say 10–20%) of σ_y, confirming that buckling governs well before crushing. Why Other Options Are Wrong: “Equal to” applies only to a special transition slenderness, not generally for long columns. “More than” contradicts the inverse square dependence on Le. “Cannot be compared” ignores established formulas. Common Pitfalls: Using material compressive strength alone; neglecting effective length factor; overlooking imperfections that further reduce actual buckling load. Final Answer: less than

Question 17

Definition check: A simply supported beam is one which is supported on more than two supports. Is this statement true?

Accepted Answer

Introduction / Context: Beam support conditions define statical determinacy and influence internal forces and deflections. Understanding the standard meaning of a simply supported beam avoids confusion with continuous beams and other boundary conditions. Given Data / Assumptions: Classical definitions from structural analysis. Supports are idealized as pin (hinge) and roller providing reactions without moment restraint. Concept / Approach: A simply supported beam is supported at its ends by a pin and a roller (two supports) allowing rotation and, in the case of the roller, horizontal movement. A beam supported at more than two points is a continuous beam and is generally statically indeterminate to a higher degree. Step-by-Step Solution: Define simply supported beam: two end supports, no end fixity.Compare with continuous beams: three or more supports → indeterminacy.Therefore, the statement claiming “more than two supports” is false. Verification / Alternative check: Textbook free-body diagrams and determinacy counts confirm that simply supported beams have two reaction points yielding three static unknowns (two verticals and possibly one horizontal), solvable by equilibrium alone if needed. Why Other Options Are Wrong: “True” contradicts the standard definition. “True only for continuous beams” mislabels a different category. Determinacy depends on constraints, not just number of supports; loading does not change the support definition. Common Pitfalls: Confusing “simply supported” with “continuous”; assuming any beam with no end fixity is simply supported regardless of additional intermediate supports (that makes it continuous, not simply supported). Final Answer: False

Question 18

Thick cylindrical shell under internal pressure
For a thick-walled cylinder subjected to an internal pressure p, how does the circumferential (tangential/hoop) stress vary across the wall thickness?

Accepted Answer

Introduction / Context: Design of pressure vessels and gun barrels requires understanding the stress distribution in thick cylinders. Unlike thin shells where hoop stress is assumed uniform, thick cylinders exhibit a radial variation governed by Lame’s equations. Given Data / Assumptions: Cylinder subject to internal pressure p only (no external pressure). Axisymmetric, long cylinder; plane strain or generalized plane stress idealization. Linear elastic, homogeneous, isotropic material. Concept / Approach: Lame’s theory states that for a thick cylinder the radial stress σ_r and hoop (tangential) stress σ_θ vary with radius r as σ_r = A − B/r^2 and σ_θ = A + B/r^2. Constants A and B are found from boundary conditions (σ_r = −p at inner surface, σ_r = 0 at outer surface). Step-by-Step Solution: Apply boundary conditions to determine A and B in σ_θ = A + B/r^2.Because of the +B/r^2 term, σ_θ increases as r decreases toward the inner surface.Therefore hoop stress is highest at the inner radius and reduces toward the outer radius. Verification / Alternative check: Plotting σ_θ(r) from typical dimensions confirms a monotonic decrease from inner to outer surface when only internal pressure acts. Why Other Options Are Wrong: Uniform stress is a thin-shell assumption; options implying zero hoop stress at one surface contradict equilibrium and elasticity solutions. Common Pitfalls: Confusing radial stress (which is zero at the outer surface) with hoop stress; applying thin-wall results to thick-wall situations. Final Answer: Maximum at the inner surface and minimum at the outer surface

Question 19

Riveted joint detailing – Minimum edge distance (margin)
To avoid tearing at the plate edge, what should be the minimum distance from the centre of a rivet hole to the nearest edge (margin), in terms of hole diameter d (in mm)?

Accepted Answer

Introduction / Context: Proper detailing of riveted joints ensures that plates do not tear at the edges under load. The minimum edge distance (margin) is a standard proportion in machine design and fabrication practice. Given Data / Assumptions: Unwelded, riveted lap or butt joint plate. Hole diameter d (mm) after drilling or punching. Need to prevent edge tearing and maintain manufacturability. Concept / Approach: If the rivet hole is placed too close to the edge, the ligament is weak and may tear. Empirical and codal recommendations prescribe a minimum margin to ensure adequate strength and avoid local failures. Step-by-Step Reasoning: Tearing strength along the edge ligament must exceed the force transferred by shear of rivets.A commonly adopted minimum margin for general machine design is 1.5 d, balancing strength and plate economy.Larger margins (2 d or more) may be used for thicker plates or severe service, but 1.5 d is the accepted minimum. Verification / Alternative check: Design handbooks and fabrication guides list 1.5 d as the minimum edge distance to prevent edge tearing and facilitate hole making without distortion. Why Other Options Are Wrong: d is too small and risks tearing; 2 d, 2.5 d, 3 d are conservative but not the minimum standard requirement for typical joints. Common Pitfalls: Confusing pitch/spacing with margin; using bolt diameter instead of hole diameter; not accounting for plate tolerance and hole drift. Final Answer: 1.5 d

Question 20

Combined stresses – Minimum normal (principal) stress under biaxial normal stresses σx, σy with in-plane shear τxy
When a body is subjected to direct stresses σx and σy on mutually perpendicular planes and a shear stress τxy, what is the minimum no…

Accepted Answer

Introduction / Context: Determining principal stresses under combined biaxial normal stresses with shear is fundamental for failure analysis using criteria such as maximum normal stress, maximum shear stress, or distortion energy. Given Data / Assumptions: Plane stress state with σx, σy, and τxy defined positive by standard sign convention. Linear elastic behavior; principal planes are orthogonal. Concept / Approach: Principal stresses are the extrema of normal stress on rotated planes where shear becomes zero. They are obtained from Mohr’s circle or the characteristic equation of the 2D stress tensor. Step-by-Step Solution: Principal stresses: σ1,2 = (σx + σy)/2 ± sqrt( ((σx − σy)/2)^2 + τxy^2 ).By definition, the minimum normal stress is σmin = (σx + σy)/2 − sqrt( ((σx − σy)/2)^2 + τxy^2 ).The maximum normal stress is the corresponding + sign. Verification / Alternative check: Construct Mohr’s circle with center at (σx + σy)/2 on the σ-axis and radius R = sqrt( ((σx − σy)/2)^2 + τxy^2 ). The leftmost intersection gives σmin as stated. Why Other Options Are Wrong: Option (b) is σmax; (c) is not a correct general expression; (d) and (e) are not invariant and ignore the quadratic coupling between σx, σy, and τxy. Common Pitfalls: Dropping the square on τxy; mixing up σmax and σmin; using absolute values incorrectly. Final Answer: (σx + σy)/2 − sqrt( ((σx − σy)/2)^2 + τxy^2 )

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