Question 1

Thin cylinder, open ends – longitudinal strain due to internal pressure A thin-walled cylinder of radius r and thickness t is open-ended (no end caps). It is subjected to internal pressure p. The Young’s modulus is E and Poisson’s ratio is μ. What i…

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Introduction / Context: Thin pressure vessels develop hoop and longitudinal stresses under internal pressure. Whether the ends are closed or open changes the longitudinal stress state and hence the axial strain. Understanding this distinction is essential in vessel design. Given Data / Assumptions: Thin cylinder (t ≪ r), linear elasticity. Open-ended ⇒ no net end force ⇒ longitudinal stress σ_l = 0. Hoop (circumferential) stress σ_h = p r / t for thin cylinders. Material constants: E (Young’s modulus), μ (Poisson’s ratio). Concept / Approach: Generalized Hooke’s law for axial strain in a biaxial stress state is epsilon_l = (1/E) * (σ_l − μ σ_h) (neglecting radial stress). With σ_l = 0 for open ends, the axial strain arises only from Poisson’s effect of the hoop stress and is negative (axial contraction) if μ > 0. Step-by-Step Solution: Compute hoop stress: σ_h = p r / t.Set σ_l = 0 (no end caps).Axial strain: epsilon_l = (1/E) * (0 − μ σ_h) = − μ (p r) / (E t). Verification / Alternative check: For closed ends, σ_l = p r / (2 t), giving epsilon_l = (1/E) * (σ_l − μ σ_h) = (p r / (2 t E)) − μ (p r / (E t)), different from the open-ended result, confirming the role of end restraint. Why Other Options Are Wrong: (a) would be true only for μ = 0, which is not general. (b), (d), (e) ignore the sign and Poisson coupling; they predict axial extension instead of contraction. Common Pitfalls: Assuming the same longitudinal stress for open and closed cylinders; always check end conditions before computing strains. Final Answer: − μ (p r) / (E t)

Question 2

Beams of uniform strength vs. uniform section – where is the economy? Beams designed for uniform strength (depth or width varied so that allowable stress is just reached everywhere) are generally more economical than prismatic beams (uniform section…

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Introduction / Context: A beam of uniform strength has its cross-section tailored along the span so that the permissible stress is fully utilized everywhere. This contrasts with a prismatic (uniform) beam where much of the material near the supports may be under-stressed when the maximum bending moment occurs near midspan. Given Data / Assumptions: Allowable stress in bending is the design limit. Self-weight may be relevant on long spans. Elastic behavior and small deflection theory. Concept / Approach: Because bending moment M(x) varies along the beam, section modulus Z(x) = M(x) / σ_allow should ideally vary in the same proportion to keep σ ≈ σ_allow everywhere. This redistribution of material (greater depth or width where M is larger) saves weight and cost while maintaining strength. Step-by-Step Solution: For a simply supported beam with UDL or midspan load, M(x) peaks near midspan and diminishes toward supports.Uniform strength design makes Z(x) largest where M(x) is largest and vice versa, reducing excess material near supports.Economy is greatest when M(x) varies strongly along the span—typically long spans with distributed loads. Verification / Alternative check: Weight savings are often reported by comparing volume integrals of variable vs. uniform sections for the same strength limit; the more variable M(x) is, the larger the saving. Why Other Options Are Wrong: (b) Short spans with heavy loads do not benefit much because the region of high M is small and fabrication complexity may dominate. (c) Load magnitude alone does not decide economy. (d) Near-uniform M(x) gives little advantage to varying the section. (e) Axial compression members are not bending-dominated beams. Common Pitfalls: Ignoring deflection criteria; varying depth can also improve stiffness distribution but must satisfy serviceability limits. Final Answer: Large spans where bending moment varies significantly

Question 3

Cantilever with uniformly distributed load – shape of the shear force diagram For a cantilever beam carrying a uniformly distributed load over its entire length, the shear force diagram has the shape of a:

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Introduction / Context: Drawing shear force (V) and bending moment (M) diagrams quickly from load diagrams is a core skill in structural analysis. Recognizing the characteristic shapes helps double-check calculations and speed up problem solving. Given Data / Assumptions: Cantilever beam of length L, fixed at one end and free at the other. Uniformly distributed load w (force per unit length) over the full span. Static equilibrium, small deflections. Concept / Approach: For a UDL, the shear force is the integral of load intensity with negative sign conventionally: dV/dx = −w. Hence V varies linearly with x. At the free end of a cantilever V = 0, and at the fixed end V = wL (magnitude). Therefore, the shear diagram is triangular. Step-by-Step Solution: Start from the free end: V(0) = 0.Integrate load: V(x) = −w x (sign per convention); magnitude grows linearly with x.At the fixed end x = L: |V(L)| = w L, confirming a straight line from 0 to wL. Verification / Alternative check: Differentiate the bending moment diagram M(x) = −w x^2 / 2; dM/dx = V(x) gives the same linear variation. Why Other Options Are Wrong: Rectangle implies constant shear, which corresponds to a point load, not a UDL. Parabola or cubic parabola describe M(x) shapes, not V(x) for a UDL on cantilever. Straight line with zero slope would be constant shear, again incorrect. Common Pitfalls: Mixing up signs or confusing shapes of V and M; remember: UDL ⇒ V linear, M quadratic. Final Answer: Triangle

Question 4

Internal forces in arches – direction of the shear force component In a solid arch under general loading, the internal shear force at a section acts:

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Introduction / Context: Arches carry load through a combination of normal thrust (along the arch axis) and shear (radial) components. Correctly identifying the direction of these internal forces is essential for section design and for interpreting influence lines in arch analysis. Given Data / Assumptions: Solid arch of arbitrary profile in a vertical plane. Section taken at an arbitrary point on the arch axis. Internal force components resolved relative to the local axis of the arch. Concept / Approach: At any section of an arch, the internal forces are typically expressed as: (i) normal thrust along the tangent to the axis, (ii) radial (or shear) force perpendicular to the axis, and (iii) bending moment about the out-of-plane axis. The “shear force” in arch theory is the radial component normal to the axis. Step-by-Step Solution: Resolve internal action at a section into two in-plane components: N_t (tangential, along the arch) and V_r (radial, perpendicular to the arch).By definition, V_r is the shear force in arch analysis, so it acts normal to the axis.Therefore, the correct directional description is “perpendicular to the axis of the arch”. Verification / Alternative check: Textbook free-body diagrams of arch segments consistently show normal thrust and radial shear at the cut, plus a bending moment. Why Other Options Are Wrong: (a) fixes the shear direction as vertical, which is not generally true for curved axes. (b) is the direction of normal thrust, not shear. (d) again corresponds to the tangential component (normal thrust), not shear. (e) is unnecessary because (c) is correct. Common Pitfalls: Confusing “shear” in beams (vertical) with “radial shear” in arches; always resolve relative to the local arch axis. Final Answer: Perpendicular to the axis of the arch (radial direction)

Question 5

Intermediate-roller beam with equal sagging and hogging: A beam of total length L rests on two intermediate roller supports and carries a uniformly distributed load over its entire length. The beam has equal overhangs at both ends. If the maximum sa…

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Introduction / Context: This classic continuous-beam problem tests the relationship between loading, reactions, and internal actions (sagging and hogging moments) for a beam with equal end overhangs supported on two intermediate rollers under a uniformly distributed load. The target is to find the overhang length at which the largest positive (sagging) and negative (hogging) bending moments are equal in magnitude. Given Data / Assumptions: Total beam length = L; equal overhangs = a each side. Two intermediate roller supports separated by span (L − 2a). Uniformly distributed load of constant intensity across the full length. Prismatic member; small deflection; linear elastic behavior. Concept / Approach: The internal moments arise from the supported span (hogging over the supports) and the overhanging cantilevers (sagging near free ends). By writing equilibrium and influence relationships (or using standard results/influence lines), one can determine the overhang a/L that equalizes the peak positive and peak negative moments. Step-by-Step Solution: Let overhangs = a; supported span = L − 2a.Under full-length UDL, reactions distribute so that hogging peaks over the supports and sagging peaks in the overhangs.Setting |M_hog|max = |M_sag|max and solving the resulting equilibrium/compatibility equation yields a/L ≈ 0.207.Therefore, each overhang length a = 0.207 L. Verification / Alternative check: This fraction is a well-known optimum for equalization of moments on a two-span-with-overhang arrangement under uniform load. Numerical checks via influence lines or structural analysis software reproduce a/L ≈ 0.207. Why Other Options Are Wrong: 0.107 L, 0.307 L, 0.407 L, 0.500 L: these do not equalize the peak hogging and sagging; analysis shows the equality occurs close to 0.207 L. Common Pitfalls: Confusing the position of maximum sagging (in the overhang) with that within the supported span, or assuming symmetry automatically implies a = L/4. Final Answer: 0.207 L.

Question 6

Compression members terminology in structural engineering: A long vertical member subjected to an axial compressive load is called:

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Introduction / Context: Naming compression members correctly is fundamental in structural design. Different terms are used depending on orientation and usage, although all share the common feature of carrying compressive force. Given Data / Assumptions: Member is long and vertical. Loading is primarily axial compression. Standard building/bridge terminology is assumed. Concept / Approach: A vertical compression member in buildings is classically called a “column.” The broader term “strut” refers to any compression member (inclined, horizontal, or vertical) in trusses and frames. “Tie” refers to a tension member and is therefore incorrect here. “Stanchion” is often used synonymously with column, particularly in steelwork, but the most general and precise term for a vertical compressive member is “column.” Step-by-Step Solution: Identify orientation and loading: vertical + axial compression → column.Exclude tie: carries tension, not compression.Strut is generic for compression but not specifically a long vertical building member.Stanchion is a steelwork term but “column” is the canonical answer. Verification / Alternative check: Textbooks consistently define a building’s vertical compression member as a column; strut is more general across orientations. Why Other Options Are Wrong: Strut: not specific to vertical members.Tie: tension member.Stanchion: acceptable in some contexts, but the standard generic term is column.All of the above: incorrect due to inclusion of the tie. Common Pitfalls: Confusing struts in trusses with columns in frames, or assuming “all of the above.” Final Answer: A column.

Question 7

Cantilever with a pure end couple: For a cantilever beam whose free end is subjected to a constant bending moment (pure couple) and no transverse load, the bending-moment diagram along the span will be a:

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Introduction / Context: Knowing the shapes of shear-force and bending-moment (BM) diagrams for standard load cases is essential for quick checks and sizing. A pure couple applied at the free end of a cantilever creates a distinctive internal action distribution. Given Data / Assumptions: Cantilever, fixed at one end, free at the other. Applied at the free end: a constant bending moment M only; no point load or UDL. Linear elastic behavior, prismatic member. Concept / Approach: With no transverse load, the internal shear force is zero everywhere. The internal bending moment equals the applied couple everywhere (equilibrium of each section), hence it is constant along the span. Step-by-Step Solution: Transverse load w(x) = 0 ⇒ shear V(x) = 0 (constant zero).dM/dx = V(x) ⇒ dM/dx = 0 ⇒ M(x) is constant.Therefore, the BM diagram is a rectangle of height M extending over the full length. Verification / Alternative check: Section any location: the resisting end moment at the cut must balance the applied couple, giving the same magnitude everywhere. Why Other Options Are Wrong: Triangle/Parabola/Cubic parabola: these arise with point loads or distributed loads, not with a pure moment.Semicircle: not a standard BM shape from statics. Common Pitfalls: Assuming a linearly varying moment because that holds for a point load; here, with zero shear, the moment cannot vary. Final Answer: Rectangle.

Question 8

Three-hinged arch under an eccentric point load: A three-hinged arch of span 10 m has the crown 1 m above the springings. An isolated load of 1000 kg acts 2.5 m horizontally from the crown. Compute the horizontal thrust.

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Introduction / Context: Three-hinged arches are statically determinate. The horizontal thrust H can be obtained from moment equilibrium at the crown hinge by using the bending moment there as if for a simply supported beam and dividing by the rise. Given Data / Assumptions: Span = 10 m; rise at crown h = 1 m. Point load P = 1000 kg located 2.5 m from crown (toward the right). Supports at the same level. Small deflection, linear elasticity, determinate analysis. Concept / Approach: For a three-hinged arch, the bending moment at the crown hinge is zero. Taking the structure as a simply supported beam to compute the “beam moment” M_beam at the crown section due to external loads, the horizontal thrust satisfies H * h = M_beam, hence H = M_beam / h. Step-by-Step Solution: Location of load from left support: 5 m (to crown) + 2.5 m = 7.5 m.Simply-supported reactions: R_A = P * (L − a) / L = 1000 * (10 − 7.5) / 10 = 250 kg.Bending moment at crown section (x = 5 m from left, load to the right): M_beam = R_A * 5 = 250 * 5 = 1250 kg·m.Horizontal thrust: H = M_beam / h = 1250 / 1 = 1250 kg. Verification / Alternative check: Symmetry about midspan is broken by the eccentric load; using the right reaction would give the same crown moment via the right segment. Why Other Options Are Wrong: 125 kg and 750 kg underpredict; 2325 kg and 2500 kg overpredict; they do not satisfy M = H h with the correct beam crown moment. Common Pitfalls: Including the load contribution in the left free-body for x = 5 m (it lies to the right and should be excluded in that cut), or using the full span moment instead of the crown section moment. Final Answer: 1250 kg.

Question 9

Shear deflection of a cantilever under uniformly distributed load: For a cantilever of length L, cross-sectional area A, and shear modulus G, carrying w (force per unit length) uniformly over its entire length, what is the vertical deflection at the…

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Introduction / Context: Total deflection of beams consists of bending (flexural) and shear components. For short/deep members, shear deflection may be significant. This question isolates the shear part for a cantilever with uniformly distributed load (UDL). Given Data / Assumptions: Cantilever length L; constant area A and shear modulus G. UDL intensity = w (force per unit length). Shear strain γ(x) = V(x) / (A G); shear-correction factor neglected as per the statement. Concept / Approach: For shear deflection, the incremental rotation due to shear is γ(x) = V(x)/(A G). The deflection at the free end equals the integral of γ(x) along the beam from the fixed end to the free end. Step-by-Step Solution: Shear force along the cantilever: V(x) = w (L − x).Shear slope: γ(x) = V(x) / (A G) = w (L − x) / (A G).Shear deflection at free end: δ_s = ∫_0^L γ(x) dx = (1 / (A G)) ∫_0^L w (L − x) dx.Evaluate integral: ∫_0^L (L − x) dx = [L x − x^2/2]_0^L = L^2 − L^2/2 = L^2/2.Therefore, δ_s = w L^2 / (2 A G). Verification / Alternative check: Dimensional check: w has units force/length; dividing by A G (force/area) yields length; multiplying by L^2 gives length → consistent. Why Other Options Are Wrong: w L^3 / (3 E I): that is a bending-deflection form (flexural, not shear).Other choices misplace coefficients or powers of L. Common Pitfalls: Mixing shear deflection with flexural deflection, or forgetting the linear variation of V(x) in a cantilever under UDL. Final Answer: w L^2 / (2 A G).

Question 10

Closely coiled helical spring deflection: A closely coiled helical spring of mean coil radius R has n turns and is subjected to an axial load W. If the wire radius is r and the material has shear modulus C, what is the axial deflection of the spring?

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Introduction / Context: Close-coiled helical springs store energy primarily through torsion of the wire. The load–deflection relation connects axial deflection to coil geometry and material stiffness. Given Data / Assumptions: Mean coil radius = R; number of active turns = n. Wire radius = r (so wire diameter d = 2 r). Shear modulus = C. Small deflection; close-coil assumption (helix angle small). Concept / Approach: For a close-coiled spring under axial load W, the torque in the wire is T = W R for each turn. The twist per turn is θ_turn = T l_w / (J C), but standard derivation yields the familiar formula δ = 64 W R^3 n / (C d^4). Substituting d = 2 r gives δ = 64 W R^3 n / (C (2 r)^4) = 64 W R^3 n / (C * 16 r^4) = 4 W R^3 n / (C r^4). Step-by-Step Solution: Start with δ = 64 W R^3 n / (C d^4).Replace d with 2 r → d^4 = 16 r^4.Hence δ = (64 / 16) * (W R^3 n) / (C r^4) = 4 W R^3 n / (C r^4). Verification / Alternative check: Dimensional check: numerator has force * length^3; denominator C r^4 has force/area * length^4 = force * length^2; ratio gives length as required. Why Other Options Are Wrong: Options with 8 or 64 numerators misuse diameter vs radius or alter powers erroneously.Expressions involving r^3, or R^2 instead of R^3, are dimensionally inconsistent. Common Pitfalls: Forgetting to convert from wire diameter to radius; mixing up shear modulus symbols (G vs C). Final Answer: δ = (4 W R^3 n) / (C r^4).

Question 11

Torsion distribution in a prismatic shaft fixed against twist at both ends: A 9 m long shaft is restrained at both ends and a torque of 30 tonne·metre is applied at a point 3 m from the nearer end (therefore 6 m from the far end). What is the reacti…

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Introduction / Context: When a torque is applied at an intermediate section of a uniform circular shaft whose ends are fixed against rotation, equal and opposite reaction torques develop at the ends. These reactions distribute in proportion to the lengths of the shaft segments on either side to satisfy compatibility of twist. Given Data / Assumptions: Total length L = 9 m; distances from torque point to ends: a = 3 m (near end), b = 6 m (far end). Applied torque T = 30 tonne·metre. Uniform torsional rigidity J G along the shaft; ends fixed (zero rotation). Concept / Approach: Compatibility requires the angle of twist of each segment (relative to the torque point) to be equal and opposite so that the net rotation at the load point is consistent. For a uniform shaft, reactions are proportional to the opposite segment lengths: T_near = T * b / L and T_far = T * a / L, with T_near + T_far = T. Step-by-Step Solution: Compute T_near: T_near = T * b / L = 30 * 6 / 9 = 20 tonne·metre.Compute T_far: T_far = T * a / L = 30 * 3 / 9 = 10 tonne·metre.Check: T_near + T_far = 20 + 10 = 30 tonne·metre (balances applied torque). Verification / Alternative check: Angles of twist: θ_near ∝ T_near * a and θ_far ∝ T_far * b; substituting shows equal and opposite rotations, confirming compatibility. Why Other Options Are Wrong: 5, 10, 15 tonne·metre do not satisfy both equilibrium and compatibility with given distances; 30 tonne·metre ignores the share at the far end. Common Pitfalls: Allocating reactions proportional to the same-side length (should be opposite-side length), or forgetting the sum must equal the applied torque. Final Answer: 20 tonne·metre.

Question 12

Flexural strength comparison of two square beams: One beam has a square cross-section of side a oriented with sides horizontal; the other identical square is rotated so that a diagonal is vertical (diamond). What is the ratio of their flexural stren…

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Introduction / Context: Flexural strength under elastic bending is proportional to the section modulus Z = I / y_max. For the same material and allowable stress, comparing section moduli reveals which orientation is stronger in bending. Given Data / Assumptions: Both sections are identical squares of side a. Comparison is about the horizontal neutral axis for vertical bending. Linear elasticity and small deflection; no stability effects. Concept / Approach: For a square with sides horizontal: I = a^4 / 12 about the centroidal horizontal axis; y_max = a/2; thus Z_sides = (a^4 / 12) / (a/2) = a^3 / 6. For the same square rotated 45° (diagonal vertical): the second moment of area about any centroidal axis remains I = a^4 / 12 (shape is the same), but the half-depth becomes y_max = (a√2)/2 = a / √2, giving Z_diagonal = (a^4 / 12) / (a/√2) = (a^3 √2)/12. Step-by-Step Solution: Z_sides = a^3 / 6.Z_diagonal = (a^3 √2) / 12.Ratio = Z_sides / Z_diagonal = (a^3 / 6) / ((a^3 √2)/12) = 12 / (6 √2) = 2 / √2 = √2 ≈ 1.414. Verification / Alternative check: Numerical substitution, e.g., a = 100 mm, reproduces the same ratio independent of size. Why Other Options Are Wrong: 1 and 1/√2: would imply no change or the diamond being stronger; both are incorrect.2 and 3/2: overestimate the advantage; the exact factor is √2. Common Pitfalls: Assuming I changes with rotation (it does not for the centroidal value of the square), or forgetting that y_max increases to a/√2 when the diagonal is vertical. Final Answer: √2 (approximately 1.414).

Question 13

Static equilibrium for support reactions: The reactions at the supports of a structure can be determined by enforcing which equilibrium conditions?

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Introduction / Context: Determining support reactions is the first step in analyzing statically determinate structures. Reactions follow directly from the equations of static equilibrium. Given Data / Assumptions: Planar structure; loads and reactions lie in a plane. Body is rigid; supports are idealized as pins/rollers/fixed as appropriate. No acceleration (static case). Concept / Approach: For planar statics, three independent equilibrium equations are available: ΣF_x = 0, ΣF_y = 0, and ΣM_O = 0. Solving these yields the unknown reaction components and moments for determinate systems. Step-by-Step Solution: Write ΣF_x = 0 to relate horizontal reaction components and applied horizontal loads.Write ΣF_y = 0 to relate vertical reaction components and applied vertical loads.Write ΣM_O = 0 (about any convenient point) to eliminate one or more unknowns and solve. Verification / Alternative check: Choosing a different moment center gives the same reactions if calculations are correct, providing a natural check. Why Other Options Are Wrong: Each single equation alone is insufficient for a general 2D support-reaction problem; all three together are required for complete solution. Common Pitfalls: Incorrect sign conventions, neglecting couple moments, or forgetting inclined loads have both x and y components. Final Answer: All of the above.

Question 14

Materials property under tension: The property by which a material can be drawn out into a smaller cross-section (for example, into wires) under tensile force is called:

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Introduction / Context: Different material properties describe behavior under various loadings. When a metal is drawn into wires or elongated plastically under tension, a specific property is being referenced. Given Data / Assumptions: Loading is tensile, producing permanent reduction in section when desired. The property is concerned with the ability to undergo substantial plastic deformation before fracture. Concept / Approach: Ductility is the capacity of a material to deform plastically in tension, permitting drawing into wires or small sections. It is often measured by percentage elongation and reduction of area in a tensile test. By contrast, malleability relates to plastic deformation under compression (e.g., rolling into sheets). Step-by-Step Solution: Identify the mode: tension with reduction of cross-section.Match the property: ductility corresponds to tensile plasticity and drawability.Exclude other properties based on definitions. Verification / Alternative check: Common ductile materials (e.g., copper, mild steel, aluminum) can be drawn into wires; malleable materials (e.g., gold, lead) form sheets under compression. Why Other Options Are Wrong: Plasticity: general ability to undergo permanent deformation (tension or compression) without specifying drawability.Elasticity: ability to recover shape once load is removed; not plastic deformation.Malleability: plastic deformation in compression (rolling, hammering), not drawing in tension.Brittleness: little plastic deformation before fracture. Common Pitfalls: Confusing malleability (compression) with ductility (tension) because both involve plastic flow. Final Answer: Ductility.

Question 15

Simply supported beam with a central point load W For a prismatic simply supported beam of span L carrying a single concentrated load W at midspan, what is the maximum vertical deflection δ_max at the centre (assume linear elastic behaviour with mod…

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Introduction / Context: This question checks recall and correct application of the classic deflection formula for a simply supported beam subjected to a single central point load. Such closed-form results are used constantly in preliminary sizing and quick serviceability checks for beams in civil and mechanical engineering. Given Data / Assumptions: Span between simple supports = L. Point load W acts at midspan. Material is linearly elastic with Young’s modulus E. Second moment of area about the bending axis = I (constant along the span). Small deflection (Euler–Bernoulli) theory; plane sections remain plane. Concept / Approach: The maximum deflection for this loading case occurs at midspan. It can be obtained from double integration of the elastic curve equation E I d²y/dx² = M(x), area-moment method, or standard tables. All methods give the same closed form: δ_max = W L^3 / (48 E I). Step-by-Step Solution: Reactions at supports = W/2 each by symmetry.Bending moment diagram is triangular, peak M_max = W L / 4 at midspan.Using the standard result (from integration or tables): δ_max = W L^3 / (48 E I). Verification / Alternative check: Area-moment method: slope discontinuity areas lead to the same coefficient 1/48. Numerical checks for typical values confirm midspan deflection scales with L^3 and inversely with E I as expected physically. Why Other Options Are Wrong: W L^3/(3 E I) and W L^3/(16 E I) greatly overestimate deflection; wrong coefficients. W L^2/(8 E I) has wrong length power (should be L^3). W L^4/(384 E I) corresponds to a different boundary condition (cantilever/UDL mix) and wrong dimension. Common Pitfalls: Forgetting the 1/48 coefficient or mixing with UDL formula δ_max = 5 w L^4/(384 E I). Using section modulus instead of second moment of area for deflection problems. Final Answer: δ_max = W L^3 / (48 E I).

Question 16

Column behaviour terminology The ratio of the effective length of a column to the minimum radius of gyration of its cross-section is known as:

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Introduction / Context: Columns fail predominantly by buckling rather than crushing when they are slender. Quantifying slenderness is therefore central to column design in steel, reinforced concrete, timber, and aluminum structures. The slenderness ratio provides a non-dimensional measure linking member length and section stiffness distribution. Given Data / Assumptions: Effective length kL reflects end restraints (pinned, fixed, etc.). Minimum radius of gyration r_min = sqrt(I_min/A) where I_min is the least second moment of area. Elastic buckling governed by Euler theory for ideal columns; design codes modify with imperfections and inelasticity. Concept / Approach: The slenderness ratio is defined as λ = (effective length)/(minimum radius of gyration) = kL / r_min. It captures both the global instability tendency (longer members buckle more easily) and cross-sectional stiffness (larger r resists curvature). Design curves and interaction checks are parameterized by λ. Step-by-Step Solution: Identify the property being asked: ratio (effective length)/(minimum r).By definition, this ratio is the slenderness ratio.Therefore, the correct choice is “slenderness ratio”. Verification / Alternative check: Euler critical stress σ_cr = π^2 E / (λ^2) (when expressed with λ = kL/r); this shows buckling sensitivity increases rapidly with λ, confirming why the term is central in column stability. Why Other Options Are Wrong: Buckling factor and crippling factor are not standard definitions for this ratio. “Stability index” is used in frame stability (second-order analysis), not for the simple ratio asked. “None of these” is incorrect since a precise, named term exists. Common Pitfalls: Using r about the wrong axis; always take the minimum radius of gyration. Ignoring end conditions; k often differs from 1.0. Final Answer: slenderness ratio.

Question 17

Modes of failure in riveted joints Riveted joints can fail in multiple ways under load. Which of the following failure modes are relevant?

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Introduction / Context: Riveted and bolted joints transfer forces through a combination of bearing, shear, and tension in the connected plates. Understanding all possible failure modes is essential for safe detailing and for verifying that the joint has balanced strength in each path. Given Data / Assumptions: Lap or butt riveted joint in ductile plate material. Static loading along the joint line producing direct shear and bearing. Holes properly drilled; edge distance and pitch influence plate tearing. Concept / Approach: Four classical limit states are checked: tearing of plate at edge (insufficient edge distance), tearing between rivets (insufficient pitch or high net-section stress), shearing of rivets (insufficient shear area), and crushing (bearing) of rivet against the plate (insufficient projected bearing area). Step-by-Step Solution: Evaluate net section across a potential tear path → plate tearing modes (options a and b).Check shear capacity of one or two shear planes per rivet → rivet shearing (option c).Check bearing stress p = P/(d*t) against allowable → crushing/bearing (option d).All are genuine and must be satisfied simultaneously. Verification / Alternative check: Design codes for steelwork explicitly require checking net-section tension, bolt/rivet shear, and bearing (sometimes block shear). This confirms that multiple distinct limit states exist and the design must satisfy the minimum. Why Other Options Are Wrong (as answers): Selecting only one of (a)–(d) ignores other governing possibilities; the correct comprehensive answer is “All the above”. Common Pitfalls: Insufficient edge distance leading to premature end tearing. Too small pitch causing net-section fracture between holes. Overlooking bearing limits when rivet diameter is increased without thickening the plate. Final Answer: All the above.

Question 18

Stress distribution in a transversely loaded rectangular beam When a straight, prismatic rectangular beam is subjected to sagging (positive) bending under transverse loading, where does the maximum compressive normal stress occur?

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Introduction / Context: Bending of beams produces linearly varying normal stress across the depth when Euler–Bernoulli assumptions hold. Recognizing where compression and tension occur is vital for placing reinforcement and checking extreme fiber stresses relative to material limits. Given Data / Assumptions: Prismatic rectangular cross-section, homogeneous isotropic material. Sagging (positive) bending moment: concave upward elastic curve in the span. Plane sections remain plane; neutral axis passes through the centroid. Concept / Approach: Normal stress due to bending is σ = M y / I, where y is measured from the neutral axis. For sagging moment in the usual sign convention, fibers above the neutral axis are in compression and those below are in tension. The magnitude is proportional to the distance |y|, so extremes happen at the outermost fibers. Step-by-Step Solution: Identify sign: sagging → top compression, bottom tension (with conventional axis).Maximum |σ| occurs where |y| is largest, i.e., at the extreme fibers.Therefore, maximum compressive stress is at the top fiber; maximum tensile stress at the bottom fiber. Verification / Alternative check: Mohr’s circle for normal stress on fibers across depth shows maxima at the extreme distances from the neutral axis, consistent with σ = M y / I. Why Other Options Are Wrong: Bottom fibre: in sagging, this is tensile, not compressive. Neutral axis: σ = 0 by definition. “Same at every point”: false for bending; stress varies linearly with y. Centroid only: also lies on the neutral axis, so stress there is zero. Common Pitfalls: Reversing the sign convention (for hogging, the answer would flip). Confusing shear stress distribution (parabolic) with bending normal stress (linear). Final Answer: Top fibre.

Question 19

Simple bending (flexure) formula Identify the correct simple bending equation relating bending moment M, section modulus terms, curvature, and stress for a prismatic, linearly elastic beam under pure bending.

Accepted Answer

Introduction / Context: The simple bending (flexure) formula links internal bending moment, section geometry, material stiffness, and curvature for beams obeying Euler–Bernoulli assumptions. It is the cornerstone for sizing beams and checking stresses under service and ultimate loads. Given Data / Assumptions: Homogeneous, isotropic material; linear elasticity. Plane sections remain plane; no significant shear deformation. Pure bending region with constant M and zero shear, or general bending where the relation still holds locally. Concept / Approach: The derivation shows neutral axis through the centroid and linear strain distribution ε = y/R. Hooke’s law gives σ = E ε = E y / R. Equilibrium of internal stresses with external moment yields M/I = σ/y = E/R, where I is the second moment of area about the neutral axis, y is distance from the neutral axis, and R is radius of curvature. Step-by-Step Solution: Strain compatibility: ε = y/R.Constitutive law: σ = E ε = E y / R.Moment equilibrium: ∫ σ y dA = M ⇒ (E/R) ∫ y² dA = M ⇒ (E/R) I = M.Rearrange: M/I = σ/y = E/R. Verification / Alternative check: Specializing to the extreme fiber at y = c gives σ_max = M c / I and section modulus Z = I / c, both standard design quantities, confirming internal consistency. Why Other Options Are Wrong: Options (b), (c), (d) scramble variables or dimensions, violating the derived proportionalities. “None of these” is incorrect since the standard identity exists and is well-known. Common Pitfalls: Using section modulus Z in place of I without tracking c correctly. Applying the formula outside its assumptions (deep beams with significant shear deformation). Final Answer: M / I = σ / y = E / R.

Question 20

Locating the neutral axis from extreme fiber stresses In a rectangular beam of total depth 10 cm, the maximum compressive stress at the top is 1600 kg/cm² and the corresponding tensile stress at the bottom is 400 kg/cm² under the same bending moment…

Accepted Answer

Introduction / Context: Under pure bending with linear elastic behaviour, normal stress varies linearly with distance from the neutral axis (N.A.). Knowing the extreme fiber stresses allows you to back-compute the N.A. position in a homogeneous rectangular section — an essential diagnostic skill when checking combined or unsymmetrical situations. Given Data / Assumptions: Total depth d = 10 cm. Top compressive stress = 1600 kg/cm². Bottom tensile stress = 400 kg/cm². Homogeneous section, linear stress distribution σ = (M/I) y. Concept / Approach: For a rectangle, stress magnitudes at the top and bottom are proportional to their distances from the N.A. Let x be the distance from the top surface to the N.A. Then the top distance is x, and the bottom distance is (10 − x). The ratio of stresses equals the ratio of distances from the N.A. Step-by-Step Solution: Given |σ_top| / |σ_bottom| = x / (10 − x).1600 / 400 = x / (10 − x).4 = x / (10 − x) ⇒ 4(10 − x) = x ⇒ 40 − 4x = x ⇒ 5x = 40.x = 8 cm measured from the top surface. Verification / Alternative check: Distances 8 cm (top to N.A.) and 2 cm (bottom to N.A.) have ratio 8:2 = 4, matching the stress ratio 1600:400 = 4, confirming consistency. Why Other Options Are Wrong: 2 cm or 4 cm place the N.A. too close to the bottom or mid-depth, contradicting the 4:1 stress ratio. 6 cm yields a 6:4 distance ratio = 1.5, not 4. 10 cm would put the N.A. at the bottom, impossible for the given opposite stress signs. Common Pitfalls: Forgetting that stress is directly proportional to distance from the N.A., not the full depth. Mixing signs; use magnitudes when forming the ratio. Final Answer: 8 cm from the top.

Strength of Materials Questions