Question 1

Euler buckling – effect of end conditions on critical load For columns of the same length, material, and cross-section, which end condition yields the highest Euler buckling load (i.e., the strongest column against buckling)?

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Introduction / Context:Euler's critical load shows how end restraints influence a column's stability. Understanding end conditions helps designers choose supports that maximize buckling resistance without unnecessary material.Given Data / Assumptions:Same material (modulus E) and cross-section (second moment I).Same clear length L.Elastic buckling, slender column (Euler regime).Concept / Approach:Euler's formula: P_cr = (pi^2 * E * I) / (L_e)^2, where L_e is the effective length depending on end conditions. Smaller L_e gives larger P_cr. Effective length factors K are: fixed–fixed K = 0.5; fixed–hinged K ≈ 0.699; hinged–hinged K = 1.0; fixed–free (cantilever) K = 2.0.Step-by-Step Solution:Compute L_e = K * L for each case.Compare K: the smallest K is 0.5 (fixed–fixed).Therefore, fixed–fixed columns have the largest P_cr for identical L, E, I.Verification / Alternative check:Ratios: P_cr(fixed–fixed) : P_cr(hinged–hinged) = (1/0.5^2) : (1/1^2) = 4 : 1 → fixed–fixed supports quadruple Euler capacity relative to pin–pin for the same column.Why Other Options Are Wrong:Fixed–hinged and hinged–hinged have larger K than fixed–fixed, giving lower P_cr. Fixed–free has the largest K and the smallest P_cr.Common Pitfalls:Confusing effective length K values; forgetting that stiffer end fixity reduces effective length and increases buckling load.Final Answer:Both ends fixed (fixed–fixed)

Question 2

Section modulus of a rectangular section – proportional dependence For a rectangular cross-section of width b and overall depth d, the elastic section modulus Z about the centroidal horizontal axis is proportional to which of the following?

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Introduction / Context:The elastic section modulus Z connects bending moment to extreme fibre stress via sigma_max = M / Z. For rectangular sections used in beams, knowing how Z scales helps with rapid sizing.Given Data / Assumptions:Rectangular section with width b and depth d.Bending about the centroidal horizontal axis (through mid-depth).Linear elastic behavior.Concept / Approach:Moment of inertia about the centroidal horizontal axis is I = b d^3 / 12. Distance to the extreme fibre is y_max = d / 2. Hence section modulus Z = I / y_max = (b d^3 / 12) / (d/2) = b d^2 / 6. Since area A = b d, it follows Z ∝ A * d.Step-by-Step Solution:I = b d^3 / 12.y_max = d / 2.Z = I / y_max = (b d^3 / 12) / (d/2) = b d^2 / 6.Express in terms of area: A = b d → Z = (A d) / 6 → proportional to A * d.Verification / Alternative check:Dimensional check: Z has units of length^3; A * d has units of length^3, confirming consistency.Why Other Options Are Wrong:Area only or area squared give incorrect scaling. Area × width becomes A * b = b^2 d, not proportional to b d^2. Half of I is not Z; Z depends on I divided by y_max.Common Pitfalls:Using I directly instead of Z; forgetting to divide by the extreme fibre distance.Final Answer:Product of the area and the depth (b d × d)

Question 3

Fillet weld design – maximum direct load on a single fillet weld of length L A flat plate is joined by a single fillet weld (size = leg length). If the permissible shear stress in the weld metal is S, what is the maximum direct load the fillet joint…

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Introduction / Context:Design of fillet welds in steel structures often requires calculating the direct load a weld can carry. A fillet weld develops shear on its effective throat, not on the full leg size. Recognizing the relationship between leg size and throat is critical for accurate and safe design.Given Data / Assumptions:Single straight fillet weld of length L.Fillet weld leg length (commonly called the “size”).Permissible (allowable) shear stress in weld metal = S.Load is in-plane and concentric; no bending or eccentricity.Concept / Approach:The effective throat thickness t_throat of a 45° equal-leg fillet weld equals 0.707 × (fillet size). The shear capacity equals shear stress × resisting area. Resisting area for a straight weld line is (throat) × (length L). Hence, capacity = S × (0.707 × size) × L.Step-by-Step Solution:Identify effective throat: t_throat = 0.707 × (fillet size).Resisting area = t_throat × L.Capacity = S × t_throat × L = S × 0.707 × (fillet size) × L.Therefore, the correct expression is 0.707 × S × (fillet size) × L.Verification / Alternative check:Many steel codes and handbooks tabulate fillet weld throat as 0.7 times the leg size. Multiplying by allowable shear stress and weld length directly yields the simple design formula.Why Other Options Are Wrong:(b) and (d) use arbitrary multipliers unrelated to the throat relation.(c) ignores the 0.707 factor and overestimates capacity.(e) is incorrect since a valid option exists.Common Pitfalls:Confusing leg size with throat thickness; forgetting to reduce for returns or end crater. For intermittent or staggered welds, use the effective total length.Final Answer:0.707 × S × (fillet size) × L

Question 4

Floating beam analogy – maximum shear force in a floating rectangular log with a central load A rectangular wooden log is floating in still water and a 100 N load is applied at its midspan (center). Treating buoyant support as continuous, what is th…

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Introduction / Context:A floating log subjected to a concentrated load behaves, in a first approximation, like a simply supported beam with equal reactions provided by buoyancy on either side of the load. Recognizing the reaction distribution allows quick identification of shear force magnitudes.Given Data / Assumptions:Applied point load W = 100 N at the midspan.Hydrostatic buoyancy adjusts to provide static equilibrium.Neglect wave action and dynamic effects; treat as quasi-static.Concept / Approach:For a simply supported beam with a central point load W, the end reactions are equal: R_A = R_B = W/2. The maximum positive/negative shear magnitude equals the reaction magnitude and occurs just to the left/right of the midspan or at the supports in the shear force diagram.Step-by-Step Solution:W = 100 N at midspan → reactions R_A = R_B = W/2 = 50 N.Shear diagram jumps from +50 N near the left support to −50 N near the right support at the load location.Maximum absolute shear magnitude = 50 N at each support region.Verification / Alternative check:Equilibrium: ΣV = R_A + R_B − W = 50 + 50 − 100 = 0; ΣM about any support confirms R = W/2.Why Other Options Are Wrong:(b) Shear at the centre is zero immediately under a central point load for an idealized SFD, not 50 N.(c) 100 N at centre violates equilibrium distribution.(d) Not applicable; a correct option exists.Common Pitfalls:Confusing support shear with bending moment; mixing total load with reaction magnitudes.Final Answer:50 N at each end

Question 5

Beam of uniform strength with constant depth – how should the width vary? For a prismatic beam designed for uniform strength while keeping its depth constant, the required width along the span must vary in proportion to which quantity?

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Introduction / Context:“Uniform strength” beams are shaped so that the extreme-fibre stress remains approximately constant along the span. When depth is held constant, width must be adjusted according to how internal bending moment varies with position.Given Data / Assumptions:Rectangular section with constant depth d.Allowable bending stress σ_allow is constant along the span.Small deflection theory; material remains elastic.Concept / Approach:For a rectangular beam, section modulus Z = b d^2 / 6. Bending stress is σ = M / Z = 6M / (b d^2). To keep σ = σ_allow, the width b must be adjusted so that b ∝ M.Step-by-Step Solution:σ_allow = 6M / (b d^2)Rearrange → b = 6M / (σ_allow d^2)Hence, b varies directly with M(x), the bending moment diagram.Verification / Alternative check:For a simply supported beam with a central point load, M is triangular from support to midspan, and so is the required b(x) when d is held constant.Why Other Options Are Wrong:Shear force primarily governs web/web-like regions; it does not determine bending stress shape for uniform strength.M^2 leads to unnecessary over-variation.Deflection depends on EI and load distribution; it is not the basis for stress-uniformity shaping.Common Pitfalls:Confusing uniform strength with uniform stiffness; the former targets constant stress, the latter targets constant EI.Final Answer:Bending moment M(x)

Question 6

Uniform-strength beam (constant depth d) under a central load – required width as a function of distance x A simply supported beam of span L carries a central point load W. If the beam is to be of uniform strength while keeping depth d constant, wha…

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Introduction / Context:Shaping a beam for uniform strength keeps the extreme-fibre bending stress constant along the span. For a rectangular beam with constant depth, the width must vary to match the bending moment diagram.Given Data / Assumptions:S.S. beam, span L, central point load W.Rectangular section, constant depth d.Allowable bending stress = σ_allow.Elastic behavior; small deflection theory.Concept / Approach:For 0 ≤ x ≤ L/2, bending moment is M(x) = R × x = (W/2) × x. For a rectangular section, Z = b d^2 / 6. Uniform strength requires σ_allow = M / Z = 6M / (b d^2) → b(x) = 6M / (σ_allow d^2).Step-by-Step Solution:M(x) = (W/2) x.b(x) = 6 × [(W/2) x] / (σ_allow d^2) = (3 W x) / (σ_allow d^2).At midspan (x = L/2), b_max = (3 W L) / (2 σ_allow d^2).Verification / Alternative check:The required b(x) is linear in x (triangular distribution), matching the triangular moment diagram for a central point load.Why Other Options Are Wrong:(b) Double counts; would imply σ varying with x.(c) Wrong dimensions and not a function of x.(d) Incorrectly suggests constant width.(e) Quadratic in x, inconsistent with linear M(x).Common Pitfalls:Mixing up allowable stress with material strength; always use design (allowable) stress for sizing.Final Answer:b(x) = (3 W x) / (σ_allow d^2)

Question 7

Cantilever deflection comparison – point load at free end vs. uniformly distributed load (same total load) For a cantilever of length L and flexural rigidity EI, compare the maximum deflections when (i) a point load W acts at the free end and (ii) a…

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Introduction / Context:Deflection formulas for common load cases are frequently compared to understand serviceability. For a cantilever, the end deflection depends strongly on how the load is distributed.Given Data / Assumptions:Cantilever length L, flexural rigidity EI.Case (i): point load W at the free end.Case (ii): UDL intensity w with total load W = wL (same total load as case i).Concept / Approach:Standard results: δ_point = W L^3 / (3 E I). For UDL with intensity w, δ_udl = w L^4 / (8 E I). With W = wL, substitute to compare.Step-by-Step Solution:δ_point = W L^3 / (3 E I).δ_udl = w L^4 / (8 E I) = (W/L) L^4 / (8 E I) = W L^3 / (8 E I).Ratio δ_point / δ_udl = (W L^3 / 3 E I) / (W L^3 / 8 E I) = 8/3.Verification / Alternative check:Since a point load is more severe than the same total UDL on a cantilever tip deflection, the ratio exceeding 1 (8/3 ≈ 2.67) is reasonable.Why Other Options Are Wrong:3/8 inverts the ratio.2 and 4/3 do not match the derived value.1 would imply equal deflections, which is false.Common Pitfalls:Comparing for equal intensities rather than equal total load; always align the basis of comparison.Final Answer:8/3

Question 8

Cantilever with uniformly distributed load over the entire length – maximum fixed-end moment A cantilever of length L carries a uniformly distributed load over the whole span. If W denotes the total load on the cantilever, what is the maximum bendin…

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Introduction / Context:For a cantilever, the critical bending moment occurs at the fixed support. When the loading is a uniform load distributed along the span, the fixed-end moment can be expressed either in terms of intensity w (per unit length) or in terms of the total load W = wL.Given Data / Assumptions:Cantilever length L.UDL over the entire length.W denotes the total load = wL.Concept / Approach:Standard result: with intensity w, the fixed-end moment M_max = w L^2 / 2. Substitute w = W / L to express the moment in terms of W.Step-by-Step Solution:M_max = (w L^2) / 2.Since W = wL, w = W / L.M_max = (W / L) × L^2 / 2 = (1/2) W L.Verification / Alternative check:Dimensional check: bending moment has units of force × length; (1/2) W L is consistent.Why Other Options Are Wrong:W L doubles the correct value.(1/3) W L is not the standard coefficient for this case.W L^2 has wrong dimensions if W is total load.Common Pitfalls:Confusing W (total load) with w (load intensity). Always confirm symbols used in the question.Final Answer:(1/2) W L

Question 9

Best rectangle from a circular log – section for maximum bending strength From a circular wooden log of diameter 100 cm, a rectangular beam is cut. What breadth b and depth d (in cm) maximize the bending strength about the strong axis?

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Introduction / Context:When cutting a rectangular section from a given circular log, the goal is often to maximize bending strength (section modulus) about the strong axis. This is a classical optimization problem with a geometric constraint.Given Data / Assumptions:Circular log diameter D = 100 cm.Rectangle breadth b and depth d inscribed in the circle → b^2 + d^2 = D^2.Strength ∝ section modulus about strong axis, Z = b d^2 / 6 for a rectangle.Concept / Approach:Maximize Z = b d^2 / 6 subject to the constraint b^2 + d^2 = D^2. Use calculus (substitute b = √(D^2 − d^2), differentiate Z with respect to d, set derivative to zero) to find the optimal ratio.Step-by-Step Solution:Constraint: b = √(D^2 − d^2).Z(d) ∝ d^2 √(D^2 − d^2).dZ/dd = 0 → 2D^2 − 3d^2 = 0 → d = (√6 / 3) D ≈ 0.8165 D.Then b = √(D^2 − d^2) = (√3 / 3) D ≈ 0.5774 D.For D = 100 cm: d ≈ 81.65 cm and b ≈ 57.73 cm.Verification / Alternative check:The ratio b : d = √3 : √6 = 1 : √2, a well-known result for the strongest rectangle from a circle.Why Other Options Are Wrong:(b) swaps the axes; yields smaller Z about the strong axis.(c), (d), (e) do not satisfy the optimal ratio and give lower section modulus.Common Pitfalls:Maximizing area instead of section modulus; ignoring the circle-inscription constraint b^2 + d^2 = D^2.Final Answer:b = 57.73 cm, d = 81.65 cm

Question 10

Yield phenomenon – stress to start yielding vs. to continue yielding In a ductile metal showing upper and lower yield points, how does the stress required to initiate yielding compare with the stress required to continue yielding?

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Introduction / Context:Many low-carbon steels display a yield point phenomenon: an upper yield point where yielding initiates and a lower yield point where plastic flow continues at a reduced stress. Understanding this behavior is important for interpreting tensile tests and designing against plastic deformation.Given Data / Assumptions:Ductile metal with a Lüders-type yield point (upper and lower yield).Monotonic tension test.Concept / Approach:At the onset of yielding, dislocation sources activate and break away from Cottrell atmospheres, causing a temporary peak (upper yield). Once plastic flow starts, fewer obstacles oppose dislocation motion, so a lower stress (lower yield) sustains continued yielding.Step-by-Step Solution:Observation: σ_upper > σ_lower for materials showing a yield point plateau.Therefore, stress to initiate yielding (at the upper yield point) exceeds the stress to continue plastic flow (lower yield point).Verification / Alternative check:Tensile test curves of mild steel typically show a drop from the upper yield to a lower, nearly constant stress as Lüders bands propagate.Why Other Options Are Wrong:(b) contradicts test evidence.(c) and (d) refer to “stop yielding,” which is not a standard descriptor in this context.(e) confuses yield with ultimate strength.Common Pitfalls:Equating 0.2% proof stress behavior (smooth curve alloys) with discrete yield point materials; they are not identical phenomena.Final Answer:It is more than that necessary to continue yielding

Question 11

Length effect on cantilever deflection – point load at free end For a cantilever beam carrying a point load W at its free end, if the length of the cantilever is doubled (other parameters unchanged), by what factor does the free-end deflection chang…

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Introduction / Context:Deflection sensitivity to span length is especially pronounced in cantilevers. Recognizing the power-law dependence helps engineers anticipate serviceability issues when spans change.Given Data / Assumptions:Cantilever with end point load W.Original length L becomes 2L; material (E) and section (I) unchanged.Concept / Approach:For a cantilever with an end point load, the free-end deflection is δ = W L^3 / (3 E I). Because δ ∝ L^3, a proportional change in length produces a cubic change in deflection.Step-by-Step Solution:Original: δ_1 = W L^3 / (3 E I).New length 2L: δ_2 = W (2L)^3 / (3 E I) = W (8 L^3) / (3 E I) = 8 δ_1.Hence, the deflection increases eightfold.Verification / Alternative check:Dimensional reasoning confirms cubic dependence on length for this loading condition; many design tables list the same relationship.Why Other Options Are Wrong:1/8 and 1/3 suggest decreases, which is incorrect.2 and 3 underestimate the cubic effect.Common Pitfalls:Assuming linear dependence on L; for beams under bending, deflections commonly scale with L^3 for simple cases.Final Answer:8

Question 12

Column effective length: For a compression member with one end fixed and the other end free (a classical cantilever column), what is the equivalent length used in Euler's buckling formula, expressed as a multiple of the actual length l?

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Introduction / Context:In structural engineering, the slenderness and end restraints of a compression member control its elastic buckling load. Euler's theory uses an effective (equivalent) length factor to model different boundary conditions. A column that is fixed at one end and free at the other is the most flexible common case.Given Data / Assumptions:Prismatic column, homogeneous, linear elastic.One end perfectly fixed (zero rotation and zero translation), the other end perfectly free (no restraint).Euler buckling (small deflection, elastic instability) is applicable.Concept / Approach:Euler buckling load P_cr = (π^2 * E * I) / (L_e)^2, where L_e is the effective length. Different end conditions are captured by L_e = K * l with a coefficient K. For fixed–free, K is largest (most flexible), so the critical load is the smallest among common cases.Step-by-Step Solution:End condition: fixed–free ⇒ effective length factor K = 2.0.Therefore, L_e = K * l = 2 * l.Thus the equivalent length is 2 l.Verification / Alternative check:Classical tables (or solving the eigenvalue problem y'' + (P/(E I)) y = 0 with the appropriate boundary conditions) yield the first buckling mode giving K = 2.0 for fixed–free.Why Other Options Are Wrong:0.5 l and 0.7 l correspond to much stiffer end conditions (e.g., fixed–fixed or fixed–pinned), not fixed–free.l corresponds to pinned–pinned.1.5 l is not a standard Euler end condition.Common Pitfalls:Confusing effective length (a stability concept) with actual length, and mixing up end-condition factors K.Final Answer:2 l.

Question 13

Hooke’s law in mechanics of materials: Within the elastic limit of a homogeneous material, the normal stress is ______ to the corresponding normal strain.

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Introduction / Context:Hooke’s law is the foundational linear relationship used in elasticity and structural analysis. It links stress and strain for materials behaving elastically and underlies the use of Young’s modulus in design calculations.Given Data / Assumptions:Material behaves linearly within its elastic (proportional) limit.Small deformations; unloading follows loading without permanent set.Uniaxial normal stress–strain state is considered.Concept / Approach:Hooke’s law states: σ = E * ε, where σ is normal stress, ε is normal strain, and E is Young’s modulus (a constant for a given material and temperature, within the elastic range). This is a direct proportionality: doubling strain doubles stress.Step-by-Step Solution:Identify regime: elastic and proportional region.Apply linear law: σ = E * ε.Conclude: stress is directly proportional to strain.Verification / Alternative check:Tensile test curves show an initial straight line whose slope is E. The relationship deviates after the proportional limit as plasticity begins.Why Other Options Are Wrong:Inversely proportional/Curvilinear: contradict the linear segment of a stress–strain curve in the elastic region.Not related/Depends on time: ignores elastic material behavior (viscoelastic exceptions are beyond the simple Hooke’s law).Common Pitfalls:Extending Hooke’s law beyond the elastic limit or confusing elastic modulus with yield strength.Final Answer:Directly proportional.

Question 14

Effect of changing section depth on deflection: A simply supported beam carries a single concentrated load at midspan. If only the depth is doubled (breadth unchanged), by what factor does the midspan deflection change?

Accepted Answer

Introduction / Context:Deflection control often governs beam sizing. For rectangular sections, the second moment of area is very sensitive to depth, so changing depth can dramatically reduce deflections without changing material or span.Given Data / Assumptions:Simply supported beam, point load P at midspan.Span L, material modulus E unchanged.Breadth b unchanged; depth h doubled to 2h.Concept / Approach:For a central point load, elastic deflection at midspan is δ = (P * L^3) / (48 * E * I). For a rectangular section, I = b * h^3 / 12. If h → 2h, then I_new = b * (2h)^3 / 12 = 8 * (b * h^3 / 12) = 8 I.Step-by-Step Solution:Original deflection: δ_1 = (P L^3) / (48 E I).New second moment: I_2 = 8 I.New deflection: δ_2 = (P L^3) / (48 E I_2) = δ_1 / 8.Hence, the deflection becomes one-eighth of the original value.Verification / Alternative check:Dimensional reasoning confirms that deflection scales inversely with I, and I scales with h^3; doubling h scales deflection by 1/2^3 = 1/8.Why Other Options Are Wrong:1/2 and 1/4 would correspond to linear or quadratic sensitivity; the correct sensitivity is cubic in depth.2, 4, 8 imply deflection increases with depth, which contradicts beam theory.Common Pitfalls:Using section modulus instead of second moment for deflection, or changing both breadth and depth accidentally when only depth is varied.Final Answer:1/8 (deflection becomes one-eighth).

Question 15

Uniform strength concept: A beam is said to be “of uniform strength” if which condition holds along its length under bending?

Accepted Answer

Introduction / Context:The design idea of a “beam of uniform strength” aims to utilize material efficiently by tailoring the section so that the extreme-fiber bending stress is constant along the entire span, rather than overdesigning where bending moments are small.Given Data / Assumptions:Elastic bending (σ = M y / I).Beam geometry may vary along the span to satisfy a target stress distribution.Focus is on bending stress, not necessarily shear or deflection.Concept / Approach:In regions of high bending moment, section modulus Z = I / y must be larger to keep σ = M / Z constant. In regions of smaller bending moment, Z can be reduced. The idealized concept results in a beam where maximum fiber stress is uniform along the length.Step-by-Step Solution:Target: σ_max(x) = constant = σ_allow.Use σ = M(x) / Z(x) ⇒ choose Z(x) ∝ M(x).Thus, by varying depth/width to achieve Z(x) matching M(x), the beam attains “uniform strength.”Verification / Alternative check:Classic tapered beams (e.g., parabolic depth variation under UDL) illustrate this principle where material use aligns with bending moment distribution.Why Other Options Are Wrong:Equal bending moment, shear stress, deflection, or constant curvature are not the defining criteria and are rarely true for real loadings.Common Pitfalls:Confusing “uniform strength” with “constant section” or with uniform deflection or shear.Final Answer:Bending stress is the same at every section along the longitudinal axis.

Question 16

Column effective length: For a column fixed at both ends (both ends fully restrained against rotation and translation), the equivalent length used in Euler’s formula is:

Accepted Answer

Introduction / Context:End conditions drastically affect a column’s critical buckling load. A column with both ends fixed is significantly stiffer against buckling than one that is pinned or free.Given Data / Assumptions:Prismatic, elastic column.Both ends fixed (zero slope and zero translation at ends).Euler elastic buckling governs.Concept / Approach:Euler's critical load P_cr = (π^2 E I)/(L_e)^2. The effective length L_e = K * l captures end conditions via K. For fixed–fixed, K = 0.5, giving the highest P_cr among the common ideal end restraints.Step-by-Step Solution:End condition: fixed–fixed ⇒ K = 0.5.L_e = K * l = 0.5 l.Verification / Alternative check:Mode-shape solutions to the column differential equation yield K = 0.5 for the lowest buckling mode with both ends fixed.Why Other Options Are Wrong:0.7 l is not a standard ideal end factor.l is for pinned–pinned; 2 l is for fixed–free; 1.5 l is nonstandard in Euler's ideal set.Common Pitfalls:Using pinned–pinned formulas or forgetting the strong influence of rotational fixity.Final Answer:0.5 l.

Question 17

Arching action in beams and slabs: What is the principal effect of arching (developing compressive arch action with horizontal thrust) on the bending moment distribution along a member?

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Introduction / Context:When structural systems can develop horizontal thrust (e.g., due to restrained supports or compressive membrane action), they exhibit arching. This phenomenon alters the internal force distribution compared with pure flexure.Given Data / Assumptions:Member capable of sustaining compression and developing thrust (e.g., masonry lintel, slab with edge restraint).Loads predominantly vertical; supports provide some horizontal restraint.Elastic behavior with modified load path due to arching.Concept / Approach:Arching shifts part of the load-carrying mechanism from bending to axial compression. The thrust line within the depth reduces flexural demands; if the line remains within the kern, tensile stresses reduce and bending moments lower overall.Step-by-Step Solution:In pure bending: internal couple M resists load effects.With arching: horizontal reactions develop; a compressive thrust T carries part of the load via a funicular path.Result: required bending moments M_arch are smaller than M_pure for the same external loads due to the axial component assisting load transfer.Verification / Alternative check:Comparisons of simply supported beams vs. two-pinned arches under UDL show lower midspan moments for arches due to thrust. Lintel arching over openings in masonry similarly reduces sagging moments.Why Other Options Are Wrong:Increase/no effect: contradict the fundamental mechanism of load sharing by axial compression and thrust line action."All of the above"/"Only changes shear": incorrect generalizations.Common Pitfalls:Confusing camber (geometric curvature) with arching action; ignoring the need for horizontal restraint to develop thrust.Final Answer:To reduce the bending moment throughout.

Question 18

Cantilever with triangular (linearly varying) load: A cantilever beam carries a load that varies linearly from zero at the free end to a maximum at the fixed end. The shape of the bending-moment diagram is a:

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Introduction / Context:Recognizing shear-force and bending-moment (BM) diagram shapes for standard load cases enables quick checks on analysis and design.Given Data / Assumptions:Cantilever fixed at one end, free at the other.Load intensity w(x) increases linearly from 0 at the free end to w_max at the fixed end (triangular load).Prismatic member, small deflection, linear elasticity.Concept / Approach:The mathematical orders relate as: load w(x) is the derivative of shear V(x) with a sign, and shear is the derivative of moment M(x) with a sign. If w(x) is linear in x, then V(x) is quadratic, and M(x) is cubic in x.Step-by-Step Solution:Let x be the distance from the fixed end.Given w(x) ∝ x (linear) ⇒ V(x) = −∫ w(x) dx ⇒ V(x) is quadratic in x.M(x) = −∫ V(x) dx ⇒ M(x) is cubic in x.Therefore, the BM diagram is a cubic parabola.Verification / Alternative check:For UDL (constant w), BM is parabolic; increasing the order of w by one (to linear) increases the order of M by one (to cubic), confirming the shape.Why Other Options Are Wrong:Triangle/Rectangle correspond to shear diagrams in some cases, not BM here.Parabola applies to UDL, not linearly varying load.Common Pitfalls:Mixing up the roles of w, V, and M in differential relationships or reading the diagram orientation incorrectly.Final Answer:Cubic parabola.

Question 19

Volumetric strain for a circular (isotropic) bar: If a circular bar experiences a longitudinal (axial) strain ε_long and an equal lateral strain ε_lat in each of the two transverse directions, what is the volumetric strain?

Accepted Answer

Introduction / Context:Volumetric strain quantifies the fractional change in volume of a 3D body due to deformation. For bars under uniaxial loading, the axial strain combines with lateral strains (due to Poisson effect) to produce the overall volume change.Given Data / Assumptions:Cylindrical (circular) prismatic bar, isotropic material.Small strains; principal (mutually orthogonal) strains are: ε_x = ε_long, ε_y = ε_lat, ε_z = ε_lat.Lateral strains are equal in the two transverse directions.Concept / Approach:For small strains, the volumetric strain ε_v equals the sum of the normal strains in three orthogonal directions: ε_v = ε_x + ε_y + ε_z. Under uniaxial loading with symmetry, this becomes ε_v = ε_long + 2 ε_lat.Step-by-Step Solution:Write principal strains: ε_x = ε_long; ε_y = ε_lat; ε_z = ε_lat.Sum them: ε_v = ε_x + ε_y + ε_z.Therefore, ε_v = ε_long + 2 ε_lat.Verification / Alternative check:If Poisson’s ratio ν is known under axial stress σ, then ε_lat = −ν ε_long, giving ε_v = ε_long (1 − 2 ν). This well-known relation confirms the form above.Why Other Options Are Wrong:Permutations with minus signs or different coefficients do not represent the sum of three orthogonal normal strains for an isotropic bar.Common Pitfalls:Forgetting there are two equal lateral directions, or confusing engineering strain sign convention (tension positive, lateral contraction negative).Final Answer:ε_long + 2 ε_lat.

Question 20

Self-weight scaling of a bar: If all linear dimensions of a prismatic bar are scaled up by a factor n (n > 1), in what ratio does the maximum axial stress produced by its own weight increase?

Accepted Answer

Introduction / Context:Dimensional scaling arguments help predict how stresses change when structures are built larger or smaller but with similar proportions. For vertical members carrying only their own weight, the governing parameter is length, not cross-sectional area.Given Data / Assumptions:Geometrically similar scaling: length, breadth, and depth multiply by n.Material properties (density ρ and modulus) unchanged.Bar is vertical; stress of interest is the maximum axial stress due to self-weight.Concept / Approach:For a vertical prismatic bar of constant area A and length L, the maximum axial stress due to its own weight is σ_max = ρ * g * L (at the top). This expression is independent of A because weight increases with A but so does the load-carrying area, canceling out.Step-by-Step Solution:Original: σ_1 = ρ g L.After scaling: L_2 = n L; area scales as A_2 = n^2 A; weight per unit length scales as ρ g A_2.However, stress at the top remains σ_2 = ρ g L_2 = ρ g (n L) = n (ρ g L) = n σ_1.Hence, the stress increases in the ratio n : 1.Verification / Alternative check:Dimensional analysis confirms σ scales with length only for self-weight in a prismatic bar; finite-element or calculus derivations of σ(y) = ρ g (L − y) lead to the same conclusion.Why Other Options Are Wrong:1 : n or 1 : n^2 imply stress decreases with size, which is incorrect for self-weight in prismatic bars.n^2 : 1 exaggerates the scaling (stress does not depend on area).Remains unchanged contradicts the direct proportionality to length.Common Pitfalls:Assuming stress scales with total weight without accounting for simultaneous scaling of area, or overlooking that σ_max occurs at the top and equals ρ g L.Final Answer:n : 1.

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