Question 1

For a prismatic beam of length L and second moment of area I subjected to a constant bending moment M along its span, determine the total strain energy stored due to bending (express your answer in terms of M, L, E, and I).

Accepted Answer

Introduction / Context: Strain energy methods are powerful for deflection calculations (Castigliano's theorem) and for understanding how members store elastic energy under load. For a constant bending moment along a span (e.g., pure bending region), the energy expression simplifies neatly. Given Data / Assumptions: Constant bending moment M over the full length L. Beam has constant E and I; linear elastic behavior. No shear deformation energy is considered (pure bending). Concept / Approach: The bending strain energy density per unit length is (M^2) / (2 E I). For variable moment, one would integrate this along x. With M constant, the integral reduces to multiplication by L. Step-by-Step Solution: 1) General formula: U = ∫ (M(x)^2 / (2 E I)) dx.2) For constant M: U = (M^2 / (2 E I)) ∫ dx from 0 to L.3) Evaluate the integral: U = (M^2 / (2 E I)) * L = M^2 L / (2 E I). Verification / Alternative check: Dimensional check: M^2 has units of (force*length)^2; dividing by E I (force/area * length^4) and multiplying by length yields force*length (work), consistent with energy. Why Other Options Are Wrong: U = M^2 L / (E I): Missing the factor 1/2; overestimates energy by 2×. U = M^2 L / (4 E I): Underestimates by 2×. U = 2 M^2 L / (E I): Incorrect scaling. Common Pitfalls: Forgetting the 1/2 factor that arises from integrating stress–strain energy density. Applying this constant-moment formula to non-uniform bending without integration. Final Answer: U = M^2 L / (2 E I).

Question 2

In structural analysis, the point of contraflexure along a beam is defined as the location where which characteristic of the bending moment (B.M.) diagram occurs?

Accepted Answer

Introduction / Context: The point of contraflexure (also called the inflection point) is central to understanding beam behavior, especially in indeterminate structures or members with variable loading. It indicates a change in curvature and is relevant for reinforcement detailing and splicing decisions. Given Data / Assumptions: We consider typical beam behavior with continuous or varying loads. The bending moment diagram (B.M. diagram) may pass through zero between positive and negative regions. Concept / Approach: Beam curvature is proportional to bending moment (for prismatic, linear-elastic beams: curvature κ ≈ M / (E I)). A change in sign of M implies a change in curvature direction (sagging to hogging or vice versa). The location where M = 0 within the span (not at a free end with zero moment by boundary) is the point of contraflexure. Step-by-Step Solution: 1) Plot or envision the B.M. diagram for the given loading.2) Identify where M crosses zero within the member length.3) Mark that location as the point of contraflexure.4) Recognize that curvature changes sign there, indicating a shift from sagging to hogging or vice versa. Verification / Alternative check: In continuous beams over multiple supports, points of contraflexure commonly occur between supports where the B.M. diagram transitions through zero between negative and positive peaks. Why Other Options Are Wrong: B.M. maximum/minimum: These are extremum points where shear force is zero, not necessarily points of sign change. S.F. is zero: Zero shear indicates a B.M. extremum, not inherently a sign change. Common Pitfalls: Confusing zero shear (extreme B.M.) with zero moment (contraflexure). Marking support ends (where M = 0 by boundary) as contraflexure without a sign change in the span. Final Answer: B.M. changes sign.

Question 3

Axial deformation and Young’s Modulus for a prismatic rod A straight rod of uniform cross-sectional area A and length L is subjected to an axial force P and undergoes an elastic deformation δ. What is the Young’s Modulus E of the material?

Accepted Answer

Introduction / Context: In strength of materials, axial deformation of a prismatic (uniform) rod under axial load provides a direct route to determine Young’s Modulus E when the load, geometry, and measured elongation are known. This test principle underlies tensile testing and field evaluations of stiffness. Given Data / Assumptions: Rod length = L; uniform area = A. Axial force = P (within elastic limit). Measured elastic extension = δ. Homogeneous, isotropic, linear-elastic behaviour is assumed. Concept / Approach: Axial stress σ = P / A. Axial strain ε = δ / L. By definition, Young’s Modulus E = σ / ε. Substituting yields E = (P / A) / (δ / L) = (P * L) / (A * δ). This relation is foundational for elastic analysis of bars and for calibration of material properties. Step-by-Step Solution: Compute stress: σ = P / A.Compute strain: ε = δ / L.Apply definition: E = σ / ε = (P / A) / (δ / L).Simplify: E = (P * L) / (A * δ). Verification / Alternative check: Dimension check: [E] = force/area divided by deformation/length = (N/m^2) / (m/m) = N/m^2, consistent with modulus units (Pa). Why Other Options Are Wrong: (A * δ) / (P * L) and others: These invert the correct relationship or scramble variables; they would not pass a simple dimensional or limiting-case check (e.g., larger P should increase E, not decrease it). Common Pitfalls: Confusing engineering strain (δ/L) with percentage elongation (100 * δ/L). Using plastic deformation values; the formula is valid only in the elastic range. Final Answer: E = (P * L) / (A * δ)

Question 4

Geometry of reactions in a two-hinged semicircular arch For a two-hinged semicircular arch under variable loading, the locus of the resultant reaction at a support traces which curve in a plane diagram of reactions?

Accepted Answer

Introduction / Context: Arches carry loads by combined axial compression and bending; reactions depend on the load position and magnitude. For certain canonical arch geometries, elegant geometric properties emerge. A semicircular two-hinged arch is one such case where the reaction locus exhibits a simple, recognizable curve. Given Data / Assumptions: Arch is semicircular with two supports (hinged at ends). Reactions are examined as loads vary in position along the span. Plane of analysis; small deflection theory and classical statics apply. Concept / Approach: In a semicircular arch, symmetry and constant radius lead to special relationships between horizontal thrust and vertical reaction as load moves. The vectorial sum at a support (resultant reaction) varies in magnitude and direction but maintains a constant relation that plots as a circle in the reaction diagram. Step-by-Step Solution: Consider unit and moving loads to generate envelopes of H (horizontal thrust) and V (vertical reaction).Combine H and V vectorially to get the resultant R at a support.Show that as the load position changes, the tip of vector R traces a circle due to the semicircular geometry and constant-radius relationships. Verification / Alternative check: Classical arch theory texts present derivations or Mohr diagrams of reactions for circular arches, where the locus of resultant reactions forms a circle under varying loading conditions. Why Other Options Are Wrong: Parabola/Hyperbola/Straight line: Do not match the geometric relation arising from the semicircular radius and symmetric boundary conditions. Common Pitfalls: Confusing the locus of the line of thrust within the arch ring with the locus of support reactions in the reaction plane. Assuming the same locus shape for noncircular arches; it is specific to semicircular geometry here. Final Answer: circle

Question 5

Analyzing forces in simple trusses Which set of methods may be used to determine member forces in a statically determinate, simple truss under given loads and supports?

Accepted Answer

Introduction / Context: Simple planar trusses are extensively used in roofs and bridges. To compute internal axial forces in members, classical statics offers several complementary techniques. Choosing the right method speeds analysis and improves accuracy. Given Data / Assumptions: Truss is statically determinate and planar. Loads are applied at joints; members carry axial forces only. Supports provide determinacy (e.g., pin and roller). Concept / Approach: The method of joints solves force equilibrium at each joint (sum of forces = 0). The method of sections cuts through the truss to analyze a subset rapidly (sum of forces and moments = 0). Graphical methods (e.g., Cremona diagram) construct force polygons to determine member forces visually, useful for quick checks and conceptual understanding. Step-by-Step Solution: Start with global reactions using static equilibrium.Use method of joints for joint-by-joint resolution in small/regular trusses.Apply method of sections to jump directly to critical members (efficient for larger trusses).Use graphical diagrams as a check or for preliminary sizing. Verification / Alternative check: All three methods are standard in structural analysis curricula and yield identical results for determinate trusses when applied correctly. Why Other Options Are Wrong: Any single-method-only option is incomplete; multiple valid methods exist and are routinely used. Common Pitfalls: Forgetting sign conventions and tension/compression identification. Cutting more than three unknowns with method of sections, which breaks determinacy of the single step. Final Answer: all the above (graphical method, method of joints, and method of sections)

Question 6

In structural analysis of determinate structures, if we apply the three static equilibrium equations (ΣH = 0, ΣV = 0, and ΣM = 0), what quantities can be fully determined?

Accepted Answer

Introduction / Context: In structural engineering, a statically determinate structure is one in which all unknowns can be found solely from the equations of static equilibrium. Knowing what can be evaluated using ΣH = 0, ΣV = 0, and ΣM = 0 helps decide the analysis path and whether additional compatibility relations are needed. Given Data / Assumptions: The structure considered is statically determinate (no redundancy). Material behavior and deformations do not need to be used to close the system of equations. Loads are known and applied quasi-statically. Concept / Approach: For a determinate structure, the static equations suffice to find: (1) support reactions, (2) internal shear forces, and (3) internal bending moments (and axial forces where relevant). Once reactions are known, section cuts combined with ΣH, ΣV, and ΣM at each cut give internal force diagrams throughout the member. Step-by-Step Solution: Use ΣH = 0 and ΣV = 0 to solve for unknown horizontal and vertical support reactions.Use ΣM = 0 about a convenient point to solve for remaining reactions.Make a cut at any section; apply ΣV = 0 and ΣM = 0 to find shear force and bending moment at that section (and ΣH = 0 for axial force).Repeat along the span to build complete internal force diagrams. Verification / Alternative check: Determinate beams, trusses, and arches yield unique internal force results independent of stiffness or deflection analysis. This confirms that equilibrium alone is sufficient. Why Other Options Are Wrong: Supporting reactions only: too narrow; internal forces are also obtained. Shear forces only / Bending moments only / Internal forces only: each ignores that reactions are also found by the same equations. Common Pitfalls: Confusing determinate with indeterminate systems where compatibility or flexibility/stiffness methods are required in addition to equilibrium. Forgetting axial forces when frames have inclined members. Final Answer: All of the above.

Question 7

Yield moment of a cross-section: the bending moment that just brings which fibre of the section to the yield stress?

Accepted Answer

Introduction / Context: Yield moment is a fundamental term in flexural design and plastic analysis. It marks the boundary between elastic behavior and the onset of yielding in bending members such as beams and slabs. Given Data / Assumptions: Material follows elastic–perfectly plastic idealization in bending. Plane sections remain plane so strain varies linearly across the depth. Yield stress is known (e.g., steel yield strength). Concept / Approach: Under pure bending, extreme fibres (furthest from the neutral axis) experience the maximum tensile and compressive strains. As bending moment increases from zero, the first location to reach the yield stress is the outermost fibre. The moment at which this first yield occurs is the yield moment, often denoted My. Step-by-Step Solution: Relate stress to curvature: σ = E * κ * y, where y is the distance from the neutral axis.Maximum |σ| occurs at |y| = c, the extreme fibre distance.Set |σ| = yield stress to find κ corresponding to first yield.Convert curvature to moment using section flexural rigidity to obtain the yield moment. Verification / Alternative check: Strain distribution linearity ensures that yielding must initiate at the largest |y|. This remains true regardless of section shape (rectangular, I-section, T-section) as long as plane sections remain plane. Why Other Options Are Wrong: Innermost fibre / neutral fibre: these are closer to the neutral axis and reach yield later. Every fibre simultaneously: that corresponds to a fully plastic state (plastic moment), not first yield. Common Pitfalls: Confusing yield moment (first yield at extremes) with plastic moment (full plastic stress block across the section yielding). Final Answer: The outermost fibre of the section.

Question 8

Flat spiral springs: which of the following statements correctly describe their construction and use?

Accepted Answer

Introduction / Context: Flat spiral springs are energy-storage devices used in timing mechanisms, measuring instruments, and recoil systems. Understanding their geometry and support/torque application helps in correct specification and design. Given Data / Assumptions: The spring is a uniform thin strip (constant thickness and width) wound into a planar spiral. One end is fixed (often the outer), the other is attached to a hub or arbor to receive torque. Applications include clocks, watches, spring balances, and meters. Concept / Approach: Under applied torque, the spiral stores strain energy by bending along its length. Because the strip is thin, bending dominates over torsion, enabling a compact form that delivers nearly constant torque over limited rotations. Step-by-Step Solution: Identify the element: a thin rectangular strip forms the spring.Fixity/support: typically the outer end is anchored to the casing; the inner end connects to a rotating arbor.Load application: a torque winds or unwinds the spiral, bending the strip and storing energy.Use cases: timekeeping mechanisms and measuring devices rely on predictable torque–deflection behavior. Verification / Alternative check: Device teardown or catalog drawings for watches and gauges consistently show a flat spiral strip with outer-end anchorage and inner-end arbor connection. Why Other Options Are Wrong: Options (a), (b), and (c) are each correct; hence the correct combined choice is 'All of the above'. Common Pitfalls: Confusing flat spiral springs with helical torsion or clock springs having different winding and support arrangements. Final Answer: All of the above.

Question 9

Euler/column basics: The equivalent (effective) length of a column of actual length L with both ends hinged is equal to:

Accepted Answer

Introduction / Context: Effective length is the notional length of a column used in buckling calculations to account for end restraints. Recognizing effective length factors is essential for critical load predictions using Euler or design codes. Given Data / Assumptions: Prismatic column of length L. Both ends hinged (pinned), allowing rotation but no translation. Elastic buckling considered (no inelastic effects). Concept / Approach: For a pin–pin column, the effective length factor K = 1.0. Thus the effective length Le = K * L = L. This is the reference case in most code tables, with other end conditions expressed relative to it (e.g., fixed–fixed K ≈ 0.5, fixed–free K ≈ 2.0, fixed–pinned K ≈ 0.7). Step-by-Step Solution: Select end condition: both ends hinged.Use K = 1.0 for pin–pin supports.Compute effective length: Le = K * L = L. Verification / Alternative check: Euler buckling mode shape for pin–pin is a half-sine wave over the full column length, confirming that the effective buckling half-wave equals the actual length. Why Other Options Are Wrong: 2L and L/2 correspond to fixed–free and fixed–fixed analogs, not pin–pin. L/√2 is not a standard effective length for column end conditions. Common Pitfalls: Mixing up end-condition factors, especially when lateral bracing exists at intermediate points. Final Answer: L.

Question 10

Beams of uniform strength: If the beam depth is kept constant along the span, how should the width b vary with the local bending moment M to maintain constant extreme-fibre stress?

Accepted Answer

Introduction / Context: A beam of uniform strength maintains the same allowable bending stress at every section under the applied loading. If depth cannot be varied (architectural constraint), designers can tailor the width b to keep bending stress constant. Given Data / Assumptions: Beam depth d is constant along the span. Allowable bending stress σ_allow is fixed. Material remains linear-elastic. Concept / Approach: Bending stress at the extreme fibre is σ = M / Z. For a rectangular section, section modulus Z = b d^2 / 6. Holding σ = σ_allow constant gives M / (b d^2 / 6) = σ_allow, thus b ∝ M when d is constant. Step-by-Step Solution: Start with σ = M / Z.Use Z_rect = b d^2 / 6.Set σ = constant ⇒ M / (b d^2 / 6) = constant.Rearrange ⇒ b ∝ M, since d is constant. Verification / Alternative check: If M doubles near midspan compared to ends, maintaining the same σ requires doubling b at midspan when depth is fixed, matching intuitive expectations for a uniform-strength design. Why Other Options Are Wrong: b ∝ 1/M or √M or M^2: these do not satisfy σ = constant with fixed depth. b constant: would cause σ to track M, defeating uniform strength. Common Pitfalls: Using moment of inertia I instead of section modulus Z directly; Z is the correct geometric property for bending stress at the extreme fibre. Final Answer: b ∝ M.

Question 11

For a triangular cross-section, what is the ratio of its second moment of area about the base to that about a centroidal axis parallel to the base?

Accepted Answer

Introduction / Context: Second moments of area (area moments of inertia) are key for calculating bending stresses and deflections. For standard shapes, knowing ratios between base and centroidal axes helps quick checks and derivations. Given Data / Assumptions: Right triangle with base b and altitude h. Axes considered: about the base, and through centroid parallel to the base. Concept / Approach: For a triangle, the formulae are I_base = b h^3 / 12 and I_centroidal = b h^3 / 36 for the axis parallel to the base through the centroid. The ratio I_base / I_centroidal = (1/12) / (1/36) = 3.0. Step-by-Step Solution: Recall I_base = b h^3 / 12.Recall I_centroidal = b h^3 / 36.Compute ratio ⇒ (b h^3 / 12) / (b h^3 / 36) = 36 / 12 = 3.0. Verification / Alternative check: Use the parallel-axis theorem: I_base = I_centroidal + A d^2 with d = h/3. Substituting A = b h / 2 gives I_base = b h^3 / 36 + (b h / 2) * (h/3)^2 = b h^3 / 36 + b h^3 / 18 = b h^3 / 12, confirming the ratio 3.0. Why Other Options Are Wrong: 1.0, 1.5, 2.0, 2.5 do not satisfy the known formulae for triangular sections. Common Pitfalls: Mixing up centroid location (h/3 from the base) leading to incorrect parallel-axis distances. Final Answer: 3.0.

Question 12

Eccentrically loaded rectangular column: If a load P has eccentricities e_x and e_y along the X- and Y-axes, what is the normal stress at a general point (x, y) on the cross-section?

Accepted Answer

Introduction / Context: Columns and pedestal footings often carry loads that are not applied through the centroid, creating additional bending about principal axes. Determining the combined stress distribution is essential to check tension zones and permissible compressive stress. Given Data / Assumptions: Rectangular cross-section with principal centroidal axes X and Y. Load P applied with eccentricities e_x (along X) and e_y (along Y) from the centroid. Second moments of area I_x (about X) and I_y (about Y) are known. Concept / Approach: An eccentric load creates resultant bending moments M_x and M_y in addition to direct compression P/A. If e_x is measured along the X-axis, it produces a moment about the Y-axis equal to M_y = P e_x; similarly, e_y produces M_x = P e_y about the X-axis. The stress at (x, y) superposes direct and bending components: σ = P/A + M_y * x / I_y + M_x * y / I_x, which becomes σ(x,y) = P/A + (P e_x / I_y) x + (P e_y / I_x) y. Step-by-Step Solution: Write direct stress: σ_d = P/A.Take moments: M_y = P e_x and M_x = P e_y.Bending stresses: σ_b,y = M_y x / I_y; σ_b,x = M_x y / I_x.Superpose: σ = σ_d + σ_b,y + σ_b,x = P/A + (P e_x / I_y) x + (P e_y / I_x) y. Verification / Alternative check: Check symmetry: if e_x = e_y = 0, the expression reduces to σ = P/A as expected. If only e_x ≠ 0, stress varies linearly with x, consistent with bending about the Y-axis. Why Other Options Are Wrong: Misplaced I_x and I_y or swapping x and y leads to incorrect stress distribution. Neglecting bending (P/A only) ignores eccentricity effects. Common Pitfalls: Sign convention: tension/compression signs depend on the chosen positive x–y directions; here a general algebraic form is provided. Final Answer: σ(x,y) = P/A + (P e_x / I_y) x + (P e_y / I_x) y.

Question 13

Axially loaded bar (gradual loading): A bar of length l and cross-sectional area A is subjected to a gradually applied tensile load W. What is the strain energy stored?

Accepted Answer

Introduction / Context: Strain energy is the elastic energy stored in a member due to deformation. For axial members, it is a fundamental quantity used in energy methods (Castigliano’s theorem) and in estimating resilience and impact effects. Given Data / Assumptions: Prismatic bar: length l, area A. Linearly elastic material with Young’s modulus E. Load W is applied gradually from 0 to W (not suddenly or with impact). Concept / Approach: Under axial loading, extension δ = W l / (A E). For a gradually applied load, the work done (and stored as strain energy) equals the area under the load–deflection curve, which is a triangle: U = (1/2) * W * δ. Substituting δ gives U = (1/2) * W * (W l / (A E)) = W^2 l / (2 A E). Step-by-Step Solution: Compute axial extension: δ = W l / (A E).Use gradual loading energy: U = (1/2) W δ.Substitute δ ⇒ U = (1/2) W * (W l / (A E)) = W^2 l / (2 A E). Verification / Alternative check: Compare with sudden loading (no impact): energy is U = W^2 l / (2 A E) for gradual loading, while the maximum deflection under sudden loading is twice the gradual case, changing energy relations accordingly. This confirms the role of the loading path. Why Other Options Are Wrong: W l / (A E) is the deflection, not energy. W^2 l / (A E) misses the 1/2 factor from the triangular load–deflection relationship. W^2 / (2 A E l) has incorrect dimensionality. (1/2) A E l is unrelated to the applied load W. Common Pitfalls: Confusing gradual and sudden loading; energy depends on the load application path. Final Answer: U = W^2 l / (2 A E).

Question 14

For typical beam loading cases, which of the following statements are correct about the shapes of shear force (SF) and bending moment (BM) diagrams?

Accepted Answer

Introduction / Context: Recognizing the qualitative shapes of shear force (SF) and bending moment (BM) diagrams is a core skill in structural analysis. It speeds up checking, helps catch modeling errors, and guides efficient design decisions. Given Data / Assumptions: Beam theory with small deflections and linear elasticity. Distributed loads expressed per unit length. Sign conventions are consistent along the beam. Concept / Approach: The governing relationships are: dV/dx = −w(x) and dM/dx = V(x), where w(x) is the load intensity. Integrating these relations establishes how the curve shapes evolve with the load pattern. Step-by-Step Solution: Uniformly distributed load w = constant: dV/dx = −w ⇒ V is linear; integrating again gives M as a quadratic function, i.e., a parabola.Linearly varying load w = kx: dV/dx = −kx ⇒ V is quadratic ⇒ M is cubic upon integration, i.e., a cubic parabola.Point loads produce jumps in V and linear segments in M, consistent with the derivative relations. Verification / Alternative check: Dimensional and boundary checks (zero moments at simple supports for simply supported beams) align with the predicted polynomial orders. Numerical examples reproduce these curve shapes exactly. Why Other Options Are Wrong: Each individual statement (a)–(d) is correct; hence the combined correct choice is 'All of the above'. Common Pitfalls: Mistaking the order of the BM curve, especially under triangular loading where M becomes cubic. Forgetting that jumps in V correspond to point loads and that M remains continuous at those points. Final Answer: All of the above.

Question 15

Compound bar under axial compression: Two bars of equal length, one steel (cross-section 3500 mm²) and one brass (cross-section 3000 mm²), are rigidly connected to act as a compound bar. The assembly is subjected to a total compressive load of 100,0…

Accepted Answer

Introduction / Context: This problem tests compound (composite) bars in axial loading. When different materials are rigidly connected and loaded axially, compatibility of deformation requires equal strain in each material. Load sharing therefore depends on both area and modulus of elasticity. Given Data / Assumptions: Steel area As = 3500 mm², brass area Ab = 3000 mm². Total load P = 100,000 N (compression). Elastic moduli: Es = 0.2 MN/mm², Eb = 0.1 MN/mm². Bars have equal length and are perfectly composite (same strain), linear elasticity holds. Concept / Approach: Equal strain implies σs/Es = σb/Eb. Hence σs = (Es/Eb)σb. Force equilibrium requires σs As + σb Ab = P. Solve these two relations for σs and σb. Step-by-Step Solution: Es/Eb = 0.2 / 0.1 = 2, so σs = 2 σb.Force balance: (2 σb)(3500) + (σb)(3000) = 100,000.Compute: 7000 σb + 3000 σb = 10,000 σb = 100,000 ⇒ σb = 10 N/mm².Thus σs = 2 σb = 20 N/mm². Verification / Alternative check: Check load share: Steel carries 203500 = 70,000 N; Brass carries 10*3000 = 30,000 N; Sum = 100,000 N ✔. Steel takes more load due to higher Es and larger area. Why Other Options Are Wrong: Other pairs do not satisfy both compatibility (ratio 2:1) and total load 100,000 N simultaneously. Common Pitfalls: Using area ratio only (ignoring modulus), or summing stresses instead of forces. Always apply both compatibility and equilibrium. Final Answer: σb = 10 N/mm², σs = 20 N/mm²

Question 16

Core (kernel) of a rectangular column: For a prismatic rectangular column of cross-sectional area A, what is the area of its core (also called kernel) within which the resultant compressive load must fall to avoid tension anywhere on the section?

Accepted Answer

Introduction / Context: The core or kernel of a section is the locus of points where a resultant compressive load may be applied so that resultant stress over the entire section remains compressive (no tension). For a rectangular column, this is a smaller concentric rectangle controlling eccentric load placement in design of foundations and columns. Given Data / Assumptions: Section is rectangular, prismatic, homogeneous, and elastic. Total section area is A = b * d. Only axial compression with possible eccentricity; no shear. Concept / Approach: For a rectangle of width b and depth d, the core dimensions are b/3 by d/3. Hence the core area equals (b/3) * (d/3) = (b d)/9 = A/9. This arises from the straight-line stress distribution condition ensuring that extreme fiber stress does not change sign when the load resultant lies within the central kernel. Step-by-Step Solution: Rectangular cross-section: kernel rectangle dimensions = b/3 and d/3.Core area = (b/3)*(d/3) = (b d)/9.Since A = b d, core area = A/9. Verification / Alternative check: For a square (b = d), kernel is an inner square with side b/3, again giving area A/9, confirming the general rectangular result. Why Other Options Are Wrong: A/4 and A/6 are too large; A/16 is the circular section core area (r/4 radius), not applicable to rectangles; A/12 is incorrect for rectangles. Common Pitfalls: Confusing rectangular and circular cores; mixing up kernel dimensions (b/3, d/3) with mid-third rule for retaining walls. Final Answer: A/9

Question 17

Thin-walled circular tube in torsion: For permissible shear stress f_s, a thin tube of mean diameter D and uniform wall thickness t transmits a torque T equal to which of the following?

Accepted Answer

Introduction / Context: Thin-walled closed sections (like tubes) resist torsion primarily through a uniform shear flow around the median line. For circular thin tubes, this gives a compact closed-form torque formula in terms of mean diameter D, wall thickness t, and allowable shear stress f_s. Given Data / Assumptions: Thin-walled circular tube; thickness t ≪ D. Uniform shear stress around the median line at value f_s. Saint-Venant torsion, elastic range. Concept / Approach: For thin closed sections, T = 2 A_m q, where A_m is the area enclosed by the median line and q is the shear flow. For a wall of thickness t, q = τ t = f_s t. For a circle, A_m = π (D/2)² = π D² / 4. Substitute to obtain the torque capacity. Step-by-Step Solution: Shear flow: q = f_s t.Median area for a circle: A_m = π D² / 4.Torque: T = 2 A_m q = 2 * (π D² / 4) * (f_s t) = (π D² t f_s) / 2. Verification / Alternative check: Units: D² t f_s has dimensions of (length³ * stress) = force * length, i.e., torque. Numerical checks against thin-tube torsion formula confirm the 1/2 factor. Why Other Options Are Wrong: Option (d) misses the 1/2 factor (overestimates T). Options (a), (b), (e) are not the correct thin-tube torsion expressions for a circular section. Common Pitfalls: Using solid-shaft formula T/J = τ/R for thin tubes; mixing polar moment J for solid sections with thin-wall torsion theory. Final Answer: T = (π D² t f_s) / 2

Question 18

Three-hinged parabolic arch (span l, rise h) under a uniformly distributed load w per unit span over the entire horizontal span: identify the correct statements about internal actions.

Accepted Answer

Introduction / Context: For a three-hinged parabolic arch carrying a uniform load over the span, the arch shape can be chosen as the funicular (thrust) line for that loading. When the geometry matches the funicular for the given loading, bending moments vanish everywhere along the arch rib, leaving only axial thrust and shear at sections. Given Data / Assumptions: Three-hinged parabolic arch with hinges at the two springings and at the crown. Uniformly distributed load w per unit horizontal span over the full span l. Rise at crown = h; small-deflection, elastic line analysis with standard arch statics. Concept / Approach: At any section x from the left springing, beam moment due to UDL on the simply supported equivalent is M_beam(x) = w x (l − x) / 2. Arch bending moment is M_arch(x) = M_beam(x) − H y(x). For a parabolic arch, ordinate y(x) = 4 h x (l − x) / l². Choosing H so that M_arch(x) = 0 for all x gives the required horizontal thrust and proves zero bending moment everywhere. Step-by-Step Solution: Set M_arch(x) = w x (l − x)/2 − H * [4 h x (l − x)/l²] = 0.Cancel common factor x(l − x) ≠ 0 in the interior to get w/2 − H * (4 h / l²) = 0.Solve for H: H = (w l²) / (8 h).With this H, M_arch(x) = 0 for all sections; however, shear force is not identically zero. Verification / Alternative check: Arch handbooks list H = w l² / (8 h) for a three-hinged parabola under UDL, with zero bending moment throughout because the rib follows the funicular polygon for that load. Why Other Options Are Wrong: Shear force is generally non-zero (option b false). Option e claims all statements are correct and is therefore wrong. Common Pitfalls: Assuming “three-hinged” alone implies zero moment; it is the match of parabolic shape to UDL that causes M = 0 everywhere. Final Answer: Both (a) and (c) are correct

Question 19

Thermal stress in a restrained steel tie: A 5 m long steel rod of cross-sectional area 250 mm² is fixed rigidly between two parallel walls. The end nuts were tightened at 100 °C. Given α_steel = 0.000012 per °C and E_steel = 0.2 MN/mm², find the ten…

Accepted Answer

Introduction / Context: When a member is fully restrained against free thermal expansion or contraction, temperature change induces internal stress. For linear elastic behavior, the thermal stress equals E * α * ΔT in magnitude, tension when cooling and compression when heating (in a restrained bar). Given Data / Assumptions: Length L = 5 m (length cancels in stress calculation for full restraint). Area A = 250 mm² (not required for stress magnitude). Coefficient of thermal expansion α = 0.000012 / °C. Elastic modulus E = 0.2 MN/mm² = 200,000 N/mm². Temperature drop ΔT = 50 − 100 = −50 °C (cooling). Ends fully restrained, no slip; linear elasticity. Concept / Approach: For a fully restrained bar, thermal strain α ΔT is prevented, producing mechanical strain ε = −α ΔT. Stress σ = E ε = −E α ΔT. Sign indicates tension for cooling (since free contraction is prevented). Step-by-Step Solution: Compute α |ΔT| = 0.000012 * 50 = 0.0006.σ = E * α * |ΔT| = 200,000 * 0.0006 = 120 N/mm².Nature: Cooling with restraint ⇒ tensile stress of 120 N/mm². Verification / Alternative check: If needed, the corresponding force would be F = σ A = 120 * 250 = 30,000 N tension, consistent with restraint forces. Why Other Options Are Wrong: 80, 100, 150 N/mm² correspond to incorrect ΔT or arithmetic; 60 N/mm² would follow if α or ΔT were halved. Common Pitfalls: Using negative sign to select a “compression” answer; here, cooling in a fixed tie produces tension. Also, mixing MPa and N/mm² units. Final Answer: 120 N/mm²

Question 20

Euler’s buckling (crippling) load forms: Identify the correct Euler critical load expression P_cr for a prismatic column (modulus E, second moment I, actual length L) under different end conditions, using the effective length concept.

Accepted Answer

Introduction / Context: Euler’s column theory gives the elastic buckling (critical) load for slender, perfectly straight columns with ideal end conditions. The formula uses an effective length L_e that depends on end restraints. Greater end fixity reduces L_e and increases the buckling load. Given Data / Assumptions: P_cr = π² E I / L_e². Idealized end conditions; column remains elastic up to buckling. No initial imperfections or eccentricity (theoretical model). Concept / Approach: Effective length L_e embodies end restraint: hinged-hinged L_e = L; fixed-fixed L_e = L/2; fixed-free (cantilever) L_e = 2 L; fixed-hinged L_e = L/√2. Substituting each L_e into P_cr = π² E I / L_e² yields the respective expressions. Step-by-Step Solution: Both ends fixed: L_e = L/2 ⇒ P_cr = π² E I / (L/2)².Both ends hinged: L_e = L ⇒ P_cr = π² E I / L².One end fixed, other free: L_e = 2 L ⇒ P_cr = π² E I / (2 L)².One end fixed, other hinged: L_e = L/√2 ⇒ P_cr = π² E I / (L/√2)². Verification / Alternative check: Ordering by restraint: fixed-fixed has 4× the hinged-hinged capacity; fixed-hinged has 2×; fixed-free has 1/4×, consistent with classical tables. Why Other Options Are Wrong: Each individual option is correct for its stated end condition; only the combined choice confirms all correctly. Common Pitfalls: Mixing L with L_e; forgetting that the cantilever (fixed-free) is the weakest case (largest effective length). Final Answer: All of the above (each matches its end condition)

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