Question 1

Choose the correct definition of engineering stress: stress is defined as force divided by the loaded area (standard SI units N/m^2 or Pa).

Accepted Answer

Introduction / Context: Stress is a primary concept in strength of materials. Correctly identifying its definition is fundamental before applying any failure criteria or deformation formulas. Given Data / Assumptions: Axial loading concept; stress symbol σ (normal) or τ (shear). Engineering (nominal) stress uses original cross-sectional area unless otherwise specified. Concept / Approach: Normal stress σ = P / A (axial force divided by area normal to the load). Shear stress τ = V / A_s (shear force divided by the relevant shear area). Units: N/m^2 (Pa) or N/mm^2 (MPa). Length or volume denominators do not represent stress. Step-by-Step Solution: 1) Identify the loaded area resisting the applied force.2) Compute stress as σ = Force / Area.3) Recognize that other denominators (length, volume) correspond to different physical quantities, not stress. Verification / Alternative check: Dimensional analysis: [Force]/[Area] = N/m^2; neither N/m nor N/m^3 aligns with stress dimensions. Why Other Options Are Wrong: Force per unit length: Relates to line load intensity, not stress. Force per unit volume: Relates to body force density (e.g., weight density), not stress. None of these: Incorrect, since force per unit area is correct. Common Pitfalls: Mixing engineering stress with true stress (which uses instantaneous area). Using average area where net area is required (e.g., bolted plates with holes). Final Answer: force per unit area.

Question 2

Combined bending and torsion: identify the correct statements about equivalent bending moment and equivalent torque for a circular shaft.

Accepted Answer

Introduction / Context: Design of circular shafts under simultaneous bending moment M and torque T often uses equivalent single-effect measures to compare against allowable stresses. The two common equivalents are equivalent bending moment (matching maximum principal stress) and equivalent torque (matching maximum shear stress). Given Data / Assumptions: Solid circular shaft; outer-fiber stresses govern. Elastic behavior with σ = 32M/(π d^3) and τ = 16T/(π d^3). Maximum principal stress and maximum shear stress from Mohr’s circle are used. Concept / Approach: For combined bending and torsion, at the outer surface: principal stress σ_1 = (σ/2) + sqrt( (σ/2)^2 + τ^2 ), and maximum shear τ_max = sqrt( (σ/2)^2 + τ^2 ). Define M_eq so that σ_1 = 32 M_eq/(π d^3), and T_eq so that τ_max = 16 T_eq/(π d^3). This yields closed-form expressions for the equivalents in terms of M and T. Step-by-Step Solution: 1) Put σ = k M and τ = (k/2) T with k = 32/(π d^3).2) σ_1 = (kM)/2 + sqrt( (kM/2)^2 + (kT/2)^2 ) = (k/2)[ M + sqrt( M^2 + T^2 ) ].3) Set σ_1 = k M_eq → M_eq = 0.5 * [ M + sqrt( M^2 + T^2 ) ].4) τ_max = sqrt( (kM/2)^2 + (kT/2)^2 ) = (k/2) sqrt( M^2 + T^2 ).5) Set τ_max = (k/2) T_eq → T_eq = sqrt( M^2 + T^2 ). Verification / Alternative check: Special cases: (i) T = 0 → M_eq = M, T_eq = |M| (incorrect dimension), but note T_eq compares shear; with T = 0, τ_max = |σ|/2 so T_eq = |M| (in the normalized formulation above) matches the definition via τ_max; (ii) M = 0 → M_eq = 0.5*|T|, T_eq = |T|, consistent with pure torsion. Why Other Options Are Wrong: Here, A and C are definitions; B presents the correct closed forms. Choosing only one omits key information—hence the inclusive statement is appropriate. Common Pitfalls: Using T_eq = sqrt( M^2 + (K T)^2 ) with an arbitrary factor K; for circular sections under the maximum shear stress criterion, K = 1. Forgetting the factor 0.5 in M_eq. Final Answer: All the above.

Question 3

For standard rolled I-beam (rolled steel beam) sections used in practice, the plastic shape factor (Mp/My) generally lies within which range?

Accepted Answer

Introduction / Context: The plastic shape factor S = Mp/My measures the reserve bending capacity beyond first yield. For rolled I-beams (compact proportions without local buckling), S is only modestly above 1.0 because flanges carry much of the plasticity while the web contributes less than in a rectangular section. Given Data / Assumptions: Standard rolled I-beam shapes with usual proportions. Elastic–perfectly plastic idealization; no premature local buckling. Concept / Approach: Typical tabulated values show S ≈ 1.10–1.20 for common I-sections, compared to S = 1.50 for rectangles. The exact value depends on flange and web dimensions, but remains within this narrow band for standard sections. Step-by-Step Solution: 1) Define S = Z_p / Z (plastic to elastic section modulus).2) Recognize I-beams have flanges far from the neutral axis (efficient elastically), but plastic neutral axis movement does not raise capacity as much as in rectangles.3) Thus, S clusters around 1.1–1.2 in common tables. Verification / Alternative check: Design manuals (steel codes) consistently list S in this range for rolled I-sections; values beyond 1.3 are atypical for standard rolled beams. Why Other Options Are Wrong: 1.20–1.30, 1.30–1.40, 1.40–1.50: Too high for standard rolled I-beams; such factors correspond to other shapes or special proportions. Common Pitfalls: Misapplying the rectangular section value (1.5) to I-beams. Ignoring slenderness/local buckling, which can reduce plastic capacity. Final Answer: 1.10 to 1.20.

Question 4

Under a uniaxial direct stress, the tangential (shear) stress on a plane inclined at angle θ to the direction of the load equals the normal stress on that plane multiplied by which factor?

Accepted Answer

Introduction / Context: Stress transformation on inclined planes under uniaxial loading is a staple of Mohr’s circle and elementary mechanics of materials. The relationship between the normal and shear components on an inclined plane is frequently tested in exams. Given Data / Assumptions: Uniaxial direct stress σ acts along one direction. Consider a plane inclined at angle θ to the load direction. Concept / Approach: For uniaxial stress, the normal stress on the inclined plane is σ_n = σ cos^2 θ, and the shear (tangential) stress is τ = σ sin θ cos θ. Eliminating σ gives a relation between τ and σ_n: τ = σ_n * tan θ. Step-by-Step Solution: 1) σ_n = σ cos^2 θ.2) τ = σ sin θ cos θ.3) Divide: τ / σ_n = (σ sin θ cos θ) / (σ cos^2 θ) = tan θ.4) Therefore, τ = σ_n * tan θ. Verification / Alternative check: Mohr’s circle also yields τ = (σ/2) sin 2θ and σ_n = (σ/2)(1 + cos 2θ). Their ratio simplifies to tan θ, consistent with the above derivation. Why Other Options Are Wrong: sin θ, cos θ: Incomplete; they relate to σ components relative to σ, not to σ_n. sin θ cos θ and sin 2θ: These multiply σ directly, not the normal stress on the plane. Common Pitfalls: Confusing the relation between τ and σ (gives sin θ cos θ) with the relation between τ and σ_n (gives tan θ). Final Answer: tan θ.

Question 5

For a rectangular cross-section under a transverse shear force F, determine the maximum shear stress at the neutral axis in terms of F and the cross-sectional area A.

Accepted Answer

Introduction / Context: The shear stress distribution in a rectangular section is parabolic, with the maximum at the neutral axis. Knowing the factor relating τ_max to the average shear stress F/A is important for beam design under shear. Given Data / Assumptions: Cross-section: rectangle of breadth b and depth d (area A = b*d). Shear force: F at the section. Small-deflection, linear elasticity. Concept / Approach: The general shear formula is τ = V*Q / (I*b), where V = F, Q is the first moment of area about the neutral axis for the portion above (or below) the level considered, I is the second moment of area, and b is the width at that level. At the neutral axis of a rectangle, Q = (b*(d/2)*(d/4)) = b*d^2/8, I = b*d^3/12, and b is the full breadth. Step-by-Step Solution: 1) For a rectangle: I = b*d^3/12, A = b*d.2) At the neutral axis: Q = b*(d/2)*(d/4) = b*d^2/8.3) τ_max = V*Q / (I*b) = F*(b*d^2/8) / ( (b*d^3/12)*b )? Use the correct formula τ = F*Q/(I*b) with b as breadth at NA = b.4) Simplify: τ_max = F * (b*d^2/8) / ( (b*d^3/12) * 1 ) = F * (12 / 8) * (1 / d*b) = (3/2) * F / (b*d) = 1.5 * (F/A). Verification / Alternative check: Average shear = F/A; the parabolic distribution in a rectangle peaks at 1.5*(F/A) at the center and is zero at the top/bottom fibers. Why Other Options Are Wrong: F/A: This is the average, not the maximum. 2*(F/A): Overestimates by 33%. 0.5*(F/A): Underestimates; corresponds to no standard location. Common Pitfalls: Using τ = F/A as a design check in rectangular beams; must use τ_max = 1.5*(F/A). Forgetting that τ varies parabolically, not uniformly. Final Answer: 1.5 * (F / A).

Question 6

For a planar truss with j joints and m members, identify the condition for a simple (statically determinate and stable) truss.

Accepted Answer

Introduction / Context: Truss classification (simple, compound, complex) and determinacy are foundational in structural analysis. A simple plane truss can be built by starting with a basic triangle (j = 3, m = 3) and adding two members and one joint at a time while maintaining determinacy and stability. Given Data / Assumptions: Planar truss; pin joints; straight members carrying only axial forces. External reactions are determinate (typically three for a planar pin-supported truss). Concept / Approach: The necessary (though not always sufficient) count condition for a simple, statically determinate plane truss is m = 2j − 3. This arises from equilibrium equations: each joint contributes two scalar equations; three overall are consumed by support reactions, leaving 2j − 3 independent member force unknowns for determinacy. Step-by-Step Solution: 1) Begin with a triangle: j = 3, m = 3 → m = 2j − 3 holds (3 = 6 − 3).2) Add a new joint with two members to existing joints → j → j+1, m → m+2 → relation preserved.3) By induction, the relation holds for any simple truss formed this way. Verification / Alternative check: Compare with method-of-joints determinacy: unknown member forces = m; available independent equations = 2j − r (r = 3 external reaction components). For a internally determinate truss with determinate supports, m = 2j − 3. Why Other Options Are Wrong: j = 2m − 3, m = 3j − 2, j = 3m − 2: Do not match the equilibrium/determinacy logic for plane trusses. Common Pitfalls: Assuming m = 2j − 3 guarantees stability in every geometry; special mechanisms can exist if members align unfavorably. Final Answer: m = 2j - 3.

Question 7

A vertical load of 1960 N is lifted at the free end of a straight steel wire. What is the minimum wire diameter (in mm) required so that the tensile stress does not exceed 100 N/mm^2?

Accepted Answer

Introduction / Context: When a straight steel wire is subjected to a tensile load, the induced normal stress equals the load divided by the cross-sectional area. Designing the minimum diameter against an allowable stress is a fundamental sizing task in strength of materials and structural design. Given Data / Assumptions: Tensile load, P = 1960 N. Allowable tensile stress, sigma_allow = 100 N/mm^2. Wire is prismatic, circular, and behaves elastically under the given load. Self-weight and stress concentrations are neglected; static loading only. Concept / Approach: Tensile stress sigma = P / A must not exceed sigma_allow. For a circular section, A = (pi * d^2) / 4. Solve for the minimum diameter d such that P / A <= 100 N/mm^2. Rearranging gives d = sqrt(4 * P / (pi * sigma_allow)). Choose the next available size that is equal to or greater than the computed minimum to ensure stress does not exceed the limit. Step-by-Step Solution: Step 1: Compute required area A_req = P / sigma_allow = 1960 / 100 = 19.6 mm^2.Step 2: Relate area to diameter: A = (pi * d^2) / 4 ⇒ d^2 = 4 * A / pi.Step 3: Substitute A_req: d = sqrt(4 * 19.6 / pi).Step 4: Numerically: 4 * 19.6 = 78.4; 78.4 / pi ≈ 24.96; sqrt(24.96) ≈ 4.996 mm.Step 5: Select a listed diameter ≥ 4.996 mm → 5.0 mm. Verification / Alternative check: Check stress with d = 5.0 mm: A = (pi * 25) / 4 ≈ 19.635 mm^2; sigma = 1960 / 19.635 ≈ 99.8 N/mm^2, which is within the 100 N/mm^2 limit. Why Other Options Are Wrong: 4.0 mm, 4.5 mm: Areas too small, producing sigma > 100 N/mm^2. 5.5 mm, 6.0 mm: Safe but not the minimum required by the stress criterion. Common Pitfalls: Using area in m^2 instead of mm^2; forgetting to check the next higher standard size; confusing tensile stress limits with yield or ultimate strengths without safety factors. Final Answer: 5.0 mm

Question 8

For a rigid body in static equilibrium under a system of forces, which complete set of equilibrium conditions must be satisfied?

Accepted Answer

Introduction / Context: Statics establishes the conditions under which a rigid body remains at rest. For planar force systems, both translational and rotational equilibria must hold. This question checks recognition of the full set of equilibrium equations needed in two dimensions. Given Data / Assumptions: Planar (2D) rigid body system. Resultant acceleration is zero. Moments are considered about any arbitrary point in the plane. Concept / Approach: Equilibrium requires zero net force and zero net moment. In 2D, the vector equations expand into ΣH = 0 for horizontal components, ΣV = 0 for vertical components, and ΣM = 0 for the sum of moments about any point. These three scalar equations are necessary and sufficient for planar rigid-body equilibrium. Step-by-Step Solution: Write force equilibrium: ΣF = 0 ⇒ ΣH = 0 and ΣV = 0.Write moment equilibrium: ΣM = 0 about any point (choice does not change truth).Recognize that all three must simultaneously hold for static equilibrium. Verification / Alternative check: Choosing different moment centers yields consistent results when ΣH and ΣV are also zero; otherwise contradictions arise. Problems with three unknown reactions often rely on these three equations. Why Other Options Are Wrong: Any single or pair (ΣH = 0 or ΣV = 0 or ΣM = 0 alone) is insufficient to ensure full equilibrium. none of these: Incorrect because all listed conditions are required. Common Pitfalls: Forgetting to enforce moment equilibrium; taking moments about a point that eliminates all unknowns of interest; sign mistakes in component resolution. Final Answer: all the above.

Question 9

In plastic analysis of beams, identify the correct statements regarding yield moment, plastic moment, plastic neutral axis, and shape factor.

Accepted Answer

Introduction / Context: Plastic analysis extends beyond elastic behavior by allowing stress redistribution after yielding. The key milestones are the yield moment (first yielding) and the plastic moment (full plasticity). The shape factor quantifies reserve strength from elastic to plastic states and depends on cross-section shape. Given Data / Assumptions: Material follows an ideal elastic–perfectly plastic model for bending. Plane sections remain plane (Bernoulli hypothesis). Longitudinal shear permits redistribution as plastic hinges form. Concept / Approach: At yield moment, only extreme fibers reach yield stress; stresses elsewhere remain below yield. At plastic moment, compressive and tensile blocks each attain yield stress across their half-areas, and the plastic neutral axis (PNA) lies so the two blocks have equal area. Shape factor = M_p / M_y captures how much higher the full plastic capacity is compared with first yield and is purely geometric (independent of the yield stress value). Step-by-Step Solution: Identify yield moment: first reaching of sigma_y at extreme fibers.Identify plastic moment: full-depth yielding with rectangular stress blocks at ±sigma_y.Recognize PNA property: equal tension and compression areas at the plastic state.Define shape factor: M_p / M_y.Since all statements align with theory, select the comprehensive option. Verification / Alternative check: For common shapes: rectangle f = 1.5, circle f ≈ 1.7, triangle f ≈ 2.34, confirming geometric dependence of shape factor. Why Other Options Are Wrong: Choosing any single statement omits the full set of correct fundamentals. Common Pitfalls: Confusing elastic neutral axis with plastic neutral axis; assuming shape factor depends on material rather than geometry. Final Answer: All the above.

Question 10

For a circular beam in transverse shear, what is the ratio of maximum shear stress to average shear stress (τ_max / τ_avg)?

Accepted Answer

Introduction / Context: Transverse shear stress distribution in beams depends on the cross-sectional shape. Designers often need the ratio between the maximum shear stress and the average shear stress to convert between nominal and peak values. This ratio is section-dependent and is used in checking shear capacity and service behavior. Given Data / Assumptions: Prismatic beam with a solid circular cross-section of diameter D. Subjected to a transverse shear force V. Linear elastic behavior; Saint-Venant assumptions apply. Concept / Approach: Average shear stress τ_avg = V / A = V / (pi * D^2 / 4). The exact shear distribution for a solid circle yields a parabolic variation with the maximum value at the neutral axis equal to τ_max = (4/3) * τ_avg. This contrasts with a solid rectangle where τ_max = (3/2) * τ_avg. Step-by-Step Solution: Step 1: Compute τ_avg = V / A.Step 2: Use the known closed-form result for a circular section: τ_max = 4/3 * τ_avg.Step 3: Hence, the ratio τ_max / τ_avg = 4/3. Verification / Alternative check: Deriving from τ = V * Q / (I * b(y)) with the circular-section expression for first moment Q confirms the 4/3 multiplier at the neutral axis. Why Other Options Are Wrong: 3/2: Applies to a solid rectangular section, not a circle. 5/4, 2, 1: Do not match the exact solution for a circular section. Common Pitfalls: Memorizing the rectangular ratio (3/2) and applying it to all shapes; using τ_avg directly to check peak stresses. Final Answer: 4/3

Question 11

In plastic analysis, what is the shape factor (M_p / M_y) for a solid circular cross-section?

Accepted Answer

Introduction / Context: The shape factor f = M_p / M_y indicates the reserve plastic strength beyond first yield and depends only on the cross-section's geometry. For a solid circle, f is higher than for a rectangle because more material lies away from the neutral axis and can contribute under plastic redistribution. Given Data / Assumptions: Elastic–perfectly plastic bending model. Solid circular cross-section. Plane sections remain plane; no shear yield limitation. Concept / Approach: Yield moment M_y is based on the elastic section modulus; plastic moment M_p is computed from equal-area compression and tension blocks at yield stress. For a circle, the resulting ratio is approximately 1.70 (commonly rounded to 1.7) and is a standard handbook value used in plastic design. Step-by-Step Solution: Step 1: Recognize f = M_p / M_y is geometric.Step 2: Use the known result for a solid circle: f ≈ 1.698 → 1.7.Step 3: Select 1.7 from the options. Verification / Alternative check: Reference tables list f_rect = 1.5, f_circle ≈ 1.7, f_triangle ≈ 2.34, validating the chosen value. Why Other Options Are Wrong: 1.5: For a rectangle. 1.6, 1.75: Near but not equal to the standard circular value. Common Pitfalls: Assuming the same shape factor across different section shapes; mixing up rectangle and circle values. Final Answer: 1.7

Question 12

A force of magnitude F is inclined at an angle θ to a given reference line. The normal (perpendicular) component of the force relative to that line equals:

Accepted Answer

Introduction / Context: Resolving a force into components along and normal to a reference direction is a basic operation in statics and dynamics. Correct trigonometric association with the angle is essential for accurate equilibrium and motion analyses. Given Data / Assumptions: A single force F making angle θ with the normal direction (or equivalently θ with the force's line to the reference axis). Right-triangle decomposition applies. Angle θ is measured from the normal to the force direction or vice versa as stated. Concept / Approach: Let the chosen axis be the normal direction. The projection of F onto that axis is F * cos θ when θ is the angle between F and the normal. If θ is defined from the tangential axis, then the normal component is F * sin θ; however, by the statement here (force inclined through θ° to the reference line with normal component needed), the standard interpretation yields the cosine factor for the component perpendicular to the line. Step-by-Step Solution: Construct a right triangle with hypotenuse F and included angle θ to the normal.Normal component = adjacent side = F * cos θ.Tangential component = opposite side = F * sin θ. Verification / Alternative check: At θ = 0°, the entire force is normal: F * cos 0° = F, consistent. At θ = 90°, normal component is zero: F * cos 90° = 0, also consistent. Why Other Options Are Wrong: F * sin θ: This is the tangential component (if θ is to the normal). F * tan θ, F * sin θ * cos θ, F * sin^2 θ: Not simple projections; dimensionally or conceptually incorrect as a component magnitude. Common Pitfalls: Mixing angle definitions; forgetting which side of the right triangle corresponds to the component sought. Final Answer: F * cos θ

Question 13

For estimating the allowable compressive stress of long (slender) columns, the commonly used empirical relation is known as:

Accepted Answer

Introduction / Context: Column design relies on empirical or semi-empirical formulas that account for instability (buckling) and material strength. Different formulas are favored for different slenderness regimes. The question asks for the named empirical relation typically associated with long columns (high slenderness ratio). Given Data / Assumptions: Long columns: high slenderness, buckling-dominated behavior. Allowable stress derived by reducing Euler buckling stress with safety margins. Empirical naming conventions as used in traditional design texts. Concept / Approach: For long columns, the allowable stress is often approximated by a straight-line relationship versus slenderness ratio, termed the Straight line formula. By contrast, Johnson's parabolic (Parabolic) addresses intermediate columns, Rankine's formula blends crushing and buckling for all ranges, and Perry's formula is another widely used intermediate-range empirical relation. Step-by-Step Solution: Identify regime: long (slender) columns → buckling governs.Associate empirical naming: Straight line formula for long columns.Recognize other formulas target other regimes or provide unified blends. Verification / Alternative check: Classical design charts show linear decay of allowable stress with 1/(slenderness)^2 for long columns, consistent with the straight-line approximation on the Rankine–Gordon/Euler asymptote. Why Other Options Are Wrong: Parabolic formula: Suits intermediate columns (Johnson). Perry's formula: An intermediate/unified approach, not specifically long-only. Rankine's formula: General formula across ranges; not the specific long-column empirical straight-line name. Common Pitfalls: Assuming one formula fits all slenderness without considering regime; confusing the names of Johnson (parabolic) and the straight-line approximation. Final Answer: Straight line formula

Question 14

For a close-coiled helical spring (stiffness calculation), which expression correctly gives the axial stiffness k = W/δ in terms of material and geometric parameters?

Accepted Answer

Introduction / Context: Close-coiled helical springs under axial load primarily deform by torsion of the wire. The spring stiffness (load per unit deflection) depends on the shear modulus and the spring’s wire/coil geometry. This relationship is essential for designing springs to meet load–deflection requirements. Given Data / Assumptions: Close-coiled spring; helix angle is small. Axial load W produces wire torsion dominated deformation. Material characterized by shear modulus G (not Young’s modulus E). Symbols: d = wire diameter, D = mean coil diameter, n = number of active coils. Concept / Approach: The classic formula for axial deflection δ of a close-coiled spring is δ = (8 * W * D^3 * n) / (G * d^4). Inverting gives stiffness k = W / δ = (G * d^4) / (8 * n * D^3). Note the strong dependence on d (to the 4th power) and the inverse cubic dependence on D, reflecting torsional compliance of the wire around the coil diameter. Step-by-Step Solution: Start from δ = (8 * W * D^3 * n) / (G * d^4).Rearrange: k = W / δ = (G * d^4) / (8 * n * D^3).Confirm units: G has units of stress; geometric ratios render k in force/length. Verification / Alternative check: Compare with design handbooks; expression matches standard spring design equations for close-coiled springs. Why Other Options Are Wrong: Using E instead of G: Spring twist depends on shear modulus G. Other exponents or constants (D^4, 16, D^2) do not match the derived relation. Common Pitfalls: Substituting E for G; forgetting that only active coils count in n; ignoring effects of end conditions on effective coil count. Final Answer: k = (G * d^4) / (8 * n * D^3)

Question 15

For a cantilever of length l and constant flexural rigidity E*I carrying a uniformly distributed load w per unit length over the entire span, what is the maximum deflection?

Accepted Answer

Introduction / Context: Deflection formulas for standard load cases enable rapid checks of serviceability and stiffness. For a cantilever with full-span UDL, the free-end deflection is the maximum and serves as a benchmark for preliminary sizing and compliance with deflection limits. Given Data / Assumptions: Cantilever length l, uniform load intensity w along 0 to l. Constant flexural rigidity E*I. Small deflections; linear elastic behavior. Concept / Approach: The classic closed-form result for a cantilever under full UDL is maximum deflection at the free end equal to δ_max = w * l^4 / (8 * E * I). This follows from integrating the moment-curvature relationship M / (E * I) = d^2y/dx^2 with boundary conditions y(0) = 0 and dy/dx(0) = 0 at the fixed end. Step-by-Step Solution: Moment at a section x from the fixed end: M(x) = w * (l - x)^2 / 2 (taking sign convention).Integrate twice to obtain slope and deflection, applying fixed-end boundary conditions.Evaluate y at x = l to get y(l) = w * l^4 / (8 * E * I). Verification / Alternative check: Compare with standard tables: simply supported with UDL gives 5 * w * l^4 / (384 * E * I), confirming that the cantilever case is much larger and specifically equals w * l^4 / (8 * E * I). Why Other Options Are Wrong: 1/384 factors relate to simply supported beams, not cantilevers. w * l^3 / (3 * E * I) is a slope-like dimension or end-load deflection form, not correct for full UDL. 1/48 corresponds to different loading/support conditions. Common Pitfalls: Mixing up simply supported vs. cantilever results; using end-point load formulas for UDL cases; missing the l^4 dependence. Final Answer: w * l^4 / (8 * E * I)

Question 16

In plastic analysis, what is the shape factor (M_p / M_y) for a triangular cross-section (about its centroidal axis in bending)?

Accepted Answer

Introduction / Context: The shape factor indicates how much plastic moment capacity exceeds the elastic yield moment for a given section shape. Triangular sections have a larger shape factor than rectangles and circles because a greater proportion of the area lies away from the neutral axis, enabling more plastic redistribution after first yield. Given Data / Assumptions: Elastic–perfectly plastic material in bending. Triangular cross-section, bending about a centroidal axis parallel to the base. Plane sections remain plane; full plasticity possible. Concept / Approach: M_y depends on elastic section modulus, while M_p is obtained by balancing equal tension and compression areas at yield stress. For a triangular section, the standard result is f = M_p / M_y ≈ 2.34, significantly higher than rectangular (1.5) or circular (≈1.7) sections. Step-by-Step Solution: Step 1: Recall definition of shape factor f = M_p / M_y.Step 2: Use known tabulated values for common shapes.Step 3: For triangle, f ≈ 2.34; select 2.34. Verification / Alternative check: Plastic design references consistently list triangular section shape factor close to 2.34, validating the choice. Why Other Options Are Wrong: 1.5: Rectangular section value. 1.34: Not a standard shape factor for a triangle. 2.5: Higher than accepted value for a triangle. Common Pitfalls: Misremembering the ordering (triangle > circle > rectangle); using material-dependent values rather than geometric constants. Final Answer: 2.34

Question 17

A steel rod that is 1 metre long with a square cross-section is subjected to a direct tensile load of 8 tonnes and undergoes an extension of 1 mm. Given Esteel = 2 × 10^6 kg/cm^2, determine the side length of the square cross-section (express the fi…

Accepted Answer

Introduction / Context: This problem applies axial deformation relations to a prismatic steel bar under a tensile load. It tests your ability to combine stress–strain definitions with unit consistency to compute an unknown cross-sectional dimension for a given elongation. The bar is square, so once the required area is found, the side length follows directly from area = side^2. Given Data / Assumptions: Length L = 1 metre = 100 cm. Axial load P = 8 tonnes = 8000 kgf. Extension δ = 1 mm = 0.1 cm. Young's modulus for steel, E = 2 × 10^6 kg/cm^2. Square cross-section with side a (cm), area A = a^2. Linear elasticity, small strain, uniform stress state. Concept / Approach: Axial stress σ = P / A. Axial strain ε = δ / L. Hooke's law for linear elastic behavior gives σ = E * ε. Combining these relations eliminates σ and ε and allows solving for A. Finally, take the square root to obtain the side a of the square cross-section. Step-by-Step Solution: 1) Compute strain: ε = δ / L = 0.1 / 100 = 0.001.2) Compute stress from Hooke's law: σ = E * ε = (2 × 10^6) * 0.001 = 2,000 kg/cm^2.3) Relate stress and area: σ = P / A ⇒ A = P / σ = 8000 / 2000 = 4 cm^2.4) For a square section, a = sqrt(A) = sqrt(4) = 2 cm. Verification / Alternative check: Dimensional check: P in kgf and E in kg/cm^2 are consistent with area in cm^2. Recomputing σ = 8000 / 4 = 2000 kg/cm^2 and ε = σ / E = 2000 / (2 × 10^6) = 0.001 gives δ = εL = 0.001 × 100 cm = 0.1 cm = 1 mm, verifying the result. Why Other Options Are Wrong: 1 cm, 1.5 cm: These give areas (1 or 2.25 cm^2) too small, producing stresses above the calculated 2000 kg/cm^2 and extensions larger than 1 mm. 2.5 cm: Area 6.25 cm^2 reduces stress too much and would predict an extension less than 1 mm. Common Pitfalls: Mistaking millimetres for centimetres when computing strain, which changes the answer by a factor of 10. Forgetting that the area for a square is a^2 (not 4a, etc.). Final Answer: 2 cm.

Question 18

For a thin-walled cylindrical tank subjected to internal liquid pressure, identify the correct statements about hoop strain, longitudinal strain, and volumetric change of the cylinder.

Accepted Answer

Introduction / Context: When a thin-walled cylindrical vessel is subjected to internal liquid pressure, it develops circumferential (hoop) and longitudinal stresses. These lead to corresponding strains and an overall change in volume. This question checks conceptual understanding of the deformation components under thin-cylinder theory. Given Data / Assumptions: Thin-walled cylinder, wall thickness t is small compared to diameter. Internal pressure p produces hoop stress σ_h and longitudinal stress σ_l. Material is linearly elastic with Young's modulus E and Poisson's ratio ν. Concept / Approach: Under thin-cylinder theory, σ_h = pD / (2t) and σ_l = pD / (4t). Strains arise as ε_h = (σ_h / E) − ν (σ_l / E) and ε_l = (σ_l / E) − ν (σ_h / E). Both are non-zero, causing increases in circumference and length. Consequently, the internal volume changes due to these dimensional increases. Step-by-Step Solution: 1) Recognize two principal stresses: hoop and longitudinal.2) Compute corresponding strains using linear elasticity and Poisson effect.3) Non-zero ε_h implies circumferential expansion (hoop strain).4) Non-zero ε_l implies axial elongation (longitudinal strain).5) Combined dimensional changes imply volumetric change of the cylinder. Verification / Alternative check: For p > 0, both σ_h and σ_l are positive, so both ε_h and ε_l are positive for typical ν (0 < ν < 0.5), confirming expansion in both directions and a corresponding volume increase. Why Other Options Are Wrong: Each of A, B, and C captures a real effect; excluding any one would give an incomplete picture. Common Pitfalls: Assuming only hoop effects are present and neglecting longitudinal strain. Ignoring Poisson's coupling between orthogonal directions. Final Answer: All the above.

Question 19

In stress analysis, principal planes are defined as those material planes on which which type(s) of stress act at a point under a general state of stress?

Accepted Answer

Introduction / Context: Principal stresses and principal planes simplify a complex, combined stress state into a convenient set of orthogonal directions with no shear. Recognizing what happens on principal planes is foundational for Mohr’s circle, failure theories, and fatigue assessments. Given Data / Assumptions: A general 2D or 3D stress state exists at a material point. We consider rotations of the reference axes to find extremal normal stresses. Concept / Approach: Principal planes are oriented such that the shear stress on them is zero. On these planes, only normal stresses act, and those normal stresses are the principal stresses (maximum or minimum among all orientations). This follows from the transformation equations or Mohr’s circle, where principal stresses lie at points where the shear ordinate is zero. Step-by-Step Solution: 1) Write the plane stress transformation formulas or use Mohr’s circle.2) The condition for principal planes is τ = 0.3) At these orientations, the normal stresses equal σ_1 and σ_2 (or σ_3 in 3D).4) Conclude: principal planes carry normal stresses only and no tangential (shear) stress. Verification / Alternative check: In Mohr’s circle, principal stresses lie at the intersections with the σ-axis, where τ = 0 by construction, confirming the absence of shear on principal planes. Why Other Options Are Wrong: Tangential only: contradicts the definition; shear vanishes on principal planes. Normal plus tangential: also incorrect; shear is zero by definition. None of these: invalid because the correct definition is known. Common Pitfalls: Confusing principal planes (τ = 0) with planes of maximum shear (τ = τ_max). Assuming principal directions align with geometry without checking stress resultants. Final Answer: normal stresses only.

Question 20

In plastic design and collapse analysis, if Q denotes the load factor, S is the plastic shape factor, and F is the factor of safety used in elastic design (based on permissible stress), identify the correct relationship among these quantities.

Accepted Answer

Introduction / Context: Plastic design compares ultimate (collapse) capacity to service (working) levels via a load factor. The reserve strength from plastic redistribution is captured by the shape factor S = Mp / My, while the elastic factor of safety F = σ_y / σ_perm links yield stress to permissible working stress. Given Data / Assumptions: Shape factor S = Mp / My (plastic to yield moment). Elastic factor of safety F = σ_y / σ_perm. Load factor Q = Collapse load / Working load. Concept / Approach: At collapse, Mp governs. At first yield, capacity is My; at working level, allowable stress is σ_perm, so the working moment is limited to My / F. Therefore, Q = (Mp) / (My / F) = (Mp / My) * F = S * F. Step-by-Step Solution: 1) Write S = Mp / My.2) Working-limit capacity = My / F (elastic permissible level).3) Load factor Q = Collapse / Working = Mp / (My / F) = S * F. Verification / Alternative check: Dimensional check is trivial (dimensionless ratios). The relation is quoted in plastic design texts for sections without premature local buckling. Why Other Options Are Wrong: Additive or subtractive relations (S ± F) have no theoretical basis in collapse design. Q = F − S also lacks physical meaning. Common Pitfalls: Confusing factor of safety F with load factor Q. Applying S of one shape (e.g., rectangle = 1.5) to another (I-beams ≈ 1.1–1.2) without checking. Final Answer: Q = S x F.

Theory of Structures Questions