Question 1

Failure theories at elastic limit (ductile materials): Which named theory corresponds to the “maximum shear stress” criterion for failure (also called the Coulomb–Tresca or Guest–Tresca criterion)?

Accepted Answer

Introduction / Context: Failure theories map multiaxial stress states to an equivalent uniaxial condition to predict yielding. For ductile materials like mild steel, shear-based criteria are widely used because yielding initiates when shear reaches a critical value related to the material’s uniaxial yield strength. Given Data / Assumptions: Elastic behavior up to yield; isotropic material. Multiaxial stress state with principal stresses σ1 ≥ σ2 ≥ σ3. No strain-rate or temperature effects considered. Concept / Approach: The maximum shear stress criterion (Guest–Tresca) states yielding begins when τ_max = (σ1 − σ3)/2 equals the uniaxial shear yield value (σ_y/2). It constructs a hexagonal yield locus in σ1–σ2 space and is conservative compared with the smooth elliptical Von Mises surface. Step-by-Step Solution: 1) Compute principal stresses (σ1, σ2, σ3).2) Evaluate τ_max = (σ1 − σ3)/2.3) Compare with shear yield threshold: yielding when τ_max ≥ σ_y/2.4) Map to named theory: this is Guest–Tresca (maximum shear stress). Verification / Alternative check: For uniaxial tension (σ1 = σ_y, σ2 = σ3 = 0), τ_max = σ_y/2, matching the Tresca yield onset condition, validating the criterion. Why Other Options Are Wrong: St. Venant (maximum principal strain) is not a standard modern failure criterion for yielding. Rankine uses maximum principal stress, more suited to brittle materials. Haigh relates to total strain energy and is seldom used for ductile yield prediction. Von Mises uses distortion energy and predicts yielding when sqrt( ( (σ1−σ2)^2 + (σ2−σ3)^2 + (σ3−σ1)^2 ) / 2 ) = σ_y, not maximum shear. Common Pitfalls: Applying Von Mises and Tresca interchangeably without noting Tresca is slightly more conservative. Using these criteria for brittle materials where Rankine or Mohr–Coulomb may be more appropriate. Final Answer: Guest’s or Tresca’s theory (maximum shear stress).

Question 2

Polar moment of inertia of a hollow circular shaft: If D and d are the external and internal diameters of a circular shaft (D > d), write the polar moment of inertia J about its centroidal axis in terms of D and d.

Accepted Answer

Introduction / Context: The polar moment of inertia J of a circular shaft governs its torsional stiffness (θ = T * L / (G * J)) and shear stress distribution (τ_max = T * R / J). Accurate J is essential for shaft strength and twist calculations. Given Data / Assumptions: Hollow circular cross-section with outer diameter D and inner diameter d. Centroidal (axisymmetric) torsion; linear elastic material. No warping or noncircular effects (Saint-Venant torsion for circular sections). Concept / Approach: For a circular section, J equals the sum of second moments about two orthogonal centroidal axes: J = I_x + I_y. For a solid circle of diameter D, J_solid = (π/32) * D^4. A hollow circle is the difference between the outer solid and the inner void. Step-by-Step Solution: 1) Outer solid: J_out = (π/32) * D^4.2) Inner void: J_in = (π/32) * d^4.3) Hollow section: J = J_out − J_in = (π/32) * (D^4 − d^4). Verification / Alternative check: Setting d = 0 recovers the solid-shaft formula J = (π/32) * D^4, confirming consistency. As d → D, J → 0, which also makes physical sense. Why Other Options Are Wrong: (π/64) * (D^4 − d^4) underestimates stiffness by a factor of 2. (π/32) * (D^2 − d^2) has wrong exponent; J scales with diameter^4. (π/16) * (D^4 − d^4) and (π/8) * (D^3 − d^3) are incorrect constants or exponents. Common Pitfalls: Using section modulus or area in place of polar moment for torsion problems. Mixing radius and diameter; remember D = 2R so J = (π/2) * R^4 = (π/32) * D^4 for solid shafts. Final Answer: J = (π/32) * (D^4 − d^4).

Question 3

Truss analysis by method of sections for member AB A straight cut passes through members AB, AD, and ED of the given truss. To isolate the internal force in member AB using a single equilibrium equation, the moments should be taken about the joint w…

Accepted Answer

Introduction / Context: Truss members are commonly analysed by the method of sections to find internal axial forces quickly. The key idea is to cut through up to three members and write equilibrium for one side of the cut. Choosing a smart moment center can eliminate unknowns and leave one target force solvable in a single step. Given Data / Assumptions: Section passes through members AB, AD, and ED. We need the force in member AB only. Planar truss, pin joints, axial forces in members, static equilibrium applies. Concept / Approach: Pick a moment center at the intersection of two cut members. The lines of action of their axial forces pass through that joint, creating zero moment. This removes two unknowns from the moment equation, leaving the third member force as the only unknown. Step-by-Step Solution: 1) Cut the truss through AB, AD, and ED.2) Choose joint D as the moment center since members AD and ED meet at D.3) Forces in AD and ED pass through D, so their moments about D are zero.4) Sum of moments about D of the cut-free body involves only the force in AB (and external loads/reactions), enabling direct computation of force in AB. Verification / Alternative check: Had we taken moments about A or E, at least two cut-member forces would contribute to the moment equation, complicating the solution. About D, exactly two vanish, which is optimal. Why Other Options Are Wrong: joint A: Only AD passes near A; AB and ED would still contribute moments. joint B: AB passes through B, but AD and ED would still add unknown moments. joint C: Not an intersection of the cut members; no special elimination happens. None of these: Incorrect because joint D works. Common Pitfalls: Forgetting that axial forces act along member centrelines and thus pass through their joint intersections; choosing a moment center that does not eliminate two of the three cut forces. Final Answer: joint D

Question 4

Short RCC column under axial load — working stress method A 200 mm × 200 mm reinforced concrete column is provided with four steel bars having a total steel area of 1200 mm². If the permissible stress in concrete is 5 N/mm² and the modular ratio m =…

Accepted Answer

Introduction / Context: This problem checks the working stress method for short axially loaded reinforced concrete columns. Under axial compression, the steel is transformed into equivalent concrete using the modular ratio, and the safe load is obtained by multiplying the permissible concrete stress by the transformed area. Given Data / Assumptions: Column size = 200 mm × 200 mm → gross area Ag = 40000 mm². Total steel area Ast = 1200 mm² (4 bars combined). Permissible concrete stress σc,perm = 5 N/mm². Modular ratio m = Es/Ec = 15. Short, concentrically loaded column; eccentricity neglected. Concept / Approach: Working stress method uses transformed section area: Aeq = Ac + m * Ast, where Ac = Ag − Ast. Safe axial load P = σc,perm * Aeq. Step-by-Step Solution: 1) Compute concrete area Ac = Ag − Ast = 40000 − 1200 = 38800 mm².2) Transformed area Aeq = Ac + m * Ast = 38800 + 15 * 1200 = 38800 + 18000 = 56800 mm².3) Safe load P = σc,perm * Aeq = 5 * 56800 = 284000 N = 284 kN. Verification / Alternative check: You may also check using percentage steel p = Ast/Ag = 1200/40000 = 3 percent (approx) and ensure that the assumed short-column, concentric load assumptions hold. The value aligns with standard transformed area computations. Why Other Options Are Wrong: 264 kN, 274 kN, 294 kN: These arise from using wrong gross area, ignoring modular ratio, or arithmetic slips. None of these: Not correct because 284 kN matches the correct calculation. Common Pitfalls: Using gross area Ag instead of Ac; forgetting to subtract steel area before adding m * Ast; mixing units (N vs kN). Final Answer: 284 kN

Question 5

Definition check — gradually applied static loads For gradually applied static loads that do not vary with time, identify which characteristic remains constant during application and sustained action.

Accepted Answer

Introduction / Context: Statics distinguishes between different types of loads. Gradually applied static loads are increased from zero to a specified level without impact and then held constant. Understanding which attributes remain unchanged helps prevent confusion between static and dynamic effects. Given Data / Assumptions: Load is applied gradually (no impact). After reaching the target, the load is maintained constant with respect to time. System is analyzed under static equilibrium; time-dependent effects are neglected. Concept / Approach: Under a static load that does not vary with time, the essential characteristics—magnitude, direction, and point of application—are fixed for the purposes of analysis. In contrast, dynamic or moving loads can vary any of these with time or position, requiring different formulations. Step-by-Step Solution: 1) Identify the loading description: “gradually applied static” implies no rapid application or vibration.2) Once the target value is reached, the magnitude is constant for the analysis period.3) The direction of a static load (e.g., vertical gravity) is fixed.4) The point of application (node, joint, area centroid) is fixed. Verification / Alternative check: Compare with a moving wheel load where the point of application changes along a beam, or with harmonic excitation where magnitude varies with time; those would not be static in this sense. Why Other Options Are Wrong: magnitude only: Incomplete; ignores invariance in direction and point of application. direction only: Incomplete. point of application only: Incomplete. None of these: False, because all listed attributes remain constant in this context. Common Pitfalls: Confusing “gradually applied” with “suddenly applied” or “impact” loading; assuming time variation exists when the statement explicitly rules it out. Final Answer: all the above

Question 6

Flexure formula relationship at a beam section Let M be the bending moment, I the second moment of area about the neutral axis, R the radius of curvature, E the modulus of elasticity, F the bending stress at a fibre, and Y the distance of that fibre…

Accepted Answer

Introduction / Context: The Euler–Bernoulli beam theory establishes the flexure formula linking bending moment, curvature, elastic modulus, section stiffness, and fibre stress. Recognizing the exact proportionalities is fundamental to beam design and stress calculations. Given Data / Assumptions: Linear elastic material behavior (Hooke’s law). Plane sections remain plane and normal to the neutral axis after bending. Small deflections and small strains. Concept / Approach: The curvature κ is 1/R and, for small deflection linear theory, κ = M / (E * I). Fibre stress varies linearly with distance Y from the neutral axis: F = (M * Y) / I. Combining these gives the standard identity: M / I = E / R = F / Y. Step-by-Step Solution: 1) Curvature: κ = 1 / R.2) From beam theory: κ = M / (E * I) ⇒ E / R = M / I.3) Stress distribution: F = (M * Y) / I ⇒ F / Y = M / I.4) Therefore, M / I = E / R = F / Y. Verification / Alternative check: Dimensional consistency: M/I has units of stress per length, equal to E/R (stress per length), and equals F/Y (stress per length), confirming the proportionality. Why Other Options Are Wrong: M / I = R / E = Y / F: Inverts E and R incorrectly. M * I = E * R = F * Y: Incorrect products; the theory gives ratios, not products. M / I = E * R = F / Y: E * R has wrong dimensions and relation. None of these: Incorrect, since option A is the standard relation. Common Pitfalls: Confusing curvature definitions, mixing up numerator and denominator, and forgetting that stress varies linearly with Y only under Euler–Bernoulli assumptions. Final Answer: M / I = E / R = F / Y

Question 7

Combined loading of a circular shaft A solid circular shaft is simultaneously subjected to a bending moment M and a torque T. Identify the correct expressions for the maximum bending stress and the maximum shear stress in terms of diameter d.

Accepted Answer

Introduction / Context: Machine shafts often experience simultaneous bending and torsion. Correct formulae for extreme normal and shear stresses at the outer fibre are essential for strength checks (e.g., maximum principal stress, maximum shear stress, or distortion-energy criteria). Given Data / Assumptions: Solid circular shaft, diameter d. Linear elastic behavior, Saint-Venant torsion theory. Small deformations, homogeneous isotropic material. Concept / Approach: For a circular section: section modulus in bending Z = π * d^3 / 32, polar section modulus in torsion Zp = J / r = π * d^3 / 16. Hence, extreme bending stress and torsional shear stress follow directly by dividing the external actions by the corresponding moduli. Step-by-Step Solution: 1) Bending: σmax = M / Z where Z = π * d^3 / 32 ⇒ σmax = 32 * M / (π * d^3).2) Torsion: τmax = T / Zp where Zp = π * d^3 / 16 ⇒ τmax = 16 * T / (π * d^3).3) Under combined loading, at the surface both σ and τ coexist and can be combined to find principal stresses if needed. Verification / Alternative check: Check limiting cases: with T = 0, only bending stress remains; with M = 0, only torsional shear remains. Dimensions are consistent (stress units for both expressions). Why Other Options Are Wrong: maximum bending stress = 32 * M / (π * d^3) alone: Incomplete for combined loading. maximum shear stress = 16 * T / (π * d^3) alone: Incomplete for combined loading. neither (a) nor (b): Both expressions are standard and correct. None of these: Incorrect because (a) and (b) are right. Common Pitfalls: Using the wrong modulus (confusing Z with J or Zp), forgetting the factor 32 vs 16, or applying hollow-shaft formulae to solid shafts. Final Answer: both (a) and (b)

Question 8

Determinacy of a compound truss from two simple trusses Two individually simple and rigid trusses have m1 and m2 members, respectively. If they are connected to form a compound truss, choose the condition on total members m for the overall frame to …

Accepted Answer

Introduction / Context: Compound trusses are built by connecting two or more simple rigid trusses. To remain statically determinate and rigid, a specific number of linking bars must be added so that the combined frame has exactly the necessary constraints without redundancy. Given Data / Assumptions: Each component truss is simple and individually rigid. The connection is made using straight bar members. Planar truss assumptions (pin-jointed, axial-force members). Concept / Approach: To properly constrain the relative motion between two rigid trusses in a plane, three independent constraints are needed (two translations and one rotation). Therefore, three appropriately placed bars are required to make the compound system rigid and statically determinate. Step-by-Step Solution: 1) Identify degrees of freedom between two rigid bodies in a plane: three (Ux, Uy, rotation).2) Each independent connecting bar removes one relative degree of freedom.3) Hence, three bars are necessary and sufficient for determinacy without redundancy.4) Total members m = m1 + m2 + 3. Verification / Alternative check: If fewer than three bars are used, the assembly will have mechanisms; if more than three, it becomes statically indeterminate due to redundant constraints. Why Other Options Are Wrong: m = m1 + m2, +1, +2: Insufficient constraints leave mechanisms. None of these: Incorrect because +3 is the established condition. Common Pitfalls: Using three bars that are concurrent or parallel so that constraints are not independent; bars must be placed to remove all three relative motions. Final Answer: m = m1 + m2 + 3

Question 9

How to connect simple trusses to form a compound truss A compound truss is formed by joining two simple, individually rigid truss frames. Select the correct way to connect them so that the combined system is rigid and statically determinate.

Accepted Answer

Introduction / Context: When two simple trusses are combined, the objective is to eliminate relative motion between them without introducing redundancies. The connection strategy must provide three independent constraints in the plane to prevent translations and rotation. Given Data / Assumptions: Two planar, simple, and rigid trusses are to be connected. Pin-jointed members; axial forces only. Determinacy is desired (no redundant members). Concept / Approach: Two rigid bodies in a plane have three relative degrees of freedom. Therefore, three independent bars are required to constrain these motions fully and exactly, yielding a rigid and statically determinate compound truss. Step-by-Step Solution: 1) Count relative DOF: 3 (two translations and one rotation).2) Each bar removes one relative DOF.3) Use three bars with non-parallel, non-concurrent arrangement to ensure independence.4) The result is a rigid, determinately connected compound truss. Verification / Alternative check: Structural determinacy can be checked by ensuring the three connections are not all parallel or all meeting at a single point, which would fail to constrain all motions independently. Why Other Options Are Wrong: two bars: Leaves one relative DOF; mechanism persists. three parallel bars: Constraints are not independent; rotation may remain. three bars intersecting at a single point: Also not independent; translations can remain unrestrained. None of these: Incorrect since three appropriately placed bars are correct. Common Pitfalls: Placing the three bars so that they are concurrent or parallel; miscounting DOF; assuming that any three bars will suffice without checking independence. Final Answer: three bars

Question 10

Strength of Materials – Radius of Gyration for a Rectangle A rectangular cross-section has depth D and width B. About the centroidal axis that is parallel to the width (i.e., the axis with bending across the depth), what is the correct expression fo…

Accepted Answer

Introduction / Context: In strength of materials, the radius of gyration k of a section connects its moment of inertia I to its area A using k = √(I/A). For a rectangle, different centroidal axes produce different values of I; here we focus on the centroidal axis parallel to the width B (bending about the axis across the depth D). Given Data / Assumptions: Section: rectangle with width B and depth D. Axis: centroidal axis parallel to B (i.e., the x-axis across the depth). Area: A = B * D. Concept / Approach: The moment of inertia of a rectangle about its centroidal axis parallel to the width is I = (B * D^3) / 12. Then k = √(I/A). Step-by-Step Solution: I = (B * D^3) / 12A = B * Dk = √(I / A) = √( (B * D^3 / 12) / (B * D) )k = √( D^2 / 12 )k = D / √12 Verification / Alternative check: As the axis takes bending across depth, I scales with D^3; dividing by area BD yields D^2, giving k proportional to D. The √12 denominator is the standard centroidal result for rectangles about this axis. Why Other Options Are Wrong: D / √3 and D / √8: incorrect constants; they do not match the rectangle's centroidal inertia form. D / 2 and D / (2√3): values are too large for the centroidal axis result. Common Pitfalls: Confusing the axis orientation (parallel to width vs parallel to depth) or using the wrong I formula (e.g., (D * B^3)/12) leads to incorrect k. Final Answer: D / √12

Question 11

Section Modulus Comparison – Square vs Circle In strength of materials and structural analysis, find the ratio of the section modulus of a square (side B) to that of a circle (diameter D), both about their centroidal axes.

Accepted Answer

Introduction / Context: Section modulus Z measures flexural strength: Z = I / y_max, where I is the second moment of area and y_max is the distance to the extreme fiber. We compare Z for square and circular sections about centroidal axes. Given Data / Assumptions: Square: side B. Circle: diameter D. Both about centroidal axes with maximum stress at the outermost fiber. Concept / Approach: Use standard formulas: I_square = B^4 / 12, y_max = B / 2; I_circle = π D^4 / 64, y_max = D / 2. Then compute Z for each and take the ratio. Step-by-Step Solution: Z_square = (B^4 / 12) / (B / 2) = B^3 / 6Z_circle = (π D^4 / 64) / (D / 2) = π D^3 / 32Required ratio = (B^3 / 6) : (π D^3 / 32) Verification / Alternative check: Dimensional check shows Z scales with length^3; both expressions have B^3 or D^3, confirming consistency. Why Other Options Are Wrong: (B^3 / 4) or (B^3 / 12): arise from mixing up I or y_max. (π D^3 / 64) is incorrect for the circle's Z; it belongs to I's denominator before dividing by D/2. Common Pitfalls: Confusing I with Z or using radius instead of diameter in circle formulas causes factor-of-2 errors. Final Answer: (B^3 / 6) : (π * D^3 / 32)

Question 12

Plastic Hinge Development in a Simply Supported I-Section A simply supported I-section beam of span L carries a central load W. In plastic analysis, what is the approximate length of the elasto-plastic zone associated with the plastic hinge that for…

Accepted Answer

Introduction / Context: When a prismatic I-section beam under bending approaches plastic collapse, a plastic hinge forms at the critical section. The hinge region is not a mathematical point; it spreads over a finite length where fibers transition from elastic to fully plastic — the elasto-plastic zone. Designers often use practical approximations for its length. Given Data / Assumptions: Simply supported beam, span L. Central concentrated load W creating maximum moment at mid-span. Prismatic I-section of overall depth D. Typical steel behavior with strain hardening neglected in first-order approximation. Concept / Approach: The plastic hinge length l_p is commonly estimated using simple rules of thumb tied to the member depth. For rolled I-sections in bending, an accepted approximation in plastic design is l_p ≈ D, representing the region over which curvature is high and stresses redistribute from elastic to plastic states. Step-by-Step Solution: At mid-span, moment is maximum and a hinge initiates.Curvature localizes over a zone scaling with section depth D.Adopting standard plastic design practice: l_p ≈ D. Verification / Alternative check: Refined models relate l_p to material strain capacity and curvature distribution, yielding values between about 0.5 * D and 2 * D. The commonly adopted design approximation is l_p ≈ D for steel I-sections, balancing safety and realism. Why Other Options Are Wrong: 0.5 * D and 2 * D: lower/upper bounds sometimes cited but not the usual design default. 1.5 * D: larger than the standard approximation for typical rolled sections. Depends only on yield stress: hinge length also depends on section geometry; depth is influential. Common Pitfalls: Treating the hinge as a zero-length point or ignoring section geometry can misestimate rotation capacity in plastic analysis. Final Answer: D

Question 13

Strongest Rectangular Beam from a Circular Log A rectangular beam is to be cut from a circular log. For maximum flexural strength (maximum section modulus), what should be the ratio of width to depth (width/depth)?

Accepted Answer

Introduction / Context: When a rectangle is inscribed in a given circle (e.g., a beam cut from a circular log), its dimensions are constrained by the circle's diameter. To maximize flexural strength, we maximize the section modulus Z of the rectangle subject to the circular constraint. Given Data / Assumptions: Rectangle of width b and depth d fits inside a circle of fixed diameter. For a rectangular section, Z ∝ b * d^2 about the centroidal axis parallel to the width. Geometric constraint: b^2 + d^2 = constant^2 (the circle). Concept / Approach: We maximize f = b * d^2 subject to b^2 + d^2 = C^2. Using optimization (e.g., Lagrange multiplier) gives a fixed ratio between b and d for the optimum. Step-by-Step Solution: Let f = b * d^2, g = b^2 + d^2 - C^2 = 0∂f/∂b = d^2, ∂f/∂d = 2 b d∂g/∂b = 2 b, ∂g/∂d = 2 dSet gradients proportional: d^2 = 2 λ b and 2 b d = 2 λ dFrom the second: b = λ; from the first: d^2 = 2 b^2 ⇒ d = √2 * bTherefore width/depth = b/d = 1/√2 ≈ 0.707 Verification / Alternative check: The result is classical: the depth should be √2 times the width to maximize Z, giving b/d = 1/√2 ≈ 0.707. Why Other Options Are Wrong: 0.303, 0.404, 0.505, 0.606: do not satisfy the optimization condition b/d = 1/√2. Common Pitfalls: Maximizing area b * d instead of section modulus b * d^2 leads to a different ratio and an incorrect answer for strength design. Final Answer: 0.707

Question 14

Principal Axes of an Area – Property of Product of Inertia For the principal centroidal axes of a plane section, which statement about the second moments is correct?

Accepted Answer

Introduction / Context: Principal axes are the orthogonal centroidal axes for which the product of inertia vanishes. They align with directions where the area's inertia matrix is diagonal, simplifying bending and torsion analysis. Given Data / Assumptions: Plane area with centroidal axes x and y. Principal axes are obtained by rotating axes so that coupling terms disappear. Concept / Approach: The product of inertia I_xy measures coupling of bending about x and y. On principal axes, the inertia matrix is diagonal, thus I_xy = 0, while I_x and I_y are generally nonzero and unequal unless the shape is isotropic. Step-by-Step Solution: Inertia matrix about centroid: [[I_x, -I_xy], [-I_xy, I_y]]Rotate axes to eliminate off-diagonal termsOn principal axes: I_xy = 0; I_x and I_y are extremal Verification / Alternative check: Mohr's circle of area moments shows that principal directions occur where the circle intersects the horizontal axis, i.e., I_xy = 0. Why Other Options Are Wrong: Sum or difference zero: not generally true; depends on geometry. None of these / both product and sum zero: contradicts standard inertia properties. Common Pitfalls: Confusing principal axes (I_xy = 0) with axes of symmetry (which may coincide but are not guaranteed). Final Answer: Product of moment of inertia is zero

Question 15

Columns – Empirical Formula for Permissible Stress The empirical expression commonly used to estimate the permissible compressive stress in a column as a function of slenderness ratio is known as:

Accepted Answer

Introduction / Context: Design codes historically used empirical relationships to estimate permissible compressive stress in columns considering slenderness. Perry’s formula is a well-known empirical relation bridging short and intermediate columns for allowable stress design. Given Data / Assumptions: Axially loaded column. Working-stress (allowable stress) context. Slenderness effects reduce permissible stress below yield. Concept / Approach: Perry’s formula relates permissible stress to slenderness ratio using a form that penalizes increasing slenderness, reflecting instability effects prior to Euler buckling. It differs from ultimate load formulas (e.g., Rankine) that predict failure loads rather than allowable stresses. Step-by-Step Solution: Identify the target: permissible stress, not ultimate load.Among the listed relations, Perry’s is specifically used to compute allowable stress with slenderness corrections.Therefore, the named empirical formula is Perry’s formula. Verification / Alternative check: Rankine’s expression combines crushing and Euler loads to give ultimate capacity, not directly an allowable stress formula. Straight-line/parabolic items may refer to steel or concrete stress-strain idealizations and not the classic column permissible stress relation. Why Other Options Are Wrong: Straight line / Parabolic formula: generic names without specific column permissible stress context. Rankine’s formula: predicts ultimate load, not allowable stress. Johnson’s parabolic formula: a capacity formula for columns rather than an allowable stress rule. Common Pitfalls: Confusing ultimate capacity formulas with allowable stress prescriptions can lead to unsafe designs. Final Answer: Perry's formula

Question 16

Strain Energy in Volumetric Compression For a body subjected to uniform hydrostatic pressure p producing volumetric strain, how does the strain energy due to volumetric strain vary with volume, pressure, and bulk modulus?

Accepted Answer

Introduction / Context: Strain energy is the elastic energy stored in a body due to deformation. Under uniform hydrostatic pressure, the energy associated with volumetric strain depends on pressure, bulk modulus, and the volume of the body. Given Data / Assumptions: Hydrostatic pressure p. Bulk modulus K of the material. Body volume V. Concept / Approach: For volumetric compression, the classic result is U_v = (p^2 * V) / (2 * K), assuming linear elasticity and small strains. This expression reveals the proportionalities directly. Step-by-Step Solution: Start from U = 1/2 * stress * strain * volume for linear elastic response.For hydrostatic loading: volumetric strain = p / K.Thus U_v = 1/2 * p * (p / K) * V = (p^2 * V) / (2 * K). Verification / Alternative check: Dimensional analysis confirms energy dimensions are consistent when combining p^2, V, and K in this form. Why Other Options Are Wrong: Each individual statement a–c is true; the most complete option is “All of the above”. Common Pitfalls: Mixing up bulk modulus in numerator rather than denominator or forgetting the square on pressure leads to incorrect trends. Final Answer: All of the above

Question 17

Section Modulus Definition at a Beam Section At any point along a beam, the section modulus Z is defined as the moment of inertia I of the cross-section divided by which geometric quantity?

Accepted Answer

Introduction / Context: The section modulus Z is a geometric property used to relate bending moment to extreme-fiber stress in flexure: σ_max = M / Z. It depends on the second moment of area and the distance to the most distant fiber. Given Data / Assumptions: Beam bending about a principal centroidal axis. Linear elastic stress distribution. Concept / Approach: By definition, Z = I / y_max, where y_max is the distance from the neutral axis to the extreme fiber (top or bottom). None of the listed options a–d correctly state “distance to the extreme fiber from the neutral axis.” Step-by-Step Solution: Z = I / y_maxy_max is not “depth of the section” (that would be overall depth).y_max is not “depth of the neutral axis.”Z is not computed by dividing by stresses (options c and d). Verification / Alternative check: For a rectangle, I = b * d^3 / 12 and y_max = d / 2, hence Z = (b * d^3 / 12) / (d / 2) = b * d^2 / 6, consistent with the definition. Why Other Options Are Wrong: a: overall depth is not the same as y_max. b: the neutral axis depth is ambiguous and not the required y_max. c and d: stress quantities are results, not geometric divisors. Common Pitfalls: Equating y_max to total depth or mixing geometric definitions with stress values yields incorrect formulas. Final Answer: None of these

Question 18

Strength of Materials – Cantilever with End Load A prismatic cantilever beam of length L, flexural rigidity EI, carries a concentrated load W at the free end. What is the maximum vertical deflection at the free end, expressed in terms of W, L, E, an…

Accepted Answer

Introduction / Context: In strength of materials, classical Euler–Bernoulli beam theory provides closed-form formulas for deflections under standard load cases. A cantilever with a point load W at its free end is a fundamental case used to validate understanding of curvature–slope–deflection relations and boundary conditions. Given Data / Assumptions: Beam type: cantilever, fixed at one end and free at the other. Length = L, flexural rigidity = E I, end load = W. Small deflection theory, linear elastic material, prismatic section, plane sections remain plane, constant E and I. Concept / Approach: The governing relation is E I * d^2y/dx^2 = M(x). Integrate twice to obtain slope and deflection, applying cantilever boundary conditions at the fixed end: y = 0 and dy/dx = 0 at x = 0. Maximum deflection occurs at the free end x = L. Step-by-Step Solution: Bending moment at a section x from the fixed end: M(x) = - W * (L - x)E I * d^2y/dx^2 = - W * (L - x)Integrate: E I * dy/dx = - W (L x - x^2/2) + C1Apply slope at fixed end: dy/dx = 0 at x = 0 gives C1 = 0Integrate: E I * y = - W (L x^2/2 - x^3/6) + C2Apply deflection at fixed end: y = 0 at x = 0 gives C2 = 0Deflection at the free end: y(L) = - W (L * L^2/2 - L^3/6) / (E I)Simplify magnitude: delta_max = W L^3 / (3 E I) Verification / Alternative check: Matches standard tables for end-loaded cantilevers and energy methods using Castigliano theorem. Why Other Options Are Wrong: W L^3 / (8 E I) and W L^3 / (48 E I) correspond to other loading and support cases. W L^2 / (2 E I) has wrong dimension. 2 W L^3 / (3 E I) overestimates by a factor of 2. Common Pitfalls: Mixing simply supported and cantilever formulas, dropping boundary conditions, or misplacing origin for M(x). Final Answer: W L^3 / (3 E I)

Question 19

Simply Supported Beam – Central Point Load A simply supported beam of span L and constant flexural rigidity E I carries a single concentrated load W at midspan. What is the maximum deflection at the centre in terms of W, L, E, and I?

Accepted Answer

Introduction / Context: Deflection of simply supported beams under common loads is a staple result in structural analysis. The central point load case yields a widely used benchmark for serviceability checks and for verifying software and hand calculations. Given Data / Assumptions: Simply supported beam, span L. Flexural rigidity E I is constant. Single point load W at midspan. Small deflections, linear elasticity, plane sections remain plane. Concept / Approach: By symmetry, reactions are W/2 at each support. Use the relation E I * d^2y/dx^2 = M(x) with a piecewise moment function or use standard integration for symmetric loading. Maximum deflection occurs at x = L/2. Step-by-Step Solution: Take origin at left support, 0 ≤ x ≤ L/2: M(x) = (W/2) xE I * d^2y/dx^2 = (W/2) xIntegrate twice and enforce y = 0 at x = 0 and continuity of slope and deflection at x = L/2Resulting centre deflection: delta_max = W L^3 / (48 E I) Verification / Alternative check: Matches tabulated formula and Castigliano theorem. Numerical check with w distribution equivalence confirms the magnitude. Why Other Options Are Wrong: W L^3 / (3 E I) is for a cantilever with end load. 5 W L^3 / (384 E I) is for a uniform load case with total load W. W L^2 / (8 E I) has incorrect dimension. W L^3 / (96 E I) is half the correct value. Common Pitfalls: Applying cantilever formulas, missing symmetry, or mixing total load with intensity. Final Answer: W L^3 / (48 E I)

Question 20

Dynamics of Lifting – Stress in the Rope A lift of weight W is hoisted upward with constant acceleration f. The supporting rope has uniform cross-sectional area A. Neglect rope mass. What is the tensile stress developed in the rope?

Accepted Answer

Introduction / Context: When accelerating a suspended load, the rope must provide both the weight support and the additional force required for acceleration. This problem connects Newton second law with normal stress definition. Given Data / Assumptions: Lift weight (force) = W. Upward acceleration = f, gravitational acceleration = g. Rope mass negligible, cross-sectional area = A. One-dimensional vertical motion, no damping or pulley friction. Concept / Approach: Let T be rope tension. For upward acceleration, net upward force equals mass times acceleration. Stress sigma equals tension divided by area. Step-by-Step Solution: Mass of the lift: m = W / gApply Newton second law upward: T - W = m * fSubstitute m: T - W = (W / g) * fSolve for tension: T = W * (1 + f / g)Stress in the rope: sigma = T / A = (W / A) * (1 + f / g) Verification / Alternative check: For f = 0, sigma reduces to W / A, as expected for static lifting. For downward acceleration of magnitude f, the sign changes to 1 - f/g. Why Other Options Are Wrong: (W / A) * (1 - f / g) is for downward acceleration. W / A ignores inertia. (W / A) * (g / f) and (W / A) * (1 + g / f) are dimensionally inconsistent for small f and unphysical at f approaching zero. Common Pitfalls: Confusing weight W with mass, forgetting to divide by g, or using downward sign convention incorrectly. Final Answer: (W / A) * (1 + f / g)

Theory of Structures Questions