Question 1

Parabolic arch geometry: Write the standard equation of a parabolic arch of span l and rise h with the origin at the left support, x measured along the span, and y the ordinate from the chord.

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Introduction / Context:Parabolic arches are widely used because the funicular shape for uniform loading minimizes bending. Describing the arch geometry with a simple quadratic helps in analysis for thrust and bending under various loadings.Given Data / Assumptions:Span = l (distance between springings).Rise = h (ordinate at crown relative to springing chord).Coordinate system: x from left springing to right, y vertical ordinate from the chord.Concept / Approach:A parabola symmetric about mid-span can be written as y = k x (l − x). Enforcing the crown ordinate y(l/2) = h determines k. Substitute x = l/2 into the general form and solve for k to get the standard parabolic equation.Step-by-Step Solution:1) Assume y = k x (l − x).2) At crown x = l/2, y = h ⇒ h = k * (l/2) * (l/2) = k * l^2 / 4.3) Solve for k: k = 4 h / l^2.4) Therefore y = (4 h / l^2) * x (l − x) = 4 h x (l − x) / l^2.Verification / Alternative check:At x = 0 and x = l, y = 0; at x = l/2, y = h. The parabola is symmetric and satisfies boundary conditions.Why Other Options Are Wrong:Linear or pure quadratic in x/l do not satisfy both end boundary conditions and crown rise simultaneously for an arch spanning 0 to l.Common Pitfalls:Forgetting the factor 4 in 4 h x (l − x) / l^2.Using x measured from the crown rather than from a springing without adjusting the equation.Final Answer:y = 4 h x (l − x) / l^2

Question 2

Cantilever of uniform strength under uniformly distributed load: keeping the breadth b constant, how should the depth d(x) vary along a cantilever of length l (measured from the free end) so that the extreme fiber stress remains constant?

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Introduction / Context:A member of “uniform strength” has constant extreme fiber stress along its length. For a cantilever with uniformly distributed load (UDL), the bending moment varies with the square of the distance from the free end. To maintain constant stress while keeping breadth b constant, the depth must vary appropriately.Given Data / Assumptions:Cantilever, span l, fixed at one end, free at the other.UDL intensity w (force per unit length).Breadth b remains constant; depth d(x) varies with position x from the free end (x = 0 at free end).Elastic behavior; rectangular section modulus Z = b * d^2 / 6.Concept / Approach:At distance x from the free end, the cantilever bending moment under UDL is:M(x) = w * x^2 / 2For uniform strength, extreme fiber stress σ = M / Z must be constant, so M(x) / [b d(x)^2 / 6] = constant ⇒ d(x)^2 ∝ M(x) ⇒ d(x) ∝ √M(x). Because M(x) ∝ x^2, we obtain d(x) ∝ x.Step-by-Step Solution:1) Write M(x) = w x^2 / 2.2) Require σ = constant ⇒ M(x) / Z = constant.3) With Z = b d(x)^2 / 6 and b constant ⇒ d(x)^2 ∝ M(x).4) Therefore d(x) ∝ √M(x) ∝ √(x^2) ⇒ d(x) ∝ x.5) Hence d = 0 at x = 0 (free end) and increases linearly to maximum at the fixed end.Verification / Alternative check:If d doubles when x doubles, the section modulus increases by a factor of 4 while M increases by a factor of 4, keeping σ uniform, confirming linear variation suffices.Why Other Options Are Wrong:Constant depth: would produce maximum stress at the fixed end and lower stress elsewhere.d ∝ x^2 or d ∝ √x: mismatch the required square relationship between Z and M.Maximum at free end: opposite to bending-moment distribution.Common Pitfalls:Measuring x from the fixed end instead of the free end when using M(x) = w x^2 / 2.Using Z ∝ d instead of Z ∝ d^2 for a rectangular section.Final Answer:d(x) is proportional to x (zero at free end and maximum at fixed end)

Question 3

Close-coiled helical spring under an applied couple M about the helix axis: what happens to the internal action and geometry?

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Introduction / Context:A close-coiled helical spring can be loaded either by an axial force (causing twisting of the wire) or by a torque/couple about the spring’s axis (causing bending of the wire). This question distinguishes these two classic loading cases in strength of materials and spring design.Given Data / Assumptions:Close-coiled helical spring (helix angle small).An external couple M is applied about the helix (spring) axis, not an axial force.Linear elastic behavior, small deformations, uniform wire and constant coil diameter.Concept / Approach:For close-coiled springs: an axial force P produces torsion of the wire (primary) and slight direct shear, while a torque M applied about the helix axis produces pure bending of the wire (primary). The distinction follows from resolving the external load into actions along the wire centerline: torque about the spring axis leads to a uniform bending moment in the wire’s local plane; no twisting moment acts about the wire’s own axis in this idealization.Step-by-Step Solution:Identify loading: couple M about spring axis ⇒ bending of the wire.Internal action: wire fibers experience tensile/compressive bending stress σ = M_w * y / I_w (locally), where M_w is the bending moment in the wire, y is fiber distance, I_w is wire section second moment.Geometric effects like change of mean diameter or number of coils are negligible in the ideal linear theory and are not fundamental consequences of M.Verification / Alternative check:Textbook spring theory tables list: “Axial force → torsion of wire; Axial couple → bending of wire.” This is a standard result used in spiral spring and clock spring calculations.Why Other Options Are Wrong:Mean diameter decrease (option b) and “increase in number of coils” (option c) are not intrinsic outcomes of the elastic analysis; they are misleading. “All of the above” (option d) is therefore incorrect.Common Pitfalls:Confusing the axial force case (wire in torsion) with the axial couple case (wire in bending).Final Answer:It is subjected to pure bending of the wire

Question 4

Polar moment of inertia of a plane section: If I_x and I_y are the second moments of area about the orthogonal centroidal X and Y axes, what is the polar moment of inertia J about the centroid?

Accepted Answer

Introduction / Context:The polar moment of inertia J of a plane area about a point (usually the centroid) is fundamental in torsion (thin sections) and in combined bending problems. It relates to resistance to rotation and to principal moments via the perpendicular axis theorem for flat areas.Given Data / Assumptions:Flat lamina (plane section).Centroidal orthogonal axes X and Y lying in the plane of the section.Polar axis Z is perpendicular to the section through the same centroid.Concept / Approach:The perpendicular axis theorem for plane areas states that the polar second moment of area about Z through the centroid equals the sum of the second moments about any two orthogonal in-plane centroidal axes: J_z = I_x + I_y.Step-by-Step Solution:Identify axes: X and Y in plane, Z out of plane.Apply perpendicular axis theorem: J = I_x + I_y.No further computation is necessary once I_x and I_y are known.Verification / Alternative check:For a circle of radius R, I_x = I_y = (π R^4)/4 and J = (π R^4)/2. Indeed, J = I_x + I_y confirms the same result.Why Other Options Are Wrong:Differences, ratios, or root-sum-squares have no basis in the perpendicular axis theorem. Option (e) holds only if I_x = I_y, but is not the general identity.Common Pitfalls:Confusing moments of inertia (area MOI) with polar mass moment of inertia; mixing centroidal with base axes.Final Answer:J = I_x + I_y

Question 5

Failure theory identification: The maximum principal strain criterion for elastic failure is also known by which classical name?

Accepted Answer

Introduction / Context:Multiple failure theories are used in design under multiaxial stress states. Each theory proposes a critical quantity (stress, strain, or energy) that governs yielding or failure. Recognizing their classical names is essential for exam and design problems.Given Data / Assumptions:Elastic limit behavior is considered.Multiaxial stress state possible.Classical nomenclature is used.Concept / Approach:The maximum principal strain theory asserts failure when the largest tensile principal strain reaches the uniaxial failure strain. Historically, this has been attributed to St. Venant. In contrast, Rankine uses maximum principal stress; Tresca/Guest uses maximum shear stress; Haigh (Beltrami-Haigh) uses total strain energy; Von Mises uses distortion energy.Step-by-Step Solution:Identify the criterion: “maximum strain” ⇒ maximum principal strain.Classical name mapping: maximum principal strain ⇒ St. Venant’s theory.Verification / Alternative check:Handbooks list: Rankine (max σ1), St. Venant (max ε1), Guest/Tresca (max τ_max), Beltrami-Haigh (total strain energy), Von Mises (distortion energy).Why Other Options Are Wrong:Each alternative refers to a different governing quantity and is not the “maximum principal strain” theory.Common Pitfalls:Confusing “maximum strain” with “maximum stress” (Rankine) or with “maximum shear stress” (Tresca/Guest).Final Answer:St. Venant's theory (maximum principal strain)

Question 6

Sizing a rectangular beam when breadth b is fixed: how does the required depth d vary with the design bending moment M to satisfy a given allowable bending stress?

Accepted Answer

Introduction / Context:Preliminary beam sizing commonly uses the bending stress formula M = σ_allow * Z. For rectangular sections with fixed breadth b, designers adjust depth d to meet moment capacity. Understanding the scaling relation between d and M fast-tracks early sizing.Given Data / Assumptions:Rectangular section, breadth b fixed.Allowable bending stress σ_allow is fixed by material and code.Elastic bending theory applies; serviceability not considered here.Concept / Approach:Section modulus for a rectangular section about its strong axis is Z = b d^2 / 6. Required moment capacity is M = σ_allow * Z. Solving for d gives the scaling with M for constant b and σ_allow.Step-by-Step Solution:Z = b d^2 / 6.M = σ_allow * Z = σ_allow * (b d^2 / 6).Rearrange: d^2 = (6 M) / (σ_allow b) ⇒ d = √[ (6 M) / (σ_allow b) ].Therefore, d ∝ √M for fixed b and σ_allow.Verification / Alternative check:Doubling the design moment M requires depth increase by √2, consistent with the square-root relation.Why Other Options Are Wrong:Linear, cube-root, or quadratic scaling does not follow from Z = b d^2 / 6 when b is constant. d cannot be independent of M.Common Pitfalls:Using I instead of Z directly; forgetting that Z ∝ d^2 for rectangles, leading to the square-root dependence.Final Answer:d ∝ M^0.5 (square-root relation)

Question 7

Euler buckling comparison: What is the ratio of the Euler crippling load of a column with both ends fixed to that of an identical column with both ends hinged?

Accepted Answer

Introduction / Context:Euler’s critical load P_cr for elastic buckling depends on the effective length L_e determined by end restraint. Increased fixity shortens L_e and raises P_cr. This question asks for the ratio of capacities for the two classic end conditions.Given Data / Assumptions:Same column (E, I, length L) in both cases.Euler formula applies: slender, straight, elastic, ideal end conditions.End conditions: fixed–fixed versus hinged–hinged.Concept / Approach:P_cr = π^2 E I / L_e^2. For hinged–hinged, L_e = L. For fixed–fixed, L_e = L/2. Ratio is set by the square of L_e.Step-by-Step Solution:Hinged–hinged: P_h = π^2 E I / L^2.Fixed–fixed: P_f = π^2 E I / (L/2)^2 = 4 π^2 E I / L^2.Ratio: P_f / P_h = 4.Verification / Alternative check:Effective length factors: K = 1.0 (hinged–hinged), K = 0.5 (fixed–fixed). Since P_cr ∝ 1/K^2, ratio = (1/0.5^2) = 4.Why Other Options Are Wrong:Other ratios do not match the effective length relationship.Common Pitfalls:Confusing fixed–free (cantilever, weakest) with fixed–fixed (strongest).Final Answer:4

Question 8

Power transmitted by a rotating shaft: A shaft runs at N revolutions per minute (rpm) under a steady torque T. What is the power P transmitted?

Accepted Answer

Introduction / Context:Power in a rotating shaft equals torque times angular speed. Converting rotational speed in rpm to angular velocity in rad/s is the key step. This relation is used widely in machine design and power transmission calculations.Given Data / Assumptions:Torque T is constant.Shaft rotational speed N in rpm.Neglect losses so mechanical power equals T times angular velocity.Concept / Approach:Angular speed ω (rad/s) = 2 π N / 60. Then power P (watts) = T * ω = T * (2 π N / 60).Step-by-Step Solution:Convert N rpm to rad/s: ω = 2 π N / 60.Power: P = T * ω = 2 π N T / 60.Units: T (N·m) * ω (rad/s) ⇒ N·m/s (W).Verification / Alternative check:If N = 60 rpm (1 rps), then P = 2 π T, matching T times 2 π rad/s.Why Other Options Are Wrong:They have missing or extra factors of 2 or 60, or incorrect dimensional form.Common Pitfalls:Forgetting to convert rpm to rad/s; mixing minutes and seconds leading to a factor-of-60 error.Final Answer:P = 2 π N T / 60

Question 9

Maximum bending moment for a simply supported beam carrying a uniformly distributed load w per unit length over the entire span l: select the correct expression.

Accepted Answer

Introduction / Context:Beams under uniformly distributed load (UDL) are fundamental in structural analysis. The maximum bending moment for a simply supported beam under full-span UDL occurs at midspan, and its value is a standard result used for sizing and checking stresses/deflections.Given Data / Assumptions:Simply supported beam, span l.Uniformly distributed load w per unit length over entire span.Linear elastic behavior.Concept / Approach:The support reactions are each R = w l / 2. The bending moment diagram is parabolic, peaking at midspan. Using statics or standard formulas yields the maximum value M_max at the center.Step-by-Step Solution:Reaction at each support: R = w l / 2.Bending moment at a section x from the left: M(x) = R x − w x^2 / 2.Set dM/dx = R − w x = 0 ⇒ x = R / w = (w l / 2)/w = l/2 (midspan).Evaluate M_max: M(l/2) = (w l / 2)(l/2) − w (l/2)^2 / 2 = w l^2 / 8.Verification / Alternative check:Area under shear diagram (triangle) from support to midspan equals w l^2 / 8, matching the computed maximum moment.Why Other Options Are Wrong:Other expressions give incorrect numerical constants, locations, or dimensions (e.g., w l^3 / 24 is dimensionally inconsistent for bending moment).Common Pitfalls:Forgetting that maximum bending moment occurs where shear is zero; misplacing the location at quarter points (that is for point loads or other cases).Final Answer:M_max = w l^2 / 8 at midspan

Question 10

Empirical column formula identification: Name the empirical formula commonly used to calculate the allowable (permissible) stress for long columns in structural design.

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Introduction / Context:Real columns deviate from Euler’s ideal; empirical or semi-empirical formulas are used to compute allowable stresses, especially for intermediate and long columns in practice. Recognizing which expression is traditionally adopted is key for code-level calculations.Given Data / Assumptions:Design for allowable stress, not ultimate load.Column may be intermediate/long with imperfections and residuals.Elastic–inelastic transition captured empirically.Concept / Approach:The Rankine–Gordon formula blends crushing (short-column) and Euler buckling (long-column) effects into one relation for permissible stress or load. It is widely cited as an empirical basis for allowable stress in long columns, predating modern code curves and Perry–Robertson refinements.Step-by-Step Solution:Identify empirical approach for long columns → Rankine–Gordon.Alternatives: Johnson’s straight-line or parabolic are approximations valid over limited slenderness ranges; Perry’s is semi-empirical.Verification / Alternative check:Handbooks present Rankine–Gordon in the form 1/P_allow = 1/P_crush + 1/P_Euler, or equivalent stress forms, illustrating the blend.Why Other Options Are Wrong:Johnson straight-line/parabolic apply primarily to intermediate columns; Perry’s is another method but Rankine–Gordon is the classic empirical name historically associated with long columns in exams.Common Pitfalls:Assuming Euler alone for design; ignoring empirical limits that account for imperfections and non-ideal behavior.Final Answer:Rankine–Gordon formula

Question 11

Compression Members: Identify the correct terminology and behavior for vertical members under axial compression

Accepted Answer

Introduction:In structural engineering, compression members are fundamental elements that resist loads pushing toward the member's centroid. This question checks whether you know the standard terminology—struts, columns (stanchions), and the instability phenomenon called buckling—used for members under axial compression.Given Data / Assumptions:Member is subjected primarily to axial compression.Terminology is based on conventional structural engineering usage.Behavior under compression may include instability (buckling) depending on slenderness.Concept / Approach:Short, stocky members mainly fail by crushing, while long, slender members can become unstable and buckle before material failure. Names also depend on orientation and usage: the generic term “strut” for compression members; “column” or “stanchion” when vertical and load-carrying in buildings/frames.Step-by-Step Solution:Step 1: Recognize naming—any member in compression can be called a strut.Step 2: When the compression member is vertical in a structure, standard terms are column or stanchion.Step 3: For sufficiently slender members, lateral deflection under compression is buckling, a stability failure mode.Step 4: Since all three statements are correct, the inclusive option is the right choice.Verification / Alternative check:Textbooks (strength of materials, structural analysis) consistently define vertical compression members as columns/stanchions and compression members more generally as struts. Euler's analysis predicts buckling in slender columns, confirming statement (c).Why Other Options Are Wrong:a: Correct but incomplete alone.b: Correct but incomplete alone.c: Correct but incomplete alone.e: Incorrect because multiple statements are correct.Common Pitfalls:Confusing crushing with buckling; assuming only very tall building members are “columns”; ignoring that “strut” applies regardless of orientation.Final Answer:All the above.

Question 12

Helical Springs: Identify the correct combined statement about form, pitch, and angle of helix

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Introduction:This question tests core terminology for helical springs: the geometric form, the concept of pitch, and the angle of helix. Understanding these basics is essential for analyzing stiffness, energy storage, and stress distribution in springs used in machines and structures.Given Data / Assumptions:Standard mechanical design definitions are used.“Close-coil” means adjacent turns are nearly touching (small pitch).“Open-coil” means turns are more widely spaced (larger pitch and helix angle).Concept / Approach:A helical spring is produced by winding a wire into a helix. Pitch is the axial distance between corresponding points on adjacent coils. The angle of helix is the angle between the helical wire path and a plane perpendicular (or parallel, by consistent convention) to the spring axis; practically, a “larger” helix angle corresponds to more open spacing and different load-deflection behavior.Step-by-Step Solution:Step 1: Definition—wound wire in spiral form is a helical spring (correct).Step 2: Close-coil springs have very small pitch; turns are nearly contiguous (correct).Step 3: The coil’s incline relative to a reference plane is the angle of helix (terminology point, correct).Step 4: Open-coil springs have a comparatively large helix angle (correct).Step 5: Since all are correct, select the comprehensive option.Verification / Alternative check:Mechanical design texts present the same definitions and note that open-coil behavior differs (torsion + bending) due to larger helix angle.Why Other Options Are Wrong:a–d: Each is correct in isolation but not complete; the best answer is the combined choice.e: Correct because it aggregates a–d.Common Pitfalls:Confusing pitch with wire diameter; assuming close-coil and open-coil formulas are interchangeable; mixing up reference planes for helix angle but keeping the concept consistent.Final Answer:All the above.

Question 13

Bimetallic (Composite) Beam of Brass and Steel: Effect of uniform temperature rise on internal forces and deformation

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Introduction:A composite (bimetallic) beam made of brass and steel experiences internal restraint when temperature changes. Because brass has a higher coefficient of thermal expansion than steel, a uniform temperature rise creates incompatible free expansions, leading to self-equilibrating internal forces and bending. This question checks understanding of thermal strain compatibility and induced internal actions.Given Data / Assumptions:Two equal strips: one brass, one steel, perfectly bonded.Uniform temperature increase across the beam.Material properties: alpha_brass > alpha_steel; linear elastic range.Concept / Approach:Free thermal strain = alpha * delta_T. Since alpha_brass > alpha_steel, brass wants to elongate more. Bonding enforces equal overall extension at the interface, inducing tension in the less-expanding material (steel) and compression in the more-expanding material (brass). The force couple generated by these equal and opposite forces produces curvature (bending) of the composite strip.Step-by-Step Solution:Step 1: Identify mismatch in free thermal strains: brass larger than steel.Step 2: Compatibility requires equal actual extension, so internal forces arise.Step 3: Steel is pulled (tension) to match brass; brass is pushed (compression) to match steel.Step 4: These opposite forces at different layers form a couple, causing bending (curvature).Step 5: Therefore all statements hold simultaneously.Verification / Alternative check:Classical bimetallic strip behavior (used in thermostats) confirms curvature toward the material with the lower expansion (steel side becomes concave).Why Other Options Are Wrong:a–d: Each is individually correct, but the question seeks the combined effect; the comprehensive option is correct.Common Pitfalls:Assuming equal expansions because strips are equal in size; forgetting that perfect bonding enforces compatibility; thinking temperature rise only changes length without causing internal forces.Final Answer:All the above.

Question 14

Short Rectangular Column (30 cm × 20 cm): Two opposite eccentric loads along a principal section parallel to the longer side produce the same extreme stress—find P1 : P2 when e1 = 4 cm and e2 = 8 cm

Accepted Answer

Introduction:This problem uses the superposition principle for combined axial load and bending in a short rectangular column. Two concentric-but-opposite eccentricities along the same principal axis create counteracting moments. The question asks for the ratio P1 : P2 such that the maximum extreme fiber stress on each side is the same in magnitude.Given Data / Assumptions:Section: 30 cm × 20 cm (dimensions ensure “short column” assumptions).Principal section considered is parallel to the longer side (30 cm).Eccentricities: e1 = 4 cm on one side; e2 = 8 cm on the opposite side.Linear elastic behavior; superposition valid; loads are axial with small eccentricities.Concept / Approach:Resultant stress at an extreme fiber is: sigma_extreme = (P/A) ± (M/Z). Here, M = Σ(Pi * ei) with sign based on sides. If the “extreme intensity on either side is the same,” the magnitudes of stress at the two opposite extremes must be equal. This condition is met when the net bending moment is zero, i.e., the opposite moments cancel each other.Step-by-Step Solution:Step 1: Write the moment balance about the centroidal axis along the given principal direction.Step 2: For equal extreme stresses on opposite faces, require net bending moment M = P1e1 − P2e2 = 0 (opposite sides give opposite signs).Step 3: Solve for ratio: P1e1 = P2e2 ⇒ P1/P2 = e2/e1.Step 4: Substitute e2 = 8 cm and e1 = 4 cm ⇒ P1/P2 = 8/4 = 2.Step 5: Therefore, P1 : P2 = 2 : 1.Verification / Alternative check:If M = 0, the section carries uniform axial stress only, so both extremes have the same intensity, satisfying the condition without needing Z or the exact dimensions.Why Other Options Are Wrong:1 : 1 — Would only work if e1 = e2.1 : 2 — Reverses the needed proportion for cancellation.4 : 1 and 1 : 4 — Over/under-compensate the moment, leaving unequal extremes.Common Pitfalls:Trying to compute with section modulus; it cancels for the equality condition. Forgetting sign convention for opposite sides. Mixing up which load should be larger.Final Answer:2 : 1.

Question 15

Hollow Circular Column (D = 25 cm, d = 15 cm): Maximum eccentricity e to avoid tension under an eccentric load of 100 t

Accepted Answer

Introduction:To prevent tension anywhere on a compression member under eccentric loading, the load’s line of action must lie within the “kern.” For circular and annular (hollow circular) sections, the kern is a concentric circle of radius r_k. This question asks for the limiting eccentricity e for a hollow circular column so that no tensile stress develops.Given Data / Assumptions:Outer diameter D = 25 cm; inner diameter d = 15 cm.Eccentric compressive load P = 100 t (magnitude does not affect kern radius).Linear elastic behavior; classical kern criterion: e ≤ r_k.Concept / Approach:The kern radius is r_k = I / (A * c), where I is the second moment of area about any centroidal axis in the plane, A is area, and c is the distance from centroid to the extreme fiber in the load direction (for a ring, c = D/2). For a hollow circle: I = (pi/64) * (D^4 − d^4), A = (pi/4) * (D^2 − d^2). Simplifying yields a compact formula for the ring: r_k = (1/8) * (D^2 + d^2) / D.Step-by-Step Solution:Step 1: Write I = (pi/64) * (D^4 − d^4).Step 2: Write A = (pi/4) * (D^2 − d^2).Step 3: Take c = D/2.Step 4: Compute r_k = I / (A * c) = [(pi/64)(D^4 − d^4)] / {[(pi/4)(D^2 − d^2)] * (D/2)}.Step 5: Cancel pi and (D^2 − d^2); use D^4 − d^4 = (D^2 − d^2)(D^2 + d^2) to get r_k = (1/8) * (D^2 + d^2) / D.Step 6: Substitute D = 25 cm, d = 15 cm ⇒ r_k = (1/8) * (625 + 225) / 25 = (1/8) * 850 / 25 = (1/8) * 34 = 4.25 cm.Verification / Alternative check:A solid circle gives r_k = D/8; putting d = 0 in the ring formula reduces to the same result, confirming consistency.Why Other Options Are Wrong:2.75, 3.00, 3.50 cm — All are less than the derived kern radius; they are not the maximum permissible eccentricity.5.0 cm — Exceeds the kern; placing the load beyond this would induce tension.Common Pitfalls:Using the solid-circle kern radius D/8 for a hollow section; forgetting that P does not affect r_k; mixing diameter and radius in c.Final Answer:4.25 cm.

Question 16

Failure Theories at Elastic Limit: The maximum principal stress theory is known as

Accepted Answer

Introduction:Failure theories predict yielding or fracture under multiaxial stress states. Among classical criteria, each theory emphasizes a different stress or energy measure. This question asks you to correctly name the theory corresponding to the maximum principal stress criterion at the elastic limit.Given Data / Assumptions:Material is ductile or brittle within elastic range for theory comparison.Principal stresses are well-defined at the point of interest.We are identifying the name attached to the maximum principal stress theory.Concept / Approach:The maximum principal stress theory states failure occurs when the largest principal stress equals the material's uniaxial failure stress. Historically, this is called Rankine's theory and is especially used for brittle materials. Other named theories correspond to different criteria (shear stress, strain energy, etc.).Step-by-Step Solution:Step 1: Recall mapping: maximum principal stress → Rankine.Step 2: Distinguish from Tresca (maximum shear stress) and Von Mises (distortion energy).Step 3: Confirm St. Venant and Haig relate to other strain/energy concepts, not maximum principal stress naming.Step 4: Select Rankine's theory as the correct naming.Verification / Alternative check:Design handbooks list Rankine as the principal stress criterion, suitable for brittle materials where tensile strength governs crack initiation.Why Other Options Are Wrong:Guest's/Tresca's — Maximum shear stress criterion, not principal stress.St. Venant's — Associated with principal strain or energy-based ideas, not the maximum principal stress naming.Haig's — Related to strain energy approaches (e.g., octahedral shear), not Rankine.Von Mises's — Distortion energy criterion, not principal stress criterion.Common Pitfalls:Confusing principal stress (Rankine) with shear-based (Tresca) or energy-based (Von Mises) criteria; assuming all names are interchangeable.Final Answer:Rankine's theory.

Question 17

Slenderness ratio of a long column (structural mechanics): Among the following definitions, identify the correct engineering definition of “slenderness ratio” used in Euler/Rankine buckling design. It should express how long (L) a member is relative…

Accepted Answer

Introduction / Context:Columns fail primarily by buckling rather than crushing when they are thin and long. To quantify how “long and thin” a column is, engineers use the slenderness ratio. This ratio is central to Euler and Rankine formulas and guides whether elastic buckling or inelastic behavior governs design.Given Data / Assumptions:Column length considered between effective end conditions is L.Radius of gyration about an axis is k = sqrt(I/A), where I is the second moment of area and A is the cross-sectional area.Least radius of gyration k_min controls buckling since buckling occurs about the weakest (least stiff) axis.Concept / Approach:The slenderness ratio quantifies susceptibility to buckling and is defined as L/k. For safety we use the smallest k, hence L/k_min. Using k_min ensures we check buckling about the axis with minimum stiffness where buckling is most likely.Step-by-Step Solution:1) Identify k about principal axes: k_x and k_y.2) Determine k_min = min(k_x, k_y).3) Compute slenderness ratio: SR = L / k_min.4) Select the definition that matches SR = L / k_min.Verification / Alternative check:When k decreases (section more “slender” about that axis), L/k increases, which correctly predicts a greater tendency to buckle. This aligns with Euler's critical load proportional to 1/L^2 and inversely to k^2 via I = A k^2.Why Other Options Are Wrong:Area divided by k or least k is dimensionally inconsistent with a ratio used for buckling.k divided by area has no direct buckling meaning.Using “any axis” can miss the weakest axis; buckling checks must use k_min.Common Pitfalls:Using overall length instead of effective length that accounts for end conditions.Checking only the strong axis and ignoring the weak axis (k_min).Final Answer:Length of column divided by least radius of gyration.

Question 18

Second moment of area of a rectangle about centroidal axis (parallel to width): For a rectangular cross-section of width B and depth D, find the moment of inertia about the centroidal axis that is parallel to the width (i.e., the horizontal centroid…

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Introduction / Context:Design against bending requires the second moment of area (also called area moment of inertia) about the relevant centroidal axis. For a rectangle, the standard formulas are frequently used in beam deflection and bending stress calculations (sigma = M * y / I).Given Data / Assumptions:Rectangular section with width B and depth D.Axis passes through centroid and is parallel to the width, i.e., a horizontal centroidal axis.Linear elastic behavior; small deflections; standard geometric properties apply.Concept / Approach:The second moment of area about a centroidal axis parallel to the width depends on the dimension perpendicular to that axis. When the axis is parallel to the width (horizontal), the dimension contributing with a power of 3 is the depth D, leading to a D^3 dependence.Step-by-Step Solution:1) Identify the correct centroidal axis: horizontal through the centroid.2) Recall the standard formula: I_centroidal,horizontal = (B * D^3) / 12.3) Confirm dimensionality: units of I are length^4, which matches B * D^3.Verification / Alternative check:Swapping axis orientation swaps which dimension is cubed. About a vertical centroidal axis (parallel to depth), the formula becomes (D * B^3)/12, confirming orientation sensitivity.Why Other Options Are Wrong:(B^3 * D)/12 corresponds to the vertical centroidal axis, not the horizontal one.(B * D^3)/3 and (B * D^3)/36 have incorrect constants; 1/12 is standard for centroidal axis.(B^3 * D)/3 has both wrong axis and constant.Common Pitfalls:Confusing which dimension is cubed—always the dimension perpendicular to the axis.Using base about base axis (B * D^3 / 3) instead of centroidal axis (B * D^3 / 12).Final Answer:I = (B * D^3) / 12.

Question 19

Normal stress on an inclined plane under biaxial normal stresses: A prismatic bar is subjected to a longitudinal normal stress σ1 and a transverse normal stress σ2. What is the normal component of stress on a plane inclined at an angle θ to the long…

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Introduction / Context:Stress transformation is fundamental in solid mechanics. When a member carries normal stresses in two perpendicular directions (σ1 longitudinal, σ2 transverse), the stress on an arbitrarily inclined plane changes with the angle. Engineers need the transformed normal stress to assess failure on critical planes.Given Data / Assumptions:Only normal stresses σ1 (longitudinal) and σ2 (transverse) act; shear stress τ12 is zero.The plane of interest is inclined at angle θ to the longitudinal axis of σ1.Small deformation, linear elasticity, continuum assumptions.Concept / Approach:Using plane stress transformation, if the plane itself is at angle θ to the σ1 direction, its normal is at (90° − θ). The general normal stress transformation (with τ = 0) reduces to σ_n = σ1 cos^2 φ + σ2 sin^2 φ where φ is the angle between the plane's normal and σ1. Substituting φ = 90° − θ gives σ_n = σ1 sin^2 θ + σ2 cos^2 θ.Step-by-Step Solution:1) Start from σ_n = σ1 cos^2 φ + σ2 sin^2 φ for τ = 0.2) Recognize φ = 90° − θ (plane normal vs plane angle).3) Use identities: cos(90° − θ) = sin θ and sin(90° − θ) = cos θ.4) Substitute to obtain σ_n = σ1 sin^2 θ + σ2 cos^2 θ.Verification / Alternative check:At θ = 0°, the plane is parallel to σ1 (normal is along σ2), giving σ_n = σ2, which matches the formula. At θ = 90°, σ_n = σ1, again consistent.Why Other Options Are Wrong:σ1 sin θ * σ2 cos θ mixes stresses multiplicatively and has wrong dimensions.σ1 cos^2 θ + σ2 sin^2 θ corresponds to the case when θ is the angle between normal and σ1, not the plane and σ1.(σ1+σ2)/2 + ... cos 2θ is correct if θ denotes the normal's angle to σ1; the problem defines θ for the plane, not the normal.σ1 cos θ + σ2 sin θ is a simple projection, not the transformation law.Common Pitfalls:Confusing the plane angle with the normal angle.Forgetting that shear is zero only in the original axes, not necessarily on the inclined plane.Final Answer:σ1 sin^2 θ + σ2 cos^2 θ.

Question 20

Sudden vs. gradual loading on a bar (impact factor concept): For the same final load W on a linear elastic bar, compare the maximum stress produced when the load is applied suddenly (without initial velocity) versus when it is applied gradually from…

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Introduction / Context:Load application rate strongly affects peak stresses in elastic members. Under impact or sudden loading, the bar stores more strain energy at the same final load level compared with a slowly applied (quasi-static) load. This concept is often introduced using energy methods in strength of materials.Given Data / Assumptions:Linear elastic, homogeneous bar obeying Hooke’’s law.Axial loading only; no damping.Sudden load means load W is applied instantaneously from zero to W with no initial kinetic energy.Gradual load means load increases slowly from 0 to W (static application).Concept / Approach:Use strain energy equivalence. Under gradual loading, the strain energy stored is (1/2) * W * δ, where δ is the static extension under W. Under sudden loading, the bar oscillates and its first peak force equals 2W, producing a peak stress twice the static stress for the same stiffness.Step-by-Step Solution:1) Static (gradual) case: σ_static = W / A. Energy = 0.5 * W * δ.2) Sudden case: initial energy input W * δ_static equals maximum strain energy at first peak: 0.5 * (2W) * δ_static.3) Therefore peak force = 2W and peak stress = 2 * (W/A) = 2 * σ_static.Verification / Alternative check:From vibration solution of a single-degree-of-freedom system with step load, the displacement overshoots to 2δ_static at first peak, confirming a factor of 2 in stress.Why Other Options Are Wrong:0.5 or 1 underestimates dynamic amplification.3 or 4 overestimates; requires additional kinetic energy (true impact with velocity), not a mere sudden step.Common Pitfalls:Confusing “sudden” with “impact with velocity”; the factor exceeds 2 only when kinetic energy is added.Comparing average stresses instead of peak stresses.Final Answer:2.

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