Question 1

Euler's column theory — identify the foundational assumptions: Which of the following are assumed in Euler's elastic buckling theory for slender columns?

Accepted Answer

Introduction: Euler's theory predicts the critical load for ideal slender columns. Knowing its assumptions clarifies when the formula Pcr = π^2EI/Le^2 applies. Given Data / Assumptions: Ideal geometry: straight column, prismatic section. Axial load through centroid with negligible initial curvature. Linear elasticity up to buckling. Concept / Approach: The theory presumes elastic behavior and buckling as the governing limit state, valid for high slenderness ratios where inelastic yielding is not reached before instability. Step-by-Step Solution: Recognize slenderness: length >> section sizeAdopt Hooke’s law: stress proportional to strainLimit state: failure by buckling, not crushing or yielding Verification / Alternative check: Columns with low slenderness do not satisfy Euler's assumptions; inelastic or empirical formulas (e.g., Rankine-Gordon) are then used, confirming the boundary of applicability. Why Other Options Are Wrong: “Load acts with large eccentricity”: That violates the central-loading assumption in the classical derivation. Common Pitfalls: Applying Euler's formula to stocky columns; ignoring end conditions when computing effective length Le; assuming plasticity while using elastic theory. Final Answer: all of the above

Question 2

Thin cylindrical shell under internal pressure — stress ratio: For a thin cylinder, what is the ratio of longitudinal stress to hoop (circumferential) stress under pressure p?

Accepted Answer

Introduction: Thin-cylinder theory yields two primary membrane stresses: hoop and longitudinal. Remembering their magnitudes is essential for shell and pressure vessel design. Given Data / Assumptions: Thin shell: t/D small, membrane action. Internal pressure p; diameter D; thickness t. Concept / Approach: Standard thin-cylinder formulas are: σ_hoop = pD/(2t) and σ_long = pD/(4t). The ratio σ_long/σ_hoop = (pD/(4t)) / (pD/(2t)) = 1/2. Step-by-Step Solution: Write hoop stress: σ_h = pD/(2t)Write longitudinal stress: σ_l = pD/(4t)Form ratio: σ_l/σ_h = (pD/(4t)) / (pD/(2t)) = 1/2 Verification / Alternative check: Equilibrium of a longitudinally cut cylinder and a transversely cut cylinder independently lead to the two expressions above, confirming the 1/2 ratio. Why Other Options Are Wrong: 3/4, 1, 1.5, 2: Do not follow from thin-shell equilibrium relations. Common Pitfalls: Confusing which stress is larger; hoop stress is twice the longitudinal stress in thin cylinders. Final Answer: 1/2

Question 3

Mechanical testing — choose the primary material type for compression testing: Compression tests are typically performed on which class of materials?

Accepted Answer

Introduction: Compression and tension tests reveal different failure behaviors. Understanding which materials are best assessed in compression is key to proper characterization. Given Data / Assumptions: Room-temperature, quasistatic tests. Standard specimen geometries. Concept / Approach: Brittle materials (e.g., cast iron, concrete, stone, ceramics) have low tensile ductility and often fail suddenly in tension. Their compressive strength can be significantly higher and measured reliably without the complications of necking that ductile materials exhibit in tension. Step-by-Step Solution: Identify class with limited tensile strain capacity: brittleSelect compression as suitable to avoid tensile cracking and to obtain meaningful strengthChoose “brittle” as the material category primarily tested in compression Verification / Alternative check: Concrete testing protocols (e.g., cube or cylinder compressive tests) exemplify the practice of evaluating brittle materials in compression as a standard index of quality. Why Other Options Are Wrong: Ductile/malleable/plastic: Commonly characterized via tension yielding and elongation; compression may cause barreling and friction effects. Elastomeric: Characterized by large elastic deformations and different test setups. Common Pitfalls: Assuming all materials are best characterized in tension; specimen-end friction and barreling must be controlled in compression tests. Final Answer: brittle

Question 4

Plane stress relations — maximum shear stress: The maximum shear stress in plane stress equals what fraction of the algebraic difference between maximum and minimum normal stresses?

Accepted Answer

Introduction: Mohr's circle for plane stress provides direct relations among principal stresses and maximum shear stress. This question checks your command of that geometric interpretation. Given Data / Assumptions: Plane stress with principal stresses σ1 and σ2. Linear elasticity and small deformations. Concept / Approach: On Mohr's circle, the radius equals (σ1 − σ2)/2 and represents the maximum shear stress τ_max in plane stress. Hence τ_max = (σ_max − σ_min)/2. Step-by-Step Solution: Define principal stresses: σ_max = σ1, σ_min = σ2Mohr’s circle radius: R = (σ1 − σ2)/2Therefore τ_max = R = (σ_max − σ_min)/2 Verification / Alternative check: Stress transformation equations also yield the same result when maximizing shear with respect to angle, confirming the half-difference relationship. Why Other Options Are Wrong: Equal to or twice: Overestimate τ_max. One-fourth or three-fourths: Arbitrary fractions not supported by the transformation equations. Common Pitfalls: Confusing average stress (σ_avg) with radius; mixing up shear in 3D with plane-stress results. Final Answer: one-half

Question 5

Reinforced concrete flexure — over-reinforced section: For an over-reinforced RC beam, is the depth of the actual neutral axis equal to the critical neutral axis depth?

Accepted Answer

Introduction: Neutral axis depth governs strain compatibility and failure mode in RC beams. Over-reinforcement leads to compression-controlled behavior rather than tension steel yielding first. Given Data / Assumptions: Rectangular RC section under flexure. Definitions: critical neutral axis depth x_c separates tension-controlled from compression-controlled behavior. Concept / Approach: In an over-reinforced beam, the actual neutral axis lies deeper (nearer to the compression face) than the critical depth. This leads to crushing of concrete before yielding of tension steel, which is brittle and undesirable per modern codes. Step-by-Step Solution: Define x_c as demarcation based on balanced strain conditionsOver-reinforced: steel area larger than balanced ⇒ NA shifts deeperHence actual neutral axis depth > critical neutral axis depth Verification / Alternative check: Strain diagrams for over-reinforced sections show higher compression block depth before steel yield, confirming a larger NA depth than x_c. Why Other Options Are Wrong: True / equal: Contradicts the definition of over-reinforced behavior. “Actual less than critical”: That corresponds to under-reinforced sections. Insufficient data: Standard definitions are enough to conclude. Common Pitfalls: Confusing under- vs over-reinforced; thinking NA position is fixed regardless of reinforcement amount. Final Answer: False

Question 6

Bending sense and fiber stresses — rectangular beam under positive (sagging) bending: With a sagging bending moment, is the upper layer of the beam in tension?

Accepted Answer

Introduction: Understanding the sign convention and where tension/compression occur under bending is fundamental to beam design and detailing. Given Data / Assumptions: Rectangular prismatic beam. Sagging (positive) bending moment convention. Concept / Approach: For a positive (sagging) moment, the beam curves concave upward. The top fibers shorten (compression) while the bottom fibers lengthen (tension). For a negative (hogging) moment, this reverses: top in tension, bottom in compression. Step-by-Step Solution: Adopt sign convention: sagging positiveFiber strain distribution: linear through depth with neutral axis at centroid for elastic bendingConclusion: upper layer is in compression under sagging; statement is false Verification / Alternative check: Standard bending stress relation σ = M*y/I shows σ sign flips across the neutral axis; with y positive upward and M sagging, σ at top (positive y) is compressive. Why Other Options Are Wrong: True: Contradicts sagging behavior. “True only for hogging”: Correct concept but the statement asked about sagging; thus answer remains false. Depends on shape only / cantilevers only: Fiber stress sense is governed by moment sign, not just section type. Common Pitfalls: Mixing up sagging vs hogging; confusing maximum deflection location (midspan) with the sign of fiber stress at the top surface. Final Answer: False

Question 7

Helical spring stiffness after cutting — closely coiled spring: A closely-coiled helical spring has stiffness k. It is cut into n equal parts (each part having 1/n of the original coils). What is the stiffness of each part?

Accepted Answer

Introduction: Spring stiffness depends strongly on the number of active coils. Cutting a spring changes its coil count and thus its force–deflection characteristics. This is frequently used to tune stiffness in mechanical assemblies. Given Data / Assumptions: Closely coiled helical spring, small pitch angle. Original stiffness: k; total active coils: N. Spring is cut into n equal parts, each with N/n coils. Concept / Approach: For closely coiled springs, stiffness k is inversely proportional to the number of active coils: k ∝ 1/N (with k = Gd^4 / (8D^3N) for round wire, where d is wire diameter, D is mean coil diameter, and G is shear modulus). Reducing N increases k proportionally. Step-by-Step Solution: Original: k ∝ 1/NEach part: N_part = N/n ⇒ k_part ∝ 1/(N/n) = n/NTherefore k_part = nk (since original constant factors are unchanged) Verification / Alternative check: If n = 2, halving coil count doubles stiffness; if n = 4, quarter of coils gives 4k. This matches practical experience and the k ∝ 1/N law. Why Other Options Are Wrong: k/n and k/n^2: Predict decreased stiffness after cutting, contrary to 1/N dependence. n^2k: Overstates the increase; proportionality is linear in n, not quadratic. k: Ignores coil-count effect. Common Pitfalls: Confusing series vs parallel combination ideas with coil-count changes; mixing up wire diameter or coil diameter effects, which remain constant when simply cutting length. Final Answer: n*k

Question 8

Support Conditions and End Moments in Beams Consider a prismatic, linearly elastic, simply supported beam without end restraints (i.e., pin–roller or pin–pin). State whether the bending moment at each support (beam ends) is zero under external trans…

Accepted Answer

Introduction: In elementary structural analysis, the relationship between boundary conditions and internal actions is fundamental. A simply supported beam has supports that do not restrain rotation; therefore, the end moments are generally expected to be zero. This item checks your understanding of how ideal support types (pins/rollers) influence bending moments at the boundaries. Given Data / Assumptions: Beam behavior is linear elastic, small deflection. Supports are ideal: no rotational restraint (pin and/or roller). Loads are transverse (point loads, uniformly distributed loads, etc.). No externally applied end couple at the supports. Concept / Approach: For a simply supported beam, rotations at the supports are free. The internal bending moment is related to curvature and rotational restraint. With free rotation, the boundary bending moment is zero unless an explicit couple is applied at the end. Shear forces at the supports can be non-zero, but bending moment values at those points vanish for the ideal model. Step-by-Step Solution: 1) Identify support type: pin/roller implies zero resisting moment.2) Draw bending moment diagram for common loads (point, UDL): the diagram starts and ends at zero.3) Recognize exceptions only arise if an external end couple is applied (not part of typical “simply supported” definition). Verification / Alternative check: From equilibrium and compatibility, the boundary condition for a free-rotation support is M = 0 at that node, which is consistent with classical solutions and textbook bending moment diagrams. Why Other Options Are Wrong: “No” is incorrect because ideal simple supports cannot sustain end moment. “Only under UDL” is false; point-load cases also show zero end moments. “Only when span is symmetric” is irrelevant; symmetry is not required. “Only when deflection is zero at ends” confuses displacement with rotational restraint. Common Pitfalls: Mixing up shear (which is often maximum at supports) with bending moment, and assuming end fixity where there is none. Final Answer: Yes

Question 9

Kern (Core) of a Circular Section — No-Tension Condition For a short column with a circular cross-section, to avoid tensile stress anywhere at the base, the line of action of the resultant load must lie within a concentric “kern” circle whose diamet…

Accepted Answer

Introduction: The “kern” (or core) of a section is the locus of points through which a compressive resultant can act without causing tension anywhere on the base. For masonry and short columns, keeping the load within the kern ensures uniform compression. This question focuses on the kern size for a circular section. Given Data / Assumptions: Short column (no significant slenderness effects). Linear elastic stress distribution under eccentric compression. Circular cross-section of radius r (main diameter = 2r). Concept / Approach: For a circle, the kern is a concentric circle of radius r/4. If the load resultant stays within this radius from the centroid, the entire base remains in compression. Therefore, the kern diameter equals 2 * (r/4) = r/2. Step-by-Step Solution: Let main radius = r ⇒ main diameter = 2r.Kern radius for a circle = r/4.Kern diameter = 2 * (r/4) = r/2.Fraction of main diameter = (r/2) / (2r) = 1/4.Hence, kern diameter = one-fourth of the main diameter. Verification / Alternative check: Classical results for kerns: rectangle → middle third; circle → concentric circle radius r/4; triangle → centroidal regions defined by medians. The circular case consistently yields r/4 radius. Why Other Options Are Wrong: one-half: This is the kern diameter in absolute units (r/2) but expressed as a fraction of the main diameter (2r) it becomes 1/4, not 1/2. one-third, one-eighth, one-fifth: Do not match the r/4 kern radius result. Common Pitfalls: Confusing kern diameter with kern radius or mixing absolute dimensions with fractional comparisons. Final Answer: one-fourth

Question 10

Plane Stress — Orientation for Maximum Normal Stress under Uniaxial Tension A body is subjected to a direct tensile stress σ in one plane only (σy = 0, τxy = 0). The section on which the maximum normal stress acts is inclined at how many degrees to …

Accepted Answer

Introduction: Understanding stress transformation is critical for failure analysis. Under uniaxial tension, principal stresses occur on specific planes. This question asks for the orientation that produces the maximum normal stress when only one normal stress is present. Given Data / Assumptions: Plane stress, σx = σ (tension), σy = 0, τxy = 0. Linear elasticity and homogeneous material. Concept / Approach: For a plane at angle θ (measured from the x-plane normal), the transformed normal stress is: σn = σx * cos^2θ With no other stresses, σn is maximized when cos^2θ = 1, i.e., θ = 0°, meaning the plane whose normal aligns with the applied stress direction (principal plane). Step-by-Step Solution: Start with σn = σ * cos^2θ.Derivative wrt θ shows maximum at θ = 0°; physically, a plane perpendicular to the load carries full σ.Thus, maximum normal stress occurs on the plane whose normal is colinear with the load, i.e., θ = 0°. Verification / Alternative check: Mohr’s circle degenerates to a point at σ, confirming the single principal stress appears on planes at θ = 0° and 90° (maximum and minimum, respectively). Why Other Options Are Wrong: 45°: Associated with maximum shear under pure shear or combined states, not here. 30°, 60°: No extremum occurs at these angles for σn. 90°: Gives σn = 0 in this case (plane parallel to load). Common Pitfalls: Confusing the angle for maximum shear (often 45°) with the angle for maximum normal stress (0° here). Final Answer: 0°

Question 11

Reinforced Concrete Beams — Under-Reinforced Section and Neutral Axis For an under-reinforced reinforced concrete (RC) beam section, how does the depth of the actual (neutral axis at failure) compare with the critical (balanced) neutral axis depth?

Accepted Answer

Introduction: RC beam behavior depends on the reinforcement ratio relative to a balanced (critical) state. Under-reinforced beams are preferred in design because they provide ductile tension failure. This question probes the relative position of the neutral axis in such sections. Given Data / Assumptions: Beam is singly reinforced and behaves per standard RC theory. Plane sections remain plane; concrete carries no tension at ultimate. Balanced (critical) section has simultaneous concrete compression and steel tension at their limiting strains. Concept / Approach: In an under-reinforced section, steel yields before concrete reaches its crushing strain. To satisfy strain compatibility, the neutral axis lies shallower (closer to the tension face) than in the balanced case, meaning its depth is less than the critical neutral axis depth. Step-by-Step Solution: 1) Define balanced depth: both materials at limit strains.2) With less steel than balanced, tensile steel yields first.3) Strain profile requires a smaller compression block depth.4) Therefore, actual neutral axis depth < critical neutral axis depth. Verification / Alternative check: Design codes and textbook stress–strain blocks show shallower neutral axis for under-reinforced sections, matching observed ductile failure modes. Why Other Options Are Wrong: same as / greater than: Contradict the fundamental definition of under-reinforcement. depends only on cover / indeterminate: Neutral axis location depends on reinforcement ratio and section geometry, not just cover. Common Pitfalls: Confusing under-reinforced with over-reinforced (which has deeper neutral axis and brittle compression failure). Final Answer: less than

Question 12

Beam Classification — Member Extending Beyond Support Lines A beam that projects beyond one or both of its supports (i.e., the span continues past a support) is called:

Accepted Answer

Introduction: Correct terminology for beam types is essential for selecting analysis models and detailing. This question differentiates among common classifications based on support configuration and span geometry. Given Data / Assumptions: Structural member behaves as a beam under transverse loads. Part of the member extends past a support line. Concept / Approach: An overhanging beam is a simply supported beam with one or both ends projecting beyond the supports. This differs from a cantilever (fixed at one end, free at the other) and from a continuous beam (more than two supports within the clear span). Step-by-Step Solution: Identify support types: at least one simple support exists within the member length.Check geometry: the physical length exceeds the distance between the outer supports.Conclude: beam is “overhanging”. Verification / Alternative check: Classical beam catalogs and bending moment diagram shapes for overhangs (with negative end regions) corroborate the definition. Why Other Options Are Wrong: simply supported: does not necessarily project beyond supports. fixed beam: both ends fixed; no overhang implied. cantilever: fixed-free configuration, not supported inside the span. continuous beam: has multiple interior supports, not merely an overhang. Common Pitfalls: Confusing the negative bending region near the overhang with a cantilever; the presence of an internal support distinguishes the two. Final Answer: overhanging beam

Question 13

Transverse Shear in Circular Sections — Maximum vs Average For a beam with a solid circular cross-section under transverse shear, the maximum shear stress is how many times the average shear stress over the section?

Accepted Answer

Introduction: Design under transverse shear requires knowledge of how shear stress distributes across various cross-sections. This question targets the ratio between maximum and average shear stress for a solid circular section. Given Data / Assumptions: Solid circular cross-section of diameter D. Beam is prismatic and linearly elastic. Uniform shear force V across the section. Concept / Approach: The average shear stress is V / A, where A is the cross-sectional area. For a solid circle, the shear stress distribution is parabolic, and the closed-form solution gives τmax = (4/3) * τavg. Step-by-Step Solution: Average shear: τavg = V / A = V / (π * D^2 / 4).Using τ = V * Q / (I * b) with circular geometry yields τmax = 4/3 * τavg.Therefore, ratio = 4/3. Verification / Alternative check: This ratio is a standard result alongside the rectangular section result (τmax = 1.5 * τavg) and thin-walled torsion approximations. Why Other Options Are Wrong: equal to / twice: Do not match the parabolic distribution result. 1.5 times / 3/2 times: This belongs to rectangular sections, not solid circles. Common Pitfalls: Interchanging the circular and rectangular ratios or assuming a uniform shear distribution. Final Answer: 4/3 times

Question 14

Flitched (Composite Timber–Steel) Beams — Purpose in Practice A flitched beam is formed by fastening steel plates to the sides of a timber beam to act compositely. The primary purpose of using a flitched beam is to:

Accepted Answer

Introduction: Flitched beams combine materials (typically timber and steel) so that they share bending stresses. Understanding their purpose helps in economical and practical strengthening of existing members without increasing depth substantially. Given Data / Assumptions: Steel plates are rigidly connected to timber so strains are compatible. Composite action is achieved (no slip at the interface). Analysis uses modular ratio to transform areas. Concept / Approach: By adding high-modulus steel plates, the effective section modulus and second moment of area increase. In transformed section terms, this is equivalent to increasing the beam’s effective cross-section, enhancing bending capacity and reducing deflection for the same overall depth. Step-by-Step Solution: 1) Compute transformed steel area = n * A_steel (n = Es / Etimber).2) Add to timber area to get A_eff and I_eff.3) Increased I reduces curvature (M / E I) and deflection; increased Z = I / y enhances bending strength. Verification / Alternative check: Strengthening projects often use side plates or flitch plates to meet moment/deflection limits when depth increase is constrained. Why Other Options Are Wrong: change the shape: incidental, not the goal. saving in material: capacity gain, not primarily less material. equalise tension/compression strength: not the standard objective in timber–steel composites. reduce self-weight: steel increases weight slightly; effect is not significant. Common Pitfalls: Assuming flitching is only for tension side; in practice, symmetric side plates are common to maximize stiffness. Final Answer: increase the cross-section of the beam

Question 15

Shear Force at the Free End — Cantilever with Uniformly Distributed Load A cantilever beam of length l carries a uniformly distributed load w per unit length over the entire span. The shear force at the free end is:

Accepted Answer

Introduction: Shear force and bending moment diagrams for basic loading cases are core knowledge in strength of materials. For a cantilever with a uniformly distributed load (UDL), recognizing boundary values helps draw accurate diagrams quickly. Given Data / Assumptions: Cantilever span length = l, fixed at one end and free at the other. Uniformly distributed load w acts over the entire length. Linearly elastic, small deflection assumptions. Concept / Approach: Shear force is the integral of load intensity, and the bending moment gradient equals shear. At the free end of a cantilever, there is no support reaction to carry shear; thus the internal shear at that section must be zero. Maximum shear occurs at the fixed support where reactions develop. Step-by-Step Solution: 1) Resultant of UDL = w * l acting at midspan from the free end.2) Shear diagram for a cantilever with UDL starts at V = w * l at the fixed end and linearly decreases to V = 0 at the free end.3) Therefore, shear at the free end is zero. Verification / Alternative check: From boundary conditions, the free end cannot develop internal shear or moment; both must be zero at that free boundary. Why Other Options Are Wrong: w l, w l / 2, w l / 4, 2 w l: These are nonzero values that contradict the cantilever free-end condition. Common Pitfalls: Confusing the fixed-end values (nonzero) with the free-end values (zero for both V and M). Final Answer: zero

Question 16

In mechanical design of suspension systems, a semi-elliptic leaf spring (laminated spring) is ordinarily arranged as a simply supported member. Identify how it is supported and where the service load is applied in standard automotive and machine app…

Accepted Answer

Introduction: Leaf springs (semi-elliptic/laminated) are widely used in vehicle suspensions to absorb shocks and support loads. Understanding the support and loading configuration is fundamental to predicting bending moments, stresses, and deflections for safe design. Given Data / Assumptions: A semi-elliptic leaf spring consisting of stacked leaves. Standard road vehicle arrangement. Small deflection theory and linear elastic behavior for analysis. Load comes from the axle or shackle assembly. Concept / Approach: The spring acts like a simply supported beam. The two eyes at the ends are connected to the chassis via hangers/shackles (supports), while the axle load is transferred near the mid-span through a U-bolt or seat. This support–load configuration dictates a triangular bending-moment diagram with maximum moment at mid-span. Step-by-Step Solution: 1) Identify supports: the two eyes form the supports at the ends.2) Identify load application: axle clamp/U-bolt applies load near the geometric centre.3) Structural model: simply supported beam with a central point load.4) Result: maximum bending at the mid-span, consistent with standard leaf spring stress formulae. Verification / Alternative check: Vehicle manuals and standard machine design texts model the semi-elliptic leaf spring as simply supported at the eyes with the axle load at the centre; derived formulas for stress and deflection assume this configuration. Why Other Options Are Wrong: Centre and loaded at the ends: reverses the actual arrangement. Ends and loaded anywhere: design assumes a defined mid-span load for predictable stress. Centre and loaded anywhere: physically incorrect for standard mounts. Cantilevered at one end: describes quarter-elliptic springs, not semi-elliptic. Common Pitfalls: Confusing semi-elliptic with quarter-elliptic/cantilever arrangements; assuming distributed loads instead of the central axle load in first-order models. Final Answer: ends and loaded at the centre

Question 17

Two prismatic beams of equal cross-sectional area—one with a circular cross-section and the other with a square cross-section—are subjected to the same bending conditions. Considering section modulus (economy in bending), which beam is more economic…

Accepted Answer

Introduction: Economy in bending for a given cross-sectional area is judged by the section modulus, Z = I / y_max. A larger Z yields lower bending stress for the same bending moment. This question compares circular and square cross-sections of equal area to determine which is more efficient in bending. Given Data / Assumptions: Equal cross-sectional areas: A_square = A_circle. Homogeneous, isotropic, linearly elastic material. Same loading and span so bending moment demand is identical. Concept / Approach: For a square of side a: I = a^4 / 12 and y_max = a / 2, hence Z_square = a^3 / 6. For a circle of diameter d: I = (π * d^4) / 64 and y_max = d / 2, hence Z_circle = (π * d^3) / 32. With equal areas, a^2 = (π * d^2) / 4, so a = (d * √π) / 2. Substituting shows Z_square > Z_circle, i.e., the square section is more economical in bending. Step-by-Step Solution: 1) Area equality: a^2 = π d^2 / 4 ⇒ a = (d √π) / 2.2) Z_square = a^3 / 6 = ((d √π) / 2)^3 / 6.3) Z_circle = (π d^3) / 32.4) Ratio Z_square / Z_circle > 1 for all positive d, confirming the square is more efficient. Verification / Alternative check: Numerical substitution (e.g., d = 1) yields Z_square ≈ 1.182 * Z_circle, validating the theoretical comparison. Why Other Options Are Wrong: Circular more economical: contradicted by computed Z. Equally economical / none / cannot be compared: incorrect because Z permits a direct, objective comparison when areas are equal. Common Pitfalls: Comparing moments of inertia alone without dividing by y_max; ignoring the equal-area constraint. Final Answer: square beam is more economical

Question 18

In a reinforced cement concrete (R.C.C.) beam subjected to positive bending (sagging), where are the main longitudinal steel bars embedded within the beam depth for efficient resistance to tension?

Accepted Answer

Introduction: R.C.C. beams rely on composite action between concrete (strong in compression) and steel (strong in tension). For sagging bending, tension occurs at the beam's soffit, dictating the placement of main reinforcement. Given Data / Assumptions: Simply supported or continuous beam segments in positive bending. Plane sections remain plane; perfect bond between steel and concrete. Service loading within elastic behavior for basic placement rationale. Concept / Approach: Under sagging, the bottom fibers are in tension and the top fibers in compression. Concrete is weak in tension and is therefore relieved by embedding tensile steel near the tension zone (bottom). Compression is resisted primarily by concrete at the top, sometimes aided by compression steel in special cases. Step-by-Step Solution: 1) Determine bending sign: sagging ⇒ tension at bottom.2) Provide main longitudinal steel near the bottom cover region.3) Use stirrups for shear and to hold longitudinal bars; compression zone near the top resists compressive stresses.4) Ensure adequate cover for durability and fire resistance per code. Verification / Alternative check: Negative bending (over supports) reverses stress zones; then top steel becomes main tension reinforcement. This confirms placement follows the tension zone. Why Other Options Are Wrong: Near the top: correct only for negative bending zones. Centre / any position / random: ineffective in tension control and violates code-based detailing practices. Common Pitfalls: Placing bars symmetrically without considering bending sign; ignoring cover and anchorage requirements. Final Answer: near the bottom

Question 19

A long, slender column under axial compressive load tends to buckle, generating significant bending stresses. Compare the magnitude of direct (uniform) compressive stress with bending stress at critical sections for a long column.

Accepted Answer

Introduction: Column behavior transitions from primarily direct compression in short columns to flexural (buckling) behavior in long, slender columns. This question examines the comparative significance of direct compressive stress versus bending stress in long columns near buckling conditions. Given Data / Assumptions: High slenderness ratio; Euler-type behavior. Axial load P acting through a small eccentricity or induced by imperfections. Linear elastic analysis near critical load for conceptual comparison. Concept / Approach: Total stress at any fiber is sigma = P/A ± My/I. In long columns approaching buckling, the bending moment M from lateral deflection grows rapidly compared to the uniform term P/A, making the bending component dominant in extreme fibers. Thus, direct stress becomes relatively insignificant (negligible) compared to bending stress at critical locations. Step-by-Step Solution: 1) Write sigma_total = P/A ± My/I.2) For long columns, lateral deflection increases ⇒ M increases sharply.3) Near failure, My/I » P/A ⇒ bending governs design and stress checks.4) Hence direct stress is negligible when assessing extreme fiber stresses. Verification / Alternative check: Euler theory P_cr = π^2 * E * I / L_e^2 shows sensitivity to L_e; large L_e implies low critical load and large curvature, reinforcing bending dominance. Why Other Options Are Wrong: Less/same/more: do not capture the typical dominance of bending in long columns. Cannot be compared: they can be compared through sigma = P/A ± My/I. Common Pitfalls: Applying short-column formulas to slender columns; ignoring initial imperfections that trigger significant bending moments. Final Answer: negligible

Question 20

In working-stress/limit-state theory for reinforced cement concrete (R.C.C.) beams, which basic assumptions are adopted for analysis and design of a singly reinforced, balanced section?

Accepted Answer

Introduction: Design of R.C.C. beams rests on standard assumptions that enable a simple yet reliable stress–strain model for composite action. Recognizing these assumptions is essential for computing neutral axis depth, internal forces, and moment of resistance. Given Data / Assumptions: Plane sections remain plane after bending. Perfect bond between steel and concrete. Concrete resists compression; steel resists tension. Working-stress or compatible limit-state approach within design limits. Concept / Approach: In singly reinforced beams under sagging, concrete's tensile capacity is neglected; tension is carried by steel. Adequate bond ensures strain compatibility, allowing steel and adjacent concrete to have the same strain at a level. Within the design range, stresses are kept within permissible/limit values to ensure serviceability and safety. Step-by-Step Solution: 1) Assume zero tensile strength of concrete in bending → tension carried by steel.2) Assume perfect bond → linear strain distribution; use modular ratio or code-defined stress blocks.3) Compute internal compression in concrete and tension in steel, enforce C = T.4) Determine lever arm and moment capacity accordingly. Verification / Alternative check: Design results align with code provisions (e.g., stress blocks, strain limits) that inherently rely on these assumptions. Why Other Options Are Wrong: Any single statement alone is incomplete; R.C.C. analysis needs all three simultaneously. None of the above contradicts established R.C.C. theory. Common Pitfalls: Counting on concrete tension in service; ignoring bond quality and strain compatibility; exceeding strain limits. Final Answer: all of the above

Strength of Materials Questions