Question 1

Simply Supported Rectangular Beam — Deflection under Central Point Load A rectangular beam of span l, simply supported at both ends, carries a central point load W. Where along the beam does the maximum deflection occur?

Accepted Answer

Introduction: Deflection shapes provide insight into stiffness and serviceability. For statically determinate beams with symmetric loading, the location of maximum deflection can often be inferred from symmetry and slope conditions without solving the full differential equation. Given Data / Assumptions: Simply supported beam, span l. Central point load W. Prismatic beam; linear elastic bending; small deflection theory. Concept / Approach: The loading and boundary conditions are symmetric about midspan, so the deflected shape is symmetric and slope is zero at the center. The point of zero slope is typically an extremum of deflection, giving the location of maximum deflection at the center. Step-by-Step Solution: Write standard result for midspan deflection: y_max = (W * l^3) / (48 * E * I).Symmetry dictates dy/dx = 0 at x = l / 2.Boundary conditions y(0) = y(l) = 0 at the supports; thus the interior extremum occurs at midspan. Verification / Alternative check: Area-moment method or conjugate-beam method both yield the same midspan location and value. Why Other Options Are Wrong: at the ends: deflection is constrained to zero at simple supports.at l / 3: not an extremum for this loading; applies to other cases (e.g., certain distributed loads) only under special conditions.none of these: incorrect since midspan is correct. Common Pitfalls: Confusing the location of maximum moment (also at midspan here) with other load cases where it shifts. Final Answer: at the centre

Question 2

Helical Springs — Effect of Cutting into Two Halves A closely coiled helical spring is cut into two equal halves. How does the stiffness (spring constant) of each resulting spring compare to the original?

Accepted Answer

Introduction: Spring stiffness determines load–deflection behavior. Understanding how stiffness changes when altering the number of active coils helps when adapting springs to new applications by cutting or combining coils. Given Data / Assumptions: Closely coiled helical spring with wire diameter d, mean coil diameter D, and N active turns. Material shear modulus G unchanged by cutting. Ends of the new springs are finished so that the number of active turns halves approximately. Concept / Approach: The stiffness k of a closely coiled helical spring is k = (G * d^4) / (8 * D^3 * N). With all other parameters unchanged, k is inversely proportional to the number of active turns N. Cutting the spring into two halves halves N and thus doubles k. Step-by-Step Solution: Original stiffness: k_0 = (G * d^4) / (8 * D^3 * N).After cutting: N_new = N / 2.New stiffness: k_new = (G * d^4) / (8 * D^3 * (N / 2)) = 2 * k_0. Verification / Alternative check: Series/parallel analogy: two identical springs in series halve stiffness; conversely, halving coil count in a single spring doubles stiffness. Why Other Options Are Wrong: same: would require N unchanged, which is not the case.half: contradicts inverse proportionality to N.one-fourth: would occur only if N increased fourfold, not halved. Common Pitfalls: Forgetting that wire and mean diameters remain the same after cutting, so only N changes. Final Answer: double

Question 3

Modulus of elasticity (Young's modulus) for mild steel — choose the closest standard value. Give your answer in kN/mm^2 (note: 1 kN/mm^2 = 1 GPa).

Accepted Answer

Introduction: Mild steel (low-carbon steel) is a standard structural material. This question tests recall of its elastic stiffness, commonly used in beam, column, shaft, and plate calculations in civil and mechanical engineering. Given Data / Assumptions: Material: mild (low-carbon) steel. Small-strain, linear-elastic range. Units: kN/mm^2 where 1 kN/mm^2 = 1 GPa. Concept / Approach: Young's modulus E links normal stress and normal strain by Hooke's law: σ = Eε. For steels, E is nearly constant across grades and heat treatments within engineering accuracy, typically around 200 to 210 GPa (i.e., 200–210 kN/mm^2). Step-by-Step Solution: Recognize standard value: E_steel ≈ 200–210 GPaConvert understanding of units: 1 GPa = 1 kN/mm^2Select the closest choice: 210 kN/mm^2 Verification / Alternative check: Handbooks and codes often tabulate E_steel = 2.010^5 to 2.1*10^5 N/mm^2, which equals 200–210 kN/mm^2, confirming the chosen answer. Why Other Options Are Wrong: 10, 25, 80, 100 kN/mm^2: Far lower than accepted values for steel; such stiffnesses are more typical of polymers (very low) or some non-ferrous alloys (still higher than listed low values). Common Pitfalls: Confusing strength (e.g., fy) with stiffness (E); mixing up units such as MPa, GPa, and N/mm^2. Remember: E for steels is relatively invariant compared to strength grades. Final Answer: 210 kN/mm^2

Question 4

Steam generation and pressure vessels — evaluate the statement: “A thick pressure vessel is always used for generating steam because it can withstand high pressures.”

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Introduction: This item checks conceptual understanding of thin vs thick pressure vessel design and the realities of steam generation equipment (boilers and drums). Not every steam generator uses a thick shell; selection depends on diameter, pressure level, and practical fabrication/heat-transfer constraints. Given Data / Assumptions: Comparing “thick” vs “thin” shell criteria. Typical steam drums and boilers operate with diameters where t/D is small. Design must satisfy stress limits and heat-transfer needs. Concept / Approach: Classification often uses t/D thresholds (e.g., thin when t/D is small). Many boiler drums and shells are analyzed with thin-shell formulas because their walls, though substantial in mm, are still “thin” relative to large diameters. Thick cylinders are favored when high pressure and small diameter push t/D into the “thick” regime and stress varies significantly through the wall thickness. Step-by-Step Solution: Identify claim: “always” thick for steam generationCounterpoint: Many boilers use thin-shell design with appropriate thickness, materials, and code safety factorsConclusion: The statement is false because selection depends on t/D and design codes, not the fact of generating steam Verification / Alternative check: Typical water-tube boilers employ drums with large diameters and wall thicknesses small relative to diameter, validating thin-shell assumptions at code-checked thicknesses. Why Other Options Are Wrong: True variants: Overgeneralize; thickness choice is not universal for steam generation. “Thin shells always regardless of pressure”: Also incorrect; very high pressures or small diameters may require thick-cylinder analysis. Common Pitfalls: Equating “withstand high pressure” with “must be thick.” Thickness is set by code equations, diameter, material strength, corrosion allowance, and fabrication—sometimes thin-shell theory remains valid. Final Answer: False

Question 5

Beam classification by supports — name the member: A beam resting on more than two supports over its span is called what?

Accepted Answer

Introduction: Support conditions define common beam types and their analysis methods. Recognizing a beam spanning over multiple supports is foundational for statically indeterminate analysis. Given Data / Assumptions: More than two supports along the beam's length. Standard textbook nomenclature. Concept / Approach: A beam that extends over three or more supports is termed a continuous beam. Internal moments and shears depend on continuity across supports and require compatibility along with equilibrium for solution. Step-by-Step Solution: Identify number of supports: more than twoMatch to standard term: “continuous beam”Conclude correct classification Verification / Alternative check: Indeterminacy check: each additional support beyond two generally introduces indeterminacy requiring slope-deflection, moment distribution, or matrix methods. Why Other Options Are Wrong: Simply supported: exactly two supports without fixity. Fixed beam: both ends fixed, but still two supports. Overhanging: extensions beyond supports, not necessarily >2 supports. Cantilever: fixed at one end only. Common Pitfalls: Confusing “overhanging” with “continuous.” Overhang concerns extension beyond a support, not the number of supports. Final Answer: continuous beam

Question 6

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 7

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 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

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 16

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 17

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 18

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 19

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 20

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

Strength of Materials Questions