Question 1

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…

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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 2

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…

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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 3

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?

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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 4

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.

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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 5

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?

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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

Question 6

For ductile engineering materials (e.g., mild steel), compare the tensile strength with the compressive strength under standard testing conditions.

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Introduction: Ductile materials exhibit significant plastic deformation before fracture. A common design heuristic is that, unlike brittle materials, ductile metals show approximately similar resistance in tension and compression at ultimate strength levels, which informs failure theories and safety checks. Given Data / Assumptions: Material class: ductile metals (e.g., low-carbon/mild steel). Standardized coupon/compression tests without instability issues. Homogeneous, isotropic behavior for first-order comparison. Concept / Approach: For ductile metals, yielding is governed by shear-dominated criteria (e.g., distortion energy), and ultimate strengths in tension and compression are of similar magnitude. In practice, compressive strength can be slightly higher due to suppression of necking and different instability modes, but the widely taught simplification is that they are taken as approximately equal for basic comparisons. Step-by-Step Solution: 1) Review tensile test results: UTS determined from maximum load/area.2) Review compression test: plastic flow without a brittle break; large strains.3) Compare: for ductile metals, ultimate strengths are close; design texts often treat them as equal in concept questions.4) Conclude: tensile strength ≈ compressive strength for ductile materials. Verification / Alternative check: Reference failure theories (e.g., von Mises) emphasize shear energy, supporting near-symmetry of strength in tension and compression for ductile metals. Why Other Options Are Wrong: Less than / greater than: oversimplify and can mislead; equality is the standard teaching assumption. Cannot be generalized / depends only on geometry: material behavior, not just geometry, controls strength. Common Pitfalls: Confusing ductile with brittle behavior (e.g., concrete, cast iron), where compressive strength is far greater than tensile strength. Final Answer: equal to

Question 7

A thin cylindrical shell provided with hemispherical end-caps is subjected to the same internal pressure and material allowables as the spherical ends. Compare the required wall thickness of the cylindrical shell with that of the hemispherical ends.

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Introduction: Pressure vessels commonly use cylindrical shells with hemispherical ends. For the same internal pressure and allowable stress, the required thickness differs for cylinder versus sphere due to different membrane stress distributions. Given Data / Assumptions: Same internal pressure p and radius r. Same allowable tensile stress sigma_t. Thin-shell (membrane) theory; negligible bending. Concept / Approach: For a thin cylinder: hoop stress sigma_h = p * r / t ⇒ t_cyl = p * r / sigma_t. For a thin sphere: membrane stress sigma_sph = p * r / (2 * t) ⇒ t_sph = p * r / (2 * sigma_t). Thus, for identical p, r, and sigma_t, t_cyl = 2 * t_sph. Hence, the cylindrical shell requires more thickness than the hemispherical ends. Step-by-Step Solution: 1) Write cylinder thickness: t_cyl = p * r / sigma_t.2) Write sphere thickness: t_sph = p * r / (2 * sigma_t).3) Compare: t_cyl / t_sph = 2 ⇒ t_cyl is double.4) Conclude: cylinder requires more thickness. Verification / Alternative check: Design codes and handbooks consistently show spherical shells as the most material-efficient shape for internal pressure, validating the relation. Why Other Options Are Wrong: Equal to / less than: contradict membrane theory. Indeterminate / depends only on joint efficiency: primary dependence is on geometry and membrane equations; joint efficiency modifies required t but does not invert the basic relation. Common Pitfalls: Using longitudinal stress instead of hoop stress for cylinders; overlooking the factor of 2 between cylinder and sphere equations. Final Answer: more than

Question 8

The shear strain energy stored in a solid body under a state of pure shear is to be expressed in terms of shear stress τ, shear modulus C (modulus of rigidity, G), and volume V. Select the correct expression for the total strain energy due to shear.

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Introduction: Strain energy represents recoverable elastic energy stored in a body under loading. Under pure shear, this energy depends on shear stress, shear strain, and the material's shear modulus. The correct expression is essential for impact, resilience, and torsion problems. Given Data / Assumptions: Uniform shear stress τ in the body. Linear elastic behavior with shear modulus C (G). Body volume V. Concept / Approach: For linear elasticity, energy density u = 0.5 * stress * strain. In shear: u = 0.5 * τ * γ. Since γ = τ / C, substitute to get u = 0.5 * τ * (τ / C) = τ^2 / (2C). Total energy U = u * V = (τ^2 * V) / (2C). Step-by-Step Solution: 1) Start with u = 0.5 * τ * γ.2) Use γ = τ / C for linear shear.3) Obtain u = τ^2 / (2C).4) Multiply by volume: U = u * V = (τ^2 * V) / (2 * C). Verification / Alternative check: Dimensional check: τ in N/m^2, C in N/m^2, so τ^2/C has units of N/m^2; multiplying by V (m^3) gives N·m (energy), which is consistent. Why Other Options Are Wrong: (τ * V) / C and (τ * V) / (2C): miss the quadratic dependence on τ for energy. (τ^2 * V) / C: missing the 1/2 factor required by energy derivation. (τ^2) / (2 C V): incorrect placement of V in the denominator. Common Pitfalls: Forgetting that energy scales with the square of stress in linear elasticity; omitting the 1/2 factor. Final Answer: U = (τ^2 * V) / (2 * C)

Question 9

State the limit up to which Hooke's law (stress is directly proportional to strain) is valid for a typical ductile metal tested in tension under standard laboratory conditions.

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Introduction: Hooke's law states that stress is proportional to strain (σ ∝ ε) in the initial portion of a material's stress–strain curve. Knowing the boundary of this linear relationship is crucial for elastic analysis and modulus determination. Given Data / Assumptions: Metallic specimen tested in monotonic tension. Quasi-static loading; room temperature. Linearly elastic behavior at small strains. Concept / Approach: The linear proportionality holds only up to the material's proportional limit. In many educational contexts and basic MCQs, the proportional limit is taken as very close to the elastic limit; hence Hooke's law is said to be valid up to the elastic limit for practical purposes. Beyond this, nonlinearity and yielding begin, invalidating σ ∝ ε. Step-by-Step Solution: 1) Identify linear region on the stress–strain curve.2) Note the end of linearity ≈ proportional limit.3) Elastic limit closely coincides with proportional limit in many metals; MCQ convention uses "elastic limit".4) Therefore, Hooke's law validity is taken up to the elastic limit for typical problems. Verification / Alternative check: Laboratory curves show deviation from linearity just before yield; the elastic domain ends at the elastic limit, consistent with Hooke's law applicability. Why Other Options Are Wrong: Yield point: plastic flow begins; σ–ε is no longer linear. Plastic limit / breaking point: far beyond elastic behavior. Creep limit: relates to time-dependent deformation, not monotonic linearity. Common Pitfalls: Confusing proportional limit with yield point; using Hooke's law in plastic range. Final Answer: elastic limit

Question 10

Identify the correct S.I. unit for normal stress (and shear stress) used in solid mechanics and materials engineering.

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Introduction: Stress is force per unit area. In the International System of Units (S.I.), it must be expressed using S.I. coherent units to ensure consistency in calculations, reporting, and code compliance. Given Data / Assumptions: Stress σ or τ defined as Force / Area. Force in Newtons (N); area in square metres (m^2). Use of S.I. coherent units. Concept / Approach: The S.I. coherent unit of stress is N/m^2, named the Pascal (Pa). Engineering practice often uses derived multiples such as MPa where 1 MPa = 1 N/mm^2 = 10^6 N/m^2. However, the base S.I. expression remains N/m^2 (Pa). Step-by-Step Solution: 1) Formula: stress = Force / Area.2) Substitute S.I. units: N / m^2 = N/m^2.3) Recognize naming: 1 Pa = 1 N/m^2; 1 MPa = 10^6 Pa = 1 N/mm^2.4) Conclude the correct S.I. unit is N/m^2 (Pa). Verification / Alternative check: Dimensional analysis confirms: [N] = kg·m/s^2; dividing by m^2 gives kg/(m·s^2), consistent with pressure/stress units. Why Other Options Are Wrong: N/mm^2: acceptable engineering unit (MPa) but not the base S.I. expression. kN/mm^2: equals GPa order (10^9 N/m^2) and is rarely used; not the base S.I. unit. Any one of these: incorrect because only N/m^2 (Pa) is the S.I. coherent unit. Common Pitfalls: Mixing base S.I. units with convenient engineering multiples; misreporting units in calculations. Final Answer: N/m2

Question 11

Shear force diagram (SFD): identifying load types by diagram jumps When there is a sudden increase or decrease (a vertical jump) in the SFD at a section between two adjacent points along a beam, it indicates the presence of what kind of loading at t…

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Introduction / Context: Engineers diagnose the type and location of loads on a beam by interpreting the shear force diagram (SFD) and bending moment diagram (BMD). A key signature is how the SFD changes from point to point: vertical jumps, straight sloping lines, or curves each correspond to different loading conditions. Given Data / Assumptions: Prismatic beam under transverse loading. Small deflection theory and linear elasticity. Standard sign conventions for shear and moment. Concept / Approach: Differential relations link load intensity w(x), shear V(x), and bending moment M(x): dV/dx = -w(x) and dM/dx = V(x). A concentrated point load is mathematically represented by an impulse (Dirac delta) in w(x), which produces a finite jump in V but leaves M continuous (unless a couple is applied). Step-by-Step Solution: If w(x) = 0 over an interval, then dV/dx = 0 ⇒ V is constant (a horizontal segment in SFD).If w(x) = constant ≠ 0 (UDL), then V varies linearly (a straight sloped line in SFD).If w(x) varies linearly (UVL), then V is quadratic (a curve in SFD).If a point load P acts at x = a, V has a sudden jump of magnitude P at x = a, while M remains continuous there.Therefore, a vertical jump between two adjacent plotted points of the SFD indicates a point load at that section. Verification / Alternative check: Integrate w(x) over an infinitesimal interval around the suspected section: ∫ w dx = P produces ΔV = -P (sign per convention). That finite ΔV confirms a point load. Why Other Options Are Wrong: No loading gives a flat SFD segment, not a jump. UDL gives linear variation, not a sudden jump. UVL gives a curved SFD. A pure couple (moment) causes a jump in the BMD, not in the SFD. Common Pitfalls: Confusing the jump in SFD (point load) with the jump in BMD (applied couple); misreading plotted points spaced too far apart and missing the exact load location. Final Answer: A concentrated (point) load acting at that section

Question 12

Axial elongation: uniformly tapering circular bar vs uniform bar The extension of a circular bar tapering uniformly from diameter d1 at one end to diameter d2 at the other end, under axial load P, is ______ the extension of a uniform circular bar ha…

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Introduction / Context: Bars with varying cross-section are common in weight-optimized designs. For an axially loaded, uniformly tapering circular bar, engineers often compare its elongation with that of a uniform bar to develop quick checks and design equivalences. Given Data / Assumptions: Prismatic bar of length L with circular cross-section tapering linearly from d1 to d2. Axial load P within elastic limit; Young’s modulus E is constant. Comparison bar has constant diameter equal to √(d1 * d2) and the same length L and load P. Concept / Approach: Differential elongation is dδ = (P dx)/(E A(x)). For a taper, area A(x) varies with diameter d(x). Integrating along the length yields a closed-form expression. A well-known result is that a linearly-tapered circular bar elongates by the same amount as a uniform bar whose diameter equals the geometric mean of the end diameters. Step-by-Step Solution: For taper: d(x) varies linearly, so A(x) = π [d(x)]^2 / 4.δ_taper = ∫[0→L] P / (E A(x)) dx = (4 P / (π E)) ∫[0→L] 1 / [d(x)]^2 dx.Carrying out the integral for linear diameter variation gives δ_taper = (4 P L) / (π E d1 d2).For uniform bar with diameter d_u = √(d1 d2): δ_uniform = (4 P L) / (π E d_u^2) = (4 P L) / (π E d1 d2).Therefore δ_taper = δ_uniform, i.e., the extensions are equal. Verification / Alternative check: Dimensional consistency and limiting checks (d1 = d2 ⇒ both reduce to the same uniform-bar formula) confirm the equality. Why Other Options Are Wrong: Less than/greater than contradict the closed-form integral. “Indeterminate” is incorrect because sufficient information is given. “Zero” would require P = 0 or infinite stiffness, neither applicable. Common Pitfalls: Using arithmetic mean instead of geometric mean; integrating area rather than its reciprocal; forgetting that E cancels in the comparison. Final Answer: equal to

Question 13

Leaf springs: do all leaves have the same length? In a typical semi-elliptic (laminated) leaf spring, are all leaves equal in length?

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Introduction / Context: Laminated (semi-elliptic) leaf springs consist of multiple steel plates (leaves) stacked with a central band or clamp. Their geometry and length distribution are chosen to manage stress uniformity, deflection, and packaging constraints in vehicles and machinery. Given Data / Assumptions: Semi-elliptic leaf spring assembly with a master (full-length) leaf and several shorter leaves. Leaves are pre-cambered and clamped together. Concept / Approach: To achieve near-uniform stress and practical assembly, a semi-elliptic spring uses one or two full-length leaves and multiple graduated leaves of decreasing length toward the ends. This graded length distribution helps distribute load and reduces stress concentration at eye ends. Step-by-Step Solution: Identify the master leaf (full length with eyes).Recognize additional full-length (secondary) leaf may be included for strength.Remaining leaves are progressively shorter (graduated), not equal in length.Thus the statement “all leaves are equal in length” is false for standard laminated springs. Verification / Alternative check: Manufacturer drawings show staggered leaf tips; textbooks depict graduated leaves creating a near-parabolic thickness profile. Why Other Options Are Wrong: “True” contradicts common construction; the qualifiers are either unrelated or already restate the falsehood. The more precise reason is that graduated leaves are shorter. Common Pitfalls: Confusing mono-leaf composite springs (single plate) with laminated designs; assuming equal-length leaves due to simplified sketches. Final Answer: False

Question 14

Material properties under deformation The ratio of lateral strain to linear (longitudinal) strain for an axially loaded specimen is called:

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Introduction / Context: When a bar is stretched axially, it becomes longer in the loading direction and thinner in the lateral directions. The coupling between these strains is captured by a fundamental material constant widely used in stress analysis and finite element modeling. Given Data / Assumptions: Homogeneous, isotropic, linear elastic material. Small strains so linear relations apply. Concept / Approach: Poisson’s ratio, denoted by ν (nu), is defined as ν = |lateral strain| / |longitudinal strain| (signs differ but magnitude is used in many definitions). It ties together axial and transverse deformations and appears in relationships among E (modulus of elasticity), G (modulus of rigidity), and K (bulk modulus). Step-by-Step Solution: Under uniaxial stress, longitudinal strain ε_l = σ / E.Transverse (lateral) strain ε_t = -ν * σ / E.Therefore |ε_t| / |ε_l| = ν, which is Poisson’s ratio. Verification / Alternative check: Use constitutive relation: E = 2 G (1 + ν) and K = E / [3 (1 - 2 ν)], confirming ν’s central role. Why Other Options Are Wrong: Modulus of elasticity measures stiffness in tension/compression; modulus of rigidity pertains to shear; bulk modulus to volumetric stiffness; thermal expansion coefficient links strain to temperature, not axial-lateral coupling. Common Pitfalls: Ignoring negative sign for lateral strain (contraction under tension); assuming ν is constant for all states (it is material dependent, may vary with plasticity). Final Answer: Poisson's ratio

Question 15

Structural members – what loading defines a strut? In a frame structure, a strut is a load-carrying member that is primarily subjected to which type of loading?

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Introduction / Context: Terminology distinguishes structural members by their primary internal force resultants: ties resist tension, struts resist compression, beams resist bending, and shafts resist torsion. Correct identification is essential for design checks such as buckling versus yielding. Given Data / Assumptions: Member is straight and slender compared to its cross-section. Ends provide axial alignment; eccentricities are small. Loads are primarily along the member’s longitudinal axis. Concept / Approach: A strut is a compression member that carries load principally through axial compressive stress. In slender struts (columns), stability (buckling) often governs design, whereas in stocky struts, material crushing may control. The term applies across materials: steel, timber, reinforced concrete, aluminum, and composites. Step-by-Step Solution: Identify the defining internal force → axial compression.Exclude other primary actions: tension (tie), bending (beam), torsion (shaft).Thus, “axial compressive loads” is the correct characterization. Verification / Alternative check: Design codes classify compression members (struts/columns) and prescribe slenderness limits, effective length factors, and buckling curves specifically for axial compression. Why Other Options Are Wrong: Tension defines a tie/rod, not a strut. Torsion and transverse loads describe shafts or beams, not struts in their primary role. Combined actions may occur in practice, but the defining action of a strut is axial compression. Common Pitfalls: Ignoring small eccentricities that can introduce bending; in design, check combined compression and bending as required, but the member is still termed a strut. Final Answer: Axial compressive loads

Question 16

Axial deformation of a steel rod – compute the required tensile force A steel rod 1000 mm long and 12 mm in diameter must elongate by 0.045 mm. If E = 2 × 10^6 kg/cm^2, what tensile force is required?

Accepted Answer

Introduction / Context: Axial deformation under load for a prismatic bar is a fundamental strength-of-materials calculation. It is frequently used to size rods, bolts, and tendons for specified elongations or to check stresses from measured strains. Given Data / Assumptions: Length L = 1000 mm = 100 cm. Diameter d = 12 mm = 1.2 cm. Required elongation Δ = 0.045 mm = 0.0045 cm. Modulus of elasticity E = 2 × 10^6 kg/cm^2. Linear elastic behavior; uniform cross-section and uniaxial loading. Concept / Approach: For an axially loaded bar, the relation between load P and elongation Δ is Δ = (P * L) / (A * E) where A is the cross-sectional area. Solve for P after converting all quantities into consistent units (centimeters here). Step-by-Step Solution: Compute area A = (π/4) * d^2 = (π/4) * (1.2)^2 cm^2.A = π * 0.36 ≈ 1.131 cm^2.Rearrange formula: P = (Δ * A * E) / L.Insert values: P = (0.0045 cm * 1.131 cm^2 * 2 × 10^6 kg/cm^2) / 100 cm.Evaluate numerator: 1.131 * 2 × 10^6 = 2.2619 × 10^6; multiply by 0.0045 → ≈ 10,178.8.Divide by 100 → P ≈ 101.8 kg (round to 102 kg). Verification / Alternative check: Back-calculate Δ using P = 102 kg: Δ ≈ (102 * 100) / (1.131 * 2 × 10^6) ≈ 0.0045 cm = 0.045 mm, confirming consistency. Why Other Options Are Wrong: 166 kg and 204 kg overpredict the load (would produce larger elongations). 74 kg underpredicts the load. 128 kg is also inconsistent with the computed requirement. Common Pitfalls: Mixing units (mm vs cm) or using area in mm^2 with E in kg/cm^2; always convert to a consistent set before substitution. Final Answer: 102 kg

Question 17

Time-dependent deformation under sustained load – what is it called? The slow, continuous extension of a material under a constant applied load, increasing with time, is known as:

Accepted Answer

Introduction / Context: Many engineering materials exhibit time-dependent deformation under sustained stress or load. Recognizing the correct term and its implications is essential for long-term serviceability checks in structures, pavements, pressure vessels, and high-temperature components. Given Data / Assumptions: Load is essentially constant over time (dead load). Temperature and moisture may influence the rate but not the basic definition. Elastic strain occurs immediately; additional strain accumulates gradually. Concept / Approach: Creep is the progressive, time-dependent increase in strain at a constant stress level. It occurs at room temperature in materials like concrete, polymers, and wood, and at elevated temperatures in metals (e.g., turbines, boilers). Yielding is an instantaneous plastic deformation when stress exceeds yield strength; fracture is sudden failure; stress relaxation is the decrease of stress under constant strain; work hardening refers to strengthening due to plastic deformation. Step-by-Step Solution: Identify the loading condition: constant load → constant stress (approximately) in many applications.Observe deformation behavior: strain increases with time under unchanged load → definition of creep.Match with terminology: select “Creep”. Verification / Alternative check: Concrete codes include creep coefficients to predict long-term deflection; metal creep curves show primary, secondary, and tertiary creep stages under constant stress and temperature. Why Other Options Are Wrong: Yielding is immediate under rising load, not time-dependent at constant load. Fracture is a failure mode, not necessarily time-dependent under constant sub-yield load. Stress relaxation assumes constant strain, not constant stress. Work hardening describes material evolution under plastic strain cycling, not time-dependent extension under constant load. Common Pitfalls: Confusing creep with shrinkage (moisture-related volume change) in concrete; creep requires sustained stress, while shrinkage occurs without load. Final Answer: Creep

Question 18

In strength of materials: Which named law states that, within the elastic limit of a material, the strain produced is directly proportional to the stress producing it (i.e., stress ∝ strain)?

Accepted Answer

Introduction / Context: Hooke's law is a foundational concept in strength of materials and structural engineering. It links stress (internal force per unit area) to strain (deformation per unit length) for materials behaving elastically. Engineers use this relationship to size members, estimate deflections, and ensure serviceability before yielding occurs. Given Data / Assumptions: Material behavior considered is within the elastic limit. Loading is such that linear elastic response is valid. Temperature and time-dependent (creep/relaxation) effects are neglected. Concept / Approach: Hooke's law states stress = E * strain for axial loading, where E is Young's modulus. In shear, tau = G * gamma, with G as shear modulus. The proportionality is valid only up to the proportional limit; beyond this, nonlinearities and plasticity occur. Step-by-Step Solution: Identify the statement: “within elastic limits, strain is proportional to stress.”Recall the named law matching this statement: Hooke's law.Therefore, select “Hooke's law.” Verification / Alternative check: Tensile test curves show a straight initial segment where stress/strain is constant (slope = E). This region corresponds to Hookean behavior. Why Other Options Are Wrong: Poisson's law: relates lateral and longitudinal strains via Poisson's ratio, not stress-proportionality.Bernoulli's law: fluid dynamics principle, not solids.“Stress law”/“None”: not standard names for this relationship. Common Pitfalls: Assuming proportionality holds beyond yield or during large deformations; ignoring effects like temperature, strain-rate, or viscoelasticity. Final Answer: Hooke's law.

Question 19

Axial tension and inclined plane: A prismatic member carries tensile force P over a normal cross-sectional area A (uniform). What is the normal stress on a plane inclined at angle θ to the transverse (normal) cross-section?

Accepted Answer

Introduction / Context: Stress transformation on inclined planes is a key topic in mechanics of materials. Even under simple axial tension, planes cut at an angle experience different normal and shear components of stress than the basic P/A on the transverse plane. Given Data / Assumptions: Member under axial tensile force P. Uniform normal cross-sectional area A. Plane considered is inclined at angle θ to the transverse (perpendicular-to-axis) plane. Concept / Approach: The force resultant across the inclined section equals P (equilibrium). The area of the inclined section increases as A_inclined = A / cos θ. The traction vector on the plane resolves into normal and shear components governed by geometry. The well-known results for axial tension are: σ_n = (P/A) * cos^2 θ and τ = (P/A) * sin θ * cos θ. Step-by-Step Solution: Compute inclined area: A_inclined = A / cos θ.Traction magnitude on the plane: p = P / A_inclined = (P/A) * cos θ.Normal component: σ_n = p * cos θ = (P/A) * cos^2 θ. Verification / Alternative check: At θ = 0°, σ_n = P/A (as expected). At θ = 90°, σ_n = 0 (the plane parallels the load and carries no normal stress). Why Other Options Are Wrong: (P/A) * sin^2 θ: gives zero at θ = 0°, contradicting basic axial stress.(P/A) * tan θ or * sec^2 θ: not dimensionally or physically consistent for σ_n.Zero: only at θ = 90°, not general. Common Pitfalls: Confusing angle definitions (to the axis vs to the transverse plane) and mixing normal stress with shear stress expressions. Final Answer: (P / A) * cos^2 θ.

Question 20

Support reactions: For a frame or beam resting on which type of support will the reaction always be normal (perpendicular) to the plane of the support?

Accepted Answer

Introduction / Context: Choosing correct boundary conditions is essential in structural analysis. Different supports impose different reaction directions and constraints, changing internal forces and deflections. Given Data / Assumptions: Idealized supports are used (roller, pin, fixed). Plane of the support surface is well-defined. No friction is assumed for the roller (classical idealization). Concept / Approach: A frictionless roller can only transmit a force normal to its contact surface. Therefore, the reaction is always perpendicular to the plane of the support. Hinges can resist forces in two orthogonal in-plane directions; fixed supports can also resist moments. Step-by-Step Solution: Identify role of a roller: allows translation along the plane; resists motion normal to the plane.Hence the reaction must be normal to that plane.Therefore, select “Roller support.” Verification / Alternative check: Free-body diagrams for beams on rollers show a single reaction line perpendicular to the surface at the contact point. Why Other Options Are Wrong: Hinged: reaction components act in predefined axes, not necessarily normal to a plane surface.Fixed: also develops moment; direction is not constrained to be normal to a plane.Free: no support reaction by definition.Cable: provides axial tension along the cable, not normal to a plane. Common Pitfalls: Assuming friction at the roller, which would add tangential reaction; the ideal roller is frictionless. Final Answer: Roller support.

Strength of Materials Questions