Question 1

Type of riveted joint pattern
When rivets in adjacent rows are directly opposite each other, how is the riveted joint described?

Accepted Answer

Introduction / Context: Riveted joints are classified by the relative position of rivets in successive rows. Correct terminology is essential for fabrication drawings and strength calculations. Given Data / Assumptions: Multiple rows of rivets connecting plates. Interest is only in the positional pattern of rivets. Concept / Approach: In chain riveting, rivets in adjacent rows are aligned directly opposite, forming straight transverse lines. In zig-zag riveting, rivets in one row are staggered relative to the next, forming a zig-zag path. Step-by-Step Identification: Observe the rows: if rivets line up across rows, it is chain riveting.If rivets alternate positions between rows, it is zig-zag (staggered). Verification / Alternative check: Handbooks and codes use these standard definitions for layout and pitch computations. Why Other Options Are Wrong: “Zig-zag” implies stagger; “Diamond” is a special pattern; “Staggered lap” is a descriptive phrase, not the standard name; “Cruciform” refers to a type of joint configuration, not this pattern. Common Pitfalls: Confusing chain and zig-zag when pitches are unequal; misreading drawings. Final Answer: Chain riveted

Question 2

Definition of thin shell in pressure vessel design
Evaluate the statement: “A pressure vessel is called a thin shell when it is made of thin sheets.”

Accepted Answer

Introduction / Context: The classification of pressure vessels as thin or thick shells determines the stress formulas used in design. It is based on geometry, not merely on how the vessel is fabricated. Given Data / Assumptions: Cylindrical shell of diameter d and wall thickness t. Internal pressure p, elastic material. Concept / Approach: A shell is considered thin when the wall thickness is small compared to its radius/diameter, so that stress through the thickness is nearly uniform and radial stress is negligible. A common criterion is t/d ≤ 1/20 (or t/r ≤ 1/10). Step-by-Step Reasoning: Check the geometric ratio t/d.If t/d ≤ about 1/20, thin-shell (use membrane formulas).If thicker, use Lame’s thick-cylinder theory with radial stress variation. Verification / Alternative check: Thin-shell formulas give hoop stress σ_h = p d / (2 t) and longitudinal stress σ_l = p d / (4 t), which assume negligible radial stress compared to σ_h and σ_l. Why Other Options Are Wrong: Fabrication by “thin sheets” is not the deciding factor; operating pressure and diameter alone do not define thinness without t/d. Common Pitfalls: Equating “sheet metal” with thin-shell theory; ignoring diameter’s role in t/d. Final Answer: Disagree

Question 3

Definition – Principal stress
Is the direct normal stress acting across a principal plane known as the principal stress?

Accepted Answer

Introduction / Context: Principal stresses and planes are fundamental in failure analysis and stress transformation. On a principal plane, the shear stress is zero and the normal stress attains an extremum value. Given Data / Assumptions: Plane stress or 3D stress state in a linear elastic continuum. Principal planes are mutually orthogonal. Concept / Approach: By definition, a principal plane has zero shear stress. The corresponding normal stress on that plane is a principal stress (maximum, middle, or minimum depending on order). Step-by-Step Clarification: Find planes where τ = 0 via Mohr’s circle or eigenvalue problem.The normal stresses on these planes are σ1, σ2 (and σ3 in 3D).Therefore the direct normal stress across a principal plane is indeed the principal stress. Verification / Alternative check: Eigenvectors of the stress tensor give principal directions; eigenvalues are principal stresses—purely normal, no shear. Why Other Options Are Wrong: Presence of shear contradicts the definition; restriction to uniaxial or ductile materials is unnecessary. Common Pitfalls: Confusing maximum principal stress with Von Mises equivalent stress; mixing plane of maximum shear with principal plane. Final Answer: Yes

Question 4

Beam classification by end fixity
A beam that is built-in (encastré) at both ends is referred to as which type?

Accepted Answer

Introduction / Context: Beam behavior depends strongly on end restraints. Knowing the correct classification is essential for selecting bending moment and deflection formulas. Given Data / Assumptions: Both ends are encastré (fully fixed: zero rotation and zero translation at supports). Linearly elastic, prismatic beam. Concept / Approach: A beam with both ends fixed is a fixed beam. By contrast, a simply supported beam has pin/roller supports allowing rotation; a cantilever is fixed at one end and free at the other; a continuous beam spans over more than two supports. Step-by-Step Reasoning: Identify boundary conditions: rotations at both ends are restrained.Thus, end moments develop even under symmetric loads, characteristic of a fixed beam. Verification / Alternative check: Deflection curves and moment diagrams of fixed beams show end negative moments compared to simply supported counterparts. Why Other Options Are Wrong: They do not match the described end fixity; continuous beams require three or more supports. Common Pitfalls: Confusing propped cantilever (fixed one end, simple support at the other) with a fully fixed beam. Final Answer: Fixed beam

Question 5

Thin cylinder under internal pressure – Ratio of longitudinal strain to hoop strain
A thin cylindrical shell of diameter d, thickness t, and length l is subjected to internal pressure p. If ν is Poisson’s ratio, what is the ratio of longitudinal st…

Accepted Answer

Introduction / Context: Thin-cylinder membrane theory gives closed-form expressions for stresses and strains under internal pressure. The ratio of longitudinal to hoop strain is important for estimating changes in length and diameter. Given Data / Assumptions: Thin shell: t/d ≤ about 1/20, radial stress negligible compared to membrane stresses. Hoop stress σ_h = p d / (2 t); longitudinal stress σ_l = p d / (4 t). Linear elastic isotropic material with Young’s modulus E and Poisson’s ratio ν. Concept / Approach: Strains considering Poisson coupling are ε_h = (σ_h/E) − ν (σ_l/E) and ε_l = (σ_l/E) − ν (σ_h/E). Form the ratio ε_l / ε_h. Step-by-Step Solution: Let σ_l = 0.5 σ_h (from thin-cylinder stresses).ε_l = (σ_l − ν σ_h)/E = (0.5 σ_h − ν σ_h)/E = σ_h (0.5 − ν)/E.ε_h = (σ_h − ν σ_l)/E = (σ_h − ν * 0.5 σ_h)/E = σ_h (1 − 0.5 ν)/E.Therefore, ε_l / ε_h = (0.5 − ν) / (1 − 0.5 ν). Verification / Alternative check: For ν = 0.3, ε_l / ε_h = (0.2) / (0.85) ≈ 0.235, consistent with the fact that hoop strain exceeds longitudinal strain. Why Other Options Are Wrong: Inversions or ν-independent values contradict the derived dependence; expressions with (1 ± ν) are for other relationships (e.g., plane stress/plane strain conversions). Common Pitfalls: Ignoring Poisson’s effect and assuming ε_l/ε_h = σ_l/σ_h = 0.5; mixing thin and thick cylinder assumptions. Final Answer: (0.5 − ν) / (1 − 0.5 ν)

Question 6

Euler’s column theory – Critical load for a pin-ended (hinged-hinged) column
According to Euler, what is the crippling (critical) load for a column of length l with both ends hinged?

Accepted Answer

Introduction / Context: Euler’s buckling formula gives the elastic critical load for long, slender columns. The effective length depends on end conditions. Given Data / Assumptions: Slender prismatic column, linear elastic material. Both ends hinged (pin–pin), no end moment restraint. Initial imperfections small; loading concentric. Concept / Approach: The Euler load is P_cr = π^2 * E * I / (L_e^2), where L_e is the effective length. For hinged–hinged ends, L_e = l. Step-by-Step Solution: Identify end condition factor K = 1.0 for hinged–hinged.Compute L_e = K * l = l.Therefore P_cr = π^2 * E * I / l^2. Verification / Alternative check: Comparing with other end conditions: fixed–fixed has L_e = l/2 (four times higher P_cr), fixed–free has L_e = 2l (one-quarter P_cr). Why Other Options Are Wrong: They use incorrect multipliers or dimensions; the only correct expression for pin–pin is π^2 E I / l^2. Common Pitfalls: Confusing effective length factors; mixing units of E, I, and length. Final Answer: P_cr = π^2 * E * I / l^2

Question 7

Elastic constants – quick evaluation:
For an isotropic material with Poisson’s ratio v = 0.25, what is the ratio of bulk modulus to Young’s modulus (K / E)?

Accepted Answer

Introduction / Context: Bulk modulus K and Young’s modulus E are two fundamental elastic constants for isotropic materials. Knowing how they relate through Poisson’s ratio v allows quick conversions between laboratory test data and design parameters. This question checks your fluency with the standard isotropic elasticity relations. Given Data / Assumptions: Material is homogeneous and isotropic. Poisson’s ratio v = 0.25 (dimensionless). Small-strain linear elasticity applies. Concept / Approach: The well-known relation between E, K, and v in isotropy is K = E / (3 * (1 − 2v)). Rearranging gives K / E = 1 / (3 * (1 − 2v)). Substituting v immediately yields the numerical ratio without unit conversions, because both K and E have identical dimensions (stress). Step-by-Step Solution: Start with K = E / (3 * (1 − 2v)).Form the ratio: K / E = 1 / (3 * (1 − 2v)).Plug in v = 0.25 → 1 − 2v = 1 − 0.5 = 0.5.Therefore K / E = 1 / (3 * 0.5) = 1 / 1.5 = 2 / 3. Verification / Alternative check: Use the companion identity E = 3 * K * (1 − 2v). Solving for K / E provides the same expression. For v in the common metallic range (0.25 to 0.35), K is typically of the same order as E, which is consistent with 2/3 for v = 0.25. Why Other Options Are Wrong: 1/3 or 1/2: arise from omitting the factor (1 − 2v) or misapplying the 3. 1: would require v = 0, which is not the case. 3/2: inverted relationship; not physically consistent. Common Pitfalls: Confusing the E–K relation with E–G (E = 2 * G * (1 + v)). Always check which modulus is in the formula and keep track of the 3 and (1 − 2v) factors. Final Answer: 2/3

Question 8

Eccentrically loaded columns – nature of stress:
When a column carries an eccentric load, what kinds of stresses are induced in the cross-section?

Accepted Answer

Introduction / Context: Columns are often loaded away from the centroidal axis due to construction tolerances or intentional eccentricity (e.g., brackets). Such eccentricity changes the internal stress state from pure compression to a combination that designers must assess for safety against crushing and buckling. Given Data / Assumptions: Eccentric axial load P acting at an eccentricity e from the centroidal axis. Prismatic, homogeneous column section. Small deflection and linear elastic behavior for section stress evaluation. Concept / Approach: An eccentric axial load is statically equivalent to a concentric load P (causing uniform direct compressive stress) plus a bending moment M = P * e (causing a linear compressive–tensile stress distribution). The resulting stress at a fibre is the algebraic sum of direct and bending components. Step-by-Step Solution: Direct stress: sigma_d = P / A.Bending moment: M = P * e.Bending stress at a fibre: sigma_b = M * y / I.Resultant stress: sigma = sigma_d ± sigma_b = (P / A) ± (P * e * y / I). Verification / Alternative check: At the neutral axis y = 0, only direct stress remains. At extreme fibres, bending may reduce or add to direct compression, potentially producing tension if (P * e * y / I) > (P / A). Why Other Options Are Wrong: Direct only / bending only: omit half of the statically equivalent effects. Shear only: eccentric axial loading does not create a primary shear resultant in uniform prismatic members. Direct and torsional: torsion requires a moment about the longitudinal axis, which is not produced by simple planar eccentricity. Common Pitfalls: Ignoring tension possibility on one side, or using net section area after cracking without justification. Always check maximum and minimum fibre stresses. Final Answer: Direct and bending stress both

Question 9

Riveted joints – shear condition:
In a double-strap butt joint with equal cover straps on both sides, the rivets are in which shear condition?

Accepted Answer

Introduction / Context: Riveted butt joints use cover plates (straps) to connect the main plates. The number and placement of straps control the load path through rivets and thus whether the rivets experience single shear or double shear. Recognizing this is essential for calculating joint efficiency and rivet strength. Given Data / Assumptions: Butt joint with two equal cover straps, one on each side of the main plates. Load is axial, transmitted symmetrically through straps. Neglect secondary bending due to eccentricity of load path. Concept / Approach: With two cover straps, load from one main plate to the other splits into two parallel shear planes across each rivet shank. The rivet is thus cut by two shear planes (one to each strap), placing it in double shear. This doubles the nominal shear area compared to single-strap configurations, increasing rivet shear capacity. Step-by-Step Solution: Identify load path: plate → upper strap via rivets, and plate → lower strap via rivets.Each rivet has two shear planes: one between the first plate and near strap, another between the second plate and far strap.Conclusively, the rivet experiences double shear under symmetric loading. Verification / Alternative check: Compare to a single-strap butt joint: there is only one strap and one shear plane through each rivet, resulting in single shear. Adding the second strap creates the second shear plane. Why Other Options Are Wrong: Single shear / either-or: contradict the symmetric two-strap configuration. Never in shear / torsion: rivets primarily carry shear in these joints; torsion is not a basic design action here. Common Pitfalls: Forgetting eccentric load distribution if straps are unequal; with equal straps and proper fit, double shear is the governing idealization. Final Answer: Always in double shear

Question 10

Material idealization of RCC members:
A reinforced cement concrete (RCC) beam is best considered as which kind of material system for analysis and design?

Accepted Answer

Introduction / Context: Reinforced cement concrete (RCC) members combine two distinct materials—concrete and steel—to exploit their complementary strengths. Understanding the correct idealization informs how we distribute stresses and calculate capacities for bending, shear, and serviceability. Given Data / Assumptions: RCC beam with steel reinforcement embedded in concrete. Perfect bond assumed between steel and concrete within the elastic range. Modular ratio or transformed section concept used in elastic analysis. Concept / Approach: Concrete is strong in compression but weak in tension, whereas steel excels in tension and has ductility. Together they form a composite system where different constituents share load. Analysis often transforms steel area into an equivalent concrete area using the modular ratio n = E_s / E_c, explicitly acknowledging composite behavior. Step-by-Step Solution: Recognize constituents: concrete (matrix) + steel (reinforcement).Assume strain compatibility due to bond: epsilon_concrete = epsilon_steel at a given level.Use transformed section method: A_s(eq) = n * A_s to compute neutral axis and stresses.Conclude that RCC behaves as a composite section, not as a single homogeneous or purely isotropic material. Verification / Alternative check: Experimental strain profiles in RCC beams show consistent strain compatibility and distinct stress sharing, validating the composite model. Design codes (e.g., working stress or limit state methods) are built on this notion. Why Other Options Are Wrong: Homogeneous / isotropic: would imply identical material properties throughout, which is not true. “Hetrogeneous” alone: although constituents differ, the engineering idealization that drives analysis is specifically a composite section with bonded interaction. Viscoelastic: time-dependent effects exist, but the primary classification for structural analysis is composite. Common Pitfalls: Ignoring bond slip or cracking effects; at ultimate states, cracked section analysis and tension stiffening concepts refine the composite model. Final Answer: Composite material

Question 11

Mohr’s circle – scope of application:
Mohr’s circle can be used to find stresses on an oblique plane of a body subjected to which of the following loading combinations?

Accepted Answer

Introduction / Context: Mohr’s circle is a powerful graphical tool for 2D stress transformation. It provides principal stresses, maximum shear stress, and normal and shear stresses on any oblique plane from a single diagram, unifying different combined-stress scenarios. Given Data / Assumptions: Plane stress condition (sigma_x, sigma_y, tau_xy) acting on a differential element. Linear elastic continuum; sign conventions consistent. Interest is in stresses on a plane at angle theta from the reference axis. Concept / Approach: Mohr’s circle plots normal stress on the abscissa and shear stress on the ordinate. For any given 2D state, the center C is at ( (sigma_x + sigma_y)/2, 0 ) and the radius R = sqrt( ((sigma_x − sigma_y)/2)^2 + tau_xy^2 ). Rotating 2 * theta around the circle yields the transformed stresses on a plane at angle theta. Step-by-Step Solution: Construct circle with center and radius from the given sigma_x, sigma_y, tau_xy.A pure uniaxial case with shear (option a) maps to a circle offset from zero with nonzero tau.Two perpendicular direct stresses without shear (option b) map to a circle with tau_xy = 0.The most general 2D case (option c) is directly represented and solved. Verification / Alternative check: The analytical transformation equations sigma_theta = (sigma_x + sigma_y)/2 + ((sigma_x − sigma_y)/2) * cos(2theta) + tau_xy * sin(2theta) and tau_theta = −((sigma_x − sigma_y)/2) * sin(2theta) + tau_xy * cos(2theta) yield identical results to the circle construction, confirming its universality in plane stress. Why Other Options Are Wrong: Each individual option (a), (b), or (c) is a subset; the comprehensive answer is that Mohr’s circle handles all of them. “None” contradicts standard strength-of-materials practice. Common Pitfalls: Misinterpreting the 2*theta rotation rule; forgetting sign conventions for shear; or applying plane-stress Mohr’s circle to 3D states without modification. Final Answer: All of the above

Question 12

Neutral axis and centroid in symmetric beams:
State whether the following is true or false: “For a symmetrical, homogeneous beam section under pure bending, the neutral axis does not pass through the centroid.”

Accepted Answer

Introduction / Context: The neutral axis (NA) is the locus of zero normal stress in bending. For many common engineering sections, quickly locating the NA is essential for stress calculation using the flexure formula. This item checks conceptual understanding of where the NA lies relative to the centroid for symmetric, homogeneous sections. Given Data / Assumptions: Homogeneous, isotropic material. Symmetric prismatic cross-section. Pure bending without shear. Concept / Approach: Under Euler–Bernoulli assumptions, plane sections remain plane and strain varies linearly with distance y from the NA: epsilon = y / rho. Enforcing zero resultant normal force over the cross-section in pure bending makes the NA pass through the centroid for homogeneous sections. Hence, for symmetric homogeneous beams, the NA coincides with the centroidal axis. Step-by-Step Solution: Flexure formula: sigma = M * y / I.Zero net normal force requires integral of sigma over area to be zero.That condition is satisfied when the NA passes through the centroid (first moment about NA equals zero).Therefore, the quoted statement is false for symmetrical homogeneous beams. Verification / Alternative check: Standard shapes (rectangular, circular, I-section with symmetric flanges/web) have NA through centroid in homogeneous materials. Composite or transformed sections are the cases where NA shifts away from geometric centroid. Why Other Options Are Wrong: Agree: contradicts beam theory fundamentals for homogeneous, symmetric sections. Conditional options apply to unsymmetrical or composite sections, not to the stated case. Common Pitfalls: Confusing centroidal axis with NA in cracked or composite sections, where transformed area methods are needed. Final Answer: Disagree

Question 13

Shear in UDL case:
For a simply supported beam carrying a uniformly distributed load of intensity w per unit length over the full span, what is the shear force at the midspan (centre) of the beam?

Accepted Answer

Introduction / Context: Shear force and bending moment distributions under simple load cases are staple knowledge in structural analysis. For a uniformly distributed load (UDL), recognizing the symmetry of shear lets you answer midspan values instantly—useful for quick checks in design and exams. Given Data / Assumptions: Simply supported beam with span l. UDL of intensity w acts along the entire span. Static, linear elastic behavior; supports provide vertical reactions only. Concept / Approach: For a symmetric UDL, reactions at supports are equal: R_A = R_B = w l / 2. Shear force V(x) at a distance x from the left support equals R_A − w x, a straight line decreasing from +w l / 2 to −w l / 2. The point where V = 0 is the midspan. Step-by-Step Solution: Compute reactions: R_A = R_B = w l / 2.Write V(x) = R_A − w x.At midspan x = l / 2: V(l/2) = (w l / 2) − w (l / 2) = 0.Hence the shear force at the centre is zero. Verification / Alternative check: The shear diagram is a straight line crossing zero at midspan. The maximum bending moment occurs where shear is zero, i.e., at the midspan, confirming internal consistency (M_max = w l^2 / 8). Why Other Options Are Wrong: w l^2 / 2, w l^2 / 4, w l^2 / 8 are bending-moment-like terms (units of force*length), not shear. w l / 2 is the end shear (support reaction), not the midspan shear. Common Pitfalls: Mixing up shear and moment formulas; always check dimensional consistency to avoid such errors. Final Answer: Zero

Question 14

Flexure formula insight:
When a beam is subjected to bending moment, the normal stress at any point varies how with the distance of that point from the neutral axis?

Accepted Answer

Introduction / Context: The classic flexure formula sigma = M * y / I underpins most beam stress calculations. It tells us exactly how normal stress varies from the neutral axis to the extreme fibres and is key to sizing sections efficiently. Given Data / Assumptions: Euler–Bernoulli beam theory (plane sections remain plane). Homogeneous, isotropic material; linear elasticity. Pure bending or regions where shear is negligible. Concept / Approach: The strain distribution is linear: epsilon = y / rho. Using Hooke’s law, sigma = E * epsilon = E * y / rho. Combining with the curvature relation M / I = E / rho yields sigma = (M / I) * y, explicitly showing direct proportionality between stress magnitude and distance from the neutral axis. Step-by-Step Solution: Start: epsilon = y / rho.Apply Hooke’s law: sigma = E * y / rho.Use M / I = E / rho → sigma = (M / I) * y.Therefore, sigma increases linearly with |y| and changes sign across the neutral axis. Verification / Alternative check: Maximum stress occurs at extreme fibres where |y| is largest; zero stress at y = 0 (neutral axis). This matches both theoretical derivation and experimental strain gauge data. Why Other Options Are Wrong: Equal to / independent of distance: contradict linear variation. Inverse or quadratic dependence: not supported by the flexure formula under standard assumptions. Common Pitfalls: For composite or cracked sections, the neutral axis location changes; still, within each transformed section, the variation remains linear with local y. Final Answer: Directly proportional to the distance

Question 15

Types of strain – basic definition:
When a change in length occurs in a member under load, what is the appropriate name of the strain developed?

Accepted Answer

Introduction / Context: Strain describes deformation normalized by original dimensions. Correctly naming the type of strain is important for applying the right constitutive relations and for interpreting test data such as tensile tests on bars or coupons. Given Data / Assumptions: A prismatic bar changes its length under axial loading. Small deformations so engineering strain definitions apply. Isotropic homogeneous material (not essential for naming). Concept / Approach: Linear (or longitudinal) strain is the ratio of change in length to original length along the direction of the applied load. If the bar elongates by delta L from an original length L, linear strain epsilon = delta L / L. Lateral strain refers to the concurrent contraction/expansion perpendicular to the load due to Poisson’s effect. Volumetric strain sums the three principal linear strains. Step-by-Step Solution: Identify deformation mode: change in length along the load direction.Use definition: epsilon_linear = delta L / L.Recognize accompanying lateral strain may exist, but the question asks specifically about strain due to change in length.Therefore, the correct term is linear strain. Verification / Alternative check: Tensile test stress–strain curves plot axial stress against axial (linear) strain; this is the primary measure reported for material properties such as Young’s modulus E. Why Other Options Are Wrong: Lateral strain: transverse response, not the axial change described. Volumetric strain: sum of principal linear strains; more general measure. Shear strain: associated with angular distortion, not axial change. Thermal strain: cause-based descriptor; the question is geometric, not causal. Common Pitfalls: Confusing cause and effect (thermal vs mechanical). Regardless of cause, if the measured deformation is axial length change, the strain type is linear. Final Answer: Linear strain

Question 16

Riveted joints – definition of joint strength:
The strength of a riveted joint (per pitch length) is defined as which of the following?

Accepted Answer

Introduction / Context: Designing riveted (and bolted) joints requires identifying the weakest failure mode. Each pitch length of a joint can fail by plate tearing, rivet shear, or bearing (crushing). The safe joint capacity equals the lowest resistance among these competing modes. Given Data / Assumptions: P_t = resistance against plate tearing across the net section per pitch. P_s = total shear resistance of all rivets in that pitch length. P_c = total bearing (crushing) resistance of rivets and plate in that pitch. Load is axial and uniformly distributed over the considered pitch. Concept / Approach: Structural capacity is governed by the first mode to reach its limit. Therefore, the joint strength equals the minimum of the three resistances. This definition aligns with limit-state design and classical working-stress design philosophies where the weakest link dictates overall strength. Step-by-Step Solution: Compute P_t from net plate area: P_t = sigma_t_allow * (p − d) * t (for single row, representative form).Compute P_s from rivet shear capacity: P_s = tau_allow * (number of shear planes) * (pi * d^2 / 4).Compute P_c from bearing capacity: P_c = sigma_b_allow * d * t * (number of rivets).Joint strength = min(P_t, P_s, P_c). Verification / Alternative check: Inspection of failed joints typically shows a clear governing mode (net-section fracture, sheared shanks, or bearing deformation), confirming that the minimum resistance controls. Why Other Options Are Wrong: Any single mode (P_t, P_s, or P_c) may overestimate capacity if another mode is weaker. Summing the capacities is non-physical; failure occurs by one mechanism, not all simultaneously in a load-sharing manner. Common Pitfalls: Ignoring staggered (diagonal) tearing paths, multi-row effects, or double shear where applicable; always evaluate all relevant modes for the given joint configuration. Final Answer: Minimum of P_t, P_s, or P_c

Question 17

In strength of materials, what is the relationship between the maximum shear stress and the average shear stress for a rectangular beam cross-section?

Accepted Answer

Introduction / Context: In beam theory, shear stress is not uniformly distributed across a cross-section. For a rectangular section (common in mechanical and civil engineering), the maximum shear stress at the neutral axis differs from the cross-section's average shear stress. This question checks your recall of the exact ratio used in design and checks. Given Data / Assumptions: Beam with a rectangular cross-section of breadth b and depth h. Transverse shear force V acts on the section. Material is linearly elastic and plane sections remain plane. Concept / Approach: The average shear stress is V / A where A = b * h. The distribution for a rectangle is parabolic, with the maximum at the neutral axis. Using the shear formula, tau = V * Q / (I * b), we compare peak to average to obtain the ratio. Step-by-Step Solution: Average shear stress: tau_avg = V / (b * h)Second moment of area: I = b * h^3 / 12First moment at NA: Q = (b * h/2) * (h/4) = b * h^2 / 8Maximum shear stress at NA: tau_max = V * Q / (I * b) = V * (b * h^2 / 8) / ( (b * h^3 / 12) * b ) * b = 3/2 * (V / (b * h))Therefore, tau_max = 1.5 * tau_avg Verification / Alternative check: Standard beam tables list tau_max for a rectangle as 3/2 times the average, confirming the derivation. Why Other Options Are Wrong: equal to, 4/3 times, twice, 0.866 times: do not match the rectangular section result derived from tau = V * Q / (I * b). Common Pitfalls: Assuming uniform shear distribution; confusing the 3/2 factor with bending stress relations; applying the rectangular result to non-rectangular shapes. Final Answer: 1.5 times

Question 18

Simply supported beam with a central point load: Does the maximum bending moment occur at the point of loading (i.e., at midspan)?

Accepted Answer

Introduction / Context: For a simply supported beam under a single central point load, identifying where the maximum bending moment occurs is fundamental. Designers use this to select section size, reinforcement, and to estimate deflection and stresses safely. Given Data / Assumptions: Beam is simply supported at both ends. A single point load P acts at the midspan (center). Material is linearly elastic; self-weight ignored for clarity. Concept / Approach: The shear force diagram changes sign at the location of maximum bending moment. For a symmetric beam and centrally placed load, the symmetry dictates that the maximum moment occurs at the midspan under the load. Step-by-Step Solution: Support reactions: R_A = R_B = P/2Shear just left of midspan: V = +P/2; just right: V = −P/2Bending moment at a distance x from left: M(x) = R_A * x for x < L/2At midspan: M_max = R_A * (L/2) = (P/2) * (L/2) = P * L / 4This is the maximum because the shear changes sign there. Verification / Alternative check: From calculus, dM/dx = V. Setting V = 0 gives the extremum at midspan; checking neighbouring values confirms a maximum. Why Other Options Are Wrong: False or conditional statements contradict the shear sign-change rule and symmetry for a central point load on a simply supported beam. Common Pitfalls: Confusing with uniformly distributed loads (UDL) or off-center point loads; mixing simply supported with fixed-ended boundary conditions. Final Answer: True

Question 19

When a bending moment acts on a beam section, internal stresses develop to resist it. What is the name of this internal resistance to bending?

Accepted Answer

Introduction / Context: Beams under transverse loads experience internal stress distributions. Distinguishing bending stress from shear stress is essential for sizing members and checking combined stress states. Given Data / Assumptions: A prismatic beam resists an external bending moment M. Linear elastic material; plane sections remain plane. Neutral axis passes through the centroid of the section for homogeneous members. Concept / Approach: Bending stress is the normal stress that varies linearly with distance from the neutral axis due to bending moment. The classic relationship is sigma = M * y / I, where y is the distance from the neutral axis and I is the second moment of area. Step-by-Step Solution: Apply external bending moment M.Internal stress distribution develops: compression on one side, tension on the opposite, zero at the neutral axis.Magnitude at a fiber distance y: sigma(y) = M * y / I (linear variation).These normal stresses together provide the resisting couple equal to M. Verification / Alternative check: Integrate sigma(y) over the cross-section to recover the internal moment: ∫ sigma * y dA = M, confirming that bending stress is the resisting mechanism. Why Other Options Are Wrong: Compressive or shear stress alone are incomplete descriptions; bending generates both tension and compression varying with y, not purely shear. Elastic modulus is a material property, not a stress. Common Pitfalls: Confusing shear stress (from transverse shear force) with bending stress (from moment); assuming uniform stress rather than linear distribution. Final Answer: bending stress

Question 20

Classification of pressure vessels: If the diameter D is 15 times the wall thickness t (i.e., D/t = 15), should the vessel be treated as a thick shell?

Accepted Answer

Introduction / Context: Pressure vessel analysis uses either thin-shell or thick-cylinder theory. Selecting the correct model is crucial because stress formulas and safety margins differ significantly between the two. Given Data / Assumptions: Diameter-to-thickness ratio D/t = 15. Isotropic, elastic material; cylindrical vessel. Internal pressure acts; external pressure assumed negligible. Concept / Approach: Rule of thumb: use thin-shell theory when t ≤ D/20 (equivalently D/t ≥ 20). If the wall is relatively thicker (D/t less than about 20), through-thickness stress variation is non-negligible and thick-cylinder (Lame) theory should be applied. Step-by-Step Solution: Given: D/t = 15 → t = D/15.Compare with thin-wall criterion: thin if t ≤ D/20 → D/15 > D/20, so wall is thicker than the thin-wall limit.Therefore, treat as a thick shell; use Lame’s equations for radial and hoop stresses. Verification / Alternative check: Check relative radial stress variation: in thick shells, sigma_r varies from −p at the inner surface to near zero at the outer surface, which thin formulas neglect; for D/t = 15 this variation is significant. Why Other Options Are Wrong: Disagree contradicts the widely used D/t ≈ 20 threshold.Dependence on pressure magnitude or material brittleness does not change the geometric classification.Length-to-diameter ratio affects end conditions, not thin/thick wall selection. Common Pitfalls: Applying thin-wall hoop stress sigma_h = p D / (2 t) when D/t is too small; ignoring radial stress in thick members. Final Answer: Agree

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