Question 1

Serviceability limit for filler joists: For filler joists supporting a structural concrete slab, the ratio of span L to the depth d (measured from the underside of the joist to the top of the structural concrete) should not exceed which limiting val…

Accepted Answer

Introduction / Context: Limiting span-to-depth ratios provide a quick serviceability check to control deflection and vibration in joist-and-slab floor systems. For traditional steel filler joists with a concrete topping, commonly used empirical limits help ensure satisfactory stiffness without detailed deflection calculations for preliminary design. Given Data / Assumptions: Filler joists are simply supported or lightly continuous with a concrete slab. Depth d is measured from the underside of the joist to the top of structural concrete (i.e., composite overall depth for serviceability). Standard occupancy loading and normal deflection criteria are implied. Concept / Approach: Span-to-depth limits are calibrated so that the elastic deflection under typical service loads remains within acceptable fractions of span (such as L/325 to L/360 for floors). For filler joists with slab, a frequently used quick limit is L/d ≤ 45. This balances economy (shallower sections) with serviceability (acceptable deflection). Step-by-Step Solution: Identify member type: filler joist with concrete slab.Apply the empirical serviceability limit L/d ≤ 45.Compare with the listed options and select 45. Verification / Alternative check: Detailed elastic deflection y_max can be estimated using y_max = 5 * w * L^4 / (384 * E * I) for uniform loading or y_max = W * L^3 / (48 * E * I) for point loads. Back-calculations typically align with L/d limits near 45 for common spans and loads. Why Other Options Are Wrong: 60: often too flexible for floor comfort and finishes. 35, 25, 20: overly conservative for typical filler-joist floors, inflating steel tonnage. Common Pitfalls: Using beam limits for composite members without recognizing the slab contribution. Ignoring continuity or partition loads that can alter deflection behavior. Final Answer: 45

Question 2

Vertical stiffener spacing in plate girders (without horizontal stiffeners): If d is the distance between the flange angles (i.e., clear depth between vertical legs), the minimum spacing of vertical stiffeners should not be less than which fraction …

Accepted Answer

Introduction / Context: Vertical stiffeners prevent web buckling and help distribute concentrated loads in plate girders. Codes and handbooks prescribe limits on the maximum and minimum spacing, often expressed as fractions of the clear web depth d (between inner faces of flange angles) when horizontal stiffeners are absent. Given Data / Assumptions: Plate girder without horizontal stiffeners. Clear depth between vertical legs of flange angles is d. We seek the minimum recommended spacing between consecutive vertical stiffeners. Concept / Approach: Too-close stiffeners are uneconomical and can cause fabrication congestion, while too-wide spacing risks web instability. Traditional guidance sets a lower bound around one-third of the clear depth, i.e., ≥ 0.33 d, to ensure a rational panel aspect ratio and practical detailing. Step-by-Step Solution: Identify the reference dimension: d (clear depth).Apply the empirical minimum spacing criterion: s_min ≈ 0.33 d.Choose the listed value 0.33 d. Verification / Alternative check: Sample designs using typical web slenderness limits show vertical panel widths between about 0.33 d and 1.5 d depending on shear demand and web thickness, consistent with the stated minimum spacing. Why Other Options Are Wrong: 0.15 d and 0.22 d: unnecessarily close spacing; impractical and uneconomical for standard girders. 0.44 d: acceptable but not the minimum commonly cited; the question asks for the minimum lower bound. Common Pitfalls: Confusing minimum spacing with maximum spacing; both limits exist but serve different stability and economy concerns. Using overall depth instead of clear depth when computing spacing ratios. Final Answer: 0.33 d

Question 3

Accounting for deviations – minimum factor of safety: If the maximum deviations in the computed load P and the computed strength E are each taken as 25% (i.e., ΔP = 0.25 P and ΔE = 0.25 E), what is the minimum factor of safety recommended to ensure …

Accepted Answer

Introduction / Context: Engineering design often addresses uncertainties in both actions (loads) and resistances (strength). A practical way to represent this in an allowable-stress context is to select a factor of safety (FOS) that maintains a margin when both the applied load is higher than computed and the available strength is lower than computed by specified percentages. Given Data / Assumptions: Computed load = P; possible positive deviation ΔP = 0.25 P (applied load may be 1.25 P). Computed strength = E; possible negative deviation ΔE = 0.25 E (available strength may be 0.75 E). We seek a single minimum factor of safety such that demand ≤ capacity even in this worst case. Concept / Approach: Define factor of safety F such that allowable load = E / F. Under worst deviations, the demand becomes 1.25 P and the capacity becomes 0.75 E. Safety requires 1.25 P ≤ 0.75 E / F, which implies F ≤ 0.75 E / (1.25 P) = 0.6 * (E / P). To guarantee safety irrespective of E/P in the original computation (i.e., when E/P ≈ F in the base design), Rearrangement shows the minimum F consistent with these deviations is about 1 / 0.6 ≈ 1.67. Step-by-Step Solution: Worst applied load = 1.25 P.Worst available strength = 0.75 E.Require: 1.25 P ≤ (0.75 E) / F.Solve for F: F ≥ (0.75 E) / (1.25 P) = 0.6 * (E / P).For a base design where E/P ≈ F, we get F ≈ 1 / 0.6 = 1.67 (rounded). Verification / Alternative check: If F = 1.67, allowable = E / 1.67 ≈ 0.60 E. Under worst deviations, demand = 1.25 P, capacity = 0.75 E / 1.67 ≈ 0.449 E. Ensuring initial E/P comfortably exceeds F provides adequate margin; the 1.67 value is a widely taught minimum in classic problems for ±25% deviations. Why Other Options Are Wrong: 1.00: no margin for uncertainty. 0.67: nonsensical for a factor of safety (less than unity). 2.67: overly conservative for the stipulated ±25% deviations. Common Pitfalls: Treating deviations as mutually exclusive rather than compounding worst-case simultaneously. Confusing partial safety factors (limit-state) with a single global FOS (allowable stress method). Final Answer: 1.67

Question 4

Steel fasteners – minimum tensile strength requirement for mild steel bolts and nuts For structural connections, the specified ultimate tensile strength of mild steel used for bolts and nuts should not be less than:

Accepted Answer

Introduction / Context: In steel design, the material grade of bolts and nuts governs connection capacity. Traditional working-stress-method tables (in kg-based units) specify a minimum ultimate tensile strength for mild steel fasteners to ensure predictable shear and bearing behavior in riveted/bolted joints. Given Data / Assumptions: Mild steel (MS) bolts and nuts used in structural steelwork. Question asks for the minimum ultimate tensile strength (not yield strength). Legacy units are kg/mm² consistent with classic Indian practice. Concept / Approach: Mild steel fasteners historically correspond to a minimum ultimate tensile strength near 400 MPa. Converting 400 MPa gives approximately 40 kg/mm², which is the commonly cited minimum for MS bolts/nuts in older tables. Step-by-Step Solution: Relate typical MS ultimate strength: 400 MPa ≈ 40 kg/mm².Select the nearest standardized minimum → 40 kg/mm². Verification / Alternative check: Modern property classes (e.g., 4.6) also imply an ultimate strength near 400 MPa, aligning with the 40 kg/mm² benchmark in legacy units. Why Other Options Are Wrong: 32, 36 kg/mm²: too low for structural-grade MS fasteners. 44 kg/mm²: higher than the usual minimum for MS; aligns more with stronger grades. Common Pitfalls: Confusing yield strength with ultimate tensile strength; the question explicitly concerns tensile (ultimate) strength. Final Answer: 40 kg/mm²

Question 5

Allowable bending stress for rolled I-sections and channels (extreme fibres on the effective section) What is the permissible tensile/compressive bending stress at extreme fibres for rolled I- and channel sections when using the effective section co…

Accepted Answer

Introduction / Context: Traditional steel design using the working-stress method specifies permissible bending stresses for standard rolled sections. For mild steel (yield ≈ 2500 kg/cm²), the allowable bending stress is typically a fraction of the yield strength to ensure adequate margin. Given Data / Assumptions: Rolled I-sections and channels. Effective section considered for bending. Working-stress philosophy with legacy units. Concept / Approach: A common allowable bending stress for MS sections is about 0.66 of the yield strength. With fy ≈ 2500 kg/cm² (≈ 250 MPa), the permissible stress is about 0.66 * 2500 ≈ 1650 kg/cm². Step-by-Step Solution: Assume fy ≈ 2500 kg/cm².Compute 0.66 * fy ≈ 0.66 * 2500 = 1650 kg/cm².Select 1650 kg/cm² as the allowable bending stress. Verification / Alternative check: Design handbooks from the working-stress era list 1650 kg/cm² (tension/compression at extreme fibres) for standard sections in bending. Why Other Options Are Wrong: 1500 or 1420 kg/cm²: conservative but not the standard tabulated limit. 2285 kg/cm²: too high, approaching yield without adequate safety margin. Common Pitfalls: Mixing limit-state design (factored) values with working-stress allowable stresses; the question clearly uses allowable-stress numbers in kg/cm². Final Answer: 1650 kg/cm²

Question 6

Double-angle tension members connected on one side of a gusset plate For double angles placed back-to-back and connected to one side of a gusset plate, the allowable working stress corresponding to a given slenderness ratio is reduced to what percen…

Accepted Answer

Introduction / Context: When double angles are connected to a gusset by only one leg (i.e., both angles connected on the same side), shear lag and eccentricity effects reduce the effective capacity. Classic specifications apply a reduction factor to the allowable stress. Given Data / Assumptions: Double-angle member, back-to-back configuration. Connected by one leg only to one side of a gusset. Working-stress design context. Concept / Approach: The reduction accounts for uneven stress distribution and shear lag in outstanding legs. A commonly adopted allowance is to reduce the allowable stress to about 80% of the value applicable to fully balanced connections. Step-by-Step Solution: Identify connection eccentricity/shear lag → stress reduction needed.Apply standard reduction → about 80% of the tabulated allowable stress. Verification / Alternative check: Design handbooks show factors (≈0.8) for single-sided connections of double angles to reflect reduced effectiveness of outstanding legs. Why Other Options Are Wrong: 60–70%: overly severe for standard double-angle details. 50% or 40%: excessively conservative and not typical of code guidance. Common Pitfalls: Neglecting reduction for single-sided connections can unconservatively overestimate tension capacity. Final Answer: 80%

Question 7

Eccentrically loaded columns – defining an equivalent axial load for design How is the equivalent axial load commonly determined for an eccentrically loaded column to reflect the effect of the produced bending moment?

Accepted Answer

Introduction / Context: Eccentric axial load P acting at an eccentricity e produces a combined effect: direct compression and bending with moment M = P * e. Some design approaches convert this combined action into an “equivalent axial load” using a bending factor to simplify checks against allowable stress. Given Data / Assumptions: Column subjected to axial load P with eccentricity e. Bending moment M = P * e considered. Working-stress context using equivalent load concepts. Concept / Approach: The extreme-fibre stress is usually checked as σ = P/A ± M/Z. Equivalently, some handbooks propose a transformed axial load by adding a term proportional to M (via a bending factor) to P to maintain a single check against allowable compressive stress. Step-by-Step Solution: Compute M = P * e.Introduce a bending factor α reflecting section properties and stress distribution.Define P_eq ≈ P + α * M to represent the combined effect conservatively. Verification / Alternative check: Direct stress interaction (P/A + M/Z ≤ σ_allow) provides the exact check; the equivalent-load approach with α * M is an expedient that parallels interaction. Why Other Options Are Wrong: Subtracting or dividing by M and a factor has no rational basis for a conservative equivalent load. Hence option (a) is the only reasonable description. Common Pitfalls: Relying only on “equivalent load” without verifying combined-stress interaction can miss critical tension on one face. Final Answer: By adding the axial load and the contribution due to eccentricity, i.e., axial load plus the product of bending moment from eccentricity and an appropriate bending factor

Question 8

Double-angle tension member connected by one leg each – effective net area to be taken for design A pair of equal angles placed back-to-back in tension is connected to a gusset by one leg of each angle (single-sided connection). What should be taken…

Accepted Answer

Introduction / Context: Shear lag reduces the effectiveness of outstanding legs in angle tension members. When two angles are placed back-to-back and each connects by only one leg to one side of a gusset, the load transfer path is primarily through the connected legs. Given Data / Assumptions: Two equal angles, back-to-back. Each angle is connected by one leg to the gusset on the same side. Tension member design under working-stress concepts. Concept / Approach: Only the connected legs have direct, efficient load transfer. Outstanding legs are prone to shear lag and are generally ignored in basic effective-area evaluation, unless a refined factor is applied. The simplest conservative approach is to use the sum of net areas of the two connected legs only. Step-by-Step Solution: Identify load path → through connected legs adjacent to the gusset.Compute net area of each connected leg (deduct hole areas).Add the two net connected-leg areas; ignore outstanding legs. Verification / Alternative check: Advanced provisions may allow a shear-lag factor on outstanding legs, but baseline conservative practice is to neglect them for design simplicity and safety. Why Other Options Are Wrong: Gross area or all-leg area overestimates capacity by ignoring shear lag. Using only one leg is too conservative for a double-angle member. Common Pitfalls: Forgetting to deduct bolt/rivet-hole areas when computing the net area of the connected legs. Final Answer: Sum of the net areas of the two connected legs only; outstanding legs neglected

Question 9

Effective length of a double-angle strut (back-to-back) connected on both sides of a gusset For a double-angle strut with angles placed back-to-back and connected to both sides of a gusset plate by at least two rivets at each end, the effective leng…

Accepted Answer

Introduction / Context: Effective length reflects end restraint and the ability of the connection to reduce buckling length. For short, gusseted angle struts with adequate end fasteners, classical guides recommend an effective length factor below 1.0. Given Data / Assumptions: Double angles back-to-back. Connected to both sides of a gusset plate. At least two rivets at each end provide fixity against rotation and translation. Concept / Approach: Gusseted struts gain partial end restraint, typically modeled by an effective length about 0.85 L for design checks, reflecting improved stability relative to pin–pin length L. Step-by-Step Solution: Recognize improved end restraint due to both-side connection.Adopt recommended factor → Le ≈ 0.85 L. Verification / Alternative check: Coefficient tables for gusseted compression members list 0.85 L for well-connected ends, matching empirical performance. Why Other Options Are Wrong: L (1.0 L) ignores added restraint. 0.50 L or 0.67 L are too low unless full fixity/strut framing warrants it. 2 L is non-physical here. Common Pitfalls: Applying 0.85 L without verifying end connection quality (minimum fasteners, detailing) can be unsafe. Final Answer: 0.85 L

Question 10

Sizing riveted connections – determining the number of rivets required In routine working-stress design, the number of rivets to be provided in a joint is determined by which basic relation?

Accepted Answer

Introduction / Context: Connection design ensures that the shear transfer between members is adequate. In riveted (or bolted) joints, a simple capacity check is used to size the number of fasteners based on the demand and the strength of a single fastener. Given Data / Assumptions: Total load to be carried by the joint is known. Design shear (or bearing) capacity of one rivet is known for the selected diameter and materials. Governing failure mode for a single rivet is identified (shear or bearing). Concept / Approach: The basic sizing rule is demand over capacity, then round up. Distribution and eccentricities may require additional checks, but the initial count is controlled by the one-rivet design capacity in the governing mode. Step-by-Step Solution: Find single-rivet design strength (shear or bearing).Compute required count = total joint load / single-rivet strength.Round up to the next whole number; then check spacing, edge distance, and secondary effects. Verification / Alternative check: Where eccentric loading exists, rivet group analysis (resultant shear, polar moment) supplements the initial count. Why Other Options Are Wrong: Relations using plate thickness, pitch, or tributary area do not directly determine strength. Bending moment divided by pitch is not a capacity-based count. Common Pitfalls: Not checking edge distance/pitch limits or bearing on plates after counting rivets can cause detailing failures. Final Answer: Number of rivets = factored load on the joint divided by the design shear strength of one rivet in the governing failure mode, rounded up to the next whole number

Question 11

Column design – factor of safety parameter “m” in the safe compressive stress formula In the classic working-stress formula for safe compressive stress of a column, the factor of safety parameter m is typically taken as:

Accepted Answer

Introduction / Context: Older column design under working-stress method used empirical/analytical formulae (e.g., Rankine or secant-type relations) containing a factor of safety parameter denoted by m. This factor provides a margin between elastic instability/yield and permissible stress. Given Data / Assumptions: Working-stress design approach. Steel columns of common structural grades. Parameter m appears in the safe compressive stress expression. Concept / Approach: Design aids and classic codes often adopt m ≈ 1.68 for steel columns, balancing safety and economy in allowable compressive stress calculations across practical slenderness ranges. Step-by-Step Solution: Identify the parameter used in the safe stress formula.Select the standard value m = 1.68 for steel columns in WSM tables. Verification / Alternative check: Tabulated allowable compressive stresses reverse-engineer to an m around 1.68 when matched to experimental buckling data and customary safety margins. Why Other Options Are Wrong: 1.50 or 1.60: used in other contexts; not the cited standard for this formula. 1.88: more conservative than usual practice for ordinary structural steel columns. Common Pitfalls: Mixing limit-state safety factors with working-stress parameter m; they are not interchangeable. Final Answer: 1.68

Question 12

Bolt holes – standard clearance over nominal diameter For structural bolting, the diameter of a bolt hole is generally taken as the nominal bolt diameter plus:

Accepted Answer

Introduction / Context: Standard bolt holes provide installation clearance. The hole diameter exceeds the nominal bolt diameter by a small allowance to accommodate fabrication tolerances and ease of erection. Given Data / Assumptions: Common structural bolts and shop/site drilling practice. Standard holes (not oversized or slotted) are considered. Concept / Approach: Conventional practice uses a clearance of about 1.5 mm for standard holes in common bolt size ranges, ensuring fit-up without excessive slop that would compromise connection stiffness. Step-by-Step Solution: Hole diameter = nominal bolt diameter + clearance.Adopt clearance ≈ 1.5 mm → hole = d + 1.5 mm. Verification / Alternative check: Design standards list standard, oversize, and short/long slotted holes. The 1.5 mm increment corresponds to the “standard” category for many bolt sizes. Why Other Options Are Wrong: 1.0–1.4 mm: may be insufficient for erection tolerance. 1.6 mm: close but not the conventional standard value. Common Pitfalls: Confusing standard holes with oversize or slotted holes, which use larger allowances and require special checks. Final Answer: 1.5 mm

Question 13

Laterally unsupported cantilever – effective projecting length for lateral–torsional buckling A cantilever beam of length L is built-in at the support and is restrained against torsion at the free end. What effective projecting length l should be us…

Accepted Answer

Introduction / Context: Lateral–torsional buckling (LTB) depends on the unbraced length and boundary restraints. For cantilevers, restraining the free end against torsion (and sometimes warping) reduces the effective buckling length compared to a fully free tip. Given Data / Assumptions: Cantilever length L, built-in at one end. Free end is restrained against torsion (lateral restraint present). LTB effective length factor required for design. Concept / Approach: Coefficient tables for LTB provide factors to convert the physical length into an effective length for buckling. With torsional restraint at the free end, the effective projecting length is reduced to about 0.70 L, reflecting improved stability compared to an unrestrained free tip (which would be ≈ L). Step-by-Step Solution: Identify boundary conditions → built-in at one end, torsion-restrained at the other.Adopt recommended factor for this case → l ≈ 0.70 L. Verification / Alternative check: Handbooks of stability coefficients list 0.7 L for cantilevers with torsional restraint at the tip, consistent with test data and theory. Why Other Options Are Wrong: 0.75 L or 0.85 L: less beneficial than typical tabulated restraint effect for this condition. 0.50 L: too optimistic for a cantilever. L: corresponds to an unrestrained free end. Common Pitfalls: Assuming restraint at the tip without detailing an actual torsion/lateral brace; the factor is valid only if the restraint is truly provided. Final Answer: l = 0.70 L

Question 14

In structural engineering practice, how can the vertical deflection of a prismatic beam be most effectively reduced (keeping material and span otherwise similar)?

Accepted Answer

Introduction / Context: Beam deflection control is a core serviceability requirement in civil and structural engineering. Designers frequently tune member proportions to limit sag under working loads without excessive material use. Given Data / Assumptions: Prismatic (constant cross-section) beam under typical service loads. Material and modulus of elasticity E unchanged. We compare the qualitative effect of changing section dimensions or span. Concept / Approach: For a rectangular section, second moment of area I = b * d^3 / 12. Deflection Δ is inversely proportional to E * I for a given loading case (e.g., Δmax ∝ 1 / (E * I)). Because I scales with d^3, increasing depth produces a cubic gain in stiffness. Increasing width b helps only linearly and is therefore less effective for the same material. Step-by-Step Solution: Recognize Δ ∝ 1 / I for a given load and E.For rectangular sections: I ∝ d^3 (depth dominates).Therefore, increasing depth substantially increases stiffness and reduces deflection.Increasing span L increases Δ strongly (e.g., L^4 for UDL on simply supported beams), hence is unfavorable. Verification / Alternative check: Compare two sections of same area: making one deeper and narrower will typically reduce Δ more than making one wider and shallower, confirming the cubic sensitivity to depth. Why Other Options Are Wrong: Increasing span raises deflection; decreasing depth reduces stiffness; increasing only width slightly gives limited benefit compared to depth for the same mass; “None” is incorrect because depth increase is effective. Common Pitfalls: Focusing only on ultimate strength and neglecting serviceability; assuming width increase equals depth increase in impact; forgetting load case and support conditions also influence Δ. Final Answer: Increasing the depth (overall height) of the beam section

Question 15

Tacking rivets for plates exposed to weather: what is the maximum permitted pitch along a line, expressed as a multiple of the outside plate thickness t?

Accepted Answer

Introduction / Context: Tacking rivets (or bolts) are auxiliary fasteners used to keep built-up elements together between primary connections. Exposure to weather demands closer spacing to control plate separation and corrosion-induced issues. Given Data / Assumptions: Plates are in compression and exposed to weather. t = thickness of the outside plate. Pitch refers to spacing of tacking rivets along a line. Concept / Approach: Legacy steel design practice and many exam standards limit the tacking pitch to the lesser of an absolute distance and a multiple of t. For weather-exposed plates, a conservative multiple of 16 t is adopted to avoid plate opening and local buckling. Step-by-Step Solution: Identify exposure category: weather-exposed.Apply the weather-exposed limit: maximum pitch = 16 * t.Also verify any absolute cap (e.g., a specific millimeter limit) in detailed design. Verification / Alternative check: Compare to non-exposed guidance where larger multiples (e.g., 32 t) are allowed; exposure necessitates tighter spacing. Why Other Options Are Wrong: Values like 24 t or 32 t are too large for weather-exposed conditions; 8 t is overly conservative beyond common practice; 48 t is unsafe for exposed plates. Common Pitfalls: Using non-exposed limits on exposed work; neglecting the smaller governing absolute distance when 16 t is large in millimeters. Final Answer: 16 t

Question 16

In a plate girder, why are stiffeners provided on or along the web plate?

Accepted Answer

Introduction / Context: Plate girders achieve economy by using thin webs with flanges carrying most bending. Thin webs are susceptible to shear buckling and local crushing near reactions; stiffeners mitigate these issues. Given Data / Assumptions: Thin web plate under shear/bearing demands. Transverse (vertical) and bearing stiffeners may be used. Longitudinal stiffeners may also be used in deep webs. Concept / Approach: Web panels buckle in shear if h/t is large. Transverse stiffeners subdivide the web into smaller panels to raise critical shear buckling stress. Bearing (end) stiffeners distribute concentrated reactions into the web and flanges, preventing local crushing. Step-by-Step Solution: Identify web panel demand (shear and bearing).Provide vertical stiffeners at code-limited spacings to prevent shear buckling.Provide bearing stiffeners under concentrated loads and at supports. Verification / Alternative check: Check web slenderness and panel proportions; confirm shear buckling capacity with and without stiffeners per design provisions. Why Other Options Are Wrong: Flange compressive stress is governed by flange area, not web stiffeners; average shear stress is not reduced (capacity is increased); bearing-only statement is incomplete. Common Pitfalls: Omitting bearing stiffeners at heavy reactions; misplacing stiffeners relative to loads; ignoring weld details causing web distortion. Final Answer: To avoid buckling (bulking) of the web plate and provide bearing where needed

Question 17

Filler (secondary) joists in steel floors: which statement is correct regarding their detailing and support?

Accepted Answer

Introduction / Context: In steel floor systems, main beams carry secondary members called filler (or jack) joists. Understanding their connections and support conditions is important for layout and constructability. Given Data / Assumptions: Typical multi-beam floor framing. Secondary joists span between main beams/girders. Conventional riveted/bolted or welded detailing with cleat angles. Concept / Approach: Filler joists are commonly detailed as continuous across multiple supports when practical and are connected to main beams using cleat angles that transfer shear and provide lateral restraint. The main beams provide vertical support to the filler joists. Step-by-Step Solution: Recognize that filler joists sit on or frame into main beams.Note that cleat angles are a standard shear connection detail for such joists.Hence statements (c) and (d) together depict correct practice. Verification / Alternative check: Standard steel detailing manuals show joist-to-beam connections via cleat angles or shear tabs; continuity depends on layout and is not limited strictly to three supports. Why Other Options Are Wrong: (a) confuses reinforced concrete with steel beams in PCC; (b) is overly restrictive—continuity depends on design and layout; selecting only (c) or only (d) ignores the paired correctness. Common Pitfalls: Assuming moment connections where only shear is intended; overlooking bracing or deck diaphragm roles. Final Answer: Both (c) and (d) are correct

Question 18

Column bases in steel structures: which of the following statements are correct about their function and design intent?

Accepted Answer

Introduction / Context: Column bases form the interface between steel columns and concrete foundations or pedestals. They ensure safe transfer of axial load, moments, and shear, while limiting bearing stresses in concrete. Given Data / Assumptions: Steel column with base plate and anchors resting on concrete. Service loads within permissible bearing of concrete and plate bending capacity. Concept / Approach: The base plate distributes concentrated column reactions across a sufficient concrete area such that bearing pressure is acceptable. Simplified design commonly assumes an approximately uniform pressure distribution (with checks for moments and shear as required) and designs plate thickness and anchors accordingly. Step-by-Step Solution: Provide a plate of adequate plan dimensions based on allowable concrete bearing.Design plate bending between column footprint and plate edges.Detail anchors and shear keys if moments or lateral loads exist. Verification / Alternative check: Check concrete bearing stress = Load / Bearing area and confirm within limits; cross-check with more refined pressure distribution if significant moments exist. Why Other Options Are Wrong: Options (a)–(d) are all correct, so choosing a subset would be incomplete. Common Pitfalls: Ignoring eccentricity and resultant moments; undersized plate thickness leading to excessive plate bending; neglecting grout and levelness. Final Answer: All the above

Question 19

Tension member with rivet/bolt holes not in a single line: the governing strength of the net section depends on which geometric parameters?

Accepted Answer

Introduction / Context: When holes do not lie on a single straight line, the critical rupture path can be zig-zag. Net-section strength must therefore consider both longitudinal and transverse spacing as well as hole sizes. Given Data / Assumptions: Tension member with multiple fastener lines. Potential straight and zig-zag failure paths. Standard deductions for holes and stagger corrections. Concept / Approach: Net width is computed by deducting hole diameters along the path, but zig-zag paths allow a stagger correction that adds back an amount proportional to gauge^2 / pitch for diagonal segments. Hence pitch, gauge, and hole diameters all influence the least net area. Step-by-Step Solution: Identify all plausible failure paths (straight and zig-zag).For each path, compute net width: b_net = b_gross − Σ(d_h) + Σ(stagger correction).Select the least net area A_net, then strength = A_net * f_t (or design value). Verification / Alternative check: Check block shear if applicable; confirm that the chosen path indeed gives the smallest net area. Why Other Options Are Wrong: Each single parameter alone is insufficient; the interaction among pitch, gauge, and hole size governs the net section. Common Pitfalls: Using nominal hole size instead of standard hole diameter; forgetting stagger correction; ignoring minimum pitch/gauge requirements causing tear-out. Final Answer: All the above

Question 20

A thin-walled cylindrical vessel subjected to internal fluid pressure develops which principal membrane stress in the circumferential direction?

Accepted Answer

Introduction / Context: Thin pressure vessels are classic subjects in mechanics of materials. Under internal pressure, the cylinder wall carries membrane stresses in two perpendicular directions: circumferential (hoop) and longitudinal. Given Data / Assumptions: Thin cylinder: wall thickness t is much smaller than diameter. Uniform internal pressure p. Neglect radial stress across thickness (thin-wall assumption). Concept / Approach: For a thin cylinder, hoop stress sigma_h = p * d / (2 * t) acts around the circumference, and longitudinal stress sigma_l = p * d / (4 * t) along the axis. Both are tensile under internal pressure, with hoop stress being twice the longitudinal stress. Step-by-Step Solution: Resolve internal pressure against a longitudinally cut section to derive hoop tension.Obtain sigma_h = p * d / (2 * t) (tension).Confirm longitudinal stress sigma_l = p * d / (4 * t) (also tension, but smaller). Verification / Alternative check: Check equilibrium of half cylinder in circumferential direction; stresses must balance the pressure resultant on the projected area. Why Other Options Are Wrong: Compression is incorrect for internal pressure; shear options do not represent the principal membrane state; “no stress” contradicts pressure vessel theory. Common Pitfalls: Confusing thin-wall with thick-wall formulas; mixing units; using radius instead of diameter without adjusting constants. Final Answer: Hoop tension (circumferential tension)

Steel Structure Design Questions