Question 1

Serviceability limit for sway: the maximum horizontal deflection of a column of actual length L under lateral forces may be permitted up to what limit (order-of-magnitude design guideline)?

Accepted Answer

Introduction / Context: Aside from strength and stability checks (like buckling), columns and frames must satisfy serviceability limits on lateral deflection (sway) to ensure functionality, comfort, and to protect non-structural components. Many design guides and exam syllabi use simple span-based ratios as permissible lateral deflection limits when detailed drift checks are not performed. Given Data / Assumptions: Column of actual length L is subjected to lateral forces (e.g., wind or seismic service-level loads). We seek a commonly used order-of-magnitude limit for maximum lateral deflection at the top. Detailed code-specific drift limits can vary by material, occupancy, and load case; here a typical exam value is intended. Concept / Approach: Deflection limits help control cracking, prevent damage to finishes, and maintain occupant comfort. For many building systems, lateral drift limits around L/500 at service loads are widely cited as a target value in educational and exam contexts. This sits between looser (L/300) and tighter (L/600) criteria sometimes adopted in special cases. Step-by-Step Solution: Identify that we need a serviceability drift ratio, not a strength check.Recall typical textbook/exam guidance: permissible sway ≈ L/500 for columns/frames at service conditions.Select L/500 as the acceptable upper bound in this context. Verification / Alternative check: Cross-check with common practice: many references present drift limits between 0.002 to 0.0033 times the height (i.e., H/500 to H/300) depending on usage; educational questions often adopt the stricter L/500 value. Why Other Options Are Wrong: L/200 and L/300: generally too large a sway for serviceability in typical buildings.L/400: sometimes used, but the question expects the more conservative educational standard.L/600: stricter than commonly required; acceptable but not the typical exam key. Common Pitfalls: Confusing ultimate drift limits with serviceability limits.Ignoring that code-specific values may differ; here a standard exam convention is used. Final Answer: L/500

Question 2

Concrete statistics: if Δ denotes the deviation of each individual test strength from the sample mean for n results, choose the correct expression for the sample standard deviation.

Accepted Answer

Introduction / Context: Quality control of concrete relies on statistical treatment of cube/cylinder strengths. The standard deviation gauges dispersion around the mean and feeds directly into characteristic strength calculations and compliance criteria. Using the correct formula (with the appropriate degrees-of-freedom correction) is essential, especially when the population variance is unknown and must be estimated from sample data. Given Data / Assumptions: We have n strength test results with mean x̄. For each result, Δ = (x_i − x̄). We seek the sample standard deviation s applicable to concrete control per common practice (e.g., IS/ASTM approaches). Concept / Approach: The unbiased estimator of variance uses (n − 1) in the denominator because one degree of freedom is consumed in calculating the sample mean. Thus, sample variance s^2 = ΣΔ^2 / (n − 1) and sample standard deviation s = sqrt(ΣΔ^2 / (n − 1)). Using /n would bias the estimate low for small samples and is typically reserved for population variance when the true mean is known a priori. Step-by-Step Solution: Define deviations: Δ_i = x_i − x̄.Compute sum of squares: ΣΔ^2 = Σ(x_i − x̄)^2.Divide by (n − 1) to obtain unbiased variance: s^2 = ΣΔ^2 / (n − 1).Take square root for s: s = sqrt(ΣΔ^2 / (n − 1)). Verification / Alternative check: Check a simple case (n = 2): using /n would understate spread; / (n − 1) correctly scales dispersion. Why Other Options Are Wrong: Using /n yields the population standard deviation, not the sample estimator used in control charts.ΣΔ alone ignores squaring and will cancel to ~0; not a spread measure.Linear forms (ΣΔ)/(n − 1) or (ΣΔ^2)/n without square root are not standard deviation. Common Pitfalls: Forgetting the square root step; mixing variance with standard deviation.Using biased /n denominator in small-sample concrete control. Final Answer: s = sqrt( (ΣΔ^2) / (n - 1) )

Question 3

Reinforced concrete detailing: the cover to the main reinforcement may be reduced by up to one-third of the specified cover but not less than which absolute minimum?

Accepted Answer

Introduction / Context: Concrete cover protects reinforcement against corrosion and fire and ensures adequate bond. Tolerances are allowed during construction, but codes limit how much the actual cover can deviate below the specified value. This question targets the commonly cited absolute minimum reduction permitted when a fractional reduction (one-third) would otherwise reduce cover too much. Given Data / Assumptions: Specified nominal cover is determined by exposure, bar diameter, and fire rating. Construction tolerances permit reducing cover up to one-third. An absolute floor (minimum) applies: the reduced cover must not be less than a fixed millimetre value. Concept / Approach: While one-third reduction accounts for practical placement variability, a hard minimum ensures durability is not compromised for small nominal covers. Many exam-oriented references and detailing handbooks state that even after reduction, cover should not fall below 5 mm for typical reinforced members, balancing constructability with durability needs in benign exposures. Designers must still check exposure-specific minimum covers: severe or marine environments may require larger minimums irrespective of tolerance rules. Step-by-Step Solution: Compute allowed reduction = min(one-third of specified cover, absolute minimum threshold).Apply absolute floor: actual cover ≥ 5 mm even if one-third would drop below this value.Therefore select 5 mm as the not-less-than limit. Verification / Alternative check: Cross-compare with detailing guides: minimum practical cover thresholds prevent bars becoming exposed due to minor surface tolerances or finishing. Why Other Options Are Wrong: 7 mm / 10 mm / 15 mm / 20 mm: larger than the commonly cited absolute minimum for general conditions in exam literature; the question’s key expects the smallest hard limit. Common Pitfalls: Assuming tolerance can reduce cover below durability requirements for aggressive exposures; always check code tables for environmental class.Confusing nominal cover with effective cover (to bar centroid). Final Answer: 5 mm

Question 4

For overall stability against overturning in structural engineering (e.g., retaining walls, shear walls, chimneys), the resisting weight or anchorage should provide a minimum restoring moment equal to what multiple of the overturning moment due to d…

Accepted Answer

Introduction / Context: Designing for stability against overturning requires a safety margin between the restoring moment (from self-weight and anchorage) and the overturning moment (from dead, live, wind, or seismic actions). Codes and classical design practice prescribe minimum factors so that minor uncertainties in loading, construction tolerances, or deterioration do not compromise safety. Given Data / Assumptions: We compare restoring (stabilizing) moment to the overturning moment due to dead load. We seek the minimum multiple that the resisting system (weight/anchorage) should provide. Target is a conservative, code-consistent value used in preliminary checks. Concept / Approach: A common benchmark for stability checks is to ensure that the available restoring moment comfortably exceeds the destabilizing effects. When evaluating only dead-load overturning, practice typically requires a restoring moment at least 1.4 times the dead-load overturning component, giving a margin above unity. Step-by-Step Solution: Define M_o,DL = overturning moment due to dead load.Define M_r = restoring (resisting) moment from weight/anchorage.Stability criterion (dead-load case): M_r ≥ 1.4 * M_o,DL.Select the option that states 1.4 times as the minimum multiple. Verification / Alternative check: Even when fuller factored load combinations are used, a separate serviceability-style stability check often preserves a margin >1.0. The 1.4 multiple aligns with conservative practice for dead-load-only comparisons. Why Other Options Are Wrong: 1.2 times: Less conservative; smaller safety margin. Sum of (a) and (b): Not meaningful dimensionally for a single multiple. None of these / 1.0 times: Offers no safety margin; unacceptable. Common Pitfalls: Using equality (M_r = M_o) with no reserve capacity. Mixing factored and unfactored moments inconsistently in a single check. Final Answer: 1.4 times the minimum overturning moment due to dead load

Question 5

For plain (unreinforced) concrete in sea water or in structures directly exposed to sea water (severe to very severe exposure), which minimum grade should be specified?

Accepted Answer

Introduction / Context: Durability in marine environments is governed by chloride ingress, wetting–drying cycles, and sulphate attack. Codes prescribe minimum grades to achieve lower water–cement ratios and denser, more impermeable concrete, especially for plain concrete lacking crack control from reinforcement. Given Data / Assumptions: Environment: sea water exposure or direct sea spray (severe to very severe). Material: plain (unreinforced) concrete. Objective: choose the minimum grade that is typically mandated for durability. Concept / Approach: Higher concrete grades correlate with lower permeability and better durability. For severe to very severe exposures (typical of marine conditions), plain concrete usually has a minimum grade requirement around M 20 to limit ingress and surface deterioration. Step-by-Step Solution: Classify exposure: sea water → severe/very severe.Associate typical minimum grade for plain concrete under such exposure → M 20.Select the matching option. Verification / Alternative check: Practical specifications often combine minimum grade with maximum w/c ratio (≈0.45–0.50) and minimum cement content to achieve required durability performance in marine zones. Why Other Options Are Wrong: M 10/M 15: Too low for marine durability; higher permeability. M 30: Overly high for plain concrete minimum; may be specified for special performance but not as the base minimum. M 12.5: Not a standard grade for structural durability prescriptions. Common Pitfalls: Assuming the same minimum grade for both reinforced and plain concrete; RC often requires even higher grades. Ignoring cover, w/c ratio, and curing, which are also critical for durability. Final Answer: M 20

Question 6

For reinforced concrete members immersed in sea water or exposed to heavy sea spray (adverse marine conditions), what is the maximum permissible nominal cover to reinforcement?

Accepted Answer

Introduction / Context: Nominal cover ensures steel reinforcement is protected from corrosion and provides fire resistance. Marine exposure accelerates chloride-induced corrosion, so codes increase the required cover. In adverse conditions such as immersion and severe sea spray, the largest standard nominal cover category is typically invoked. Given Data / Assumptions: Environment: sea water immersion or severe spray. Member type: reinforced concrete (RC). We seek the maximum nominal cover commonly prescribed for such exposure. Concept / Approach: As exposure severity increases from moderate to extreme, nominal cover rises stepwise (e.g., ≈20–50 mm for normal conditions; up to ≈75 mm in extreme/marine). The intent is to slow chloride ingress and delay corrosion onset in reinforcement. Step-by-Step Solution: Identify exposure: adverse marine (severe/very severe/extreme).Select the highest standard cover used in such cases → 75 mm.Confirm suitability for immersion/splash zones. Verification / Alternative check: Marine projects often pair 75 mm cover with dense concrete (e.g., M 30+ for RC), low w/c ratio, and supplementary cementitious materials (fly ash/slag) to enhance durability. Why Other Options Are Wrong: 40–70 mm: Typical for lower exposure classes; may be insufficient for harsh marine conditions. Common Pitfalls: Using nominal cover values meant for moderate exposure in aggressive chloride environments. Neglecting construction tolerances; inadequate cover on site defeats the design intent. Final Answer: 75 mm

Question 7

For webs of standard steel sections, identify the incorrect statement about gross web area definitions for rolled I-beams, channels, and plate girders.

Accepted Answer

Introduction / Context: In steel design, quick estimates of web area use simple geometric products of web depth and web thickness. These gross areas support preliminary shear and buckling checks before refined deductions for fillets, holes, or cope details are considered. Given Data / Assumptions: We are dealing with gross (not net) web areas. Rolled I-beams and channels have defined web thicknesses and overall depths. Plate girders have separate web plates with explicit plate depth and thickness. Concept / Approach: Gross web area is taken as depth * thickness for the web element. For rolled shapes, the “depth” used for gross approximations is the overall section depth; for plate girders it is the clear web plate dimension. Net deductions (e.g., for copes/holes) are handled later if required. Step-by-Step Solution: I-beam web (gross) ≈ overall depth * web thickness → acceptable for first-order checks.Channel web (gross) ≈ overall depth * web thickness → same rationale.Plate girder web (gross) = web plate depth * web plate thickness → direct definition.Thus, none of A–C is incorrect in the stated gross sense. Verification / Alternative check: Design manuals may refine “effective” web depth for buckling or shear lag, but that does not invalidate the gross definition used for initial sizing. Why Other Options Are Wrong: A–C: Each reflects the standard gross-area estimate. E: True in many preliminary calculations and does not contradict A–C. Common Pitfalls: Confusing gross and net section properties; deductions apply to net, not gross. Mixing up clear web depth (between fillets) with overall depth without noting which idealization is used. Final Answer: None of these.

Question 8

A solid slab has overall depth 20 cm and effective depth 15 cm. What is the maximum permissible horizontal spacing between parallel main reinforcement bars in the slab?

Accepted Answer

Introduction / Context: Bar spacing limits in slabs control crack widths and ensure adequate distribution of reinforcement. Codes cap spacing as a function of effective depth to keep flexural crack spacing within acceptable bounds and to promote load sharing between bars. Given Data / Assumptions: Overall depth = 20 cm; effective depth d = 15 cm (i.e., 150 mm). Main reinforcement under consideration (tension zone). Use standard spacing rule for slabs: maximum of 3d or 300 mm, whichever is smaller. Concept / Approach: For slabs, the maximum spacing S_max is min(3d, 300 mm). This ensures crack control independent of absolute slab thickness beyond a threshold. Step-by-Step Solution: Compute 3d = 3 * 150 mm = 450 mm.Compare with 300 mm → take the smaller: 300 mm.Convert to centimetres: 300 mm = 30 cm.Therefore, the maximum horizontal spacing of main bars = 30 cm. Verification / Alternative check: Even if the slab were thicker, the 300 mm cap would still govern once 3d exceeds 300 mm, maintaining consistent crack control. Why Other Options Are Wrong: 40, 45, 60 cm: Exceed the 300 mm cap; would risk excessive crack spacing. 35 cm: Still greater than 30 cm limit in this case. Common Pitfalls: Using overall depth instead of effective depth in 3*d calculation. Confusing slab rules with beam spacing provisions, which differ. Final Answer: 30 cm

Question 9

For steel compression members, the maximum permissible slenderness ratio for single-angle discontinuous struts connected by a single rivet or bolt is limited to which value?

Accepted Answer

Introduction / Context: Slenderness ratio (effective length/radius of gyration) limits prevent excessive buckling susceptibility and serviceability problems. Discontinuous single-angle struts connected by a single fastener behave less effectively in compression, so codes specify lower slenderness limits than for well-braced members. Given Data / Assumptions: Member type: single angle, discontinuous strut. Connection: single rivet/bolt (less restraint). We seek the upper limit of slenderness ratio for such a member class. Concept / Approach: Poor end restraint and eccentricity in single-angle connections reduce compression capacity and stiffness. Hence the allowable slenderness is capped at a more stringent value (≈130) compared to more robust compression members. Step-by-Step Solution: Classify the member as a discontinuous single-angle strut with minimal connection restraint.Recall the typical limit adopted for such members → 130.Select 130 from the options. Verification / Alternative check: Design handbooks list varying limits based on member use; values around 120–130 are commonly associated with lightly connected discontinuous struts, lower than 150–180 used elsewhere. Why Other Options Are Wrong: 110: Unduly restrictive relative to standard allowances. 150 or 180: Too permissive for single-angle, single-fastener struts. 95: Far too low and impractical for typical steelwork. Common Pitfalls: Applying limits for main compression members to lightly connected angle struts without adjustment. Ignoring eccentric fastener effects which reduce effective buckling resistance. Final Answer: 130

Question 10

A 24-hour load test is performed on a flexural member. If l is the effective span in metres and D is the overall depth in mm, acceptance requires the measured maximum deflection (in mm) to be less than which expression?

Accepted Answer

Introduction / Context: Load testing verifies in-situ stiffness and serviceability of flexural members. Beyond general span/ratio limits, some acceptance clauses relate permissible deflection to both span and section depth, recognizing that deeper sections deflect less under the same span and load pattern. Given Data / Assumptions: Effective span l (metres) and overall depth D (mm) are given. We seek the acceptance criterion expressed directly in millimetres of deflection. 24-hour maintained load test is considered (serviceability verification). Concept / Approach: An empirical acceptance expression ties allowable deflection to span squared divided by depth times a constant. This reflects classic elastic beam behavior (deflection ∝ l^2/EI for uniform loads) and introduces D in the denominator as a proxy for I. Step-by-Step Solution: Set allowable deflection δ_allow (mm) = l^2 / (D * K).For standard acceptance in 24-h tests, K ≈ 9000.Thus δ_allow = l^2 / (D * 9000) with l in metres and D in mm.Compare options and select the matching expression. Verification / Alternative check: While simple span/250 rules exist, the l^2/(DK) form better captures the influence of section depth. Residual deflection recovery criteria may also be checked (e.g., recovery fraction after unloading), but the question asks only the deflection bound. Why Other Options Are Wrong: l/250, l/360: Span-ratio limits not expressed with D; also dimensional inconsistency unless l is in mm. l^2/(D6000): More permissive (larger deflection) than the typical acceptance, not conservative. 0.003*l: Generic limit; does not reflect depth influence and unit handling is ambiguous. Common Pitfalls: Forgetting unit consistency (l in m and D in mm) leads to wrong magnitudes. Confusing total maximum deflection with residual deflection after unloading. Final Answer: l^2 / (D * 9000)

Question 11

In reinforced concrete beams, bars are not to be bundled in direct contact if the bar diameter exceeds which limit?

Accepted Answer

Introduction / Context: Bundling reinforcement bars in contact can ease congestion and placement, but beyond a certain diameter it impedes concrete flow and cover reliability, increasing risks of voids, honeycombing, and poor bond. Hence, codes limit the diameter for which in-contact bundling is permitted. Given Data / Assumptions: Member: RC beam. Practice: bundling several tension bars in contact. We seek the maximum diameter at which this is allowed. Concept / Approach: Larger single bars have bigger ribs and greater spacing demands. When bundled in contact, the interstices can obstruct aggregate and cement paste flow, degrading compaction. A conventional upper limit around 36 mm is therefore enforced for in-contact bundling. Step-by-Step Solution: Identify the relevant practice limit for in-contact bundling.Recall the maximum diameter commonly permitted → 36 mm.Choose 36 mm as the correct option. Verification / Alternative check: Alternative detailing strategies for larger reinforcement include using multiple smaller bars, staggering layers, or increasing member width to maintain spacing and cover, avoiding the need for oversized bundles. Why Other Options Are Wrong: 12/20/30/35 mm: Either too restrictive or slightly below the conventional cap; 36 mm is the recognized threshold. Common Pitfalls: Bundling large bars in narrow beams leading to compaction defects. Ignoring minimum clear spacing and cover when forming bundles. Final Answer: 36 mm

Question 12

For 15 cm cube compressive strength at 7 days, what should be the minimum ratio f7(M 20) : f7(M 40) expected from standard curing and testing?

Accepted Answer

Introduction / Context: Early-age compressive strength (e.g., at 7 days) is often used to track quality and predict 28-day results. For normal concretes cured under standard conditions, strength tends to scale proportionally across grades at the same age, allowing simple ratios to be inferred between different characteristic grades. Given Data / Assumptions: Compare 7-day strengths of M 20 and M 40 (same curing regime). Assume typical maturity/strength-gain curves (similar cement type, no special admixture effects). Seek a minimum ratio reflecting proportional scaling across grades. Concept / Approach: At any given early age, strengths of different grades are approximately proportional to their 28-day characteristics if mix design methodology is consistent. Hence, f7(M 20) : f7(M 40) ≈ 20 : 40 = 1 : 2 → a ratio of 1/2 as a reasonable lower bound. Step-by-Step Solution: Assume proportionality: f7 ∝ f28 for comparable mixes.Compute ratio: f7(M20)/f7(M40) ≈ 20/40 = 1/2.Adopt 1/2 as the minimum expected ratio. Verification / Alternative check: Empirical curves often show f7 ≈ 0.60–0.70 of f28 for OPC systems; if both grades follow similar curves, the ratio between grades at 7 days mirrors the 28-day ratio, supporting 1/2. Why Other Options Are Wrong: 1/3: Too low; implies unusually poor early strength for M 20 relative to M 40. 3/2 or 2.0: Inverted; would suggest M 20 stronger than M 40 at 7 days. 3/4: Overestimates relative M 20 strength compared to M 40 at early age. Common Pitfalls: Comparing cubes with different curing or cement types, which skews ratios. Ignoring that absolute early-age strengths depend on temperature and admixtures. Final Answer: 1/2

Question 13

For exposed concrete surfaces, what is the minimum duration of moist curing typically specified to achieve adequate early hydration and durability?

Accepted Answer

Introduction / Context: Proper curing maintains moisture and temperature to support cement hydration, reduce plastic shrinkage cracking, and improve durability. For exposed surfaces, inadequate curing leads to a weak, porous skin and higher permeability, undermining long-term performance. Given Data / Assumptions: Surface type: exposed concrete (not permanently protected). Objective: select the minimum moist curing period commonly required. Assume ordinary Portland cement without special accelerators or hot-weather adjustments. Concept / Approach: Baseline prescriptions for OPC concretes specify not less than about 7 days of continuous moist curing for exposed surfaces. Longer periods (10–14 days) are recommended for blended cements or harsh climates, but the minimum benchmark remains 7 days for standard conditions. Step-by-Step Solution: Identify standard minimum curing for OPC → 7 days.Recognize that supplementary cementitious materials may warrant ≥10 days; however, the question asks the minimum.Choose 7 days accordingly. Verification / Alternative check: Field QA often checks curing records and surface temperature/humidity; achieving 7 continuous days significantly improves near-surface strength and permeability versus shorter durations. Why Other Options Are Wrong: 3 or 5 days: Too short for reliable hydration and durability gains. 14 days: Beneficial but beyond the asked minimum. 10 days: Typical for blended cements; not the baseline minimum for OPC. Common Pitfalls: Stopping curing early due to schedule pressure, leading to dusting or map cracking. Counting intermittent wetting as continuous curing; continuity matters. Final Answer: 7 days

Question 14

In reinforced concrete compression members (columns), what is the minimum lap splice length for a bar, expressed as a multiple of bar diameter φ, as per standard detailing practice?

Accepted Answer

Introduction / Context: Lap splices are used to provide continuity of reinforcing bars when individual bar lengths are insufficient. In compression members (such as reinforced concrete columns), codal provisions specify a minimum lap length to ensure safe transfer of compressive forces between lapped bars without premature slip or crushing of surrounding concrete. Given Data / Assumptions: Member type: reinforced concrete compression member (column). Quantity asked: minimum lap length as a multiple of bar diameter φ. Design context: basic codal/detaling rule of thumb for compression splices (no special confinement or mechanical couplers considered). Concept / Approach: The lap splice in compression must be long enough to develop the required compressive force in the bar through bond with the surrounding concrete. While exact development length depends on concrete grade, bar type, and bond conditions, codes provide a minimum lap length for compression to avoid unsafe reductions. A widely adopted minimum is 24 φ for bars in compression, ensuring reliable force transfer even under minor construction deviations. Step-by-Step Solution: Identify member action: bars are in compression → use compression splice rule.Recall minimum lap length requirement in compression: not less than 24 φ.Compare with options → select 24 φ as the smallest acceptable multiple that meets the rule. Verification / Alternative check: For tension splices, a larger limit (often 30 φ or more, depending on development length) is typical; the question specifically targets compression, where 24 φ suffices as the minimum. Why Other Options Are Wrong: 12 φ / 16 φ / 20 φ: too short for a reliable compression splice in standard detailing.30 φ: conservative but not the minimum; exceeds the requirement for compression splices. Common Pitfalls: Confusing tension and compression lap requirements; applying development-length equations without observing codal minimums.Ignoring bar location and confinement, which affect required lengths but do not reduce the codal minimum. Final Answer: 24 φ

Question 15

In concrete mix classification for reinforced concrete design, which grade is generally not used for reinforced concrete (R.C.C.) structural members?

Accepted Answer

Introduction / Context: Concrete grades (M10, M15, M20, M25, M40, etc.) denote the specified characteristic compressive strength in N/mm^2 at 28 days. In reinforced concrete design, minimum grades are stipulated to ensure adequate strength, durability, and bond with reinforcement. Lower grades are typically restricted to plain concrete works such as blinding, levelling courses, and non-structural applications. Given Data / Assumptions: We are selecting the grade that is generally not used for R.C.C. members. Typical practice requires at least M20 for R.C.C. in many modern codes for durability and strength. Lower grades like M10 are used for plain cement concrete (P.C.C.), not for reinforced members. Concept / Approach: Reinforced members must meet minimum compressive strength to develop rebar bond and sustain design actions with adequate safety and serviceability. Therefore, grades below the threshold (e.g., M10) are unsuitable for R.C.C. and are reserved for non-reinforced works. Higher grades such as M25 and M40 are common for beams, slabs, columns, and high-performance elements, respectively. Step-by-Step Solution: Identify the minimum accepted R.C.C. grade in typical practice (≥ M20).Recognize M10 as a low-strength grade used for P.C.C. and non-structural layers.Conclude that M10 is generally not used for R.C.C. members. Verification / Alternative check: Check standard detailing/design guides: R.C.C. often starts from M20 upward due to strength and durability requirements. Why Other Options Are Wrong: M15: sometimes used historically or for minor R.C.C., but modern practice prefers ≥ M20.M20 / M25 / M40: regularly used in R.C.C. structural elements depending on performance needs. Common Pitfalls: Confusing plain concrete (P.C.C.) grades with reinforced concrete requirements.Selecting higher grades as “not used,” which contradicts common structural practice. Final Answer: M 10

Question 16

For reinforced concrete columns under axial load with inevitable eccentricity, the design eccentricity shall be taken as the distance from the assumed load application point to the centroid of the section, subject to a minimum of how many millimetre…

Accepted Answer

Introduction / Context: Columns are rarely loaded with perfectly concentric axial forces due to construction tolerances, load transfer paths, and second-order effects. Design standards therefore enforce a minimum eccentricity to avoid unconservative assessments of strength and to capture realistic bending moments even when axial load is intended to be concentric. Given Data / Assumptions: Member: reinforced concrete column. Quantity: minimum design eccentricity to be assumed. Definition: distance from the assumed load application point to the centroid of the cross-section. Concept / Approach: Minimum eccentricity protects against idealized zero-moment assumptions. Codes commonly prescribe e(min) based on overall dimensions (e.g., l/500 + D/30) but not less than a fixed floor value to account for practical misalignments. The fixed floor is widely taken as 20 mm to ensure a minimum bending moment is always present in design checks, enhancing safety against brittle compression failures. Step-by-Step Solution: Recognize that actual columns experience unintentional eccentricity due to tolerances and load path.Apply codal rule: compute e = geometric expression, but enforce a minimum of 20 mm.Select 20 mm as the required not-less-than limit from the options. Verification / Alternative check: Design moment can be taken as P * e(min), guaranteeing non-zero bending even if the load is nominally axial. Why Other Options Are Wrong: 10 mm: too small to cover practical eccentricities.30 / 40 / 50 mm: larger than the standard minimum; these may arise in specific geometries but are not the universal floor. Common Pitfalls: Ignoring minimum eccentricity and designing purely for axial compression.Confusing the geometric formula for eccentricity with the absolute minimum constraint. Final Answer: 20 mm

Question 17

In steel fabrication using fillet welds, the permissible gap between flange plates and web plates should not exceed which maximum value?

Accepted Answer

Introduction / Context: When connecting flange plates to web plates with fillet welds (e.g., built-up plate girders), the fit-up gap affects weld quality and load transfer. Excessive separation can reduce throat fusion, promote defects, and necessitate oversizing beyond what was designed. Hence codes and good practice set a maximum permissible gap for fillet-welded connections. Given Data / Assumptions: Connection: fillet welds between web and flange plates. Quantity: maximum allowable gap between mating surfaces. Standard practice allows small separations with limits; beyond that, corrective measures are required. Concept / Approach: Fillet welds are sized based on required throat thickness. If the parts are separated, the effective throat is compromised. Up to a small limit (commonly taken as 1.5 mm), the joint may still be accepted or the weld size marginally increased; larger gaps call for additional measures such as backing bars, surfacing, or re-fitting to restore proper contact. Step-by-Step Solution: Identify the interface: flange-to-web plate junction in a welded girder.Apply the fabrication tolerance for fit-up: maximum permitted gap = 1.5 mm.Select 1.5 mm from the options as the limiting value. Verification / Alternative check: Workshop quality control typically flags larger separations for rework; WPS/PQR documents often note gap tolerances and remedies. Why Other Options Are Wrong: 0.5 / 0.75 / 1.0 / 1.25 mm: stricter than necessary; while acceptable, they are not the maximum permissible gap widely adopted for fillet welds in such joints. Common Pitfalls: Assuming any visible gap is acceptable without adjusting weld size or procedure.Ignoring inspection criteria related to root opening and heat input, which affect weld integrity. Final Answer: 1.5 mm

Question 18

In steel beam behavior, the dispersion of load from the flange into the web is commonly assumed to spread uniformly at what angle θ to the flange plane?

Accepted Answer

Introduction / Context: When loads are applied to a steel beam flange (e.g., via floor systems or brackets), they transfer into the web through a dispersion mechanism. For simplified design checks—such as bearing, web crippling, and local web compressive checks—an assumed uniform dispersion angle is used to estimate load distribution from flange to web. Given Data / Assumptions: Member: rolled or built-up I-section. Load path: from loaded flange into web. Assumption: uniform dispersion at a standard angle θ to the flange plane. Concept / Approach: Design aids often take a conservative, codified dispersion angle to convert concentrated flange loads into an equivalent bearing width in the web, simplifying local checks. A commonly adopted value in traditional exam literature is θ = 20°, which leads to manageable bearing lengths without overestimating web capacity. This is an approximation for preliminary or code-aligned checks rather than a substitute for detailed finite element modeling. Step-by-Step Solution: Identify the need to spread load from flange into the web to avoid unrealistic point load effects.Apply the canonical assumption: dispersion at 20°.Use the corresponding geometry to compute the effective spread length for bearing/web checks as needed. Verification / Alternative check: Compare with other simplified methods (e.g., 1:2 or 1:3 spreads); 20° falls within customary conservative approximations for flange-to-web transfer. Why Other Options Are Wrong: 10° / 15°: too shallow; overestimates spread length and may unconservatively reduce local stresses.25° / 30°: steeper than typical assumption in this context, potentially underestimating spread length. Common Pitfalls: Treating the assumed angle as universal for all details; local stiffeners, fillets, or welds can alter real behavior.Ignoring web slenderness and crippling checks after dispersion. Final Answer: 20°

Question 19

For a simply supported reinforced concrete beam where the ratio of effective span l to overall depth D is ≥ 1 and ≤ 2, what lever-arm approximation should be used for design?

Accepted Answer

Introduction / Context: Short or deep members (including pile caps and beams with small span-to-depth ratios) are often designed using strut-and-tie or truss analogies. In such cases, simplified lever-arm expressions relate the resultant compressive strut location to the tensile tie (reinforcement), enabling rapid sizing of steel and verification of shear and anchorage, especially when beam theory assumptions are not strictly valid. Given Data / Assumptions: Member: simply supported beam behaving as a deep or short member. Span-to-depth ratio: l/D is between 1 and 2 (inclusive). Quantity: lever arm z to be used in design expressions for internal forces. Concept / Approach: For low l/D, the internal force path is better represented by a tied arch or truss model. Educational and handbook formulas provide lever-arm approximations as functions of l and D. For 1 ≤ l/D ≤ 2, a commonly recommended expression is z = 0.2 (l + 2D), balancing the influence of span and overall depth when the compression strut is steep and the tension tie is near the bottom reinforcement. Step-by-Step Solution: Recognize that standard beam theory (z ≈ 0.9 d) is less appropriate for deep members.Adopt the truss/strut-and-tie inspired approximation valid for 1 ≤ l/D ≤ 2.Select z = 0.2 (l + 2D) from the given options. Verification / Alternative check: Check limiting values: as l/D approaches 1, z tends to a function dominated by D, consistent with a deep tie; as l/D approaches 2, z scales with both l and D, reflecting intermediate behavior. Why Other Options Are Wrong: 0.3(l + 3D) and 0.3(l + 2D): generally overpredict the lever arm for this range, potentially underestimating steel demand.0.2(l + 3D): misaligned with standard guidance for the specified l/D band.0.25(l + 2D): not a recognized approximation for this l/D range. Common Pitfalls: Applying slender-beam formulas to deep members; ignoring anchorage and shear capacity under truss action.Using an l/D-specific expression outside its validity range. Final Answer: 0.2 (l + 2D)

Question 20

For a riveted joint not exposed to weather, what is the recommended distance from the rivet line to the nearest plate edge, expressed as a multiple of the thickness t of the thinner outside plate?

Accepted Answer

Introduction / Context: In plate fabrication with riveted or bolted joints, layout dimensions such as edge distance and pitch are governed by practical fabrication limits and plate buckling considerations. For joints not exposed to weather (i.e., not required to be watertight), the spacing from the rivet line to the plate edge helps prevent edge tearing and local buckling while keeping the connection economical. Given Data / Assumptions: Joint type: riveted, not exposed to weather. Parameter: distance from rivet line to the nearest plate edge. Reference thickness: t is the thickness of the thinner outside plate in the joint. Concept / Approach: The distance from the line of fasteners to the free edge must be sufficient to limit out-of-plane deformation between the fastener row and the edge and to provide adequate net section for force transfer. Traditional detailing rules specify multiples of plate thickness to maintain stiffness and fabrication tolerance, with a commonly accepted value for non-watertight joints being 12 t. Step-by-Step Solution: Identify the joint category: not required to be weather-tight; hence caulking requirements do not govern.Apply standard detailing guidance: edge distance from rivet line to edge = 12 t.Select 12 t from the given choices. Verification / Alternative check: Compare with rules for watertight joints, which may prescribe different distances to accommodate caulking and seal integrity. Why Other Options Are Wrong: 6 t / 8 t / 10 t: may be insufficient to prevent local buckling or edge tearing in standard detailing.16 t: conservative but not the common requirement for non-weather joints, potentially uneconomical. Common Pitfalls: Confusing center-to-edge minimums (margins) with line-to-edge distances specified as multiples of t.Applying watertight-joint criteria to ordinary structural joints unnecessarily increasing plate widths. Final Answer: 12 t

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