Question 1

Minimum edge thickness of reinforced concrete pile caps / footings on piles (as per classic IS-style working stress practice): According to traditional guidance aligned with IS:456-era practice, the thickness of RCC footing on piles at its edges sho…

Accepted Answer

Introduction / Context:Pile caps and footings on piles require a minimum edge thickness to ensure adequate cover, shear capacity, and constructability. Historical working-stress-oriented guidance provides clear minimums for preliminary detailing.Given Data / Assumptions:RCC footing supported on piles.Edge thickness requirement in centimeters.Classic IS:456-era working-stress guidance context.Concept / Approach:The minimum edge thickness prevents spalling, provides room for reinforcement and cover, and aids in distributing concentrated pile reactions. A value of not less than 150 mm (15 cm) is conventionally adopted in many textbooks and practice notes for preliminary sizing.Step-by-Step Solution:Identify minimum edge thickness requirement for pile-supported footing: ≥ 15 cm.Select the correct option corresponding to 15 cm.Note that actual depth may be governed by punching shear and strut-and-tie action near piles.Verification / Alternative check:Checking punching shear around pile heads or groups and ensuring development length for bars may lead to greater practical thicknesses; however, the minimum edge thickness baseline remains about 15 cm for preliminary checks.Why Other Options Are Wrong:5 cm or 10 cm: Too thin for cover, shear, and reinforcement detailing.20 cm or 25 cm: Acceptable but not the minimum; these may be required by strength checks, not by the basic minimum edge thickness rule.Common Pitfalls:Using the minimum blindly without verifying punching shear and reinforcement anchorage; ignoring aggressive exposure that could demand larger covers and hence greater thickness.Final Answer:15 cm

Question 2

In reinforced cement concrete (R.C.C.) construction, which grade/type of steel is most commonly used for reinforcement in general building practice?

Accepted Answer

Introduction / Context:R.C.C. (reinforced cement concrete) combines concrete in compression with steel in tension. Examinations often ask which steel is generally used in everyday building work, focusing on practicality, economy, and code traditions rather than specialty applications.Given Data / Assumptions:Question targets “generally used” reinforcement for common R.C.C. work.Special steels like stainless or very high-carbon types are available but expensive or unsuitable for routine use.Answer is framed for typical building structures under standard codes and budgets.Concept / Approach:Mild steel (including deformed mild-steel-based bars in older practice) has long been the staple reinforcement owing to its ductility, weldability, and cost-effectiveness. Although modern practice frequently uses high-yield strength deformed (HYSD) bars (often low-alloy, not the “high-carbon” option shown), traditional objective-type questions usually mark “mild steel” as the general-purpose answer for R.C.C. reinforcement in buildings.Step-by-Step Solution:List options and identify the one most associated with everyday, economical R.C.C.: mild steel.Exclude specialty or unsuitable steels for routine work (stainless/high-carbon).Recognize that “high-tensile alloy steel” is used in prestressing, not ordinary R.C.C. bars.Select: mild steel.Verification / Alternative check:Textbook summaries of traditional R.C.C. practice emphasize mild steel (and later HYSD) as the common reinforcement choice for slabs, beams, and columns in buildings, due to cost and ductility advantages.Why Other Options Are Wrong:Stainless steel: corrosion-resistant but too expensive for general use.High-carbon steel: too brittle; poor ductility for R.C.C. detailing.High-tensile alloy steel: typical of prestressed concrete tendons, not ordinary R.C.C. reinforcement.Common Pitfalls:Confusing prestressing steel with reinforcement for conventional R.C.C.; selecting an exotic material based on corrosion resistance without considering cost/practice.Final Answer:Mild steel

Question 3

Continuous floor slab layout: If the interior (intermediate) span of a continuous reinforced concrete slab is 5 m, what length is commonly adopted for each end span to obtain practical continuity and economy?

Accepted Answer

Introduction / Context:During framing layout of continuous slabs/beams, end spans are often selected slightly shorter than interior spans to control negative moments at supports, improve serviceability, and fit architectural grids. This rule-of-thumb planning concept is frequently examined in building design.Given Data / Assumptions:Interior (intermediate) span = 5 m.End span is commonly set as a fraction of interior span for practical continuity.Typical building design guidance is assumed.Concept / Approach:A widely used planning guide is to take end spans at about 0.9 of the interior span in continuous systems. This leads to smoother moment distribution, ease of detailing, and better control of deflection and cracking at exterior supports.Step-by-Step Solution:Compute 0.9 × 5 m = 4.5 m.Compare with offered choices.Select 4.5 m as the adopted end-span length.Verification / Alternative check:Design handbooks and practice notes commonly reflect the 0.9L guideline for preliminary layout; final design still requires rigorous analysis.Why Other Options Are Wrong:4.0 m, 3.5 m, 3.0 m: excessive reduction, not a common planning proportion for continuous slabs.None of these: incorrect because 4.5 m is a recognized practice value.Common Pitfalls:Treating preliminary span proportions as a substitute for full structural analysis; ignoring actual support conditions and loading variations.Final Answer:4.5 m

Question 4

Mohr’s circle relationship for plane stress: If p1 and p2 are mutually perpendicular principal stresses in a soil mass, the normal stress σ on a plane at angle θ to the p1-plane is given by which expression?

Accepted Answer

Introduction / Context:Stress transformation in soils and solids is routinely performed using Mohr’s circle or equivalent equations. Knowing the normal stress on a plane inclined to principal directions is essential for earth pressure, bearing capacity, and slope stability problems.Given Data / Assumptions:Principal stresses p1 (major) and p2 (minor) act on mutually perpendicular planes.Angle θ is measured from the plane carrying p1.Plane stress condition.Concept / Approach:For any plane at angle θ from the p1-plane, the normal stress is obtained from standard transformation equations. The fundamental identity is σ = p1 cos^2θ + p2 sin^2θ. Equivalent forms using double-angle identities also exist but must be used carefully with correct trigonometric factors.Step-by-Step Solution:Start with transformation: σ = p1 cos^2θ + p2 sin^2θ.Recognize that shear stress transformation involves sin 2θ terms, but the normal stress here uses squared trigs.Confirm angle is referenced properly to p1-plane as stated.Select the matching expression (option a).Verification / Alternative check:Using double-angle identity: σ = (p1 + p2)/2 + (p1 − p2)/2 cos 2θ, which is algebraically identical to option (a). Among choices, (d) is the double-angle equivalent; since the stem defines angle from the p1-plane, either (a) or the correct double-angle form works. The explicit squared-trig version is unambiguous, hence chosen.Why Other Options Are Wrong:(b) swaps sin^2θ and cos^2θ, not matching the reference plane stated.(c) uses sin 2θ for normal stress, which is incorrect; sin 2θ appears in shear stress.Common Pitfalls:Confusing sign conventions and angle reference; mixing up the normal and shear transformation formulas.Final Answer:σ = p1 cos^2θ + p2 sin^2θ

Question 5

In reinforced concrete construction, when bars need to be continued beyond standard stock length, which type(s) of splice connection are commonly used to obtain the required greater length?

Accepted Answer

Introduction / Context:Reinforcing bars are supplied in standard lengths. On site, members often exceed these lengths, so bars must be spliced to continue reinforcement safely and effectively. Understanding splice types is vital for detailing and construction sequences.Given Data / Assumptions:Need to extend bar length beyond standard stock.Common field methods are acceptable under code provisions.Quality of lap lengths, hooks, and dowels is properly ensured.Concept / Approach:There are three widely recognized splice categories for tension/compression bars in R.C.C.: straight lap splices, hooked splices, and dowel splices. The choice depends on bar diameter, space constraints, required development length, and location (tension or compression zone).Step-by-Step Solution:Enumerate typical splice forms: straight lap, hooked, dowel.Check compatibility with bar size, cover, and detailing rules for development length.Conclude that all listed methods are used in practice when properly designed.Verification / Alternative check:Codes and detailing manuals contain lap splice requirements, hook dimensions, and dowel splice practices that confirm all three are legitimate techniques.Why Other Options Are Wrong:Picking only one splice type ignores site constraints and varied detailing situations.Common Pitfalls:Insufficient lap length; placing splices at sections with highest tension; ignoring confinement ties for large bar splices.Final Answer:All of the above

Question 6

Stair design and user comfort: Which type of stair is generally considered the most comfortable for everyday use in buildings due to uniform goings, easy orientation, and straightforward handrail arrangement?

Accepted Answer

Introduction / Context:Comfort in stair design depends on uniform treads/risers, clear sight lines, and simple movement patterns. For everyday buildings, ease of ascent and descent across all ages is important.Given Data / Assumptions:Standard building usage (residential/office).Uniform riser and going dimensions per codes.No architectural constraints forcing curved or winders.Concept / Approach:Straight stairs provide the simplest movement: users proceed in a single direction with uniform goings and risers, and a continuous handrail. There are no triangular winders or turning landings that force gait adjustment, making them the most comfortable for most users.Step-by-Step Solution:Compare stair types by uniformity and gait consistency.Identify that straight stairs avoid variable tread widths common in winders.Conclude that straight stairs generally offer the best comfort.Verification / Alternative check:Ergonomic guidance and many architectural texts recommend straight stairs for maximum comfort and safety, provided space permits.Why Other Options Are Wrong:Dog-legged: requires a 180° turn; can interrupt rhythm though still common.Geometrical: includes winders/curved treads with variable going; less comfortable.Open newel: often includes large voids and turns, not inherently more comfortable.Common Pitfalls:Equating architectural elegance with comfort; neglecting consistent riser/going proportions and adequate handrail design.Final Answer:Straight stair

Question 7

Working-stress design relation: If C is the permissible compressive stress in concrete (kg/cm²), the modular ratio m used in working-stress R.C.C. design is commonly taken as which of the following?

Accepted Answer

Introduction / Context:In classic working-stress design (WSD), a modular ratio m converts concrete area to an equivalent steel area. Many objective questions test the empirical relation connecting m with the permissible compressive stress in concrete.Given Data / Assumptions:Permissible compressive stress in concrete = C (kg/cm²).WSD philosophy with long-term effects implicitly included in m.Concept / Approach:A standard empirical expression frequently adopted is m = 280 / (3 C). This accounts in a simplified manner for creep and long-term effects by reducing the effective stiffness contribution of concrete when transformed to steel.Step-by-Step Solution:Recall WSD relation: m = 280 / (3 C).Check units and that C is the permissible compressive stress (not characteristic strength).Select option matching the standard formula.Verification / Alternative check:Several traditional design manuals use 280/(3C); some variations exist historically, but this expression remains a common exam value.Why Other Options Are Wrong:280/C: ignores the one-third factor used in the conventional empirical relation.350/(3C) or 230/(3C): nonstandard constants for this context.Common Pitfalls:Confusing permissible stress C with characteristic strength; applying this WSD relation in limit-state design without proper conversion.Final Answer:m = 280 / (3 C)

Question 8

Main reinforcement spacing in slabs: For a simply supported reinforced concrete slab, the centre-to-centre spacing (pitch) of main steel bars shall not exceed how many times the slab’s effective depth?

Accepted Answer

Introduction / Context:Bar spacing limits in slabs prevent wide crack widths and ensure proper distribution of stress. Codes specify maximum spacing as a multiple of the effective depth and as an absolute value, whichever is smaller.Given Data / Assumptions:Simply supported slab with main tension reinforcement.Standard code-type spacing rules applied.Concept / Approach:For main reinforcement in slabs, a common limit is: spacing ≤ 3 × effective depth (and also ≤ 300 mm as an absolute limit in many codes). These limits ensure adequate crack control and structural performance.Step-by-Step Solution:Identify the governing rule: spacing ≤ 3d (and ≤ 300 mm).Since the question asks “not exceed its effective depth by how many times,” choose 3 times.Hence, select “Three times the effective depth.”Verification / Alternative check:Typical slab detailing examples confirm that the 3d limit is standard for main bars in tension zones.Why Other Options Are Wrong:4d, 5d, 6d, 7d: exceed the usual code-prescribed limit for main reinforcement spacing.Common Pitfalls:Forgetting the absolute limit (e.g., 300 mm) in addition to the 3d rule; mixing distribution and main bar limits.Final Answer:Three times the effective depth

Question 9

Prestressed rectangular beam under bending: A tendon with prestressing force P acts on a rectangular section of area A and section modulus Z. If the beam is subjected to a maximum bending moment M, what is the minimum extreme-fiber stress f (conside…

Accepted Answer

Introduction / Context:Precompression from prestress counteracts tensile stresses from external bending moments. For a rectangular section with tendon near the centroid, the resultant stress at an extreme fiber equals the algebraic sum of uniform prestress P/A and bending stress M/Z with appropriate sign.Given Data / Assumptions:P = prestressing force in the tendon.A = gross cross-sectional area.Z = section modulus about the relevant axis.M = maximum external bending moment (sagging). Tendon at or near centroid so eccentricity is negligible for this expression.Concept / Approach:Stress at an extreme fiber = direct stress (from P) ± bending stress (from M). For the tension-critical extreme fiber under sagging bending, bending stress is tensile (positive tension). Prestress P/A provides uniform compression, thus algebraic combination is f = P/A − M/Z at the fiber where the bending stress would otherwise be tensile.Step-by-Step Solution:Compute uniform compressive stress due to prestress: σp = P/A.Compute bending stress at extreme fiber: σb = M/Z (tension under sagging at soffit).Combine algebraically for the tension-critical fiber: f = σp − σb = P/A − M/Z.This gives the minimum (most negative) stress among fibers.Verification / Alternative check:If M/Z > P/A, net tension occurs; otherwise the fiber remains in compression. This check matches standard prestress serviceability calculations.Why Other Options Are Wrong:P/A + M/Z: would increase compression and does not represent the tension-critical fiber under sagging.M/Z − P/A: same magnitude but sign-reversed; not the “minimum” stress at tension-critical side.(P/A)(M/Z): incorrect dimensional form.Common Pitfalls:Forgetting the sign convention; ignoring eccentricity when it is actually significant (then add ± P*e/Z term).Final Answer:f = P/A - M/Z

Question 10

Balanced reinforced concrete section: What do we call the cross-section of an R.C.C. beam in which the extreme concrete in compression and the tension steel simultaneously reach their respective permissible stresses at the same load level?

Accepted Answer

Introduction / Context:Design philosophies distinguish between under-reinforced, balanced, and over-reinforced sections. Understanding “balanced section” helps clarify ductility and failure mode expectations for safe design.Given Data / Assumptions:R.C.C. beam under bending.Concrete in compression and steel in tension have specified permissible stresses (working-stress) or design limits (limit-state).Concept / Approach:In a balanced section, the chosen reinforcement ratio is such that, at the limiting state, concrete in compression and steel in tension both reach their allowable/limit stresses simultaneously. This condition defines the balanced steel percentage and neutral axis depth for that material set and stress limits.Step-by-Step Solution:Define balanced: concrete and steel reach permissible stresses together.Contrast with under-reinforced: steel yields first (ductile).Contrast with over-reinforced: concrete crushes first (brittle).Hence, the named condition is “balanced section.”Verification / Alternative check:Textbook stress–strain diagrams and rectangular stress block analyses confirm the balanced condition at a specific reinforcement ratio for a given concrete and steel grade.Why Other Options Are Wrong:Economic section: refers to minimal cost/weight for required capacity, not necessarily balanced.Critical section: generic phrase; not the definition of balanced condition.All of the above: only one is the precise definition.Common Pitfalls:Assuming balanced sections are always desirable; in modern limit-state design, under-reinforced (ductile) behavior is generally preferred.Final Answer:Balanced section

Question 11

Reinforced concrete ribbed slabs (narrow ribs): What is the maximum permissible nominal size of coarse aggregate to be used when casting the ribs of a slab, considering the narrow rib width and the need for proper compaction and cover?

Accepted Answer

Introduction / Context:In reinforced concrete ribbed (joist) slabs, the ribs are relatively narrow. Selecting an appropriate nominal maximum size of coarse aggregate is essential to ensure good workability, proper compaction, and adequate cover around reinforcement without honeycombing.Given Data / Assumptions:Narrow rib width typical of ribbed slabs.Standard cover and bar spacing must be maintained.Concrete must flow and compact properly within the rib.Concept / Approach:Code-based and good-practice limits state that the nominal maximum aggregate size should be less than the minimum clear distance between bars, less than three-fourths of the clear cover, and usually not more than about one-third of the section depth in very thin members. For narrow ribs, a 10 mm aggregate is widely adopted to avoid blocking and to improve consolidation quality.Step-by-Step Solution:Check rib width and reinforcement layout to identify clear spacing.Apply the rule: max size ≤ min(clear spacing, 3/4 cover, practical thin-section limit).For narrow ribs, a 10 mm nominal size satisfies these constraints and improves placement quality.Verification / Alternative check:Site trials and slump tests often show that larger aggregates (15–20 mm) increase the risk of voids and segregation in tight ribs. Reducing to 10 mm improves passing ability around bars and corners and reduces vibration effort.Why Other Options Are Wrong:5 mm and 7.5 mm: These are closer to fine aggregate/very small coarse aggregate; unnecessarily small for economy and strength in most ribs.15 mm and 20 mm: Too large for narrow ribs; higher risk of honeycombing and cover violations.Common Pitfalls:Ignoring the bar spacing/cover check and choosing aggregate solely by availability.Using high slump to compensate for large aggregate instead of selecting the right size.Final Answer:10 mm

Question 12

One-way (simply supported) slabs — bar curtailment rule: In a simply supported slab, at what fraction of the span are alternate tension bars commonly curtailed (cut off) from the midspan side towards the support under standard detailing practice?

Accepted Answer

Introduction / Context:Curtailment of reinforcement in slabs and beams is used to reduce congestion and improve economy while maintaining structural safety. In simply supported slabs, alternate bars may be curtailed because positive bending moments are highest near midspan and reduce towards supports.Given Data / Assumptions:One-way slab, simply supported.Uniform loading scenario typical of floors/roofs.Standard detailing practices for bar extension and curtailment.Concept / Approach:The intent is to keep full reinforcement where moments are largest and reduce steel where moments diminish. A conventional detailing guideline adopted in many practice-oriented question banks is to curtail alternate bars at approximately 1/7 of the span from the support side, while ensuring minimum anchorage/development lengths are satisfied.Step-by-Step Solution:Identify maximum positive moment region at midspan (keep full bars continuous there).Towards supports, moment reduces; alternate bars can be terminated safely at a codified fraction of span while meeting development length requirements.Adopt the common objective-test guideline distance = L/7 from support for curtailment of alternate bars in simply supported slabs.Verification / Alternative check:A more rigorous approach would compute actual cut-off points using moment diagrams and development length checks. Nevertheless, the L/7 rule is a frequently cited conservative detailing cue for exam-style problems.Why Other Options Are Wrong:L/4, L/5, L/6: These do not reflect the conventional slab-curtailment rule used in such questions; also may not provide the same balance between economy and anchorage.None of these: Incorrect because a commonly accepted proportion exists for such MCQs.Common Pitfalls:Ignoring development length; bars must not be cut where anchorage is insufficient.Cutting too close to regions with still-appreciable bending demand.Final Answer:1/7 of the span

Question 13

Minimum diameter of main longitudinal bars in RCC columns: What is the minimum nominal bar diameter permitted for the main (longitudinal) reinforcement in a reinforced concrete column?

Accepted Answer

Introduction / Context:Main longitudinal reinforcement in columns must satisfy minimum size requirements to ensure ductility, robustness during handling, and adequate bond with surrounding concrete. Very small bars are prone to buckling between ties and offer poor confinement compatibility.Given Data / Assumptions:Reinforced concrete column (short or long).Main bars refer to the longitudinal steel, distinct from lateral ties.Standard building-code practice is assumed.Concept / Approach:Common RC codes specify a minimum nominal diameter for main column bars. A widely adopted limit in building practice is 12 mm minimum for longitudinal bars in columns, which balances constructability, stiffness, and bond while avoiding excessive bar count.Step-by-Step Reasoning:Check code minimum diameter for longitudinal column bars.Select the smallest diameter satisfying the requirement: 12 mm.Ensure tie spacing and bar arrangement are then designed for confinement and buckling control.Verification / Alternative check:Practical detailing handbooks and code commentaries reinforce the 12 mm minimum as a robust lower bound, especially under construction tolerances and the need for adequate development length.Why Other Options Are Wrong:6, 8, 10 mm: Too small for main column bars; risk of instability and poor bond.Common Pitfalls:Confusing main bars with confinement ties (which can be smaller).Ignoring clear cover and spacing constraints when increasing bar count.Final Answer:12 mm

Question 14

Nominal shear reinforcement in beams: In a reinforced concrete beam zone where the nominal shear stress is less than 5 kg/cm^2, nominal shear reinforcement (stirrups) is provided at a pitch equal to which fraction of the lever arm (z) of the section?

Accepted Answer

Introduction / Context:Even when calculated shear demand is small, nominal shear reinforcement is mandated to control diagonal cracking, provide ductility, and confine longitudinal steel. The pitch (spacing) of stirrups is often expressed as a fraction of the effective depth or lever arm z of the section.Given Data / Assumptions:Beam zone with nominal shear stress Lever arm z approximately equal to 0.9 d (where d is effective depth), used for practical spacing rules.Concept / Approach:Where shear is low, codes permit larger stirrup spacings but still require a minimum amount. A common rule-of-thumb (reflected in many objective-type questions) is to limit the pitch to around 0.5 z in such zones, subject to absolute maximum spacing limits and minimum shear steel requirements.Step-by-Step Reasoning:Identify that only nominal shear reinforcement is required.Adopt the guideline pitch ≈ 0.5 z (also cross-check with code max spacing limits such as a multiple of bar diameter or 0.75 d).Provide closed stirrups with adequate anchorage and hook details.Verification / Alternative check:Design charts and code clauses for minimum shear reinforcement will generally lead to a similar or slightly more conservative spacing once maximum spacing caps (e.g., 300 mm or 0.75 d, whichever is less) are applied.Why Other Options Are Wrong:One-third z: Overly conservative for nominal zones.Full z or 1.5 z: Excessive spacing; risks diagonal cracking and does not meet minimum reinforcement philosophy.Common Pitfalls:Not checking absolute maximum spacing limits in the governing code.Ignoring anchorage/hook requirements of stirrups at the compression face.Final Answer:one-half of the lever arm

Question 15

Slab–column interaction (transverse bending zone): The plan extent of the zone where significant transverse bending of a slab is likely near a column face can be obtained by drawing lines from the column faces at what inclination θ to the horizontal?

Accepted Answer

Introduction / Context:Near slab–column connections, load transfer and bending are three-dimensional. Designers often identify an influence or dispersion zone emanating from the column faces to approximate where higher bending effects occur and to guide reinforcement placement, punching checks, and detailing.Given Data / Assumptions:Flat slab or slab supported on columns.Identification of a simplified dispersion zone.Guidance for preliminary detailing and checking.Concept / Approach:A commonly used dispersion angle in practice-oriented problems is 45°. Projecting 45° lines from column faces provides a practical estimate of the area influenced by high transverse bending and shear, aiding placement of top steel around columns and informing punching shear perimeters (which themselves are defined by code, e.g., at a distance of d/2 from the loaded area).Step-by-Step Reasoning:Mark the column perimeter in plan.From each face, project lines at 45° to identify the critical bending region.Ensure adequate top reinforcement and anchorage within this zone and check punching shear independently.Verification / Alternative check:Advanced analysis (e.g., finite-element plate models) yields a similar qualitative influence region radiating from column faces; 45° is a reasonable simplified envelope for routine detailing guidance questions.Why Other Options Are Wrong:30° or 60°: These are not the standard simplified dispersion angles used in such MCQs.None of these: A simple and widely recognized angle exists for this purpose.Common Pitfalls:Confusing this detailing cue with the exact code-defined punching critical perimeter (at distance d/2).Neglecting to provide sufficient top steel cages around columns.Final Answer:45°

Question 16

Short RC column (load factor method): According to the load factor method, the permissible axial load W on a short reinforced concrete column with longitudinal bars and lateral stirrups is taken as which of the following?

Accepted Answer

Introduction / Context:Axial load capacity of short columns in reinforced concrete arises from both concrete and longitudinal steel. The load factor (working stress–type) approach sums the contributions of each material at their permissible stresses to estimate allowable axial load.Given Data / Assumptions:Short column (buckling effects neglected).Longitudinal bars with lateral ties (stirrups) for confinement.Use of permissible (allowable) stresses for concrete and steel.Concept / Approach:The basic expression is W_allowable = σ_c,perm * A_c + σ_s,perm * A_s, where σ_c,perm and σ_s,perm are the permissible compressive stresses for concrete and steel respectively, A_c is net concrete area (gross minus steel), and A_s is area of longitudinal steel. Reduction factors for slenderness or eccentricity apply as required by design provisions.Step-by-Step Solution:Identify material permissible stresses.Compute areas A_c and A_s.Sum contributions: W = σ_c,perm * A_c + σ_s,perm * A_s.Verification / Alternative check:Limit state design (ultimate strength method) uses factored loads and strengths with different equations; however, under the load factor / working stress framing of this question, the linear addition of permissible stress contributions is correct for short columns under concentric load.Why Other Options Are Wrong:Options a and b consider only one material, underestimating capacity.Option d dismisses the established expression.Common Pitfalls:Forgetting to reduce A_c by A_s when computing concrete contribution.Ignoring eccentricity, which reduces permissible load.Final Answer:Stress in concrete × Area of concrete + Stress in steel × Area of steel

Question 17

End cover for reinforcing bars (bar termination): The minimum thickness of cover at the end (terminating face) of a reinforcing bar should be at least twice the bar diameter, subject to a lower bound not less than which absolute value?

Accepted Answer

Introduction / Context:Concrete cover protects reinforcement from corrosion and fire and provides adequate anchorage. At the bar termination (ends), cover must be sufficient to prevent splitting and spalling and to allow effective transfer of bond stresses.Given Data / Assumptions:Cover at the end face where a reinforcing bar terminates.Rule: cover ≥ 2 × bar diameter with a minimum absolute value.Typical building exposure conditions assumed.Concept / Approach:Many detailing guides prescribe that end cover should not be less than twice the bar diameter and, additionally, should not be less than an absolute minimum such as 25 mm. This ensures that the concrete cone around the bar end is robust and bond transfer is reliable.Step-by-Step Reasoning:Compute 2d for the bar in question.Compare with the absolute minimum (25 mm) and adopt the larger value.Provide proper compaction and avoid cutting bars too close to the end face.Verification / Alternative check:Project specifications may increase cover for harsh exposures; however, the 25 mm absolute minimum is a widely cited baseline in many exam contexts for normal exposures.Why Other Options Are Wrong:10, 15, 20 mm: Too small as an absolute minimum for bar ends in typical structural members.30 mm: May be specified for severe exposure, but the baseline minimum used in such questions is 25 mm.Common Pitfalls:Confusing side cover, top/bottom cover, and end cover requirements.Not accounting for concrete tolerance when detailing end cover.Final Answer:25 mm

Question 18

Combined footing depth governed by punching shear: Based on punching shear considerations under column A, the overall depth of a combined footing should be chosen so that which condition is satisfied?

Accepted Answer

Introduction / Context:Combined footings support two or more columns and are designed for bending, one-way shear, and two-way (punching) shear. Of these, punching shear near a column can govern the required thickness, especially under high concentrated loads.Given Data / Assumptions:Combined footing supporting at least one column A.Critical section for punching shear taken around the column.Allowable shear stress of concrete considered without shear reinforcement (unless provided).Concept / Approach:The critical perimeter for punching is typically located at a distance d/2 from the loaded area (column face) all around. The design condition is: V_u / (b_0 * d) ≤ τ_c,allowable, where V_u is factored shear, b_0 is critical perimeter length, and d is effective depth. Satisfying this often controls the overall depth.Step-by-Step Solution:Determine column load and footing reaction near the column.Compute V_u at the critical section located d/2 from the column face.Choose depth so that V_u / (b_0 * d) ≤ τ_c,allowable; increase overall depth if needed.Verification / Alternative check:After meeting punching, check one-way shear along a line at distance d from column face, and bending moments along the footing length. Adjust bar sizes and layout accordingly.Why Other Options Are Wrong:Bending stress limit (option b) controls steel area, not necessarily overall depth.Average soil pressure (option c) relates to sizing and proportioning plan area, not thickness.None of these: The punching criterion is the standard depth-governing check.Common Pitfalls:Using the wrong punching perimeter location (not at d/2).Ignoring load eccentricity and column moments that increase V_u.Final Answer:The nominal punching shear stress at the critical perimeter located at a distance d/2 from the column face does not exceed the allowable shear stress of concrete

Question 19

Loads on stairs — conversion from slope area to horizontal plan area: If a stair has tread T and rise R, and carries a uniformly distributed load w per square metre on its sloping surface, what is the equivalent load intensity per square metre when …

Accepted Answer

Introduction / Context:Stairs are often loaded and detailed on their sloping surface, but many design calculations and drawings refer loads to the horizontal plan area. A clear conversion between sloping-surface UDL and plan UDL is needed for consistent load take-off and design of supporting members.Given Data / Assumptions:Tread = T (horizontal going per step).Rise = R (vertical per step).Uniform load on slope = w (kN/m^2 or similar).Concept / Approach:For one step, the sloping length equals √(T^2 + R^2). The plan (horizontal) projection for that step is T. Load per plan area equals load per slope area multiplied by the ratio of slope area to plan area, which reduces to the ratio of sloping length to tread for unit width.Step-by-Step Solution:Slope area per unit width of one step = √(T^2 + R^2).Plan area per unit width of one step = T.Equivalent plan UDL = w × (slope area / plan area) = w × √(T^2 + R^2) / T.Verification / Alternative check:Dimensional consistency holds: ratio is nondimensional, so the units remain as UDL. For small R (gentle stair), √(T^2 + R^2) ≈ T, giving plan load ≈ w, as expected. For steep stairs (larger R), plan intensity increases appropriately.Why Other Options Are Wrong:w × T / √(T^2 + R^2) and w × R / √(T^2 + R^2): These invert the correct ratio.w × T / R or w × √(T^2 + R^2) / R: Do not represent the projection relationship.Common Pitfalls:Confusing plan load with line load on stringers; the latter requires separate derivation.Mixing total flight loads with per-step geometry.Final Answer:w × √(T^2 + R^2) / T

Question 20

Precast concrete piles — commonly adopted cross-section: Which cross-sectional form is most commonly used for precast concrete piles to minimize edge damage during handling and driving?

Accepted Answer

Introduction / Context:Precast concrete piles must withstand handling, transportation, and driving stresses. Cross-sectional shape influences durability of corners, ease of reinforcement placement, and driving performance in varying soil conditions.Given Data / Assumptions:Precast (not cast-in-situ) piles.Driven or jacked into ground; risk of edge spalling exists.Standard reinforcement cages and covers considered.Concept / Approach:A plain square section has vulnerable sharp corners that can spall under impact. Chamfering the corners distributes stress and reduces chipping during handling and driving. This detailing choice is widely used in practice for durability and constructability while keeping formwork simple.Step-by-Step Reasoning:Select a section that is easy to fabricate and reinforce.Modify corners to reduce stress concentration and spalling risk.Adopt a square cross-section with chamfered corners as the most common compromise.Verification / Alternative check:Octagonal and circular sections also mitigate corner damage, but they can be more complex or costly to form. For many projects, the chamfered square gives the best balance of performance and economy.Why Other Options Are Wrong:Circular: Good for driving but usually more expensive formwork and reinforcement cages.Square: Vulnerable sharp corners.Octagonal: Better than square but less common and more complex than chamfered square.Common Pitfalls:Insufficient cover at corners even after chamfering.Ignoring reinforcement layout needed to control handling and driving stresses.Final Answer:square with corners chamfered

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