Question 1

A simply supported R.C.C. beam, span 6 m, effective depth d = 50 cm, carries a uniformly distributed load of 2400 kg/m (including self-weight). With lever arm factor j = 0.85 and permissible steel stress σst = 1400 kg/cm², compute the required steel…

Accepted Answer

Introduction / Context: Sizing tension steel for a simply supported beam under UDL uses the bending moment at mid-span and the internal lever arm between compression and tension resultants. Working-stress design relates moment capacity to steel stress and lever arm. Given Data / Assumptions: Span L = 6 m. Effective depth d = 50 cm. UDL w = 2400 kg/m (kgf units). Lever arm factor j = 0.85 → z = jd. Permissible steel stress σst = 1400 kg/cm². Concept / Approach: For simply supported beam under UDL, maximum bending moment M = wL²/8. Moment resistance in working-stress: M = Tz where T = Astσst and z = jd. Step-by-Step Solution: Compute M: M = 2400 * 6² / 8 = 2400 * 36 / 8 = 2400 * 4.5 = 10800 kg·m.Convert to kg·cm: 10800 * 100 = 1,080,000 kg·cm.Lever arm z = jd = 0.85 * 50 = 42.5 cm.Ast = M / (σst * z) = 1,080,000 / (1400 * 42.5) = 1,080,000 / 59,500 ≈ 18.15 cm².Choose the nearest provided value: 18 cm². Verification / Alternative check: Back-compute capacity with 18 cm²: T = 18*1400 = 25,200 kg; Mres = 25,200 * 42.5 = 1,071,000 kg·cm, close to demand (difference due to rounding). Slight conservatism can be handled with bar selection. Why Other Options Are Wrong: 14–17 cm²: Under-reinforced for the given stress limit and lever arm; moment capacity would be insufficient. Common Pitfalls: Mixing N and kgf; forgetting to convert m to cm; using overall depth instead of effective depth; misusing j. Final Answer: 18 cm²

Question 2

Flat slab (beam-less) construction: which of the following practical advantages and performance statements are true?

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Introduction / Context: Flat slabs eliminate downstand beams, yielding a flat soffit and transferring loads directly from slab to columns via column heads and drop panels. This has architectural, services, and structural implications in building design. Given Data / Assumptions: Flat slab system with appropriate column heads and/or drop panels where required. Normal building occupancy and service routing. Concept / Approach: The flat soffit provides uniform light diffusion, reduces deep beam shadows, and eases suspended ceiling coordination. Absence of beams simplifies routing of ducts, pipes, and conduits. Structurally, shear (punching) at columns is addressed with drops/heads so that significant concentrated loads can be managed within code limits. Step-by-Step Solution: Assess architectural benefit: plain ceiling → better light distribution, spacious feel.Assess services benefit: no beams → easier MEP (mechanical, electrical, plumbing) layout and installation.Assess maintenance benefit: flat surfaces → easier painting and access.Assess structural performance: drops/column capitals enhance punching shear and negative moment capacity, enabling concentrated loads within design checks. Verification / Alternative check: Design guides show increased punching shear capacity with adequate drop panels/stirrups; many commercial buildings employ flat slabs for these combined benefits. Why Other Options Are Wrong: Each individual statement (a–d) is correct; therefore the combined option “All the above” is the most accurate. Common Pitfalls: Ignoring punching shear checks; omitting drop panels where needed; assuming flat slabs suit very long spans without supplementary measures (post-tensioning or column layout optimization). Final Answer: All the above

Question 3

Two-way slab bending: If a slab is simply supported on all four edges and spans in two directions with equal side lengths (a square slab), by what factor is the maximum bending moment in a one-way slab reduced due to two-way action? Select the stand…

Accepted Answer

Introduction / Context: When a reinforced concrete slab is supported on all four sides and the side lengths are equal (a square panel), the load is carried in two directions. This two-way action reduces the bending moment in each direction compared to a one-way slab spanning in a single direction only. Given Data / Assumptions: Simply supported slab on all four edges. Equal sides, so span along x equals span along y (square slab). Uniformly distributed load is considered for basic design comparison. Concept / Approach: In a one-way simply supported slab, the maximum midspan moment is M = w * l^2 / 8. For a square, simply supported two-way slab under uniform load, classical plate theory and standard design tables indicate that the load is shared by orthogonal strips. Each direction carries approximately half the one-way moment, giving a multiplying factor of 0.5 for design moments in each direction. Step-by-Step Solution: Identify base moment (one-way): M1 = w * l^2 / 8.Recognize two-way sharing for a square slab: moments distribute equally in both directions.Adopt the standard factor: Mx = My = 0.5 * (w * l^2 / 8).Hence the multiplying factor is 0.5. Verification / Alternative check: Design handbooks and code commentaries present coefficient tables for two-way slabs; for a square, simply supported slab, the coefficients in both directions are equal and sum to the one-way value, validating the 0.5 factor per direction. Why Other Options Are Wrong: 0.2, 0.3, 0.4: Underestimate moments and would be unsafe. 0.7: Overestimates per-direction moment for a square panel; not economical. Common Pitfalls: Confusing total plate moment distribution with per-direction design moments; using one-way moments for a square two-way panel leads to overdesign. Final Answer: 0.5

Question 4

Stair design: For stairs spanning horizontally (waist slab spanning between supports), what is the minimum waist (minimum structural thickness along the slope) generally provided for safe and serviceable construction?

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Introduction / Context: The waist of a stair slab is the minimum structural thickness measured perpendicular to the soffit along the slope. Providing an adequate minimum waist ensures strength, stiffness, durability, and sufficient cover to reinforcement for stairs spanning horizontally between supports. Given Data / Assumptions: Reinforced concrete stair flight spanning horizontally. Conventional residential or institutional usage with standard loads. Minimum practical thickness targets based on common code-aligned practice. Concept / Approach: Practical design balances structural capacity with constructability. Too small a waist can lead to inadequate cover, difficulty in placing bars, excessive deflection, and vibration. Common practice adopts a minimum waist of about 80 mm for general stairs; higher values (100–120 mm) are used for longer spans, heavier loads, or vibration control. Step-by-Step Solution: Identify the design goal: safe, serviceable minimum thickness.Review typical code guidance and practice: minimum waist about 8 cm for standard spans and loads.Select the smallest value satisfying cover, bar placement, and stiffness requirements: 8 cm. Verification / Alternative check: Serviceability checks on deflection and vibration for typical tread-riser geometry indicate 8–10 cm waist is adequate for common spans; 8 cm is widely accepted as a baseline minimum. Why Other Options Are Wrong: 4 cm, 6 cm: Insufficient for cover and stiffness; impractical to reinforce. 10 cm, 12 cm: Safe but not the minimum; chosen when spans/loads demand extra stiffness. Common Pitfalls: Ignoring cover requirements and bar diameters; using too thin a waist can compromise durability and cause cracking. Final Answer: 8 cm

Question 5

Foundation detailing: In cantilever retaining wall foundations, the toe projection of the base slab (distance from stem face to front edge) is commonly taken as which proportion of the total base width for preliminary sizing?

Accepted Answer

Introduction / Context: Preliminary proportioning of cantilever retaining walls includes selecting a sensible toe projection to satisfy stability (overturning, sliding, and bearing) and facilitate economical reinforcement layouts. A practical starting fraction of the base width helps streamline early design before detailed checks. Given Data / Assumptions: Conventional cantilever retaining wall with base slab divided into toe and heel. Preliminary sizing stage (detailed stability checks to follow). Typical soil parameters and surcharge; no extreme conditions assumed. Concept / Approach: A common rule of thumb is to start with a toe projection about one third of the total base width, with the remaining two thirds toward the heel under the backfill. This initial proportion generally leads to acceptable bearing pressure distribution and adequate lever arm for resisting overturning, subject to refinement by calculations. Step-by-Step Solution: Adopt a preliminary base width B from empirical relations (often 0.5H–0.7H).Propose toe ≈ B/3 and heel ≈ 2B/3 for initial layout.Perform stability checks and adjust toe/heel as needed to satisfy safety factors and serviceability. Verification / Alternative check: Trial sections with toe near B/3 frequently yield bearing pressure diagrams with the resultant kept within the middle third, facilitating economical reinforcement. Why Other Options Are Wrong: One sixth of overall height: Not a standard proportioning rule for toe projection. Equal to heel slab: The heel is commonly larger than the toe. Below ground surface / variable regardless of stability: Not meaningful sizing guidance. Common Pitfalls: Fixing toe width without running stability checks; final dimensions must be validated for overturning, sliding, and bearing stress limits. Final Answer: as one third of the base

Question 6

Column ties: The diameter of transverse reinforcement (lateral ties) in reinforced concrete columns should be at least one-fourth of the diameter of the largest longitudinal bar, but not less than what absolute minimum diameter?

Accepted Answer

Introduction / Context: Transverse reinforcement (ties) confines concrete and prevents buckling of the longitudinal bars in columns. Codes prescribe minimum diameters to ensure adequate stiffness and robustness of ties during construction and service. Given Data / Assumptions: Tie diameter must be ≥ one-fourth of the largest longitudinal bar diameter. An absolute minimum tie diameter also applies to avoid flimsy ties. Ordinary reinforced concrete columns in buildings. Concept / Approach: Even when one-fourth of the largest bar would permit a very small tie, practice imposes a floor on tie size. Commonly adopted absolute minimum is 6 mm for mild steel ties (and often 8 mm for high-strength deformed, depending on code/project). Thus, when one-fourth rule yields less than 6 mm, the minimum 6 mm governs. Step-by-Step Solution: Check one-fourth of largest longitudinal bar.Compare with absolute minimum permitted for ties.Adopt the greater value; for small bars, this is typically 6 mm. Verification / Alternative check: Design examples and detailing guides routinely specify 6 mm or 8 mm ties; 6 mm is a widespread minimum for conventional building columns with smaller longitudinal bars. Why Other Options Are Wrong: 4 mm, 5 mm: Too slender and difficult to handle; below common minimum. 7 mm, 8 mm: Acceptable but not the minimum in the general case posed. Common Pitfalls: Using only the fractional rule and forgetting the absolute minimum leads to under-sized, flexible ties that are hard to place and insufficiently confining. Final Answer: 6 mm

Question 7

Bond (development) length: For a reinforcing bar of diameter d subjected to allowable tensile stress ft in concrete with allowable bond stress fb, what is the required development/bond length Lb?

Accepted Answer

Introduction / Context: The bond length (development length) is the minimum embedment required so that the stress in a reinforcing bar can be safely transferred to surrounding concrete through bond. It is a fundamental detailing parameter in reinforced concrete design. Given Data / Assumptions: Bar diameter = d. Allowable tensile stress in steel = ft. Allowable bond stress in concrete = fb. Straight bar in tension; standard bond conditions. Concept / Approach: Equating the tensile force in steel to the bond resistance along the embedded length gives the basic formula. Tensile force = area * stress = (π d^2 / 4) * ft. Bond resistance = perimeter * length * bond stress = (π d) * Lb * fb. Solving for Lb yields the classical expression. Step-by-Step Solution: Tension in steel: T = (π d^2 / 4) * ft.Bond capacity: R = (π d) * Lb * fb.Equate T = R and simplify: (π d^2 / 4) * ft = (π d) * Lb * fb.Cancel π d and solve: Lb = (d * ft) / (4 * fb). Verification / Alternative check: The derived expression matches standard RC design texts for straight tension bars under allowable stress design assumptions. Why Other Options Are Wrong: Forms with 2fb or 3fb in denominator mis-state the equilibrium constants. Inverse or squared-diameter forms do not come from basic force equilibrium of bar area vs. bond perimeter. Common Pitfalls: Forgetting to multiply perimeter by length, or using area times bond stress; mixing design stress formats (working vs. ultimate) without adjusting fb values. Final Answer: Lb = (d * ft) / (4 * fb)

Question 8

Retaining wall sizing: For a retaining wall of overall height H that retains soil with surcharge, what is the usual preliminary proportion adopted for the total base slab width (B) before detailed stability checks?

Accepted Answer

Introduction / Context: Preliminary sizing of cantilever retaining walls is commonly done using proportioning rules that are later refined by detailed checks for overturning, sliding, and allowable bearing pressure. When surcharge exists on the backfill, a wider base is typically required than for the no-surcharge case. Given Data / Assumptions: Overall wall height = H. Soil retained with surcharge loading. Conventional cantilever RC retaining wall on spread footing. Concept / Approach: Rules of thumb often choose total base width B between 0.5 H and 0.7 H depending on soil properties and surcharge. With surcharge present, a value around 0.6 H is a widely used starting point that provides a reasonable lever arm for resisting overturning while keeping bearing pressures within limits after optimization. Step-by-Step Solution: Select B preliminarily as 0.6 H to account for surcharge effects.Divide B into toe and heel (e.g., toe ≈ B/3, heel ≈ 2B/3 for trial).Carry out stability and bearing checks; adjust B as needed. Verification / Alternative check: Experience-based design charts and example problems commonly converge to B in the 0.55 H–0.65 H range when surcharge is present, making 0.6 H a practical initial choice. Why Other Options Are Wrong: 0.3 H, 0.4 H: Typically too small under surcharge; likely to fail overturning/bearing checks. 0.5 H: Common without surcharge; may be marginal with surcharge. 0.7 H: Safe but often uneconomical as a starting point. Common Pitfalls: Treating preliminary proportions as final; ignoring soil strength and groundwater can lead to unsafe or overly conservative designs. Final Answer: 0.6 H

Question 9

RCC slab design (working stress approach): If the maximum bending moment for a simply supported slab (per metre width) is M kg·cm and the material strength factor (moment of resistance factor) is Q, what is the required effective depth d of the slab?

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Introduction / Context: This question checks your understanding of the working-stress design relation for slabs, where the ultimate (allowable) moment capacity per metre width is expressed using a material constant Q. Slabs are normally designed per 1 m width, so breadth b = 100 cm when using kg·cm units. Given Data / Assumptions: Simply supported slab designed by working-stress method. Maximum bending moment per metre width = M kg·cm. Moment of resistance factor (material factor) = Q. Per metre width: b = 100 cm. Concept / Approach: For slabs in WSM, the limiting (safe) moment of resistance is M = Q * b * d^2. Here Q groups material and section constants. Rearranging gives the effective depth d needed for the demand moment M. Step-by-Step Solution: Start with M = Q * b * d^2.Use b = 100 cm for one-metre strip of slab.Rearrange: d^2 = M / (Q * b) = M / (Q * 100).Therefore d = sqrt(M / (Q * 100)). Verification / Alternative check: Dimensional check: M (kg·cm) divided by Q (kg/cm^2) and by b (cm) leaves cm^2; the square root gives cm, as required for d. Why Other Options Are Wrong: d = sqrt(M / Q): Ignores b = 100 cm and overestimates depth.d = M / (Q * 100): Linear instead of square-root dependence.d = sqrt(Q / M) or d = M / Q: Dimensionally inconsistent. Common Pitfalls: Forgetting to use b = 100 cm for a 1 m strip. Mixing SI (N·mm) with kg·cm units; be consistent. Final Answer: d = sqrt(M / (Q * 100))

Question 10

Lever arm in a singly reinforced RCC beam: Given permissible stresses σ_st = 1400 kg/cm^2 (steel), σ_cbc = 80 kg/cm^2 (concrete), modular ratio m = 18, beam width b = 30 cm, and effective depth d = 46 cm, determine the lever arm z of the section (wo…

Accepted Answer

Introduction / Context: The lever arm z in a singly reinforced beam (working-stress method) is the distance between the compressive resultant in concrete and the tensile force in steel. At the balanced condition, both concrete and steel reach their permissible stresses simultaneously, enabling a direct geometric solution for the neutral axis and lever arm. Given Data / Assumptions: σ_st = 1400 kg/cm^2, σ_cbc = 80 kg/cm^2, modular ratio m = 18. Rectangular section with effective depth d = 46 cm (tension steel depth from top fibre). Working-stress theory, linear stress distribution. Concept / Approach: For a balanced section in WSM: σ_st / σ_cbc = m * (d − x) / x, where x is the neutral axis depth from the top. Once x is known, the lever arm z is taken as the distance from the compressive resultant (at x/3 from the top for a triangular stress block) to the tension steel at depth d, i.e., z = d − x/3. Step-by-Step Solution: Compute ratio: σ_st / (m * σ_cbc) = 1400 / (18 * 80) = 1400 / 1440 ≈ 0.9722.Solve for neutral axis: (d − x) / x = 0.9722 ⇒ x = d / (1 + 0.9722) = 46 / 1.9722 ≈ 23.33 cm.Lever arm: z = d − x/3 = 46 − 23.33/3 ≈ 46 − 7.78 ≈ 38.22 cm.Rounded practical value ≈ 38 cm. Verification / Alternative check: Using the same inputs with more precise arithmetic gives z within a few millimetres of 38 cm; design rounding to whole centimetres is customary at preliminary stages. Why Other Options Are Wrong: 37 cm / 39 cm / 40 cm / 36 cm: Do not match the computed balanced-section lever arm with given stresses and modular ratio. Common Pitfalls: Using z = j * d from reinforced concrete constants without confirming balanced condition. Forgetting that the concrete resultant for a triangular stress block is at x/3 from the top. Final Answer: 38 cm

Question 11

Fundamentals of prestressed concrete: A prestressed concrete member is best described as which of the following?

Accepted Answer

Introduction / Context: Prestressed concrete (PSC) is a structural material in which internal stresses are deliberately introduced to counteract tensile stresses produced by service loads. This enables slimmer, longer spans with reduced cracking and deflection compared to ordinary RC. Given Data / Assumptions: Member is a concrete element with high-strength steel tendons. Stressing may be pre-tensioning or post-tensioning. Concept / Approach: The defining characteristic of PSC is the presence of initial internal compressive stress in concrete caused by tensioned steel. Whether the stressing is before (pre-tensioning) or after (post-tensioning) casting, the key feature is the engineered internal stress state, not merely the material type. Step-by-Step Explanation: Tendons are tensioned relative to the concrete.Anchorage/bond transfers this force, compressing the concrete.Service tensile stresses are reduced or eliminated, improving crack control and deflection. Verification / Alternative check: Pre-tensioned pretensioning beds and post-tensioned ducts both create internal compressive stress; this shared outcome distinguishes PSC from ordinary RC. Why Other Options Are Wrong: is made of concrete / is made of reinforced concrete: Too generic; RC can exist without prestress.is stressed after casting: Only post-tensioning; PSC also includes pre-tensioning.is free of any initial stress: Opposite of PSC definition. Common Pitfalls: Confusing PSC with high-strength RC without deliberate prestress. Thinking only post-tensioned systems count as PSC. Final Answer: possesses internal stresses

Question 12

Singly reinforced beam behavior: If, in a singly reinforced RCC beam, the permissible stress in concrete is reached earlier than the permissible stress in steel, the section is called a(n)…

Accepted Answer

Introduction / Context: Classifying a singly reinforced beam as under-reinforced, balanced, or over-reinforced is essential to predict failure mode and ductility. The sequence in which concrete and steel reach their permissible or ultimate stresses dictates this classification. Given Data / Assumptions: Singly reinforced rectangular section. Comparison is based on permissible (working) stresses or limit state strains. Concept / Approach: Under-reinforced: steel reaches its limit first → tension steel yields before concrete crushes → ductile behavior. Balanced: steel and concrete reach limits simultaneously. Over-reinforced: concrete reaches its limit first → crushing in compression zone governs → brittle behavior. Step-by-Step Reasoning: Given that concrete reaches permissible stress earlier than steel, the compression zone limits govern.Hence the section has more tensile reinforcement relative to balanced, i.e., it is over-reinforced. Verification / Alternative check: Over-reinforced sections are generally avoided in modern limit-state design because they lack ductility and warning before failure. Why Other Options Are Wrong: Under-reinforced: Opposite condition (steel would govern).Economic / critical: Not standard classifications for this context.Balanced: Implies simultaneous attainment, not earlier concrete control. Common Pitfalls: Assuming more steel always improves safety; it can reduce ductility. Final Answer: over reinforced section

Question 13

Shear safety check for a simply supported RCC beam: A beam of width b = 25 cm, clear span = 5 m carries a uniformly distributed load w = 2000 kg/m (inclusive of self-weight). If the lever arm j·d = 45 cm, comment on shear safety near the supports ba…

Accepted Answer

Introduction / Context: Near beam supports, the shear force is maximum. Checking nominal shear stress against permissible values helps decide whether the section is adequate in shear or whether shear reinforcement or section revision is required. Given Data / Assumptions: Width b = 25 cm. Clear span L = 5 m; simply supported; UDL w = 2000 kg/m (includes self-weight). Lever arm j·d = 45 cm (effective shear area taken as b * j * d). Typical permissible concrete shear stress τ_c for common RC percentages is higher than about 4.5 kg/cm^2 (varies with % steel). Concept / Approach: Support shear V = w * L / 2. Nominal shear stress τ_v = V / (b * j * d). Compare τ_v with τ_c (from code tables depending on percentage of steel). If τ_v ≤ τ_c, concrete alone is adequate in shear (nominal links still provided). Step-by-Step Solution: Compute shear: V = 2000 * 5 / 2 = 5000 kgf.Effective area: A_v = b * j * d = 25 * 45 = 1125 cm^2.Nominal shear stress: τ_v = 5000 / 1125 ≈ 4.44 kg/cm^2.Compare with typical τ_c (generally ≥ about 5 kg/cm^2 for usual reinforcement ratios): τ_v ≤ τ_c → section safe in shear. Verification / Alternative check: Using b * d instead of b * j * d gives an even lower τ_v; thus the conclusion remains conservative that the section is safe in shear. Why Other Options Are Wrong: is safe with stirrups / with inclined members: Unnecessary here since τ_v does not exceed τ_c.needs revision: Not warranted by the calculated τ_v.shear cannot be assessed: All required inputs are available. Common Pitfalls: Forgetting that τ_c depends on percentage tension steel; always consult the relevant code table. Mixing units; keep kgf and cm consistent. Final Answer: safe in shear

Question 14

Bar spacing rule (different diameters): For horizontal parallel reinforcement bars of different diameters placed side-by-side, the minimum clear spacing should not be less than _____.

Accepted Answer

Introduction / Context: Minimum clear spacing between parallel bars ensures adequate concrete flow, aggregate accommodation, and bond. When the adjacent bars are of different sizes, the code-based limit is governed by the larger diameter bar. Given Data / Assumptions: Two or more horizontal bars placed in the same layer. Bars of unequal diameters are adjacent. Concept / Approach: Codes specify that the clear spacing shall not be less than the bar diameter (for equal bars). For unequal bars, the controlling dimension is the larger diameter bar. Additional aggregate-based limits may also apply (e.g., not less than 5 mm more than nominal maximum aggregate size), but the bar-diameter criterion here is decisive. Step-by-Step Reasoning: Identify the larger of the adjacent bar diameters.Set minimum clear spacing ≥ diameter of the thicker bar (and also ≥ aggregate-based limit as applicable). Verification / Alternative check: This rule promotes proper compaction and bond; it is consistent with widely used RC design standards for bar spacing. Why Other Options Are Wrong: One diameter of thinner bar / half diameter: May be too small, risking honeycombing and poor bond.Sum or twice thinner bar: Not a general minimum requirement and may be overly conservative or arbitrary. Common Pitfalls: Ignoring aggregate size limits which may govern in some mixes. Confusing clear spacing with centre-to-centre spacing. Final Answer: one diameter of thicker bar

Question 15

Anchorage value of a standard hook in tension reinforcement: If the bar diameter is d, what is the anchorage value contributed by a standard hook?

Accepted Answer

Introduction / Context: Anchorage length provides the necessary bond to develop the bar's tensile force. Standard bends and hooks add equivalent straight length, reducing the straight embedment otherwise required at bar terminations. Given Data / Assumptions: Plain or deformed bar with a standard 180° hook (common detailing at ends). Bar diameter = d. Concept / Approach: Design standards assign equivalent anchorage values to bends and hooks. A standard hook typically contributes an anchorage value of about 16d for a 180° hook in tension, which can be added to the straight development length to meet the required L_d. Step-by-Step Reasoning: Determine required development length from bond considerations.Account for end hook: add 16d as anchorage contribution of the hook. Verification / Alternative check: Many codes tabulate equivalent lengths: 90° bends ≈ 8d, 135° hooks ≈ 12d, and 180° hooks ≈ 16d (exact values may vary slightly by code and bar type). Why Other Options Are Wrong: 4d / 8d / 12d: Correspond to smaller bend angles; insufficient for a standard full hook.20d: More than typical requirement, not standard. Common Pitfalls: Confusing added anchorage value with the total development length needed. Not checking cover and bend radii that affect practicality of detailing. Final Answer: 16d

Question 16

Soil mechanics definition: The angle of internal friction φ of a soil mass is the angle whose…

Accepted Answer

Introduction / Context: In shear strength theory for soils, the angle of internal friction φ characterizes frictional resistance between particles. It appears in the Mohr-Coulomb criterion, relating shear stress to normal stress on a potential failure plane. Given Data / Assumptions: Soil is considered under drained, frictional conditions (c ≈ 0 for clean sands). Maximum resistance refers to limiting shear at failure. Concept / Approach: At failure, for cohesionless soils, τ = σ * tan φ on the critical plane, where τ is shear stress and σ is the normal effective stress. Hence tan φ is the ratio of limiting shear resistance to normal pressure on that plane. Step-by-Step Explanation: Identify the failure condition: shear/normal ratio reaches its maximum.Write τ / σ = tan φ → tan φ represents the ratio of maximum resistance to normal pressure. Verification / Alternative check: For granular soils, direct shear and triaxial tests yield approximately linear envelopes in τ–σ space, with slope equal to tan φ. Why Other Options Are Wrong: Using sine or cosine would not match the Mohr-Coulomb frictional definition.Cotangent statement is not the conventional definition and inverts the intended ratio. Common Pitfalls: Confusing φ with the angle of repose; they are related but not always identical under all conditions. Final Answer: tangent is equal to the ratio of maximum resistance to sliding on an internal plane to the normal pressure on that plane

Question 17

Foundation classification: A foundation is called shallow if its depth is less than or equal to its width. Which of the following depth choices therefore qualify as shallow foundations?

Accepted Answer

Introduction / Context: Foundations are grouped into shallow and deep types based on the embedment depth relative to foundation width (or least lateral dimension). This classification guides analysis methods and construction techniques. Given Data / Assumptions: Shallow foundation definition: depth D_f ≤ width B (or least lateral dimension). Depth options provided are fractions up to and including equality with width. Concept / Approach: Shallow foundations transfer loads to the near-surface soils; typical forms include isolated footings, strip footings, and raft foundations. All the listed depths are ≤ the width, so each option satisfies the criterion for being shallow. Step-by-Step Reasoning: Check each option against D_f ≤ B.1/4 B, 1/2 B, 3/4 B, and D_f = B all satisfy D_f ≤ B.Therefore, all listed choices are shallow. Verification / Alternative check: In contrast, deep foundations (e.g., piles, drilled shafts) typically have D_f >> B and transfer load by end bearing and/or skin friction at deeper strata. Why Other Options Are Wrong: Individual selections are not wrong, but the comprehensive correct choice is the option that includes all qualifying depths. Common Pitfalls: Assuming a single fixed ratio (e.g., D_f = B/2) defines shallow; the accepted definition is D_f ≤ B. Final Answer: all the above

Question 18

In reinforced concrete (R.C.C.) slab design, when should a roof slab be treated as a two-way slab rather than a one-way slab? Consider the practical rule that compares the longer span to the shorter span.

Accepted Answer

Introduction / Context: Slab behavior depends on how bending moments distribute across two orthogonal directions. Distinguishing between one-way and two-way action is essential for choosing correct reinforcement and estimating moments and deflections accurately. Given Data / Assumptions: A rectangular slab with different spans in two perpendicular directions. Support conditions are comparable on all edges (typical room slabs). Rule of thumb for classification is based on span ratio. Concept / Approach: When the longer span (Ly) is not much larger than the shorter span (Lx), load distributes significantly in both directions. A common practical criterion is: if Ly/Lx < 2, the slab is analyzed as a two-way slab; otherwise, for Ly/Lx ≥ 2, it is treated as one-way spanning along the shorter direction. Step-by-Step Solution: Identify Lx (short span) and Ly (long span).Compute span ratio R = Ly / Lx.If R < 2 → adopt two-way slab design (reinforcement in both directions for flexure).If R ≥ 2 → adopt one-way slab design (main steel along short span, distribution steel orthogonal). Verification / Alternative check: Moment coefficients and strip methods in common design aids reflect significantly higher two-directional moments when Ly/Lx is below 2, which validates the practical rule used by designers for preliminary categorization. Why Other Options Are Wrong: It supports live loads in both directions: All slabs carry area loads; this statement does not define two-way action. Slab continuous over two supports: Continuity impacts end moments but not the one-way vs two-way classification. Slab discontinuous at edges: Edge condition affects distribution but is not the primary criterion. Common Pitfalls: Confusing continuity with two-way action, or using panel aspect ratio without checking support conditions. Also, ignoring torsional reinforcement at discontinuous edges can lead to cracking even in two-way panels. Final Answer: When the ratio of the longer span to the shorter span is less than 2.

Question 19

Flexural resistance of a singly reinforced concrete beam: If At is the area of tension steel, f is the design stress in that steel at the limiting state, d is the effective depth, and y is the depth to the resultant compressive force in concrete, wh…

Accepted Answer

Introduction / Context: In reinforced concrete flexure, internal compression in concrete and tension in steel form a couple. The moment of resistance equals one of the forces times the lever arm between the lines of action. For singly reinforced beams, the tension side is governed by steel stress at the design limit state. Given Data / Assumptions: At = area of tension reinforcement. f = design stress in tension steel at the limit state. d = effective depth to tension steel centroid. y = depth to the resultant compressive force in concrete from the extreme compression fiber (so lever arm is approximately d - y). Concept / Approach: Equilibrium requires compressive force C = tensile force T. With steel governing: T = At * f. The internal moment is T multiplied by the lever arm z = d - y, giving M = At * f * (d - y). Step-by-Step Solution: Compute tensile force: T = At * f.Determine lever arm: z = d - y.Moment of resistance: M = T * z = At * f * (d - y). Verification / Alternative check: Using compressive block: C = 0.36 * fck * b * x (code-based) with its centroid at y; matching C = T gives the same lever arm to evaluate M, consistent with M = T * z. Why Other Options Are Wrong: M = At * (d - y): missing the steel stress f. M = At * f * (d + y): lever arm is not d + y. M = (At / f) * (d - y): dimensionally inconsistent. None of these: incorrect because a correct expression is available. Common Pitfalls: Confusing y with neutral axis depth or using effective depth incorrectly; mixing working-stress and limit-state symbols can also cause errors. Final Answer: M = At * f * (d - y).

Question 20

In a combined footing for two columns spaced L apart, the top (negative) reinforcement must extend into the projecting cantilever portions beyond each column face. What practical detailing rule is commonly adopted for this extension?

Accepted Answer

Introduction / Context: Combined footings support two columns on a single base when individual isolated footings would overlap or when soil pressure distribution must be controlled. Top bars resist negative bending in the cantilevered projections beyond the column faces and must be adequately developed for moment transfer. Given Data / Assumptions: Two columns separated by distance L on a common footing. Top reinforcement over columns must be extended into the projecting regions. Need a practical rule for minimum extension to ensure anchorage and moment capacity. Concept / Approach: Under column loads, the footing slab behaves as a beam-slab with negative bending over columns and positive bending in mid-panel. To prevent premature cracking at the column face and to ensure full development of negative moment capacity, top bars are commonly extended a fraction of the column spacing from the column’s centre line; a widely used detailing guide adopts 0.1L beyond the column centre line into the projection. Step-by-Step Solution: Identify column spacing L and projection lengths beyond faces.Provide top bars continuous across both columns.Extend the bars into each projection by about 0.1L measured from the centre line of the column. Verification / Alternative check: Checking against development length requirements shows that 0.1L typically exceeds minimum bar anchorage in many practical layouts, providing redundancy against local stress concentrations near column faces. Why Other Options Are Wrong: 0.1L from the outer edge: not the conventional reference point used in detailing rules. Half or one-fourth projection: may be insufficient or inconsistent across varying geometries. No extension: unsafe and contrary to sound practice. Common Pitfalls: Stopping bars close to the column face without adequate anchorage; neglecting bar hooks or bends where cover is limited; ignoring development length checks in thinner footings. Final Answer: Extend 0.1 L from the centre line of each column.

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