Question 1

For a simply supported T-beam in reinforced concrete design, what is the maximum permissible ratio of effective span to overall depth that is typically used for preliminary sizing to control deflection?

Accepted Answer

Introduction / Context:In reinforced concrete beam and T-beam design, span/depth limits are used at the preliminary stage to keep service deflections within acceptable limits without doing full deflection calculations. For simply supported members, the basic limit is the most restrictive because end fixity is low and midspan deflections are relatively larger.Given Data / Assumptions:Member behaves as a simply supported T-beam under service loads.Basic span/depth ratio is sought for preliminary sizing (before modification factors for tension steel, compression steel, and finishes).Normal exposure and common material grades.Concept / Approach:The span/depth ratio is a serviceability control. Codes and standard practice provide “basic” ratios for different support conditions. For simply supported members, the basic ratio is smallest (more conservative) because the support restraint is minimal. The T-beam flange enhances compression capacity but does not typically change the basic starting ratio for deflection control at the sizing stage; refinement comes later with modification factors and detailed checks.Step-by-Step Solution:Identify the support condition: simply supported.Adopt the basic span/depth ratio for preliminary sizing = 20.After preliminary sizing, apply modification factors as permitted (based on percentage of steel, stress levels, compression reinforcement, and finishes) to fine-tune the depth.Carry out a detailed deflection check at the final design stage if the member is sensitive (e.g., brittle finishes or long spans).Verification / Alternative check:Compare with other support conditions: continuous beams often use a higher basic ratio (less restrictive), while cantilevers use a much lower ratio. This confirms that 20 is the correct baseline for simply supported members.Why Other Options Are Wrong:10 and 15 are overly conservative and lead to uneconomical depths.25 and 30 are more appropriate to stiffer, continuous systems, risking excessive deflection if used for simple spans.Common Pitfalls:Confusing beam and slab limits, forgetting to apply modification factors, or skipping a final deflection check when spans are long or finishes are crack-sensitive.Final Answer:20.

Question 2

Design assumptions for prestressed concrete members: which set of statements appropriately reflects the standard working assumptions used in analysis and service-level design?

Accepted Answer

Introduction / Context:Prestressed concrete (PSC) analysis commonly uses a set of simplifying assumptions at the serviceability stage. These assumptions allow the designer to apply linear elastic methods, distribute stresses, and estimate losses with reasonable accuracy for everyday bridge and building members.Given Data / Assumptions:Service-level behavior is considered (pre-cracking or limited cracking conditions).Members may be pretensioned or post-tensioned; bonded systems are typical in building/bridge practice.Small strain, linear elasticity is assumed at service to enable transformed-section analysis.Concept / Approach:The core assumptions are: (1) plane sections remain plane, so strain varies linearly over depth; (2) within working ranges, concrete and steel are treated as linearly elastic (Hook’s law); (3) in fully prestressed, bonded members, changes in tendon stress under modest service load variations are limited, enabling simplified checks that emphasize concrete service stresses and crack control rather than precise tendon stress increments.Step-by-Step Solution:Adopt the plane-sections assumption to relate curvature to strain distribution.Use elastic moduli to transform steel to an equivalent concrete area (modular ratio) for service stress calculations.For bonded systems, treat tendon stress changes at service as modest; focus on concrete fiber stresses and allowable limits in compression/tension.Verification / Alternative check:Where higher accuracy is required (e.g., ultimate strength or unbonded tendons), employ non-linear or refined methods that capture tendon stress increments and cracked-section behavior; this confirms the service assumptions are a reasonable simplification, not a universal rule.Why Other Options Are Wrong:Each of options a–c is valid in the stated service context; hence “All the above” is correct.“None of these” contradicts standard service design assumptions in PSC.Common Pitfalls:Applying these assumptions at ultimate limit states without appropriate modifications, ignoring time-dependent effects, or forgetting that unbonded tendons can experience larger stress variations under load changes.Final Answer:All the above.

Question 3

Singly reinforced rectangular beam (working stress): A 25 cm wide beam with effective depth 70 cm is provided with 18.75 cm² tension steel. If the modular ratio m = 15, determine the depth of the neutral axis.

Accepted Answer

Introduction / Context:Under the working stress method, neutral axis depth in a singly reinforced rectangular beam is obtained using transformed-section analysis. The position of the neutral axis governs the stress distribution in concrete and steel and is a key input for service stress checks and moment capacity calculations within the elastic range.Given Data / Assumptions:Beam width b = 25 cm; effective depth d = 70 cm.Area of tension steel Ast = 18.75 cm².Modular ratio m = Es/Ec = 15.Working stress assumptions: plane sections remain plane; linear elastic behavior; concrete in tension ignored.Concept / Approach:For transformed-section analysis, replace steel by an equivalent concrete area m*Ast located at the steel level. The neutral axis depth n (from top) follows by force equilibrium in the transformed section:m * Ast * (d - n) = (b * n²) / 2Step-by-Step Solution:Insert values: 15 * 18.75 * (70 - n) = (25 * n²) / 2Compute constants: 281.25 * (70 - n) = 12.5 * n²Expand: 19687.5 - 281.25 n = 12.5 n²Rearrange: 12.5 n² + 281.25 n - 19687.5 = 0Divide by 12.5: n² + 22.5 n - 1575 = 0Solve: Discriminant = 22.5² + 4 * 1575 = 6806.25; sqrt = 82.5n = (-22.5 + 82.5) / 2 = 30 cmVerification / Alternative check:With n = 30 cm, n Why Other Options Are Wrong:20 cm, 25 cm, 35 cm, 40 cm do not satisfy the transformed equilibrium equation for the given b, d, Ast, and m.Common Pitfalls:Using gross area instead of transformed steel area, forgetting to square n in the compression term, or mixing units (cm vs. mm).Final Answer:30 cm.

Question 4

Foundations for a line of multiple columns constructed in a straight row: which type is most appropriate when loads are distributed along a line rather than concentrated at a single point?

Accepted Answer

Introduction / Context:When columns are arranged in a straight line and carry similar loads, it is often economical and practical to use a continuous foundation that supports the entire line. This spreads the load uniformly and simplifies construction, especially on soils of moderate bearing capacity.Given Data / Assumptions:Multiple columns aligned in a single row.Loads are relatively uniform along the line.Soil bearing conditions allow a shallow continuous footing.Concept / Approach:A strip footing (continuous footing) is a long footing that runs beneath a line of columns or a load-bearing wall. It distributes loads along length and is ideal where the structural layout presents a linear load path. Alternatives such as isolated pads or rafts are selected for different layouts (point loads scattered or whole-plan coverage).Step-by-Step Solution:Identify the load path: a linear sequence of column reactions.Select a footing that integrates these reactions along one axis: a strip footing.Size width based on allowable soil bearing pressure and service load combinations; check bearing, sliding, settlement, and shear/bending within the footing strip.Verification / Alternative check:Compare with a raft when coverage over a large area or poor soils is needed; isolated pads if columns are far apart; strap footings for eccentric cases combining adjacent pads. The linear column layout favors a strip footing in common practice.Why Other Options Are Wrong:Isolated pads: suited to discrete column loads with separation.Raft: used when the entire building footprint needs support or soil is very soft; not just a single row.Strap: connects two pads to control eccentricity, not for a continuous row.“Combination of the above for a single column” does not fit the scenario.Common Pitfalls:Ignoring differential settlement along the strip, omitting temperature/shrinkage steel, or failing to check punching or two-way shear under columns on the strip.Final Answer:Strip footing (continuous footing under the row).

Question 5

Effect of increasing steel percentage in a singly reinforced concrete section: how does the depth of the neutral axis and the lever arm respond as steel percentage rises (keeping section size similar)?

Accepted Answer

Introduction / Context:The depth of the neutral axis (NA) in a reinforced concrete flexural section shifts with the relative proportions of concrete compression and steel tension. Understanding how NA moves as steel percentage changes is vital for service stress calculations, curvature, ductility expectations, and ultimate strength behavior.Given Data / Assumptions:Rectangular, singly reinforced section under positive bending.Concrete in tension is neglected in service calculations.Linear strain distribution (plane sections remain plane).Concept / Approach:When the percentage of steel increases, the tensile force capacity on the tension side increases. To maintain force equilibrium, a larger compressive force in the concrete is required. This is achieved by deepening the compression block, which means the neutral axis shifts downward (measured from the compression face), i.e., the depth of NA increases. With greater NA depth, the internal lever arm z between resultant compression and tension forces tends to reduce slightly for a given overall depth, affecting moment capacity and curvature.Step-by-Step Solution:Increase Ast (tension steel area).Equilibrium requires C = T; with higher T, the compression block must grow, pushing NA deeper.Result: neutral axis depth increases; lever arm generally decreases modestly because the compression resultant moves farther from the tension steel.Verification / Alternative check:Transformed-section equations show n increases as m*Ast increases; typical design charts also reflect a deeper NA with higher reinforcement ratios.Why Other Options Are Wrong:Decrease in NA depth contradicts equilibrium for higher steel tension.Increase in lever arm is not typical; with deeper NA, z tends to reduce slightly.“None of these” is incorrect because the correct trend is known.Common Pitfalls:Assuming lever arm always increases with more steel; misinterpreting ultimate stress block behavior as service behavior; neglecting ductility trade-offs with very high steel ratios.Final Answer:Depth of neutral axis increases.

Question 6

Parabolic tendon in a prestressed beam: If the tendon carries prestressing force P, has maximum dip h at midspan, and the effective span is L, what is the equivalent upward uniform load produced by the tendon curvature?

Accepted Answer

Introduction / Context:Curved prestressing tendons exert an equivalent transverse load on the concrete member. For a parabolic profile (common in simply supported members), this load can be represented as a uniform upward load that counterbalances part of the downward external loads, reducing midspan moments and improving service performance.Given Data / Assumptions:Parabolic tendon with maximum eccentricity (dip) h at midspan.Prestressing force P (assumed constant along the span, neglecting friction/anchorage for simplicity).Effective span L; small deflection theory applies.Concept / Approach:The equivalent load per unit length induced by a tendon profile y(x) is w = P * d²y/dx². For a symmetric parabola with zero eccentricity at supports and maximum h at midspan, y(x) = 4h * x * (L - x) / L². The second derivative is constant, giving a uniform equivalent load.Step-by-Step Solution:Let y(x) = 4h * x * (L - x) / L².Compute dy/dx and then d²y/dx²; result: d²y/dx² = 8h / L².Equivalent upward load: w = P * d²y/dx² = 8 * P * h / L².Verification / Alternative check:Check units: P has units of force, h/L² yields 1/length, so w has force/length as required. Also, the total equivalent upward force w * L equals 8Ph/L, which, combined with the parabolic distribution, reproduces the same internal effect as the curved tendon.Why Other Options Are Wrong:4Ph/L² and 6Ph/L² underpredict the curvature effect.P/(hL) and 2P/L have incorrect functional dependence or units.Common Pitfalls:Using first derivative instead of second derivative, applying the formula to non-parabolic profiles, or forgetting that friction and anchorage can change P and thus w along the span.Final Answer:w = 8 * P * h / L^2.

Question 7

Angle of repose of soil: Which reference line is used to define the maximum angle that the natural slope (outer face) of a soil mass makes under its own weight without lateral support?

Accepted Answer

Introduction / Context:The angle of repose is a fundamental geotechnical property indicating the steepest angle at which a loose granular material remains stable without lateral support. It controls natural slopes, embankments, storage piles, and influences earth pressure on retaining structures.Given Data / Assumptions:Dry or near-dry granular soil considered for the classical definition.No cementation or apparent cohesion.Mass forms a self-supported slope under gravity alone.Concept / Approach:The angle of repose, typically denoted by phi_r (sometimes approximating the friction angle for dry, clean sands), is defined with respect to the horizontal plane. It is the maximum angle at which grains interlock and frictional resistance is sufficient to prevent sliding or rolling. Moisture, particle shape, and gradation alter this angle in practice.Step-by-Step Solution:Visualize a soil heap formed by pouring granular material.Measure the slope angle at which sliding begins; the greatest stable angle is the angle of repose.Reference is the horizontal plane because gravity acts vertically and slope stability is judged relative to level ground.Verification / Alternative check:Laboratory tilt box tests or simple pile tests give angles commonly ranging from about 28° to 40° for sands and gravels depending on particle shape and moisture content.Why Other Options Are Wrong:“With the vertical” and “with a line perpendicular to the slope” are unconventional references and would confuse standard definitions.“None of these” is incorrect since the standard reference is clearly the horizontal.Common Pitfalls:Assuming angle of repose equals the friction angle for all soils; cohesive or wet materials can show different behavior (e.g., apparent cohesion raises temporary slopes above the dry angle of repose).Final Answer:With the horizontal.

Question 8

Reinforced concrete beam (b = 25 cm, effective depth d = 70 cm): The extreme-fiber stresses are 62.5 kg/cm² in concrete (compression) and 250 kg/cm² in steel (tension). Using modular ratio m = 15 and elastic theory, determine the depth of the neutra…

Accepted Answer

Introduction / Context:Neutral axis depth in a singly reinforced concrete beam under working-stress (elastic) theory is obtained from strain compatibility between concrete and steel using the modular ratio m. This problem tests the direct use of the steel-to-concrete stress ratio to find the neutral axis without needing the steel area.Given Data / Assumptions:Beam breadth b = 25 cm, effective depth d = 70 cm.Concrete stress at top σc = 62.5 kg/cm² (compression).Steel stress σs = 250 kg/cm² (tension).Modular ratio m = 15.Elastic theory (linear strain distribution) and concentric bending; concrete in tension ignored.Concept / Approach:From similar triangles in the strain diagram and Hooke’s law: σs/σc = m * ( (d − x) / x ), where x is the neutral-axis (NA) depth from the top compression face. Rearranging gives x in terms of m, d, and the stress ratio σs/σc.Step-by-Step Solution:Compute stress ratio: σs/σc = 250 / 62.5 = 4.Use compatibility: 4 = m * (d − x) / x = 15 * (d − x) / x.Solve: (d − x) / x = 4 / 15 → d = x * (1 + 4/15) = x * (19/15).Therefore x = (15/19) * d = (15/19) * 70 ≈ 55.3 cm.Verification / Alternative check:You can also write x = m d / ( (σs/σc) + m ) = 15 * 70 / (4 + 15) = 1050 / 19 ≈ 55.3 cm, confirming the result.Why Other Options Are Wrong:25–40 cm and 30–35 cm are far smaller than the computed 55.3 cm using the given stresses and m; they contradict the compatibility equation.Common Pitfalls:Mixing effective depth with overall depth; forgetting to apply m; using force equilibrium that requires As (not provided) instead of direct strain compatibility.Final Answer:55 cm (approximately)

Question 9

R.C.C. slab detailing: For a slab with main reinforcement diameter 16 mm, select the minimum nominal clear cover to main bars as per common building practice.

Accepted Answer

Introduction / Context:Clear cover in slabs protects steel from corrosion and fire and ensures proper bond. Many design practices (e.g., IS 456 guidance) adopt a minimum nominal cover of 15 mm or the bar diameter, whichever is greater, for mild exposure slabs.Given Data / Assumptions:Main bar diameter = 16 mm.Slab in typical building exposure (not aggressive).Adopt rule: cover = max(15 mm, bar diameter).Concept / Approach:Choose the higher of 15 mm and 16 mm to satisfy durability and cover tolerances in slabs.Step-by-Step Solution:Compute max(15, 16) = 16 mm.Verification / Alternative check:If exposure is severe, higher cover may be required; for normal exposure, 16 mm satisfies the common minimum requirement.Why Other Options Are Wrong:12–15 mm are less than the bar diameter; they do not satisfy the max(15 mm, bar dia) rule for a 16 mm bar.Common Pitfalls:Confusing slab cover with beam cover; ignoring exposure class adjustments.Final Answer:16 mm

Question 10

Ribbed (joist) slab: For a rib containing two longitudinal bars, the practical breadth (web width) should be within which range for good detailing and placement?

Accepted Answer

Introduction / Context:Ribbed slabs have narrow ribs supporting a thin topping slab. The rib breadth must be sufficient for bar placement, concrete flow, cover, and shear capacity. With two longitudinal bars, typical practice keeps the rib width around 100–120 mm to avoid congestion and ensure proper compaction.Given Data / Assumptions:Two longitudinal bars in the rib.Standard bar spacing, concrete cover, and link (stirrup) detailing constraints.Ordinary building-grade concrete and construction methods.Concept / Approach:Minimum width must accommodate 2 bars plus clear spacing and cover, while meeting shear web requirements. Hence, 10–12 cm is a widely used practical range.Step-by-Step Solution:Check bar diameters + clear cover + spacing → implies ≥ about 100 mm.Adopt a practical range of 10–12 cm to satisfy constructability and shear.Verification / Alternative check:Project detailing manuals and site experience confirm that ribs narrower than ~100 mm cause congestion and poor compaction.Why Other Options Are Wrong:6–10 cm too narrow for two bars and links; 12–15 cm is workable but often uneconomical for typical rib spacing; the question targets the commonly recommended band.Common Pitfalls:Ignoring link width and nominal cover; choosing too-narrow ribs leading to honeycombing.Final Answer:10 cm to 12 cm

Question 11

Two-way slab simply supported on all four edges carrying uniformly distributed load w: If l1 and l2 are the longer and shorter spans respectively, the load split into w1 on strips parallel to l2 and w2 on strips parallel to l1 is in the ratio ?

Accepted Answer

Introduction / Context:In classic plate theory for simply supported rectangular slabs under uniform load, the load distribution to orthogonal strip actions relates to the spans to the fourth power. This gives a convenient way to apportion load effects for preliminary design or educational problems.Given Data / Assumptions:Two-way slab, simply supported on all sides.Uniformly distributed load w.Strips considered parallel to l2 (carrying w1) and parallel to l1 (carrying w2).Concept / Approach:Deflection and stiffness of plate action lead to a fourth-power span dependency. Hence the load split is proportional to the fourth power of the orthogonal spans.Step-by-Step Solution:Use standard result: w1 : w2 = l2^4 : l1^4.Therefore, w1/w2 = l2^4 / l1^4.Verification / Alternative check:Many textbooks present the same fourth-power law for load apportionment in simply supported two-way plates under UDL.Why Other Options Are Wrong:Ratios with 2nd power, 1st power, or inverted fourth power contradict the plate-theory result used for such educational distribution.Common Pitfalls:Confusing which span is associated with which strip; swapping l1 and l2 leads to the reciprocal.Final Answer:w1/w2 = l2^4 / l1^4

Question 12

Working-stress (elastic) theory for a cracked rectangular beam: If m is the modular ratio, r = As/(b d) is the steel ratio, and d is the overall effective depth, the neutral axis depth x from the top is given by which expression?

Accepted Answer

Introduction / Context:For a cracked elastic section (concrete in tension neglected), the neutral axis of a singly reinforced beam is obtained using the transformed-section method. The neutral axis passes through the centroid of the transformed section (compression concrete + transformed steel area).Given Data / Assumptions:Modular ratio m = Es/Ec.Steel ratio r = As/(b d).Rectangular section; concrete tension ignored, linear elasticity.Concept / Approach:Locate the centroid (neutral axis) of the transformed section by equating moments of areas about the neutral axis: (b * x) * (x/2) = m As (d − x). This yields a quadratic in k = x/d.Step-by-Step Solution:Write (b/2) x^2 = m As (d − x).Divide by b d^2 and set r = As/(b d), k = x/d → (1/2) k^2 = m r (1 − k).Rearrange: k^2 + 2 m r k − 2 m r = 0.Solve for k: k = - m r + sqrt( m^2 r^2 + 2 m r ).Hence x = d * [ - m r + sqrt( m^2 r^2 + 2 m r ) ].Verification / Alternative check:Check limiting cases: as r → 0, k → 0; as r increases, k grows sublinearly, consistent with mechanics.Why Other Options Are Wrong:Options (b), (c), (d), and (e) do not satisfy the derived quadratic or have incorrect functional forms.Common Pitfalls:Using area equilibrium instead of first moment of area; forgetting to neglect concrete in tension when cracked; algebraic sign errors.Final Answer:x = d * [ - m r + sqrt( m^2 r^2 + 2 m r ) ]

Question 13

Prestressed concrete per IS 1343: The total shrinkage strain to be considered for a pretensioned member is approximately equal to which value?

Accepted Answer

Introduction / Context:Shrinkage of concrete shortens prestressed members and reduces effective prestress via time-dependent losses. IS 1343 recommends typical values of total shrinkage strain for design purposes in the absence of detailed testing.Given Data / Assumptions:Pretensioned prestressed concrete member.Use standard code-recommended total shrinkage strain.Concept / Approach:Total shrinkage strain is an order of 10^-4 for normal concretes; adopting 3.0 × 10^-4 is a common design value for pretensioned members as per IS practice.Step-by-Step Solution:Select from given magnitudes: 10^-2 and 10^-3 are too large; 10^-5 too small.Adopt 3.0 × 10^-4 as the representative value.Verification / Alternative check:Comparisons with code tables and standard textbooks confirm this order of magnitude for total shrinkage strain.Why Other Options Are Wrong:10^-2 or 10^-3 would imply unrealistically huge shortening; 10^-5 values underestimate shrinkage and associated prestress loss.Common Pitfalls:Confusing shrinkage with creep; mixing units of strain and stress.Final Answer:3.0 × 10^-4

Question 14

Material property recall: The Young’s modulus of elasticity of structural steel is typically taken as which standard value?

Accepted Answer

Introduction / Context:Young’s modulus (E) for steel is a fundamental material constant used in deflection, stress, and stability calculations across steel and reinforced concrete design (for modular ratio).Given Data / Assumptions:Structural carbon steel of ordinary grade.Standard room temperature conditions.Concept / Approach:Conventional design adopts E_steel ≈ 2.0 × 10^5 N/mm², which is 200 kN/mm². This approximation is sufficiently accurate for most building and bridge calculations.Step-by-Step Solution:Recognize standard constant: E ≈ 200 kN/mm².Verification / Alternative check:Many codes and textbooks list the same nominal value; small variations (±5%) do not change the standard selection for exam purposes.Why Other Options Are Wrong:150, 250, and 275 kN/mm² deviate significantly from the accepted nominal constant for structural steel.Common Pitfalls:Confusing units (N/mm² vs kN/mm²) or using stainless/temperature-dependent values.Final Answer:200 kN/mm²

Question 15

Heel slab design of a cantilever retaining wall: The maximum design bending moment in the heel is considered from which combination of actions?

Accepted Answer

Introduction / Context:Heel slabs of cantilever retaining walls act as cantilevers from the stem. Accurate bending moment evaluation must include all vertical loads and upward soil pressure acting on the heel area.Given Data / Assumptions:Heel slab with retained soil above it.Possible surcharge on backfill.Upward soil reaction distributed under the base.Concept / Approach:Bending results from downward loads (self-weight of concrete, weight of backfill, surcharge) balanced by upward soil reaction. Maximum heel moment occurs from their combined effect and must satisfy strength and serviceability.Step-by-Step Solution:Compute downward loads: W_conc + W_soil + W_surcharge.Determine base pressure distribution and resultant upward reaction under the heel.Evaluate net load diagram over the heel and compute maximum moment at the stem face.Verification / Alternative check:Check with stability analysis (sliding, overturning, bearing) to ensure reactions used are consistent with overall equilibrium.Why Other Options Are Wrong:Each single effect alone is incomplete; the heel must be designed for the combined action to be safe.Common Pitfalls:Omitting surcharge; ignoring heel self-weight; using uniform soil reaction regardless of resultant location.Final Answer:All of the above acting together

Question 16

Foundation pressure distribution: For a perfectly rigid foundation base resting on a homogeneous elastic soil and subjected to a concentric vertical load, how does the contact pressure vary?

Accepted Answer

Introduction / Context:Contact-pressure assumptions depend on the relative rigidity of the footing and the soil. For a perfectly rigid footing on homogeneous soil under concentric vertical load, the conventional assumption is a uniform pressure distribution.Given Data / Assumptions:Rigid foundation base (no differential deformation).Homogeneous supporting soil.Concentric vertical loading (no overturning moment).Concept / Approach:Rigid base implies plane sections remain plane; with symmetry and no moments, the soil reaction is uniform to balance the applied load.Step-by-Step Solution:Apply equilibrium: resultant reaction = applied vertical load.With no net moment and rigidity, reaction intensity = constant = Load / Area.Verification / Alternative check:Flexible bases show non-uniform distributions (often more at center); presence of moments shifts maxima to edges. The stated ideal leads to uniform pressure.Why Other Options Are Wrong:Edges higher or center zero require eccentricity or flexibility; arbitrary shapes contradict concentric, rigid conditions.Common Pitfalls:Confusing rigid with flexible base assumptions; ignoring effect of eccentric loading and rocking.Final Answer:Uniform over the base

Question 17

Retaining wall stability: If W is the self-weight of the wall and P is the horizontal active earth pressure, what minimum factor of safety against sliding is generally adopted at working loads?

Accepted Answer

Introduction / Context:Sliding resistance of retaining walls is checked by comparing available resisting forces with the driving horizontal earth pressure. Design guides commonly recommend a minimum factor of safety (FS) at service loads.Given Data / Assumptions:Retaining wall on soil or rock foundation.Service load combinations (working state).FS_sliding = (frictional resistance + passive resistance, if any) / P.Concept / Approach:A typical target is FS ≈ 1.5 for sliding under service loads, unless otherwise required by project codes or seismic provisions.Step-by-Step Solution:Compute resisting force = μ (W + any vertical surcharge on base) + passive (if mobilized).Compute FS = Resisting / P and check FS ≥ 1.5.Verification / Alternative check:Many texts align on 1.5 for sliding and 1.5–2.0 for overturning (varies by standard). Ultimate limit-state methods convert to partial factors instead.Why Other Options Are Wrong:1.0 and 1.25 are too low; 2.0–2.5 may be overly conservative for routine static cases unless specified by the client or code.Common Pitfalls:Ignoring base key contribution; overestimating passive resistance without proper compaction; neglecting uplift water pressure.Final Answer:1.5

Question 18

Retaining wall earth pressure (active case): For a vertical retaining wall of height h (per unit length out of plane) retaining horizontal backfill with unit weight w, and active earth pressure coefficient Ka, what is the total thrust exerted by the…

Accepted Answer

Introduction / Context:This problem tests Rankine’s active earth pressure theory for a vertical retaining wall with level backfill. Knowing the total lateral thrust and its dependence on the active pressure coefficient Ka is fundamental for sizing retaining walls and checking sliding, overturning, and base pressures.Given Data / Assumptions:Wall height = h (per unit run along the wall).Backfill is horizontal, dry, and homogeneous with unit weight w.Active earth pressure condition with coefficient Ka (soil yields away from the wall).No surcharge load and no water table.Concept / Approach:Under Rankine active conditions, the lateral pressure at a depth z is pa(z) = Ka * w * z (a linear distribution from zero at the top to Ka * w * h at the base). The total active thrust is the area of this triangular pressure diagram times unit length of wall.Step-by-Step Solution:At depth z: pa(z) = Ka * w * z.Resultant thrust Pa = area of triangle = (1/2) * base * height = (1/2) * (Ka * w * h) * h.Therefore, Pa = (1/2) * Ka * w * h^2.Point of action is at h/3 above the base (for a triangular distribution), though the question asks only for the total magnitude.Verification / Alternative check:Dimensional check: w has units of force/volume; multiplying by h^2 (area) and by Ka (dimensionless) gives force per unit run—consistent with thrust per unit length of wall.Why Other Options Are Wrong:Ka * w * h: Misses the triangular integration (factor 1/2 and h^2).(1/3) * Ka * w * h^3: Wrong power of h; thrust scales with h^2, not h^3.(1/2) * w * h^2: Ignores soil friction angle effect embedded in Ka.w * h^2: Also ignores triangular distribution and Ka.Common Pitfalls:Forgetting the factor of 1/2 from the triangular diagram, neglecting Ka, or confusing active with at-rest (K0) or passive (Kp) conditions.Final Answer:(1/2) * Ka * w * h^2

Question 19

Beam detailing per IS practice: Side-face reinforcement shall be provided in a beam when the overall web depth exceeds which threshold?

Accepted Answer

Introduction / Context:Side-face reinforcement prevents surface cracking and improves crack control in deep beams where the depth of web is large. Codes prescribe a depth threshold beyond which such reinforcement is obligatory.Given Data / Assumptions:Reinforced concrete beam with large overall depth.IS-style practice for side-face reinforcement.Question asks the threshold web depth.Concept / Approach:As per standard Indian practice, side-face reinforcement is provided when the overall depth exceeds approximately 750 mm. This steel is placed on the side faces of the beam, distributed uniformly, mainly for crack control due to shrinkage and temperature, and to improve ductility.Step-by-Step Solution:Compare given thresholds with code trigger: 750 mm = 75 cm.When depth > 75 cm, provide side-face reinforcement, generally not less than the prescribed minimum percentage, distributed equally on two faces.Therefore, the correct option is 75 cm.Verification / Alternative check:Practical detailing guides and handbooks reiterate the 750 mm criterion and specify minimum steel (e.g., 0.1% of the web area, split equally on both faces) as a typical guideline.Why Other Options Are Wrong:50 cm: Too small; routine beams would all require side-face steel unnecessarily.100 cm and 120 cm: Too lenient; important crack control would be missed for deep beams between 75–100 cm.Common Pitfalls:Confusing side-face reinforcement with shear stirrups; omitting it in deep transfer girders; not distributing bars evenly or providing insufficient cover.Final Answer:75 cm

Question 20

Effective flange width of an intermediate T-beam in a floor system: A hall measures 12 m × 7 m. Intermediate T-beams are spaced at 3 m centre-to-centre. One such beam is 20 cm wide (web) and cast monolithically with a slab; clear cover to tensile st…

Accepted Answer

Introduction / Context:The effective flange width b_f for a T-beam limits how much of the slab participates in compression. IS-style rules define b_f as the least of certain quantities related to span, slab thickness, and beam spacing.Given Data / Assumptions:Hall size = 12 m × 7 m.Intermediate beam spacing = 3 m (centre-to-centre).Beam web (bw) = 20 cm.Flange thickness (slab over the web) ≈ 13 cm.Beam likely spans the 12 m direction (common in such problems).Concept / Approach:IS-style criterion for T-beams (typical): b_f is the least of1) l0/6 + bw + 6Df,2) centre-to-centre spacing of beams,3) actual flange width available. Here, Df is effective flange (slab) thickness over the beam.Step-by-Step Solution:Take l0 ≈ 12 m = 1200 cm.Compute l0/6 + bw + 6Df = 1200/6 + 20 + 613 = 200 + 20 + 78 = 298 cm.Centre-to-centre spacing = 300 cm.Actual available is not less than spacing in this intermediate beam case.Therefore, b_f = least of (298, 300, actual) ≈ 298 cm; selecting a practical rounded choice from options gives 300 cm.Verification / Alternative check:298 cm is within typical rounding to 3.0 m in many hand-calculation contexts; conservative designers may keep 2.98 m explicitly.Why Other Options Are Wrong:233, 176, 236, 255 cm: All are below the computed limit and do not reflect the codal calculation for the given geometry.Common Pitfalls:Using clear span instead of effective span for l0; forgetting to limit by beam spacing; misreading slab thickness; using gross slab width rather than effective width.Final Answer:300 cm

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