CELE Reinforced & Prestressed Concrete — Reinforced Concrete Footings, Bond and DevelopmentExam Answer Templates

Marks

1

Topic

Footing Sizing

Difficulty

easy

Template Id

T1

Examiner Tip

This is a pure conceptual recall question. Examiners want the exact terms 'service' and 'factored' paired correctly with 'sizing' and 'structural design.' One mismatched pairing loses the mark.

Model Answer

Service (unfactored) loads are used to size the footing plan area against allowable soil bearing capacity. Factored (ultimate) loads are used for the structural design of shear and flexure.

Question Type

very_short_answer

Answer Structure

• Line 1: State 'service load' for sizing against q_a [0.5 mark]
• Line 2: State 'factored load' for structural design [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'ultimate load' for sizing — this is a conceptual error that loses the full mark
• Vague answer such as 'dead and live load' without specifying service vs. factored

Key Phrases To Include

• service load
• allowable soil bearing capacity
• factored load
• structural design

Marks

1

Topic

Two-Way (Punching) Shear

Difficulty

easy

Template Id

T2

Examiner Tip

Board exams often ask this as a one-liner. Write the location first, then the formula — examiners score both sub-parts.

Model Answer

The critical perimeter for punching shear is located at a distance d/2 from each face of the column. For a square column of side c: b_o = 4(c + d).

Question Type

very_short_answer

Answer Structure

• Line 1: State location — at d/2 from column face [0.5 mark]
• Line 2: Write formula b_o = 4(c + d) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating d instead of d/2 from column face — loses 0.5 mark
• Writing b_o = 4c + d instead of 4(c + d)

Key Phrases To Include

• d/2 from the column face
• b_o = 4(c + d)
• critical perimeter

Marks

1

Topic

Two-Way (Punching) Shear

Difficulty

easy

Template Id

T3

Examiner Tip

Memorize both coefficients: 0.33 for punching, 0.17 for one-way beam shear. Confusing them is the top error in shear formula questions.

Model Answer

For a square interior column (where the governing expression applies): V_c = 0.33 λ √f'c · b_o · d, where λ = 1.0 for normal-weight concrete, f'c is in MPa, b_o is the critical perimeter in mm, and d is the effective depth in mm.

Question Type

very_short_answer

Answer Structure

• Line 1: Write the formula V_c = 0.33 λ √f'c b_o d [0.5 mark]
• Line 2: Define all symbols with units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using 0.17 (which is the beam shear coefficient) — loses the mark
• Omitting λ or defining it incorrectly

Key Phrases To Include

• 0.33
• λ
• √f'c
• b_o
• d
• normal-weight concrete

Marks

2

Topic

Footing Sizing

Difficulty

easy

Template Id

T4

Examiner Tip

Always round UP the footing dimension and verify — examiners explicitly check that actual area ≥ required area as a safety statement.

Model Answer

Given: P_service = 900 kN, q_a = 150 kPa Required: Side B of square footing Step 1 — Required area: A_req = P_service / q_a = 900 / 150 = 6.0 m² Step 2 — Side length: B = √A_req = √6.0 = 2.449 m Provide: Use B = 2.5 m × 2.5 m (rounded up to the next 0.1 m increment) Actual area = 6.25 m² > 6.0 m² ✓

Question Type

numerical

Answer Structure

• Step 1: A_req = P/q_a with substitution [1 mark]
• Step 2: B = √A_req, rounded up to a practical dimension with check [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using factored load instead of service load for sizing — loses 1 mark immediately
• Rounding down the footing size (unconservative) — loses 0.5 mark
• No verification that actual area ≥ required area

Key Phrases To Include

• A_req = P_service / q_a
• service load
• rounded up
• actual area > required area

Marks

3

Topic

One-Way (Beam) Shear

Difficulty

medium

Template Id

T5

Examiner Tip

The critical-section location is different for each shear mode: d/2 for punching, d for beam shear. Write both in your Given section if doing both checks — examiners reward clarity.

Model Answer

Given: B = 3.0 m, c = 500 mm, d = 550 mm, P_u = 2200 kN, f'c = 28 MPa, λ = 1.0, ϕ = 0.75 Required: One-way shear check Step 1 — Net upward pressure: q_u = P_u / A = 2200 / (3.0)² = 2200 / 9.0 = 244.4 kPa Step 2 — Shear demand (critical section at d from column face): Projecting length from column face to footing edge = (3000 − 500)/2 = 1250 mm Critical section distance from footing edge = 1250 − 550 = 700 mm V_u = q_u × B × (1250 − 550) / 1000 = 244.4 × 3.0 × 0.700 = 513.2 kN Step 3 — Nominal and design shear capacity (NSCP, one-way): V_c = 0.17 λ √f'c · B · d = 0.17 × 1.0 × √28 × 3000 × 550 = 0.17 × 5.292 × 1 650 000 = 1 484 190 N = 1484.2 kN ϕV_c = 0.75 × 1484.2 = 1113.2 kN Conclusion: ϕV_c = 1113.2 kN > V_u = 513.2 kN → One-way shear is ADEQUATE.

Question Type

numerical

Answer Structure

• Step 1: Compute q_u = P_u/A [1 mark]
• Step 2: Identify critical section at d from column face; compute V_u [1 mark]
• Step 3: Apply V_c = 0.17λ√f'c·B·d, multiply by ϕ = 0.75, compare and conclude [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using critical section at d/2 instead of d from column face (confusing punching with beam shear) — loses 1 mark
• Using 0.33 coefficient instead of 0.17 for one-way shear
• Forgetting to apply ϕ = 0.75 — loses 0.5 mark
• No explicit conclusion statement

Key Phrases To Include

• q_u = P_u / A
• critical section at d from column face
• V_c = 0.17 λ √f'c · B · d
• ϕ = 0.75
• ϕV_c > V_u
• ADEQUATE

Marks

2

Topic

Shear and Flexure

Difficulty

easy

Template Id

T6

Examiner Tip

Always provide the 'why' after the location — examiners allocate partial marks to the reason even if the location is wrong, so never omit the explanation.

Model Answer

(a) Two-way (punching) shear — critical section at d/2 from each face of the column, forming a closed perimeter around the column. This location is used because test data show the diagonal crack initiates at approximately d/2 from the column face. (b) Flexure — critical section at the face of the column (for concrete columns and walls). The footing behaves as a cantilever projecting from the column face, so the maximum bending moment occurs at this section.

Question Type

short_answer

Answer Structure

• Part (a): State d/2 from column face for punching, with brief reason [1 mark]
• Part (b): State face of column for flexure, with cantilever reason [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating 'd from column face' for punching instead of d/2
• Stating critical flexure section is at mid-span of footing

Key Phrases To Include

• d/2 from column face
• closed perimeter
• face of the column
• cantilever

Marks

3

Topic

Two-Way (Punching) Shear

Difficulty

medium

Template Id

T7

Examiner Tip

The most common numerical error is using the full 6.25 m² instead of 5.44 m² for V_u. Write the subtraction explicitly to show the examiner you understand the mechanics.

Model Answer

Given: B = 2.5 m, c = 400 mm, d = 500 mm, P_u = 1700 kN, f'c = 28 MPa, λ = 1.0, ϕ = 0.75 Required: Punching shear check Step 1 — Net upward factored pressure: q_u = P_u / A = 1700 / (2.5)² = 1700 / 6.25 = 272.0 kPa Step 2 — Critical perimeter and shear demand: b_o = 4(c + d) = 4(400 + 500) = 4(900) = 3600 mm Area inside perimeter = (c + d)² = (900)² = 810 000 mm² = 0.81 m² V_u = q_u × (A_footing − A_inside) = 272.0 × (6.25 − 0.81) = 272.0 × 5.44 = 1479.7 kN Step 3 — Nominal punching shear capacity (governing expression for square column): V_c = 0.33 λ √f'c · b_o · d = 0.33 × 1.0 × √28 × 3600 × 500 = 0.33 × 5.292 × 1 800 000 = 3 143 412 N = 3143.4 kN ϕV_c = 0.75 × 3143.4 = 2357.6 kN Conclusion: ϕV_c = 2357.6 kN > V_u = 1479.7 kN → Punching shear is ADEQUATE.

Question Type

numerical

Answer Structure

• Step 1: Compute q_u = P_u/A [0.5 mark]
• Step 2: Compute b_o = 4(c+d); compute area inside perimeter; compute V_u using net area [1 mark]
• Step 3: Apply V_c = 0.33λ√f'c·b_o·d; apply ϕ = 0.75; write explicit conclusion [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using total footing area instead of net area for V_u (subtracting the punched-out zone) — loses 1 mark
• Using 0.17 instead of 0.33 for punching
• b_o calculated as 4c instead of 4(c+d)
• No conclusion or using wrong inequality direction

Key Phrases To Include

• b_o = 4(c + d)
• net area = A_footing − (c+d)²
• V_c = 0.33 λ √f'c b_o d
• ϕ = 0.75
• ϕV_c > V_u

Marks

2

Topic

Flexure

Difficulty

medium

Template Id

T8

Examiner Tip

Sketch the footing in cross-section showing the cantilever length ℓ — it takes 10 seconds and guarantees you set up the formula correctly.

Model Answer

Given: B = 2.5 m, q_u = 272 kPa, c = 400 mm = 0.4 m Required: M_u at column face Projecting cantilever length from column face to footing edge: ℓ = (B − c) / 2 = (2.5 − 0.4) / 2 = 2.1 / 2 = 1.05 m Factored moment per unit width then integrated over width B: M_u = q_u × B × ℓ² / 2 = 272 × 2.5 × (1.05)² / 2 = 272 × 2.5 × 1.1025 / 2 = 272 × 1.378 = 374.9 kN·m ∴ M_u ≈ 375 kN·m at the face of the column.

Question Type

numerical

Answer Structure

• Step 1: Compute projecting length ℓ = (B − c)/2 [1 mark]
• Step 2: Apply M_u = q_u · B · ℓ²/2 with substitution and unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using full footing width B as the lever arm instead of (B−c)/2 — loses 1 mark
• Forgetting to multiply by B (treating as per-metre strip only)

Key Phrases To Include

• critical section at face of column
• ℓ = (B − c)/2
• M_u = q_u · B · ℓ²/2
• cantilever

Marks

2

Topic

Development Length

Difficulty

medium

Template Id

T9

Examiner Tip

Write the bar size check explicitly: '25 mm > 20 mm → use 1.7'. This single line earns partial credit even if you make a numerical error.

Model Answer

Given: d_b = 25 mm (> 20 mm → use 1.7 coefficient), f_y = 415 MPa, f'c = 28 MPa, λ = 1.0, ψ_t = ψ_e = 1.0 Required: ℓ_d Formula (NSCP/ACI, large bars — favorable spacing and cover): ℓ_d = [f_y · ψ_t · ψ_e / (1.7 · λ · √f'c)] · d_b = [415 × 1.0 × 1.0 / (1.7 × 1.0 × √28)] × 25 = [415 / (1.7 × 5.292)] × 25 = [415 / 8.996] × 25 = 46.13 × 25 = 1153 mm Check minimum: ℓ_d = 1153 mm > 300 mm ✓ ∴ ℓ_d = 1150 mm (use 1200 mm in practice).

Question Type

numerical

Answer Structure

• Step 1: Identify bar size category (> 20 mm → coefficient 1.7) and state formula [0.5 mark]
• Step 2: Substitute all values correctly [1 mark]
• Step 3: Compute result and check 300 mm minimum [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using coefficient 2.1 for a 25 mm bar (which applies only to ≤20 mm bars) — loses 1 mark
• Forgetting to multiply by d_b at the end
• Not checking the 300 mm minimum

Key Phrases To Include

• db > 20 mm
• coefficient 1.7
• ℓ_d = f_y ψ_t ψ_e / (1.7 λ √f'c) · db
• minimum 300 mm

Marks

3

Topic

Development Length — Modification Factors

Difficulty

hard

Template Id

T10

Examiner Tip

The ψ_t = 1.3 multiplier for top bars (horizontal bars with more than 300 mm of fresh concrete cast below) is frequently tested. Always explain why ψ_t ≠ 1.0 — the reason is worth partial credit.

Model Answer

Given: d_b = 20 mm (≤ 20 mm → use 2.1 coefficient), f_y = 415 MPa, f'c = 35 MPa, λ = 1.0, ψ_t = 1.3 (top bar: more than 300 mm of concrete cast below), ψ_e = 1.0 Required: ℓ_d Formula (NSCP/ACI, bars ≤ 20 mm with favorable spacing/cover): ℓ_d = [f_y · ψ_t · ψ_e / (2.1 · λ · √f'c)] · d_b Step 1 — Denominator: 2.1 × 1.0 × √35 = 2.1 × 5.916 = 12.424 Step 2 — Numerator: 415 × 1.3 × 1.0 = 539.5 Step 3 — Development length: ℓ_d = (539.5 / 12.424) × 20 = 43.43 × 20 = 868.6 mm Check minimum: 868.6 mm > 300 mm ✓ ∴ ℓ_d ≈ 870 mm

Question Type

numerical

Answer Structure

• Step 1: Identify 20 mm bar category (≤20 mm → 2.1 coefficient); state ψ_t = 1.3 reason [0.5 mark]
• Step 2: Compute denominator 2.1 × λ × √f'c [0.5 mark]
• Step 3: Compute numerator f_y × ψ_t × ψ_e [0.5 mark]
• Step 4: Divide and multiply by d_b; check minimum 300 mm [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using ψ_t = 1.0 for a top bar — loses 1 mark
• Using 1.7 coefficient for a 20 mm bar (boundary case: 20 mm uses 2.1)
• Arithmetic errors in √35

Key Phrases To Include

• d_b ≤ 20 mm → coefficient 2.1
• ψ_t = 1.3 (top bar, more than 300 mm concrete cast below)
• minimum ℓ_d = 300 mm

Marks

2

Topic

Bond and Development

Difficulty

easy

Template Id

T11

Examiner Tip

Use the word 'composite action' — it is the professional engineering term and signals to the examiner that you understand why bond matters at a systems level.

Model Answer

Bond stress is the interface shear stress developed between a reinforcing bar and the surrounding concrete that transfers tensile (or compressive) force from the bar to the concrete. It arises from mechanical bearing of concrete against the deformations (lugs) of deformed bars plus adhesion. Bond is essential for structural integrity because reinforced concrete relies on the two materials acting compositely. If bond is inadequate, bars slip through the concrete without reaching yield stress f_y, causing sudden loss of capacity — a brittle failure mode. This is why a minimum development length ℓ_d must be provided for every bar.

Question Type

short_answer

Answer Structure

• Sentence 1–2: Define bond stress as interface shear transferred via lugs/adhesion [1 mark]
• Sentence 3–4: Explain consequence of bond failure — slip, loss of composite action, brittle failure [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining bond only as 'stickiness' without mentioning mechanical bearing on deformations
• Not linking inadequate bond to structural failure mode

Key Phrases To Include

• interface shear stress
• deformations/lugs
• composite action
• development length
• brittle failure

Marks

5

Topic

Complete Footing Design Check

Difficulty

hard

Template Id

T12

Examiner Tip

Label each part clearly (a), (b), (c), (d) and start each with a sub-heading like 'TWO-WAY SHEAR:'. In a 5-mark question, examiners award marks part-by-part — clear labeling ensures you get credit for every part you complete correctly, even if one part has an error.

Model Answer

Given: B = 2.5 m, c = 400 mm = 0.4 m, d = 500 mm = 0.5 m f'c = 28 MPa, f_y = 415 MPa, λ = 1.0, ϕ_shear = 0.75 P_service = 1200 kN, q_a = 200 kPa, P_u = 1700 kN ───────────────────────────────────── (a) FOOTING SIZE — SERVICE LOAD CHECK ───────────────────────────────────── A_req = P_service / q_a = 1200 / 200 = 6.0 m² Actual area = 2.5 × 2.5 = 6.25 m² > 6.0 m² ✓ Size is adequate. ───────────────────────────────────── (b) TWO-WAY (PUNCHING) SHEAR — FACTORED LOAD ───────────────────────────────────── q_u = P_u / A = 1700 / 6.25 = 272.0 kPa b_o = 4(c + d) = 4(400 + 500) = 3600 mm V_u = q_u × [A − (c + d)²] = 272.0 × [6.25 − (0.9)²] = 272.0 × [6.25 − 0.81] = 272.0 × 5.44 = 1479.7 kN V_c = 0.33 × 1.0 × √28 × 3600 × 500 = 0.33 × 5.292 × 1 800 000 = 3 143 412 N = 3143.4 kN ϕV_c = 0.75 × 3143.4 = 2357.6 kN ϕV_c = 2357.6 kN > V_u = 1479.7 kN ✓ Punching shear ADEQUATE. ───────────────────────────────────── (c) ONE-WAY (BEAM) SHEAR — FACTORED LOAD ───────────────────────────────────── Projection from column face to edge = (2500 − 400)/2 = 1050 mm Critical section at d = 500 mm from column face: Net strip = 1050 − 500 = 550 mm = 0.55 m V_u1 = q_u × B × (0.55) = 272.0 × 2.5 × 0.55 = 374.0 kN V_c = 0.17 × 1.0 × √28 × 2500 × 500 = 0.17 × 5.292 × 1 250 000 = 1 124 550 N = 1124.6 kN ϕV_c = 0.75 × 1124.6 = 843.4 kN ϕV_c = 843.4 kN > V_u1 = 374.0 kN ✓ One-way shear ADEQUATE. ───────────────────────────────────── (d) FACTORED MOMENT AT COLUMN FACE ───────────────────────────────────── ℓ = (B − c)/2 = (2.5 − 0.4)/2 = 1.05 m M_u = q_u × B × ℓ²/2 = 272.0 × 2.5 × (1.05)²/2 = 272.0 × 2.5 × 0.5513 = 374.9 kN·m ∴ M_u ≈ 375 kN·m — use this to design flexural steel A_s.

Question Type

long_answer

Answer Structure

• Part (a): A_req = P_service/q_a; compare with actual area [1 mark]
• Part (b): Compute q_u; b_o; net V_u; V_c with 0.33 coefficient; ϕV_c vs V_u; conclusion [1.5 marks]
• Part (c): Locate critical section at d from face; compute V_u; V_c with 0.17 coefficient; ϕV_c; conclusion [1.5 marks]
• Part (d): Compute ℓ = (B−c)/2; M_u = q_u·B·ℓ²/2 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using factored load for part (a) — loses 1 mark
• Using full footing area instead of net area in punching V_u
• Swapping 0.33 and 0.17 coefficients between shear types
• Using wrong critical section for beam shear (d/2 instead of d)
• Omitting ϕ = 0.75 in either shear check
• No explicit ADEQUATE/INADEQUATE conclusion for each check

Key Phrases To Include

• service load for sizing
• factored load for design
• b_o = 4(c + d)
• net area = A_footing − (c+d)²
• V_c = 0.33 λ √f'c b_o d
• V_c = 0.17 λ √f'c B d
• ϕ = 0.75
• critical section at d from column face (beam shear)
• M_u = q_u · B · ℓ²/2

Marks

1

Topic

Development Length

Difficulty

easy

Template Id

T13

Examiner Tip

A memory aid: smaller bars bond better (more perimeter per area) → larger coefficient → 2.1. Larger bars → 1.7.

Model Answer

Minimum: ℓ_d ≥ 300 mm in all cases. Coefficient 2.1 applies to bars of 20 mm diameter and smaller; coefficient 1.7 applies to bars larger than 20 mm.

Question Type

very_short_answer

Answer Structure

• Part 1: State 300 mm minimum [0.5 mark]
• Part 2: State 2.1 for ≤20 mm and 1.7 for >20 mm [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Assigning coefficients in reverse (2.1 to large bars, 1.7 to small bars)
• Not stating the 300 mm floor

Key Phrases To Include

• 300 mm minimum
• 2.1 for ≤ 20 mm
• 1.7 for > 20 mm

Marks

2

Topic

Hooks and Anchorage

Difficulty

medium

Template Id

T14

Examiner Tip

Draw a tiny sketch of each hook with the extension dimension labeled — it earns full marks for the hook description even with minimal text.

Model Answer

A standard hook is used when the available straight embedment length (from the critical section to the edge of concrete minus cover) is insufficient to develop the full ℓ_d for straight bars. Hooks develop the bar in a shorter length by providing a mechanical anchorage in addition to bond. The two standard hook configurations per NSCP/ACI are: 1. 90° hook — the bar is bent 90° with an extension of 12d_b beyond the bend. 2. 180° hook — the bar is bent 180° (hairpin) with an extension of 4d_b (minimum 65 mm) beyond the bend.

Question Type

short_answer

Answer Structure

• Sentence 1: Reason for hook use — insufficient straight development length [0.5 mark]
• Sentence 2: How hooks work — mechanical anchorage [0.5 mark]
• Points 1 & 2: 90° hook with 12d_b extension; 180° hook with 4d_b extension [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating hook extension as 6d_b for 90° (incorrect) — loses 0.5 mark
• Confusing 90° and 180° extensions

Key Phrases To Include

• insufficient straight embedment
• mechanical anchorage
• 90° hook with 12d_b extension
• 180° hook with 4d_b extension

Marks

2

Topic

Development Length — Modification Factors

Difficulty

hard

Template Id

T15

Examiner Tip

Board exams often give you a footing detail and ask you to justify which coefficient to use. Always state the specific conditions: 'Clear spacing = __ mm, d_b = __ mm, therefore favorable/less-favorable.'

Model Answer

Favorable conditions exist when clear spacing between bars ≥ d_b AND clear cover ≥ d_b AND minimum transverse reinforcement (stirrups) is provided. Under these conditions, concrete confinement around the bar is adequate, and the higher coefficients (2.1 for ≤20 mm bars; 1.7 for >20 mm) apply, giving a shorter ℓ_d. Less-favorable (or other) conditions occur when spacing or cover is tighter. The coefficients drop to 1.4 (for ≤20 mm) and 1.1 (for >20 mm), resulting in a longer ℓ_d — reflecting higher splitting-crack risk around closely spaced or poorly covered bars.

Question Type

short_answer

Answer Structure

• Part 1: Define favorable condition (clear spacing ≥ d_b, cover ≥ d_b, stirrups) and higher coefficients 2.1/1.7 [1 mark]
• Part 2: Define less-favorable condition and lower coefficients 1.4/1.1 with longer ℓ_d explanation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing which condition gives the larger coefficient
• Not mentioning the physical reason (splitting crack risk)

Key Phrases To Include

• clear spacing ≥ d_b
• clear cover ≥ d_b
• coefficients 2.1 / 1.7 (favorable)
• coefficients 1.4 / 1.1 (less-favorable)
• splitting failure