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CELE Reinforced & Prestressed Concrete — Reinforced Concrete SlabsExam Answer Templates

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Exam answer templates for Reinforced Concrete Slabs in CELE Reinforced & Prestressed Concrete. These are the response frameworks that consistently earn full marks on Professional Regulation Commission (PRC) — Board of Civil Engineering's questions. Each template is tuned to a specific question type — learn them all and your CELE 2026 performance will reflect it.

Exam context

Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Reinforced & Prestressed Concrete section sits under a "Core" weighting, and Reinforced Concrete Slabs is the 5th chapter in the 7-chapter CELE Reinforced & Prestressed Concrete rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Reinforced & Prestressed Concrete.

Reinforced Concrete Slabs - Exam Answer Templates

Proper answer writing is the single most controllable factor in your board exam score. In the PRC Civil Engineer Licensure Examination, partial credit is awarded based on demonstrated understanding, correct use of code provisions, proper formula citation, and logical solution flow — not just the final numerical answer. These templates show you exactly how a top-scoring examinee writes answers: which formulas to cite first, how to show units at every step, when to invoke NSCP 2015 provisions, and how to present conclusions clearly. Follow these structures and you eliminate the most common sources of mark deduction: missing units, skipped steps, wrong code reference, and absent minimum-steel checks. Study each template until you can reproduce its structure from memory.

Templates

Classify each slab panel as one-way or two-way: (a) 3.0 m × 7.0 m panel; (b) 5.5 m × 6.0 m panel.

Marks

Topic

One-Way vs. Two-Way Classification

Difficulty

easy

Template Id

Examiner Tip

Examiners reward the ratio computation even if the final label is correct — show the number.

Model Answer

One-way vs. two-way criterion (NSCP 2015): if L_long/L_short ≥ 2 → one-way; if < 2 → two-way. (a) L_long/L_short = 7.0/3.0 = 2.33 ≥ 2 → ONE-WAY slab. (b) L_long/L_short = 6.0/5.5 = 1.09 < 2 → TWO-WAY slab.

Question Type

very_short_answer

Answer Structure

Line 1: State the criterion with the threshold value of 2 [0.5 mark]
Line 2: Compute ratio for each panel and state the classification [0.5 mark]

Scoring Breakdown

Marks

Criteria

Correct classification of BOTH panels with the ratio computed or implied

Common Mark Deductions

Inverting the ratio (computing short/long instead of long/short)
Classifying the 5.5 m × 6.0 m panel as one-way without computing the ratio

Key Phrases To Include

L_long/L_short
≥ 2 → one-way
< 2 → two-way

State the minimum slab thickness formula for a one-way simply supported slab under NSCP 2015 and list all four support conditions with their corresponding fractions.

Marks

Topic

Minimum Slab Thickness

Difficulty

easy

Template Id

Examiner Tip

A neat table format signals organized thinking and earns the full mark faster than a prose list — use it whenever you list four or more items.

Model Answer

Under NSCP 2015 (ACI 318-based provisions for deflection control), the minimum thickness of a non-prestressed one-way solid slab for f_y = 420 MPa is: | Support Condition | Minimum h | |-------------------------|------------| | Simply supported | L / 20 | | One end continuous | L / 24 | | Both ends continuous | L / 28 | | Cantilever | L / 10 | For f_y ≠ 420 MPa, multiply by the factor: (0.4 + f_y/700).

Question Type

very_short_answer

Answer Structure

Line 1: Cite NSCP 2015 / ACI 318 basis and state f_y = 420 MPa reference condition [0.5 mark]
Lines 2–5: List all four support conditions with correct fractions in tabular or list form [1 mark]
Line 6: State the f_y correction factor [0.5 mark]

Scoring Breakdown

Marks

Criteria

All four fractions correctly stated for the four support conditions

Marks

Criteria

f_y correction factor (0.4 + f_y/700) stated with the condition that it applies when f_y ≠ 420 MPa

Common Mark Deductions

Mixing up L/24 and L/28 for the two continuous conditions
Forgetting the cantilever value (L/10)
Omitting the f_y correction factor entirely

Key Phrases To Include

L/20, L/24, L/28, L/10
f_y = 420 MPa
(0.4 + f_y/700)
NSCP 2015

Define shrinkage and temperature reinforcement in concrete slabs. Give the minimum steel ratio for f_y = 415 MPa and the maximum spacing limit.

Marks

Topic

Shrinkage and Temperature Steel

Difficulty

easy

Template Id

Examiner Tip

The examiner specifically checks whether you use h (total thickness) and not d (effective depth) in the A_s formula for temperature steel.

Model Answer

Shrinkage and temperature (S&T) reinforcement is steel placed perpendicular to the main flexural bars to control crack widths caused by concrete shrinkage and thermal expansion or contraction — not to resist flexure. Minimum steel ratio (NSCP 2015 / ACI 318-14 §24.4.3): ρ_temp = 0.0018 (for f_y = 415–420 MPa) ρ_temp = 0.0020 (for f_y = 275 MPa) Required area per metre: A_s,temp = 0.0018 × b × h Maximum spacing limit for temperature steel: s_max = min(5h, 450 mm)

Question Type

short_answer

Answer Structure

Sentence 1: Define S&T steel — purpose and direction relative to main bars [0.5 mark]
Line 2: State ρ_temp = 0.0018 for f_y = 415–420 MPa [0.5 mark]
Line 3: Write the area formula A_s,temp = 0.0018 × b × h [0.5 mark]
Line 4: State the spacing limit min(5h, 450 mm) [0.5 mark]

Scoring Breakdown

Marks

Criteria

Correct definition including direction (perpendicular to main bars) and purpose (crack control)

Marks

Criteria

Correct ρ = 0.0018, formula A_s = 0.0018bh, and spacing limit min(5h, 450 mm)

Common Mark Deductions

Using ρ = 0.0018 on effective depth d instead of total thickness h
Quoting the main-steel spacing limit (3h) instead of the temperature-steel limit (5h)
Confusing S&T steel direction — writing 'parallel to main bars'

Key Phrases To Include

perpendicular to main bars
crack control
ρ_temp = 0.0018
min(5h, 450 mm)

A one-way slab is continuous at both ends with a clear span of 4.2 m. Steel yield strength f_y = 275 MPa. Determine the minimum required slab thickness.

Marks

Topic

Minimum Slab Thickness

Difficulty

easy

Template Id

Examiner Tip

Always round slab thickness UP — a slab thinner than the computed minimum violates the deflection provision.

Model Answer

Given: L = 4.2 m = 4200 mm; f_y = 275 MPa; both ends continuous. Step 1 – Base formula (NSCP 2015, both ends continuous): h_base = L/28 = 4200/28 = 150 mm Step 2 – Apply f_y correction factor: Factor = 0.4 + f_y/700 = 0.4 + 275/700 = 0.4 + 0.393 = 0.793 Step 3 – Minimum thickness: h_min = 150 × 0.793 = 118.9 mm ∴ Use h = 120 mm (rounded up to nearest 5 mm).

Question Type

numerical

Answer Structure

Line 1: List given data with units [implied mark]
Line 2: Identify support condition and write base formula h = L/28 [0.5 mark]
Line 3: Compute and write the f_y correction factor [0.5 mark]
Line 4: Multiply and state h_min rounded up [1 mark total for T4 at 2 marks]

Scoring Breakdown

Marks

Criteria

Correct base value L/28 computed as 150 mm

Marks

Criteria

Correct factor (0.793) applied and final answer rounded up (≈120 mm)

Common Mark Deductions

Using L/20 (simply supported) instead of L/28 (both ends continuous)
Forgetting the f_y correction factor because f_y = 275 ≠ 420
Rounding DOWN instead of up

Key Phrases To Include

h = L/28
0.4 + f_y/700
0.793
round up

A 200 mm thick one-way slab uses f_y = 415 MPa for all steel. Determine the required area and spacing of 12 mm diameter temperature bars per metre width.

Marks

Topic

Shrinkage and Temperature Steel

Difficulty

medium

Template Id

Examiner Tip

Always adopt a practical spacing (e.g., 250, 300 mm) and verify the provided area exceeds the required area — this shows professional judgment.

Model Answer

Given: h = 200 mm; f_y = 415 MPa; temperature bars: φ12 mm (A_b = 113 mm²). Step 1 – Minimum temperature steel ratio (NSCP 2015 / ACI 318): ρ_temp = 0.0018 (since f_y = 415 MPa) Step 2 – Required area per metre width: A_s,temp = ρ_temp × b × h = 0.0018 × 1000 × 200 = 360 mm²/m Step 3 – Bar spacing: s = (A_b × 1000) / A_s,temp = (113 × 1000) / 360 = 314 mm Step 4 – Check maximum spacing limit: s_max = min(5h, 450 mm) = min(5×200, 450) = min(1000, 450) = 450 mm Since 314 mm < 450 mm → OK. ∴ Provide φ12 mm temperature bars at s = 300 mm (round down to standard spacing) → A_s provided = 113×1000/300 = 377 mm²/m > 360 mm²/m ✓

Question Type

numerical

Answer Structure

Line 1–2: State ρ_temp = 0.0018 citing NSCP 2015 [1 mark]
Line 3–4: Compute A_s,temp = 0.0018 × 1000 × 200 = 360 mm²/m [1 mark]
Line 5–6: Compute spacing s = 113×1000/360 = 314 mm [0.5 mark]
Line 7–8: Check s_max = min(5h,450) = 450 mm; confirm 314 < 450 → adopt 300 mm [0.5 mark]

Scoring Breakdown

Marks

Criteria

Correct ρ = 0.0018 and A_s,temp = 360 mm²/m

Marks

Criteria

Correct bar spacing formula and computed s = 314 mm

Marks

Criteria

Spacing limit check min(5h, 450) and adoption of practical spacing (300 mm)

Common Mark Deductions

Using d (effective depth) instead of h in A_s,temp formula
Using spacing limit min(3h, 450) — which applies to MAIN steel, not temperature steel
Not rounding spacing to a practical value or not verifying the provided area

Key Phrases To Include

ρ_temp = 0.0018
A_s = ρ × b × h
s = A_b × 1000 / A_s
min(5h, 450 mm)

A simply supported one-way slab with a span of 3.5 m is to be used as a roof slab. Given f_y = 415 MPa, determine the minimum slab thickness to satisfy NSCP 2015 deflection requirements.

Marks

Topic

Minimum Slab Thickness

Difficulty

medium

Template Id

Examiner Tip

The f_y factor must be applied whenever f_y ≠ 420 MPa — the 5 MPa difference between 415 and 420 is not negligible in code compliance.

Model Answer

Given: L = 3.5 m = 3500 mm; f_y = 415 MPa; simply supported; roof slab (non-prestressed). Step 1 – Base minimum thickness (NSCP 2015, simply supported): h_base = L/20 = 3500/20 = 175 mm Step 2 – f_y correction factor (since f_y = 415 ≠ 420 MPa): Factor = 0.4 + f_y/700 = 0.4 + 415/700 = 0.4 + 0.5929 = 0.9929 Step 3 – Adjusted minimum thickness: h_min = 175 × 0.9929 = 173.7 mm → round up to 175 mm ∴ Minimum slab thickness h = 175 mm. (Note: Rounding to 175 mm also satisfies the 174 mm computed value; practical minimum is 150 mm for roofs in most Philippine practice, but code minimum governs here.)

Question Type

numerical

Answer Structure

Line 1: Identify support condition; write h_base = L/20 = 175 mm [1 mark]
Line 2: Write and compute f_y factor = 0.4 + 415/700 = 0.993 [1 mark]
Line 3: Multiply h_base × factor = 174 mm; round up to 175 mm [1 mark]

Scoring Breakdown

Marks

Criteria

Correct base formula L/20 and base value 175 mm

Marks

Criteria

Correct f_y factor = 0.4 + 415/700 evaluated numerically

Marks

Criteria

Final h rounded up to 175 mm with clear conclusion statement

Common Mark Deductions

Not applying the f_y factor because 415 is 'close to' 420
Rounding to 170 mm (rounding down)
Using L/24 or L/28 for a simply supported slab

Key Phrases To Include

h = L/20
0.4 + f_y/700
round up
175 mm

A one-way slab strip (b = 1000 mm, d = 150 mm, h = 175 mm, f'_c = 28 MPa, f_y = 415 MPa) carries a factored moment M_u = 15 kN·m/m. Determine the required main flexural steel area per metre and check against the minimum.

Marks

Topic

Flexural Design of One-Way Slabs

Difficulty

hard

Template Id

Examiner Tip

The minimum-steel check is the most frequently missed step in board exam slab problems — it is always worth a dedicated mark. Write the comparison explicitly: 272 < 315, ∴ minimum governs.

Model Answer

Given: b = 1000 mm; d = 150 mm; h = 175 mm; f'_c = 28 MPa; f_y = 415 MPa; M_u = 15 kN·m/m; φ = 0.90 (flexure, NSCP 2015). Step 1 – Compute nominal resistance coefficient R_n: R_n = M_u / (φ b d²) = (15 × 10⁶) / (0.90 × 1000 × 150²) = 15,000,000 / 20,250,000 = 0.741 MPa Step 2 – Compute β₁ (NSCP 2015; f'_c = 28 MPa ≤ 28 MPa → β₁ = 0.85): β₁ = 0.85 Step 3 – Compute steel ratio ρ: ρ = (0.85 f'_c / f_y) × [1 − √(1 − 2R_n / (0.85 f'_c))] = (0.85 × 28 / 415) × [1 − √(1 − 2(0.741)/(0.85×28))] = (0.05735) × [1 − √(1 − 1.482/23.8)] = (0.05735) × [1 − √(1 − 0.06227)] = (0.05735) × [1 − √(0.93773)] = (0.05735) × [1 − 0.96836] = (0.05735) × (0.03164) = 0.001814 Step 4 – Compute required A_s: A_s,calc = ρ × b × d = 0.001814 × 1000 × 150 = 272 mm²/m Step 5 – Check minimum steel (temperature/shrinkage governs for thin slabs): A_s,min = 0.0018 × b × h = 0.0018 × 1000 × 175 = 315 mm²/m Step 6 – Compare and govern: A_s,calc = 272 mm²/m < A_s,min = 315 mm²/m ∴ Minimum steel GOVERNS. ∴ Provide A_s = 315 mm²/m (e.g., φ12 mm at s = 113×1000/315 = 359 mm → use s = 350 mm; check: 350 mm < min(3h,450) = min(525,450) = 450 mm ✓)

Question Type

numerical

Answer Structure

Step 1: Write R_n formula; substitute correctly; compute R_n = 0.741 MPa [1 mark]
Step 2: State β₁ = 0.85 for f'_c = 28 MPa (optional but shows code awareness)
Step 3: Write ρ formula with √ term; substitute and compute ρ = 0.001814 [1 mark]
Step 4: Compute A_s,calc = ρ × b × d = 272 mm²/m [1 mark]
Step 5: Compute A_s,min = 0.0018 × b × h = 315 mm²/m [1 mark]
Step 6: Compare, state minimum governs, adopt 315 mm²/m; convert to spacing [1 mark]

Scoring Breakdown

Marks

Criteria

R_n correctly computed as 0.741 MPa with φ = 0.90

Marks

Criteria

ρ formula written and evaluated to approximately 0.00181

Marks

Criteria

A_s,calc = 272 mm²/m correctly computed

Marks

Criteria

A_s,min = 315 mm²/m correctly computed using h (not d)

Marks

Criteria

Comparison made, minimum governs stated, practical spacing provided with limit check

Common Mark Deductions

Using b×d² but forgetting φ = 0.90 in the denominator of R_n
Computing A_s,min with d instead of h
Failing to compare A_s,calc with A_s,min (omitting the minimum check)
Not converting A_s to a practical bar spacing with a limit check

Key Phrases To Include

R_n = M_u / (φ b d²)
ρ = (0.85 f'_c / f_y)(1 − √(1 − 2R_n / 0.85f'_c))
A_s,min = 0.0018 × b × h
minimum governs
min(3h, 450 mm)

State the total static moment formula used in the Direct Design Method (DDM) for two-way slabs and define each term.

Marks

Topic

Two-Way Slabs — Direct Design Method

Difficulty

medium

Template Id

Examiner Tip

The most common error is using center-to-center span instead of clear span for L_n — always distinguish these in your answer.

Model Answer

For two-way slabs analyzed by the Direct Design Method (DDM) per NSCP 2015 (ACI 318): M_o = w_u × L₂ × L_n² / 8 Where: M_o = total factored static moment in the design strip (kN·m) w_u = factored uniformly distributed load per unit area (kN/m²) L₂ = transverse span length center-to-center of supports (m) L_n = clear span in the direction of analysis, measured face-to-face of supports (m); L_n ≥ 0.65 L₁ This total moment M_o is then distributed between the column strip and middle strip as negative and positive moments using DDM distribution factors.

Question Type

short_answer

Answer Structure

Line 1: Write the formula M_o = w_u L₂ L_n² / 8 [1 mark]
Lines 2–5: Define all four variables with units [1 mark]

Scoring Breakdown

Marks

Criteria

Formula written correctly with 8 in denominator and L_n² (not L₁² or L₂²)

Marks

Criteria

All four terms defined with correct descriptions (especially L_n as clear span, not center-to-center)

Common Mark Deductions

Using L₁ (center-to-center span) instead of L_n (clear span) in the formula
Swapping L₁ and L₂ definitions
Writing L_n instead of L_n² (forgetting the square)

Key Phrases To Include

M_o = w_u L₂ L_n² / 8
clear span L_n
transverse span L₂
Direct Design Method

What is the maximum allowable bar spacing for (a) main flexural steel and (b) shrinkage/temperature steel in a one-way slab of thickness h?

Marks

Topic

Spacing Limits — Slab Reinforcement

Difficulty

easy

Template Id

Examiner Tip

The number is easy to remember: MAIN steel gets the SMALLER multiplier (3h) because it carries loads; temperature steel gets the LARGER multiplier (5h) because it only controls cracks.

Model Answer

Per NSCP 2015: (a) Main flexural steel: s_max = min(3h, 450 mm) (b) Shrinkage/temperature steel: s_max = min(5h, 450 mm)

Question Type

very_short_answer

Answer Structure

One line each for (a) and (b) with the correct multiplier (3 or 5) and the 450 mm cap

Scoring Breakdown

Marks

Criteria

Both limits stated correctly: 3h vs 5h with 450 mm cap for each

Common Mark Deductions

Swapping the multipliers — writing 5h for main steel and 3h for temperature steel
Omitting the 450 mm absolute cap

Key Phrases To Include

min(3h, 450 mm)
min(5h, 450 mm)

A two-way slab panel measuring 6 m × 7.5 m (center-to-center) carries a factored load w_u = 12 kN/m². The clear span in the 6 m direction is 5.6 m. Compute the total static moment M_o for the design strip spanning in the 6 m direction.

Marks

Topic

Two-Way Slabs — Direct Design Method

Difficulty

medium

Template Id

T10

Examiner Tip

L_n is always the CLEAR span (face-to-face of supports) — not the center-to-center span. Substituting the wrong value is the single most common error in DDM problems.

Model Answer

Given: L₁ = 6.0 m (span direction); L₂ = 7.5 m (transverse span, c/c); L_n = 5.6 m (clear span in L₁ direction); w_u = 12 kN/m². Verify DDM applicability — classify slab: L_long/L_short = 7.5/6.0 = 1.25 < 2 → two-way slab ✓ Total static moment (DDM, NSCP 2015): M_o = w_u × L₂ × L_n² / 8 = 12 × 7.5 × (5.6)² / 8 = 12 × 7.5 × 31.36 / 8 = 2822.4 / 8 = 352.8 kN·m ∴ M_o = 352.8 kN·m for the design strip in the 6 m direction.

Question Type

numerical

Answer Structure

Line 1: Classify panel as two-way (ratio = 1.25 < 2) [0.5 mark]
Line 2: Write the DDM formula M_o = w_u L₂ L_n² / 8 [0.5 mark]
Line 3: Substitute values correctly [1 mark]
Line 4: Compute and state final answer with units [1 mark]

Scoring Breakdown

Marks

Criteria

Formula written and L_n (5.6 m, clear span) correctly used rather than L₁ (6.0 m)

Marks

Criteria

Numerical computation correct: 12 × 7.5 × 31.36 / 8

Marks

Criteria

Final answer M_o = 352.8 kN·m with units

Common Mark Deductions

Using L₁ = 6.0 m instead of L_n = 5.6 m for the clear span
Using L₁ in place of L₂ (transverse span) in the formula
Omitting units from the final answer

Key Phrases To Include

M_o = w_u L₂ L_n² / 8
L_n = 5.6 m (clear span)
L₂ = 7.5 m
352.8 kN·m

Design the main flexural steel for a one-way slab with the following data: M_u = 22 kN·m/m; d = 160 mm; h = 190 mm; f'_c = 28 MPa; f_y = 415 MPa; b = 1000 mm. Select φ12 mm bars (A_b = 113 mm²) and provide the required spacing.

Marks

Topic

Flexural Design of One-Way Slabs

Difficulty

hard

Template Id

T11

Examiner Tip

Show both A_s values side by side before declaring the governing value — examiners award a dedicated mark for the explicit comparison.

Model Answer

Given: b = 1000 mm; d = 160 mm; h = 190 mm; f'_c = 28 MPa; f_y = 415 MPa; M_u = 22 kN·m/m; φ = 0.90. Step 1 – Resistance coefficient: R_n = M_u / (φ b d²) = (22 × 10⁶) / (0.90 × 1000 × 160²) = 22,000,000 / 23,040,000 = 0.955 MPa Step 2 – Steel ratio: ρ = (0.85 f'_c / f_y) [1 − √(1 − 2R_n / (0.85 f'_c))] = (0.85 × 28 / 415) [1 − √(1 − 2(0.955)/(0.85×28))] = (0.05735) [1 − √(1 − 0.08025)] = (0.05735) [1 − √(0.91975)] = (0.05735) [1 − 0.95903] = (0.05735)(0.04097) = 0.002350 Step 3 – Required A_s from analysis: A_s,calc = ρ b d = 0.002350 × 1000 × 160 = 376 mm²/m Step 4 – Minimum A_s check: A_s,min = 0.0018 × b × h = 0.0018 × 1000 × 190 = 342 mm²/m Step 5 – Governing A_s: A_s,calc = 376 mm²/m > A_s,min = 342 mm²/m ∴ Analysis value GOVERNS → A_s = 376 mm²/m Step 6 – Bar spacing for φ12 mm bars: s = A_b × 1000 / A_s = 113 × 1000 / 376 = 300 mm Step 7 – Check spacing limit: s_max = min(3h, 450 mm) = min(3×190, 450) = min(570, 450) = 450 mm 300 mm < 450 mm ✓ ∴ Provide φ12 mm main bars at s = 300 mm (A_s provided = 113×1000/300 = 377 mm²/m > 376 mm²/m ✓)

Question Type

numerical

Answer Structure

Step 1: R_n = 0.955 MPa [1 mark]
Step 2: ρ computed = 0.002350 [1 mark]
Step 3: A_s,calc = 376 mm²/m [0.5 mark]
Step 4–5: A_s,min = 342 mm²/m; analysis governs [1 mark]
Step 6–7: s = 300 mm; spacing limit check 300 < 450 ✓ [0.5 mark]

Scoring Breakdown

Marks

Criteria

R_n correctly computed as 0.955 MPa

Marks

Criteria

ρ formula applied and ρ ≈ 0.00235 obtained

Marks

Criteria

A_s,calc and A_s,min both computed; correct value identified as governing

Marks

Criteria

Bar spacing s = 300 mm computed from s = A_b × 1000 / A_s

Marks

Criteria

Spacing limit min(3h, 450) checked and satisfied; final adoption stated

Common Mark Deductions

Forgetting φ = 0.90 in R_n denominator
Using h instead of d in A_s,calc = ρ b d
Incorrectly identifying which A_s governs
Not performing the spacing limit check after computing s

Key Phrases To Include

R_n = M_u / (φ b d²)
ρ formula with √ term
A_s,min = 0.0018 × b × h
analysis governs
s = A_b × 1000 / A_s
min(3h, 450 mm)

What is the minimum slab thickness h for a two-way slab (flat plate) supported on columns under NSCP 2015, for a 6 m × 6 m panel with f_y = 415 MPa? State the applicable code limit and formula.

Marks

Topic

Two-Way Slabs — Minimum Thickness

Difficulty

medium

Template Id

T12

Examiner Tip

Know that the NSCP minimum thickness provisions are different for flat plates (L_n/33) versus two-way slabs with beams — the coefficient changes based on the beam stiffness ratio α_fm.

Model Answer

For two-way flat plates (slabs without beams on column lines), NSCP 2015 (based on ACI 318-14 §8.3.1) provides minimum thickness to control deflection: h_min = L_n / 33 (for slabs with α_fm ≤ 0.2, i.e., flat plates, f_y = 420 MPa) For f_y = 415 MPa (≈ 420 MPa), the difference is negligible and the same formula applies: h_min = L_n / 33 For a 6 m × 6 m panel, assuming column size ~500 mm: L_n ≈ 6000 − 500 = 5500 mm h_min = 5500 / 33 = 167 mm → use h = 170 mm (round up) Absolute minimum for flat plates: h ≥ 125 mm (NSCP 2015). ∴ Minimum h = 170 mm.

Question Type

short_answer

Answer Structure

Line 1: Identify slab type (flat plate, α_fm ≤ 0.2) and cite NSCP 2015 [0.5 mark]
Line 2: State formula h_min = L_n/33 [0.5 mark]
Line 3: Compute L_n and h_min; round up [1 mark]

Scoring Breakdown

Marks

Criteria

Correct formula h_min = L_n/33 cited with NSCP/ACI basis

Marks

Criteria

Numerical evaluation with correct L_n and rounded-up result

Common Mark Deductions

Applying one-way slab formulas (L/20, L/24) to a flat plate
Using center-to-center span instead of clear span for L_n
Not mentioning the 125 mm absolute minimum

Key Phrases To Include

flat plate
h_min = L_n/33
α_fm ≤ 0.2
125 mm absolute minimum

A slab is supported on stiff beams on all four sides. The panel dimensions are 4 m × 6 m. Using the coefficient method, if the coefficient C = 0.048 and w_u = 10 kN/m², compute the bending moment in the short-span direction.

Marks

Topic

Two-Way Slabs — Coefficient Method

Difficulty

medium

Template Id

T13

Examiner Tip

In the coefficient method, always identify which span (short or long) the coefficient and moment apply to — the problem statement specifies this; do not assume.

Model Answer

Given: Short span L_s = 4 m; C = 0.048; w_u = 10 kN/m². Verify two-way: L_long/L_short = 6/4 = 1.5 < 2 → two-way slab ✓ Coefficient method moment (short-span direction): M = C × w_u × L_s² = 0.048 × 10 × (4)² = 0.048 × 10 × 16 = 7.68 kN·m/m ∴ Bending moment in short-span direction M = 7.68 kN·m/m.

Question Type

numerical

Answer Structure

Line 1: Verify two-way classification (ratio = 1.5 < 2) [implied]
Line 2: Write the coefficient method formula M = C w_u L_s² [1 mark]
Line 3: Substitute and compute M = 7.68 kN·m/m [1 mark]

Scoring Breakdown

Marks

Criteria

Formula M = C × w_u × L_s² correctly applied using the SHORT span

Marks

Criteria

Correct numerical result 7.68 kN·m/m with units

Common Mark Deductions

Using the long span (6 m) instead of the short span (4 m)
Omitting kN·m/m units

Key Phrases To Include

M = C × w_u × L_s²
short span L_s = 4 m
coefficient method

A one-way cantilever slab has a span of 2.0 m and f_y = 420 MPa. (a) What is the minimum thickness? (b) Why is the cantilever minimum (L/10) larger than the simply supported minimum (L/20)?

Marks

Topic

Minimum Slab Thickness

Difficulty

medium

Template Id

T14

Examiner Tip

For concept-explanation parts, always anchor your answer to a physical phenomenon (deflection magnitude) and the code's intent (avoid explicit deflection calculation) — not just 'the code says so.'

Model Answer

(a) Minimum thickness for a cantilever (NSCP 2015, f_y = 420 MPa — no correction factor needed): h_min = L/10 = 2000/10 = 200 mm (No f_y correction since f_y = 420 MPa → factor = 0.4 + 420/700 = 1.0) ∴ h_min = 200 mm (b) Conceptual reason: A cantilever slab experiences maximum moment and maximum deflection at the fixed support, with zero restraint at the free tip. For a given span, the computed deflection of a cantilever under uniform load is approximately five times greater than that of a simply supported beam of the same span and stiffness. To limit this larger deflection to acceptable levels without explicit calculation, the code requires a proportionally thicker slab — hence L/10 (twice the L/20 minimum for simply supported), reflecting the higher deflection demand and the absence of continuity relief.

Question Type

short_answer

Answer Structure

Part (a): Write h = L/10; note f_y = 420 so factor = 1.0; compute 200 mm [1.5 marks]
Part (b): State deflection is ~5× larger for cantilever; no continuity relief; code uses L/10 to compensate [1.5 marks]

Scoring Breakdown

Marks

Criteria

Correct formula L/10 and computed value h = 200 mm

Marks

Criteria

f_y factor noted as 1.0 (no penalty if not stated, but shows awareness)

Marks

Criteria

Valid conceptual explanation: cantilever deflects more; greater minimum thickness needed; reference to deflection being larger at free end

Common Mark Deductions

Applying f_y correction unnecessarily when f_y = 420 MPa exactly
Part (b): Giving a vague answer like 'because it is cantilever' without explaining the deflection basis

Key Phrases To Include

h = L/10
200 mm
f_y = 420 → factor = 1.0
deflection is greater in cantilever
no continuity

Explain the per-metre strip method for designing one-way slab flexural steel. Why is b set to 1000 mm and how is A_s converted to bar spacing?

Marks

Topic

Flexural Design of One-Way Slabs

Difficulty

medium

Template Id

T15

Examiner Tip

Concept questions expect a brief worked example to earn the full mark — show the formula and then plug in numbers even if the question does not ask for computation.

Model Answer

The per-metre strip method treats a one-way slab as a series of identical rectangular beam strips, each of width b = 1000 mm and depth equal to the slab thickness h, running across the short span. Why b = 1000 mm: Since the slab is uniform across its width, any 1-metre length perpendicular to the span represents a representative beam. Using b = 1000 mm allows the designer to express steel requirements directly in mm²/m (area per metre of slab width) — a practical unit for detailing. Design steps: 1. Compute R_n = M_u / (φ × 1000 × d²) 2. Compute ρ and then A_s = ρ × 1000 × d [result in mm²/m] Converting A_s to bar spacing: For a chosen bar diameter with single-bar area A_b (mm²), bars placed at spacing s (mm) provide: A_s,provided = A_b × 1000 / s → s = A_b × 1000 / A_s Example: A_s = 315 mm²/m; φ10 mm bars (A_b = 78.5 mm²): s = 78.5 × 1000 / 315 = 249 mm → use 200 mm (round down for conservatism). Finally, verify s ≤ min(3h, 450 mm).

Question Type

short_answer

Answer Structure

Paragraph 1: Define the strip and explain b = 1000 mm [1 mark]
Paragraph 2: Write the A_s formula and state units mm²/m [1 mark]
Paragraph 3: Write and explain s = A_b × 1000 / A_s; include limit check [1 mark]

Scoring Breakdown

Marks

Criteria

Clear explanation of why b = 1000 mm — representative strip, units in mm²/m

Marks

Criteria

Correct formula for A_s and units clearly stated

Marks

Criteria

Spacing formula s = A_b × 1000 / A_s shown with a worked example or statement of limit check

Common Mark Deductions

Not explaining WHY b = 1000 mm is used
Inverting the spacing formula: writing s = A_s / A_b instead of s = A_b × 1000 / A_s
Omitting the spacing limit check

Key Phrases To Include

representative 1-metre strip
b = 1000 mm
mm²/m
s = A_b × 1000 / A_s
min(3h, 450 mm)

Mark Wise Strategy

Dos

State the answer directly in the first line
Include the threshold value (e.g., ≥ 2 for one-way) when classifying
Write units even for short answers
Use the exact NSCP code notation for ratios and formulas

Donts

Do not write lengthy introductory sentences — go straight to the answer
Do not compute if only a formula or classification is asked
Do not leave the ratio unevaluated — always compute the decimal value

Marks

Strategy

Deliver one crisp, complete sentence or a single formula with the correct numerical value. For classification questions, always show the ratio computed. For formula-recall questions, write the formula and the variable it applies to.

Expected Length

1–2 lines or a single equation

Time Allocation

1–2 minutes

Dos

Structure as Step 1 / Step 2 explicitly
Include the code reference (NSCP 2015, ACI 318) at least once
Show units at each step
Round final answers to the appropriate precision (nearest mm or 2 decimal places for MPa)

Donts

Do not combine both steps into one line — examiners need to see each mark earned separately
Do not skip intermediate calculations for 2-mark numerical problems
Do not forget the f_y correction factor for minimum thickness

Marks

Strategy

Show two distinct, marked steps. For numerical problems: Step 1 = formula, Step 2 = substitution and result. For concept questions: definition + code reference or formula + spacing/limit rule.

Expected Length

3–5 lines or 2 clear steps

Time Allocation

3–4 minutes

Dos

Number each step explicitly (Step 1, Step 2, Step 3)
Perform the minimum-steel check as a dedicated step — it is almost always worth 1 mark
Draw a small table or list spacing limits clearly
State which value governs with a comparison (e.g., 272 < 315 ∴ minimum governs)

Donts

Do not bury the minimum-steel check inside another step — give it its own line
Do not forget to check spacing limits after computing s
Do not write paragraphs — use step-by-step format

Marks

Strategy

Structure as three clearly separated steps — each earning one mark. For design problems: formula → compute required A_s → minimum check and spacing. For concept problems: definition → code provision → worked numerical example. Always end with a boxed or underlined conclusion.

Expected Length

6–10 lines or 3–4 numbered steps

Time Allocation

5–8 minutes

Dos

Begin with a 'Given' block listing all data with units
Write the formula on one line, then substitute on the next — never combine
Perform and show the minimum-steel check explicitly as a dedicated step
End with a professional conclusion: 'Provide φXX mm bars at s = YYY mm'
Verify the selected spacing against both the computed spacing and the code limit

Donts

Do not skip the β₁ check or the φ factor — these show code compliance
Do not use d in place of h for minimum temperature steel area
Do not present a long answer without clear step numbering — disorganized answers lose marks even if computations are correct
Do not forget to verify that provided A_s ≥ required A_s after selecting the practical spacing

Marks

Strategy

Full design solution format: list all given data → write all formulas before substituting → compute step by step → perform minimum-steel check → select bar size and spacing → verify spacing limit → state conclusion. Each major computation earns one mark. Clarity and organization are rewarded.

Expected Length

15–25 lines with 5–7 numbered steps and a conclusion

Time Allocation

10–15 minutes

General Answer Writing Tips

Always state the governing code provision (e.g., NSCP 2015 Section 406) before applying a formula — examiners award marks for code awareness, not just arithmetic.
Write units at every intermediate step; a correct numerical answer with missing units can lose marks in the scoring rubric.
For slab problems, always state whether the slab is one-way or two-way by citing the span ratio L_long/L_short and comparing it to 2.0 — this is the mandatory first step.
Design per 1-metre strip: explicitly write 'b = 1000 mm' at the start of every slab flexural design to signal to the examiner you are applying the per-metre convention.
Always perform the minimum-steel check (temperature steel A_s,min = 0.0018bh for f_y = 415 MPa) and explicitly state which value governs — many students lose marks by omitting this comparison.
Convert A_s (mm²/m) to bar spacing using s = A_b × 1000 / A_s and then check against the spacing limit min(3h, 450 mm) for main steel or min(5h, 450 mm) for temperature steel.
For minimum thickness problems, always write the base formula first (e.g., h = L/20), then apply the f_y correction factor (0.4 + f_y/700), and round UP to the nearest 5 mm or 25 mm increment.
Box or underline your final answers with proper units and a brief conclusion statement (e.g., 'Use h = 175 mm') — examiners scanning answer sheets reward clearly identified final answers.

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