CELE Reinforced & Prestressed Concrete — Reinforced Concrete Beams: FlexureExam Answer Templates

Marks

1

Topic

Basic Concepts — Singly Reinforced Beam

Difficulty

easy

Template Id

T1

Examiner Tip

Both endpoints of the measurement must be named correctly. A partial definition (only one endpoint) earns zero on most rubrics.

Model Answer

The effective depth d is the distance from the extreme compression fiber to the centroid of the tension reinforcement.

Question Type

very_short_answer

Answer Structure

• One complete sentence: state what d is measured from AND to [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'bottom of beam' instead of 'centroid of tension steel' — loses the mark
• Confusing total depth h with effective depth d
• Omitting 'centroid' — just writing 'tension steel' is incomplete

Key Phrases To Include

• extreme compression fiber
• centroid of tension reinforcement
• effective depth d

Marks

1

Topic

Tension-Controlled Sections — Ductility

Difficulty

easy

Template Id

T2

Examiner Tip

Board exams often give 0.5 mark each for the strain value and the φ factor. Never state one without the other.

Model Answer

A section is tension-controlled when the net tensile strain in the extreme tension steel at nominal strength is εt ≥ 0.005, permitting the use of the strength-reduction factor φ = 0.90 for flexure (NSCP 2015 / ACI 318-19 Table 21.2.2).

Question Type

very_short_answer

Answer Structure

• State the numerical strain limit: εt ≥ 0.005 [0.5 mark]
• State the resulting φ factor: φ = 0.90 [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing εt ≥ 0.004 (the transition zone boundary) instead of 0.005
• Stating φ = 0.85 — incorrect value for flexure
• Omitting the φ value entirely

Key Phrases To Include

• net tensile strain
• εt ≥ 0.005
• φ = 0.90
• tension-controlled

Marks

2

Topic

Equivalent Stress Block — Singly Reinforced Beam

Difficulty

easy

Template Id

T3

Examiner Tip

The factor 0.85 is the single most tested constant in RC flexure. Always write it explicitly — never absorb it into f'c.

Model Answer

The depth of the equivalent rectangular stress block is: a = (As · fy) / (0.85 · f'c · b) Where: As = area of tension steel (mm²) fy = yield strength of steel (MPa) f'c = specified compressive strength of concrete (MPa) b = width of the compression zone (mm) 0.85 = Whitney stress block intensity factor (NSCP 2015 / ACI 318-19 Section 22.2.2)

Question Type

short_answer

Answer Structure

• Line 1: Write the formula correctly [1 mark]
• Lines 2–6: Define all variables with correct units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing f'c instead of 0.85 f'c in the denominator — loses the formula mark
• Omitting units from variable definitions
• Confusing b (width) with d (effective depth) in the formula

Key Phrases To Include

• 0.85 f'c
• tension reinforcement
• compression zone width b
• Whitney stress block

Marks

3

Topic

Steel Ratio Limits — ρ_b and ρ_max

Difficulty

medium

Template Id

T4

Examiner Tip

Board examiners reward derivation logic, not just formulas. Show the strain triangle step even in a 3-mark question — it earns the process mark.

Model Answer

Balanced Steel Ratio ρ_b At balanced failure, concrete reaches εcu = 0.003 simultaneously as steel reaches εy = fy/Es. By similar triangles on the strain diagram: c_b / d = 0.003 / (0.003 + fy/Es) With Es = 200,000 MPa: c_b = d · [600 / (600 + fy)] Since a_b = β₁ c_b and the stress-block equilibrium gives ρ_b = 0.85 β₁ (f'c/fy)(c_b/d): ρ_b = 0.85 β₁ (f'c/fy) · [600 / (600 + fy)] …(1) Maximum Steel Ratio ρ_max (Tension-Controlled Limit) NSCP 2015 (ACI 318-19 Table 21.2.2) requires εt ≥ 0.005 for φ = 0.90. At εt = 0.005: c/d = 0.003 / (0.003 + 0.005) = 0.003/0.008 = 0.375 Therefore: ρ_max = 0.85 β₁ (f'c/fy) · (0.375) …(2) Note: Equation (2) supersedes the older 0.75ρ_b limit in NSCP 2001; the two are numerically close but not identical.

Question Type

short_answer

Answer Structure

• Part 1 — ρ_b: Strain compatibility diagram statement → c_b expression → final ρ_b formula [1.5 marks]
• Part 2 — ρ_max: State εt = 0.005 requirement → derive c/d = 0.375 → state ρ_max formula [1 mark]
• Code citation and comparison note [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating ρ_max = 0.75ρ_b without noting it is the older NSCP 2001 expression
• Omitting the derivation and just writing the formula — no derivation marks awarded
• Wrong sign or arrangement in strain-triangle ratio

Key Phrases To Include

• strain compatibility
• εcu = 0.003
• β₁
• 0.375
• εt ≥ 0.005
• NSCP 2015

Marks

3

Topic

Analysis — Singly Reinforced Rectangular Beam

Difficulty

medium

Template Id

T5

Examiner Tip

The ductility check step is worth its own mark in most rubrics. Write it as a separate numbered step even if it seems obvious.

Model Answer

Given: b = 300 mm, d = 500 mm, As = 1 473 mm², f'c = 28 MPa, fy = 415 MPa β₁ = 0.85 (since f'c = 28 MPa ≤ 28 MPa → β₁ = 0.85, NSCP 2015 Section 422.2.2.4.3) Step 1 — Stress-block depth: a = As·fy / (0.85·f'c·b) a = (1 473 × 415) / (0.85 × 28 × 300) a = 611 295 / 7 140 a = 85.6 mm Step 2 — Neutral-axis depth: c = a / β₁ = 85.6 / 0.85 = 100.7 mm Step 3 — Net tensile strain (ductility check): εt = 0.003 × (d − c) / c εt = 0.003 × (500 − 100.7) / 100.7 εt = 0.003 × 3.963 = 0.01189 > 0.005 ✓ Tension-controlled → φ = 0.90 Step 4 — Nominal moment: Mn = As·fy·(d − a/2) Mn = 1 473 × 415 × (500 − 42.8) Mn = 611 295 × 457.2 Mn = 279.5 × 10⁶ N·mm = 279.5 kN·m Step 5 — Design capacity: φMn = 0.90 × 279.5 = 251.5 kN·m Answer: φMn = 251.5 kN·m

Question Type

numerical

Answer Structure

• Given block with all data [0.5 mark]
• Formula and calculation of a [0.5 mark]
• Tension-controlled check (εt vs 0.005) with φ stated [1 mark]
• Mn and φMn with correct units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Skipping the εt check and assuming φ = 0.90 without proof
• Using full beam depth h instead of effective depth d
• Forgetting to divide 'a' by 2 in the moment-arm expression
• Leaving the answer in N·mm instead of converting to kN·m

Key Phrases To Include

• a = As fy / (0.85 f'c b)
• tension-controlled
• εt > 0.005
• φ = 0.90
• φMn

Marks

3

Topic

Steel Ratio Limits — Verification

Difficulty

medium

Template Id

T6

Examiner Tip

ρ_min uses the MAXIMUM of two expressions — this is a classic trap. Write both values and explicitly pick the larger one.

Model Answer

Given: b = 300 mm, d = 500 mm, As = 1 473 mm², f'c = 28 MPa, fy = 415 MPa, β₁ = 0.85 Actual steel ratio: ρ = As / (b·d) = 1 473 / (300 × 500) = 0.009 82 Minimum steel ratio (NSCP 2015 Section 409.6.1.1 / ACI 318-19 Section 9.6.1.2): ρ_min = max(1.4/fy , √f'c / (4fy)) ρ_min = max(1.4/415 , √28 / (4×415)) ρ_min = max(0.003 37 , 5.292/1 660) ρ_min = max(0.003 37 , 0.003 19) ρ_min = 0.003 37 ← governs Maximum steel ratio (tension-controlled, NSCP 2015): ρ_max = 0.85 β₁ (f'c/fy)(0.375) ρ_max = 0.85 × 0.85 × (28/415) × 0.375 ρ_max = 0.7225 × 0.067 47 × 0.375 ρ_max = 0.01828 Check: ρ_min = 0.003 37 ≤ ρ = 0.009 82 ≤ ρ_max = 0.018 3 ✓ Conclusion: The section is under-reinforced and tension-controlled. Both NSCP 2015 steel ratio limits are satisfied.

Question Type

numerical

Answer Structure

• Compute actual ρ [0.5 mark]
• Compute ρ_min correctly using both expressions and taking the larger [1 mark]
• Compute ρ_max using 0.375 factor [1 mark]
• State comparison and conclusion [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Taking the smaller of the two ρ_min expressions instead of the larger
• Using ρ_max = 0.75ρ_b without justification (old code value)
• Not stating the comparison conclusion

Key Phrases To Include

• ρ_min = max(1.4/fy, √f'c/4fy)
• ρ_max = 0.85β₁(f'c/fy)(0.375)
• under-reinforced
• tension-controlled

Marks

5

Topic

Design — Singly Reinforced Rectangular Beam

Difficulty

medium

Template Id

T7

Examiner Tip

In a 5-mark design problem, every step is worth approximately 1 mark. A complete 7-step solution as shown above hits every rubric point. Never compress multiple steps into one line.

Model Answer

Given: Mu = 200 kN·m = 200 × 10⁶ N·mm, b = 300 mm, d = 450 mm f'c = 28 MPa, fy = 415 MPa, β₁ = 0.85 Step 1 — Assume tension-controlled (φ = 0.90); verify later. Step 2 — Coefficient of resistance Rn: Rn = Mu / (φ · b · d²) Rn = (200 × 10⁶) / (0.90 × 300 × 450²) Rn = 200 × 10⁶ / 54 675 000 Rn = 3.658 MPa Step 3 — Required steel ratio: ρ = (0.85 f'c / fy) · [1 − √(1 − 2Rn/(0.85 f'c))] 0.85 f'c = 0.85 × 28 = 23.8 MPa 2Rn/(0.85 f'c) = 2(3.658)/23.8 = 0.3074 ρ = (23.8/415) · [1 − √(1 − 0.3074)] ρ = 0.05735 · [1 − √0.6926] ρ = 0.05735 · [1 − 0.8322] ρ = 0.05735 × 0.1678 = 0.009 622 Step 4 — Required steel area: As = ρ · b · d = 0.009 622 × 300 × 450 As = 1 299 mm² Step 5 — Check ρ limits: ρ_min = 1.4/415 = 0.003 37 ρ_max = 0.85(0.85)(28/415)(0.375) = 0.01828 0.003 37 ≤ 0.009 62 ≤ 0.018 3 ✓ Step 6 — Verify ductility (confirm φ = 0.90): a = As·fy / (0.85 f'c·b) = 1 299×415/(0.85×28×300) = 539 085/7 140 = 75.5 mm c = 75.5/0.85 = 88.8 mm εt = 0.003(450 − 88.8)/88.8 = 0.003 × 4.068 = 0.01220 > 0.005 ✓ φ = 0.90 confirmed Step 7 — Bar selection: Use 3 – 25 mm bars: As,provided = 3 × 490.9 = 1 473 mm² > 1 299 mm² ✓ Answer: Provide 3 – 25 mm bars (As = 1 473 mm²).

Question Type

numerical

Answer Structure

• Given block with unit conversion of Mu [0.5 mark]
• Rn formula and calculation [1 mark]
• ρ formula and calculation [1 mark]
• As computed and bar selection [0.5 mark]
• ρ_min ≤ ρ ≤ ρ_max check with values [1 mark]
• ductility verification (εt check, φ confirmed) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using Mu in kN·m without converting to N·mm before computing Rn
• Not verifying ductility after design — most rubrics give a dedicated mark for this
• Selecting a bar size without confirming As,provided ≥ As,required
• Skipping ρ_min check — an examiner will deduct if omitted

Key Phrases To Include

• Rn = Mu/(φbd²)
• ρ = (0.85f'c/fy)[1 − √(1 − 2Rn/0.85f'c)]
• As = ρbd
• tension-controlled
• εt > 0.005

Marks

2

Topic

Over-Reinforced Sections — Ductility

Difficulty

medium

Template Id

T8

Examiner Tip

Always link the strain condition (εt < 0.005) to the φ value when discussing over-reinforced sections. This shows code awareness.

Model Answer

An over-reinforced beam has excess tension steel so the concrete compression zone crushes (εc = 0.003) before the steel reaches its yield strain (εt < εy). The failure is sudden and brittle — there is no ductile warning (large deflections or cracking) before collapse, which is unsafe for occupants. Because εt < 0.005 (and possibly εt < 0.002), the section falls into the transition or compression-controlled zone. NSCP 2015 / ACI 318-19 reduces φ below 0.90, approaching φ = 0.65 for compression-controlled sections, thereby increasing the required nominal strength and making the design less efficient. NSCP 2015 prohibits the use of over-reinforced sections for new design by enforcing ρ ≤ ρ_max.

Question Type

short_answer

Answer Structure

• Sentence 1–2: Describe the failure mechanism (concrete crushes first, no steel yielding, brittle failure) [1 mark]
• Sentence 3–4: Effect on φ — reduced below 0.90, approaching 0.65; code prohibits such sections [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying the beam is 'stronger' because it has more steel — factually wrong; capacity does not increase proportionally
• Not mentioning the effect on φ — this is specifically asked
• Confusing over-reinforced with doubly reinforced

Key Phrases To Include

• concrete crushes first
• steel does not yield
• brittle failure
• no ductile warning
• φ reduced
• compression-controlled

Marks

2

Topic

Minimum Steel Ratio — Code Provisions

Difficulty

easy

Template Id

T9

Examiner Tip

Write the keyword 'max' in the formula explicitly. If you write 'min' by accident, it signals you do not understand the purpose and loses the mark.

Model Answer

Per NSCP 2015 Section 409.6.1.1 (ACI 318-19 Section 9.6.1.2), the minimum steel ratio is: ρ_min = max( 1.4/fy , √f'c / (4fy) ) The first expression (1.4/fy) governs for low-strength concrete (f'c ≤ 31.4 MPa for fy = 415 MPa). The second expression (√f'c / 4fy) becomes larger for higher-strength concrete and controls in those cases. Both must be evaluated because ρ_min must be sufficient to prevent a sudden brittle failure immediately after the concrete cracks — the cracking moment must not exceed the section's nominal moment capacity. Using only one expression may yield an unsafe (too low) minimum for a particular concrete strength.

Question Type

short_answer

Answer Structure

• Write both formulas with code citation [1 mark]
• Explain why 'max' is used — one expression governs depending on f'c [1 mark]

Scoring Breakdown

Common Mark Deductions

• Taking the smaller (minimum) of the two expressions — this is the most common mistake
• Citing ρ_min = 1.4/fy alone without acknowledging the second expression
• No explanation of why ρ_min is required (purpose mark lost)

Key Phrases To Include

• ρ_min = max(1.4/fy, √f'c/4fy)
• NSCP 2015 Section 409.6.1.1
• post-cracking brittle failure
• governs

Marks

5

Topic

T-Beam Analysis — Flange vs Web

Difficulty

hard

Template Id

T10

Examiner Tip

The first and most critical step for any T-beam problem is to compute a with bf and compare to tf. State the comparison explicitly — it is its own rubric mark.

Model Answer

Given: bf = 800 mm, tf = 100 mm, bw = 300 mm, d = 500 mm As = 2 500 mm², f'c = 28 MPa, fy = 415 MPa, β₁ = 0.85 Step 1 — Assume stress block in flange only (rectangular section of width bf): a = As·fy / (0.85·f'c·bf) a = (2 500 × 415) / (0.85 × 28 × 800) a = 1 037 500 / 18 960 a = 54.7 mm Step 2 — Check flange containment: a = 54.7 mm < tf = 100 mm ✓ The stress block lies entirely within the flange. Therefore, treat the section as a rectangular beam of width bf = 800 mm. Step 3 — Neutral axis and ductility: c = a / β₁ = 54.7 / 0.85 = 64.4 mm εt = 0.003 × (500 − 64.4) / 64.4 = 0.003 × 6.763 = 0.02029 > 0.005 ✓ Section is tension-controlled → φ = 0.90 Step 4 — Nominal moment: Mn = As·fy·(d − a/2) Mn = 2 500 × 415 × (500 − 27.35) Mn = 1 037 500 × 472.65 Mn = 490.4 × 10⁶ N·mm = 490.4 kN·m Step 5 — Design capacity: φMn = 0.90 × 490.4 = 441.4 kN·m Answer: a = 54.7 mm < tf = 100 mm → stress block in flange only; φMn = 441.4 kN·m.

Question Type

numerical

Answer Structure

• Compute a using full flange width bf [1 mark]
• Compare a to tf and state conclusion (rectangular or T-beam analysis) [1 mark]
• Ductility check with εt value and φ = 0.90 [1 mark]
• Mn computation [1 mark]
• φMn with correct units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using bw instead of bf to compute a — this gives the wrong a and a wrong conclusion
• Not comparing a to tf before choosing the analysis approach
• If a > tf: not accounting for the T-section web contribution (different analysis needed)

Key Phrases To Include

• a < tf
• treat as rectangular beam of width bf
• stress block within flange
• φ = 0.90
• tension-controlled

Marks

1

Topic

Doubly Reinforced Beams — When to Use

Difficulty

easy

Template Id

T11

Examiner Tip

The key trigger is 'Mu > φMn,max of singly reinforced section at ρ_max.' State this quantitatively even in a 1-mark answer.

Model Answer

A doubly reinforced beam (with both tension and compression steel) is used when the factored moment Mu exceeds the maximum tension-controlled moment capacity φMn,max of the singly reinforced section with the same b and d, or when section dimensions are architecturally fixed and cannot be increased.

Question Type

very_short_answer

Answer Structure

• One complete sentence: when Mu > φMn,max of the singly reinforced section OR when dimensions are fixed [1 mark]

Scoring Breakdown

Common Mark Deductions

• Vague answer such as 'when the beam is too weak' without quantifying the trigger
• Saying doubly reinforced is used for all beams

Key Phrases To Include

• Mu exceeds φMn,max
• fixed section dimensions
• compression steel
• ρ_max

Marks

2

Topic

T-Beam — Effective Flange Width

Difficulty

medium

Template Id

T12

Examiner Tip

Citing NSCP 2015 Section 406.3.2 (or ACI 318-19 Section 6.3.2) earns the code-awareness mark that separates boarders from passers.

Model Answer

The effective flange width bf is the portion of the slab that is assumed to act compositely with the beam web in resisting compression during bending. NSCP 2015 Section 406.3.2 (ACI 318-19 Section 6.3.2) limits bf based on span length, slab thickness, and beam spacing to account for shear lag — the phenomenon where stress in the slab decreases with distance from the web. Using bf rather than the full slab width gives a conservative estimate of the compression area actually engaged. As long as the equivalent stress block depth a ≤ tf (flange thickness), the T-beam is analyzed as a rectangular beam of width bf, simplifying computation.

Question Type

short_answer

Answer Structure

• Define bf and cite the code limit [1 mark]
• Explain why it is used — shear lag, conservative compression area, simplification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining bf as 'the full slab width' — incorrect
• Not mentioning shear lag as the physical reason for the limit

Key Phrases To Include

• effective flange width
• shear lag
• NSCP 2015 Section 406.3.2
• composite action
• compression flange

Marks

1

Topic

Whitney Stress Block — β₁

Difficulty

easy

Template Id

T13

Examiner Tip

Memorize β₁ = 0.85 − 0.05(f'c − 28)/7 for f'c > 28 MPa. Board exams frequently use f'c = 35, 40, or 42 MPa to test this reduction.

Model Answer

β₁ is the ratio of the equivalent rectangular stress-block depth a to the neutral-axis depth c (i.e., a = β₁c). Per NSCP 2015 Section 422.2.2.4.3 (ACI 318-19), β₁ = 0.85 for f'c ≤ 28 MPa and decreases by 0.05 for each 7 MPa above 28 MPa, with a minimum of 0.65. For f'c = 35 MPa: β₁ = 0.85 − 0.05×(35−28)/7 = 0.85 − 0.05 = 0.80.

Question Type

very_short_answer

Answer Structure

• Define β₁ relationship (a = β₁c) and state the value for 35 MPa = 0.80 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using β₁ = 0.85 for 35 MPa — wrong; must apply the reduction
• Not knowing the minimum limit of 0.65

Key Phrases To Include

• a = β₁c
• 0.85 for f'c ≤ 28 MPa
• decreases 0.05 per 7 MPa
• minimum 0.65
• β₁ = 0.80 for 35 MPa

Marks

5

Topic

Analysis — Singly Reinforced Beam with Reduced β₁

Difficulty

hard

Template Id

T14

Examiner Tip

When f'c > 28 MPa, compute β₁ first before anything else. Write it as the very first calculation in your solution to avoid forgetting it.

Model Answer

Given: b = 350 mm, d = 600 mm, As = 4 × π(28)²/4 = 4 × 615.75 = 2 463 mm² f'c = 35 MPa, fy = 415 MPa β₁ = 0.85 − 0.05(35−28)/7 = 0.85 − 0.05 = 0.80 Step 1 — Stress-block depth: a = As·fy / (0.85·f'c·b) a = (2 463 × 415) / (0.85 × 35 × 350) a = 1 022 145 / 10 412.5 a = 98.2 mm Step 2 — Neutral-axis depth: c = a / β₁ = 98.2 / 0.80 = 122.7 mm Step 3 — Net tensile strain (ductility check): εt = 0.003 × (d − c) / c εt = 0.003 × (600 − 122.7) / 122.7 εt = 0.003 × 3.889 = 0.01167 > 0.005 ✓ → φ = 0.90 Step 4 — Nominal moment: Mn = As·fy·(d − a/2) Mn = 2 463 × 415 × (600 − 49.1) Mn = 1 022 145 × 550.9 Mn = 563.1 × 10⁶ N·mm = 563.1 kN·m Step 5 — Design capacity: φMn = 0.90 × 563.1 = 506.8 kN·m Answer: φMn = 506.8 kN·m

Question Type

numerical

Answer Structure

• β₁ computed correctly for f'c = 35 MPa [0.5 mark]
• a computed = 98.2 mm [1 mark]
• c computed and εt check with conclusion φ = 0.90 [1.5 marks]
• Mn computed = 563.1 kN·m [1 mark]
• φMn = 506.8 kN·m with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using β₁ = 0.85 for f'c = 35 MPa instead of applying the reduction to 0.80
• Incorrect bar area — must use π(28)²/4 per bar, not πd (diameter formula error)
• Not showing the ductility check step

Key Phrases To Include

• β₁ = 0.80
• a = 98.2 mm
• εt = 0.01167
• tension-controlled
• φMn = 506.8 kN·m

Marks

2

Topic

Design Procedure — Coefficient of Resistance

Difficulty

easy

Template Id

T15

Examiner Tip

Always include φ in the Rn formula. Its absence is the surest sign of a common student error and costs the formula mark.

Model Answer

Rn is the required flexural resistance per unit area of the beam section (units: MPa or N/mm²). It condenses the factored moment demand Mu, the strength-reduction factor φ, and the section geometry (b and d) into a single parameter that can be directly mapped to a steel ratio ρ through the design equation. Rn = Mu / (φ · b · d²) Where Mu is in N·mm, b and d in mm, giving Rn in MPa. A higher Rn means the section is more heavily demanded and will require more steel.

Question Type

short_answer

Answer Structure

• Physical meaning: resistance per unit area of section; links Mu to ρ [1 mark]
• Formula written with variables defined and units stated [1 mark]

Scoring Breakdown

Common Mark Deductions

• Leaving out φ in the formula — one of the most common errors
• Using Mu in kN·m without converting, giving inconsistent units

Key Phrases To Include

• Rn = Mu/(φbd²)
• MPa
• coefficient of resistance
• maps to steel ratio ρ