CELE Reinforced & Prestressed Concrete — Reinforced Concrete ColumnsExam Answer Templates

Marks

1

Topic

Reinforcement Limits

Difficulty

easy

Template Id

T1

Examiner Tip

A one-mark VSA item expects a single complete statement. Do not pad it with unnecessary explanation — just define ρg and state both bounds.

Model Answer

Under NSCP 2015 (consistent with ACI 318), the longitudinal steel ratio ρg = Ast/Ag must satisfy 0.01 ≤ ρg ≤ 0.08.

Question Type

very_short_answer

Answer Structure

• Line 1: State both limits with the correct symbol in one concise sentence [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating only one limit (minimum or maximum) instead of both
• Confusing gross area Ag with core area Ach
• Writing 1% and 8% without showing the ratio expression ρg = Ast/Ag

Key Phrases To Include

• ρg = Ast/Ag
• 0.01
• 0.08
• NSCP 2015

Marks

2

Topic

Axial Capacity of Short Columns

Difficulty

easy

Template Id

T2

Examiner Tip

Examiners specifically look for the cap factor AND the φ factor written separately before they are combined. Show both multipliers clearly.

Model Answer

For a short tied column (φ = 0.65): φPn,max = 0.80 φ Po = 0.80(0.65) Po = 0.52 Po For a short spiral column (φ = 0.75): φPn,max = 0.85 φ Po = 0.85(0.75) Po = 0.6375 Po where Po = 0.85 f′c (Ag − Ast) + fy Ast is the pure-axial nominal strength.

Question Type

short_answer

Answer Structure

• Line 1: Write the tied column formula with correct cap (0.80) and φ (0.65) [1 mark]
• Line 2: Write the spiral column formula with correct cap (0.85) and φ (0.75), and define Po [1 mark]

Scoring Breakdown

Common Mark Deductions

• Swapping the cap factors (using 0.85 for tied and 0.80 for spiral)
• Using φ = 0.90 (flexure φ) instead of 0.65 or 0.75
• Omitting the definition of Po

Key Phrases To Include

• 0.80
• 0.65
• 0.85
• 0.75
• Po = 0.85f′c(Ag − Ast) + fyAst
• tied
• spiral

Marks

3

Topic

Axial Capacity of Short Columns

Difficulty

easy

Template Id

T3

Examiner Tip

Part (b) is the mark most students lose. The examiner expects the specific mechanism: the spiral confines the core even after the concrete cover spalls, preventing sudden brittle failure.

Model Answer

(a) Confinement: A tied column uses discrete rectangular or square closed ties spaced along the column height. A spiral column uses a continuous helical spiral winding around the core, providing uniform confinement pressure. (b) Ductility: Spiral columns are significantly more ductile. After the outer shell spalls, the spiral continues to confine the core, allowing the column to sustain load through large deformations without sudden failure. (c) φ factor (NSCP 2015 / ACI 318): For compression-controlled members, φ = 0.65 for tied columns and φ = 0.75 for spiral columns. The higher φ for spirals reflects their superior post-peak performance and ductility.

Question Type

short_answer

Answer Structure

• Part (a): Describe tie/spiral confinement in one to two sentences [1 mark]
• Part (b): Explain why spirals are more ductile — continued confinement after shell spalling [1 mark]
• Part (c): State correct φ values for both types and justify the difference [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not mentioning core confinement continuing after shell spalls — this is the key ductility argument
• Reversing φ values (0.75 for tied and 0.65 for spiral)
• Generic statements about spirals being 'stronger' without explaining the mechanism

Key Phrases To Include

• discrete ties
• continuous helical spiral
• core confinement
• shell spalls
• ductility
• φ = 0.65
• φ = 0.75

Marks

5

Topic

Axial Capacity of Short Columns

Difficulty

medium

Template Id

T4

Examiner Tip

Show every multiplication step separately. If you commit an arithmetic error but the formula and method are correct, you can still earn 3 out of 5 marks through the follow-through principle.

Model Answer

Given: b = h = 400 mm → Ag = 400 × 400 = 160,000 mm² Ast = 3927 mm² (8-25 mm ø bars) f′c = 28 MPa, fy = 415 MPa Column type: Tied → φ = 0.65, cap = 0.80 Step 1 — Net concrete area: Ag − Ast = 160,000 − 3,927 = 156,073 mm² Step 2 — Pure-axial nominal strength: Po = 0.85 f′c (Ag − Ast) + fy Ast Po = 0.85(28)(156,073) + 415(3,927) Po = 3,714,537 + 1,629,705 Po = 5,344,242 N Step 3 — Design axial capacity (tied, NSCP 2015 Sec. 422.4.2): φPn,max = 0.80 × φ × Po φPn,max = 0.80 × 0.65 × 5,344,242 φPn,max = 0.52 × 5,344,242 ∴ φPn,max = 2,779,006 N ≈ 2779 kN

Question Type

numerical

Answer Structure

• Block 1: List all given data with units [0.5 mark implicit — sets up solution]
• Step 1: Compute Ag and net area (Ag − Ast) [1 mark]
• Step 2: Apply Po formula with correct net area and compute [2 marks]
• Step 3: Apply cap factor 0.80 and φ = 0.65 for tied column and compute final answer [2 marks]

Scoring Breakdown

Common Mark Deductions

• Using Ag instead of (Ag − Ast) in the concrete term — loses 2 marks
• Using cap = 0.85 (spiral) instead of 0.80 (tied)
• Using φ = 0.90 (flexure) instead of 0.65
• No unit on final answer
• Arithmetic error in Po without showing intermediate steps

Key Phrases To Include

• Ag − Ast
• 0.85f′c
• 156,073 mm²
• Po = 5,344,242 N
• 0.80 × 0.65
• 2779 kN

Marks

2

Topic

Reinforcement Limits

Difficulty

easy

Template Id

T5

Examiner Tip

The mark for steel-ratio verification is specifically for the explicit inequality comparison and the verdict word 'acceptable' or 'OK'. Do not skip the inequality — it is worth a dedicated mark.

Model Answer

Given: Ast = 3,927 mm², Ag = 160,000 mm² Longitudinal steel ratio: ρg = Ast / Ag = 3,927 / 160,000 = 0.0245 Code check (NSCP 2015 / ACI 318-19): 0.01 ≤ ρg ≤ 0.08 0.01 ≤ 0.0245 ≤ 0.08 ✓ ∴ The steel ratio is within the allowable range. The reinforcement is acceptable.

Question Type

numerical

Answer Structure

• Line 1: Compute ρg = Ast/Ag with substitution [1 mark]
• Line 2: Show the code inequality check and state verdict ✓ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing the ratio but not writing the inequality check — loses 1 mark
• Using net area (Ag − Ast) in the denominator instead of gross area Ag
• No conclusive statement (merely stating the number without judging acceptability)

Key Phrases To Include

• ρg = Ast/Ag
• 0.0245
• 0.01 ≤ ρg ≤ 0.08
• within code limits

Marks

3

Topic

Axial Capacity of Short Columns

Difficulty

medium

Template Id

T6

Examiner Tip

When the problem says 'how much more', always compute both absolute (kN) and percentage differences. Questions that ask for comparison almost always reward the percentage — include both to be safe.

Model Answer

Given: Po = 5,344,242 N (computed previously) Column type: Spiral → φ = 0.75, cap = 0.85 Step 1 — Design axial capacity (spiral, NSCP 2015 Sec. 422.4.2): φPn,max = 0.85 × φ × Po φPn,max = 0.85 × 0.75 × 5,344,242 φPn,max = 0.6375 × 5,344,242 φPn,max = 3,406,954 N ≈ 3407 kN Step 2 — Increase over tied column: Tied: φPn,max = 2779 kN Spiral: φPn,max = 3407 kN Increase = 3407 − 2779 = 628 kN % Increase = (628 / 2779) × 100 ≈ 22.6% ∴ The spiral column carries approximately 23% more axial load than the equivalent tied column.

Question Type

numerical

Answer Structure

• Step 1: Identify spiral cap (0.85) and φ (0.75) — not the tied values [1 mark]
• Step 2: Apply formula and compute φPn,max = 3407 kN [1 mark]
• Step 3: Compute the difference and express as percentage [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using cap = 0.80 (tied value) for the spiral column
• Not showing the percentage computation — examiner expects the arithmetic
• Omitting the comparison entirely and only computing the spiral capacity

Key Phrases To Include

• 0.85
• 0.75
• 0.6375
• 3407 kN
• 23%
• ductility

Marks

5

Topic

Axial Capacity of Short Columns

Difficulty

medium

Template Id

T7

Examiner Tip

For circular columns, always write Ag = πD²/4 explicitly. Examiners see mistakes when students confuse radius and diameter. Write D = 500 mm, r = 250 mm, then compute πr² to double-check.

Model Answer

Given: D = 500 mm → Ag = π(500)²/4 = 196,350 mm² Bars: 8-28 mm ø → Ast = 8 × 615.75 = 4,926 mm² f′c = 30 MPa, fy = 415 MPa Column type: Spiral → φ = 0.75, cap = 0.85 Step 1 — Net concrete area: Ag − Ast = 196,350 − 4,926 = 191,424 mm² Step 2 — Check steel ratio: ρg = 4,926 / 196,350 = 0.0251 → 0.01 ≤ 0.0251 ≤ 0.08 ✓ Step 3 — Pure-axial nominal strength: Po = 0.85 f′c (Ag − Ast) + fy Ast Po = 0.85(30)(191,424) + 415(4,926) Po = 4,881,312 + 2,044,290 Po = 6,925,602 N Step 4 — Design axial capacity (NSCP 2015 Sec. 422.4.2): φPn,max = 0.85 × 0.75 × Po φPn,max = 0.6375 × 6,925,602 ∴ φPn,max = 4,415,071 N ≈ 4415 kN

Question Type

numerical

Answer Structure

• Block 1: Compute Ag for circular section (π D²/4) and total Ast [1 mark]
• Step 2: Steel ratio check (optional but earns mark if shown) [0.5 mark — see note]
• Step 3: Po formula stated and correctly computed with net area [2 marks]
• Step 4: Correct spiral cap (0.85) and φ (0.75) applied; final answer in kN [2 marks]

Scoring Breakdown

Common Mark Deductions

• Using D/2 = 250 in π r² incorrectly — arithmetic errors in Ag
• Forgetting to multiply bar area by 8 (number of bars) to get total Ast
• Using concrete term 0.85f′c × Ag instead of 0.85f′c × (Ag − Ast)
• Applying tied φ = 0.65 to the spiral column

Key Phrases To Include

• πD²/4
• 196,350 mm²
• 4926 mm²
• 0.85(0.75)
• 4415 kN

Marks

3

Topic

Reinforcement Limits

Difficulty

hard

Template Id

T8

Examiner Tip

The most-tested trap in spiral problems is the core area definition. 'Ach is the area measured to the OUTSIDE of the spiral' — write this sentence first before computing Dch. It sets up the correct geometry.

Model Answer

Given: Dc = 400 mm (outside column diameter) Clear cover = 40 mm, spiral wire diameter ds = 10 mm f′c = 28 MPa (assume), fyt = 415 MPa (assume, ≤ 700 MPa limit) Step 1 — Areas: Gross area: Ag = π(400)²/4 = 125,664 mm² Core diameter (to outside of spiral): Dch = 400 − 2(40) + 2(10) = 400 − 80 + 20 = 340 mm [subtract cover on both sides, add back one spiral wire on each side] Core area: Ach = π(340)²/4 = 90,792 mm² Step 2 — Minimum spiral ratio formula (NSCP 2015 / ACI 318-19 Sec. 410.7.6): ρs,min = 0.45 (Ag/Ach − 1)(f′c/fyt) ρs,min = 0.45 (125,664/90,792 − 1)(28/415) ρs,min = 0.45 (1.384 − 1)(0.0675) ρs,min = 0.45 (0.384)(0.0675) ρs,min = 0.45 × 0.02592 ∴ ρs,min = 0.01166 ≈ 1.17%

Question Type

numerical

Answer Structure

• Step 1: Compute Ag and correctly determine Dch (column diameter minus cover each side plus spiral wire each side) [1 mark]
• Step 2: Compute Ach from Dch [0.5 mark]
• Step 3: Apply ρs formula and substitute correctly [1 mark]
• Step 4: State final ρs,min value [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using the inside of the spiral as Dch (subtracting the full spiral diameter) — most common error
• Using Ag in place of Ach or vice versa in the denominator
• Forgetting the fyt cap of 700 MPa in the formula
• Not converting the final answer to percentage when percentage is requested

Key Phrases To Include

• Ach
• Dch = D − 2(cover) + 2(spiral wire)
• 0.45
• Ag/Ach − 1
• f′c/fyt
• 700 MPa cap

Marks

2

Topic

Axial–Moment Interaction

Difficulty

medium

Template Id

T9

Examiner Tip

The word 'simultaneously' is the mark-earning word for the balanced point definition. Both strain conditions (concrete crushing AND steel yielding) must occur at the same time — without 'simultaneously', the definition is incomplete.

Model Answer

The balanced point (Pb, Mb) on an axial-moment interaction diagram is the unique combination of axial load and bending moment at which the concrete compressive strain simultaneously reaches εc = 0.003 AND the tension steel exactly reaches its yield strain εy = fy/Es. Significance: It represents the maximum moment capacity of the section. Above the balanced point, failure is compression-controlled (φ = 0.65); below it, failure is tension-controlled (φ = 0.90 for flexure). The balanced point divides these two behavioural zones and is the point most dangerous for combined loading — the column can fail in a relatively brittle manner near this point without adequate ductility.

Question Type

short_answer

Answer Structure

• Sentence 1: Define balanced point precisely — two simultaneous strain conditions [1 mark]
• Sentence 2: State its significance — maximum moment, zone boundary, φ transition [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'maximum load' instead of 'maximum moment' — the balanced point is the moment peak, not load peak
• Not mentioning the simultaneous strain condition — simply defining it as 'where the column fails' earns no marks
• Confusing balanced point with the pure-axial point at the top of the diagram

Key Phrases To Include

• εc = 0.003
• εs = εy
• simultaneously
• maximum moment
• compression-controlled
• tension-controlled
• balanced point (Pb, Mb)

Marks

3

Topic

Axial–Moment Interaction

Difficulty

medium

Template Id

T10

Examiner Tip

Draw a quick sketch of the interaction diagram even if not asked — it helps you organize the three regions correctly and examiners often award bonus credit for a correctly labeled sketch.

Model Answer

An interaction diagram has three key regions: 1. Compression-controlled region (above the balanced point): Failure is initiated by concrete crushing before the tension steel yields. The section is relatively brittle. φ = 0.65 (tied) or φ = 0.75 (spiral). Most column design falls in this region. 2. Transition zone (between balanced point and tension-controlled limit): Both concrete compression and steel tension contribute to failure, with the steel strain between yield and 0.005. φ increases linearly from 0.65 (or 0.75) to 0.90 as the strain increases. 3. Tension-controlled region (below the transition zone): Failure begins with yielding of the tension steel (εt ≥ 0.005). This is the ductile, beam-like behaviour zone. φ = 0.90 (consistent with flexure-dominant members). Note: On the interaction diagram, any factored load point (Pu, Mu) must plot INSIDE the φ-reduced boundary to be safe.

Question Type

short_answer

Answer Structure

• Region 1: Compression-controlled — definition + φ value(s) [1 mark]
• Region 2: Transition zone — definition + φ interpolation rule [1 mark]
• Region 3: Tension-controlled — definition + φ = 0.90 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing only two regions (compression and tension) — omitting the transition zone loses 1 mark
• Not stating specific φ values for each region
• Confusing which region is above vs. below the balanced point on the diagram

Key Phrases To Include

• compression-controlled
• tension-controlled
• balanced point
• εt ≥ 0.005
• φ = 0.65
• φ = 0.75
• φ = 0.90
• linear interpolation

Marks

2

Topic

Slenderness

Difficulty

medium

Template Id

T11

Examiner Tip

Memorise both slenderness limits: 22 for unbraced/sway frames and 34 − 12(M1/M2) ≤ 40 for braced/non-sway frames. The board exam often tests which limit applies to a given scenario.

Model Answer

For a column in a braced (non-sway) frame, slenderness effects must be considered when: kℓu/r > 34 − 12(M1/M2) where: k = effective length factor (≤ 1.0 for braced frame) ℓu = unsupported column length r = radius of gyration of the column cross-section (r ≈ 0.30h for rectangular, 0.25D for circular) M1/M2 = ratio of smaller to larger end moments (positive if column bends in single curvature) The expression 34 − 12(M1/M2) shall not exceed 40. If kℓu/r ≤ the limit, slenderness effects may be neglected and the column is treated as short.

Question Type

short_answer

Answer Structure

• Line 1: Write the slenderness inequality kℓu/r > 34 − 12(M1/M2) [1 mark]
• Line 2: Define all terms and state the upper limit of 40 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing the unbraced (sway) frame limit of kℓu/r > 22 for a braced frame question
• Omitting the upper limit of 40
• Not defining the sign convention for M1/M2 (positive for single curvature)

Key Phrases To Include

• kℓu/r
• 34 − 12(M1/M2)
• not exceed 40
• braced frame
• non-sway
• r = 0.30h or 0.25D

Marks

1

Topic

Reinforcement Limits

Difficulty

easy

Template Id

T12

Examiner Tip

The number of minimum bars matches the geometry: triangular = 3, rectangular = 4, spiral/circular = 6. This is a simple recall item — zero partial marks if wrong.

Model Answer

(a) Rectangular tied column: minimum 4 longitudinal bars (b) Circular spiral column: minimum 6 longitudinal bars (c) Triangular tied column: minimum 3 longitudinal bars (one at each corner)

Question Type

very_short_answer

Answer Structure

• Line 1–3: State minimum bar count for each column shape [1 mark total — all three must be correct]

Scoring Breakdown

Common Mark Deductions

• Stating 4 bars for spiral columns instead of 6
• Not knowing the triangular column minimum (3 bars)

Key Phrases To Include

• 4 bars (rectangular tied)
• 6 bars (spiral)
• 3 bars (triangular)

Marks

5

Topic

Axial Capacity of Short Columns

Difficulty

medium

Template Id

T13

Examiner Tip

In a three-part numerical question, label each part (a), (b), (c) clearly and draw a horizontal line between them. Examiners grade each part separately; a clear layout prevents mark leakage.

Model Answer

Given: b = h = 450 mm → Ag = 450 × 450 = 202,500 mm² ρg = 0.02, f′c = 35 MPa, fy = 415 MPa Column type: Tied → φ = 0.65, cap = 0.80 (a) Steel area: Ast = ρg × Ag = 0.02 × 202,500 = 4,050 mm² (b) Pure-axial nominal strength: Ag − Ast = 202,500 − 4,050 = 198,450 mm² Po = 0.85 f′c (Ag − Ast) + fy Ast Po = 0.85(35)(198,450) + 415(4,050) Po = 5,913,413 + 1,680,750 Po = 7,594,163 N (c) Design axial capacity: φPn,max = 0.80 × 0.65 × Po φPn,max = 0.52 × 7,594,163 ∴ φPn,max = 3,948,965 N ≈ 3949 kN

Question Type

numerical

Answer Structure

• Part (a): Ast = ρg × Ag — straightforward multiplication [1 mark]
• Part (b): Net area computed, Po formula stated and evaluated correctly [2 marks]
• Part (c): Correct cap 0.80 and φ 0.65 applied; final answer in kN [2 marks]

Scoring Breakdown

Common Mark Deductions

• Computing Ast correctly but then using Ag (not Ag − Ast) in the Po formula
• Using f′c = 35 directly without the 0.85 modifier in the concrete term
• Rounding Po too early before applying the cap factor — carry full precision

Key Phrases To Include

• ρg × Ag
• 4050 mm²
• 198,450 mm²
• 7,594,163 N
• 0.52
• 3949 kN

Marks

3

Topic

Slenderness

Difficulty

hard

Template Id

T14

Examiner Tip

The 0.75 in the denominator is often omitted by students. It is a stiffness reduction factor for sustained loads — write it explicitly. Examiners treat it as a required term.

Model Answer

When a column is slender (kℓu/r exceeds the short-column limit), second-order P-δ effects amplify the moments along the column length. The moment-magnifier method (NSCP 2015 / ACI 318 Sec. 406) accounts for this by multiplying the first-order factored moment by a magnification factor δns: Mc = δns × M2 where M2 is the larger factored end moment, and the non-sway (braced) moment magnifier is: δns = Cm / (1 − Pu / (0.75 Pc)) ≥ 1.0 Cm = 0.6 − 0.4(M1/M2) is the equivalent moment factor (accounts for end-moment gradient) Pc = π²EI / (kℓu)² is the Euler critical load, with EI computed using ACI stiffness approximations 0.75 accounts for the possibility that all columns in a story may be loaded simultaneously The amplified moment Mc replaces M2 in the interaction diagram check.

Question Type

short_answer

Answer Structure

• Sentence 1–2: Explain why amplification is needed (P-δ effect) [1 mark]
• Equation block: Write Mc = δns × M2 and the δns formula [1 mark]
• Definition block: Define Cm, Pc, and the 0.75 factor [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the sway (Δo) magnifier formula for a braced frame (mixing up δns and δs)
• Omitting the ≥ 1.0 condition on δns
• Not defining Cm or writing only the formula without explanation

Key Phrases To Include

• P-δ effect
• moment magnifier
• δns
• Cm = 0.6 − 0.4(M1/M2)
• Pc = π²EI/(kℓu)²
• 0.75 Pc
• Mc = δns × M2

Marks

5

Topic

Axial–Moment Interaction

Difficulty

hard

Template Id

T15

Examiner Tip

In interaction diagram adequacy questions, always state the conclusion explicitly in words: 'The load point (Pu, Mu) lies INSIDE the φ-reduced envelope; therefore the section is adequate.' One clear sentence earns the verdict mark.

Model Answer

Given: b = h = 450 mm, Ag = 202,500 mm² 8-25 mm bars → Ast = 8 × 490.9 = 3927 mm² f′c = 28 MPa, fy = 415 MPa Tied: φ = 0.65, cap = 0.80 Pu = 2500 kN, Mu = 180 kN·m Balanced point: φPb ≈ 0.65(1500) = 975 kN, φMb ≈ 0.65(320) = 208 kN·m Step 1 — Compute φPn,max (pure axial): Ag − Ast = 202,500 − 3,927 = 198,573 mm² Po = 0.85(28)(198,573) + 415(3,927) Po = 4,722,137 + 1,629,705 = 6,351,842 N φPn,max = 0.80(0.65)(6,351,842) = 3,303,158 N = 3303 kN Step 2 — Identify zone of Pu = 2500 kN: Since Pu = 2500 kN > φPb = 975 kN, the design point lies in the COMPRESSION-CONTROLLED region (above the balanced point). Step 3 — Adequacy check: Pu = 2500 kN < φPn,max = 3303 kN ✓ (within axial capacity) Mu = 180 kN·m < φMb = 208 kN·m (moment is within the compression-side envelope) The load point (2500 kN, 180 kN·m) plots inside the φ-reduced interaction diagram. ∴ The column is ADEQUATE for the given loading combination.

Question Type

case_study

Answer Structure

• Step 1: Compute φPn,max correctly (with net area) [2 marks]
• Step 2: Apply φ = 0.65 to the given balanced point and identify the control zone [1 mark]
• Step 3: Compare Pu and Mu against diagram limits and state adequacy [2 marks]

Scoring Breakdown

Common Mark Deductions

• Not applying φ = 0.65 to the given balanced point before comparing
• Comparing Pu to Po (unfactored) instead of φPn,max
• No conclusive statement — computing numbers without a verdict loses 1 mark
• Treating the problem as pure axial without discussing the moment check

Key Phrases To Include

• compression-controlled
• φPn,max = 3303 kN
• φPb
• inside the interaction diagram
• adequate