CELE Reinforced & Prestressed Concrete — Prestressed ConcreteExam Answer Templates

Marks

1

Topic

Introduction to Prestressed Concrete

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners check for two elements: (1) pre-applied compression and (2) purpose. Hit both in one sentence to guarantee the mark.

Model Answer

Prestressed concrete is a form of concrete in which a pre-applied internal compressive force (prestress) is introduced — through high-strength steel tendons — before service loads are applied, so that tensile stresses that would otherwise develop under load are reduced or eliminated.

Question Type

very_short_answer

Answer Structure

• One sentence: state 'pre-applied internal compressive force via high-strength tendons' AND purpose 'eliminate/reduce tensile stress under load' [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'reinforced concrete with steel strands' — does not capture the prestress concept.
• Omitting 'before service loads' — timing is fundamental to the definition.
• Confusing prestressing with ordinary reinforcement.

Key Phrases To Include

• pre-applied compressive force
• high-strength steel tendons
• before service loads
• reduces or eliminates tensile stress

Marks

2

Topic

Pre-tensioning vs Post-tensioning

Difficulty

easy

Template Id

T2

Examiner Tip

The key discriminator examiners look for is the force-transfer mechanism (bond vs. anchorage), not just 'before/after casting.'

Model Answer

Pre-tensioning: The tendons are tensioned against external abutments BEFORE concrete is cast. After the concrete reaches the required strength, the tendons are released and the prestress is transferred to the concrete by bond. Typical applications include factory-produced members such as piles, hollow-core slabs, and precast girders. Post-tensioning: The tendons are placed in ducts within the formwork and tensioned AFTER the concrete has cured. The prestress is transferred to the concrete through end anchorages, not bond. Typical applications include cast-in-place bridges, transfer beams, and post-tensioned flat slabs.

Question Type

short_answer

Answer Structure

• Line 1–2: Pre-tensioning — strand tensioned before casting, force transferred by bond [1 mark]
• Line 3–4: Post-tensioning — strand tensioned after casting, force transferred by end anchorage [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the definitions — saying post-tensioning uses bond or pre-tensioning uses anchorage.
• Writing only vague timing differences without mentioning the force-transfer mechanism.
• No mention of typical applications (optional but adds quality).

Key Phrases To Include

• before casting
• after curing
• transfer by bond
• end anchorage
• ducts / sheaths

Marks

2

Topic

Prestress Losses

Difficulty

easy

Template Id

T3

Examiner Tip

Grouping into 'immediate' and 'time-dependent' shows systematic thinking and may earn a bonus impression from examiners, even in a 2-mark question.

Model Answer

Prestress losses in a post-tensioned member include: Immediate losses: 1. Elastic shortening of concrete (ES) 2. Anchorage seating (draw-in) loss (ANC) 3. Friction loss along the duct (FR) Time-dependent losses: 4. Concrete creep (CR) 5. Concrete shrinkage (SH) 6. Steel relaxation (RE) (Any four of the above earns full marks.)

Question Type

short_answer

Answer Structure

• State at least 2 immediate losses with correct labels [1 mark]
• State at least 2 time-dependent losses with correct labels [1 mark]

Scoring Breakdown

Common Mark Deductions

• Listing the same loss twice with different names (e.g., 'friction' and 'wobble friction' counted as one).
• Inventing losses not in the standard classification (e.g., 'temperature loss').
• Confusing steel relaxation with concrete creep.

Key Phrases To Include

• elastic shortening
• anchorage seating
• friction
• creep
• shrinkage
• relaxation

Marks

2

Topic

Load Balancing

Difficulty

easy

Template Id

T4

Examiner Tip

Always state units alongside each variable definition — examiners deduct for dimensionally inconsistent answers.

Model Answer

For a simply supported beam with a parabolic tendon, the equivalent upward balanced load is: w_bal = 8Pe / L² Where: w_bal = equivalent uniformly distributed upward load balanced by the tendon (kN/m) P = effective prestress force in the tendon (kN) e = sag (eccentricity) of the tendon at midspan, measured from the centroidal axis (m) L = span length of the beam (m) Physical meaning: By choosing P and e, the designer can cancel ('balance') a target gravity load, leaving the beam under essentially uniform axial precompression P/A only.

Question Type

short_answer

Answer Structure

• State the formula correctly: w_bal = 8Pe/L² [1 mark]
• Define all four variables with units (w_bal, P, e, L) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing w_bal = 8Pe/L (missing the square on L).
• Not defining e clearly as midspan sag from the centroidal axis.
• Mixing units (e.g., e in mm, L in m) without conversion.

Key Phrases To Include

• w_bal = 8Pe/L²
• parabolic tendon
• sag / eccentricity at midspan
• equivalent upward load
• uniform axial precompression

Marks

2

Topic

Prestress Losses

Difficulty

easy

Template Id

T5

Examiner Tip

Show the R formula explicitly; it earns a method mark and signals understanding of the effectiveness concept.

Model Answer

Given: P_i = 1 400 kN Total losses = 18% = 0.18 Effectiveness ratio: R = 1 - 0.18 = 0.82 Effective prestress: P_e = R × P_i P_e = 0.82 × 1 400 P_e = 1 148 kN ∴ P_e = 1 148 kN and R = 0.82

Question Type

numerical

Answer Structure

• State R = 1 - loss fraction [0.5 mark]
• Compute P_e = R × P_i with correct arithmetic [1 mark]
• State final answer with units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using P_e = P_i - 18 (treating 18% as a fixed kN value instead of a percentage).
• Omitting units in the final answer.
• Reversing: computing P_e = P_i / (1 - losses).

Key Phrases To Include

• R = 1 - (loss fraction)
• P_e = R × P_i
• effectiveness ratio
• kN

Marks

3

Topic

Load Balancing

Difficulty

medium

Template Id

T6

Examiner Tip

Convert mm to m immediately when writing 'Given:' data — it prevents unit errors throughout the solution.

Model Answer

Given: P = 800 kN e = 120 mm = 0.120 m L = 8 m Formula (load-balancing concept): w_bal = 8Pe / L² Substituting: w_bal = [8 × 800 × 0.120] / (8)² w_bal = 768 / 64 w_bal = 12.0 kN/m ∴ The tendon exerts an equivalent upward distributed load of 12.0 kN/m on the beam. Under this load, the beam is under essentially uniform axial precompression P/A, with no net bending.

Question Type

numerical

Answer Structure

• State given data with unit conversion (e in m) [0.5 mark]
• Write formula w_bal = 8Pe/L² [1 mark]
• Substitute and compute correctly [1 mark]
• State final answer with unit and physical interpretation [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Leaving e = 120 mm without converting to 0.120 m — gives answer 12 000 kN/m (off by 1000).
• Using L = 8 instead of L² = 64.
• Not stating units in the final answer.

Key Phrases To Include

• w_bal = 8Pe/L²
• e = 0.120 m
• 12.0 kN/m
• upward equivalent load
• uniform axial precompression

Marks

3

Topic

Service Stresses

Difficulty

medium

Template Id

T7

Examiner Tip

An examiner's marking scheme almost always assigns one mark to transfer, one to service, and one to the explanation of the flip — structure your answer in exactly those three parts.

Model Answer

Stage 1 — Transfer Stage: Immediately after the tendons are released (pre-tensioning) or anchored (post-tensioning), the beam carries only its self-weight with the initial prestress P_i. The eccentric prestress compresses the bottom fiber heavily and relieves — or even creates tension at — the top fiber. Therefore, the critical check at transfer is the TOP fiber for excessive tension (which could crack the beam upward before service loads arrive), checked against the allowable tensile stress (ACI 318-19: 0.25√f'ci). Stage 2 — Service Stage: Under full service load, the applied sagging moment M_service creates tension in the bottom fiber. The effective prestress P_e (= P_i minus all losses) adds compression, but if M_service is large, the bottom fiber may go into tension. Therefore, the critical check in service is the BOTTOM fiber for tension, checked against the allowable tensile stress (ACI 318-19: 0.5√f'c for Class T). Conclusion: The critical fiber flips between stages because the eccentric prestress and the applied moment have opposite effects — prestress relieves bottom (and stresses top), while the service moment stresses bottom (and relieves top).

Question Type

short_answer

Answer Structure

• Describe transfer stage: P_i, self-weight only, top fiber critical for tension [1 mark]
• Describe service stage: P_e, full load, bottom fiber critical for tension [1 mark]
• Explain WHY the critical fiber flips — opposite action of prestress vs. service moment [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying the top fiber is always critical — ignores the service stage.
• Using P_e at transfer instead of P_i — the losses have not yet occurred.
• No explanation of the opposing actions of prestress and service moment.

Key Phrases To Include

• transfer stage
• P_i (initial prestress)
• top fiber tension at transfer
• service stage
• P_e (effective prestress)
• bottom fiber tension in service
• eccentric prestress
• sagging moment

Marks

3

Topic

Service Stresses

Difficulty

medium

Template Id

T8

Examiner Tip

Write the stress formula first, then substitute — this earns the method mark even if you make an arithmetic error later.

Model Answer

Given (compression positive): b = 250 mm, h = 500 mm A = 250 × 500 = 125 000 mm² I = (250)(500³)/12 = 2.604 × 10⁹ mm⁴ c_top = c_bot = 250 mm (symmetric section) P = 750 kN = 750 000 N e = 100 mm (below centroid) M = 0 Stress formula: f = P/A ± Pec/I Axial term: P/A = 750 000 / 125 000 = 6.0 MPa Eccentricity term: Pec/I = (750 000 × 100 × 250) / (2.604 × 10⁹) = 18 750 000 000 / 2 604 000 000 = 7.20 MPa Top fiber (eccentric prestress RELIEVES top): f_top = P/A − Pec/I = 6.0 − 7.20 = −1.20 MPa (tension) Bottom fiber (eccentric prestress ADDS compression at bottom): f_bot = P/A + Pec/I = 6.0 + 7.20 = 13.20 MPa (compression) ∴ f_top = −1.20 MPa (tension) ; f_bot = +13.20 MPa (compression)

Question Type

numerical

Answer Structure

• Compute section properties A and I [0.5 mark]
• Compute axial term P/A [0.5 mark]
• Compute eccentricity term Pec/I [1 mark]
• Apply formula correctly with correct signs for both fibers [1 mark]

Scoring Breakdown

Common Mark Deductions

• Wrong sign: adding Pec/I to top fiber and subtracting from bottom.
• Using c = 500 mm (full depth) instead of c = 250 mm (half depth to centroid).
• Forgetting to convert P from kN to N.

Key Phrases To Include

• f = P/A ± Pec/I
• compression positive
• f_top = P/A − Pec/I
• f_bot = P/A + Pec/I
• MPa

Marks

5

Topic

Service Stresses

Difficulty

hard

Template Id

T9

Examiner Tip

Examiners for 5-mark numericals use a step-by-step marking scheme; write each step as a clearly labeled calculation block so partial marks are clearly earned even if a later step has an error.

Model Answer

Step 1 — Section properties: b = 300 mm, h = 600 mm A = 300 × 600 = 180 000 mm² I = (300)(600³)/12 = 5.40 × 10⁹ mm⁴ c_top = c_bot = 300 mm Step 2 — Applied moment at midspan: w = 20 kN/m, L = 10 m M = wL²/8 = (20)(10²)/8 = 250 kN·m = 250 × 10⁶ N·mm Step 3 — Stress components (compression positive): P = 900 kN = 900 000 N, e = 150 mm Axial term: P/A = 900 000 / 180 000 = 5.0 MPa Eccentricity term: Pec/I = (900 000 × 150 × 300) / (5.40 × 10⁹) = 40 500 000 000 / 5 400 000 000 = 7.50 MPa Moment term: Mc/I = (250 × 10⁶ × 300) / (5.40 × 10⁹) = 75 000 000 000 / 5 400 000 000 = 13.89 MPa Step 4 — Fiber stresses: f_top = P/A − Pec/I + Mc/I = 5.0 − 7.50 + 13.89 = +11.39 MPa (compression) ✓ f_bot = P/A + Pec/I − Mc/I = 5.0 + 7.50 − 13.89 = −1.39 MPa (tension) ✗ Step 5 — Safety check: f_top = 11.39 MPa < 18 MPa ✓ (safe in compression) f_bot = −1.39 MPa < 0 MPa ✗ (tension developed — NOT safe for a no-tension allowable) ∴ The top fiber is safe; however, the bottom fiber develops 1.39 MPa tension, which exceeds the zero allowable tensile stress. The beam is NOT fully safe under the given loading — a higher prestress force, larger eccentricity, or reduced span loading is required.

Question Type

numerical

Answer Structure

• Step 1: Compute A, I, c correctly [1 mark]
• Step 2: Compute midspan moment M = wL²/8 [1 mark]
• Step 3: Compute all three stress terms (P/A, Pec/I, Mc/I) [1 mark]
• Step 4: Apply formula correctly for both fibers with proper signs [1 mark]
• Step 5: Check against allowable stresses and conclude [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to check BOTH fibers against the respective allowable (tension and compression).
• Using kN·m without converting to N·mm in the Mc/I calculation.
• Wrong sign on the Pec/I term (adding to top instead of subtracting).
• Not writing a conclusion sentence — the 'safety' question requires a verdict.

Key Phrases To Include

• f_top = P/A − Pec/I + Mc/I
• f_bot = P/A + Pec/I − Mc/I
• M = wL²/8
• compression positive
• allowable stress check
• tension in bottom fiber

Marks

3

Topic

Load Balancing

Difficulty

medium

Template Id

T10

Examiner Tip

Always sanity-check your P value — a prestress force should be in the range of hundreds to a few thousand kN for typical beams; a mega-kN answer signals a unit error.

Model Answer

Given: w_bal = 18 kN/m (load to be balanced) L = 12 m e = 200 mm = 0.200 m Formula (rearranged for P): w_bal = 8Pe / L² P = w_bal × L² / (8e) Substituting: P = (18 × 12²) / (8 × 0.200) P = (18 × 144) / 1.6 P = 2 592 / 1.6 P = 1 620 kN ∴ A tendon force of P = 1 620 kN is required to balance 18 kN/m on the 12 m span.

Question Type

numerical

Answer Structure

• State the load-balancing formula and rearrange for P [1 mark]
• Convert e to metres and substitute values [1 mark]
• Correct final answer with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using e = 200 mm (not converted) giving P = 1 620 000 kN — absurd answer.
• Forgetting to square L (using L = 12 instead of L² = 144).
• No unit on final answer.

Key Phrases To Include

• w_bal = 8Pe/L²
• P = w_bal L²/(8e)
• e = 0.200 m
• 1 620 kN

Marks

2

Topic

Prestress Losses

Difficulty

medium

Template Id

T11

Examiner Tip

Mentioning a code reference (ACI 318, NSCP 2015) in a 2-mark question is a quick way to demonstrate professional knowledge and may earn an impression mark.

Model Answer

Typical total prestress losses: • Post-tensioned members: approximately 15–20% of the jacking force. • Pre-tensioned members: approximately 18–25% of the jacking force. Pre-tensioned members typically experience higher total losses because: 1. Elastic shortening: In pre-tensioning, all strands shorten simultaneously when released, causing relatively large elastic shortening loss. In post-tensioning (multi-strand), each successive tendon shortens the concrete, but previously tensioned tendons absorb this loss differently. 2. Longer time-dependent exposure: Factory-produced pre-tensioned members are often steam-cured, affecting creep and shrinkage development. 3. No compensating re-stressing: Post-tensioned tendons can be re-tensioned to recover some losses; pre-tensioned cannot. Note: Exact losses must be calculated per ACI 318-19 Section 26.10 or NSCP 2015 for design.

Question Type

short_answer

Answer Structure

• State correct loss ranges for both types [1 mark]
• Give at least one valid reason why pre-tensioned losses are generally higher [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the ranges (saying post-tensioned loses more).
• Giving the range without any explanation — only earns partial credit.
• Quoting specific percentage without the word 'typical' or 'approximate.'

Key Phrases To Include

• 15–20% post-tensioned
• 18–25% pre-tensioned
• elastic shortening
• cannot be re-tensioned
• ACI 318 / NSCP 2015

Marks

1

Topic

Load Balancing

Difficulty

easy

Template Id

T12

Examiner Tip

The word 'uniform axial precompression' is the key examiner trigger phrase — one sentence with that phrase guarantees the mark.

Model Answer

When exactly the balanced load (w_bal = 8Pe/L²) acts, the upward force from the parabolic tendon exactly cancels the downward gravity load, producing zero bending moment in the concrete. The beam experiences only uniform axial precompression equal to P/A throughout its length — no flexural stresses, no deflection due to load.

Question Type

very_short_answer

Answer Structure

• State: zero bending moment AND uniform axial precompression P/A [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'zero stress' — the beam still has axial compression P/A.
• Not mentioning the uniform axial compression residual.

Key Phrases To Include

• zero bending moment
• uniform axial precompression
• P/A
• balanced load

Marks

5

Topic

Service Stresses & Prestress Losses

Difficulty

hard

Template Id

T13

Examiner Tip

In 5-mark questions, write 'Step 1, Step 2…' explicitly. Examiners mark by steps — a wrong number in Step 3 should not cost you Step 4 marks if your procedure is correct.

Model Answer

Step 1 — Effective prestress: R = 1 − 0.20 = 0.80 P_e = 0.80 × 1 500 = 1 200 kN = 1 200 000 N Step 2 — Stress terms: P_e/A = 1 200 000 / 200 000 = 6.0 MPa P_e × e × c / I = (1 200 000 × 180 × 350) / (8.0 × 10⁹) = 75 600 000 000 / 8 000 000 000 = 9.45 MPa M × c / I = (300 × 10⁶ × 350) / (8.0 × 10⁹) = 105 000 000 000 / 8 000 000 000 = 13.125 MPa Step 3 — Service fiber stresses (compression positive): f_top = P_e/A − P_e·e·c/I + M·c/I = 6.0 − 9.45 + 13.125 = +9.675 MPa (compression) f_bot = P_e/A + P_e·e·c/I − M·c/I = 6.0 + 9.45 − 13.125 = +2.325 MPa (compression) Step 4 — Allowable stress check: f_top = 9.675 MPa < 16.5 MPa ✓ (compression OK) f_bot = 2.325 MPa > 0 MPa (compression, no tension) ✓ Both fibers remain in compression; maximum compression (9.675 MPa) < 16.5 MPa. No tensile stress develops; tension limit of 1.5 MPa is not triggered. ∴ Both fibers are within allowable limits. The beam is safe at service.

Question Type

numerical

Answer Structure

• Step 1: Compute P_e = R × P_i = 1 200 kN [1 mark]
• Step 2: Compute all three stress terms correctly [1 mark]
• Step 3: Apply formula for both fibers with correct signs [1 mark]
• Step 4: Compare with allowable values and state verdict [1 mark]
• Overall clarity: defined steps, correct units throughout [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using P_i = 1 500 kN instead of P_e = 1 200 kN (most common 5-mark error).
• Incorrect moment arm (using h = 700 mm instead of c = 350 mm).
• Missing the conclusion sentence.
• Mixing N and kN in the same calculation.

Key Phrases To Include

• P_e = R × P_i
• f_top = P_e/A − P_e·e·c/I + Mc/I
• f_bot = P_e/A + P_e·e·c/I − Mc/I
• compression positive
• allowable stress check
• beam is safe

Marks

2

Topic

Introduction to Prestressed Concrete

Difficulty

easy

Template Id

T14

Examiner Tip

Each advantage should follow the pattern: Name → What it means → Why it matters. Three-part micro-structure earns full marks reliably.

Model Answer

1. Crack control and waterproofing: By keeping the concrete in permanent compression, prestressed concrete eliminates (or greatly reduces) tensile cracking under service loads. This makes it ideal for water-retaining structures, marine wharves, and bridge decks exposed to aggressive environments. 2. Longer spans with shallower sections: Because the full cross-section is effective (no cracked zone), prestressed members are structurally more efficient. They can span 1.5–2× farther than equivalent reinforced concrete sections of the same depth, reducing material volume and dead load.

Question Type

short_answer

Answer Structure

• Advantage 1: crack control — correct description with practical example [1 mark]
• Advantage 2: longer spans / shallower sections — correct description [1 mark]

Scoring Breakdown

Common Mark Deductions

• Listing advantages without describing them (e.g., 'crack control' alone without explanation).
• Giving the same advantage twice in different words.
• Mixing up advantages with properties of steel.

Key Phrases To Include

• permanent compression
• eliminates cracking
• longer spans
• full cross-section effective
• reduced dead load

Marks

3

Topic

Load Balancing & Prestress Losses

Difficulty

medium

Template Id

T15

Examiner Tip

Part (b) specifically tests whether you know the conceptual significance of the balanced load state — state 'no net bending, only axial precompression' before computing the number.

Model Answer

Part (a) — Effective prestress: R = 1 − 0.15 = 0.85 P_e = R × P_i = 0.85 × 1 000 = 850 kN Part (b) — Axial precompression at balanced load: Under the balanced load, the parabolic tendon forces cancel the applied gravity load exactly, leaving only uniform axial compression in the slab: f_axial = P_e / A = 850 000 N / 150 000 mm² = 5.67 MPa (compression) ∴ P_e = 850 kN ; f_axial = 5.67 MPa (uniform compression, no bending).

Question Type

numerical

Answer Structure

• Part (a): Correct R = 0.85 and P_e = 850 kN [1 mark]
• Part (b): State load-balancing condition: only P_e/A remains [1 mark]
• Part (b): Correct computation f_axial = 5.67 MPa with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using P_i = 1 000 kN for Part (b) instead of P_e = 850 kN.
• Not stating 'no bending' under balanced load — the question asks for the conceptual basis.
• Dividing by A in incorrect units (using mm² but P in kN without converting to N).

Key Phrases To Include

• R = 1 − loss fraction
• P_e = R × P_i
• balanced load condition
• f = P_e / A
• uniform axial compression
• no bending