CELE Reinforced & Prestressed Concrete — Prestressed ConcreteMisconception Buster
Avoid the most common Prestressed Concrete mistakes made by CELE reviewers. Each misconception here has been pulled from real CELE Reinforced & Prestressed Concrete questions where Professional Regulation Commission (PRC) — Board of Civil Engineering used it to separate strong reviewers from weak ones. Learn these before your next mock.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Reinforced & Prestressed Concrete subtest is marked as "Core" in the official pattern, and Prestressed Concrete appears in position 7th of 7 in the CELE Reinforced & Prestressed Concrete review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Prestressed Concrete - Misconception Buster
Prestressed concrete is one of the highest-yield topics in the PRC Civil Engineer Licensure Examination, yet it is also the chapter where many reviewees lose the most marks. The concepts seem straightforward on paper — introduce compression, apply load, check stresses — but the devil is in the details: sign conventions that flip between transfer and service, the difference between P_i and P_e, and the subtle direction of the eccentricity moment. Board exam problems are specifically designed to catch these exact errors. This guide identifies the 10 most dangerous misconceptions in prestressed concrete, explains why smart reviewees fall for them, and equips you with trap questions so you can test yourself before the exam does.
Summary
Mastering prestressed concrete for the PRC board exam requires getting five things right every single time: (1) SIGN CONVENTION — the eccentric prestress adds compression at the BOTTOM and relieves the TOP (f_bot = P/A + Pec/I, f_top = P/A − Pec/I); (2) TWO STAGES — always check BOTH transfer (using P_i, M_SW only, top fiber often critical) and service (using P_e, full M, bottom fiber often critical); (3) LOAD BALANCING DATUM — eccentricity e is measured from the CENTROIDAL AXIS, not the soffit or any other reference, and w_bal = 8Pe/L² leaves the beam in uniform axial compression P/A, not zero stress; (4) LOSSES — friction and anchorage seating are post-tensioning only; elastic shortening is smaller on average for sequentially stressed post-tensioned members; always apply loss percentage to the initial jacking force P_i; (5) CRACKING — prestressed members can be Class U, T, or C per ACI 318-19, and cracking is not automatically prohibited — it is managed through bonded reinforcement and cracked section deflection calculations. The exam is specifically designed to exploit these five areas. Every misconception in this guide corresponds to a question type that has appeared in Philippine engineering licensure examinations. Own these corrections and you own the prestressed concrete portion of the board exam.
Misconceptions
The eccentric prestress moment ALWAYS puts the top fiber in compression and the bottom in tension — just like a regular applied load.
Tags
- sign_convention
- formula_confusion
- critical_error
Topic
Service Stresses in a Prestressed Beam
Severity
critical
Exam Impact
A student with this misconception will REVERSE the sign of the Pec/I term, producing completely wrong fiber stresses. Both the top and bottom answers will be incorrect. In a 5-point board problem, this is a total loss.
The Reality
The eccentric prestress is a COMPRESSIVE force applied BELOW the centroid. This is the OPPOSITE of a downward applied load. The prestress eccentricity moment (Pe × e) causes ADDITIONAL compression at the bottom fiber and TENSION (or stress relief) at the top fiber. The formula is: f_bot = P/A + Pec/I (plus sign) and f_top = P/A - Pec/I (minus sign). An applied sagging moment does exactly the opposite: f_top = +Mc/I (adds compression at top) and f_bot = -Mc/I (relieves compression at bottom). The two moments partially cancel at the bottom — which is the entire design intent.
Trap Question
Question
A simply supported prestressed beam has an effective prestress P = 800 kN applied at e = 100 mm below the centroid. The cross-section is 250 mm × 500 mm (A = 125,000 mm², I = 2.604 × 10⁹ mm⁴, c = 250 mm). With NO applied load (M = 0), find the bottom fiber stress.
Explanation
With no applied load, the eccentric prestress alone should put the bottom fiber into EXTRA compression — that is the whole point of draping the tendon below the centroid! If your answer gives tension at the bottom with no applied load, you have the Pe sign wrong. The correct formula: f_bot = P/A + Pec_bot/I - Mc_bot/I.
Wrong Answer
f_bot = P/A - Pec/I = 800,000/125,000 - (800,000×100×250)/(2.604×10⁹) = 6.4 - 7.68 = -1.28 MPa (tension) [WRONG — student used minus for Pe at bottom]
Correct Answer
f_bot = P/A + Pec/I = 6.4 + 7.68 = +14.08 MPa (compression). The eccentric prestress ADDS compression at the bottom fiber.
Misconception Id
M1
Correct Vs Incorrect
Correct Approach
The eccentric prestress P at eccentricity e below the centroid applies a HOGGING moment (P×e) to the section. This hogging moment adds compression at the bottom and tension at the top. Therefore: f_top = P/A - Pec_top/I + Mc_top/I and f_bot = P/A + Pec_bot/I - Mc_bot/I. The Pe term and the M term have OPPOSITE signs at each fiber.
Incorrect Approach
Student writes f_top = P/A + Pec/I and f_bot = P/A - Pec/I, reasoning that the eccentric tendon below the centroid 'bends the beam like a load would.' This reverses the correct formula entirely.
Why Students Believe It
Students learn bending stress as f = Mc/I, where a sagging moment produces tension at the bottom and compression at the top. They apply the same intuition to the prestress eccentricity moment without thinking about the direction of the force. Since both the applied load and the tendon are 'below the centroid,' students assume both create the same sense of bending.
Use the EFFECTIVE prestress P_e for BOTH transfer and service stress checks.
Tags
- stage_confusion
- P_i_vs_Pe
- critical_error
Topic
Transfer vs Service Stage
Severity
critical
Exam Impact
Board exam problems often ask for stresses 'at transfer' specifically. A student using P_e instead of P_i will compute the wrong stress and select the wrong answer. This is a deliberate board exam trap — the numerical choices are designed to match both P_i and P_e results.
The Reality
There are TWO critical stages with DIFFERENT loads AND different prestress values: (1) AT TRANSFER — use P_i (initial prestress, before time-dependent losses) with only the self-weight moment M_SW. This checks whether the TOP fiber is over-tensioned or the BOTTOM is over-compressed before the full load arrives. (2) IN SERVICE — use P_e (effective prestress after all losses) with the full service moment M. Using P_e at transfer UNDERESTIMATES the prestress force and may miss a critical overstress. Using P_i in service OVERESTIMATES the prestress and gives unconservatively high compression — unsafe.
Trap Question
Question
A post-tensioned beam is jacked to P_i = 1400 kN. Total losses are 20%. At transfer, only the beam self-weight acts (M_SW = 60 kN·m). Which prestress value do you use to find the TOP fiber stress at transfer?
Explanation
Losses are NOT all instantaneous. Elastic shortening happens immediately; friction and anchorage seating happen at jacking. But creep, shrinkage, and relaxation accumulate over months and years. At the moment of transfer, the full P_i (minus elastic shortening for pre-tensioned) acts. Using P_e at transfer unconservatively underestimates the compressive stress at the bottom fiber and the potential tension at the top fiber.
Wrong Answer
P_e = 0.80 × 1400 = 1120 kN — use this value because losses have already been subtracted.
Correct Answer
P_i = 1400 kN. At the moment of transfer, the tendon has just been released (pre-tensioned) or just been anchored (post-tensioned) and time-dependent losses (creep, shrinkage, relaxation) have NOT yet occurred. Only elastic shortening loss may be subtracted for pre-tensioned members at transfer.
Misconception Id
M2
Correct Vs Incorrect
Correct Approach
At transfer: use P_i (e.g., 1200 kN) with M_SW only — check for tension at top (allowable = 0.25√f'ci per ACI 318 Section 24.5). In service: use P_e (e.g., 1000 kN) with full M — check for tension at bottom (allowable = 0.5√f'c for Class C or zero for Class U per ACI 318).
Incorrect Approach
Student uses P_e = 1000 kN for all stages. At transfer with M_SW = 80 kN·m, computes f_top = P_e/A - P_e×e×c/I + M_SW×c/I. This underestimates the prestress effect at transfer and may miss an overstress condition.
Why Students Believe It
Students learn that losses reduce the prestress to P_e and naturally use P_e as 'the' prestress force throughout all calculations. The concept of two separate stages — transfer and service — is often glossed over in quick reviews, so students apply the single value P_e everywhere.
Prestress losses only reduce the force — they have no effect on which fiber is critical.
Tags
- critical_fiber
- stage_confusion
- conceptual_gap
Topic
Transfer vs Service Stage
Severity
major
Exam Impact
A student who checks only one fiber or assumes the same fiber is always critical will miss the governing condition. Board exam questions routinely ask 'which fiber controls at transfer' — students who know the flip phenomenon answer immediately; others guess.
The Reality
At TRANSFER (high P_i, low M): The BOTTOM fiber gets extra compression from Pe and the TOP fiber may experience NET TENSION (since M_SW is small and the Pe term dominates at the top). The critical check at transfer is often TENSION at the top fiber. In SERVICE (lower P_e, high M): The BOTTOM fiber may go into tension from the large M overcoming the Pe compression. The critical check in service is often TENSION at the bottom fiber. The critical fiber FLIPS between transfer and service — this is a fundamental design principle of prestressed concrete.
Trap Question
Question
A prestressed beam with e = 180 mm below centroid has P_i = 1500 kN and M_SW = 50 kN·m. At transfer, which fiber is more likely to be in TENSION and must be checked against the allowable tension limit?
Explanation
At transfer, M is small (only self-weight) and the prestress eccentricity term dominates. The formula for the top fiber is f_top = P_i/A - P_i×e×c_top/I + M_SW×c_top/I. The minus sign of the Pe term means the top fiber is pushed toward tension. This is why post-tensioned beams sometimes need mild steel reinforcement at the top near the ends — to control transfer-stage tension cracking.
Wrong Answer
The BOTTOM fiber, because the live load always causes tension there in a simply supported beam.
Correct Answer
The TOP fiber. At transfer, the large P_i and eccentricity create a large hogging effect (-P_i×e×c_top/I) at the top fiber that the small M_SW cannot overcome, resulting in net tension at the top.
Misconception Id
M3
Correct Vs Incorrect
Correct Approach
At transfer (P_i high, M low): f_top = P_i/A - P_i×e×c_top/I + M_SW×c_top/I — this could be TENSILE (negative). Check against allowable tension at transfer = 0.25√f'ci (ACI 318-19 Table 24.5.3.2). At service (P_e lower, M high): f_bot = P_e/A + P_e×e×c_bot/I - M×c_bot/I — this could go TENSILE. Check against service tension limits.
Incorrect Approach
Student checks only the bottom fiber at both transfer and service because 'that's where tension from live load occurs in a regular beam.' Misses the potential top-fiber tension at transfer when the prestress eccentricity term dominates.
Why Students Believe It
Students think losses simply scale everything down proportionally, so the same fiber that was critical in service remains critical at transfer, just with a different number. They treat the two stages as identical in character, differing only in magnitude.
In the load-balancing formula w_bal = 8Pe/L², the eccentricity e is measured from the support to the tendon, not from the centroid to the tendon.
Tags
- formula_confusion
- datum_error
- load_balancing
Topic
Load Balancing
Severity
major
Exam Impact
Using the wrong reference for e gives a wrong w_bal value. If a student uses the tendon depth from the soffit (say 50 mm cover + 150 mm eccentricity = 200 mm depth from bottom, but only 150 mm from centroid), the computed balanced load is wrong and the net moment on the section is incorrect.
The Reality
In the load-balancing formula w_bal = 8Pe/L², the eccentricity e is the SAG of the parabolic tendon profile measured from the CENTROIDAL AXIS of the beam to the LOWEST POINT of the tendon at midspan. It is NOT measured from the soffit, the top fiber, or the support. If the tendon is concentric at the supports and drops to e = 150 mm below the centroid at midspan, that 150 mm is the sag. The upward equivalent load comes from the geometry of the parabolic profile relative to the centroid.
Trap Question
Question
A simply supported beam has a total depth of 700 mm. The centroid is at mid-depth (350 mm from soffit). The parabolic tendon has its lowest point at 100 mm from the soffit at midspan and is concentric at both supports. P = 1200 kN, L = 12 m. What is w_bal?
Explanation
The sag e in the load-balancing formula is always measured from the centroidal axis to the tendon's lowest point. Using the soffit as reference understates e and severely underestimates the balanced load. Always draw the section, locate the centroid, and measure e from there.
Wrong Answer
e = 100 mm (distance from soffit to tendon), w_bal = 8(1200)(0.1)/(12²) = 6.67 kN/m [WRONG datum]
Correct Answer
e = 350 − 100 = 250 mm = 0.25 m (below centroid), w_bal = 8(1200)(0.25)/(144) = 16.67 kN/m
Misconception Id
M4
Correct Vs Incorrect
Correct Approach
e = centroid distance from soffit − tendon distance from soffit = 300 − 80 = 220 mm. The tendon is 220 mm BELOW the centroid. Use e = 220 mm in w_bal = 8Pe/L².
Incorrect Approach
Beam is 600 mm deep; centroid at 300 mm from soffit; tendon is 80 mm from soffit at midspan. Student uses e = 600 - 80 = 520 mm (distance from top). WRONG.
Why Students Believe It
The word 'sag' is sometimes confused with the tendon profile depth from the soffit. Students sketch the beam and measure the tendon's distance from the soffit (or support level), not from the centroidal axis. They use the wrong reference datum for e.
Prestress losses of 15–25% are simply subtracted from the jacking stress — the percentage applies to the jacking STRESS, not force.
Tags
- unit_confusion
- loss_calculation
- formula_confusion
Topic
Prestress Losses
Severity
minor
Exam Impact
Board exam loss problems are usually straightforward percentage calculations, so this error loses only the final numeric mark. However, it creates cascading errors in subsequent parts of multi-part questions.
The Reality
The loss percentage applies to the JACKING PRESTRESS FORCE P_i (or equivalently to the initial jacking stress f_pj since the tendon area is constant). P_e = (1 − loss fraction) × P_i. If P_i = 1500 kN and losses = 20%, then P_e = 0.80 × 1500 = 1200 kN. In SI units, if f_pj = 1395 MPa and losses = 20%, then f_pe = 0.80 × 1395 = 1116 MPa. The key error is when a student computes losses as a fixed MPa value from one quantity and subtracts it from a different quantity (e.g., computing the loss in kN from the stress).
Trap Question
Question
A prestressing tendon with area A_ps = 1000 mm² is jacked to a stress of 1300 MPa. Total losses are 18%. What is the effective prestress FORCE P_e?
Explanation
Both approaches give the same answer when done correctly. The trap is mixing up where to apply the percentage. Always apply the loss fraction to the initial jacking stress to get f_pe, then multiply by A_ps for force. Or apply it directly to P_i. Never apply it to partial quantities.
Wrong Answer
P_i = 1300 × 1000 = 1,300,000 N = 1300 kN; Loss = 0.18 × 1300 = 234 MPa; P_e = (1300 − 234) × 1000 = 1,066,000 N = 1066 kN [correct method, but watch for students who mix units]
Correct Answer
f_pe = (1 − 0.18) × 1300 = 1066 MPa; P_e = f_pe × A_ps = 1066 × 1000 = 1,066,000 N = 1066 kN
Misconception Id
M5
Correct Vs Incorrect
Correct Approach
P_e = (1 − 0.20) × P_i = 0.80 × 1500 = 1200 kN. Equivalently, f_pe = 0.80 × 1395 = 1116 MPa. Both are consistent since A_ps is constant.
Incorrect Approach
P_i = 1500 kN, stress in tendon = 1395 MPa, 20% loss. Student computes: loss = 0.20 × 1395 = 279 MPa, then converts incorrectly to force, getting a different P_e than intended.
Why Students Believe It
Board exam loss problems often state 'losses are 20%' without specifying the reference. Students assume the percentage is universal and apply it to whatever quantity appears in the problem, sometimes applying it to the jacking stress f_pj instead of the jacking FORCE P_i or vice versa.
Pre-tensioning and post-tensioning only differ in the sequence of casting — their stress transfer mechanisms are the same.
Tags
- conceptual_gap
- loss_types
- system_distinction
Topic
Pre-tensioning vs Post-tensioning
Severity
major
Exam Impact
Board exam conceptual questions directly test this distinction. 'Which type uses end anchorages?' and 'Which type is more suitable for plant production?' are classic board items. Mixing up the mechanisms costs easy marks.
The Reality
The transfer mechanisms are FUNDAMENTALLY DIFFERENT: Pre-tensioning transfers prestress by BOND between strand and concrete along the DEVELOPMENT LENGTH — no mechanical anchor at the end is needed in service, but the transfer zone has a stress build-up region. Post-tensioning transfers prestress through MECHANICAL ANCHORAGES (wedge anchors, bearing plates) at the beam ENDS — the full prestress is delivered at the anchorage point. This difference affects: (1) where losses occur (friction + anchorage set only in post-tensioning); (2) required development length (pre-tensioning has transfer length ~50–60 strand diameters); (3) application — pre-tensioning is factory/plant (piles, hollowcore), post-tensioning is field/cast-in-place (bridges, transfer beams).
Trap Question
Question
Which of the following is a loss that occurs in POST-TENSIONING but NOT in pre-tensioning? A) Elastic shortening B) Creep C) Friction along the duct D) Steel relaxation
Explanation
Elastic shortening, creep, shrinkage, and relaxation occur in BOTH systems. Friction along the duct and anchorage seating (draw-in) losses are UNIQUE to post-tensioning. For pre-tensioned members, elastic shortening loss is calculated differently because all strands shorten simultaneously when released.
Wrong Answer
A) Elastic shortening — students think all losses are common to both systems.
Correct Answer
C) Friction along the duct. Friction loss (wobble friction and curvature friction) is unique to post-tensioning because the tendon slides inside a duct. Pre-tensioned strands are straight (or slightly deflected by hold-down points) and transfer by bond — there is no duct friction.
Misconception Id
M6
Correct Vs Incorrect
Correct Approach
Pre-tensioning: strands tensioned against external abutments BEFORE casting; after curing, strands released and force transferred by BOND (friction/adhesion/mechanical interlock) along the development length. Post-tensioning: strands in ducts tensioned AFTER curing using hydraulic jack bearing on the concrete; force transferred to concrete via MECHANICAL ANCHORAGE HARDWARE (wedge plates, barrel anchors).
Incorrect Approach
Student states: 'Both pre- and post-tensioning use end anchors to hold the strands.' This is correct only for post-tensioning. Pre-tensioned strands rely on bond — no end anchorage device is embedded in the concrete.
Why Students Believe It
Both methods ultimately result in a compressed concrete member with a tensioned tendon. Students focus on the end state (compressed beam) and overlook the fundamentally different mechanisms — bond vs mechanical anchorage — and the different types of losses each incurs.
Under the balanced load, the prestressed beam has ZERO stress everywhere — both axial and bending stresses are eliminated.
Tags
- conceptual_gap
- load_balancing
- stress_interpretation
Topic
Load Balancing
Severity
major
Exam Impact
A question asking 'What is the stress at the extreme fiber when the applied load equals w_bal?' will catch students who answer 'zero.' The correct answer is P/A (uniform compression). This is a classic board exam numerical trap.
The Reality
Under the balanced load (w = w_bal), the NET BENDING MOMENT from the combination of gravity load and the equivalent tendon load is ZERO — but the beam is still subjected to a UNIFORM AXIAL COMPRESSION of P/A throughout. This is the elegance of load balancing: the member acts as a pure compression member (like a column) under the balanced condition, with no bending. Any load BEYOND w_bal causes additional bending that must be superposed onto the P/A precompression. The beam never has ZERO stress as long as the prestress force is active.
Trap Question
Question
A prestressed beam has P_e = 900 kN, A = 180,000 mm², and a parabolic tendon. The applied uniform load exactly equals w_bal = 8P_e×e/L². What is the top fiber stress?
Explanation
Load balancing eliminates BENDING STRESS, not ALL stress. The tendon's axial component still pushes the beam together with force P_e, creating a uniform compression of P/A. This is why prestressed flat plates under balanced loading behave like a membrane in pure compression — flat and un-deflected but still stressed.
Wrong Answer
0 MPa — the load is perfectly balanced so all stresses cancel.
Correct Answer
f_top = P_e/A = 900,000/180,000 = 5.0 MPa (compression). There is still uniform axial precompression even though the bending moment is zero.
Misconception Id
M7
Correct Vs Incorrect
Correct Approach
Under balanced loading, the parabolic tendon exerts w_bal upward + P forces at the ends (horizontal). The net external effect is only the axial force P at the ends. The beam is in UNIFORM COMPRESSION = P/A. Both top and bottom fibers carry f = P/A (compression). No bending stress exists, but axial stress does.
Incorrect Approach
Student says: 'Since w_bal balances the load perfectly, all stresses are zero at the balanced condition.' Answers f_top = f_bot = 0 MPa.
Why Students Believe It
The load-balancing concept says the equivalent upward load w_bal cancels the applied gravity load, so 'everything balances' and students interpret this as no stress at all. They forget that canceling the NET MOMENT does not cancel the AXIAL precompression.
More prestress force P is always better — higher P gives more compression and thus more moment capacity.
Tags
- design_concept
- allowable_stress
- stage_confusion
Topic
Transfer vs Service Stage
Severity
minor
Exam Impact
This is more of a design-concept question than a computation trap, but boards test it in 'which of the following is TRUE' format. Reviewees who understand the feasibility zone concept answer these instantly.
The Reality
There is an UPPER BOUND to P that is set by the TRANSFER STAGE LIMIT on the BOTTOM FIBER. At transfer (maximum P, minimum M), the bottom fiber receives P/A + Pec/I — both terms are positive and can cause the bottom concrete to OVERSTRESS in compression before the design service load arrives. ACI 318-19 Section 24.5.3.1 limits the compressive stress at transfer to 0.60 f'ci. Exceeding this can crush the young concrete (typically f'ci = 0.75–0.80 f'c at transfer). Additionally, overpressing makes the BOTTOM fiber in compression and the TOP fiber in tension at transfer — creating cracking before the member even reaches service. Good design finds P within the feasible zone defined by ALL four stress limits (top and bottom at transfer; top and bottom in service).
Trap Question
Question
At transfer, the GOVERNING overstress condition for the bottom fiber of a simply supported prestressed beam is most likely:
Explanation
At transfer, before the service load arrives, the large P_i forces the bottom into high compression while the low M_SW provides little relief. The top fiber may go into tension. At service, the roles partially reverse — the bottom is relieved by the high M and may go into tension. Both stages must be checked.
Wrong Answer
Tension — because live load will cause tension at the bottom in service.
Correct Answer
Compression — at transfer, P_i is maximum and M is only the small self-weight moment. The bottom fiber has P_i/A + P_i×e×c_bot/I − M_SW×c_bot/I, which can exceed the allowable compression 0.60 f'ci if P is too large.
Misconception Id
M8
Correct Vs Incorrect
Correct Approach
P must satisfy ALL four inequalities: (1) f_top at transfer ≥ -allowable tension at transfer; (2) f_bot at transfer ≤ +allowable compression at transfer (0.60 f'ci); (3) f_top in service ≤ +allowable compression in service (0.45 f'c or 0.60 f'c); (4) f_bot in service ≥ -allowable tension in service (depends on class: U, T, or C).
Incorrect Approach
Student increases P to increase the moment capacity without checking the transfer-stage bottom fiber stress. The bottom becomes overstressed in compression at transfer, violating the allowable stress limit.
Why Students Believe It
Students correctly understand that prestress adds compression and reduces the likelihood of tension cracking. They extrapolate this to conclude: 'if some is good, more is better.' This linear thinking ignores the upper fiber overstress that develops at transfer.
Elastic shortening loss is the same for pre-tensioned and post-tensioned members with the same concrete strength and prestress level.
Tags
- elastic_shortening
- pre_vs_post
- loss_calculation
Topic
Prestress Losses
Severity
major
Exam Impact
Board problems on elastic shortening sometimes ask which system has MORE elastic shortening loss. The answer for sequential post-tensioning is LESS average loss than pre-tensioning — counterintuitive but correct. Wrong answers here reflect this misconception.
The Reality
For PRE-TENSIONED members: ALL strands are released simultaneously. The concrete shortens ONCE under the full prestress, and ALL strands lose the same elastic shortening. For POST-TENSIONED members with MULTIPLE tendons tensioned SEQUENTIALLY: the FIRST tendon tensioned suffers the MOST elastic shortening loss (all subsequent tendons shorten the concrete further), while the LAST tendon tensioned suffers ZERO elastic shortening loss (no further shortening of concrete occurs after it is anchored). The average elastic shortening loss for a post-tensioned member with n tendons is approximately 1/2 × the full pre-tensioned elastic shortening loss (or (n-1)/(2n) times, for all tendons stressed sequentially).
Trap Question
Question
A post-tensioned beam has 4 tendons stressed one at a time. Compared to a pre-tensioned beam with the same 4 strands and same prestress level, the average elastic shortening loss in the post-tensioned beam is approximately:
Explanation
When post-tensioned tendons are stressed one at a time, each successive tendon causes the concrete to shorten further, but this shortening is locked into the previously anchored tendons (as an additional loss). However, from the perspective of the last tendon stressed, the concrete has already shortened and no further shortening occurs. The average loss = (n-1)/(2n) × full elastic shortening loss.
Wrong Answer
The same — elastic shortening depends only on concrete stiffness and prestress level, not on how the tendons are stressed.
Correct Answer
About 3/8 (= (n-1)/2n = 3/8 for n=4) times the pre-tensioned elastic shortening — significantly LESS. Post-tensioning with sequential stressing averages out the elastic shortening loss.
Misconception Id
M9
Correct Vs Incorrect
Correct Approach
Pre-tensioned: Δf_ES = (E_ps/E_ci) × f_cgp (full loss for all strands). Post-tensioned (single tendon or all at once): Δf_ES ≈ 0 for the last tendon stressed. Post-tensioned (n tendons stressed sequentially): average Δf_ES = [(n−1)/(2n)] × (E_ps/E_ci) × f_cgp. For one tendon or simultaneous stressing: the concrete has already been placed and not yet loaded — first tendon: full loss; last tendon: zero loss.
Incorrect Approach
Student computes Δf_ES = (E_ps/E_ci) × f_cgp for both pre- and post-tensioned members using the same formula and the same factor, obtaining equal losses.
Why Students Believe It
Students learn that elastic shortening = P×L/(A×Ec) for any axially loaded member and apply it uniformly without considering the number of tendons tensioned and whether the concrete has already shortened when each subsequent tendon is stressed.
A prestressed concrete member never cracks — the whole point of prestressing is to eliminate cracking.
Tags
- ACI_318
- cracking_classes
- conceptual_gap
Topic
Service Stresses in a Prestressed Beam
Severity
minor
Exam Impact
Board conceptual questions test knowledge of the three ACI 318 prestressed member classes. A student who believes prestressed = never cracks will answer Class U for every member and miss Class T and Class C questions.
The Reality
ACI 318-19 classifies prestressed flexural members into THREE classes based on the maximum extreme fiber tensile stress ft in the precompressed tension zone: (1) CLASS U (Uncracked): ft ≤ 0.62√f'c — member behaves as uncracked; (2) CLASS T (Transition): 0.62√f'c < ft ≤ 1.0√f'c — intermediate behavior; (3) CLASS C (Cracked): ft > 1.0√f'c — member is treated as cracked; deflection must be computed using cracked section (Ie). Cracking is PERMITTED in Class C prestressed members — they just need sufficient bonded reinforcement to control crack widths. Not all prestressed members are designed to remain uncracked.
Trap Question
Question
A prestressed beam has an extreme fiber tensile stress of 0.85√f'c under service loads. According to ACI 318-19, this member is classified as:
Explanation
ACI 318-19 Section 24.5.2 defines the three classes based on extreme fiber tensile stress. Class U ≤ 0.62√f'c, Class T between 0.62 and 1.0√f'c, Class C > 1.0√f'c. Prestressed members CAN be in Class T or C if the design does not fully suppress tension — and this is acceptable if deflection and crack control requirements are met.
Wrong Answer
Class U — because it is prestressed, so it should be uncracked.
Correct Answer
Class T (Transition). The stress 0.85√f'c falls between 0.62√f'c and 1.0√f'c, placing it in the transition class. It is neither fully uncracked nor treated as fully cracked.
Misconception Id
M10
Correct Vs Incorrect
Correct Approach
Compute the net tensile stress at the bottom fiber under full service load: ft = |P_e/A + P_e×e×c_bot/I − M×c_bot/I|. Compare to 0.62√f'c and 1.0√f'c to determine Class U, T, or C. Class C members require cracked section Ie for deflection calculations.
Incorrect Approach
Student automatically classifies all prestressed beams as Class U and uses the uncracked section moment of inertia I_g for all deflection calculations.
Why Students Believe It
The fundamental principle of prestressing IS to keep concrete in compression under service loads, eliminating tensile cracking. Students take this to mean cracking can NEVER happen in a well-designed prestressed member.
In the stress formula f = P/A ± Pec/I ∓ Mc/I, the 'c' value is always the same for all three terms.
Tags
- asymmetric_section
- formula_confusion
- centroid_location
Topic
Service Stresses in a Prestressed Beam
Severity
major
Exam Impact
An asymmetric section problem with a single c value substituted throughout will yield wrong answers for at least one of the two fibers. This is a guaranteed wrong answer in a multi-part calculation problem.
The Reality
For a SYMMETRIC section (rectangle, I-section with equal flanges), c_top = c_bot = h/2, so one value suffices. But for ASYMMETRIC sections (T-beams, inverted T, unsymmetric I), c_top ≠ c_bot. You must use the correct c for each fiber: (1) For TOP fiber: c_top = distance from centroid to TOP fiber; (2) For BOTTOM fiber: c_bot = distance from centroid to BOTTOM fiber. Also note that c_top + c_bot = h (total depth) for any section. Board exam problems frequently use T-beam or I-sections specifically to test whether students substitute the correct c.
Trap Question
Question
A prestressed T-beam has centroid located 200 mm from the TOP fiber and 500 mm from the BOTTOM fiber (total depth 700 mm). P_e = 1000 kN, e = 200 mm below centroid, I = 1.2 × 10¹⁰ mm⁴, M = 300 kN·m. Compute the BOTTOM fiber stress.
Explanation
The section modulus S = I/c is fiber-specific. S_top = I/c_top and S_bot = I/c_bot. For symmetric sections these are equal; for asymmetric sections they differ. Always locate the centroid first, then compute c_top and c_bot separately. This is a classic board exam T-beam trap.
Wrong Answer
Using c = 700/2 = 350 mm: f_bot = 1,000,000/A + (1,000,000×200×350)/(1.2×10¹⁰) − (300×10⁶×350)/(1.2×10¹⁰) — WRONG c value for an asymmetric section.
Correct Answer
Use c_bot = 500 mm: f_bot = P_e/A + P_e×e×c_bot/I − M×c_bot/I. Each term at the bottom fiber uses c_bot = 500 mm, not the half-depth.
Misconception Id
M11
Correct Vs Incorrect
Correct Approach
f_top = P/A − Pec_top/I + Mc_top/I using c_top = 250 mm. f_bot = P/A + Pec_bot/I − Mc_bot/I using c_bot = 450 mm. Each term uses the distance from the centroid to the fiber being checked.
Incorrect Approach
T-beam: total depth 700 mm, flange at top, centroid at 250 mm from TOP (c_top = 250 mm, c_bot = 450 mm). Student uses c = 350 mm (half of 700) for both fibers. Both computed stresses are wrong.
Why Students Believe It
The formula looks symmetric and students memorize it as a single equation, assuming one value of c applies throughout. They use c = h/2 for a rectangular section for ALL terms without thinking about which fiber they are computing.
The load-balancing method gives the exact bending moment at every section — it is a complete analysis method, not an approximation.
Tags
- load_balancing
- formula_limits
- conceptual_gap
Topic
Load Balancing
Severity
minor
Exam Impact
Board exam questions on load balancing in the licensure exam almost always involve simply supported beams with parabolic tendons, so this misconception rarely directly costs marks. However, conceptual questions about the limitations of the method test this understanding.
The Reality
The formula w_bal = 8Pe/L² is EXACT only for a SIMPLY SUPPORTED beam with a PERFECTLY PARABOLIC tendon profile and CONCENTRIC tendon at both ends. For continuous beams, the tendon drape and the equivalent load distribution are more complex — the tendon exerts both distributed transverse loads AND concentrated moments at the supports (if the tendon changes slope). For non-parabolic profiles or variable P (due to friction losses in post-tensioning), the equivalent load must be computed from first principles (d²y/dx² of the tendon profile). Load balancing is a design TOOL for choosing P and e, not a universal analysis method.
Trap Question
Question
A parabolic tendon in a simply supported beam has sag e = 200 mm at midspan and is concentric at both supports. If P = 1000 kN and L = 10 m, what is the equivalent upward uniform load?
Explanation
The formula w_bal = 8Pe/L² is derived from the curvature of a parabola: d²y/dx² = 8e/L² for a parabola with sag e. For this specific geometry (simply supported, parabolic profile, concentric ends), it is exact. Students who add 'approximately' to the answer lose the conceptual point that the formula is exact for this case.
Wrong Answer
The load-balancing formula is only approximate, so the answer is approximately 8(1000)(0.2)/100 ≈ 16 kN/m.
Correct Answer
For a simply supported beam with a parabolic tendon concentric at both supports, w_bal = 8Pe/L² is EXACT: w_bal = 8(1000)(0.2)/(10²) = 16 kN/m exactly — not approximately.
Misconception Id
M12
Correct Vs Incorrect
Correct Approach
For simply supported beam with parabolic tendon: w_bal = 8Pe/L² exactly. For continuous beams: the equivalent load must be computed for each span considering the tendon profile geometry, concentrated moments at anchorage points, and variable P due to friction. The load-balancing concept still applies but the formula changes.
Incorrect Approach
Student applies w_bal = 8Pe/L² to a two-span continuous beam using the full span L without considering the different parabolic profiles in each span or the tendon slope change over the interior support.
Why Students Believe It
The load-balancing formula w_bal = 8Pe/L² is exact for a parabolic tendon in a simply supported beam. Students extend this exactness to all configurations — continuous beams, non-parabolic profiles, variable eccentricity — without realizing the limits of the simple formula.
Quick Self Check
The Pec/I term adds compression at the BOTTOM fiber (same direction as the axial P/A term) and RELIEVES compression at the TOP fiber (opposite direction). The eccentric prestress has a hogging moment effect: f_top = P/A − Pec_top/I, f_bot = P/A + Pec_bot/I.
Statement
In the service stress formula for a prestressed beam, the term Pec/I has the same sign (adds compression) at BOTH the top and bottom fibers.
At transfer, P_i is large and the moment M_SW is small. The Pec/I term dominates at the top fiber, creating net tension: f_top = P_i/A − P_i×e×c_top/I + M_SW×c_top/I. This can be tensile (negative) and must be checked against the allowable tension at transfer (ACI 318: 0.25√f'ci for pre-tensioned, 0.50√f'ci for post-tensioned without bonded reinforcement — noting current ACI 318-19 specifies 0.25√f'ci for normal-weight Class U).
Statement
At transfer, the critical allowable stress check is typically TENSION at the TOP fiber for a simply supported prestressed beam with the tendon below the centroid.
Under the balanced load, the bending stress is zero (bending moment from gravity plus equivalent tendon load = 0), but the beam still carries a uniform axial precompression of P_e/A. Both top and bottom fibers have stress = P_e/A (compression), not zero.
Statement
Under the balanced load condition (w = w_bal), the extreme fiber stresses in a prestressed beam are ZERO.
Friction loss (curvature friction and wobble friction) occurs because the post-tensioning tendon slides inside a duct and friction develops between the strand and the duct wall. Pre-tensioned strands are tensioned BEFORE casting against external abutments — there is no duct involved — so friction loss does not apply to pre-tensioning.
Statement
Friction loss along the duct is a prestress loss type that occurs in POST-TENSIONED members only.
For a T-beam (wide flange at top), the centroid is pulled upward — closer to the top fiber. Therefore c_top (centroid to top) < c_bot (centroid to bottom). The section modulus for the bottom fiber S_bot = I/c_bot is SMALLER than S_top = I/c_top, meaning the bottom fiber has higher stress for the same moment. This is why T-beams are efficient in hogging situations (negative moment) but the prestress helps manage the sagging service loads.
Statement
For a T-beam prestressed section with the centroid closer to the top flange, c_top is LESS than c_bot.
R = P_e/P_i = 1 − (loss fraction) = 1 − 0.22 = 0.78. The effective prestress is 78% of the initial jacking force. This ratio is used in simplified calculations when losses are given as a total percentage.
Statement
If total prestress losses are 22%, then the effectiveness ratio R = P_e/P_i = 0.78.
Class C is defined as members where the extreme fiber tensile stress EXCEEDS 1.0√f'c. This does NOT mean ACI 318 allows unlimited tension — Class C members must have sufficient bonded reinforcement to control cracking, and the cracked section transformed properties must be used for deflection calculations. Tensile stress beyond 1.0√f'c is PERMITTED in Class C if these conditions are met, but it is not unlimited.
Statement
ACI 318-19 allows tensile stresses in prestressed flexural members — Class C members may have tensile stress exceeding 1.0√f'c at the extreme fiber.
Unit consistency is critical. If P is in kN and e is in mm, you must convert e to m (divide by 1000) before using L in m. Formula: w_bal = 8×P(kN)×e(m)/[L(m)]². Using e in mm gives kN/m only if L² is in mm² — a dimensional inconsistency. Always use consistent SI units: P in kN, e in m, L in m → w_bal in kN/m. This is one of the most common arithmetic errors in board exam calculations.
Statement
In the load-balancing formula w_bal = 8Pe/L², using P in kN, e in mm, and L in m will give w_bal in kN/m without any unit conversion.
Ready to practise for the CELE 2026?
Super Tutor's AI review plan adapts to your weak areas and builds a weekly practice schedule around your target CELE exam date.