Detailed ExplanationCELE · Reinforced & Prestressed ConcreteReal content

CELE Reinforced & Prestressed Concrete — Prestressed ConcreteDetailed Explanation

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Detailed explanations for CELE Reinforced & Prestressed Concrete — Prestressed Concrete. This page treats you like a serious reviewer: we unpack the concepts thoroughly, show worked examples of how Professional Regulation Commission (PRC) — Board of Civil Engineering frames Prestressed Concrete questions, and explain the underlying reasoning that gets you to the right answer every time.

Exam context

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

Prestressed Concrete - Detailed Explanation

Prestressed concrete is one of the most frequently tested topics in the PRC Civil Engineer Licensure Examination under Reinforced and Prestressed Concrete Design. Unlike ordinary reinforced concrete, which relies on steel to resist tension after cracking, prestressed concrete introduces a deliberate compressive force into the member before service loads arrive. This precompression either eliminates tensile stresses entirely or reduces them below the modulus of rupture, keeping the section uncracked, deflection-controlled, and structurally efficient. The result is a member that can span greater distances with shallower depth — the reason why prestressed concrete dominates Philippine long-span bridges (e.g., Estrella–Pantaleon Bridge) and precast infrastructure. This chapter covers the four pillars of board-exam PSC problems: (1) the two prestressing methods (pre- vs. post-tensioning), (2) service-stress computation using the combined-stress formula, (3) prestress losses, and (4) the load-balancing concept. Mastery of these four areas, together with careful sign-convention discipline, will allow you to solve any board-type PSC problem confidently and correctly.

Concepts

Principle of Prestressing and the Two Methods

Prestressing works by applying a compressive force P to the concrete section before external loads act. Because concrete is strong in compression but weak in tension (tensile strength ≈ 0.62√f'c MPa per ACI 318-19 §19.2.3), prestressing ensures that service loads merely reduce the existing compression rather than inducing net tension — preventing cracking. The net section stress at any fiber is the algebraic sum of the axial prestress term (P/A), the eccentricity-bending term (±Pec/I), and the applied-moment term (∓Mc/I). Two methods deliver this precompression: PRE-TENSIONING: High-strength strands (typically 1860 MPa low-relaxation, ASTM A416) are stretched between external abutments in a casting yard before concrete is placed. After the concrete achieves the required transfer strength (commonly 28 MPa or 0.75f'c, per ACI 318 §25.5), the strands are cut. The strand tries to shorten, but the bond with the surrounding concrete transfers the compressive force into the member. The transfer length (≈ 50–60 strand diameters) defines the zone where prestress builds up from zero to full value. Pre-tensioning is ideal for mass-produced precast elements: piles, prestressed concrete hollow-core slabs, I-girders, and double-tee beams. POST-TENSIONING: Here, ducts (galvanized metal or HDPE) are cast into the hardened member. After concrete reaches adequate strength, high-strength tendons are threaded through the ducts and stressed against the hardened concrete ends using hydraulic jacks, then anchored with wedge-type or plate-type anchors. The jacking force is recorded via a calibrated pressure gauge and a strand elongation check (elongation theory: ΔL = PL/ApsEps). Ducts may be grouted (bonded) for corrosion protection or left ungrouted (unbonded) for flat-slab systems. Post-tensioning is used in cast-in-place bridges, transfer beams, elevated slabs, and segmental bridge construction. Key distinction for board exams: pre-tensioning transfers by BOND; post-tensioning transfers by END ANCHORAGE.

Examples

Pre-tensioned members require adequate transfer length for the bond to develop fully. Within the transfer length, the prestress increases from zero at the cut end to the full value Pi. Board exams may ask you to identify the transfer length or recognize that the critical section is at Lt from the end, not at the face of support.

Scenario

A precast prestressed pile is pre-tensioned with 12 strands of 12.7 mm diameter (Aps = 98.7 mm² each). The strands are jacked to 0.75fpu = 0.75 × 1860 = 1395 MPa. Identify the method and the mechanism of force transfer.

Solution

Method: Pre-tensioning (strands stressed before casting). Jacking force per strand = 1395 × 98.7 = 137,717 N ≈ 137.7 kN. Total jacking force Pi = 12 × 137.7 = 1652 kN. Force transfer mechanism: BOND between the strand surface (seven-wire helical surface) and the surrounding hardened concrete along the transfer length Lt ≈ 50db = 50 × 12.7 = 635 mm ≈ 640 mm.

Even without curvature, long post-tensioned tendons lose prestress to wobble (unintended angular deviation). This is why long post-tensioned members are often stressed from both ends, and intermediate re-stressing anchors are provided in very long structures.

Scenario

A post-tensioned transfer beam has a 50 m tendon. The friction coefficient μ = 0.25 and wobble coefficient K = 0.0015/m. Compute the friction loss at the far end if the tendon is straight (angular change α = 0).

Solution

Friction loss formula: ΔPf = Pi(1 − e^(−μα − KL)). For a straight tendon, α = 0: ΔPf = Pi(1 − e^(−0.0015 × 50)) = Pi(1 − e^(−0.075)) = Pi(1 − 0.9277) = 0.0723Pi. So about 7.2% of the jacking force is lost to wobble friction alone over 50 m.

Applications

Pre-tensioned: precast piles (PCCP piles used extensively in Philippine port and bridge foundations), hollow-core planks, prestressed I-girders for short-span bridges.
Post-tensioned: cast-in-place bridge decks, transfer beams in high-rise buildings, post-tensioned flat plates in condominium construction, segmental box-girder bridges.
The Skyway Stage 3 in Metro Manila uses post-tensioned segmental box girders — a landmark Philippine PSC application.

Misconceptions

MISCONCEPTION: Pre-tensioning and post-tensioning differ only in timing — the stress in the concrete is the same. FACT: The mechanism of force transfer is fundamentally different (bond vs. anchorage), which affects the prestress profile, end-zone bursting stresses, and the type of losses that occur.
MISCONCEPTION: Any steel grade can be used for prestressing. FACT: Only high-strength steel (fpu ≥ 1725 MPa) is practical because losses of ~150–300 MPa must be a small fraction of the initial stress.
MISCONCEPTION: Post-tensioning always means unbonded. FACT: Post-tensioning can be bonded (grouted ducts) or unbonded; bonded is required in most structural bridge applications.

Related Concepts

Service stress computation
Prestress losses
Load balancing
Transfer length and development length
Anchorage zone design (strut-and-tie)

Common Exam Questions

Example

Which prestressing method uses end anchorage to transfer force to the concrete? Answer: Post-tensioning.

Approach

Identify the method (pre vs. post) based on sequence of operations: strands stressed before or after casting? Force transferred by bond or by anchorage?

Question Type

Identification/Conceptual

Example

8 strands, Aps = 98.7 mm² each, fjack = 0.75 × 1860 = 1395 MPa. Pi = 8 × 98.7 × 1395 = 1,101,492 N ≈ 1101.5 kN.

Approach

Given Aps and jacking stress (0.70–0.80 fpu per ACI 318 Table 26.10.1), compute Pi = Aps × fjack.

Question Type

Computation — Jacking Force

Key Points To Remember

Prestressing keeps concrete in compression (or low tension) under service loads — the fundamental purpose.
Pre-tensioning: strands stressed BEFORE casting; force transferred by BOND after release.
Post-tensioning: tendons stressed AFTER casting; force transferred by END ANCHORAGE.
High-strength steel (fpu = 1860 MPa) is required because total losses (~200 MPa) are a small fraction of the large initial stress; ordinary rebar at 415 MPa would lose all prestress to creep and shrinkage alone.
Transfer strength must be attained before releasing pre-tensioned strands (ACI 318 §26.10.2).
Bonded post-tensioning (grouted ducts) is required for most bridge applications in the Philippines per DPWH bridge design standards.

Service Stresses in a Prestressed Beam

The fundamental tool for PSC board problems is the combined stress equation. At any cross-section, three stress contributions are superposed. Using compression as POSITIVE (the standard PSC sign convention): For a tendon at eccentricity e BELOW the centroid (the normal, downward-draping case) under a sagging moment M: f_top = P/A − Pec_top/I + Mc_top/I f_bot = P/A + Pec_bot/I − Mc_bot/I where: P = effective prestress force (kN → convert to N for MPa results) A = gross cross-sectional area (mm²) e = eccentricity of tendon below centroid (mm), always taken as positive c_top = distance from centroid to top fiber (mm) c_bot = distance from centroid to bottom fiber (mm) I = moment of inertia of gross section about centroidal axis (mm⁴) M = applied bending moment (kN·m → N·mm for consistency) PHYSICAL INTERPRETATION OF EACH TERM: P/A — uniform axial compression from the prestress (always compressive, adds to both fibers) Pec/I — hogging moment of the eccentric prestress (compresses bottom, relieves top) Mc/I — sagging moment of applied load (compresses top, tensions bottom) DESIGN STAGES — TWO CRITICAL CHECKS: 1. AT TRANSFER (t = 0, initial prestress Pi, dead load only, no live load): Top fiber is critical for tension (the Pec/I effect dominates). Use Pi = jacking force minus immediate losses (elastic shortening, anchorage set). Check: f_top ≥ −allowable tension (ACI 318 Table 24.5.3.1 gives −0.25√f'ci for Class C or zero for uncracked). 2. IN SERVICE (t = ∞, effective prestress Pe after all losses, full dead + live load): Bottom fiber is critical for tension. Check: f_bot ≥ −allowable tension; top fiber for compression: f_top ≤ 0.45f'c (sustained) or 0.60f'c (total), per ACI 318 §24.5.3. SECTION MODULI SHORTCUT: For a rectangular section b × h: A = bh, I = bh³/12, c_top = c_bot = h/2 S_top = S_bot = bh²/6 Then: f = P/A ± Pe/S ∓ M/S

Examples

Notice that even with a relatively large applied moment (187.5 kN·m), both fibers stay in compression because of the eccentric prestress. If this were an ordinary RC beam, the bottom fiber would be cracked under 187.5 kN·m unless heavily reinforced. This illustrates the efficiency of prestressing.

Scenario

BOARD-TYPE PROBLEM: A simply supported rectangular prestressed beam has b = 300 mm, h = 600 mm, span L = 10 m. The effective prestress is Pe = 900 kN at eccentricity e = 150 mm below the centroid. The beam carries a superimposed uniform load of w = 15 kN/m (including self-weight). Compute the stress at the top and bottom fibers at midspan.

Solution

Step 1 — Section properties: A = 300 × 600 = 180,000 mm² I = 300 × 600³/12 = 5.40 × 10⁹ mm⁴ c_top = c_bot = 300 mm S = I/c = 5.40 × 10⁹/300 = 18 × 10⁶ mm³ Step 2 — Applied moment at midspan: M = wL²/8 = 15 × 10²/8 = 187.5 kN·m = 187.5 × 10⁶ N·mm Step 3 — Stress terms (P in N): P/A = 900,000/180,000 = 5.00 MPa Pe/S = 900,000 × 150/(18 × 10⁶) = 7.50 MPa M/S = 187.5 × 10⁶/(18 × 10⁶) = 10.42 MPa Step 4 — Fiber stresses: f_top = P/A − Pe/S + M/S = 5.00 − 7.50 + 10.42 = +7.92 MPa (compression) ✓ f_bot = P/A + Pe/S − M/S = 5.00 + 7.50 − 10.42 = +2.08 MPa (compression) ✓ Both fibers remain in compression — the beam is uncracked under service load.

At transfer, the large Pi and small M_sw combine to produce high compression at the bottom (12 MPa) and near-zero (or tension) at the top. Board problems often ask you to verify whether the top fiber exceeds the allowable tension at transfer, which per ACI 318 is limited to 0.25√f'ci (for normal-weight concrete, Class C) or 0.5√f'ci (with bonded reinforcement to control crack width).

Scenario

TRANSFER CHECK: For the same beam, at transfer Pi = 1080 kN (before losses), the only moment is the beam self-weight: w_sw = 0.30 × 0.60 × 24 = 4.32 kN/m, M_sw = 4.32 × 10²/8 = 54 kN·m. Find fiber stresses at transfer.

Solution

P/A = 1,080,000/180,000 = 6.00 MPa Pe/S = 1,080,000 × 150/(18 × 10⁶) = 9.00 MPa M/S = 54 × 10⁶/(18 × 10⁶) = 3.00 MPa f_top = 6.00 − 9.00 + 3.00 = 0.00 MPa → exactly zero tension f_bot = 6.00 + 9.00 − 3.00 = 12.00 MPa (compression) The top fiber is at zero stress — it has used up all the tensile capacity. If the eccentricity were slightly larger, the top would crack at transfer. This confirms that the transfer stage controls the top fiber, while service controls the bottom.

Applications

Design of prestressed bridge girders: compute stresses at midspan (maximum moment) and at the support (maximum shear; eccentricity may reduce to zero at the anchorage).
Checking adequacy of existing prestressed members when subjected to increased live loads (load rating).
Proportioning the tendon eccentricity and effective force to satisfy both transfer and service stress limits simultaneously (the 'magnel diagram' or stress-envelope approach).
Verification of flat-slab post-tensioned systems: uniform flat-plate stresses under balanced and unbalanced loads.

Misconceptions

MISCONCEPTION: The Pec/I term always increases top fiber stress. FACT: For a below-centroid tendon with compression-positive convention, Pec_top/I is SUBTRACTED from the top fiber (eccentricity creates a hogging moment that relieves top compression).
MISCONCEPTION: You only need to check service stresses. FACT: The TRANSFER stage (Pi, minimal moment) can be more critical for the top fiber tension than the service stage. Board problems that omit the transfer check are testing this exact pitfall.
MISCONCEPTION: M in the formula is the total load moment. FACT: At TRANSFER, M includes only loads present at that time (usually self-weight only). At SERVICE, M includes all superimposed dead + live loads plus self-weight.
MISCONCEPTION: For a symmetric section, c_top = c_bot always. FACT: True only for doubly-symmetric sections (rectangle, I-section). For T-sections and asymmetric sections, c_top ≠ c_bot, and you must use the correct distance for each fiber.

Related Concepts

Prestress losses (affects Pi vs. Pe)
Load balancing (an alternative approach to stress computation)
Cracking moment and ultimate flexural strength of PSC members
Section modulus and centroid location for T and I sections
Allowable stress design limits (ACI 318 §24.5)

Common Exam Questions

Example

A 250 × 500 mm PSC beam carries P = 600 kN at e = 100 mm below centroid and M = 80 kN·m. S = 250 × 500²/6 = 10.417 × 10⁶ mm³. A = 125,000 mm². f_top = 4.80 − 5.76 + 7.68 = +6.72 MPa; f_bot = 4.80 + 5.76 − 7.68 = +2.88 MPa.

Approach

Apply f = P/A ± Pe/S ∓ M/S directly. For non-rectangular sections, compute I and c separately. Always state the sign of each fiber stress (+ = compression, − = tension).

Question Type

Computation — Find fiber stresses given P, e, M, and section

Example

If f_bot ≥ 0 (no tension allowed): P/A + Pe/S ≥ M/S. Rearrange: P(1/A + e/S) ≥ M/S. Solve for P_min.

Approach

Set the critical fiber stress equal to the allowable (e.g., zero tension at bottom in service). Solve algebraically for P or e.

Question Type

Computation — Find P or e to limit a specific fiber stress

Example

At transfer, the top fiber stress of a simply supported beam with a below-centroid tendon is: f_top = Pi/A − Pie/S + M_sw/S. If M_sw is small, f_top can be tensile — this is the critical transfer condition.

Approach

Transfer: top fiber (Pec/I relieves top; only small self-weight moment restores compression to top). Service: bottom fiber (large moment reduces Pe's bottom compression).

Question Type

Identification — Which fiber is critical at transfer vs. service?

Key Points To Remember

Sign convention: compression POSITIVE throughout — be consistent, or you will get the wrong answer.
Eccentric prestress below centroid: ADDS compression to bottom (Pec_bot/I has + sign), RELIEVES compression from top (−Pec_top/I).
Applied sagging moment: ADDS compression to top (+Mc_top/I), REDUCES compression at bottom (−Mc_bot/I).
Two checks required: TRANSFER (Pi, minimal moment → top fiber critical for tension) and SERVICE (Pe, full moment → bottom fiber critical for tension).
For a rectangular section, use S = bh²/6 to simplify calculations.
Units must be consistent: if P is in N and dimensions in mm, stresses come out in MPa directly.
The critical fiber SHIFTS from top (at transfer) to bottom (in service) — this is a classic board-exam trap.

Prestress Losses

The prestressing force is not constant over time. It drops from the jacking force Pj to the initial prestress Pi (after immediate losses at transfer) and then further to the effective prestress Pe (after all time-dependent losses in service). The ratio R = Pe/Pi is the effectiveness ratio, typically 0.80–0.85. LOSSES ARE CLASSIFIED INTO TWO GROUPS: 1. IMMEDIATE (SHORT-TERM) LOSSES — occur at or shortly after stressing: a. ELASTIC SHORTENING (ΔPes): When the prestress is applied, the concrete shortens elastically, and the tendon shortens with it, losing some stress. For PRE-TENSIONING (all strands stressed simultaneously against external abutments and concrete cures while unstressed), the concrete shortens when strands are released: Δfes = n × fcpa, where n = Eps/Ec (modular ratio), fcpa = concrete stress at the centroid of the tendon due to Pi. For post-tensioning (each tendon stressed sequentially), only tendons already anchored lose stress when subsequent tendons are stressed. Average loss = (n−1)/(2n) × n × fcpa for n tendons stressed sequentially. b. ANCHORAGE SEATING LOSS (ΔPanch): In post-tensioning, wedge anchors seat 3–10 mm into the anchor plate when the jack is released. This shortening ΔL translates to a stress loss: Δfanch = Eps × ΔL / L_tendon. Typically 35–70 MPa. Affects more in short tendons. c. FRICTION LOSS (POST-TENSIONING ONLY): As the tendon is pulled through the duct, friction resists movement. Two components: — Curvature friction: proportional to the contact force (μ × normal force) — Wobble (unintended curvature): proportional to duct length (K × L) Combined formula: Px = Pj × e^(−μα − KL) ≈ Pj(1 − μα − KL) for small values. Where: μ = curvature friction coefficient (0.15–0.25 for grouted metal ducts), α = total angular change of tendon (rad), K = wobble coefficient (0.001–0.003/m). 2. TIME-DEPENDENT (LONG-TERM) LOSSES — develop over months and years: a. CREEP OF CONCRETE (ΔPcr): Under sustained compressive stress, concrete deforms (creeps). The strand shortens with the concrete, losing prestress. ΔPcr = n × Cu × fcs, where Cu = ultimate creep coefficient (≈ 2.0 for standard conditions) and fcs = concrete stress at tendon level due to sustained loads. b. SHRINKAGE OF CONCRETE (ΔPsh): Concrete shrinks as it cures. The tendon shortens with it. ΔPsh = Eps × εsh, where εsh ≈ 200–400 × 10⁻⁶ (ultimate shrinkage strain). Approximately ΔPsh = 0.0002 × Eps = 0.0002 × 195,000 = 39 MPa for low-relaxation strand. c. STEEL RELAXATION (ΔPrel): Under constant strain, the stress in high-strength steel decreases over time. Low-relaxation strand (the standard today) has ΔPrel ≈ 2–3% of initial prestress; stress-relieved strand ≈ 8–10%. AASHTO and ACI 318 provide relaxation tables. TYPICAL TOTAL LOSSES: — Post-tensioned: 15–20% of Pi — Pre-tensioned: 18–25% of Pi (no friction or seating loss, but elastic shortening is higher) EFFECTIVE PRESTRESS: Pe = Pi − ΔP_total = R × Pi

Examples

At service, the beam sees only 936 kN of effective prestress — not the original 1200 kN. The 264 kN difference is 'lost' to elastic shortening, creep, shrinkage, and relaxation. Designs that mistakenly use Pi instead of Pe in service will OVERESTIMATE the beneficial precompression and may underestimate tensile stresses at the bottom fiber.

Scenario

A pre-tensioned beam has Pi = 1200 kN. Total losses are estimated at 22%. Find Pe and the axial precompression stress on a section with A = 200,000 mm².

Solution

Pe = (1 − 0.22) × 1200 = 0.78 × 1200 = 936 kN Axial precompression = Pe/A = 936,000/200,000 = 4.68 MPa (compression)

The far end receives about 9.5% less force than the jacking end. In practice, engineers may stress the tendon from both ends (double-end stressing) to reduce the maximum friction loss to about half this value. Alternatively, a friction loss correction factor is applied in design. The board exam often gives you Pj and asks for Px at a specific point using this formula.

Scenario

FRICTION LOSS: A post-tensioned tendon is jacked to Pj = 1500 kN. The duct has μ = 0.20, K = 0.002/m, total length L = 20 m, and total angular change α = 0.30 rad. Find the prestress force at the far end.

Solution

Using the exponential formula: Px = Pj × e^(−μα − KL) = 1500 × e^(−0.20 × 0.30 − 0.002 × 20) = 1500 × e^(−0.06 − 0.04) = 1500 × e^(−0.10) = 1500 × 0.9048 = 1357.2 kN Friction loss = 1500 − 1357.2 = 142.8 kN (9.5% loss)

Seating loss is inversely proportional to tendon length. A 6 mm seating in a 15 m tendon causes 78 MPa loss — significant. In a 150 m tendon, the same 6 mm seating causes only 195,000 × 6/150,000 = 7.8 MPa loss — negligible. This is why seating loss is critical in short tendons but unimportant in long ones.

Scenario

ANCHORAGE SEATING LOSS: A post-tensioned tendon (Eps = 195,000 MPa, Aps = 150 mm²) has a length of 15 m. The anchor seating is 6 mm. Find the prestress force lost due to seating.

Solution

Stress loss due to seating: Δfanch = Eps × ΔL / L = 195,000 × 6/15,000 = 78 MPa Force loss = 78 × 150 = 11,700 N = 11.7 kN

Applications

Determining the design effective prestress Pe from the jacking force Pj for service-stress computations.
Specifying jacking force and over-tensioning to compensate for anticipated losses.
Selecting low-relaxation strand vs. stress-relieved strand to minimize relaxation losses.
Evaluating existing PSC structures: if Pe is lower than assumed due to higher-than-expected losses, the member may be cracked in service.
Post-tensioning shop drawings: compute expected elongation at jacking — a key quality-control check during construction.

Misconceptions

MISCONCEPTION: Friction loss affects pre-tensioned members. FACT: Pre-tensioned strands are external to the member during tensioning; they are released by cutting. There is NO friction loss or anchorage seating loss in pre-tensioning.
MISCONCEPTION: All losses happen instantaneously. FACT: Elastic shortening, anchorage seating, and friction are immediate; creep, shrinkage, and relaxation develop over months to years.
MISCONCEPTION: A higher initial jacking force always gives a higher final effective force. FACT: Higher jacking force increases all losses proportionally. The percentage loss is roughly constant; the absolute loss in kN increases with higher Pi.
MISCONCEPTION: Relaxation loss is the same for all strand types. FACT: Low-relaxation strand loses only 2–3% vs. 8–10% for stress-relieved strand — a critical distinction for modern design.

Related Concepts

Elastic shortening formula and concrete stress at tendon level
Creep coefficient and shrinkage strain (ACI 209, ACI 318)
Friction coefficient for different duct types (metal, HDPE, unsheathed)
Modular ratio n = Eps/Ec
Quality control: elongation measurement during post-tensioning

Common Exam Questions

Example

Pi = 1000 kN, losses = 18%. Pe = 0.82 × 1000 = 820 kN.

Approach

Pe = (1 − loss fraction) × Pi. Simple but must know whether Pi or Pj is the starting point.

Question Type

Computation — Effective prestress after given loss percentage

Example

Pj = 800 kN, μ = 0.25, α = 0.2 rad, K = 0.001/m, L = 30 m. Exponent = 0.25(0.2) + 0.001(30) = 0.05 + 0.03 = 0.08. Px = 800e^(−0.08) = 800 × 0.923 = 738.4 kN.

Approach

Px = Pj × e^(−μα − KL). If the exponent is small (<0.2), approximate as Px ≈ Pj(1 − μα − KL).

Question Type

Computation — Friction loss at a given point

Example

If n = 6.5 and fcpa = 8 MPa: Δfes = 6.5 × 8 = 52 MPa. For Aps = 800 mm²: ΔP = 52 × 800 = 41,600 N = 41.6 kN.

Approach

Δfes = n × fcpa = (Eps/Ec) × (Pi/A + Pi×e²/I). Then ΔP = Δfes × Aps.

Question Type

Computation — Elastic shortening loss (pre-tensioned)

Key Points To Remember

Immediate losses: elastic shortening, anchorage seating, friction (post-tensioning only).
Long-term losses: creep, shrinkage, relaxation (all three are time-dependent and occur in both methods).
Friction loss affects ONLY post-tensioned members; pre-tensioned members do not have friction or anchorage seating losses.
Elastic shortening affects both methods but is computed differently: full loss for pre-tensioning (all strands released at once), partial/average loss for post-tensioning (sequential stressing).
Typical R = Pe/Pi = 0.80–0.85 (use 0.83 if not given).
Total losses for post-tensioned ≈ 15–20%; for pre-tensioned ≈ 18–25%.
Shrinkage strain εsh ≈ 0.0002–0.0004 for normal-weight concrete in Philippine conditions (ACI 209 values).
Low-relaxation strand (the dominant type today) has relaxation loss ≈ 2–3% — much less than stress-relieved strand (8–10%).

Load Balancing

Load balancing, introduced by T.Y. Lin, is an elegant alternative approach to PSC design. A draped tendon exerts forces on the concrete not just at the anchorage ends but also along its length because of curvature. For a parabolic tendon with sag e (measured from chord connecting end anchorages to the lowest point of the parabola), the equivalent upward distributed load on the concrete is: w_bal = 8Pe/L² where P is the tendon force, e is the sag (in meters if P is in kN and L in meters, giving w_bal in kN/m), and L is the span. INTUITION: The tendon is like a bowstring pulling upward on the beam. The higher the sag e and the greater the force P, the stronger this upward push. At the anchorages (ends of a simply supported beam), the tendon exerts horizontal compressive forces and (if the tendon is not at the centroid at the ends) small vertical or moment reactions. UNDER THE BALANCED LOAD (w = w_bal): If the applied downward load exactly equals w_bal, the beam experiences NO bending — the tendon's upward forces perfectly cancel the gravity loads. The only stress in the beam is uniform axial compression P/A (plus friction and any end moments from non-centroidal anchorage). Deflection is theoretically zero under the balanced load. FOR THE UNBALANCED LOAD (w_net = w_total − w_bal): Any load in excess of w_bal causes bending in the section, computed by ordinary flexural theory. The precompressed section (P/A acts uniformly) then resists this net moment. Net bottom fiber tension = P/A − M_net/S ≥ −allowable. DESIGN STEPS USING LOAD BALANCING: 1. Choose what fraction of the dead load (or total gravity load) to balance: w_bal = 80–100% of DL is common. 2. From w_bal = 8Pe/L², with e and L known, solve for the required P: P = w_bal × L²/(8e). 3. Compute net unbalanced moment: M_net = (w_total − w_bal)L²/8. 4. Check fiber stresses: f = P/A ± M_net × c/I. LIMITATION: The formula w_bal = 8Pe/L² applies ONLY to a parabolic tendon in a simply supported beam. For continuous spans, each parabola segment over each span generates its own equivalent load, and the end eccentricities (at interior supports) add concentrated moments. Board exams typically use the simple-span formula.

Examples

The load-balancing approach cleanly separates the problem: the balanced portion causes only P/A (no bending), and the unbalanced portion is treated as an ordinary bending problem on a precompressed section. This is far more intuitive than superposing three stress terms directly — but both methods give the same answer.

Scenario

BOARD-TYPE PROBLEM: A post-tensioned simply supported beam spans L = 12 m. The parabolic tendon has sag e = 200 mm and effective prestress Pe = 1050 kN. The beam carries a total uniform load w = 22 kN/m. (a) Find the balanced load. (b) Find the net unbalanced moment. (c) Find fiber stresses if A = 200,000 mm² and S = 33.33 × 10⁶ mm³.

Solution

(a) Balanced load: w_bal = 8Pe/L² = 8 × 1050 × 0.20/(12²) = 1680/144 = 11.67 kN/m (b) Net unbalanced moment: w_net = 22 − 11.67 = 10.33 kN/m M_net = w_net × L²/8 = 10.33 × 144/8 = 185.9 kN·m (c) Axial precompression: f_axial = Pe/A = 1,050,000/200,000 = 5.25 MPa Net moment bending stress: M_net/S = 185.9 × 10⁶/(33.33 × 10⁶) = 5.58 MPa Top fiber: f_top = P/A + M_net/S = 5.25 + 5.58 = 10.83 MPa (compression) Bottom fiber: f_bot = P/A − M_net/S = 5.25 − 5.58 = −0.33 MPa (TENSION) The bottom fiber has 0.33 MPa tension — small, but must be checked against the allowable tension (e.g., 0.5√f'c ≈ 0.5√35 = 2.96 MPa for f'c = 35 MPa). Acceptable.

This type of problem asks you to DESIGN the tendon force rather than CHECK stresses. The load-balancing method gives a direct one-step solution: once the balanced load and sag are fixed, the required P follows immediately. Note that 'per meter width' is the unit for slab design — you would then determine the number of tendons required per meter from the force per strand.

Scenario

DESIGN PROBLEM: A simply supported post-tensioned slab with L = 8 m and e = 100 mm must balance 80% of a total gravity load of 12 kN/m. Find the required effective tendon force P.

Solution

w_bal = 0.80 × 12 = 9.6 kN/m From w_bal = 8Pe/L²: P = w_bal × L²/(8e) = 9.6 × 8²/(8 × 0.10) = 9.6 × 64/0.80 = 614.4/0.80 = 768 kN/m width So the tendon force per meter width of slab must be 768 kN.

Applications

Post-tensioned flat-plate slabs in high-rise buildings: balance dead load + 50% of live load to minimize deflections and control cracking.
Prestressed bridge girders: choose w_bal ≈ full dead load so the structure shows no net deflection under DL; only LL causes deflection.
Continuous prestressed beams: the equivalent loads from the tendon (upward UDL over spans, downward concentrated forces at interior supports due to reversed curvature) are applied as external loads and analyzed using ordinary structural analysis methods.
Preliminary design: a rapid way to select P and e before doing the full fiber-stress check.

Misconceptions

MISCONCEPTION: The formula w_bal = 8Pe/L² works for any tendon profile. FACT: It applies ONLY to a parabolic (second-degree curve) tendon profile. For a harped (straight segments) tendon, the equivalent load is concentrated forces at the harping points, not a UDL.
MISCONCEPTION: Under the balanced load, the beam has zero stress. FACT: Under the balanced load, the beam has UNIFORM AXIAL COMPRESSION P/A — not zero stress. Only bending stress is zero.
MISCONCEPTION: Load balancing and the fiber-stress formula give different results. FACT: Both are exact methods and give identical answers. Load balancing is simply a more intuitive way to organize the computation.
MISCONCEPTION: The sag e is the distance from the beam bottom to the tendon at midspan. FACT: The sag e is the distance from the CHORD (the straight line connecting the two end anchor points of the tendon) to the lowest point of the tendon. If the anchors are at the centroid, e = eccentricity at midspan. If the anchors are above the centroid, e < eccentricity at midspan.

Related Concepts

Equivalent load concept in structural analysis
Service stress formula (same result, different computation path)
Continuous prestressed beams — secondary moments from indeterminate prestress
Deflection control in prestressed members
Parabolic tendon geometry

Common Exam Questions

Example

P = 900 kN, e = 150 mm = 0.15 m, L = 10 m. w_bal = 8 × 900 × 0.15/100 = 1080/100 = 10.8 kN/m.

Approach

Direct substitution: w_bal = 8Pe/L². Convert e to meters and P to kN.

Question Type

Computation — Find balanced load given P, e, L

Example

w_bal = 14 kN/m, L = 12 m, e = 250 mm = 0.25 m. P = 14 × 144/(8 × 0.25) = 2016/2 = 1008 kN.

Approach

Rearrange: P = w_bal × L²/(8e).

Question Type

Computation — Find P to balance a given load

Example

See the worked example above (L = 12 m, P = 1050 kN, e = 200 mm, w = 22 kN/m).

Approach

Step 1: Compute w_bal. Step 2: w_net = w_total − w_bal. Step 3: M_net = w_net L²/8. Step 4: f = P/A ± M_net/S.

Question Type

Computation — Find fiber stresses using load-balancing approach

Key Points To Remember

w_bal = 8Pe/L² — applicable only to a PARABOLIC tendon in a SIMPLY SUPPORTED beam.
Under the balanced load, the beam is in UNIFORM AXIAL COMPRESSION only: f = P/A, no bending.
Sag e is measured from the chord (line connecting the two end anchor points) to the lowest point of the tendon parabola — NOT from the beam top or bottom.
If the tendon anchors are AT the centroid at both ends, then e = distance from centroid to tendon low point. If anchors are off-center, adjust accordingly.
For design: choose w_bal, then solve for P = w_bal L²/(8e). Alternatively, choose P and solve for required e.
Units: if P in kN, e in m, L in m → w_bal in kN/m. If P in N, e in mm, L in mm → w_bal in N/mm = kN/m (consistent).
Partially balanced: if w_bal < w_total, the difference w_net causes ordinary bending in the precompressed section.

Practice Problems

Key check: all terms computed in MPa (N/mm²). The bottom fiber has only 0.851 MPa compression — a relatively small margin. If the load were increased by about 7% more, the bottom would crack. This shows that the beam is well-designed but not excessively over-prestressed. Note: if f'c = 35 MPa, the allowable service compression = 0.45 × 35 = 15.75 MPa (sustained), and 0.60 × 35 = 21 MPa (total). The top fiber at 8.13 MPa is well within both limits.

Problem

PROBLEM 1 (Service Stresses — Rectangular Section): A simply supported prestressed concrete beam has a cross-section of 350 mm × 700 mm and a span of 12 m. The effective prestress is Pe = 1100 kN at an eccentricity e = 200 mm below the centroidal axis. The beam carries a superimposed uniform load of w = 18 kN/m (including self-weight). Compute: (a) the section modulus S, (b) the top fiber stress at midspan, (c) the bottom fiber stress at midspan, and (d) state whether each fiber is in tension or compression.

Solution

(a) Section properties: A = 350 × 700 = 245,000 mm² S = 350 × 700²/6 = 350 × 490,000/6 = 28.583 × 10⁶ mm³ (b)+(c) Applied moment at midspan: M = wL²/8 = 18 × 12²/8 = 18 × 144/8 = 324 kN·m = 324 × 10⁶ N·mm Stress terms: P/A = 1,100,000/245,000 = 4.490 MPa Pe/S = 1,100,000 × 200/(28.583 × 10⁶) = 220 × 10⁶/28.583 × 10⁶ = 7.697 MPa M/S = 324 × 10⁶/28.583 × 10⁶ = 11.336 MPa (b) Top fiber: f_top = P/A − Pe/S + M/S = 4.490 − 7.697 + 11.336 = +8.129 MPa → COMPRESSION (c) Bottom fiber: f_bot = P/A + Pe/S − M/S = 4.490 + 7.697 − 11.336 = +0.851 MPa → COMPRESSION (d) Both fibers are in compression — the member is uncracked. The prestress is sufficient to keep the bottom fiber in compression despite the large applied moment.

A classic two-part design problem: first find Pe from the load-balance requirement, then work backward through losses to find Pi. Board problems often stop at part (a), but the complete design requires part (b) to specify the actual jacking/transfer force. Note: the jacking force Pj would be even higher than Pi because Pi already accounts for elastic shortening and anchorage seating (immediate losses), while Pj is the raw force applied by the jack.

Problem

PROBLEM 2 (Load Balancing — Find P): A parabolic tendon in a simply supported post-tensioned beam (L = 14 m, sag e = 250 mm) must balance a uniform load of 16 kN/m. After losses of 18%, find: (a) the required effective prestress Pe, (b) the required jacking force Pi.

Solution

(a) From the load-balance equation: w_bal = 8Pe/L² Pe = w_bal × L²/(8e) = 16 × 14²/(8 × 0.25) = 16 × 196/2.0 = 3136/2 = 1568 kN (b) Losses = 18%, so Pe = (1 − 0.18)Pi = 0.82Pi: Pi = Pe/0.82 = 1568/0.82 = 1912.2 kN ≈ 1912 kN This is the force that must exist in the tendon immediately after transfer (before time-dependent losses reduce it to Pe).

This problem illustrates the critical nature of the transfer check. The top fiber at transfer is in tension (−1.375 MPa) because the large eccentricity term (10.208 MPa upward at top) overwhelms both the axial compression (5.833 MPa) and the self-weight moment relief (3.000 MPa). The allowable is exceeded — a common board-exam scenario testing whether examinees recognize that transfer governs, not service.

Problem

PROBLEM 3 (Transfer Check): A pre-tensioned beam (b = 300 mm, h = 600 mm) has Pi = 1050 kN at e = 175 mm at transfer. The only load at transfer is self-weight (unit weight of concrete = 24 kN/m³, span L = 10 m). Check whether the top fiber stress at midspan exceeds the allowable tension of 0.25√f'ci where f'ci = 28 MPa.

Solution

Section properties: A = 300 × 600 = 180,000 mm² S = 300 × 600²/6 = 18 × 10⁶ mm³ Self-weight: w_sw = 0.30 × 0.60 × 24 = 4.32 kN/m M_sw = 4.32 × 10²/8 = 54 kN·m = 54 × 10⁶ N·mm Stress terms: Pi/A = 1,050,000/180,000 = 5.833 MPa Pie/S = 1,050,000 × 175/(18 × 10⁶) = 183,750,000/18,000,000 = 10.208 MPa M_sw/S = 54 × 10⁶/18 × 10⁶ = 3.000 MPa Top fiber at transfer: f_top = Pi/A − Pie/S + M_sw/S = 5.833 − 10.208 + 3.000 = −1.375 MPa (TENSION) Allowable tension at transfer: f_t,allow = 0.25√28 = 0.25 × 5.292 = 1.323 MPa Checking: |f_top| = 1.375 MPa > 1.323 MPa → OVERSTRESS — the top fiber EXCEEDS the allowable tension at transfer by 0.052 MPa. Conclusion: The design is slightly overstressed at transfer. Corrective options: (1) reduce e slightly, (2) add bonded mild steel in the top zone to provide crack control (which raises the allowable to 0.5√f'ci per ACI 318), or (3) delay release until f'ci is higher.

Friction loss reduces the force at the far end (1176.2 kN vs. 1300 kN at jacking end). Seating loss reduces the force at the jacking end (by 62.4 kN). After both immediate losses, the force range along the tendon is from 1237.6 kN (at jacking end, after seating) to 1176.2 kN (at far end, after friction). Long-term creep, shrinkage, and relaxation will further reduce both values. This is why engineers average the friction and seating losses or use a sophisticated calculation at multiple points along the tendon.

Problem

PROBLEM 4 (Friction and Seating Loss — Combined): A post-tensioned tendon (Aps = 200 mm², Eps = 195,000 MPa) is stressed from one end with Pj = 1300 kN. The duct is 25 m long with μ = 0.20, K = 0.002/m, and total angular change α = 0.25 rad. The anchor seating is 8 mm. Find: (a) the force at the far end after friction loss, (b) the seating loss in stress at the jacking end, and (c) the effective jacking stress after seating loss.

Solution

(a) Friction loss to far end: Px = Pj × e^(−μα − KL) = 1300 × e^(−0.20 × 0.25 − 0.002 × 25) = 1300 × e^(−0.05 − 0.05) = 1300 × e^(−0.10) = 1300 × 0.9048 = 1176.2 kN (b) Seating loss at jacking end: Δfanch = Eps × δ/L = 195,000 × 8/25,000 = 62.4 MPa ΔP_anch = 62.4 × 200 = 12,480 N = 12.48 kN (c) Effective jacking stress after seating: Initial jacking stress = Pj/Aps = 1,300,000/200 = 6500 MPa [NOTE: This exceeds fpu, so let's reframe: the jacking force Pj = 1300 kN is given.] Stress after seating at jacking end: Pj,eff = Pj − ΔP_anch = 1300 − 12.48 = 1287.5 kN fjack,eff = 1,287,500/200 = 6437.5 MPa... [Recheck: If Aps = 200 mm² and Pj = 1300 kN, then fjack = 1,300,000/200 = 6500 MPa — this is unrealistic for a single tendon. The problem intends Aps = 1000 mm² (10 strands of 100 mm² each) for typical values. Using Aps = 1000 mm²:] fjack = 1,300,000/1000 = 1300 MPa (realistic for 0.70fpu × 1860 = 1302 MPa ✓) ΔP_anch = 195,000 × 8/25,000 × 1000 = 62.4 × 1000 = 62,400 N = 62.4 kN Pj,eff at jacking end after seating = 1300 − 62.4 = 1237.6 kN Effective stress at jacking end = 1,237,600/1000 = 1237.6 MPa

This comprehensive problem demonstrates both the direct stress formula and the load-balancing approach, confirming they yield identical results. The bottom fiber is in tension (2.26 MPa) but within the allowable 3.16 MPa — the member is a Class C (ACI 318 classification) prestressed beam, meaning it is designed to permit limited tension but remain below the cracking stress. Note that 'Class U' (uncracked) would require f_bot ≥ 0 — this beam would not qualify as Class U.

Problem

PROBLEM 5 (Combined — Complete Service Check): A rectangular PSC beam (b = 400 mm, h = 800 mm) simply supported over L = 15 m has a parabolic tendon with sag e = 300 mm and Pe = 1400 kN (after all losses). The total uniform service load is w = 25 kN/m. If f'c = 40 MPa, check whether the service stresses comply with ACI 318 allowable limits (compression ≤ 0.45f'c sustained; tension ≤ 0.5√f'c for Class C).

Solution

Section properties: A = 400 × 800 = 320,000 mm² S = 400 × 800²/6 = 42.667 × 10⁶ mm³ Midspan moment: M = 25 × 15²/8 = 703.125 kN·m = 703.125 × 10⁶ N·mm Stress terms: P/A = 1,400,000/320,000 = 4.375 MPa Pe/S = 1,400,000 × 300/(42.667 × 10⁶) = 420 × 10⁶/42.667 × 10⁶ = 9.844 MPa M/S = 703.125 × 10⁶/42.667 × 10⁶ = 16.481 MPa Top fiber: f_top = 4.375 − 9.844 + 16.481 = +11.012 MPa (compression) Allowable compression = 0.45 × 40 = 18.0 MPa → 11.012 < 18.0 ✓ Bottom fiber: f_bot = 4.375 + 9.844 − 16.481 = −2.262 MPa (TENSION) Allowable tension = 0.5√40 = 0.5 × 6.325 = 3.162 MPa → |−2.262| < 3.162 ✓ Alternative check using load balancing: w_bal = 8 × 1400 × 0.30/15² = 3360/225 = 14.93 kN/m w_net = 25 − 14.93 = 10.07 kN/m M_net = 10.07 × 225/8 = 283.4 kN·m M_net/S = 283.4 × 10⁶/42.667 × 10⁶ = 6.643 MPa f_top = 4.375 + 6.643 = +11.018 MPa ✓ (same result, difference due to rounding) f_bot = 4.375 − 6.643 = −2.268 MPa ✓ (same result) CONCLUSION: Both fiber stresses comply with ACI 318 service stress limits. The member is adequate in service.

Exam Preparation Tips

MASTER THE SIGN CONVENTION FIRST: Board exam PSC problems hinge entirely on getting the signs of the three stress terms correct. Write out the formula and label each term BEFORE substituting numbers: f = P/A (always +) ± Pe/S (+ for bottom, − for top with below-centroid tendon) ∓ M/S (+ for top, − for bottom under sagging moment). Practice until this is automatic.
ALWAYS DO TWO STRESS CHECKS — TRANSFER AND SERVICE: Many board problems specify Pi and Pe separately and provide both self-weight and superimposed loads. At transfer: use Pi and M_sw only. At service: use Pe and M_total. The critical fiber is the top at transfer (tension risk) and the bottom in service (tension risk). Skipping one check is the most common exam mistake.
MEMORIZE w_bal = 8Pe/L² FOR PARABOLIC TENDON: This formula appears in virtually every PSC board problem set. Know it forward (find w_bal given P, e, L) and backward (find P given w_bal, e, L or find e given w_bal, P, L). Units trap: ensure P in kN, e in m, L in m for w in kN/m.
LOSS SEQUENCE — IMMEDIATE THEN LONG-TERM: For board problems, classify losses before computing: (a) Pre-tensioned: immediate = elastic shortening; long-term = creep + shrinkage + relaxation. No friction, no seating. (b) Post-tensioned: immediate = friction + seating + elastic shortening; long-term = creep + shrinkage + relaxation.
TYPICAL TOTAL LOSS VALUES TO MEMORIZE: Post-tensioned ≈ 15–20%; Pre-tensioned ≈ 18–25%. If not given, assume R = Pe/Pi = 0.82 (18% loss) for a conservative mid-range estimate in multiple-choice problems.
FRICTION FORMULA DRILL: Px = Pj × e^(−μα − KL). For small exponents (<0.2), use the linear approximation: Px ≈ Pj(1 − μα − KL). Always check if the exponent justifies the approximation — if μα + KL > 0.2, use the exact exponential.
SECTION MODULUS SHORTCUT FOR RECTANGULAR SECTIONS: S = bh²/6. For a 300 × 600 mm beam, S = 300 × 360,000/6 = 18 × 10⁶ mm³. Compute this immediately upon seeing a rectangular section — it simplifies all stress calculations.
UNITS DISCIPLINE: The most common arithmetic error is mixing units. Adopt a single consistent system: P in N (multiply kN × 1000), e in mm, L in mm, I in mm⁴, M in N·mm (multiply kN·m × 10⁶). This gives stresses in N/mm² = MPa directly.
KNOW THE ACI 318 ALLOWABLE STRESS LIMITS: At transfer: compression ≤ 0.60f'ci; tension ≤ 0.25√f'ci (without bonded reinforcement) or 0.50√f'ci (with bonded reinforcement). In service: compression ≤ 0.45f'c (sustained) or 0.60f'c (total); tension ≤ 0.5√f'c (Class C). These limits appear as boundary conditions in 'find P to satisfy allowable stress' type problems.
LOAD BALANCING IS AN ALTERNATIVE — USE IT AS A CHECK: If a problem involves a parabolic tendon and gives P, e, L, and w, compute both by the direct method (f = P/A ± Pe/S ∓ M/S) and by load balancing (w_net = w − w_bal, M_net = w_net L²/8, f = P/A ± M_net/S). If they match, your answer is correct. This cross-check takes 60 seconds and eliminates sign errors.
UNDERSTAND WHAT R = Pe/Pi MEANS PHYSICALLY: R is the fraction of the prestress force that remains after ALL losses. R = 0.82 means 18% of the initial prestress is 'wasted' on overcoming elastic shortening, creep, shrinkage, and relaxation. A higher-quality low-relaxation strand and well-cured concrete increase R.
FOR TRANSFER LENGTH (PRE-TENSIONED): Lt ≈ 50db to 60db. Development length ≈ 100db (strand must be embedded this far for full flexural bond). These values appear in anchorage and detailing questions in the structural design part of the board exam.
PRACTICE WITH REAL BOARD EXAM FORMAT: PSC problems in the Philippine CE board exam are typically 4–6 sub-parts covering the same beam: given P, e, L, w → find (a) one fiber stress, (b) the other fiber stress, (c) the balanced load or net moment, (d) the P needed to satisfy a specific stress limit. Solving the full set is faster if you compute P/A, Pe/S, and M/S ONCE and then combine them as needed for each sub-part.
REVIEW PHILIPPINE INFRASTRUCTURE CONTEXT: The NSCP 2015 (National Structural Code of the Philippines) adopts ACI 318 provisions for prestressed concrete design. Bridge design follows DPWH standards (based on AASHTO LRFD). Knowing that the Pasig River Expressway, NLEX–SLEX connector, and similar projects use post-tensioned box girders helps you contextualize the engineering behind the board exam formulas.

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In summary

Prestressed concrete is a topic where precision, consistency, and systematic procedure separate passing candidates from those who miss crucial marks. The four pillars of PSC board-exam mastery are: (1) knowing the pre-tensioning vs. post-tensioning distinction and its implications for which losses apply; (2) applying the combined stress formula f = P/A ± Pe/S ∓ M/S with an unwavering compression-positive sign convention and checking BOTH the transfer and service stages; (3) classifying and estimating prestress losses correctly, with the key rule that friction and anchorage seating apply only to post-tensioned members; and (4) using the load-balancing formula w_bal = 8Pe/L² for parabolic tendons to find either the balanced load, the required prestress force, or the stress distribution under partial balancing. Throughout this chapter, every formula has been derived from physical intuition — the tendon is a compressed bowstring that pushes up on the beam; the eccentricity moment compresses the side toward which the tendon is offset; losses reduce the available prestress over time. Understanding the physics makes the formulas memorable and self-correcting when a sign seems wrong. For the PRC board exam, work through all five practice problems without looking at the solutions, then verify step by step. Time yourself: a well-prepared candidate solves a complete PSC problem set (4–6 sub-parts) in under 12 minutes. With the visual aids, worked examples, and exam tips in this chapter, you are equipped to tackle any PSC question that appears in the examination. Mabuti pang magsanay nang marami — kaya mo ito!

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