07-Str-A5 · Undated paper

Question 7 of 7: C1: Prestressed beam for the loading of Figure 3

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams — 07-Str-A5 Advanced Structural Design, printed footer 07-Str-A5/May 2019. Three hours, "closed book" with handbooks and textbooks permitted. Seven questions of equal value (20 marks each) in three parts: A1–A3 (do two), B1–B3 (do two), C1 (do one) — five solutions make a complete paper. All seven are solved here, because the set is a study resource rather than an exam attempt.

Design data given on page 1. Solutions to CAN/CSA S16 (steel), CAN/CSA A23.3 (concrete) and CAN/CSA O86 (timber). All loads shown on the figures are unfactored. All structural steel is G40.21 300W — $F_y = 300$ MPa, $F_u = 450$ MPa, $E = 200\,000$ MPa, $G = 77\,000$ MPa. All reinforcement is 400W — $f_y = 400$ MPa. Concrete strength is stated per question. Steel sections are W-shapes unless noted otherwise.

Reference texts.

CSA S16:19, Design of Steel Structures — Cl. 11 (section classification), 13.3 (compression), 13.4 (shear), 13.5–13.6 (bending and lateral–torsional buckling), 13.8 (beam-columns), 13.13 (welds), 14.3.2 (web crippling), 8.4.3 (stability effects).
CSA A23.3:19, Design of Concrete Structures — Cl. 8.6 (material properties), 10 (flexure and axial load, slenderness), 11 (shear), 15 (footings), 18 (prestressed concrete).
National Building Code of Canada 2020, Part 4 — Table 4.1.3.2 load combinations, 4.1.6 snow and drift loads.
CISC, Handbook of Steel Construction, 11th ed. — W-shape dimensions and properties.
Salmon, Johnson & Malhas, Steel Structures: Design and Behavior, 5th ed. — Ch. 9 (LTB), 12 (beam-columns), 13 (connections).
MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design, Canadian ed. — Ch. 4–6 (flexure), 11 (columns), 12 (slender columns), 15 (footings).
Collins & Mitchell, Prestressed Concrete Structures — Ch. 4 (service stresses), 6 (flexural strength); PCI Design Handbook, 8th ed. — multiplier method for long-term camber.

Check — load combinations. The figures split the loads into DEAD, LIVE, SNOW and WIND, so NBCC 2020 Table 4.1.3.2 is applied literally. Throughout Part A the governing case is $1.25D + 1.5L$ (case 2 with no companion snow or wind on the figure). For Figures 3 the governing case is $1.25D + 1.5S + 0.4W$ (case 3), with $1.25D + 1.4W + 0.5S$ and $1.4D$ checked and shown not to govern. The "LIVE/SNOW" label on W1 of Figure 3 is read as a single 3 kN/m specified roof load that may be either occupancy live or balanced snow, and the triangular W2 is read as the additional snow-drift surcharge that its shape implies.

Question 7 — C1: Prestressed beam for the loading of Figure 3 (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The 4.5 m simply supported beam and loads of Figure 3 (see Question 4), now pretensioned. Shop-fabricated by a CSA-certified plant (so pretensioning with straight strands is available and $f'_{ci}$ can be taken at 30 MPa at release); $f_{pu} = 1\,860$ MPa low-relaxation strand; $f'_c = 40$ MPa.

Find. The concrete section, prestress force and strand layout; a check of stresses at transfer and at service, the ultimate moment capacity, and the long-term deflection under dead load with the method stated.

Approach. Fix a section, size the effective prestress for zero bottom tension at service, then work backwards to the strand count and check the transfer stresses at the member end (where the self-weight moment vanishes and the transfer condition is worst). Finish with strength, cracking and the PCI multiplier method for camber.

Section and section properties. Take a 250 mm × 400 mm rectangle: $$A_g = 100\,000\ \text{mm}^2,\qquad I_g = 1.333\times10^9\ \text{mm}^4,\qquad S_t = S_b = 6.667\times10^6\ \text{mm}^3$$ Self weight is 2.40 kN/m, so $M_{sw} = 2.40(4.5)^2/8 = 6.08$ kN$\cdot$m.
Service moment. Adding the 4.5 kN/m superimposed dead load and the 3.0 kN/m uniform snow to the self weight, and superposing the unfactored 6 kN/m drift, the peak service moment falls at 2.127 m from A: $$\boxed{M_T = 28.3\ \text{kN}\cdot\text{m}}$$
Prestress required for a fully compressed soffit. Requiring $f_b \le 0$ at service with eccentricity $e$, $$P_e\left(\frac{1}{A_g} + \frac{e}{S_b}\right) \ge \frac{M_T}{S_b}$$ Selecting $e = 100$ mm (strand centroid 100 mm above the soffit) gives $P_e \ge 169.6$ kN.
Choose the strands. Two 13 mm seven-wire low-relaxation strands give $A_{ps} = 2(99) = 198$ mm$^2$. Allowing 30% total loss to service and 26% at transfer, standard for a shop-cast member with low-relaxation strand, $$P_i = 0.70f_{pu}A_{ps} = 258\ \text{kN},\qquad P_e = 0.60f_{pu}A_{ps} = 221\ \text{kN} \ \ge\ 169.6\ \text{kN}\ \checkmark$$
Check stresses at transfer. The governing point is the member end, where the prestress is fully developed but the self-weight moment is zero: $$f_t = -\frac{P_i}{A_g} + \frac{P_ie}{S_t} = -2.578 + 3.867 = +1.289\ \text{MPa (tension)}$$ $$f_b = -\frac{P_i}{A_g} - \frac{P_ie}{S_b} = -6.445\ \text{MPa (compression)}$$ Against the Cl. 18.3.1 limits at $f'_{ci} = 30$ MPa — tension $0.25\sqrt{f'_{ci}} = 1.37$ MPa and compression $0.6f'_{ci} = 18$ MPa — both pass. This limit is what fixes $e$ at 100 mm rather than the 140 mm a bottom-fibre strand pattern would give.
Check stresses at service. At the critical section, $$f_b = -\frac{P_e}{A_g} - \frac{P_ee}{S_b} + \frac{M_T}{S_b} = -2.210 - 3.315 + 4.239 = \boxed{-1.28\ \text{MPa}}$$ so the soffit stays in compression — no tension anywhere, hence no cracking and no crack-width check. The top fibre reaches $-3.14$ MPa against the $0.6f'_c = 24$ MPa limit.
Ultimate moment capacity. For bonded strand, Cl. 18.6.2 with $k_p = 2(1.04 - f_{py}/f_{pu}) = 0.28$ for low-relaxation strand, $\alpha_1 = 0.79$, $\beta_1 = 0.87$ and $d_p = 300$ mm: $$c = \frac{\phi_pA_{ps}f_{pu}}{\alpha_1\phi_cf'_c\beta_1b + k_p\phi_pA_{ps}f_{pu}/d_p} = 69.4\ \text{mm},\qquad f_{pr} = f_{pu}\left(1 - \frac{k_pc}{d_p}\right) = 1\,740\ \text{MPa}$$ $$M_r = \phi_pA_{ps}f_{pr}\left(d_p - \frac{\beta_1c}{2}\right) = \boxed{83.6\ \text{kN}\cdot\text{m}}$$ The factored demand, using the same governing combination as Question 4, is $M_f = 41.7$ kN$\cdot$m — comfortably satisfied.
Check the minimum-strength rule. The cracking moment is $$M_{cr} = S_b\left(0.6\lambda\sqrt{f'_c} + \frac{P_e}{A_g} + \frac{P_ee}{S_b}\right) = 62.1\ \text{kN}\cdot\text{m}$$ and $M_r = 83.6 \ge 1.2M_{cr} = 74.6$ kN$\cdot$m, so the member will not fail suddenly on first cracking — the check that prevents an under-reinforced prestressed section from being brittle.
Immediate deflections. With $E_{ci} = 4\,500\sqrt{30} = 24\,648$ MPa at release and $E_c = 4\,500\sqrt{40} = 28\,460$ MPa at 28 days, and a straight tendon, $$\Delta_{p} = \frac{P_ieL^2}{8E_{ci}I_g} = 1.99\ \text{mm}\ \uparrow,\quad \Delta_{sw} = \frac{5w_{sw}L^4}{384E_{ci}I_g} = 0.39\ \text{mm}\ \downarrow,\quad \Delta_{sd} = \frac{5w_DL^4}{384E_cI_g} = 0.63\ \text{mm}\ \downarrow$$ so the member leaves the bed with 1.60 mm of net camber.
Long-term deflection — the method, stated. Use the PCI multiplier method (PCI Design Handbook, after Martin), which is the standard treatment for a shop-cast pretensioned member: each immediate deflection component is multiplied by an empirical factor that folds creep, shrinkage and prestress loss into a single long-term amplifier, with different factors for prestress camber (2.45 final), member self weight (2.70 final) and superimposed dead load applied after erection (3.00 final). Then $$\Delta_{LT} = 2.45(1.99) - 2.70(0.39) - 3.00(0.63) = 4.87 - 1.05 - 1.90 = \boxed{+1.91\ \text{mm (upward)}}$$ The beam therefore retains about 1.9 mm of camber under sustained dead load rather than sagging; the sustained-load limit of $L/480 = 9.4$ mm is not approached in either direction. The alternative permitted route — an age-adjusted effective-modulus calculation to A23.3 Cl. 18.3.2 — gives the same conclusion but needs creep and shrinkage coefficients that a shop with CSA A23.4 certification would supply from its own mix data.
Detail. Two 13 mm strands straight at 100 mm from the soffit; 2–10M top bars for handling and to control end-zone tension; 10M stirrups at 200 mm, closing to 75 mm over the 600 mm transfer zone at each end to resist bursting; 40 mm cover throughout.

Question 7 — results
Quantity	Value
Section	250 mm × 400 mm, $f'_c = 40$ MPa ($f'_{ci} = 30$ MPa)
Strand	2 × 13 mm low-relaxation, $A_{ps} = 198$ mm$^2$, straight at $e = 100$ mm
Prestress at transfer / effective	258 kN / 221 kN (169.6 kN required)
Peak service moment / factored moment	28.3 / 41.7 kN$\cdot$m
Transfer stresses, member end (top / bottom)	+1.29 / −6.45 MPa (limits 1.37 / 18)
Service stresses (top / bottom)	−3.14 / −1.28 MPa — no tension
Ultimate: $c$ / $f_{pr}$ / $M_r$	69.4 mm / 1 740 MPa / 83.6 kN$\cdot$m
$M_{cr}$ / $1.2M_{cr}$	62.1 / 74.6 kN$\cdot$m ($M_r$ governs)
Immediate camber / self wt / superimposed	1.99 ↑ / 0.39 ↓ / 0.63 ↓ mm
Long-term deflection (PCI multipliers)	1.91 mm upward — net camber retained

Topic. Pretensioned rectangular beam: service stress control, flexural strength and long-term camber.

Key relations. Service fibre stresses $f = -P/A \mp Pe/S \pm M/S$; A23.3 Cl. 18.3.1 transfer limits ($0.25\sqrt{f'_{ci}}$ tension, $0.6f'_{ci}$ compression); Cl. 18.6.2 bonded-strand stress $f_{pr} = f_{pu}(1 - k_pc/d_p)$; $M_r \ge 1.2M_{cr}$; PCI multipliers 2.45 (camber), 2.70 (self weight), 3.00 (superimposed dead).

Why this works. On a span this short the design is not driven by strength — $M_r$ is twice $M_f$ and the section never even cracks — but by the transfer condition at the member ends, where the full prestress acts on an unloaded section and puts the top fibre into tension. That single limit is what forces the eccentricity down to 100 mm; a designer who placed the two strands at the usual 60 mm from the soffit would see +1.93 MPa at release and would then have to debond one strand or add top prestress to fix it. Because the eccentric prestress hogs the member more than the dead load sags it, the long-term multipliers work in the beam's favour and the net movement is upward — which is exactly what the question means by "define the long-term deflection": name the method, then state the sign as well as the magnitude.

Common pitfall. Checking transfer stresses only at mid-span, where the self-weight moment relieves the top fibre by 0.91 MPa. The end of the member, not the middle, is the critical transfer section for a straight tendon. The second is treating the PCI multipliers as if they applied to the net deflection; they must be applied component by component, because camber and load deflection creep at different effective rates.

Source. CSA A23.3:19 Cl. 18.3, 18.6.2 and 18.8; Collins & Mitchell, Prestressed Concrete Structures, Ch. 4 and 6; PCI Design Handbook, 8th ed., Table 5.8.2 (multipliers).

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