07-Str-A5 · May 2015

Question 5 of 7: Prestressed concrete tee-beam

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

Notes on this paper

Paper format. National Exams, May 2015 — 07-Str-A5 Advanced Structural Design. Three hours, "closed book" (textbooks and design handbooks permitted, no notes). Seven questions; any five constitute a complete paper and all are of equal value, the printed mark split being 20 marks each. All loads shown on Figures 1–4 are unfactored. All seven questions are answered here.

Design data printed on page 1 and used throughout. Concrete $f'_c = 30\ \text{MPa}$; structural steel $F_y = 350\ \text{MPa}$; reinforcing steel $f_y = 400\ \text{MPa}$; prestressed concrete $f'_{ci} = 35\ \text{MPa}$ at transfer and $f'_c = 50\ \text{MPa}$ in service, $n = 6$; $f_{pu} = 1750\ \text{MPa}$, $f_{py} = 1450\ \text{MPa}$, $f_{p,\text{initial}} = 1200\ \text{MPa}$, losses $= 240\ \text{MPa}$, hence $f_{pe} = 960\ \text{MPa}$ and the loss ratio $\eta = 0.80$.

Check: load factors. The paper says only that "all loads shown are unfactored" and gives no load classification. Every printed load is therefore treated as a specified live load and factored at 1.5; member self weight is treated as dead and factored at 1.25 (NBCC 2020 Case 2, $1.25D + 1.5L$). Resistance factors are $\phi = 0.90$ (steel), $\phi_w = 0.67$ and the extra 0.67 of CSA S16 Cl. 13.13.2.2 (welds), $\phi_{sc} = 0.80$ (shear studs), $\phi_c = 0.65$, $\phi_s = 0.85$, $\phi_p = 0.90$ (CSA A23.3). In an examination you would state exactly this and proceed.

Reference texts for 07-Str-A5.

CSA S16:19, Design of Steel Structures, with the CISC Handbook of Steel Construction (11th ed.) — Cl. 13.3 (compression), 13.5–13.6 (flexure and lateral–torsional buckling), 13.7 (bracing at plastic hinges), 13.8 (beam–columns), 13.13 (welds), 14.3–14.6 (plate girders), Cl. 17 (composite beams).
CSA A23.3:19, Design of Concrete Structures, with the CAC Concrete Design Handbook (4th ed.) — Cl. 10 (flexure and columns), Cl. 11 (shear), Cl. 18 (prestressed concrete).
Salmon, Johnson and Malhas, Steel Structures: Design and Behavior, 5th ed. — plastic analysis (Ch. 10), plate girders (Ch. 11), composite construction (Ch. 16).
MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design, Canadian edition — frames, beam–columns, detailing.
Collins and Mitchell, Prestressed Concrete Structures — cable zone, transfer and service stress checks, harped profiles.
National Building Code of Canada 2020 for load combinations; CSA S6 (CHBDC) for the pedestrian-bridge serviceability and vibration criteria.

Question 5: Prestressed concrete tee-beam (14 + 6 = 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.

Quantity	Value
Span and loads (Figure 3)	14 m simply supported; 400 kN at 4 m and at 10 m from $A$
Concrete	$f'_{ci} = 35\ \text{MPa}$ at transfer, $f'_c = 50\ \text{MPa}$ in service
Prestressing steel	$f_{pu} = 1750$, $f_{py} = 1450$, $f_{pi} = 1200\ \text{MPa}$; losses 240 MPa, so $f_{pe} = 960\ \text{MPa}$, $\eta = 0.80$
Stress limits adopted	zero tension at both stages; $0.6f'_{ci} = 21.0\ \text{MPa}$ at transfer, $0.45f'_c = 22.5\ \text{MPa}$ in service

Find. Tee proportions that admit a workable cable zone with no tensile stress at either stage, the number of strands, and the tendon profile from end to end.

Trial tee cross-section: 1400 × 200 flange over a 350 × 1000 web.

Approach. Work in kern terms. No tension at the top at transfer puts a ceiling on the eccentricity; no tension at the soffit in service puts a floor on it. The two bounds intersect only if the prestress force is large enough, and that intersection condition sizes the strands directly. Then check the four corner stresses, the ultimate moment and the shear.

Trial section and properties. Take a 1400 × 200 flange over a 350 × 1000 web, total depth 1200 mm:
$$A = 630\,000\ \text{mm}^2 \qquad y_b = 766.7\ \text{mm} \qquad I = 86.10\times10^9\ \text{mm}^4$$ $$S_b = 112.3\times10^6,\quad S_t = 198.7\times10^6\ \text{mm}^3 \qquad k_t = \frac{S_b}{A} = 178.3,\quad k_b = \frac{S_t}{A} = 315.4\ \text{mm}$$
Self weight $w = 0.63(24) = 15.12\ \text{kN/m}$, so $M_{sw} = wL^2/8 = 370.4\ \text{kN}\cdot\text{m}$. Between the two 400 kN loads the moment is constant at $400(4) = 1600\ \text{kN}\cdot\text{m}$, giving $M_T = 1970.4\ \text{kN}\cdot\text{m}$ at midspan.
The two eccentricity bounds. Writing compression as positive, the top fibre at transfer and the bottom fibre in service give
$$e \le k_b + \frac{M_{sw}}{P_i} \qquad\text{and}\qquad e \ge \frac{M_T}{\eta P_i} - k_t$$
Minimum prestress force. The zone is non-empty only if the ceiling is at least the floor:
$$P_i \ge \frac{M_T/\eta - M_{sw}}{k_t + k_b} = \frac{(1970.4/0.80 - 370.4)\times10^6}{493.7} = 4239\ \text{kN}$$
At exactly that force the zone shrinks to the single point $e = 402.8\ \text{mm}$, which leaves no tolerance, so use a larger force.
Strands. With 15.2 mm seven-wire strands ($A = 140\ \text{mm}^2$ each) at $f_{pi} = 1200\ \text{MPa}$, 28 strands give
$$A_{ps} = 3920\ \text{mm}^2 \qquad P_i = 4704\ \text{kN} \qquad P_e = \eta P_i = 3763\ \text{kN}$$
Twenty-six strands were tried first and rejected: at 4368 kN the zone at the harp point ($\le 384.6$ mm) and at midspan ($\ge 385.6$ mm) do not overlap, so a constant eccentricity between the loads is impossible. This is the characteristic behaviour of a beam under two equal point loads — the moment diagram is flat between them, so the cable must be flat too.
Cable zone and profile. With $P_i = 4704$ kN the bounds become 327 ≤ $e$ ≤ 380 mm at the harp points and 345 ≤ $e$ ≤ 394 mm at midspan, while at the ends (where $M = 0$) the ceiling is simply $k_b = 315\ \text{mm}$. Adopt a harped profile: $e = 150\ \text{mm}$ at each end, ramping linearly to $e = 360\ \text{mm}$ at 4 m, held constant to 10 m, then mirrored. In terms of the soffit the tendon centroid runs from 617 mm at the ends to 407 mm over the middle 6 m — two ducts of 14 strands in the 350 mm web.

Elevation, permissible cable zone and adopted harped tendon profile. A single parabola cannot fit: the zone is flat between the load points because the moment diagram is.

Check all four corner stresses. Sweeping the profile along the span, the extreme values are

Stage and fibre	Governing value	Limit	Status
Transfer, top fibre (minimum)	+0.47 MPa	≥ 0 (no tension)	satisfied
Transfer, bottom fibre (maximum)	19.85 MPa	$0.6f'_{ci} = 21.0$	satisfied
Service, bottom fibre (minimum)	+0.49 MPa	≥ 0 (no tension)	satisfied
Service, top fibre (maximum)	9.07 MPa	$0.45f'_c = 22.5$	satisfied

At midspan, for instance, $P_i/A = 7.47$, $P_ie/S_t = 8.52$ and $M_{sw}/S_t = 1.86\ \text{MPa}$, so the top fibre at transfer is $7.47 - 8.52 + 1.86 = +0.81\ \text{MPa}$ — small, but compressive, which is exactly what "allowing no tension" demands.

Ultimate moment. $d_p = 1200 - 407 = 793.3\ \text{mm}$, $\rho_p = 3920/(1400 \times 793.3) = 0.00353$, and with $k_p = 2(1.04 - f_{py}/f_{pu}) = 0.4229$, $\alpha_1 = 0.775$, $\beta_1 = 0.845$ at 50 MPa,
$$f_{pr} = f_{pu}\left(1 - \frac{k_p\phi_pf_{pu}\rho_p}{\alpha_1\phi_cf'_c}\right) = 1750(1 - 0.0934) = 1587\ \text{MPa}$$ $$a = \frac{\phi_pA_{ps}f_{pr}}{\alpha_1\phi_cf'_cb} = 158.7\ \text{mm} < 200\ \text{mm} \ \ (\text{rectangular behaviour, block inside the flange})$$ $$\boxed{M_r = \phi_pA_{ps}f_{pr}\left(d_p - \frac{a}{2}\right) = 3997\ \text{kN}\cdot\text{m} \ \ge\ M_f = 1.25(370.4) + 1.5(1600) = 2863\ \text{kN}\cdot\text{m}}$$
Minimum reinforcement is satisfied too: $M_{cr} = 2502$ and $1.2M_{cr} = 3003 \le 3997\ \text{kN}\cdot\text{m}$.
Shear. At the support $V_f = 1.25(15.12)(7) + 1.5(400) = 732\ \text{kN}$. With $d_v = \max(0.9d_p, 0.72h) = 864\ \text{mm}$, $\beta = 0.18$ and the harped tendon inclined at $(360-150)/4000 = 0.0525$:
$$V_c = \phi_c\beta\sqrt{f'_c}\,b_wd_v = 250\ \text{kN} \qquad V_p = P_e(0.0525) = 198\ \text{kN} \qquad V_s = 732 - 250 - 198 = 285\ \text{kN}$$
Ten-millimetre double-leg stirrups ($A_v = 200\ \text{mm}^2$, $\theta = 35^\circ$) at 275 mm centres supply $V_s$, well inside the 600 mm maximum spacing. Harping the tendons earns roughly 27 % of the shear resistance for free.

Result	Value
Cross-section	1400 × 200 flange + 350 × 1000 web, $h = 1200$ mm, $A = 0.63$ m²
$y_b$ / $I$ / $S_b$ / $S_t$	766.7 mm / 86.10 × 10⁹ mm⁴ / 112.3 × 10⁶ / 198.7 × 10⁶ mm³
Kern distances $k_t$, $k_b$	178.3 mm, 315.4 mm
Minimum prestress force	4239 kN
Prestressing steel	28 strands of 15.2 mm, $A_{ps} = 3920$ mm²; $P_i = 4704$ kN, $P_e = 3763$ kN
Profile	harped: $e = 150$ mm at the ends, 360 mm from 4 m to 10 m (c.g.s. 617 mm → 407 mm above the soffit)
Extreme stresses	transfer +0.47 / 19.85 MPa; service +0.49 / 9.07 MPa — no tension anywhere
Ultimate	$M_r = 3997$ vs $M_f = 2863$ kN·m (0.716)
Shear reinforcement	10M stirrups at 275 mm near the supports

Topic. Zero-tension (Class U) design of a post-tensioned tee, and the cable zone as a design tool.

Key relations. Kern distances $k_t = S_b/A$, $k_b = S_t/A$; cable-zone bounds $M_T/(\eta P_i) - k_t \le e \le k_b + M_{sw}/P_i$; the intersection condition $P_i \ge (M_T/\eta - M_{sw})/(k_t + k_b)$; A23.3 Cl. 18.6.2 $f_{pr} = f_{pu}[1 - k_p\phi_pf_{pu}\rho_p/(\alpha_1\phi_cf'_c)]$ with $k_p = 2(1.04 - f_{py}/f_{pu})$; $V_r = V_c + V_s + V_p$.

Why this works. Prestressing is a self-equilibrating load, so the design question is never "how much steel resists the moment" but "where can the resultant sit". The upper bound comes from transfer, when the beam is strongest and least loaded and an over-eccentric cable would crack the top; the lower bound comes from service, when the full moment is on and an under-eccentric cable would crack the soffit. Both bounds move with $1/P_i$, so increasing the force opens the zone from both sides — which is why a force barely above the theoretical minimum leaves nothing to detail with.

Common pitfall. Reaching for a parabolic tendon out of habit. Between two equal point loads the moment is constant, so the cable zone is essentially flat there, and a parabola with its vertex at midspan sits too low at the harp points and too high at the vertex. Deepening the section does not help, because both bounds move together; only changing the profile shape does. The second pitfall is placing 28 strands at the physically lowest position the web allows, roughly $e = 614\ \text{mm}$, which cracks the top flange at transfer.

Source. Collins and Mitchell, Prestressed Concrete Structures, Ch. 4–5 (cable zone, transfer and service checks); CSA A23.3:19 Cl. 18; Lin and Burns, Design of Prestressed Concrete Structures, the classical kern equations.

← Question 4 Next: Question 6 →

Result	Value
Cross-section	1400 × 200 flange + 350 × 1000 web, \(h = 1200\) mm, \(A = 0.63\) m²
\(y_b\) / \(I\) / \(S_b\) / \(S_t\)	766.7 mm / 86.10 × 10⁹ mm⁴ / 112.3 × 10⁶ / 198.7 × 10⁶ mm³
Kern distances \(k_t\), \(k_b\)	178.3 mm, 315.4 mm
Minimum prestress force	4239 kN
Prestressing steel	28 strands of 15.2 mm, \(A_{ps} = 3920\) mm²; \(P_i = 4704\) kN, \(P_e = 3763\) kN
Profile	harped: \(e = 150\) mm at the ends, 360 mm from 4 m to 10 m (c.g.s. 617 mm → 407 mm above the soffit)
Extreme stresses	transfer +0.47 / 19.85 MPa; service +0.49 / 9.07 MPa — no tension anywhere
Ultimate	\(M_r = 3997\) vs \(M_f = 2863\) kN·m (0.716)
Shear reinforcement	10M stirrups at 275 mm near the supports