Question 5 of 7: Prestressed concrete tee-beam — section, strands and profile

Question 5: Prestressed concrete tee-beam — section, strands and profile (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. A simply supported post-tensioned tee-beam of 14 m span carrying two 400 kN loads at 4 m and 10 m from the left support, with the page-1 prestressing data.

Find. Cross-section dimensions that admit a no-tension solution at both transfer and service, the prestressing force and strand area, and the tendon profile along the span.

Approach. Combine the transfer and service stress limits at each face to get minimum section moduli, choose a tee that meets them, then solve the pair of governing inequalities simultaneously for the prestressing force and its eccentricity, convert to strands, and set the end eccentricity from the zero-moment condition at the supports. Close with the ultimate flexural and shear checks.

1. Part (a) — applied moments. The two loads are symmetric, so each reaction is 400 kN and the moment is constant between them: $$M_{live} = 400(4) = 1600\ \text{kN}\cdot\text{m}$$ For the section adopted below the self weight is 13.10 kN/m, giving $$M_i = \frac{13.10(14)^2}{8} = 321.0\ \text{kN}\cdot\text{m},\qquad M_s = 1600 + 321.0 = 1921.0\ \text{kN}\cdot\text{m}$$
2. Minimum section moduli from the stress limits. Writing the bottom-fibre stress at transfer (limited to $0.6f'_{ci}$) and in service (limited to zero tension), multiplying the first by the loss ratio $\eta = 0.80$ and subtracting eliminates the prestress terms: $$Z_b \ge \frac{M_s - \eta M_i}{\eta(0.6f'_{ci})} = \frac{(1921.0 - 0.8 \times 321.0)\times10^6}{0.8(21)} = 9.91\times10^7\ \text{mm}^3$$ The same operation on the top fibre (no tension at transfer, $0.6f'_c$ in service) gives $$Z_t \ge \frac{M_s - \eta M_i}{0.6f'_c} = \frac{1664.2\times10^6}{30} = 5.55\times10^7\ \text{mm}^3$$
3. Choose the tee. Take a flange 1000 mm wide × 180 mm deep on a 300 mm web, overall depth 1400 mm. On the gross section $$A = 546\,000\ \text{mm}^2,\qquad \bar{y}_b = 840.8\ \text{mm},\qquad I = 1.050\times10^{11}\ \text{mm}^4$$ $$Z_b = \frac{I}{\bar{y}_b} = 1.249\times10^8 \ge 9.91\times10^7\ \checkmark,\qquad Z_t = \frac{I}{h-\bar{y}_b} = 1.878\times10^8 \ge 5.55\times10^7\ \checkmark$$ Its self weight, $0.546(24) = 13.10$ kN/m, is the value assumed in step 1, so no iteration is needed. The kern distances that control the tendon are $k_t = Z_b/A = 228.7$ mm and $k_b = Z_t/A = 343.9$ mm.
Adopted cross-section — 1000 × 180 mm flange on a 300 mm web, 1400 mm overall, with the 22 strands grouped 411 mm above the soffit at mid-span.
4. Part (b) — the two governing inequalities. No tension at the bottom in service and no tension at the top at transfer give, at mid-span, $$\eta P\left(\frac{1}{A} + \frac{e}{Z_b}\right) \ge \frac{M_s}{Z_b} \;\Longrightarrow\; P \ge \frac{M_s}{\eta(k_t + e)}$$ $$\frac{P}{A} - \frac{Pe}{Z_t} + \frac{M_i}{Z_t} \ge 0 \;\Longrightarrow\; e \le k_b + \frac{M_i}{P}$$ Solving the pair simultaneously (a two-line iteration that converges in four cycles) gives $$\boxed{P_i = 3633\ \text{kN},\qquad e = 432\ \text{mm}}$$
5. Convert to strands. At an initial stress of 1200 MPa, $$A_{ps} = \frac{3633\times10^3}{1200} = 3027\ \text{mm}^2$$ Using 15.2 mm seven-wire strand at 140 mm² each, provide 22 strands ($A_{ps} = 3080\ \text{mm}^2$) in two ducts, giving $P_i = 3696$ kN and $P_e = 0.8(3696) = 2957$ kN. Rounding the eccentricity down to $e = 430$ mm keeps the slightly larger force inside the transfer limit.
6. Verify all four stresses with the strands actually provided. With $P_i/A = 6.769$, $P_ie/Z_b = 12.724$ and $P_ie/Z_t = 8.463$ MPa:

   Stage and fibre	Stress (MPa)	Limit (MPa)
   Transfer, top	+0.02 compression	≥ 0 (no tension)
   Transfer, bottom	+16.92	≤ 21.0
   Service, top	+8.87	≤ 30.0
   Service, bottom	+0.21 compression	≥ 0 (no tension)

   Every fibre is in compression at both stages, so the member is fully prestressed and uncracked as the question requires.
7. Profile along the beam. At the supports the applied moment is zero, so the no-tension condition at the top reduces to $e \le k_b = 343.9$ mm; take $e_{end} = 340$ mm. Between the two loads the moment is constant at its maximum, so the tendon is held at $e = 430$ mm there. The natural profile is therefore a harped tendon: straight from 340 mm at each end to 430 mm under the 4 m and 10 m load points, and horizontal over the central 6 m — a profile that follows the trapezoidal moment diagram exactly, which a parabola would not.
8. Ultimate flexural resistance. With the strands 411 mm above the soffit, $d_p = 989$ mm. Using CSA A23.3 Cl 18.6.2 with $k_p = 2(1.04 - f_{py}/f_{pu}) = 0.4229$, $\alpha_1 = 0.775$ and $\beta_1 = 0.845$ for $f'_c = 50$ MPa, iterating $f_{pr} = f_{pu}(1-k_pc/d_p)$ against the compression block gives $c = 208$ mm, $a = 175$ mm and $f_{pr} = 1595$ MPa, so $$M_r = \phi_pA_{ps}f_{pr}\left(d_p - \frac{a}{2}\right) = 3985\ \text{kN}\cdot\text{m} \ \ge\ M_f = 1.5(1600) + 1.25(321.0) = 2801\ \text{kN}\cdot\text{m}$$ a ratio of 0.70, so the section that serviceability demanded is comfortably strong at ultimate.
9. Shear reinforcement. At the support $V_f = 1.5(400) + 1.25(13.10)(7) = 715$ kN. With $d_v = \max(0.9d_p,\ 0.72h) = 1008$ mm and $\beta = 0.18$, $$V_c = \phi_c\lambda\beta\sqrt{f'_c}\,b_vd_v = 250\ \text{kN},\qquad V_p = P_e\frac{430-340}{4000} = 67\ \text{kN}$$ leaving $V_s = 398$ kN for the stirrups: $$s = \frac{\phi_sA_vf_yd_v\cot\theta}{V_s} = \frac{0.85(200)(400)(1008)(1.428)}{398\times10^3} = 246\ \text{mm}$$ Use 10M double-leg stirrups at 240 mm in the end quarters, relaxing to 400 mm between the loads where the shear is only the self-weight component. End-block bursting reinforcement is provided behind each anchorage.