Question 4 of 7: Prestressed concrete T-beam — no-tension design and tendon profile

Question 4: Prestressed concrete T-beam — no-tension design and tendon profile (12 + 6 + 2 = 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 design starts from a post-tensioned T-beam of 15 m clear span with a 3 m overhang, with $f'_{ci} = 35$ MPa at transfer and $f'_c = 50$ MPa in service.

Find. Cross-section dimensions, the number of strands and the effective prestress force such that no tensile stress occurs anywhere at transfer or in service, and the tendon profile that delivers it.

Approach. Compute the gross section properties and kern distances, obtain the service and self-weight moment envelopes, combine the "no tension at the bottom in service" and "no tension at the top at transfer" conditions into a single requirement on $P_e$, choose a strand count, then plot the permissible cable zone and fit a harped profile inside it.

1. Fix the section and compute its properties. Try a flange $1200 \times 200$ mm on a $350$ mm web, total depth 1400 mm ($L/h \approx 10.7$, appropriate for two 300 kN point loads). Then $$A = 660\,000\ \text{mm}^2, \quad y_t = 545.5, \quad y_b = 854.5\ \text{mm}, \quad I_g = 126.0 \times 10^9\ \text{mm}^4,$$ so $S_t = 231.1 \times 10^6$ and $S_b = 147.5 \times 10^6$ mm³, and the kern distances are $k_t = S_b/A = 223.5$ mm and $k_b = S_t/A = 350.1$ mm. Self-weight is $15.84$ kN/m.
2. Obtain the moments. The overhang produces a hogging moment at B of $M_B = -(80)(3) - 15.84(3)^2/2 = -311.3$ kN·m, of which $-71.3$ kN·m is self-weight. Carrying that end moment into the span together with the two 300 kN loads and the uniform self-weight gives a service maximum at mid-span, $$M_s = 1489.9\ \text{kN}\cdot\text{m} \quad\text{with}\quad M_{sw} = 409.9\ \text{kN}\cdot\text{m}\ \text{at the same section.}$$
3. Combine the two no-tension conditions. No tension at the bottom in service requires $P_e(k_t + e) \ge M_s$; no tension at the top at transfer requires $e \le k_b + M_{sw}/P_i$. Eliminating $e$ between them, with $P_i = (1200/960)P_e = 1.25P_e$, gives the section requirement $$P_e \ge \frac{M_s - 0.8M_{sw}}{k_t + k_b} = \frac{(1489.9 - 0.8 \times 409.9) \times 10^6}{223.5 + 350.1} = \boxed{2026\ \text{kN}}.$$ This corresponds to 15.1 strands, but at exactly that force the two bounds on $e$ coincide and there is no usable cable zone, so a margin is needed.
4. Choose the tendon. Provide $18 - 15.2$ mm strands, $A_{ps} = 2520$ mm², giving $$P_e = 2520(960) = 2419\ \text{kN}, \qquad P_i = 2520(1200) = 3024\ \text{kN},$$ grouped in three ducts of six and stressed from both ends. The cable zone at mid-span then opens to $392 \le e \le 486$ mm, a 93 mm band, which is workable.
5. Plot the cable zone along the member. Evaluating $e_{min} = M_s/P_e - k_t$ and $e_{max} = k_b + M_{sw}/P_i$ section by section gives

6. Fit a harped profile, not a parabola. Between two equal point loads the moment diagram is a straight line, so the cable zone is essentially flat there; a single parabola with its vertex at mid-span sits too low at the load points and too high at the vertex, and deepening the section does not help because both bounds move together. Adopt instead a harped profile: $e = 0$ at the free end A, rising to 150 mm at B, ramping to 430 mm at the first harp point 5 m from B, held constant to the second harp point at 11 m, then ramping back to $e = 0$ at the anchorage at C. Every ordinate in the table above falls inside its zone.
7. Verify the extreme-fibre stresses. With $P_e = 2419$ kN and $e = 430$ mm at mid-span, $$\sigma_{bot,serv} = \frac{P_e}{A} + \frac{P_ee}{S_b} - \frac{M_s}{S_b} = 3.67 + 7.05 - 10.10 = +0.62\ \text{MPa (compression)},$$ $$\sigma_{top,trans} = \frac{P_i}{A} - \frac{P_ie}{S_t} + \frac{M_{sw}}{S_t} = 4.58 - 5.63 + 1.77 = +0.73\ \text{MPa (compression)}.$$ Repeating at every section listed, and along the overhang, the extreme-fibre stress is compressive everywhere; the largest value is $\boxed{11.2\ \text{MPa}}$ at the bottom at transfer, against the limits $0.6f'_{ci} = 21$ MPa at transfer and $0.6f'_c = 30$ MPa in service. The no-tension requirement is met throughout.
8. Check the ultimate limit state. Factoring the loads gives $M_f = 2132$ kN·m at mid-span. With $d_p = y_t + e = 975.5$ mm and a stress at ultimate of about $f_{pr} = 1600$ MPa, the compression block depth is $a = 120$ mm, which stays inside the 200 mm flange, so $$M_r = \phi_p A_{ps}f_{pr}\left(d_p - \frac{a}{2}\right) = 3322\ \text{kN}\cdot\text{m} \; > \; 2132\ \text{kN}\cdot\text{m}.$$ Strength is comfortable, as it usually is when a member is proportioned for zero tension; provide nominal untensioned bars in the flange over the support to control cracking on the overhang.