Question 2 of 7: Post-tensioned T-beam with cantilever

Question 2: Post-tensioned T-beam with cantilever (10 + 5 + 2 + 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.

Find. (a) T-section dimensions, prestressing-strand area and tendon profile such that no tensile stress occurs at any fibre at transfer or in service; (b) the long-term vertical deflection of the cantilever tip \(C\).

Approach. Size a uniform T-section, compute the service moment envelope including self-weight, then impose the two no-tension conditions — bottom fibre in service and top fibre at transfer — as a cable zone \(M_s/P_e - k_b \le e \le k_t + M_{sw}/P_i\). The prestress force is the smallest value for which that zone is non-empty at every section; the profile is then any curve inside the zone. Deflection follows from the unit-load method on the gross section, amplified for creep.

1. Trial section and properties. Take a flange \(2000 \times 200\ \text{mm}\) over a \(450\ \text{mm}\) web, overall depth \(h = 1600\ \text{mm}\) (span/depth \(\approx 10.6\), normal for a heavily loaded post-tensioned beam):
   $$A = 1.030 \times 10^{6}\ \text{mm}^2, \quad y_b = 1010.7\ \text{mm}, \quad I_g = 2.608 \times 10^{11}\ \text{mm}^4$$
   $$S_b = 2.581 \times 10^{8}\ \text{mm}^3, \quad S_t = 4.426 \times 10^{8}\ \text{mm}^3$$
   The kern distances that control the no-tension checks are \(k_b = S_b/A = 250.5\ \text{mm}\) and \(k_t = S_t/A = 429.7\ \text{mm}\). The self-weight is \(w = 1.030\ \text{m}^2 \times 24 = 24.72\ \text{kN/m}\).
2. Service analysis. The beam is statically determinate, so there are no secondary prestress moments. Taking moments about \(B\) for the whole beam (including self-weight over the full 20 m),
   $$R_A = 592.99\ \text{kN}, \qquad R_B = 761.41\ \text{kN}$$
   The service moments at the sections that matter are
   $$M(6) = 3113.0, \quad M(8.5) = 3147.4, \quad M(11) = 3027.3, \quad M(B) = -291.2\ \text{kN}\cdot\text{m}$$
   with self-weight-only values \(M_{sw}(6) = 776.5\) and \(M_{sw}(8.5) = 837.4\ \text{kN}\cdot\text{m}\).
3. The two no-tension conditions. Writing the bottom-fibre stress in service and the top-fibre stress at transfer and setting each to zero:
   $$-\frac{P_e}{A} - \frac{P_e e}{S_b} + \frac{M_s}{S_b} \le 0 \;\Rightarrow\; e \ge \frac{M_s}{P_e} - k_b$$
   $$-\frac{P_i}{A} + \frac{P_i e}{S_t} - \frac{M_{sw}}{S_t} \le 0 \;\Rightarrow\; e \le k_t + \frac{M_{sw}}{P_i}$$
   The first is a lower bound that falls as \(P_e\) rises; the second is an upper bound that also rises with \(P_i\). The minimum viable force is where the two meet.
4. Prestress force and strand area. Equating the bounds at the critical section and using \(P_i = P_e (1200/960) = 1.25P_e\) gives \(P_e \approx 3.66\ \text{MN}\). Rounding up to a practical strand count, adopt 30 no. 15.2 mm seven-wire strands, \(A_{p1} = 140\ \text{mm}^2\):
   $$A_{ps} = 30(140) = 4200\ \text{mm}^2, \quad P_i = 4200(1200) = \boxed{5040\ \text{kN}}, \quad P_e = 4200(960) = 4032\ \text{kN}$$
   Detail these as five tendons of six strands in 85 mm ducts, stressed from both ends.
5. Cable zone and adopted profile. Evaluating both bounds section by section with these forces:
   $$x = 6\ \text{m}: \; 521.5 \le e \le 583.7 \qquad x = 8.5\ \text{m}: \; 530.1 \le e \le 595.8$$
   $$x = 11\ \text{m}: \; 500.3 \le e \le 577.3 \qquad x = B: \; -322.8 \le e \le 407.6\ \text{mm}$$
   The zone between the two point loads is only about 47 mm wide, and it is flat: between two equal point loads the moment diagram is a straight line, so a single parabola with its vertex at midspan sits too low under the loads and too high at the vertex. A harped profile is therefore adopted — straight from \(e = 0\) at \(A\) to \(e = 550\ \text{mm}\) at \(x = 6\ \text{m}\), constant to \(x = 11\ \text{m}\), straight back to \(e = 0\) at \(B\), and level through the cantilever. Every section from \(x = 0\) to \(x = 20\ \text{m}\) then lies inside its zone.
6. Stress check at the critical section. At \(x = 6\ \text{m}\) with \(e = 550\ \text{mm}\):
   $$\text{transfer:}\quad f_{top} = -0.38\ \text{MPa}, \qquad f_{bot} = -12.63\ \text{MPa} \;\le\; 0.60f_{ci} = 21.0\ \text{MPa}$$
   $$\text{service:}\quad f_{top} = -5.94\ \text{MPa} \;\le\; 0.45f'_c = 22.5\ \text{MPa}, \qquad f_{bot} = -0.45\ \text{MPa}$$
   Every fibre is in compression at both stages, so the no-tension requirement is met with a small margin at the two fibres that control.
7. Ultimate flexural check. The strand is unbonded-equivalent bonded post-tensioning; using A23.3 Cl. 18.6.2 with \(k_p = 2(1.04 - f_{py}/f_{pu}) = 0.4229\) and \(d_p = 1600 - (1010.7 - 550) = 1139\ \text{mm}\), equilibrium of the rectangular stress block within the flange gives \(c = 146.9\ \text{mm}\), \(a = \beta_1 c = 124.2\ \text{mm} < 200\ \text{mm}\) and \(f_{pr} = 1654.6\ \text{MPa}\), so
   $$M_r = \phi_p A_{ps} f_{pr}\left(d_p - \frac{a}{2}\right) = \boxed{6737\ \text{kN}\cdot\text{m}} \;\ge\; M_f = 1.25(776.5) + 1.5(2336.5) = 4475\ \text{kN}\cdot\text{m}$$
8. (b) Immediate deflection of C. Applying a unit downward load at \(C\) and integrating \(\int M m\,\mathrm{d}x / EI_g\) with \(E_c = 4500\sqrt{50} = 31\,820\ \text{MPa}\) and \(I_g = 2.608\times10^{11}\ \text{mm}^4\), each load case contributes
   $$\Delta_{sw} = -1.57\ \text{mm}, \quad \Delta_{loads} = -4.34\ \text{mm}, \quad \Delta_{prestress} = +4.41\ \text{mm}$$
   (negative = upward). The loads inside span \(AB\) rotate \(B\) so that the cantilever tip rises, while the prestress camber in \(AB\) rotates it the other way and pushes \(C\) down. The net immediate movement is \(\Delta_i = -1.50\ \text{mm}\), i.e. 1.5 mm upward.
9. Long-term deflection. All the loads are sustained, so the effective modulus method applies with a creep coefficient \(\varphi_{cc} = 2.0\), i.e. \(E_{eff} = E_c/(1 + \varphi_{cc})\). Since the prestress term already uses the after-loss force \(P_e\),
   $$\Delta_{LT} = (1 + \varphi_{cc})\,\Delta_i = 3.0(-1.50) = \boxed{-4.50\ \text{mm}\ \ (4.5\ \text{mm upward})}$$
   Against the A23.3 Table 9.3 limit for a cantilever, taken as twice the projection, \(2(3000)/480 = 12.5\ \text{mm}\), the tip movement is comfortably acceptable and — being upward — will not be visible as sag.