Question 2 of 7: Post-tensioned prestressed-concrete girder

Question 2: Post-tensioned prestressed-concrete girder (10 + 5 + 5 = 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. From Figure 2, a girder with a 2 m overhang and a 12 m span: the tip A is free and carries 60 kN, B is a roller 2 m from A, C is a pin 14 m from A, and a 400 kN load acts at midspan of BC (8 m from A).

Find. A gross concrete section, the strand area \(A_{ps}\) and a tendon profile such that no fibre anywhere carries tension either at transfer or in service.

Approach. Compute the service moments, then size the section from the cable-zone condition — the requirement that the band \(M_T/P_e - k_t \le e \le k_b + M_{sw}/P_i\) is not empty — because with "no tension" enforced at both stages it is the section, not the strand count, that governs; finally lay a harped tendon inside that band and confirm the ultimate capacity.

1. Service moments from the applied loads. Taking moments about B for the overhanging beam, \(R_C(12) + 60(2) = 400(6)\), so \(R_C = 190\ \text{kN}\) and \(R_B = 270\ \text{kN}\). Hence $$M_B = -60(2) = -120\ \text{kN}\cdot\text{m}, \qquad M_{\text{mid}} = R_C(6) = 190(6) = 1140\ \text{kN}\cdot\text{m}$$ The overhang puts the girder into hogging over B and sagging at midspan, so both faces must be kept in compression.
2. Choose a trial section and add its self weight. Try a solid rectangle \(450 \times 1800\): \(A = 810\,000\ \text{mm}^2\), \(S = 2.43 \times 10^{8}\ \text{mm}^3\) and, since the section is symmetric, both kern distances equal \(k = S/A = 300\ \text{mm}\). Its self weight is \(0.45(1.8)(24) = 19.44\ \text{kN/m}\), which adds \(M_{sw} = 330.5\ \text{kN}\cdot\text{m}\) at midspan and \(-38.9\ \text{kN}\cdot\text{m}\) at B. The total service moments are therefore \(M_T = 1470.5\ \text{kN}\cdot\text{m}\) at midspan and \(-158.9\ \text{kN}\cdot\text{m}\) at B.
3. Size the section from the cable zone, not from the strand count. Keeping the bottom fibre in compression in service requires \(e \ge M_T/P_e - k_t\); keeping the top fibre in compression at transfer requires \(e \le k_b + M_{sw}/P_i\). Subtracting, a usable zone exists only when $$k_t + k_b + \frac{M_{sw}}{P_i} - \frac{M_T}{\eta P_i} \;\ge\; 0 \quad\Longrightarrow\quad P_i \;\ge\; \frac{M_T/\eta - M_{sw}}{k_t + k_b} = \frac{1838.1 - 330.5}{0.600} = \boxed{2513\ \text{kN}}$$ This is the real design driver: a smaller section closes the zone entirely, and no strand count can reopen it.
4. Strands. Adopt \(n = 17\) low-relaxation strands of 15.2 mm diameter, \(A_{ps} = 17(140) = \boxed{2380\ \text{mm}^2}\), giving $$P_i = 2380(1200) = 2856\ \text{kN}, \qquad P_e = 2380(960) = 2284.8\ \text{kN}$$ At midspan the cable zone is then $$\frac{M_T}{P_e} - k_t = \frac{1470.5\times10^6}{2.285\times10^6} - 300 = 343.6\ \text{mm} \;\le\; e \;\le\; 300 + \frac{330.5\times10^6}{2.856\times10^6} = 415.7\ \text{mm}$$ so take \(e_{\text{mid}} = 400\ \text{mm}\) below the centroid, comfortably inside the band and leaving 500 mm of concrete below the duct.
5. Confirm the stresses at both stages. Using \(f = -P/A \mp Pe/S \pm M/S\) with compression negative: $$\text{transfer, midspan:}\quad f_{\text{top}} = -0.19\ \text{MPa}, \quad f_{\text{bot}} = -6.87\ \text{MPa}$$ $$\text{service, midspan:}\quad f_{\text{top}} = -5.11\ \text{MPa}, \quad f_{\text{bot}} = -0.53\ \text{MPa}$$ Every value is compressive, so the "no tension" requirement is met, and the peaks sit well below the permissible \(0.6f'_{ci} = 21\ \text{MPa}\) at transfer and \(0.45f'_c = 22.5\ \text{MPa}\) in service.
6. Profile through the overhang and the anchorages. Over B the moment is hogging, so the tendon must rise: the top fibre stays in compression only while \(e \le k + M_B/P_e = 300 - 69.5 = \boxed{230.5\ \text{mm}}\). At A and C the moment is zero, so the anchorages must lie inside the kern, \(|e| \le 300\ \text{mm}\). A harped (trapezoidal) profile therefore serves: \(e = 0\) at A, 200 mm at B, 400 mm at midspan and 0 at C. A single parabola would not, because the moment diagram of an overhanging beam with one point load is made of straight lines and a parabola cuts across them; the shaded band in the figure above shows the harped line staying inside the zone at every quarter point, for instance \(31.8 \le 300 \le 383\ \text{mm}\) at 3 m from B.
7. Ultimate flexural check. With \(k_p = 2(1.04 - f_{py}/f_{pu}) = 0.423\), \(d_p = 900 + 400 = 1300\ \text{mm}\), \(\alpha_1 = 0.775\) and \(\beta_1 = 0.845\) for \(f'_c = 50\ \text{MPa}\), equilibrium gives \(c = 347\ \text{mm}\), \(f_{pr} = 1552\ \text{MPa}\) and $$M_r = \phi_p A_{ps} f_{pr}\left(d_p - \tfrac{\beta_1 c}{2}\right) = \boxed{3835\ \text{kN}\cdot\text{m}}$$ against \(M_f = 1.5(1140) + 1.25(330.5) = 2123\ \text{kN}\cdot\text{m}\). The service no-tension rule, not strength, governs this girder — which is the usual outcome for a fully prestressed member.