Question 5 of 7: Post-tensioned girder designed for no tension

Question 5: Post-tensioned girder designed for no tension

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. Simply supported girder of Figure 3, 14 m between a pin at A and a roller at B, carrying two unfactored 450 kN loads at 4 m from each end. From page 1: $f_{ci} = 35$ MPa at transfer, $f'_c = 50$ MPa, $f_{initial} = 1200$ MPa, loss of prestress 240 MPa, so $f_{se} = 960$ MPa and the ratio of effective to initial force is $\eta = 0.80$.

Find. A cross-section, the area of post-tensioned steel, and the tendon profile, such that no tension occurs at any fibre at any stage.

Approach. Write the four fibre-stress conditions, combine them into a minimum prestress and a permissible band of eccentricity, size the section so that band is open, then choose the strand count and check the profile at every section and the ultimate strength.

1. Set out the loading. The live moment is constant between the loads at $M_L = P a = 450 \times 4 = 1800$ kN·m. Trial a symmetric I-section 1500 mm deep with 500 × 250 mm flanges and a 250 mm web: $A = 500000$ mm$^2$, $I = 1.1979 \times 10^{11}$ mm$^4$, $Z = 1.5972 \times 10^{8}$ mm$^3$ and the kern distance $Z/A = 319.4$ mm. Its self weight is 12.0 kN/m, giving $M_0 = 294.0$ kN·m at mid-span and a total service moment $M_T = 2094.0$ kN·m.
2. Write the no-tension conditions. Taking compression positive and $e$ measured downwards from the centroid, the two conditions that can produce tension are the top fibre at transfer (full $P_i$, self weight only) and the bottom fibre in service (reduced $P_e$, full moment): $$\frac{P_i}{A} - \frac{P_i e}{Z_t} + \frac{M_0}{Z_t} \ge 0, \qquad \frac{P_e}{A} + \frac{P_e e}{Z_b} - \frac{M_T}{Z_b} \ge 0$$ Rearranged, these bracket the eccentricity: $$\frac{M_T}{P_e} - \frac{Z_b}{A} \;\le\; e \;\le\; \frac{Z_t}{A} + \frac{M_0}{P_i}$$
3. Find the prestress that opens the band. The band is non-empty only when the lower limit falls below the upper one. Substituting $P_e = \eta P_i$ and collecting terms, $$P_i \ \ge \ \frac{M_T/\eta - M_0}{(Z_t + Z_b)/A} = 3637 \text{ kN}$$ This single inequality is what sizes the section: increase $Z/A$ and the required force falls. A 500 × 1500 mm solid rectangle would need over 5000 kN for the same duty, which is why the flanged section is worth the extra formwork.
4. Choose the tendons. With 15.2 mm seven-wire strand ($A = 140$ mm$^2$ each) at $f_{initial} = 1200$ MPa, each strand delivers 168 kN, so 22 strands are needed as a minimum. Adopt 24 strands, $A_{ps} = 3360$ mm$^2$, giving $$P_i = 4032 \text{ kN}, \qquad P_e = \eta P_i = 3226 \text{ kN}$$ Place them in two ducts of twelve, stacked vertically in the 250 mm web.
5. Fix the eccentricity at mid-span. Substituting these forces into the band gives $$329.7 \text{ mm} \;\le\; e \;\le\; 392.4 \text{ mm}$$ a window only 62.6 mm wide. Take $\boxed{e = 370 \text{ mm}}$, which puts the tendon centroid 380 mm above the soffit, just inside the web above the bottom flange.
6. Verify all four fibre stresses. Substituting back:

Every value is positive, so no fibre is ever in tension, and both compressive limits are met with reserve.

7. Set the profile. The limits move along the span with $M_T(x)$ and $M_0(x)$. At the supports both moments vanish and the band becomes $-319.4 \le e \le +319.4$ mm, so anchoring the tendon on the centroid is safe and is also the easiest anchorage to detail. Because the moment diagram is trapezoidal — linear over the outer 4 m, constant between the loads — a harped profile matching that shape is the natural choice: straight from $e = 0$ at each end to $e = 370$ mm under each load, constant between. Checking the band at eight sections confirms the cable stays inside it everywhere, with the tightest margin at $x = 4$ m where the lower limit is 313.0 mm.
8. Check the ultimate limit state. Factoring, $M_f = 1.5 \times 1800 + 1.25 \times 294.0 = 3067.5$ kN·m. With $d_p = 1120$ mm and $k_p = 2(1.04 - f_{py}/f_{pu}) = 0.4229$, solving $\phi_p A_{ps} f_{pr} = \alpha_1 \phi_c f'_c (\text{stress block})$ gives $c = 508.0$ mm, $f_{pr} = 1414$ MPa and $$\boxed{M_r = 4013 \text{ kN}\cdot\text{m} \; \gt \; M_f = 3067.5 \text{ kN}\cdot\text{m}}$$ with $c/d_p = 0.454$, inside the 0.5 limit. The cracking moment is 2902 kN·m, so $M_r$ also exceeds $1.2 M_{cr} = 3482$ kN·m as A23.3 Cl 18.5 requires — the member will warn before it fails.