Question 3 of 7: Post-Tensioned Girder — Section and Tendons

Question 3: Post-Tensioned Girder — Section and Tendons (12 + 4 + 4 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. Figure 2 shows a 12 m concrete girder on a roller at the left end and built in at the right (marked FIXED END), carrying unfactored point loads of 600 kN at 3 m and 400 kN at 6 m from the roller. The prestressing data from page 1 apply.

Find. A rectangular section whose extreme fibres reach — but do not exceed — the permitted tension at both the transfer and the service stage, together with the strand area and a tendon profile that stays inside the permissible zone all along the member.

Approach. Analyse the propped cantilever for service moments, size the depth from the standard section-modulus inequality, then write the two governing tension limits as equalities in the two unknowns $P_{e}/A$ and $P_{e}e/Z$ and solve them simultaneously — "maximum permitted tension" is a two-equation closed-form problem, not an iteration. Convert $P_{e}$ to strands, re-trim the eccentricity for the rounded strand count, and finish by mapping the permissible zone and drawing a parabolic profile inside it.

1. Part (a) — analyse the propped cantilever. Taking the cantilever fixed at the right end as the released structure and the prop reaction $R_{A}$ as the redundant, compatibility at the roller gives $$R_{A} = \sum \frac{P b^{2}(3L-b)}{2L^{3}} = \frac{600(9)^{2}(36-9) + 400(6)^{2}(36-6)}{2(12)^{3}} = \frac{1\,744\,200}{3456} = 504.7\ \text{kN}$$ where $b$ is measured from the built-in end. The applied-load moments follow directly: $M = 1514.1$ kN·m under the 600 kN load, $1228.1$ kN·m under the 400 kN load, and $-1743.8$ kN·m at the fixed end, with the point of contraflexure at $x = 8.48$ m.
2. Add self weight and identify the critical section. The section reached in Step 3 is $400 \times 1500$ mm, weighing $0.4(1.5)(24) = 14.4$ kN/m. For a propped cantilever under a uniform load, $M_{0} = -wL^{2}/8 = -259.2$ kN·m at the fixed end and $+129.6$ kN·m at 3 m. Superposing, $$M_{T} = 2003.0\ \text{kN}\cdot\text{m}\ \text{(hogging, fixed end)}, \qquad M_{T} = 1643.7\ \text{kN}\cdot\text{m}\ \text{(sagging, 3 m)}$$ The built-in end governs, and there the tension face is the top, so the tendon must sit above the centroid at that section.
3. Size the section. Requiring the transfer compression and the service tension limits to be simultaneously satisfiable gives the standard inequality $$Z \ge \frac{M_{T} - \eta M_{0}}{\eta\,(0.60f'_{ci}) + f_{ts}} = \frac{(2003.0 - 0.80 \times 259.2)\times 10^{6}}{0.80(21.0) + 3.536} = \frac{1795.6 \times 10^{6}}{20.34} = 88.3 \times 10^{6}\ \text{mm}^{3}$$ A $400 \times 1500$ mm rectangle supplies $Z = bh^{2}/6 = 150 \times 10^{6}\ \text{mm}^{3}$ and $A = 600\,000\ \text{mm}^{2}$, comfortably above the requirement. The surplus is deliberate: a depth of $L/8$ keeps the strand count sensible and, as Step 8 shows, keeps the ultimate-strength check under-reinforced.
4. Write the two tension limits as equalities. Let $X = P_{e}/A$ and $Y = P_{e}e/Z$, and measure both moments as magnitudes about the tension face. Service tension at the top fibre and transfer tension at the bottom fibre give $$X + Y = \frac{M_{T}}{Z} - f_{ts}, \qquad X - Y = \eta\left(-f_{ti} - \frac{M_{0}}{Z}\right)$$ Substituting $M_{T}/Z = 13.353$ MPa and $M_{0}/Z = 1.728$ MPa, $$X + Y = 9.817, \qquad X - Y = -2.565 \;\Rightarrow\; X = 3.626\ \text{MPa}, \quad Y = 6.191\ \text{MPa}$$ Both permitted tensions are reached exactly, which is what the wording of the question asks for.
5. Recover the prestressing force and eccentricity. $$\begin{gathered}P_{e} = X A = 3.626 \times 600\,000 = 2176\ \text{kN}, \\ e = \frac{Y Z}{P_{e}} = \frac{6.191 \times 150 \times 10^{6}}{2.176 \times 10^{6}} = 427\ \text{mm}\end{gathered}$$ measured above the centroid at the built-in end. With 120 mm to the strand centroid the physical limit is 630 mm, so the cable fits with room to spare.
6. Part (b) — strand area, then re-trim the eccentricity. The area follows from the effective stress: $$A_{ps} = \frac{P_{e}}{f_{se}} = \frac{2176 \times 10^{3}}{960} = 2266\ \text{mm}^{2} \;\Rightarrow\; 2266/140 = 16.2 \ \text{strands}$$ so provide 17 strands of 15.2 mm diameter ($A_{ps} = 2380\ \text{mm}^{2}$), in two ducts. Rounding up raises the force to $P_{e} = 2380(960) = 2285$ kN, which pushes the transfer stage past its limit unless the eccentricity is pulled back. Re-solving the transfer equation alone with the supplied area, $$Y = X + \eta f_{ti} + \eta\frac{M_{0}}{Z} = 3.808 + 1.183 + 1.382 = 6.373 \;\Rightarrow\; e = 418\ \text{mm}$$ The band that satisfies both stages is then 394.5 to 418.4 mm; adopt $\boxed{e = 410\ \text{mm}}$ above the centroid at the built-in end, a round buildable value inside it. Quoting the trial 427 mm with 17 strands would overshoot the transfer tension — a small slip that a marker will find.

7. Map the permissible zone and draw the profile. At every section the four stress limits (two fibres, two stages) become upper and lower bounds on the eccentricity. Evaluated with $P_{e} = 2285$ kN and $P_{i} = P_{e}/\eta = 2856$ kN, and measuring $e$ positive below the centroid, the band runs as tabulated below. A pair of parabolas that meet tangentially at $x = 4.5$ m — $e = 330\,(2x/4.5 - x^{2}/4.5^{2})$ up to the vertex, then $e = 330 - 740\,[(x-4.5)/7.5]^{2}$ beyond it — starts on the centroid at the anchorage, reaches 330 mm below the centroid at 4.5 m, crosses the axis near the point of contraflexure and finishes 410 mm above it at the built-in end. Checked at 0.25 m intervals the profile lies inside the zone everywhere, with no violation.
8. Confirm the ultimate limit state. Serviceability sized the member, but Clause 18 still requires $M_{r} \ge M_{f}$. At the built-in end $d_{p} = 750 + 410 = 1160$ mm, $\rho_{p} = 2380/(400 \times 1160) = 0.00513$ and $f_{pr} = f_{pu}(1 - 0.5\rho_{p}f_{pu}/f'_{c}) = 1593$ MPa, giving $a = 339$ mm, $c/d_{p} = 0.35$ (safely under-reinforced) and $$M_{r} = \phi_{p}A_{ps}f_{pr}\left(d_{p} - \frac{a}{2}\right) = 3380\ \text{kN}\cdot\text{m} \ge M_{f} = 1.5(1743.8) + 1.25(259.2) = 2940\ \text{kN}\cdot\text{m}$$ The sagging section at 3 m returns $M_{r} = 2955$ against $M_{f} = 2433$ kN·m. Both pass.

The band closes almost completely at the built-in end — barely 24 mm wide — which is the whole difficulty of this question. That is why the eccentricity there has to be fixed first and the rest of the profile drawn to suit, rather than the other way round.