Question 7 of 7: Post-Tensioned Prestressed Concrete T-Girder
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Civ-B2 Advanced Structural Design, National Exams May 2016 — 3 hours, closed book (design handbooks permitted). Seven design questions of equal value; any five constitute a complete paper, and all seven are solved here.
Check — support conditions for the prestressed girder. Figure 1 shows the steel beam of Question 1 built in at both ends. A post-tensioned member cannot be built into rigid abutments: the supports would resist the elastic shortening that the prestress causes, and much of the precompression would be lost into them rather than into the concrete. The girder is therefore designed as simply supported over the same 12 m span with the same loads, which is also the conservative reading — the simple-span moment of 2100 kN.m is 65 per cent larger than the 1275 kN.m a built-in span would attract. If continuity really were intended, the same section would be used with a concordant cable profile and secondary moments computed explicitly.
Given. A 12 m post-tensioned girder carrying three point loads, to be proportioned so that no fibre goes into tension at any stage.
Quantity
Value
Span, simply supported
L = 12 000 mm
Unfactored point loads
300 kN at 3 m, 400 kN at 6 m, 300 kN at 9 m
Concrete at transfer / final
f'ci = 35 MPa, f'c = 50 MPa
Strand
fpu = 1750 MPa, fpy = 1450 MPa
Initial stress / losses
1200 MPa; losses 240 MPa, so fpe = 960 MPa
Permissible stresses
no tension anywhere; 0.6f'ci = 21.0 MPa at transfer, 0.45f'c = 22.5 MPa in service
Find. T-section proportions, the area of post-tensioned strand, and the cable profile along the span.
The post-tensioned T-section and the cable profile. Holding the tendon on the centroid at the anchorages is what keeps the end sections free of tension when the prestress acts alone.
Approach. Write the two no-tension conditions — bottom fibre under full service load, top fibre at transfer under self weight alone — as limits on the eccentricity at every section. Those two curves bound a permissible zone; pick a section deep enough for the zone to exist at mid-span, read the prestress force off the service condition, then draw a cable that stays inside the zone from end to end.
Get the applied moments. The three point loads give end reactions of 500 kN, so the moment is $500x$ up to the first load, reaching $M = 500(3) = 1500$ kN.m at 3 m and $M = 500(6) - 300(3) = 2100$ kN.m at mid-span. These are service moments; no load factor is applied because "no tension" is a serviceability requirement.
Propose a T-section and compute its properties. Take a 1400 mm by 220 mm flange on a 400 mm web, 1500 mm deep overall. Then $A = 820\,000$ mm$^{2}$, the centroid lies $y_{t} = 578.3$ mm below the top and $y_{b} = 921.7$ mm above the soffit, and $$I = 1.7932\times10^{11}\ \text{mm}^{4},\qquad Z_{t} = 3.101\times10^{8},\qquad Z_{b} = 1.946\times10^{8}\ \text{mm}^{3}$$ Its self weight is $0.820\times24 = 19.68$ kN/m, adding 354.2 kN.m at mid-span, so the total service moment there is $M = 2100 + 354 = 2454.2$ kN.m.
Write the two no-tension conditions. Taking compression as negative and the eccentricity $e$ positive downwards, the bottom fibre in service and the top fibre at transfer require $$\frac{P_{e}}{A} + \frac{P_{e}e}{Z_{b}}\ \ge\ \frac{M}{Z_{b}}\qquad\text{and}\qquad \frac{P_{i}e}{Z_{t}} - \frac{P_{i}}{A} \le \frac{M_{0}}{Z_{t}}$$ Rearranged, they bound the eccentricity at every section: $$\frac{M}{P_{e}} - \frac{Z_{b}}{A}\ \le\ e\ \le\ \frac{Z_{t}}{A} + \frac{M_{0}}{P_{i}}$$ The two kern distances are $Z_{b}/A = 237.3$ mm and $Z_{t}/A = 378.2$ mm.
Solve for the prestress force. The two conditions are coupled, because raising $P$ tightens the transfer limit while relaxing the service one. Iterating at mid-span converges on $P_{e}\approx3.6$ MN, and choosing 27 strands of 15.2 mm diameter ($A_{ps} = 27\times140 = 3780$ mm$^{2}$) gives $$P_{i} = 3780\times1200 = 4536\ \text{kN},\qquad P_{e} = 3780\times960 = \boxed{3629\ \text{kN}}$$ housed in three ducts of nine strands in the 400 mm web.
Fix the mid-span eccentricity. With that force the permissible band at mid-span is $$\frac{2454.2\times10^{6}}{3.629\times10^{6}} - 237.3 = 439.1\ \text{mm}\ \le\ e\ \le\ 378.2 + \frac{354.2\times10^{6}}{4.536\times10^{6}} = 456.3\ \text{mm}$$ a window only 17 mm wide, which is exactly what "no tension at either stage" costs. Take $e = 450$ mm, giving a duct centroid 471.7 mm above the soffit — ample room for three ducts with cover.
Verify every fibre stress. At mid-span, with $P_{i}/A = 5.53$ MPa and $P_{e}/A = 4.43$ MPa: $$\text{transfer: } f_{\text{top}} = -5.53 + 6.58 - 1.14 = -0.09\ \text{MPa},\quad f_{\text{bot}} = -5.53 - 10.50 + 1.82 = -14.20\ \text{MPa}$$ $$\text{service: } f_{\text{top}} = -4.43 + 5.27 - 7.91 = -7.07\ \text{MPa},\quad f_{\text{bot}} = -4.43 - 8.40 + 12.62 = -0.20\ \text{MPa}$$ Every value is compressive, so no tension occurs, and the largest compression, 14.20 MPa at transfer, is well inside $0.6f'_{ci} = 21.0$ MPa while the service maximum of 7.07 MPa is inside $0.45f'_{c} = 22.5$ MPa.
Draw the cable profile. At the anchorages the applied moment is zero, so the tendon must lie inside the kern: $e\le Z_{t}/A = 378.2$ mm to keep the top in compression, and $e\ge -237.3$ mm for the bottom. Anchoring on the centroid, $e = 0$, satisfies both with the widest margin and is the standard detail. A parabola between the ends and mid-span, $$\boxed{e(x) = 450\left[1 - \left(\frac{x-6}{6}\right)^{2}\right]\ \text{mm}}$$ gives $e = 337.5$ mm at 3 m, where the permissible band is 249.3 to 436.7 mm, so the cable lies inside the zone at every section. A parabola is also the natural profile for a draped post-tensioned duct and produces a uniform upward balancing load.
Check the ultimate limit state. Serviceability fixed the prestress, but strength must still be confirmed. With $d_{p} = 578.3 + 450 = 1028.3$ mm and $k_{p} = 2(1.04 - f_{py}/f_{pu}) = 0.423$, iterating $f_{pr} = f_{pu}(1 - k_{p}c/d_{p})$ with the stress-block equilibrium converges on $f_{pr} = 1617$ MPa and $a = 156$ mm, which stays inside the 220 mm flange. Then $$M_{r} = \phi_{p}A_{ps}f_{pr}\left(d_{p} - \frac{a}{2}\right) = 0.9(3780)(1617)(950.3)\times10^{-6} = 5228\ \text{kN.m}$$ against $M_{f} = 1.5(2100) + 1.25(354.2) = 3593$ kN.m, a utilisation of 0.687, with $c/d_{p} = 0.180$ confirming a ductile failure.
Permissible-eccentricity envelope. The lower curve is the service no-tension requirement, the upper curve the transfer no-tension requirement; the cable must lie between them everywhere, and the chosen parabola does.
