Question 7 of 7: Post-tensioned prestressed concrete girder

Question 7: Post-tensioned prestressed concrete girder (12 + 6 + 2 = 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. Girder 17 m overall: 3 m overhang plus a 14 m span; unfactored loads 60 kN at the free tip, 250 kN at 2 m and at 12 m from the left support. Prestressed-concrete data from page 1: $f_{ci}=35$ MPa at transfer, $f'_c=50$ MPa, $n=6$, $f_{pu}=1750$ MPa, $f_{py}=1450$ MPa, $f_{p,initial}=1200$ MPa, losses $=240$ MPa (so the effective stress is 960 MPa and $\eta=P_e/P_i=0.80$). Gross-section properties may be used; no tension is permitted at any stage.

Find. (a) A rectangular cross-section for which neither fibre goes into tension at transfer or in service; (b) the area of prestressing strand and the tendon profile that achieves it.

[Figure not reproduced: Figure 3 as printed, with the draped tendon profile designed below: 3 m overhang carrying 60 kN, 14 m span with 250 kN at 2 m from each support. See the official exam paper.]

Approach. Find the service and self-weight moment diagrams; for a trial section, impose no tension at the bottom in service and no tension at the top at transfer, eliminate the eccentricity between the two to get the minimum prestress force, then re-introduce $e$, convert the force to strands, and finally check that the whole tendon profile lies inside the resulting cable zone.

(a) The cross-section

1. Try 500 × 1200 mm. $$A=6.00\times10^{5}\ \text{mm}^{2},\qquad S=\frac{bh^{2}}{6}=1.20\times10^{8}\ \text{mm}^{3},\qquad k=\frac{S}{A}=\frac{h}{6}=200\ \text{mm}$$ Self-weight $w=0.5(1.2)(24)=14.4$ kN/m.
2. Moments. With $x$ measured from the left support, the reactions are $R_1=471.5$ kN and $R_2=333.3$ kN under the full load, and $R_1=148.6$ kN, $R_2=96.2$ kN under self-weight alone. The moment diagram peaks at $$M_{max}=741.0\ \text{kN}\cdot\text{m}\ \text{at}\ x=8.21\ \text{m},\qquad M_{S1}=-244.8\ \text{kN}\cdot\text{m}$$ with $M=637.8$ kN·m still at $x=12$ m under the second 250 kN load.
3. State the two no-tension conditions. Writing $u=1/A$ and $v=e/S$, and taking compression positive, $$\text{service, bottom fibre:}\quad \eta P_i\left(u+v\right)\ \ge\ \frac{M_s}{S}$$ $$\text{transfer, top fibre:}\quad P_i\left(v-u\right)\ \le\ \frac{M_{sw}}{S}$$ The first says the prestress must be big enough (or eccentric enough) to close the bottom fibre under full load; the second says it must not be so eccentric that it cracks the top when only self-weight is present.
4. Eliminate the eccentricity. Substituting the second inequality into the first removes $v$ altogether: $$2\eta P_iu\ \ge\ \frac{M_s-\eta M_{sw}}{S}\quad\Longrightarrow\quad P_i\ \ge\ \frac{A\left(M_s-\eta M_{sw}\right)}{2\eta S}$$ This is the key result: the section is feasible only if some $P_i$ satisfies it, and the governing section is the one that maximises $M_s-\eta M_{sw}$ — not necessarily the section of maximum moment.
5. Find the governing section. Scanning the span, $\left(M_s-\eta M_{sw}\right)/S$ peaks at $x=11.79$ m with a value of 4.226 MPa, slightly ahead of the peak-moment section (4.072 MPa at $x=8.21$ m), because the self-weight moment falls away faster than the live-load moment near the right-hand support. Hence $$P_i\ \ge\ \frac{6.00\times10^{5}(4.226)}{2(0.80)}\times10^{-3}=\boxed{1585\ \text{kN}}$$
6. Confirm the section. The required $P_i$ implies a mean precompression of $1585\times10^{3}/6.00\times10^{5}=2.64$ MPa — a normal, economical figure for a post-tensioned girder, and only 7.5% of $0.6f_{ci}=21$ MPa. The 500 × 1200 mm section is therefore adopted: $h/L=1200/14\,000=1/11.7$, which is the usual proportion for a post-tensioned girder of this span. A shallower section would drive $P_i$ up sharply, because $M/S$ grows as $h^{-2}$.

(b) Strand area and tendon profile

1. Area of prestressing steel. Stressing to the given initial stress of 1200 MPa, $$A_{ps}=\frac{P_i}{f_{p,initial}}=\frac{1585\times10^{3}}{1200}=1320\ \text{mm}^{2}$$ Using 13 mm seven-wire strand ($A=98.7$ mm$^2$ each), $1320/98.7=13.4$, so provide 14 strands, $A_{ps}=1382$ mm$^2$, in two grouted ducts of seven. Then $$P_i=1382(1200)\times10^{-3}=1658\ \text{kN},\qquad P_e=1382(1200-240)\times10^{-3}=1327\ \text{kN}$$ The jacking stress $1200/1750=0.686f_{pu}$ is inside the usual $0.70f_{pu}$ ceiling for post-tensioning.
2. Build the cable zone. Rearranging the two conditions gives, at every section, an allowable band for the tendon centroid (positive $e$ = below the section centroid): $$\frac{M_s}{P_e}-k\ \le\ e\ \le\ k+\frac{M_{sw}}{P_i}$$ The lower limit keeps the bottom fibre closed in service; the upper limit keeps the top fibre closed at transfer.
3. Tabulate the zone and choose ordinates.

   $x$ (m)	$M_s$ (kN·m)	$M_{sw}$ (kN·m)	$e_{min}$ (mm)	$e_{max}$ (mm)	$e$ adopted (mm)
   −3.0 (tip)	0	0	−200	+200	0
   0.0 (S1)	−244.8	−64.8	−385	+161	0
   2.0	+463.0	+117.3	+149	+271	+200
   4.0	+613.1	+241.7	+262	+346	+290
   6.0	+705.7	+308.6	+332	+386	+345
   8.0	+740.7	+317.8	+358	+392	+375
   10.0	+718.1	+269.5	+341	+363	+360
   12.0	+637.8	+163.5	+281	+299	+290
   14.0 (S2)	0	0	−200	+200	0

   Every adopted ordinate lies inside its band, and the maximum eccentricity of 375 mm is inside the $h/2-150=450$ mm allowed by cover and duct spacing.
4. Verify the whole girder, not just the table. Sweeping the adopted profile at 10 mm intervals along all 17 m gives a minimum top-fibre stress at transfer of $+0.04$ MPa and a minimum bottom-fibre stress in service of $+0.08$ MPa — both compressive, so $$\boxed{\text{no tension occurs anywhere, at either stage}}$$ The peak compressions are 5.5 MPa at transfer (limit $0.6f_{ci}=21$ MPa) and 4.3 MPa in service (limit $0.45f'_c=22.5$ MPa), so the section is governed entirely by the no-tension requirement rather than by crushing.
5. Describe the profile. The tendon leaves the anchorage at the free tip on the section centroid, stays on the centroid at the left support (where the moment is hogging), drapes as a smooth parabola to 375 mm below the centroid at $x=8$ m, and returns to the centroid at the right-hand anchorage. Because the girder is post-tensioned, this profile is set by the duct positions in the falsework and the strands are stressed after the concrete has reached $f_{ci}=35$ MPa, then grouted; stressing from both ends halves the friction loss along the 17 m of curved duct.

Checking the assumption — ultimate strength and the cantilever

1. Ultimate flexural strength. The critical sagging section is at $x=8.39$ m, where $$M_f=1.25M_{D}+1.5M_{L}=1.25(316)+1.5(425)=1033\ \text{kN}\cdot\text{m}$$ With $d_p=600+372=972$ mm, a stress in the bonded strand at ultimate of about 1550 MPa, $\alpha_1=0.85-0.0015(50)=0.775$ and $\phi_p=0.90$, $$T_p=0.90(1382)(1550)\times10^{-3}=1928\ \text{kN},\qquad a=\frac{T_p}{\alpha_1\phi_cf'_cb}=153\ \text{mm}$$ $$M_r=1928\left(972-\frac{153}{2}\right)\times10^{-3}=1726\ \text{kN}\cdot\text{m}\ \ge\ 1033\ \text{kN}\cdot\text{m}$$ The no-tension serviceability requirement, not strength, governs the design — which is typical of a fully prestressed member and is why the question asks for the section on a stress basis.
2. The cantilever at ultimate. Over the left support the factored hogging moment is $M_f=1.25(-64.8)+1.5(-180)=-351$ kN·m, and the tendon is on the centroid there, so ordinary reinforcement must carry it: $$A_s=\frac{351\times10^{6}}{0.85(400)(0.9\times1100)}=1043\ \text{mm}^{2}\ \Rightarrow\ \textbf{4 – 20M}\ (1200\ \text{mm}^{2})\ \text{in the top}$$ extended a development length past the point of contraflexure.
3. Anchorage zone. Provide bursting reinforcement behind each anchorage — a spiral or a set of closed 10M ties over a length equal to the section depth — designed for a bursting force of roughly $0.25P_i=415$ kN, and check bearing under the anchor plates against $f_{ci}=35$ MPa. This end-block detail is where post-tensioned girders most often crack.