Question 1 of 7: Post-tensioned girder — section, strand area and cable profile (12 + 6 + 2 marks)

Question 1: Post-tensioned girder — section, strand area and cable profile (12 + 6 + 2 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. A post-tensioned girder simply supported over 16 m with a 2 m overhang, carrying three 350 kN loads at the quarter points of the span and an 80 kN load at the free end, built in 50 MPa concrete stressed at 35 MPa.

Find. A rectangular cross-section proportioned so that the extreme-fibre tensile stresses reach, but do not exceed, the permissible values at both transfer and full service load; then the strand area and the cable profile.

Approach. Size the section from the classical permissible-stress requirement on the bottom section modulus, then solve the transfer top-fibre and service bottom-fibre limits simultaneously for the prestress force and its eccentricity, so that both tension limits are reached exactly — which is what “maximum permitted tension” asks for.

1. Find the service moments from statics. Taking moments about A, the reaction at B follows from the three span loads and the overhang load: $$R_B=\frac{350(4+8+12)+80(18)}{16}=\frac{8400+1440}{16}=615\ \text{kN},\qquad R_A=3(350)+80-615=515\ \text{kN}$$ The largest sagging moment is at mid-span, and the overhang produces a small hogging moment over B: $$M_{8}=515(8)-350(4)=\boxed{2720\ \text{kN}\cdot\text{m}},\qquad M_{B}=-80(2)=-160\ \text{kN}\cdot\text{m}$$
2. Write down the permissible concrete stresses. For an uncracked (Class U) member, CSA A23.3 Clause 18.3 gives, with compression taken as positive, $$f_{ti}=0.5\sqrt{f_{ci}}=0.5\sqrt{35}=2.96\ \text{MPa},\qquad f_{ci,\text{all}}=0.60f_{ci}=21.0\ \text{MPa}$$ $$f_{ts}=0.5\sqrt{f'_c}=0.5\sqrt{50}=3.54\ \text{MPa},\qquad f_{cs,\text{all}}=0.45f'_c=22.5\ \text{MPa}$$ The prestress ratio after losses is $\eta=f_{se}/f_{pi}=960/1200=0.80$.
3. Size the section from the bottom-fibre requirement. Combining the transfer bottom-fibre compression limit with the service bottom-fibre tension limit eliminates the prestress force and leaves a pure section requirement, $$Z_b\ \ge\ \frac{M_T-\eta M_0}{\eta f_{ci,\text{all}}+f_{ts}}$$ Trying a 500 mm wide by 1500 mm deep rectangle, the self weight is $w=0.5(1.5)(24)=18.0$ kN/m, which on this span with its overhang gives $M_0=558$ kN·m and hence $M_T=2720+558=3278$ kN·m. Then $$Z_{b,\text{req}}=\frac{(3278-0.8\times558)\times10^6}{0.8(21.0)+3.54} =139.2\times10^6\ \text{mm}^3\ \le\ Z=\frac{500(1500)^2}{6} =\boxed{187.5\times10^6\ \text{mm}^3}$$ so the trial section has ample section modulus and the depth is governed by the eccentricity that can actually be accommodated, checked next.
4. Solve the two tension limits simultaneously. Writing $X=P_e/A$ and $Y=P_ee/Z$, the service bottom fibre and the transfer top fibre give $$X+Y=\frac{M_T}{Z}-f_{ts}=17.48-3.54=13.94\ \text{MPa}$$ $$X-Y=\eta\left(-f_{ti}-\frac{M_0}{Z}\right)=0.8(-2.96-2.98)=-4.75\ \text{MPa}$$ whence $X=4.60$ MPa and $Y=9.35$ MPa, i.e. $$P_e=4.60(750\,000)=\boxed{3450\ \text{kN}},\qquad e=\frac{9.35(187.5\times10^6)}{3.45\times10^6}=\boxed{508\ \text{mm}}$$
5. Convert the force to strands and re-close the eccentricity. The required strand area is $A_{ps}=P_e/f_{se}=3\,450\,000/960=3594$ mm². Using 15.2 mm seven-wire strand at 140 mm² each, $$n=\frac{3594}{140}=25.7\ \rightarrow\ \boxed{26\ \text{strands},\ A_{ps}=3640\ \text{mm}^2}$$ $$P_i=3640(1200)=4368\ \text{kN},\qquad P_e=3640(960)=3494\ \text{kN}$$ Because slightly more strand is supplied than required, the eccentricity must be trimmed so the transfer top fibre still just reaches its limit; solving that one equation gives $e=\boxed{504\ \text{mm}}$ at mid-span, i.e. the cable centroid sits 246 mm above the soffit — ample room for five 100 mm ducts.
6. Check all four extreme-fibre stresses. With $A=750\times10^3$ mm² and $Z=187.5\times10^6$ mm³, $$f=\frac{P}{A}\mp\frac{Pe}{Z}\pm\frac{M}{Z}$$ At transfer ($P_i$, $M_0$ only): top $=-2.94$ MPa (tension, limit 2.96) and bottom $=+14.59$ MPa (limit 21.0). Under full service load ($P_e$, $M_T$): top $=+12.75$ MPa (limit 22.5) and bottom $=-3.43$ MPa (tension, limit 3.54). Both tension limits are reached to within 1 per cent and neither is exceeded, so the section is a genuine maximum-permitted-tension design.
7. Confirm the ultimate flexural resistance. With the factoring convention stated at the head of this paper, $M_f=1.5(2720)+1.25(558)=4777.5$ kN·m. Taking the bonded strand at yield ($f_{pr}=f_{py}=1450$ MPa) and $\alpha_1=0.775$, $\beta_1=0.845$ for 50 MPa concrete, $$T_r=\phi_pA_{ps}f_{py}=0.9(3640)(1450)=4750\ \text{kN},\qquad a=\frac{T_r}{\alpha_1\phi_cf'_cb}=377\ \text{mm}$$ $$M_r=T_r\!\left(d_p-\frac{a}{2}\right)=4750\!\left(1254-188.6\right) =\boxed{5061\ \text{kN}\cdot\text{m}}\ \ge\ 4777.5\ \text{kN}\cdot\text{m}$$ with $c/d_p=0.356$, comfortably ductile.
8. Check the minimum-reinforcement rule. The cracking moment is $$M_{cr}=\left(f_r+\frac{P_e}{A}+\frac{P_ee}{Z}\right)Z =(4.24+4.66+9.41)(187.5\times10^6)=3430\ \text{kN}\cdot\text{m}$$ and $M_r=5061 \ge 1.2M_{cr}=4116$ kN·m, so Clause 18.8 is satisfied without supplementary mild steel.
9. Set the cable profile inside the permissible zone. At any section the eccentricity must satisfy $$\frac{M_T-\left(f_{ts}+P_e/A\right)Z}{P_e}\ \le\ e\ \le\ k_b+\frac{M_0+f_{ti}Z}{P_i}$$ At the third points the zone is [271, 474] mm at 4 m and [243, 470] mm at 12 m, while a parabola anchored on the centroid at A and B passes through 378 mm at both — inside the zone with margin at every section. Anchor the cable on the centroid at A, run a parabola to $e=504$ mm at mid-span and back to the centroid at B, then straight through the overhang to the anchorage at C, where the small hogging moment leaves both fibres in compression with $e=0$.