Question 2 of 7: Post-Tensioned Prestressed Concrete Girder

Question 2: Post-Tensioned Prestressed Concrete Girder (10 + 5 + 5 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.

Find. T-section dimensions, the effective prestressing force and its eccentricity, the strand area, and a cable profile that keeps every section free of tension.

Approach. Compute the service moments, adopt a trial T-section, then solve the two “no-tension” equalities — bottom fibre in service and top fibre at transfer — simultaneously for the force and its eccentricity, and finish with the cable-limit zone and an ultimate check.

1. Establish the service moments. The 2 m overhang hangs a hogging moment on the support: $M_B = -60(2) = -120$ kN·m. Over the simple 12 m span the 400 kN load gives $PL/4 = 1200$ kN·m, from which the support moment subtracts linearly, contributing $-120/2 = -60$ kN·m at mid-span. Adding self weight (below) at $17w$ kN·m, $$\boxed{M_{\text{mid}} = 1366.4\ \text{kN}\cdot\text{m}, \qquad M_B = -146.6\ \text{kN}\cdot\text{m}, \qquad M_o = 226.4\ \text{kN}\cdot\text{m}}$$ where $M_o$ is the self-weight moment acting at transfer.
2. Adopt a trial T-section and get its properties. Try a 1200 × 200 flange over a 350 × 900 web, total depth 1100 mm. Then $A = 555\,000$ mm², $w_{sw} = 13.32$ kN/m, and taking first moments about the soffit, $$y_b = 687.8\ \text{mm},\quad I = 63.268\times10^9\ \text{mm}^4,\quad Z_b = 91.98\times10^6,\quad Z_t = 153.50\times10^6\ \text{mm}^3$$
3. Write the two no-tension conditions. With compression positive and e measured downwards from the centroid, the bottom fibre in service and the top fibre at transfer are the two critical faces: $$\text{service, soffit:}\quad \frac{P}{A} + \frac{Pe}{Z_b} - \frac{M_s}{Z_b} \ge 0 \qquad\qquad \text{transfer, top:}\quad \frac{P_i}{A} - \frac{P_i e}{Z_t} + \frac{M_o}{Z_t} \ge 0$$ The second rearranges to a ceiling on the eccentricity, $e \le Z_t/A + M_o/P_i$, where $Z_t/A = 276.6$ mm is the upper kern distance. Because both are equalities at the economical design, they can be solved together.
4. Solve for the force and the eccentricity. Substituting $e = 276.6 + 0.8M_o/P$ (the 0.8 converts $P_i$ to $P$ at 20 % loss) into the service condition gives a single linear equation in P: $$P\left(\frac{1}{A} + \frac{276.6}{Z_b}\right) + \frac{0.8M_o}{Z_b} = \frac{M_s}{Z_b} \quad\Longrightarrow\quad \boxed{P = 2680\ \text{kN}, \qquad e = 344\ \text{mm}}$$ The tendon therefore sits 344 mm below the centroid, i.e. 344 mm above the soffit — well inside the 900 mm web with room for ducts and cover.
5. Confirm the stresses. With $P_i = P/0.8 = 3350$ kN the four corner stresses are as tabulated. Both no-tension limits are met exactly, by construction, and the compressive limits have ample margin.

   Stage	Top fibre	Soffit	Limit
   Transfer (Pi + self weight)	0.00 MPa	16.11 MPa	0.6f'ci = 21.0 MPa
   Service (Pe + full load)	7.72 MPa	0.00 MPa	0.45f'c = 22.5 MPa

6. Size the strands. The effective stress after losses is $f_{pe} = 1200 - 240 = 960$ MPa, so $$A_{ps} = \frac{P}{f_{pe}} = \frac{2680\times10^3}{960} = 2791\ \text{mm}^2$$ Use 20 – 15.2 mm seven-wire strands (140 mm² each, 2800 mm² total) in four ducts of five. At jacking this delivers $P_i = 3360$ kN and after losses $P_e = 2688$ kN, both marginally above the requirement. The jacking stress of 1200 MPa is 0.69f_ult, inside the usual 0.70 ceiling.
7. Fix the cable profile from the limit zone. At any section the same two conditions bracket the eccentricity, $-Z_b/A + M/P \le e \le Z_t/A + M/P$. At the free end A and at C, where $M = 0$, the zone is $-166 \le e \le +277$ mm; at B, where the hogging moment is $-146.6$ kN·m, it is $-220 \le e \le +222$ mm. Anchoring the tendon at the centroid (e = 0) at A, B and C therefore satisfies every requirement, and the cable is draped as a symmetric parabola from e = 0 at B down to e = 344 mm at mid-span and back to e = 0 at C, running straight at the centroid over the 2 m overhang.
8. Check the ultimate limit state. The factored moment is $M_f = 1.5(1140) + 1.25(226.4) = 1993$ kN·m. With $d_p = 756$ mm and $k_p = 2(1.04 - f_{py}/f_{pu}) = 0.423$, Clause 18.6.2 gives $f_{pr} = 1581$ MPa. The stress block $a = 132$ mm stays inside the 200 mm flange, so the section is effectively rectangular, and $$M_r = \phi_p A_{ps} f_{pr}\left(d_p - \frac{a}{2}\right) = \boxed{2751\ \text{kN}\cdot\text{m}} \;\ge\; 1993\ \text{kN}\cdot\text{m}$$ As expected for a “no tension” design, serviceability governs and strength is comfortable; a small amount of untensioned bar (say 4 – 20M top over B) handles the hogging region and handling stresses.

Final Results