Prestressed concrete girder with no tension in the cross-section: 16-Civ-B2 Advanced Structural Design, December 2017, Q5

Question 5: Prestressed concrete girder with no tension in the cross-section (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. From Figure 4: a girder pinned at A, on a roller at B 18 m away, and cantilevering a further 2 m to a free end at C. Point loads of 500 kN act at 6 m and 12 m from A, and 80 kN acts at the tip C. Page 1 gives the prestressed-concrete data.

Quantity	Symbol	Value
Span A–B / overhang B–C	—	18 m / 2 m
Point loads at 6 m and 12 m	$P$	500 kN each
Load at the tip C	$P_C$	80 kN
Concrete strength at transfer	$f_{ci}$	35 MPa
Concrete strength in service	$f'_c$	50 MPa
Initial tendon stress	$f_{pi}$	1200 MPa
Losses	$\Delta f$	240 MPa, so $f_{se} = 960$ MPa
Tendon ultimate / yield	$f_{ult}$ / $f_{py}$	1750 / 1450 MPa

Find. A concrete section, a prestressing force and a cable profile such that no fibre anywhere on the member goes into tension, either at transfer or in service, and the ultimate flexural resistance is adequate.

Figure 4: 18 m span with a 2 m overhang. The parabolic tendon is anchored on the centroid at A and at B and reaches e = 500 mm at mid-span; it runs along the centroid over the cantilever.

Approach. Work out the service and self-weight moment diagrams, use the "no tension" conditions to write the two limits on the tendon eccentricity at each section, size the section so that a permissible window exists, choose the strand count that opens it comfortably, then draw a parabola inside the window and confirm every fibre stress and the ultimate moment.

Reactions and the applied moment diagram. Taking moments about A for the point loads, $$R_B = \frac{500(6) + 500(12) + 80(20)}{18} = 588.9\ \text{kN},\qquad R_A = 1080 - 588.9 = 491.1\ \text{kN}.$$ The applied sagging moment peaks between the two 500 kN loads, and the tip load puts $-160$ kN·m of hogging at B.
Choose a trial section and add self weight. A T-section with a $1200 \times 250$ flange on a 400 mm web, 1800 mm deep, gives $A = 920\,000$ mm$^2$, $\bar y_t = 731.5$ mm, $I = 2.895\times10^{11}$ mm$^4$, hence $$\begin{gathered}Z_t = 3.957\times10^{8}, \\ Z_b = 2.709\times10^{8}\ \text{mm}^3, \\ k_t = \frac{Z_b}{A} = 294.5, \\ k_b = \frac{Z_t}{A} = 430.1\ \text{mm}.\end{gathered}$$ Its self weight is 22.08 kN/m. Superposing, the total service moment peaks at $M_T = 3795$ kN·m at $x = 8.49$ m, where the self-weight share is $M_0 = 870.5$ kN·m; at B the total is $-204.2$ kN·m.
Check the section is big enough. With $\eta = f_{se}/f_{pi} = 960/1200 = 0.80$ and zero permissible tension, the bottom fibre requires $$Z_b \;\ge\; \frac{M_T - \eta M_0}{\eta(0.6f_{ci})} = \frac{3795 - 0.8(870.5)}{0.8(21)}\times10^{3} = 1.844\times10^{8}\ \text{mm}^3,$$ comfortably below the $2.709\times10^{8}$ mm$^3$ provided.
The permissible eccentricity window. No tension at the service bottom fibre and no tension at the transfer top fibre bound $e$ from both sides: $$\underbrace{\frac{M_T}{P_e} - k_t}_{e_{min}} \;\le\; e \;\le\; \underbrace{k_b + \frac{M_0}{P_i}}_{e_{max}} .$$ A window exists only if $P_i \ge A(M_T/\eta - M_0)/(Z_t + Z_b)$, which for this section is 4282 kN — the two limits then touch, so a workable design needs more.
Choose the tendon. Take 38 – 15.2 mm strands, $A_{ps} = 5320$ mm$^2$: $$P_i = 5320(1200)/10^3 = 6384\ \text{kN},\qquad P_e = 5320(960)/10^3 = 5107\ \text{kN}.$$ At $x = 6$ m this opens the window to $$435.3\ \text{mm} \;\le\; e \;\le\; 552.3\ \text{mm},$$ and at mid-span to $448$–$567$ mm.
Draw the profile inside the window. Anchor the cable on the centroid at A, where $M = 0$ and the kern rule alone allows $-294.5 \le e \le 430.1$ mm, and again on the centroid at B, where the hogging from the cantilever pulls the upper limit down to 423 mm. Between them use the parabola $$e(x) = \frac{4e_m x(L - x)}{L^2},\qquad e_m = 500\ \text{mm},$$ which gives $e = 444.4$ mm at both 6 m and 12 m and 500 mm at mid-span. Running the check at 0.1 m intervals over the whole 20 m gives $$\boxed{\text{0 violations of the permissible window}}$$ — the cable is admissible everywhere. Over the cantilever it continues straight along the centroid, which satisfies the kern condition at C where $M = 0$.
Verify the extreme-fibre stresses. At mid-span, with $P_i$ acting on the girder self weight alone, $$f_{top} = \frac{P_i}{A} - \frac{P_i e}{Z_t} + \frac{M_0}{Z_t} = 6.94 - 8.07 + 2.20 = 1.08\ \text{MPa}\ \ (\ge 0),$$ $$f_{bot} = \frac{P_i}{A} + \frac{P_i e}{Z_b} - \frac{M_0}{Z_b} = 6.94 + 11.78 - 3.22 = 15.50\ \text{MPa} \;\le\; 0.6f_{ci} = 21\ \text{MPa}.$$ In service, with $P_e$ and the full moment, the same fibres give 8.68 MPa compression at the top ($\le 0.45f'_c = 22.5$ MPa) and 0.98 MPa compression at the bottom. At B both fibres stay in compression (5.04 MPa top, 6.30 MPa bottom) despite the cantilever hogging.
Ultimate flexural resistance. At mid-span $d_p = \bar y_t + e = 1231.5$ mm. With $\alpha_1 = 0.85 - 0.0015(50) = 0.775$ and $\phi_p = 0.90$, $$\begin{gathered}T_u = \phi_p A_{ps} f_{ult} = 0.90(5320)(1750)/10^3 = 8379\ \text{kN}, \\ a = \frac{T_u}{\alpha_1\phi_c f'_c b} = 277.2\ \text{mm},\end{gathered}$$ $$M_r = T_u\!\left(d_p - \frac{a}{2}\right) = 9158\ \text{kN}\cdot\text{m} \;\gt\; M_f = 1.5(2920) + 1.25(872) = 5470\ \text{kN}\cdot\text{m} \quad\checkmark$$ As is usual for a no-tension design, serviceability governs and the ultimate check passes with a factor of 1.67 in hand.

Mid-span section: 1200 x 250 flange on a 400 mm web, 1800 mm deep, with 38 - 15.2 mm strands at e = 500 mm below the centroid.

Quantity	Value
Section	T: 1200 × 250 flange, 400 mm web, 1800 mm deep
$A$ / $Z_t$ / $Z_b$	$0.92\times10^{6}$ mm$^2$ / $3.96\times10^{8}$ / $2.71\times10^{8}$ mm$^3$
Kern distances $k_t$ / $k_b$	294.5 / 430.1 mm
Maximum service moment	3795 kN·m at $x = 8.49$ m
Prestressing steel	38 – 15.2 mm strands ($A_{ps} = 5320$ mm$^2$)
$P_i$ / $P_e$	6384 kN / 5107 kN
Cable profile	parabola, $e = 0$ at A and B, $e_m = 500$ mm at mid-span
Transfer stresses (mid-span)	1.08 MPa top, 15.50 MPa bottom
Service stresses (mid-span)	8.68 MPa top, 0.98 MPa bottom — no tension
$M_r$ / $M_f$	9158 / 5470 kN·m

Topic: Full prestressing (no tension permitted) of a simply supported girder with an overhang, and the permissible-eccentricity window.

Key relations: $e_{min} = M_T/P_e - Z_b/A$; $e_{max} = Z_t/A + M_0/P_i$; $Z_b \ge (M_T - \eta M_0)/(\eta f_{ci,all})$ with $\eta = f_{se}/f_{pi}$; extreme-fibre stress $f = P/A \pm Pe/Z \mp M/Z$; $M_r = \phi_p A_{ps} f_{ult}(d_p - a/2)$.

Why this works: A no-tension design is really a pair of inequalities that move in opposite directions as the prestress changes. Raising $P$ lowers $e_{min}$ (less eccentricity is needed to keep the bottom in compression under full load) but also lowers $e_{max}$ (more prestress means the top fibre cracks sooner at transfer, when only self weight opposes it). The section has to be large enough that the two curves still cross; if they do not, no prestressing force whatever will work and the section must grow. Plotting the two limits along the member and fitting a parabola between them is the whole design.

Common pitfall: Checking only the mid-span service condition. The transfer condition at the ends is what usually kills a profile: with $M_0 = 0$ at a support, $e$ cannot exceed the kern distance $k_b$, which is why the tendon must be brought back to (or above) the centroid at the anchorages. A second trap here is the 2 m cantilever — it hogs at B, so the top-fibre check at B is not automatically satisfied by a mid-span design.

Source: CSA A23.3:19 Cl 18.3 and 18.6; Collins & Mitchell, Prestressed Concrete Structures, Ch. 4 (service-load design); Nawy, Prestressed Concrete, Ch. 4.

Question 5 of 7: Prestressed concrete girder with no tension in the cross-section

Question 5: Prestressed concrete girder with no tension in the cross-section (12 + 6 + 2 marks)