16-Civ-B2 Advanced Structural Design · December 2014

Question 2 of 7: Post-tensioned T-beam with cantilever

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2014 — 98-Civ-B2 Advanced Structural Design, 3 hours, closed book (handbooks and textbooks permitted). Seven design questions of equal value; any five constitute a complete paper.

Reference texts. CSA S16:19 Design of Steel Structures (Clause 13 members, Clause 14 plate girders, Clause 17 composite, Clause 21 plastic design); CISC Handbook of Steel Construction; CSA A23.3:19 Design of Concrete Structures (Clause 10 flexure and columns, Clause 11 shear, Clause 18 prestressed concrete); National Building Code of Canada 2020 for load combinations; Hibbeler, Structural Analysis, for the frame and continuous-beam analysis.

Design data printed on page 1. Concrete $f_c' = 30\ \text{MPa}$; structural steel $F_y = 350\ \text{MPa}$; reinforcing steel $f_y = 400\ \text{MPa}$. Prestressed concrete: $f_{ci}' = 35\ \text{MPa}$ at transfer, $f_c' = 50\ \text{MPa}$, modular ratio $n = 6$, $f_{ult} = 1750\ \text{MPa}$, $f_y = 1450\ \text{MPa}$, $f_{initial} = 1200\ \text{MPa}$, losses $= 240\ \text{MPa}$. Design in SI. All loads shown are unfactored.

Check — load factors. Page 1 states only that the loads shown are unfactored; it gives no dead/live split. Throughout this paper the figure loads are treated as a single variable action and factored by 1.5, while self weight introduced by the solver (the reinforced-concrete frame, the prestressed girder, the bridge deck) is factored by 1.25, per NBCC 2020 Case 2. Mechanisms, section classifications and interaction ratios are unaffected by this choice — only the magnitudes are. Serviceability checks (the no-tension prestress design of Q2 and the soil bearing of Q5) use unfactored loads, as they must.

Question 2: Post-tensioned T-beam with cantilever (10 + 5 + 2 + 3 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.

Quantity	Value
Simple span AB / cantilever BC	17.0 m / 3.0 m
Point loads on AB	400 kN at 6 m, 400 kN at 11 m from A
Point load at C	60 kN
Concrete	$f_c' = 50\ \text{MPa}$; $f_{ci}' = 35\ \text{MPa}$ at transfer
Strand stresses	$f_{initial} = 1200\ \text{MPa}$; losses 240 MPa; $f_{eff} = 960\ \text{MPa}$
Loss ratio	$\eta = 960/1200 = 0.80$
Permissible stresses	no tension at either stage; $0.6f_{ci}' = 21\ \text{MPa}$ at transfer; $0.45f_c' = 22.5\ \text{MPa}$ in service

Find. A uniform T-section and a strand area and drape such that the extreme-fibre stress is compressive everywhere at transfer and in service, and the long-term vertical movement of the cantilever tip C.

Figure 2.1 — Beam ABC with the adopted parabolic tendon profile (anchored on the centroid at A and B, straight through the cantilever), and the designed T-section.

Approach. Compute the service and transfer moment envelopes, size the T-section against the minimum section-modulus inequalities that follow from combining the transfer and service stress limits, then find the prestressing force and eccentricity from the Magnel pair of no-tension constraints at the governing section, and finally integrate $Mm/EI$ with a unit load at C to get the tip movement, multiplying by the creep factor.

Part (a) — find the moment envelope from the applied loads. Taking moments about B for the whole beam, and remembering that the 60 kN at C sits beyond B and so relieves the far reaction, $$R_A = \frac{400(11) + 400(6) - 60(3)}{17} = \frac{6620}{17} = 389.4\ \text{kN}, \qquad R_B = 860 - 389.4 = 470.6\ \text{kN}$$ The hogging moment at B is $-60(3) = -180\ \text{kN}\cdot\text{m}$.
Add the self weight of a trial section. Try a T-section with a 900 × 180 mm flange on a 450 mm web, overall depth 1700 mm. Then $A = 846\,000\ \text{mm}^2$ and $w_{sw} = 20.30\ \text{kN/m}$, whose own reaction at A is 167.2 kN. Superposing, the total service moment peaks a little away from the load point: $$\boxed{M_s = 3004\ \text{kN}\cdot\text{m} \text{ at } x = 7.71\ \text{m}}, \qquad M_t = M_{sw} = 686\ \text{kN}\cdot\text{m} \text{ there}$$ and at B the total service moment is $-271\ \text{kN}\cdot\text{m}$. The transfer condition uses $M_t$ alone, because only self weight acts when the strands are stressed.
Compute the gross section properties. From first moments about the top face, $$\bar{y}_t = 777.2\ \text{mm}, \qquad \bar{y}_b = 922.8\ \text{mm}, \qquad I_g = 226.8 \times 10^{9}\ \text{mm}^4$$ $$Z_t = 291.8\times10^{6}\ \text{mm}^3, \quad Z_b = 245.7\times10^{6}\ \text{mm}^3, \quad k_t = \frac{Z_b}{A} = 290.5\ \text{mm}, \quad k_b = \frac{Z_t}{A} = 344.9\ \text{mm}$$ The kern distances $k_t$ and $k_b$ are what turn the stress limits into a force-eccentricity pair in the next step.
Confirm the section is large enough before choosing the force. Eliminating $P$ between the transfer and service stress limits gives the two minimum section moduli $$Z_t \ge \frac{M_s - \eta M_t}{f_{cs} - \eta f_{ti}} = \frac{(3004 - 0.8 \times 686)\times10^{6}}{22.5 - 0} = 109.1\times10^{6}\ \text{mm}^3$$ $$Z_b \ge \frac{M_s - \eta M_t}{\eta f_{ci} - f_{ts}} = \frac{2455\times10^{6}}{0.8(21) - 0} = 146.2\times10^{6}\ \text{mm}^3$$ Both are met with a wide margin, so a feasible force-eccentricity pair exists and the trial section is retained. Had either inequality failed, no prestress whatever could have made the section work.
Solve for the force and eccentricity at the governing section. No tension at the bottom in service and none at the top at transfer give $$\eta P_i\left(\frac{1}{A} + \frac{e}{Z_b}\right) \ge \frac{M_s}{Z_b} \quad\text{and}\quad P_i\left(\frac{e}{Z_t} - \frac{1}{A}\right) \le \frac{M_t}{Z_t}$$ Solving the pair as equalities at $x = 7.71\ \text{m}$ gives $e \le 487\ \text{mm}$ and $P_i \approx 4830\ \text{kN}$; carrying the same check along the whole beam with a parabolic drape raises the requirement slightly to $P_i = 5110\ \text{kN}$. Hence $$A_{ps} = \frac{P_i}{f_{initial}} = \frac{5110\times10^{3}}{1200} = 4258\ \text{mm}^2$$
Select the strands. Using 15.2 mm seven-wire strand of 140 mm² nominal area, $$n = \frac{4258}{140} = 30.4 \;\rightarrow\; \boxed{31 \text{ strands}, \; A_{ps} = 4340\ \text{mm}^2}$$ $$P_i = 4340(1200) = 5208\ \text{kN}, \qquad P_e = 4340(960) = 4166\ \text{kN}$$ Arrange them in four ducts of 8-8-8-7 strands, stressed from both ends.
Fix the profile and verify the stresses along the beam. Adopt a parabola anchored on the centroid at A and at B, $$e(x) = 470\,\frac{4x(17-x)}{17^{2}}\ \text{mm} \quad (0 \le x \le 17\ \text{m}), \qquad e = 0 \text{ over the cantilever}$$ so $e = 429$ mm at both 6 m and 11 m and $e = 470$ mm at mid-span, putting the duct 453 mm above the soffit. Checking all four stress limits at 1 m intervals: $$\sigma_{top,transfer} \ge 0.12\ \text{MPa}, \quad \sigma_{bot,transfer} \le 13.3 < 21\ \text{MPa}$$ $$\sigma_{top,service} \le 9.0 < 22.5\ \text{MPa}, \quad \sigma_{bot,service} \ge 0.10\ \text{MPa}$$ Every fibre is in compression at both stages, so the no-tension requirement is satisfied. Anchoring on the centroid at A and B also keeps the end blocks simple and puts zero eccentricity where the moment is zero, which is exactly the kern requirement at a support.
Part (b) — set up the tip deflection. The beam is statically determinate (hinge, roller, free end), so prestress induces no secondary moments and the tendon moment is simply $M_p(x) = -P_e\,e(x)$. Applying a unit downward load at C gives $$m(x) = -\tfrac{3}{17}x \;(0 \le x \le 17\ \text{m}), \qquad m(x) = -(20 - x) \;(17 \le x \le 20\ \text{m})$$ and $\Delta_C = \int \frac{M\,m}{E_cI_g}\,\mathrm{d}x$ with $E_c = 4500\sqrt{50} = 31\,820\ \text{MPa}$.
Evaluate the immediate deflection. Integrating numerically over the full 20 m, $$\Delta_{loads} = -6.47\ \text{mm}, \qquad \Delta_{prestress} = +4.61\ \text{mm}, \qquad \Delta_{imm} = \boxed{-1.86\ \text{mm}\ (\text{upward})}$$ The sign is worth reading rather than assuming: the heavily loaded span sags, the tangent at B rotates, and that rotation lifts the cantilever tip far more than the 60 kN tip load pushes it down. The prestress hogs the span and therefore acts in the opposite sense at C.
Apply the long-term multiplier. Taking a creep coefficient $\varphi_{cc} = 2.0$ for post-tensioned concrete stressed at 28 days, and treating the applied loads as sustained, $$\Delta_{LT} = (1 + \varphi_{cc})\,\Delta_{imm} = 3.0 \times (-1.86) = \boxed{-5.6\ \text{mm}\ (\text{upward at C})}$$ against a cantilever limit of $L/180 = 3000/180 = 16.7\ \text{mm}$. The tip rises rather than droops, which is the usual and benign outcome for a short cantilever off a heavily prestressed span, but it must be allowed for in the expansion joint and the deck profile.

Result	Value
Reactions (service)	$R_A = 389.4$ kN, $R_B = 470.6$ kN
Governing service moment	3004 kN·m at $x = 7.71$ m
Moment at transfer (self weight)	686 kN·m
T-section	900 × 180 flange, 450 mm web, 1700 mm deep
Gross properties	$A = 846\times10^{3}$ mm², $I_g = 226.8\times10^{9}$ mm⁴
Strand area	31 × 15.2 mm strand = 4340 mm²
Prestress force	$P_i = 5208$ kN, $P_e = 4166$ kN
Profile	parabola, $e = 0$ at A and B, $e = 470$ mm at mid-span, straight over BC
Extreme stresses (transfer / service)	0.12 to 13.3 MPa / 0.10 to 9.0 MPa — all compressive
Immediate deflection at C	1.86 mm upward
Long-term deflection at C	5.6 mm upward (limit 16.7 mm)

Topic. Service-load design of a post-tensioned T-beam for zero tension, and long-term deflection of a cantilever off a prestressed span.

Key relations. Extreme-fibre stress $\sigma = P/A \pm Pe/Z \mp M/Z$; minimum section moduli $Z_t \ge (M_s - \eta M_t)/(f_{cs} - \eta f_{ti})$ and $Z_b \ge (M_s - \eta M_t)/(\eta f_{ci} - f_{ts})$; kern distances $k_t = Z_b/A$, $k_b = Z_t/A$; unit-load theorem $\Delta = \int Mm/EI\,\mathrm{d}x$; long-term multiplier $(1 + \varphi_{cc})$.

Why this works. A no-tension design is governed by two conditions acting at opposite times and opposite faces: at transfer the force is largest and the moment smallest, so the top fibre wants to crack; in service the force has shed its losses and the moment is largest, so the bottom fibre wants to crack. Eliminating the unknown force between those two inequalities produces section-modulus requirements that depend only on the moments and the loss ratio, which is why they can be tested before any strand is selected. Once the section passes, the same two inequalities become a linear pair in $P$ and $Pe$ — the Magnel diagram — whose feasible region shrinks as the section gets smaller and vanishes at exactly the point where the modulus inequalities fail.

Common pitfall. The first is designing at the largest applied-load moment rather than the largest total moment: adding self weight moves the critical section away from the point load, here from 6.0 m to 7.71 m. The second is assuming a cantilever tip must deflect downward. For a stiff, heavily loaded back span the rotation at the support dominates and the tip rises; getting the sign wrong reverses the camber allowance built into the formwork.

Source. CSA A23.3:19 Clause 18 (prestressed concrete) and Clause 9.8 (deflections); Collins and Mitchell, Prestressed Concrete Structures, Ch. 4; Nawy, Prestressed Concrete, Ch. 4 for the minimum section-modulus equations.

← Question 1 Next: Question 3 →