07-Str-A5 · May 2014

Question 3 of 7: Prestressed concrete beam — section, strands and long-term deflection

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

Notes on this paper

Paper format. 07-Str-A5 Advanced Structural Design, National Exams, May 2014 — 3 hours, closed book (handbooks and textbooks permitted, no notes on them; Casio or Sharp approved calculator). Seven questions of equal value; any five constitute a complete paper and only the first five presented are marked. All seven are solved here, because the set is intended as a study resource. All loads printed on the figures are unfactored.

Design data given on the paper (SI). Concrete $f'_c = 30$ MPa; structural steel $F_y = 350$ MPa; reinforcing steel $f_y = 400$ MPa. Prestressed concrete: $f'_{ci} = 35$ MPa at transfer, $f'_c = 50$ MPa, $n = 6$, $f_{pu} = 1750$ MPa, $f_{py} = 1450$ MPa, $f_{pi} = 1200$ MPa, losses $= 240$ MPa, hence $f_{pe} = 1200 - 240 = 960$ MPa.

Reference texts. CSA S16 Design of Steel Structures with the CISC Handbook of Steel Construction — plastic design (Cl. 8.6, 13.7), lateral–torsional buckling (Cl. 13.6), beam-columns (Cl. 13.8), plate girders (Cl. 14), composite beams (Cl. 17), welded connections (Cl. 13.13, 21); CSA A23.3 Design of Concrete Structures — flexure and shear (Cl. 10, 11), slenderness (Cl. 10.13–10.16), deflection (Cl. 9.8), footings (Cl. 15), prestressed concrete (Cl. 18); NBCC for load combinations; C. G. Salmon, J. E. Johnson & F. A. Malhas, Steel Structures: Design and Behavior; J. G. MacGregor & J. K. Wight, Reinforced Concrete: Mechanics and Design; T. Y. Lin & N. H. Burns, Design of Prestressed Concrete Structures; L. S. Beedle, Plastic Design of Steel Frames.

Check: load factors. The paper states only that the printed loads are unfactored, and gives no dead/live split. Every question below therefore treats each printed concentrated load as specified live load and factors it by $\alpha_L = 1.5$ (NBCC principal case $1.25D + 1.5L$); member self-weight, wherever it matters, is taken from the trial section and factored at $1.25$. A candidate assuming a different split obtains proportionally different sizes; the method is what is examined.

Check: section properties. Every steel section selected below is quoted by its plate dimensions (flange $b \times t$, web $h \times w$) and every property — $A$, $I_x$, $I_y$, $S$, $Z$, $J$, $C_w$, $r_x$, $r_y$ — is computed from those dimensions with root fillets neglected, which is slightly conservative. This makes every line checkable without a handbook; a rolled W-shape of equal or greater properties may be substituted directly.

Check: readings taken from the examination drawing. Three details are shown on the drawing but not stated in the question text. In Figure 1 the bases at A and D carry the fixed-support hatching and the question text says "the frame is fixed at the bases A and D", so both bases are fixed and the frame is three times statically indeterminate. In Figure 4 the 100 kN arrow at D points towards the column, i.e. to the left; A and E both carry pin symbols, making that frame indeterminate to the first degree. In Figure 5 the deck is 3 + 3 = 6 m wide over three beams spaced at 2.5 m, leaving 0.5 m overhangs — the slab-thickness annotation is not legible on the printed figure, so the deck thickness is designed in Question 7 rather than read off.

[Figure not reproduced: Figure 1 as printed on the examination paper: rectangular steel portal, fixed at A and D, beam capacity 1.5 M₊ and column capacity M₊. All loads shown are unfactored. See the official exam paper.]

Question 3: Prestressed concrete beam — section, strands and long-term deflection (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. Simply supported beam A–B, span 16 m, single 300 kN specified load at mid-span (8 m from each support). Prestressed-concrete data from page 1:

Property	Value	Property	Value
$f'_{ci}$ at transfer	35 MPa	$f_{pu}$	1750 MPa
$f'_c$ at 28 d	50 MPa	$f_{py}$	1450 MPa
modular ratio $n$	6	$f_{pi}$	1200 MPa
losses	240 MPa	$f_{pe}$	960 MPa

Find. The rectangular section, the strand area and its profile along the span, and the long-term mid-span deflection on the gross section.

Approach. Size the section so that the service stresses at mid-span stay inside the allowable envelope at transfer and in service; solve directly for the effective prestress the bottom fibre demands; convert to strands; cap the end eccentricity so the top fibre does not crack at transfer; confirm the factored moment resistance; then add the elastic camber and load deflections with PCI long-term multipliers.

Question 3: 500 × 1200 mm rectangular section with a harped tendon — e = 250 mm at the anchorages, 450 mm at mid-span, matching the triangular moment diagram of the single central load.

Part (a) — trial section and load effects. For a post-tensioned beam a span/depth ratio near 13–16 is economic; try $b = 500$, $h = 1200$ mm. Then $$A = 6.00\times10^5\ \text{mm}^2, \quad I_g = \frac{bh^3}{12} = 7.20\times10^{10}\ \text{mm}^4, \quad S_t = S_b = 1.20\times10^8\ \text{mm}^3$$ The kern distance is $k = S/A = 200$ mm each way. Self-weight is $w_g = 0.5(1.2)(24) = 14.4$ kN/m, so $$M_g = \frac{w_gL^2}{8} = \frac{14.4(16)^2}{8} = 460.8\ \text{kN}\cdot\text{m}, \qquad M_L = \frac{PL}{4} = \frac{300(16)}{4} = 1200\ \text{kN}\cdot\text{m}$$ giving a total service moment $M_T = 1660.8$ kN·m.
Allowable stresses. At transfer, $0.60f'_{ci} = 21.0$ MPa in compression and $0.25\sqrt{f'_{ci}} = 1.48$ MPa in tension. In service, $0.45f'_c = 22.5$ MPa in compression and $0.5\sqrt{f'_c} = 3.54$ MPa in tension.
Prestress demanded by the bottom fibre. Placing the tendon centroid 150 mm above the soffit gives $e = 450$ mm at mid-span. Requiring the service bottom-fibre stress to sit exactly on the tension limit, $$\frac{P_e}{A} + \frac{P_ee}{S_b} = \frac{M_T}{S_b} - f_{t,\text{allow}} = 13.84 - 3.54 = 10.30\ \text{MPa}$$ $$P_e = \frac{10.30}{1/(6\times10^5) + 450/(1.2\times10^8)} = \frac{10.30}{5.417\times10^{-6}} = 1902\ \text{kN}$$
Part (b) — strand area. With $f_{pe} = 960$ MPa the area needed is $1902\times10^3/960 = 1981$ mm². Thirteen-millimetre seven-wire strands have $A_{ps} = 99$ mm² each, so $$\boxed{21\ \text{strands} = 2079\ \text{mm}^2,\ \text{in three ducts of seven}}$$ giving $P_i = 2079(1200) = 2495$ kN at transfer and $P_e = 2079(960) = 1996$ kN in service. The initial stress is $f_{pi}/f_{pu} = 0.686 \le 0.74$, inside the CSA A23.3 Cl. 18.4 limit immediately after transfer.
Stress check at mid-span. With $f = P/A \mp Pe/S \pm M/S$ (compression positive), at transfer under $P_i$ and self-weight alone $$f_{\text{top}} = 4.16 - 9.36 + 3.84 = -1.36\ \text{MPa} \ (\ge -1.48\ \text{allowed}), \qquad f_{\text{bot}} = 4.16 + 9.36 - 3.84 = 9.67\ \text{MPa} \ (\le 21.0)$$ and in service under $P_e$ and the full moment $$\boxed{f_{\text{top}} = 9.68\ \text{MPa} \le 22.5, \qquad f_{\text{bot}} = -3.03\ \text{MPa} \ge -3.54}$$ Both states are satisfied with a little in hand, so the $500 \times 1200$ section stands.
Profile of the strands. At the anchorages the self-weight moment vanishes, so the eccentricity must be reduced or the top fibre cracks. Setting $f_{\text{top}} = -1.48$ MPa with $M = 0$, $$e_{\text{end}} \le \left(\frac{P_i}{A} + f_{ti}\right)\frac{S}{P_i} = (4.16 + 1.48)\frac{1.2\times10^8}{2.495\times10^6} = 271\ \text{mm}$$ Adopt $e = 250$ mm at each anchorage. Because a single central point load produces a triangular moment diagram, the tendon is harped — two straight runs from $e = 250$ mm at the ends to $e = 450$ mm at one harp point at mid-span — not parabolic; the straight profile follows the moment diagram exactly and keeps the cable inside its permissible zone at every section. Checking the quarter point ($e = 350$ mm, $M_g = 345.6$, $M_T = 945.6$ kN·m): top fibre $-0.24$ MPa at transfer and bottom fibre $+1.27$ MPa compression in service, both comfortable.
Factored flexural resistance. $M_f = 1.25M_g + 1.5M_L = 576 + 1800 = 2376$ kN·m. With $d_p = 600 + 450 = 1050$ mm, $\alpha_1 = 0.775$, $\beta_1 = 0.845$ for $f'_c = 50$ MPa and $k_p = 2(1.04 - f_{py}/f_{pu}) = 0.423$, equilibrium of $\alpha_1\phi_cf'_c\beta_1cb = \phi_pA_{ps}f_{pr}$ with $f_{pr} = f_{pu}(1 - k_pc/d_p)$ gives $$c = 274\ \text{mm}, \qquad f_{pr} = 1750\left(1 - 0.423\frac{274}{1050}\right) = 1557\ \text{MPa}, \qquad a = \beta_1c = 231\ \text{mm}$$ $$\boxed{M_r = \phi_pA_{ps}f_{pr}\left(d_p - \frac{a}{2}\right) = 0.9(2079)(1557)(934) = 2722\ \text{kN}\cdot\text{m} \ \ge\ 2376\ \text{kN}\cdot\text{m}}$$ and $c/d_p = 0.26$, so the section is comfortably under-reinforced and will warn before it fails.
Part (c) — elastic deflections on the gross section. $E_{ci} = 4500\sqrt{35} = 26\,622$ MPa and $E_c = 4500\sqrt{50} = 31\,820$ MPa. For a tendon that is straight at eccentricity $e_{\text{end}}$ plus a triangular drape of amplitude $e_c - e_{\text{end}}$, the upward camber is $$\Delta_p = \frac{P e_{\text{end}}L^2}{8EI} + \frac{P(e_c - e_{\text{end}})L^2}{12EI}$$ At transfer, with $P_i$ and $E_{ci}$, this gives $\Delta_p = 15.97$ mm up, against a self-weight deflection $5w_gL^4/384E_{ci}I_g = 6.41$ mm down, so the beam leaves the bed with $15.97 - 6.41 = 9.55$ mm of camber.
Long-term values. Applying the PCI multipliers for a non-composite member (2.45 on the prestress camber, 2.70 on the self-weight deflection) and adding the short-term live-load deflection $PL^3/48E_cI_g = 11.17$ mm, $$\Delta_{\text{camber}} = 2.45(15.97) = 39.1\ \text{mm}\uparrow, \qquad \Delta_{g} = 2.70(6.41) = 17.3\ \text{mm}\downarrow$$ $$\boxed{\Delta_{\text{sustained}} = 39.1 - 17.3 = 21.8\ \text{mm}\ \text{upward}; \quad \text{with full live load } 21.8 - 11.2 = 10.6\ \text{mm}\ \text{still upward}}$$ The member therefore never deflects below its casting line. The live-load component alone, 11.2 mm, is $L/1432$, far inside the usual $L/360$ limit, so serviceability is governed by the residual camber rather than by sag — which is exactly what a designer must watch when setting bearing and topping levels.

Quantity	Result
(a) Cross-section	500 mm $\times$ 1200 mm rectangular, $I_g = 72\,000 \times 10^6$ mm⁴
Effective prestress required	$P_e = 1902$ kN
(b) Strand area	21 × 13 mm strands $= 2079$ mm² ($P_i = 2495$ kN, $P_e = 1996$ kN)
(b) Profile	harped: $e = 250$ mm at anchorages, 450 mm at mid-span
Service stresses at mid-span	top $+9.68$ MPa, bottom $-3.03$ MPa
Factored resistance	$M_r = 2722$ kN·m vs $M_f = 2376$ kN·m
(c) Camber at transfer	9.6 mm upward
(c) Long-term, sustained load	21.8 mm upward
(c) Long-term with live load	10.6 mm upward (live component 11.2 mm $= L/1432$)

Topic. Service-stress design of a rectangular prestressed beam, cable profiling, and time-dependent deflection.

Key relations. Fibre stress $f = P/A \pm Pe/S \mp M/S$; kern $k = S/A$; allowable stresses $0.60f'_{ci}$, $0.25\sqrt{f'_{ci}}$, $0.45f'_c$, $0.5\sqrt{f'_c}$; CSA A23.3 Cl. 18.6.2 $f_{pr} = f_{pu}(1 - k_pc/d_p)$ with $k_p = 2(1.04 - f_{py}/f_{pu})$; camber $Pe L^2/8EI$ (straight) and $P\Delta e L^2/12EI$ (single harp point); PCI long-term multipliers 2.45 and 2.70.

Why this works. A prestressed beam is checked at two instants that pull in opposite directions. At transfer the prestress is largest and the counteracting moment smallest, so the danger is tension at the top and over-compression at the bottom; in service the prestress has shed its losses while the full moment acts, so the danger is tension at the bottom. Writing both conditions at mid-span produces two inequalities in $P$ and $e$, and the bottom-fibre service condition is almost always the binding one, which is why it is solved first for $P_e$. The end-eccentricity limit is the same top-fibre condition written where $M = 0$, and it is what forces the tendon to be draped at all. The profile shape then follows the moment diagram: a uniform load wants a parabola, a single point load wants a harp.

Common pitfall. Fitting a parabolic tendon to a beam loaded by concentrated loads. Between point loads the moment diagram is a straight line while a parabola is not, so the cable dips below the permissible zone at the loads and rises above it at the vertex — and deepening the section does not help, because both limits of the zone move together. The second pitfall is forgetting that losses have already been applied: $f_{pe} = 960$ MPa is what carries service load, while $f_{pi} = 1200$ MPa is what the section must survive at transfer.

Source. CSA A23.3 Cl. 18 (prestressed concrete); T. Y. Lin & N. H. Burns, Design of Prestressed Concrete Structures, Ch. 4–6 and 9; PCI Design Handbook, multipliers for camber and deflection.

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