Prestressed beam for the Figure 3 loading, and its long-term deflection: 16-Civ-B2 Advanced Structural Design, Undated paper, Q7

Question C1: Prestressed beam for the Figure 3 loading, and its long-term deflection (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.

Quantity	Value
Span and loads	4.5 m simply supported, loads as Figure 3 (see B1)
Concrete	$f_c' = 40\ \text{MPa}$; $f_{ci}' = 0.75f_c' = 30\ \text{MPa}$ at transfer
Moduli	$E_c = 4500\sqrt{40} = 28\,460$ MPa; $E_{ci} = 4500\sqrt{30} = 24\,648$ MPa
Strand	13 mm seven-wire low relaxation, $f_{pu} = 1860$ MPa, $A_p = 99\ \text{mm}^2$ each
Jacking and effective stress	$f_{pi} = 0.74f_{pu} = 1376$ MPa; 18 % total loss, $f_{se} = 1129$ MPa
Fabrication	CSA certified pretensioning plant, so straight bonded strand is assumed

Find. A pretensioned rectangular section, its strand pattern, verification at transfer, in service and at ultimate, and the long-term deflection under dead load.

Approach. Choose a section and strand force that keep the transfer and service fibre stresses inside the A23.3 limits, confirm the factored resistance and the minimum-reinforcement rule, then compute the immediate camber and dead-load deflection and apply long-term multipliers.

Choose a trial section. A 4.5 m span is short for prestressing, so the section is set by handling and by the strand pattern rather than by strength. Take a rectangle $b = 250\ \text{mm}$, $h = 400\ \text{mm}$: $$A_g = 100\times10^{3}\ \text{mm}^2, \quad I_g = \frac{250(400)^{3}}{12} = 1333\times10^{6}\ \text{mm}^4, \quad Z = 6.667\times10^{6}\ \text{mm}^3$$ $$w_{sw} = 0.25(0.40)(24) = 2.40\ \text{kN/m}, \quad M_0 = \frac{2.40(4.5)^{2}}{8} = 6.08\ \text{kN}\!\cdot\!\text{m}$$
Size the prestress from the service tension limit. The service moment is the same envelope computed in B1, now with the lighter self weight; the full service load (dead plus live plus triangular snow) gives $M_T = 28.26\ \text{kN}\!\cdot\!\text{m}$. Requiring the bottom fibre tension to stay under $0.5\sqrt{f_c'} = 3.16\ \text{MPa}$: $$\frac{P}{A} + \frac{Pe}{Z} \ge \frac{M_T}{Z} - 3.16 = 4.24 - 3.16 = 1.08\ \text{MPa}$$ which needs barely 30 kN of effective prestress. The design is therefore controlled at the other end — by the tension that appears at the top fibre at transfer, when only the self weight opposes the prestress.
Fix the strand pattern from the transfer condition. Try two 13 mm strands at 90 mm from the soffit, so $A_p = 198\ \text{mm}^2$, $e = 200 - 90 = 110\ \text{mm}$, $d_p = 310\ \text{mm}$: $$P_i = 198(1376) = 272.5\ \text{kN}, \quad P_e = 198(1129) = 223.5\ \text{kN}$$ At the member end, where no self-weight moment has developed, A23.3 Clause 18.3.1.1 permits $0.5\sqrt{f_{ci}'} = 2.74\ \text{MPa}$ of tension provided bonded reinforcement is available: $$f_{top,end} = -\frac{P_i}{A} + \frac{P_ie}{Z} = -2.72 + 4.50 = +1.77\ \text{MPa} \le 2.74\ \checkmark$$ At mid-span the interior limit $0.25\sqrt{f_{ci}'} = 1.37\ \text{MPa}$ applies: $$f_{top,mid} = 1.77 - \frac{M_0}{Z} = 1.77 - 0.91 = +0.86\ \text{MPa} \le 1.37\ \checkmark$$ $$f_{bot,mid} = -2.72 - 4.50 + 0.91 = -6.31\ \text{MPa}, \quad |{-6.31}| \le 0.6f_{ci}' = 18.0\ \checkmark$$ Raising the strands from 60 mm to 90 mm is what brings the end fibre inside its limit; the alternative would be to debond one strand over the end metre.
Check the service stresses. With the effective force after all losses: $$f_{top} = -\frac{P_e}{A} + \frac{P_ee}{Z} - \frac{M_T}{Z} = -2.23 + 3.69 - 4.24 = -2.79\ \text{MPa}$$ $$f_{bot} = -2.23 - 3.69 + 4.24 = -1.68\ \text{MPa}$$ $$\boxed{\text{No tension anywhere under full service load; peak compression 2.79 MPa} \ll 0.6f_c' = 24\ \text{MPa}}$$ The member is fully uncracked, which is the class of behaviour a prestressed member is normally specified to deliver.
Determine the factored moment. Re-running the B1 combinations with the 2.40 kN/m self weight, the governing case is again $1.25D + 1.4W + 0.5L$ at 15.17 kN/m uniform: $$M_f = 38.39\ \text{kN}\!\cdot\!\text{m}, \quad V_f = 35.3\ \text{kN}$$
Check the factored flexural resistance. For low-relaxation strand, Clause 18.6.2 gives $k_p = 2\left(1.04 - f_{py}/f_{pu}\right) = 0.28$ and $$f_{pr} = f_{pu}\left(1 - k_p\frac{c}{d_p}\right)$$ Solving that together with $C = \alpha_1\phi_cf_c'b\beta_1c = \phi_pA_pf_{pr}$, with $\alpha_1 = 0.79$ and $\beta_1 = 0.87$: $$c = 69.5\ \text{mm}, \quad a = 60.5\ \text{mm}, \quad f_{pr} = 1743\ \text{MPa}$$ $$\boxed{M_r = \phi_pA_pf_{pr}\left(d_p - \tfrac{a}{2}\right) = 86.9\ \text{kN}\!\cdot\!\text{m} > 38.4\ \text{kN}\!\cdot\!\text{m}}$$ The ratio $c/d_p = 0.224$ is well under 0.5, so the section is ductile.
Check the minimum-reinforcement rule, which is the real constraint. Clause 18.8.2 requires $M_r \ge 1.2M_{cr}$ so that the member does not fail the instant it cracks: $$M_{cr} = \left(f_r + \frac{P_e}{A} + \frac{P_ee}{Z}\right)Z = (3.795 + 2.235 + 3.684)(6.667\times10^{6}) = 64.8\ \text{kN}\!\cdot\!\text{m}$$ $$1.2M_{cr} = 77.7\ \text{kN}\!\cdot\!\text{m} \le M_r = 86.9\ \text{kN}\!\cdot\!\text{m}\ \checkmark$$ This governs the strand count: one strand would satisfy strength comfortably and fail this check. Shear is trivial at 35.3 kN with the axial precompression, so nominal 10M stirrups at 200 mm are provided.
Compute the immediate deflections. Camber from the prestress couple on a simply supported member with straight strand, and the downward deflection of the self weight, both at transfer with $E_{ci}$: $$\Delta_p = \frac{P_ieL^{2}}{8E_{ci}I_g} = \frac{272.5\times10^{3}(110)(4500)^{2}}{8(24\,648)(1333\times10^{6})} = 2.31\ \text{mm}\ \uparrow$$ $$\Delta_{sw} = \frac{5w_{sw}L^{4}}{384E_{ci}I_g} = 0.39\ \text{mm}\ \downarrow$$ $$\text{net camber at transfer} = 2.31 - 0.39 = 1.92\ \text{mm}\ \uparrow$$ The superimposed dead load of 4.5 kN/m is applied later, on mature concrete: $$\Delta_{SDL} = \frac{5(4.5)(4500)^{4}}{384(28\,460)(1333\times10^{6})} = 0.63\ \text{mm}\ \downarrow$$
Apply the long-term multipliers and answer the question. Using the standard PCI multipliers for a non-composite member — 2.45 on the prestress camber, 2.70 on the member self weight and 3.00 on superimposed dead load: $$\Delta_{LT} = 2.45(2.31) - 2.70(0.39) - 3.00(0.63)$$ $$\boxed{\Delta_{LT} = 5.66 - 1.05 - 1.90 = +2.71\ \text{mm upward}\ \ (L/1660)}$$ As a cross-check, applying a creep coefficient of $\varphi_{cr} = 2.0$ to the immediate deflection computed with the effective prestress gives $3.0(1.64 - 0.34 - 0.63) = +2.01\ \text{mm}$, the same answer to within the accuracy of either method. Under dead load the member does not sag at all: it cambers upward by roughly 2 to 3 mm, and that camber grows with time rather than being lost.
State the practical consequence. A 2 to 3 mm long-term camber over 4.5 m is invisible and needs no compensation, but it must be allowed for when setting screed levels on a floor made of several such units, since adjacent members with different ages at transfer will camber by different amounts. Specify a minimum concrete strength at transfer of 30 MPa and release the strands only when cylinder tests confirm it, because releasing early both raises the transfer stresses and increases the eventual camber.

Pretensioned section for C1: 250 x 400 mm, two straight 13 mm low-relaxation strands at 90 mm from the soffit.

Result	Value
Section	250 mm x 400 mm rectangular, $f_c' = 40$ MPa
Prestressing steel	2 straight 13 mm low-relaxation strands, $A_p = 198$ mm², $e = 110$ mm
Prestress force	$P_i = 272.5$ kN at transfer, $P_e = 223.5$ kN after 18 % loss
Transfer stresses (top)	+1.77 MPa at the end (limit 2.74), +0.86 MPa at mid-span (limit 1.37)
Service stresses	top −2.79 MPa, bottom −1.68 MPa — uncracked, no tension
Factored moment / resistance	38.4 / 86.9 kN·m, ratio 0.44; $c/d_p = 0.224$
Minimum reinforcement	$1.2M_{cr} = 77.7 \le M_r = 86.9$ kN·m (this governs the strand count)
Immediate camber at transfer	+1.92 mm (2.31 camber less 0.39 self weight)
Long-term deflection under dead load	+2.71 mm upward (PCI multipliers); +2.01 mm by the creep-coefficient method
Shear reinforcement	10M stirrups at 200 mm (nominal)

Topic. Design of a short pretensioned beam and the calculation of its long-term camber.

Key relations. Fibre stresses $f = -\frac{P}{A} \pm \frac{Pe}{Z} \mp \frac{M}{Z}$; A23.3 Clause 18.3 transfer limits $0.25\sqrt{f_{ci}'}$ within the span and $0.5\sqrt{f_{ci}'}$ at the ends; Clause 18.6.2 $f_{pr} = f_{pu}\left(1 - k_pc/d_p\right)$ with $k_p = 0.28$ for low relaxation; Clause 18.8.2 $M_r \ge 1.2M_{cr}$; camber $\Delta_p = P_ieL^{2}/\left(8E_{ci}I_g\right)$; PCI long-term multipliers 2.45 / 2.70 / 3.00.

Why this works. A prestressed member is checked at two quite different instants and the two pull in opposite directions. At transfer the prestress is at its largest and the only opposing moment is the self weight, so the danger is tension at the top; in service the prestress has relaxed and the applied moment is at its largest, so the danger is tension at the bottom. Between them they trap the permissible force and eccentricity in a band, and on a lightly loaded short span it is the transfer end of the band that binds — which is why the answer is a small force at a modest eccentricity rather than the largest prestress the strands could deliver. Creep then amplifies the camber and the dead-load sag alike, but the camber is larger, so the member ends up further from the soffit than it started.

Common pitfall. Sizing the prestress from the service condition alone and placing the strands as low as cover permits. On this beam that gives an end top-fibre tension of 3.0 MPa against a 2.74 MPa limit and the section cracks at transfer, before it has carried any load. The second pitfall is reporting the long-term deflection as a downward figure by treating the prestress as if it were fully lost; the correct answer here is an upward movement, and a designer expecting sag would set the soffit level the wrong way.

Source. CSA A23.3 Clauses 18.3 (permissible stresses), 18.6 (flexural resistance), 18.8 (minimum reinforcement); Collins & Mitchell, Prestressed Concrete Structures, Chapters 3 and 5; CPCI Design Manual, 5th ed., Chapter 3 and the long-term multiplier table; Nawy, Prestressed Concrete, Chapter 7 (deflection and camber).

Question 7 of 7: Prestressed beam for the Figure 3 loading, and its long-term deflection

Question C1: Prestressed beam for the Figure 3 loading, and its long-term deflection (20 marks)