Prestressed beam section for the Figure 4 loading: 16-Civ-B2 Advanced Structural Design, December 2019, Q7

Question C1: Prestressed beam section for the Figure 4 loading (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
Geometry	3.0 m simple span plus a 1.0 m cantilever overhang
Superimposed dead load	2.0 kN/m
Live load	3.0 kN/m
Tip load P1	3.0 kN dead + 2.5 kN live
Concrete at 28 days	$f'_c = 40$ MPa, $E_c = 4500\sqrt{40} = 28\,460$ MPa
Concrete at release	$f'_{ci} = 30$ MPa, $E_{ci} = 24\,648$ MPa
Strand	1860 MPa seven-wire low-relaxation, $f_{pi} = 0.70f_{pu} = 1302$ MPa
Fabrication	CSA-certified precast shop — pretensioned, not post-tensioned

Find. A pretensioned section and strand layout; the extreme-fibre stresses verified at transfer and in service against the Clause 18.3 limits; the ultimate moment resistance including the Clause 18.8.2 cracking rule; and the long-term deflection under dead load.

Approach. Recognise that the member has reversing moment — sagging in the span, hogging over the support — so place the strands symmetrically about the centroid; then compute the moments at each stage, estimate the losses component by component, verify the fibre stresses, check the ultimate limit state including $1.2M_{cr}$, and finish with the creep-amplified dead-load deflection.

Recognise the governing feature: the moment reverses. The 1.0 m cantilever puts tension in the top fibre over the support while the span puts tension in the bottom fibre. A single eccentric tendon group cannot precompress both faces, and a 4 m precast member is far too short to harp. The correct response is to place the strands symmetrically about the centroid, giving $e = 0$, so the prestress is pure uniform precompression that benefits both faces equally and produces no camber.
Choose a trial section and strand pattern. Take a $150 \times 400\ \text{mm}$ rectangle with two 9.5 mm strands 50 mm from the top and two 50 mm from the bottom: $$A = 60\,000\ \text{mm}^2, \quad Z = \frac{bh^2}{6} = 4.00\times10^6\ \text{mm}^3, \quad I_g = 8.00\times10^8\ \text{mm}^4, \quad A_p = 4(55) = 220\ \text{mm}^2$$ Two strands per row fit the 150 mm width with 25 mm cover: the 70 mm clear spacing exceeds the $4d_b = 38\ \text{mm}$ minimum of Clause 18.7. Self weight is $0.150(0.400)(24) = 1.44\ \text{kN/m}$.
Compute the prestress force. $$\begin{aligned} f_{pi} &= 0.70 f_{pu} = 0.70(1860) = 1302\ \text{MPa} \\ P_i &= f_{pi} A_p = 1302(220)/10^3 = 286.4\ \text{kN} \end{aligned}$$ and since $e = 0$ the concrete stress at the strand level is simply $f_{cir} = P_i/A = 286\,440/60\,000 = 4.77\ \text{MPa}$, uniform over the section.
Estimate the losses component by component. $$\begin{aligned} \text{elastic shortening} &: \Delta f_{ES} = \frac{E_p}{E_{ci}} f_{cir} = \frac{200\,000}{24\,648}(4.77) = 38.7\ \text{MPa} \\ \text{shrinkage} &: \Delta f_{SH} = \varepsilon_{sh} E_p = 400\times10^{-6}(200\,000) = 80.0\ \text{MPa} \\ \text{creep} &: \Delta f_{CR} = k_{cr}\frac{E_p}{E_c} f_{cir} = 2.0\frac{200\,000}{28\,460}(4.77) = 67.1\ \text{MPa} \\ \text{relaxation (low-relax)} &: \Delta f_{RE} = 0.025 f_{pi} = 32.6\ \text{MPa} \end{aligned}$$ The total is $218.4\ \text{MPa}$, i.e. 16.8 per cent of $f_{pi}$. Adopt 20 per cent as the design value, which is conservative in the direction that matters: a lower effective prestress leaves less compression to offset the service tension.
State the effective prestress. $$\begin{aligned} P_e &= 0.80 P_i = 229.2\ \text{kN} \\ f_{cpe} &= \frac{P_e}{A} = \boxed{3.82\ \text{MPa}}\ \text{uniform compression} \end{aligned}$$ with $f_{cpi} = 4.77\ \text{MPa}$ at transfer.
Compute the moments at each stage. Using $M_B = wa^2/2 + Pa$ over the support and $R_A = (wL^2/2 - M_B)/L$ with $M_{sag} = R_A^2/2w$ in the span: $$\begin{aligned} \text{transfer (self weight alone)} &: M_B = 0.72, \quad M_{sag} = 1.28 \\ \text{dead load in service} &: M_B = 4.72 \\ \text{full service load} &: M_B = 8.72, \quad M_{sag} = 3.54 \end{aligned}$$ all in kN·m. The hogging moment over B is the largest at every stage, which is unusual for a simply supported member and is entirely due to the tip load on the 1.0 m cantilever.
Verify the transfer stresses (Clause 18.3.1). At release only the self weight acts. With $e = 0$ the stress is the uniform precompression plus the flexural term $M/Z$: $$\begin{aligned} f_{top} &= -4.77 + \frac{0.72\times10^6}{4.00\times10^6} = -4.59\ \text{MPa} \\ f_{bot} &= -4.77 - 0.18 = -4.95\ \text{MPa} \end{aligned}$$ against limits of $0.25\sqrt{f'_{ci}} = 1.37\ \text{MPa}$ tension and $0.60f'_{ci} = 18.0\ \text{MPa}$ compression. Both fibres are in compression, so there is no tension to check and the compression is at 28 per cent of its limit ✓
Verify the service stresses (Clause 18.3.2). After losses, under full service load: $$\begin{aligned} \text{top fibre over B} &: -3.82 + 2.18 = -1.64\ \text{MPa} \\ \text{bottom fibre at mid-span} &: -3.82 - 0.89 = -4.70\ \text{MPa} \\ \text{top fibre at mid-span} &: -3.82 + 0.89 = -2.93\ \text{MPa} \end{aligned}$$ $$\boxed{\text{every fibre remains in compression at every stage — the member never cracks}}$$ and the peak compression is 20 per cent of the $0.60f'_c = 24\ \text{MPa}$ limit. This is exactly what the symmetric strand layout buys.
Check the ultimate limit state. $w_f = 1.25(3.44) + 1.5(3.0) = 8.80\ \text{kN/m}$ and $P_f = 1.25(3.0) + 1.5(2.5) = 7.50\ \text{kN}$, so $M_f = 8.80(0.5) + 7.50 = 11.90\ \text{kN}\cdot\text{m}$ hogging and 4.84 kN·m sagging. For the hogging case the two top strands act as tension steel at $d_p = 350\ \text{mm}$, with $f_{pr} \approx 0.9f_{pu} = 1674\ \text{MPa}$: $$\begin{aligned} T &= \phi_p A_p f_{pr} = 0.90(110)(1674) = 165.7\ \text{kN} \\ a &= \frac{165\,726}{0.79(0.65)(40)(150)} = 53.8\ \text{mm} \end{aligned}$$ $$M_r = 165.7\left(350 - \frac{53.8}{2}\right)/10^3 = 53.55\ \text{kN}\cdot\text{m} \;\Longrightarrow\; \frac{M_f}{M_r} = 0.222$$
Apply the cracking rule, which is what actually sizes the strand. Clause 18.8.2 requires the factored resistance to exceed 1.2 times the cracking moment, so that a member which cracks does not immediately fail: $$M_{cr} = \left(0.6\sqrt{f'_c} + f_{cpe}\right) Z = (3.79 + 3.82)(4.00\times10^6)/10^6 = 30.46\ \text{kN}\cdot\text{m}$$ $$1.2 M_{cr} = 36.55\ \text{kN}\cdot\text{m} \le M_r = 53.55\ \text{kN}\cdot\text{m} \ \ \checkmark$$ Since $1.2M_{cr} = 36.6$ is three times the applied $M_f = 11.9$, this rule — not the applied load — is the binding requirement on the amount of prestressing steel. Adding prestress raises $M_{cr}$ as well as $M_r$, so over-prestressing does not help; the balance found here is deliberate.
Compute the immediate dead-load deflection. The member is uncracked, so use $E_c I_g = 28\,460(8.00\times10^8) = 2.277\times10^{13}\ \text{N}\cdot\text{mm}^2$. Superposing the span UDL and the hogging moment the cantilever applies at B, the two effects almost cancel: $$\Delta_{span} = \frac{5wL^4}{384EI} - \frac{M_B L^2}{16EI} = 0.159 - 0.117 = 0.047\ \text{mm}$$ and the free end of the cantilever drops $0.100\ \text{mm}$. Because $e = 0$ there is no prestress camber to superpose.
Amplify for creep to get the long-term deflection. Only the sustained load creeps, and here all of the dead load is sustained. With a creep coefficient $\phi_{cr} = 2.0$ for a steam-cured precast member, $$\Delta_{LT} = (1 + \phi_{cr})\Delta_i = 3.0 \Delta_i$$ $$\boxed{\begin{aligned} \Delta_{LT} &= 0.14\ \text{mm at mid-span}\ (L/21\,000) \\ & 0.30\ \text{mm at the cantilever tip}\ (a/3300) \end{aligned}}$$ The PCI final multiplier of 2.70 for the self-weight component of a non-composite member gives 0.13 and 0.27 mm, so the two methods agree. Both are negligible against any serviceability criterion.
Detail for handling and for the end zone. A symmetric strand pattern has a practical advantage beyond the stress checks: the member is safe to lift or store either way up, which matters for a 4 m precast unit that will be handled several times. Provide 6 mm welded-wire stirrups at 100 mm over the transfer length (about $50d_b = 475\ \text{mm}$ from each end) to control the bursting stresses at release, and 25 mm cover to the strands, appropriate for a certified shop under controlled conditions.

Figure C1.1 — Pretensioned section: 150 × 400 mm, two 9.5 mm low-relaxation strands 50 mm from the top and two 50 mm from the bottom. The strand centroid coincides with the section centroid, so the prestress is uniform precompression and no camber develops.

Final Results

Item	Result
Cross-section	150 mm × 400 mm rectangle
Prestressing steel	4 – 9.5 mm seven-wire low-relaxation strands ($A_p = 220$ mm$^2$), two at 50 mm from the top and two at 50 mm from the bottom, $e = 0$
Force at transfer	$P_i = 286.4$ kN, $f_{cpi} = 4.77$ MPa uniform
Losses (itemised)	ES 38.7 + SH 80.0 + CR 67.1 + RE 32.6 = 218.4 MPa = 16.8 %; design value 20 %
Effective force	$P_e = 229.2$ kN, $f_{cpe} = 3.82$ MPa uniform
Transfer stresses	top −4.59 MPa, bottom −4.95 MPa (limits +1.37 / −18.0) ✓
Service stresses	top over B −1.64, span bottom −4.70, span top −2.93 MPa (limit −24.0) ✓
Tension at any stage	none — the section is fully compressed throughout
Ultimate moments	$M_f = 11.90$ kN·m hogging, 4.84 kN·m sagging
Moment resistance	$M_r = 53.55$ kN·m, utilisation 0.222
Governing check	Cl 18.8.2: $1.2M_{cr} = 36.55 \le M_r = 53.55$ kN·m
Immediate dead deflection	0.047 mm mid-span, 0.100 mm at the tip
Long-term dead deflection	0.14 mm mid-span ($L/21\,000$), 0.30 mm at the tip ($a/3300$)
Camber	nil, because the strand centroid is at the section centroid
End-zone steel	6 mm WWF stirrups at 100 mm over 475 mm from each end

Check — prestressing is not the economical choice at this load, and the section is minimum-governed. The applied moment of 11.9 kN·m would not crack an unstressed 150 × 400 mm section, so this member is prestressed because question C1 asks for a prestressed design, not because the loading demands one. Two consequences are reported rather than hidden: the ultimate utilisation is only 0.22, and the amount of strand is set by the Clause 18.8.2 cracking rule rather than by the load. In practice the same loading over a 4 m span at 2 m centres would be carried by a reinforced or precast hollow-core unit at lower cost; the prestressed solution is justified only if the member must be precast for speed or if the same section is repeated many times.

Concept & Theory

Topic. Design of a pretensioned member whose bending moment reverses, and verification of concrete stresses at transfer and in service.

Key relations. Uniform precompression $f_{cp} = P/A$ with a symmetric strand group ($e = 0$); fibre stresses $f = -P/A \pm M/Z$; A23.3 Cl 18.3.1 transfer limits $0.60f'_{ci}$ compression and $0.25\sqrt{f'_{ci}}$ tension; Cl 18.3.2 service limits $0.60f'_c$; losses $\Delta f_{ES} = (E_p/E_{ci})f_{cir}$, $\Delta f_{SH} = \varepsilon_{sh}E_p$, $\Delta f_{CR} = k_{cr}(E_p/E_c)f_{cir}$, $\Delta f_{RE} \approx 0.025f_{pi}$ for low-relaxation strand; Cl 18.8.2 $M_r \ge 1.2M_{cr}$ with $M_{cr} = (0.6\sqrt{f'_c} + f_{cpe})Z$; $\Delta_{LT} = (1 + \phi_{cr})\Delta_i$ on the uncracked section.

Why this works. Prestressing works by pre-compressing the fibre that the service load will put into tension, so a member whose moment changes sign presents a genuine conflict: an eccentricity that helps one region harms the other. Making the strand group symmetric resolves it. The eccentricity is then zero, the prestress becomes pure axial compression, and both faces gain the same margin against cracking. Two further benefits follow that are easy to overlook. Zero eccentricity means zero prestress camber, so the long-term deflection is simply the creep-amplified dead-load deflection with no upward component to subtract — a real simplification for a member with a cantilever, where camber and cantilever droop would otherwise fight each other. And it makes the unit reversible for handling. On a lightly loaded member the design then stops being governed by the applied load: the $1.2M_{cr}$ rule requires enough prestressing steel that the flexural resistance exceeds the moment at which the section cracks, which guards against the sudden failure a lightly reinforced prestressed member would otherwise suffer.

Common pitfall. Two mistakes. Placing all the strand near the bottom face by habit, which leaves the top fibre over the support with no precompression at all and cracks the member at the section carrying its largest moment. And treating $1.2M_{cr}$ as a formality: because prestress raises $M_{cr}$ at the same time as it raises $M_r$, the check does not become easier as strand is added, and on lightly loaded members it is the binding constraint. A third, quieter error is computing the long-term deflection on a cracked transformed section: a properly designed prestressed member is uncracked at service, so $I_g$ is correct.

Source. Collins & Mitchell, Prestressed Concrete Structures, Ch. 3 (stress limits and losses) and Ch. 4 (flexural strength); CPCI Design Manual, 5th ed., Ch. 3 and the multiplier tables in Ch. 4; CAC Concrete Design Handbook, 4th ed., Ch. 6 (prestressed concrete) with the Cl 18 explanatory notes.

Question 7 of 7: Prestressed beam section for the Figure 4 loading

Question C1: Prestressed beam section for the Figure 4 loading (20 marks)