Post-tensioned prestressed concrete girder (12 + 6 + 2 marks): 16-Civ-B2 Advanced Structural Design, May 2013, Q7

Question 7: Post-tensioned prestressed concrete girder (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. The post-tensioned girder of Figure 3: a 3 m cantilever carrying 60 kN at its tip, then a 14 m span between the two supports with 250 kN applied 2 m inside each support. The section is to be rectangular with no tension permitted, and the gross section may be used for the moment of inertia.

Given data — prestressed girder
Quantity	Symbol	Value
Cantilever	—	3 m, 60 kN at the tip
Span between supports	$L$	14 m
Applied loads in the span	$P$	250 kN at 2 m from each support
Concrete at transfer / at service	$f^{\prime}_{ci}$ / $f^{\prime}_{c}$	35 MPa / 50 MPa
Strand strengths	$f_{pu}$ / $f_{py}$	1750 MPa / 1450 MPa
Initial stress and losses	$f_{pi}$	1200 MPa, losses 240 MPa
Effective stress	$f_{pe}$	960 MPa

Find. (a) a rectangular cross-section that admits no tension, and (b) the area of prestressing strand required together with its profile.

Figure 3 — post-tensioned girder with the adopted parabolic tendon profile, held above the centroid over the cantilever support.

Approach. Get the service moment diagram including self weight, size the section from the no-tension condition at the critical section, then draw the cable zone — the band of eccentricity within which no fibre goes into tension — and choose the strand count that lets a practical parabolic drape fit inside it, closing with ultimate strength and shear.

Part (a) — reactions and service moments. Taking moments about the right-hand support, the applied loads give $R_{1}=322.9\ \text{kN}$ and $R_{2}=237.1\ \text{kN}$. Trying a $450\times1200$ section, its self weight of $12.96\ \text{kN/m}$ adds $R_{1}=133.8$ and $R_{2}=86.6\ \text{kN}$. Combining, the maximum sagging moment is $709.5\ \text{kN}\cdot\text{m}$ at $x=11.31\ \text{m}$ from the free end, with $238.3\ \text{kN}\cdot\text{m}$ hogging over the cantilever support.
Part (a) — the no-tension condition. With $A=540\times10^{3}\ \text{mm}^{2}$ and $S=108\times10^{6}\ \text{mm}^{3}$ the kern distance is $k=S/A=200\ \text{mm}$. Requiring the bottom fibre to stay in compression, $$\frac{P}{A}+\frac{Pe}{S}-\frac{M}{S}\ \ge\ 0\quad\Longrightarrow\quad P(k+e)\ \ge\ M$$ With 150 mm of cover to the strand centroid, $e_{\max}=600-150=450\ \text{mm}$, so the critical section alone would need only $P_{e}\ge709.5/(0.650)=1092\ \text{kN}$.
Part (b) — the cable zone is what really governs. Rearranging the same inequality at every section, and its mirror image for the top fibre, gives the band $$\frac{M}{P_{e}}-k\ \le\ e(x)\ \le\ \frac{M}{P_{e}}+k$$ a strip 400 mm wide that follows the moment diagram. A tendon can only be a smooth parabola, so it cannot chase the sharp rise in moment just past the first 250 kN load. Increasing $P_{e}$ shrinks $M/P_{e}$ and pulls the band down to meet the achievable drape, and the smallest tendon that fits is 16 strands of 13 mm, $A_{ps}=1579\ \text{mm}^{2}$.
Part (b) — forces and the adopted profile. $P_{e}=1579(960)=1516\ \text{kN}$ and $P_{i}=1579(1200)=1895\ \text{kN}$, with $f_{pi}/f_{pu}=0.686$, inside the code ceiling of 0.74 immediately after transfer. The drape runs from the centroid at the free end, rises to 150 mm above the centroid over the cantilever support to answer the hogging moment there, then falls as a parabola to a low point 450 mm below the centroid at $x=10.5\ \text{m}$ and returns to the centroid at the far anchorage.
Part (a) — check every fibre, not just the critical one. Scanning the whole girder with that profile, the minimum service stress is $2.71\ \text{MPa}$ at the top and $0.34\ \text{MPa}$ at the bottom, both compressive: $\boxed{\text{no tension anywhere under full service load}}$. The largest compression is $5.27\ \text{MPa}$, far below $0.45f^{\prime}_{c}=22.5\ \text{MPa}$.
Check the transfer condition. At transfer only self weight opposes the full $P_{i}$, so the top fibre near the low point goes into slight tension: $1.71\ \text{MPa}$, within the A23.3 allowance $0.5\sqrt{f^{\prime}_{ci}}=2.96\ \text{MPa}$, while the largest compression is $8.73\ \text{MPa}$ against $0.6f^{\prime}_{ci}=21\ \text{MPa}$. Transfer is satisfied without changing the section.
Ultimate strength. With $1.25$ on self weight and $1.5$ on the applied loads, $M_{f}=993.6\ \text{kN}\cdot\text{m}$. For bonded tendons, $k_{p}=2(1.04-f_{py}/f_{pu})=0.423$, giving $c=235.1\ \text{mm}$, $a=198.7\ \text{mm}$ and $f_{pr}=1584\ \text{MPa}$, hence $M_{r}=\phi_{p}A_{ps}f_{pr}(d_{p}-a/2)=2141\ \text{kN}\cdot\text{m}$. The minimum reinforcement rule is also met: $1.2M_{cr}=1732\ \text{kN}\cdot\text{m}\le M_{r}$, so the girder will not fail the instant it cracks.
Shear and detailing. The largest factored shear is $513\ \text{kN}$ just inside the cantilever support. There $V_{c}=\phi_{c}\beta\sqrt{f^{\prime}_{c}}b_{w}d_{v}=352\ \text{kN}$ and the drape, sloping at 0.160, adds a vertical component $V_{p}=P_{e}\sin\theta=243\ \text{kN}$, so minimum stirrups suffice: 10M closed at 400 mm, closed to 200 mm through the anchorage zone where bursting reinforcement is also required behind each anchor.

Question 7 — post-tensioned girder
Quantity	Result
Section	450 $\times$ 1200 mm rectangular, $k=S/A=200\ \text{mm}$
Service moments	$+709.5$ sagging, $-238.3$ hogging $\text{kN}\cdot\text{m}$
Effective prestress	$P_{e}=1516\ \text{kN}$ ($P_{i}=1895\ \text{kN}$)
Strand	16 of 13 mm, $A_{ps}=1579\ \text{mm}^{2}$
Maximum eccentricity	$e=450\ \text{mm}$ at $x=10.5\ \text{m}$
Eccentricity over the support	150 mm above the centroid
Service stresses	top $\ge2.71$, bottom $\ge0.34\ \text{MPa}$ (no tension)
Transfer stresses	top $-1.71\ \text{MPa}$ against a limit of $-2.96\ \text{MPa}$
Ultimate	$M_{r}=2141\ \text{kN}\cdot\text{m}$ against $M_{f}=994\ \text{kN}\cdot\text{m}$
Stirrups	10M closed at 400 mm, 200 mm in the anchorage zone

Topic. Post-tensioned concrete: the no-tension design condition and the cable zone.

Key relations. $f=\dfrac{P}{A}\pm\dfrac{Pe}{S}\mp\dfrac{M}{S}$; no bottom tension requires $P(k+e)\ge M$ with $k=S/A$; the permissible eccentricity band is $M/P-k\le e\le M/P+k$; ultimate $f_{pr}=f_{pu}(1-k_{p}c/d_{p})$ with $k_{p}=2(1.04-f_{py}/f_{pu})$.

Why this works. Prestress is a way of pre-loading the concrete so that the tension caused by the applied load only ever cancels an existing compression. Because the axial term is constant and the bending term varies along the beam, the eccentricity has to follow the moment diagram, and the cable zone is simply the statement of how much freedom the kern leaves at each section.

Common pitfall. Sizing the prestress from the critical section alone. Here that gives 1092 kN, but a tendon that can physically be built — smooth, within the cover, and anchored at the centroid — needs 1516 kN before it fits inside the cable zone at every section, 39 percent more.

Source. CSA A23.3 Clauses 18.3, 18.6 and 18.8; Collins & Mitchell, Prestressed Concrete Structures, Chapters 3 and 4; Nawy, Prestressed Concrete: A Fundamental Approach, Chapters 4 and 12.

Question 7 of 7: Post-tensioned prestressed concrete girder (12 + 6 + 2 marks)

Question 7: Post-tensioned prestressed concrete girder (12 + 6 + 2 marks)