16-Civ-B2 Advanced Structural Design · December 2013
Question 4 of 7: Prestressed-concrete T-beam — no tension permitted
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2013 — 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; CISC Handbook of Steel Construction; CSA A23.3:19 Design of Concrete Structures (Clause 18 for prestressed concrete, Clause 17 of S16 for composite members); 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 printed loads are unfactored; it gives no dead/live split, so no code load combination can be assembled uniquely. Every load printed on Figures 1–4 is therefore factored by 1.5, and any self weight the solver itself introduces (the concrete frame, the prestressed girder, the steel sections) by 1.25, consistent with NBCC 2020 Table 4.1.3.2. State this assumption on the answer paper as page-1 Note 1 invites. The choice scales the magnitudes only — the collapse mechanisms, section classifications and interaction equations below are unaffected.
Check: section properties. Rolled-shape properties cannot be reproduced from first principles, so every steel member selected here is a welded three-plate I-section whose area, $I$, $Z$, $J$ and $C_w$ are computed from its plate dimensions and are therefore fully checkable. A candidate with the CISC Handbook would instead take the lightest rolled W-shape whose $Z_x$ (or $M_r$, $C_r$) exceeds the value derived in each question; the design logic is identical.
Find. The prestressing force, the number of strands and the tendon profile such that no fibre anywhere in the member goes into tension at transfer or in service, plus a confirmation that the ultimate flexural strength is adequate.
Figure 3 as analysed, with the parabolic tendon profile superimposed.
Gross T-section used throughout, as the question directs.
Approach. Compute the gross section properties, find the service moment envelope including self weight, then choose the pair (prestressing force, eccentricity) that keeps the bottom fibre in compression under full load and the top fibre in compression at transfer, and finally check the factored moment against the A23.3 Clause 18.6 flexural resistance.
Gross section properties. For the flange $1400 \times 200$ over a $350$ wide web to a total depth of $1200\ \text{mm}$,$$A = 630\times10^3\ \text{mm}^2, \quad y_t = 433.3\ \text{mm}, \quad y_b = 766.7\ \text{mm}, \quad I = 86.10\times10^9\ \text{mm}^4$$so $Z_t = 198.7\times10^6$ and $Z_b = 112.3\times10^6\ \text{mm}^3$. The section weighs $0.63(24) = 15.12\ \text{kN/m}$. Note how much larger $Z_t$ is than $Z_b$: the flange pulls the centroid high, which is exactly what the hogging moment over the support will need.
Service moment envelope. The overhang produces a hogging moment at B of $M_B = -(80 \times 3 + 15.12 \times 3^2/2) = -308.0\ \text{kN}\cdot\text{m}$. In the span, $R_B = 559.3\ \text{kN}$ and $R_C = 392.9\ \text{kN}$; shear vanishes $4.86\ \text{m}$ beyond the first 300 kN load, so the maximum sagging moment occurs at $x = 8.86\ \text{m}$ from B:$$\boxed{M_{max} = 1485.2\ \text{kN}\cdot\text{m at }x = 8.86\ \text{m}, \qquad M_{sw} = 383.4\ \text{kN}\cdot\text{m there at transfer}}$$
Part (a) — the two no-tension conditions. Taking compression positive, the bottom fibre in service and the top fibre at transfer require$$\frac{P}{A} + \frac{Pe}{Z_b} - \frac{M}{Z_b} \ge 0, \qquad \frac{P_i}{A} - \frac{P_ie}{Z_t} + \frac{M_{sw}}{Z_t} \ge 0$$These pull in opposite directions on $e$: the first wants a large eccentricity, the second wants a small one, and $P_i/P = 1200/960 = 1.25$ makes the transfer condition the harder of the two. The design is therefore the smallest force that satisfies both at every section, not just at mid-span.
Force and strand count. Adopting a parabolic profile with $e = 0$ over both supports and a sag of $410\ \text{mm}$ at mid-span, and scanning every section of the span, the smallest satisfactory force is about $2.9\ \text{MN}$. With 15.2 mm seven-wire strand at $A_{str} = 140\ \text{mm}^2$, take$$\boxed{22\ \text{strands}: \ A_{ps} = 3080\ \text{mm}^2, \quad P_i = 3696\ \text{kN}, \quad P_{final} = 2956.8\ \text{kN}}$$The jacking stress ratio $f_{initial}/f_{ult} = 1200/1750 = 0.686$ is below the 0.74 limit of Clause 18.4, so the given initial stress is admissible.
Verify the stresses everywhere. Scanning the whole span, the worst values are$$\text{top at transfer } +0.21\ \text{MPa}, \qquad \text{bottom in service } +0.22\ \text{MPa}$$both compressive, so no tension occurs anywhere, as required. The largest compressions are $15.88\ \text{MPa}$ at transfer (limit $0.60f_{ci}' = 21\ \text{MPa}$) and $7.22\ \text{MPa}$ in service (limit $0.45f_c' = 22.5\ \text{MPa}$), both comfortable.
The support region. Over B the beam hogs at $308.0\ \text{kN}\cdot\text{m}$, which puts the top fibre in tension. With the tendon at the centroid there, the top fibre stress is$$\frac{P}{A} - \frac{M_B}{Z_t} = \frac{2956.8\times10^3}{630\,000} - \frac{308.0\times10^6}{198.7\times10^6} = 4.69 - 1.55 = +3.14\ \text{MPa}$$still compressive, so the overhang needs no separate top prestress. This is the payoff from the large $Z_t$ of a T-section noted in step 1 — an inverted-T or a rectangle of the same depth would have required either a raised tendon or extra untensioned top steel here.
Part (b) — the profile. The tendon is a single parabola through $e = 0$ at B, $e = 410\ \text{mm}$ at mid-span and $e = 0$ at C:$$e(x) = 410\,\frac{4x(15-x)}{15^2}\ \text{mm}\qquad (x \text{ in metres from B})$$giving $e = 396.6\ \text{mm}$ at the critical section $x = 8.86\ \text{m}$. A parabola is the correct shape because it produces a uniform upward balancing load $w_p = 8Pe_{sag}/L^2 = 8(2956.8)(0.410)/15^2 = 43.1\ \text{kN/m}$, which offsets nearly three times the self weight and most of the point loads' effect. Anchor both ends at the centroid so no end eccentricity moment is introduced, and carry the tendon straight through the overhang.
Ultimate flexural strength. With $\alpha_1 = 0.775$ and $\beta_1 = 0.845$ for $f_c' = 50\ \text{MPa}$, $k_p = 2(1.04 - f_{py}/f_{pu}) = 0.423$ and $d_p = y_t + e = 433.3 + 396.6 = 829.9\ \text{mm}$, iterating $f_{pr} = f_{pu}(1 - k_pc/d_p)$ with $a = \phi_pA_{ps}f_{pr}/(\alpha_1\phi_cf_c'b)$ converges to $f_{pr} = 1616\ \text{MPa}$ and $a = 127.0\ \text{mm}$, inside the 200 mm flange, so$$M_r = \phi_pA_{ps}f_{pr}\left(d_p - \tfrac{a}{2}\right) = 3432.9\ \text{kN}\cdot\text{m} > M_f = 2131.9\ \text{kN}\cdot\text{m}$$The 61 % reserve is normal for a no-tension design: forbidding tension at service is far more onerous than the ultimate limit state, which is why the serviceability criterion sized the member.
