16-Civ-B2 Advanced Structural Design · December 2013
Question 2 of 7: Reinforced-concrete design of member BC
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.
Question 2: Reinforced-concrete design of member BC (14 + 6 marks)
Find. The flexural and shear reinforcement for beam BC, and the immediate and long-term mid-span deflections compared with the A23.3 Table 9.3 limits.
Figure 2 with factored loads. The 900 kN loads act directly over the columns and pass straight into them; only the 750 kN mid-span load and the beam self weight bend member BC.
Approach. Analyse the pin-based portal by direct stiffness using the cracked-section stiffnesses of A23.3 Clause 10.14.1.2, separate the non-sway and sway moment sets, magnify the sway set through the stability index $Q$, then size the flexural steel from the rectangular stress block and the stirrups from the simplified shear method.
Part (a) — frame analysis. With a $600 \times 1500$ beam the factored self weight is $1.25 \times 0.6 \times 1.5 \times 24 = 27.0\ \text{kN/m}$. Using $0.35I_g$ for the beam and $0.70I_g$ for the columns, the gravity case and the lateral case are solved separately. The first-order sway under the 120 kN lateral load alone is $\Delta_o = 44.3\ \text{mm}$, and with $\sum P_f = 2955\ \text{kN}$ the stability index is$$Q = \frac{\sum P_f\,\Delta_o}{V_f\,l_c} = \frac{2955 \times 44.3}{120 \times 10\,000} = 0.109$$Because $0.05 < Q < 0.2$ the frame is a sway frame whose second-order effects must be included but which need not be stiffened; the sway moments are magnified by $\delta_s = 1/(1-Q) = 1.122$.
Design moments and shears in BC. Combining the non-sway set with the magnified sway set gives$$\boxed{M_f^{+} = 2660.6\ \text{kN}\cdot\text{m at mid-span}, \quad M_f^{-} = 1584.7\ \text{kN}\cdot\text{m at C}, \quad M_f^{-} = 237.9\ \text{kN}\cdot\text{m at B}}$$with end shears $V_f = 487.7\ \text{kN}$ at B and $667.3\ \text{kN}$ at C. The strong asymmetry is physical, not an error: with pinned bases the lateral load adds to the thrust in the leeward column and subtracts from the windward one, so the hogging moment at C is nearly seven times that at B.
Factored bending moment in beam BC. Mid-span sagging governs the bottom steel; the hogging peak at C governs the top steel.
Bottom steel at mid-span. With two layers of 30M bars, $d = 1400\ \text{mm}$. Trying $10 - 30\text{M}$ ($A_s = 7000\ \text{mm}^2$),$$T_r = \phi_s A_s f_y = 0.85(7000)(400) = 2380\ \text{kN}, \qquad a = \frac{T_r}{\alpha_1\phi_c f_c' b} = \frac{2.38\times10^6}{0.805(0.65)(30)(600)} = 252.7\ \text{mm}$$so $c = a/\beta_1 = 282.3\ \text{mm}$ and $c/d = 0.202$, comfortably below the $700/(700+f_y) = 0.636$ limit of Clause 10.5.2, i.e. the section is tension-controlled and will warn before it fails. Hence$$\boxed{M_r = T_r\left(d - \tfrac{a}{2}\right) = 3031.3\ \text{kN}\cdot\text{m} > M_f^{+} = 2660.6\ \text{kN}\cdot\text{m}}$$
Top steel over the supports. The same calculation with $6 - 30\text{M}$ ($A_s = 4200\ \text{mm}^2$) gives $M_r = 1890.9\ \text{kN}\cdot\text{m} > 1584.7$ at C. At B the demand is only $237.9\ \text{kN}\cdot\text{m}$, so minimum steel governs: Clause 10.5.1.2 requires $A_{s,min} = 0.2\sqrt{f_c'}\,b_th/f_y = 2465\ \text{mm}^2$, met by $4 - 30\text{M}$ ($2800\ \text{mm}^2$, $M_r = 1284.7\ \text{kN}\cdot\text{m}$). Run $4 - 30\text{M}$ continuously along the top and lap in the extra two bars over C.
Shear reinforcement. Using the simplified method with at least minimum stirrups, $\beta = 0.18$ and $d_v = \max(0.9d,\,0.72h) = 1260\ \text{mm}$:$$V_c = \phi_c\lambda\beta\sqrt{f_c'}\,b_wd_v = 0.65(1.0)(0.18)\sqrt{30}(600)(1260) = 484.5\ \text{kN}$$The balance must come from stirrups. With 10M double-leg stirrups ($A_v = 200\ \text{mm}^2$) at $s = 400\ \text{mm}$ and $\theta = 35^{\circ}$,$$V_s = \frac{\phi_sA_vf_yd_v\cot\theta}{s} = 305.9\ \text{kN}, \qquad \boxed{V_r = 484.5 + 305.9 = 790.4\ \text{kN} > V_f = 667.3\ \text{kN}}$$The spacing also satisfies the minimum-area rule ($s \le 405.7\ \text{mm}$) and the maximum-spacing rule ($s \le \min(0.7d_v, 600) = 600\ \text{mm}$), and $V_r$ is far below the crushing limit $0.25\phi_cf_c'b_wd_v = 3686\ \text{kN}$.
Beam BC reinforcement layout at mid-span.
Part (b) — cracked stiffness for deflection. The service mid-span moment is $1906.3\ \text{kN}\cdot\text{m}$, far above $M_{cr} = f_rI_g/y_t = 3.286(1.6875\times10^{11})/750 = 739.4\ \text{kN}\cdot\text{m}$, so the section is fully cracked and $I_e \to I_{cr}$. With $n = E_s/E_c = 200\,000/24\,648 = 8.11$, the transformed neutral axis solves $\tfrac{b}{2}x^2 + [nA_s + (n-1)A_s']x - [nA_sd + (n-1)A_s'd'] = 0$, giving $x = 408.2\ \text{mm}$ and$$I_{cr} = \frac{bx^3}{3} + nA_s(d-x)^2 + (n-1)A_s'(x-d')^2 = 71.49\times10^9\ \text{mm}^4$$i.e. 42 % of $I_g$ — the usual order for a heavily loaded beam.
Immediate and long-term deflection. Re-running the frame at service load with $E_cI_{cr}$ in the beam gives an immediate mid-span deflection of $18.8\ \text{mm}$, against the $L/360 = 41.7\ \text{mm}$ limit. Treating the whole load as sustained (conservative, since no dead/live split is given) and using Clause 9.8.2.5 with $\rho' = A_s'/(bd) = 0.00333$ and $s = 2.0$ for five years,$$\zeta = \frac{s}{1+50\rho'} = 1.714, \qquad \boxed{\Delta_{total} = 18.8(1 + 1.714) = 51.1\ \text{mm} < L/240 = 62.5\ \text{mm}}$$so both the immediate and the total deflection satisfy A23.3 Table 9.3. The margin on the long-term check is only 18 %, which is the honest answer for a 15 m beam carrying a 500 kN point load: it is deflection, not strength, that sets the 1500 mm depth.
