16-Civ-B2 Advanced Structural Design · December 2014

Question 6 of 7: Reinforced concrete design of member AC

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2014 — 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 (Clause 13 members, Clause 14 plate girders, Clause 17 composite, Clause 21 plastic design); CISC Handbook of Steel Construction; CSA A23.3:19 Design of Concrete Structures (Clause 10 flexure and columns, Clause 11 shear, Clause 18 prestressed concrete); 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 loads shown are unfactored; it gives no dead/live split. Throughout this paper the figure loads are treated as a single variable action and factored by 1.5, while self weight introduced by the solver (the reinforced-concrete frame, the prestressed girder, the bridge deck) is factored by 1.25, per NBCC 2020 Case 2. Mechanisms, section classifications and interaction ratios are unaffected by this choice — only the magnitudes are. Serviceability checks (the no-tension prestress design of Q2 and the soil bearing of Q5) use unfactored loads, as they must.

Question 6: Reinforced concrete design of member AC (14 + 6 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
Member AC	16.0 m, roller at A, monolithic with the column at C
Factored applied loads	300 kN at 4 m, 300 kN at 12 m, 600 kN at C; 120 kN horizontal at D
Trial sections	beam 500 × 1400 mm; column 600 × 1000 mm
Materials	$f_c' = 30$ MPa, $f_y = 400$ MPa, $\phi_c = 0.65$, $\phi_s = 0.85$
Stress block	$\alpha_1 = 0.805$, $\beta_1 = 0.895$
Effective depth	$d = 1330$ mm (40 mm cover, 10M stirrups, 30M bars)

Find. The flexural steel at the sagging peak and at the joint C, the shear reinforcement, and a bar layout that can actually be built in a 500 mm web.

Check — frame stiffnesses. The elastic redundant depends on the ratio of beam to column stiffness, which in a reinforced-concrete frame is not known until the members are sized. The frame is therefore analysed once with the gross-section stiffnesses of the trial members ($I_b/I_c = 2.29$), and the sections are confirmed against the moments that analysis returns. A second iteration using cracked-section stiffnesses would shift moment between the joint and the span, but the joint moment here is already below the minimum-steel requirement, so only the sagging design is load-bearing and it is conservative to keep it.

Approach. Re-run the one-redundant elastic analysis with the reinforced-concrete self weight and the true stiffness ratio, then design the critical sagging section as a singly reinforced rectangle, check the joint moment against minimum steel, and size the stirrups by the simplified method of Clause 11.3.

Add self weight and re-solve for the redundant. A 500 × 1400 beam weighs $16.8\ \text{kN/m}$, factored $w_f = 21.0\ \text{kN/m}$. With the flexibility coefficients weighted by $I_b/I_c = 2.29$ for the column terms, $$\delta_{11} = 1365 + 2560(2.29) = 7219, \qquad \delta_{10} = -3\,260\,763 \quad\Rightarrow\quad V_A = 451.7\ \text{kN}$$
Build the moment and shear envelope. $$M(x) = 451.7x - 300\langle x-4\rangle - 300\langle x-12\rangle - 10.5x^{2}$$ $$\boxed{M_f^{+} = 1748\ \text{kN}\cdot\text{m at } x = 7.22\ \text{m}}, \qquad M_f^{-} = 261\ \text{kN}\cdot\text{m at C}, \qquad V_f = 484\ \text{kN at C}$$ The hogging moment at the joint is modest because the 1000 mm column is far more flexible than the 1400 mm beam and rotates rather than restrains.
Design the sagging steel. Iterating on the lever arm with $a = \phi_sA_sf_y/(\alpha_1\phi_cf_c'b)$, $$A_s = \frac{M_f}{\phi_sf_y(d - a/2)} = \frac{1748\times10^{6}}{0.85(400)(1330 - 91)} = 4145\ \text{mm}^2$$ $$\text{Provide } \boxed{6 - 30\text{M} = 4200\ \text{mm}^2}, \qquad a = 181.9\ \text{mm}, \qquad M_r = 1769\ \text{kN}\cdot\text{m} \;>\; 1748$$
Confirm the section is under-reinforced. $$\frac{c}{d} = \frac{a/\beta_1}{d} = \frac{203.2}{1330} = 0.153 \;\ll\; \frac{700}{700+f_y} = 0.636$$ The steel is far past yield when the concrete crushes, so failure is ductile and the $\phi_s$ used in step 3 is legitimate.
Design the hogging steel at C. The joint moment needs only $A_s = 582\ \text{mm}^2$, well below the code minimum $$A_{s,min} = \frac{0.2\sqrt{f_c'}\,b_th}{f_y} = \frac{0.2(5.477)(500)(1400)}{400} = 1917\ \text{mm}^2$$ $$\text{Provide } \boxed{3 - 30\text{M} = 2100\ \text{mm}^2} \text{ top, continuous through the joint}, \qquad M_r = 917\ \text{kN}\cdot\text{m}$$ Minimum steel, not the analysis, governs the top face — a common and entirely proper outcome at a flexible joint.
Design the shear reinforcement. With $d_v = \max(0.9d,\ 0.72h) = 1197\ \text{mm}$ and the simplified method ($\beta = 0.18$, $\theta = 35^{\circ}$, minimum stirrups present), $$V_c = \phi_c\lambda\beta\sqrt{f_c'}\,b_wd_v = 0.65(0.18)(5.477)(500)(1197)/10^{3} = 384\ \text{kN}$$ $$V_s = 484 - 384 = 100\ \text{kN} \quad\Rightarrow\quad s = \frac{\phi_sA_vf_yd_v\cot\theta}{V_s} = 1154\ \text{mm}$$ Spacing is therefore controlled by the maximum, $\min(0.7d_v,\ 600) = 600\ \text{mm}$. Adopt 10M closed stirrups at 400 mm throughout, which also clears $A_{v,min} = 0.06\sqrt{f_c'}b_ws/f_y = 164\ \text{mm}^2 < 200\ \text{mm}^2$. The section is nowhere near the crushing limit, $0.25\phi_cf_c'b_wd_v = 2918\ \text{kN}$.
Check that the bars fit. Clear width between stirrups is $500 - 2(40) - 2(11.3) = 397\ \text{mm}$; six 30M bars occupy $6(29.9) = 179\ \text{mm}$, leaving five gaps of 43.6 mm against a minimum of $1.4d_b = 41.9\ \text{mm}$. The bottom steel fits in a single layer, which is what keeps $d$ at the assumed 1330 mm.
Set out the layout. Run 3 – 30M continuously top and bottom over the full 16 m for crack control and to anchor the stirrups. Add the extra 3 – 30M bottom bars over the middle of the span, extending them at least $d = 1330\ \text{mm}$ or $12d_b = 359\ \text{mm}$ beyond the point where they are no longer required, which places the cut-off near 2.0 m from A and 1.5 m from C. Anchor the top steel into the column with 90 degree standard hooks turned down into the joint, developing $l_{dh}$ inside the 1000 mm column depth.

Result	Value
Redundant (factored, gross stiffnesses)	$V_A = 451.7$ kN
Design moments	$+1748$ kN·m at 7.22 m; $-261$ kN·m at C
Design shear at C	484 kN
Section	500 × 1400 mm, $d = 1330$ mm
Bottom steel	6 – 30M ($A_s = 4200$ mm²), $M_r = 1769$ kN·m
Top steel	3 – 30M ($A_s = 2100$ mm²) — minimum steel governs
Neutral axis ratio $c/d$	0.153 (ductile)
Concrete shear resistance	384 kN
Stirrups	10M closed at 400 mm throughout

Figure 6.1 — Reinforcement layout for member AC: three continuous 30M bars top and bottom with three curtailed bottom bars through the sagging region, and 10M closed stirrups at 400 mm.

Topic. Limit states flexural and shear design of a reinforced concrete frame beam with an unequal-stiffness joint.

Key relations. $a = \phi_sA_sf_y/(\alpha_1\phi_cf_c'b)$; $M_r = \phi_sA_sf_y(d - a/2)$; ductility limit $c/d \le 700/(700+f_y)$; $A_{s,min} = 0.2\sqrt{f_c'}b_th/f_y$; simplified shear $V_c = \phi_c\lambda\beta\sqrt{f_c'}b_wd_v$ with $\beta = 0.18$, $\theta = 35^{\circ}$; $V_s = \phi_sA_vf_yd_v\cot\theta/s$.

Why this works. In a monolithic frame the joint moment is a stiffness question before it is a strength question: the beam and column share the joint rotation in proportion to their $I/L$, so a deep beam framing into a slender column sheds very little moment into the column and behaves close to a propped cantilever with a soft prop. That is why the analysis must be redone once the trial sizes are known, and why the answer changes materially if the column is resized. The consequence here is comfortable rather than awkward — the hogging demand falls below minimum steel, so the top mat is set by crack control and the design reduces to a single sagging calculation.

Common pitfall. Carrying the steel-frame analysis of Question 4 straight into the concrete design. The plastic redundant is wrong twice over: it belongs to a collapse state that a reinforced-concrete frame designed this way will not reach, and it ignores both the self weight (which is negligible in steel and dominant in concrete) and the stiffness ratio. The second pitfall is designing the bars without checking they fit: six 30M bars in a 500 mm web clear the spacing rule by 1.7 mm, and a seventh would force a second layer and cost 60 mm of effective depth.

Source. CSA A23.3:19 Clauses 10.1 (flexure), 10.5.1.2 (minimum reinforcement), 11.3 (simplified shear method) and 12.10 (bar cut-offs); MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design, Ch. 4, 6 and 10.

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