16-Civ-B2 Advanced Structural Design · December 2014

Question 7 of 7: Reinforced concrete design of column CE

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 7: Reinforced concrete design of column CE (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
Column CE	10.0 m overall, fixed at E, monolithic with beam AC at C
Trial section	600 mm × 1000 mm (1000 mm in the plane of bending)
Clear height	$l_u = 10\,000 - 700 = 9300$ mm
Factored axial (base, incl. self weight)	$C_f = 1264$ kN
Factored moments	261 kN·m at C, 861 kN·m at E
Materials	$f_c' = 30$ MPa, $f_y = 400$ MPa, $E_c = 24\,648$ MPa

Find. Longitudinal and tie reinforcement for a rectangular column that satisfies the axial-moment interaction after second-order effects, together with the slenderness classification the code requires.

Figure 7.1 — Column CE: factored bending moment diagram (constant from C to D, rising to the fixed base) and the designed 600 × 1000 section with 10 – 30M and 10M ties.

Approach. Classify the column for slenderness in the sway frame, magnify the base moment by the sway magnifier, then find the point on the strain-compatibility interaction diagram at the factored axial load and compare its moment with the magnified demand.

Assemble the design actions. From the Question 6 analysis, the column carries the beam reaction plus its own factored weight: $$C_f = 1200 + 21(16) - 451.7 + 18(10) = 1264\ \text{kN}$$ The moment is constant at 261 kN·m from C down to D, where the 120 kN horizontal load enters, and grows linearly to $$M_{f,E} = 261 + 120(5) = 861\ \text{kN}\cdot\text{m}$$ The eccentricity is 681 mm, so this is a moment-dominated member that behaves far more like a beam than a column.
Classify for slenderness. The frame sways (A is a roller), so the effective length must come from the sway chart. With $$\psi_{top} = \frac{I_c/l_u}{I_b/L_b} = \frac{5.0\times10^{10}/9300}{11.43\times10^{10}/16\,000} = 0.75, \qquad \psi_{bottom} = 0 \text{ (fixed)}$$ take $k = 1.30$. Then, with $r = 0.3h = 300\ \text{mm}$, $$\frac{kl_u}{r} = \frac{1.30(9300)}{300} = 40.3$$ This exceeds the sway-frame threshold of 22, so second-order effects must be included; it is well inside the limit of 100 above which Clause 10.13.2 would demand a full second-order analysis, so the moment magnifier method is admissible.
Magnify the sway moment. $$EI = \frac{0.4E_cI_g}{1+\beta_d} = 0.4(24\,648)(5.0\times10^{10}) = 4.93\times10^{14}\ \text{N}\cdot\text{mm}^2$$ $$P_c = \frac{\pi^2EI}{(kl_u)^2} = \frac{\pi^2(4.93\times10^{14})}{(12\,090)^2} = 33\,285\ \text{kN}$$ $$\delta_s = \frac{1}{1 - C_f/P_c} = \frac{1}{1 - 1264/33\,285} = 1.040 \quad\Rightarrow\quad \boxed{M_f = 861(1.040) = 895\ \text{kN}\cdot\text{m}}$$ The magnification is only four per cent because the column, though slender by the code's index, is stabilised by a very large critical load relative to the modest axial force it carries.
Set the reinforcement from the minimum ratio. Clause 10.9.1 requires $\rho \ge 0.01$ for a column, which for $A_g = 600\,000\ \text{mm}^2$ means $6000\ \text{mm}^2$. Provide $$\boxed{10 - 30\text{M} = 7000\ \text{mm}^2}, \quad \text{five per face}, \quad \rho = 1.17\ \text{per cent}$$ with $d = 930\ \text{mm}$ and $d' = 70\ \text{mm}$.
Locate the neutral axis at the design axial load. Strain compatibility with $\varepsilon_{cu} = 0.0035$ gives, for equilibrium at $P_r = C_f = 1264\ \text{kN}$, $$P_r = \alpha_1\phi_cf_c'b\beta_1c + \phi_sA_s'f_s' - \phi_sA_sf_y \quad\Rightarrow\quad c = 155.4\ \text{mm}$$ so the compression steel is at $f_s' = 383\ \text{MPa}$ (just below yield) and the tension steel is well past yield. The section is on the tension-controlled branch of the interaction diagram, which is where a column with an eccentricity of 681 mm belongs.
Compute the moment resistance at that axial load. Taking moments about mid-depth, $$M_r = C_c\left(\frac{h}{2}-\frac{a}{2}\right) + \phi_sA_s'f_s'\left(\frac{h}{2}-d'\right) + \phi_sA_sf_y\left(d-\frac{h}{2}\right) = \boxed{1568\ \text{kN}\cdot\text{m}}$$ $$\frac{M_f}{M_r} = \frac{895}{1568} = 0.57 \quad\checkmark$$ The reserve is real but not wasteful: the steel area is set by the one per cent minimum, and the concrete dimension is set by the joint stiffness assumed in Question 6, so neither can be trimmed without re-analysing the frame.
Check the minimum eccentricity and the axial cap. The code minimum $e = 15 + 0.03h = 45\ \text{mm}$ is exceeded many times over by the actual 708 mm, so no accidental-eccentricity provision applies. The pure axial capacity is $P_{r,max} = 0.8P_o = 9351\ \text{kN}$, more than seven times $C_f$ — confirming that this member is governed entirely by flexure.
Check shear and detail the ties. The column shear is 120 kN below D and zero above it. With $d_v = 837\ \text{mm}$, $$V_c = \phi_c\lambda\beta\sqrt{f_c'}\,b_wd_v = 322\ \text{kN} \;>\; 120\ \text{kN}$$ so no shear reinforcement beyond the ties is needed. Tie spacing is the least of $16d_b = 478$, $48d_{tie} = 542$ and the least column dimension 600 mm, so provide 10M ties at 450 mm, with every alternate bar restrained by a tie corner or a cross-tie and the spacing halved to 225 mm over a distance $h = 1000\ \text{mm}$ below the joint, where the column moment gradient is steepest.

Result	Value
Factored axial at base	1264 kN
First-order moments	261 kN·m at C, 861 kN·m at E
Effective length factor (sway)	$k = 1.30$, $\psi_{top} = 0.75$
Slenderness $kl_u/r$	40.3 — slender ($>22$), magnifier method valid ($<100$)
Critical load / magnifier	$P_c = 33\,285$ kN, $\delta_s = 1.040$
Magnified design moment	895 kN·m
Section	600 × 1000 mm
Longitudinal steel	10 – 30M (5 per face), $\rho = 1.17$ per cent
Neutral axis depth at $C_f$	155.4 mm (tension-controlled)
Moment resistance	1568 kN·m — utilisation 0.57
Ties	10M at 450 mm, reduced to 225 mm for 1.0 m below the joint

Topic. Limit states design of a slender reinforced concrete column in a sway frame, including second-order magnification.

Key relations. Slenderness $kl_u/r$ against the sway threshold 22 and the Clause 10.13.2 ceiling of 100; $EI = 0.4E_cI_g/(1+\beta_d)$; $P_c = \pi^2EI/(kl_u)^2$; magnifier $\delta_s = 1/(1 - \sum P_f/\sum P_c)$; strain-compatibility equilibrium $P_r = \alpha_1\phi_cf_c'b\beta_1c + \phi_sA_s'f_s' - \phi_sA_sf_y$; $\rho_{min} = 0.01$, $P_{r,max} = 0.8P_o$.

Why this works. A column carrying 1264 kN on a 600 000 mm² section is nominally under one tenth of its squash load, so its interaction diagram is being sampled far down the tension-controlled branch where added axial load actually increases the moment resistance. That is why the section looks generously reinforced and still returns a utilisation of only 0.57: the governing constraints are the one per cent minimum steel ratio and the gross dimensions needed to give the frame the joint stiffness the Question 6 analysis assumed. Recognising which constraint is binding matters, because trimming the section to raise the utilisation would invalidate the frame analysis and push moment back into the beam.

Common pitfall. Treating the frame as braced and taking $k \le 1.0$. With a roller at A there is nothing to prevent sidesway, so $k$ exceeds unity and the slenderness index roughly doubles; a column that looks stocky at $kl_u/r = 31$ is slender at 40 and the magnifier becomes mandatory. The second pitfall is magnifying the moment and then forgetting to check it against a resistance computed at the correct axial load — the interaction diagram is a curve, and reading $M_r$ at the wrong $P$ can be wrong by a factor of two on the tension-controlled branch.

Source. CSA A23.3:19 Clauses 10.9 (limits on reinforcement), 10.10 (ties), 10.13 (slenderness) and 10.16 (sway magnification); MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design, Ch. 11 and 12.

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