07-Str-A5 · December 2013

Question 3 of 7: Reinforced concrete column AB and its footing

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

Notes on this paper

Paper format. 07-Str-A5 Advanced Structural Design, National Exams, December 2013 — 3 hours, closed book (handbooks and textbooks permitted, no notes). Seven questions of equal value; any five constitute a complete paper. All seven are solved here, because the set is intended as a study resource. All loads printed on the figures are unfactored.

Design data given on the paper (SI). Concrete $f'_c = 30$ MPa; structural steel $F_y = 350$ MPa; reinforcing steel $f_y = 400$ MPa. Prestressed concrete: $f'_{ci} = 35$ MPa at transfer, $f'_c = 50$ MPa, $n = 6$, $f_{pu} = 1750$ MPa, $f_{py} = 1450$ MPa, $f_{initial} = 1200$ MPa, losses $= 240$ MPa (hence $f_{pe} = 960$ MPa).

Reference texts. CSA S16 Design of Steel Structures with the CISC Handbook of Steel Construction — plate girders (Cl. 14), lateral–torsional buckling (Cl. 13.6), plastic design (Cl. 8.6, 13.7), beam-columns (Cl. 13.8), composite beams (Cl. 17), connections (Cl. 21); CSA A23.3 Design of Concrete Structures — flexure and shear (Cl. 10, 11), slenderness (Cl. 10.13–10.16), footings (Cl. 15), prestressed concrete (Cl. 18); NBCC for load combinations; C. G. Salmon, J. E. Johnson & F. A. Malhas, Steel Structures: Design and Behavior; J. G. MacGregor & J. K. Wight, Reinforced Concrete: Mechanics and Design; T. Y. Lin & N. H. Burns, Design of Prestressed Concrete Structures; L. S. Beedle, Plastic Design of Steel Frames.

Check: load factors. The paper states only that the printed loads are unfactored, and gives no dead/live split. Every question below therefore treats each printed concentrated load as specified live load and factors it by $\alpha_L = 1.5$ (NBCC principal case $1.25D + 1.5L$); member self-weight, wherever it matters, is estimated from the trial section and factored at $1.25$. A candidate assuming a different split would obtain proportionally different sizes; the method is what is examined.

Check: section properties. All steel sections selected below are quoted by their plate dimensions (flange $b \times t$, web $h \times w$) and every property — $A$, $I_x$, $I_y$, $Z$, $J$, $C_w$, $r_x$, $r_y$ — is computed from those dimensions with root fillets neglected (slightly conservative). This makes every line checkable without a handbook; a rolled shape of equal or greater properties may be substituted directly.

Check: reading of Figures 3 and 4. On the examination drawing the dimension chain under Figure 3 reads 3 m + 4 m + 7 m + 4 m, and the overall dimension reads 15 m. Reading the drawing directly resolves this: the 15 m dimension line spans B to C ($4 + 7 + 4 = 15$ m) and the 3 m is a cantilever overhang projecting beyond support B to the free end A, which carries the 80 kN load. Similarly in Figure 4 the 10.5 m is the overall width of the floor ($3 \times 3.5$ m beam spacing), not a span; the design span is the 15 m stated in the note. Both readings are self-consistent and are used throughout.

Check: support conditions in Figure 2. The frame drawing shows identical triangle-and-roller symbols at A and D, and the frame must resist the 80 kN horizontal load, so both bases are taken as pinned. The frame is then statically indeterminate to the first degree, which is the classical form of this problem for both the elastic (Questions 2, 3) and the plastic (Questions 5, 6) treatments.

The three welded I-sections selected in this paper, drawn to a common scale.

Question 3: Reinforced concrete column AB and its footing (14 + 6 = 20 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 design starts from the same pinned-base portal analysed in Question 2: column AB is 10 m high, pinned at A, and framed rigidly into beam BC at B.

Given data
Quantity	Symbol	Value
Column height (pinned base)	$l_u$	10 000 mm
Factored axial load at A	$P_f$	1352.5 kN (1512.5 kN at D)
First-order factored moment at B	$M_{f1}$	1846.2 kN·m (load-reversal envelope)
Concrete / steel	$f'_c$ / $f_y$	30 MPa / 400 MPa
Assumed allowable bearing pressure	$q_a$	200 kPa

Find. A rectangular column section with its longitudinal and tie reinforcement, satisfying strength and slenderness, and the plan size, thickness and reinforcement of the spread footing at A.

Factored bending moments in the frame. The column moment is zero at the pinned base and equals the beam end moment at the joint.

Check: envelope for the lateral load. The 80 kN horizontal load is a lateral action and must be assumed reversible. With the load acting to the right the moments are 646 kN·m at B and 1846 kN·m at C; reversed, they swap. Column AB is therefore designed for the envelope — the larger moment, 1846 kN·m, combined with the smaller axial load, 1352.5 kN, which is the more critical pairing for a section governed by flexure. The same section then serves both columns.

Approach. Establish the effective length of the sway column, decide whether slenderness must be considered, magnify the sway moment by the A23.3 Cl. 10.16 stability index, then find the section and steel that satisfy the axial–moment interaction; finally size the footing on the assumed bearing pressure and check punching, one-way shear and flexure.

Choose a trial section and find the effective length. Take the column as $600 \times 1400$ mm, bending about the 1400 mm dimension in the plane of the frame. With A23.3 Cl. 10.14 stiffnesses ($0.70I_g$ for columns, $0.35I_g$ for beams) the joint restraint factors are $\psi_B = 3.60$ at the top and $\psi_A = 10$ at the pinned base, giving for a sway frame $$k = \sqrt{\frac{1.6\psi_A\psi_B + 4(\psi_A + \psi_B) + 7.5}{\psi_A + \psi_B + 7.5}} = 2.38.$$
Test for slenderness. With $r = 0.3h = 420$ mm, $kl_u/r = 2.38(10\,000)/420 = 56.7$. This exceeds the sway-frame threshold of 22, so second-order effects must be included.
Magnify the sway moment. Using $EI = 0.4E_cI_g/(1 + \beta_d)$ with $\beta_d = 0.35$, $$P_c = \frac{\pi^2 EI}{(kl_u)^2} = 17\,461\ \text{kN per column}, \qquad \delta_s = \frac{1}{1 - \dfrac{\sum P_f}{0.75\sum P_c}} = \frac{1}{1 - \dfrac{3369}{26\,192}} = 1.148.$$ Only the sway component of the moment is magnified. That component is the 600 kN·m produced by the factored horizontal load, so $$M_f = M_{ns} + \delta_s M_s = 1246.2 + 1.148(600) = \boxed{1935\ \text{kN}\cdot\text{m}}$$ acting together with $P_f = 1352.5$ kN.
Provide the longitudinal steel. The minimum column reinforcement of 1 % requires $0.01(600)(1400) = 8400$ mm², met exactly by $12 - 30$M arranged as six bars in each 600 mm face, with $d = 1330$ mm and $d' = 70$ mm.
Check the interaction. Strain compatibility with $\varepsilon_{cu} = 0.0035$ and the rectangular stress block gives, at the applied axial load of 1352.5 kN, a neutral axis at $c = 161.7$ mm ($a = 144.7$ mm). Both bar layers are then at yield, and taking moments about the section centroid, $$M_r = C_c\!\left(\frac{h}{2} - \frac{a}{2}\right) + C_s\!\left(\frac{h}{2} - d'\right) + T\!\left(d - \frac{h}{2}\right) = \boxed{2648\ \text{kN}\cdot\text{m}} \; > \; 1935\ \text{kN}\cdot\text{m}.$$ The column is on the tension-controlled branch of the interaction diagram — the axial load actually increases the moment resistance — which is why the low-axial, high-moment pairing was selected as critical.
Detail the ties. Tie spacing is limited to the least of $16d_b = 478$ mm, $48d_{tie} = 542$ mm and the least column dimension 600 mm. Provide 10M ties at 450 mm, closed, with cross-ties engaging every alternate longitudinal bar, and close the spacing to 150 mm over the top 1.4 m to confine the joint region where the full moment is delivered.
Size the footing. The base is pinned, so the footing carries axial load and horizontal shear but no applied moment. The service axial load is 1230.9 kN including the column self-weight, and the service base shear under the reversed lateral case is 123.1 kN. Trying a $3.0 \times 3.0 \times 0.8$ m pad founded 1.5 m down, the footing weighs 172.8 kN and the overburden 113.4 kN, so $$q_{avg} = \frac{1230.9 + 172.8 + 113.4}{9.0} = 168.6\ \text{kPa}, \qquad e = \frac{123.1 \times 0.8}{1517.1} = 65\ \text{mm},$$ $$q_{max} = q_{avg}\left(1 + \frac{6e}{B}\right) = \boxed{190\ \text{kPa} \le q_a = 200\ \text{kPa}}.$$ The resultant stays well inside the middle third, and base friction at $\mu = 0.5$ gives 758 kN against the 123 kN shear, so sliding is not critical.
Check the footing structurally. With a factored net pressure of 178.3 kPa and $d = 705$ mm, the punching perimeter is $b_o = 6820$ mm and $\beta_c = 2.33$, so $v_c = 0.353\phi_c\sqrt{f'_c} = 1.26$ MPa and $V_r = v_c b_o d = 6040$ kN against $V_f = 1115$ kN — ample. One-way shear is trivial because the critical section falls within 95 mm of the edge. The largest cantilever moment, on the 1200 mm projection, is 128.4 kN·m/m, which needs only 564 mm²/m; minimum steel $0.002bh = 1600$ mm²/m governs, so provide 20M at 180 mm each way in the bottom.

Question 3 — final results
Item	Result
Column section	$600 \times 1400$ mm rectangular
Longitudinal steel	$12 - 30$M (8400 mm², $\rho = 1.0\%$), six per 600 mm face
Ties	10M closed ties at 450 mm; 150 mm over the top 1.4 m
Effective length	$k = 2.38$, $kl_u/r = 56.7$ (slender, sway)
Moment magnifier	$\delta_s = 1.148$; $M_f = 1935$ kN·m with $P_f = 1352.5$ kN
Strength check	$M_r = 2648$ kN·m $> 1935$ kN·m ✓
Footing	$3.0 \times 3.0 \times 0.8$ m, founded 1.5 m below grade
Bearing pressure	$q_{max} = 190$ kPa vs assumed $q_a = 200$ kPa ✓
Footing steel	20M at 180 mm each way, bottom (minimum governs)

Topic. Slender reinforced concrete column in a sway frame, and the spread footing that carries it.

Key relations. $k$ from the sway alignment chart or its algebraic approximation; $P_c = \pi^2EI/(kl_u)^2$ with $EI = 0.4E_cI_g/(1+\beta_d)$; $\delta_s = 1/[1 - \sum P_f/(0.75\sum P_c)]$; interaction by strain compatibility; footing bearing $q = P/A(1 + 6e/B)$; two-way shear $v_c = \min[(1 + 2/\beta_c)0.19,\ (\alpha_s d/b_o + 0.19),\ 0.38]\lambda\phi_c\sqrt{f'_c}$.

Why this works. The column sits far up the tension-controlled side of its interaction diagram: at $e = M/P = 1430$ mm the eccentricity exceeds the section depth, so the member behaves essentially as a beam with a helpful axial thrust. That is why the lower axial load is the critical pairing, and why choosing the load case by "largest $P$" would be wrong. The pinned base is what keeps the footing simple — no applied moment means a concentric pad rather than a much larger eccentrically loaded one — but it is also what drives $k$ up to 2.38 and forces the second-order check. Those two consequences of the same detail are worth seeing together.

Common pitfall. Neglecting slenderness because the column "looks stocky" at 1400 mm deep. In a sway frame it is $kl_u/r$ with $k \approx 2.4$, not $l_u/h$, that matters, and the threshold is 22 rather than the 34 of a braced frame. A second pitfall is magnifying the whole moment instead of only the sway component, which here would overstate $M_f$ by about 8 %.

Source. CSA A23.3 Cl. 10.9, 10.13–10.16 and 15.3–15.4; MacGregor & Wight, Ch. 11 (slender columns) and Ch. 15 (footings).

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