07-Str-A5 · Undated paper

Question 6 of 7: B3: Rectangular footing and pier for column AB

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

Notes on this paper

Paper format. National Exams — 07-Str-A5 Advanced Structural Design, printed footer 07-Str-A5/May 2019. Three hours, "closed book" with handbooks and textbooks permitted. Seven questions of equal value (20 marks each) in three parts: A1–A3 (do two), B1–B3 (do two), C1 (do one) — five solutions make a complete paper. All seven are solved here, because the set is a study resource rather than an exam attempt.

Design data given on page 1. Solutions to CAN/CSA S16 (steel), CAN/CSA A23.3 (concrete) and CAN/CSA O86 (timber). All loads shown on the figures are unfactored. All structural steel is G40.21 300W — $F_y = 300$ MPa, $F_u = 450$ MPa, $E = 200\,000$ MPa, $G = 77\,000$ MPa. All reinforcement is 400W — $f_y = 400$ MPa. Concrete strength is stated per question. Steel sections are W-shapes unless noted otherwise.

Reference texts.

CSA S16:19, Design of Steel Structures — Cl. 11 (section classification), 13.3 (compression), 13.4 (shear), 13.5–13.6 (bending and lateral–torsional buckling), 13.8 (beam-columns), 13.13 (welds), 14.3.2 (web crippling), 8.4.3 (stability effects).
CSA A23.3:19, Design of Concrete Structures — Cl. 8.6 (material properties), 10 (flexure and axial load, slenderness), 11 (shear), 15 (footings), 18 (prestressed concrete).
National Building Code of Canada 2020, Part 4 — Table 4.1.3.2 load combinations, 4.1.6 snow and drift loads.
CISC, Handbook of Steel Construction, 11th ed. — W-shape dimensions and properties.
Salmon, Johnson & Malhas, Steel Structures: Design and Behavior, 5th ed. — Ch. 9 (LTB), 12 (beam-columns), 13 (connections).
MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design, Canadian ed. — Ch. 4–6 (flexure), 11 (columns), 12 (slender columns), 15 (footings).
Collins & Mitchell, Prestressed Concrete Structures — Ch. 4 (service stresses), 6 (flexural strength); PCI Design Handbook, 8th ed. — multiplier method for long-term camber.

Check — load combinations. The figures split the loads into DEAD, LIVE, SNOW and WIND, so NBCC 2020 Table 4.1.3.2 is applied literally. Throughout Part A the governing case is $1.25D + 1.5L$ (case 2 with no companion snow or wind on the figure). For Figures 3 the governing case is $1.25D + 1.5S + 0.4W$ (case 3), with $1.25D + 1.4W + 0.5S$ and $1.4D$ checked and shown not to govern. The "LIVE/SNOW" label on W1 of Figure 3 is read as a single 3 kN/m specified roof load that may be either occupancy live or balanced snow, and the triangular W2 is read as the additional snow-drift surcharge that its shape implies.

Question 6 — B3: Rectangular footing and pier for column AB (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. Column A–B of Figure 2, 600 mm square reinforced concrete (as designed in Question 5), founded on a rectangular pad whose length is three times its width. Allowable bearing 150 kPa at service, ultimate bearing 225 kPa at factored load, footing concrete $f'_c = 25$ MPa.

Given data — footing design
Quantity	Symbol	Value
Service axial load at A (D + L)	$N_{ser}$	597.5 kN
Service horizontal shear at A	$H_{ser}$	12.5 kN
Factored axial load at A	$N_f$	806.2 kN
Factored horizontal shear at A	$V_f$	18.6 kN
Base moment	$M$	0 (pinned base)
Bearing: allowable / ultimate	$q_a$ / $q_u$	150 / 225 kPa

Find. Plan dimensions, thickness, reinforcement in both directions, and a detailed pier connection consistent with the pinned base the frame analysis assumed.

Approach. Size the plan on service bearing including the footing's own weight, then set the thickness by one-way shear, check punching, and design flexure as a cantilever from the column face. The pier detail must reproduce a pin, not a fixed base.

Size the plan area on service loads. Trying $B = 1.3$ m by $L = 3B = 3.9$ m at 0.55 m thick, the footing weighs $1.3(3.9)(0.55)(24) = 66.9$ kN, so $$q_{ser} = \frac{597.5 + 66.9}{1.3(3.9)} = \frac{664.4}{5.07} = 130.9\ \text{kPa} \ \le\ 150\ \text{kPa}$$
Account for the horizontal thrust. The 12.5 kN base shear acts through the footing thickness, so $M = 12.5(0.55) = 6.87$ kN$\cdot$m and $$e = \frac{6.87}{664.4} = 10.3\ \text{mm} \ \ll\ \frac{L}{6} = 650\ \text{mm}$$ The resultant stays deep inside the kern, so bearing remains trapezoidal with $$q_{\max} = q_{ser}\left(1 + \frac{6e}{L}\right) = 130.9(1.016) = 133.0\ \text{kPa} \ \le\ 150\ \text{kPa}\ \checkmark$$ Sliding is not an issue: with a friction coefficient of 0.5 the available resistance is 332 kN against 12.5 kN.
Net factored pressure. The footing's own weight is carried directly by the soil beneath it and does not bend the pad, so structural design uses the column load alone: $$q_f = \frac{806.2}{5.07} = \boxed{158.9\ \text{kPa}} \ \le\ 225\ \text{kPa}\ \checkmark$$
Set the thickness by one-way shear. With 75 mm cover against soil and 20M bars, $d = 550 - 75 - 20 = 455$ mm and $d_v = \max(0.9d,\,0.72h) = 409.5$ mm. The critical section sits $d$ from the column face, i.e. $1.65 - 0.455 = 1.195$ m in from the end, so $V_f = 158.9(1.195)(1.3) = 246.8$ kN. With no stirrups, $\beta = 230/(1\,000 + d_v) = 0.1632$ and $$V_c = \phi_c\lambda\beta\sqrt{f'_c}\,b\,d_v = 0.65(0.1632)(5)(1\,300)(409.5)/10^3 = 282.3\ \text{kN} \ \ge\ 246.8\ \text{kN}$$ This is the check that fixes $h$: a 500 mm pad fails it.
Check punching. The critical perimeter lies $d/2$ from the column face, $b_o = 4(600 + 455) = 4\,220$ mm, and the load inside it is deducted: $$V_f = 806.2 - 158.9(1.055)^2 = 629.1\ \text{kN}$$ Of the three Cl. 13.3.4 expressions, $0.38$ governs over $(1 + 2/\beta_c)0.19 = 0.57$ and $\alpha_sd/b_o + 0.19 = 0.62$, so $v_c = 0.38\phi_c\sqrt{f'_c} = 1.235$ MPa and $$V_r = v_cb_od = 1.235(4\,220)(455)/10^3 = 2\,372\ \text{kN} \ \gg\ 629\ \text{kN}$$
Design the flexural steel, long direction. The cantilever from the column face is $(3.9 - 0.6)/2 = 1.65$ m, so $$M_f = q_fB\frac{c^2}{2} = 158.9(1.3)\frac{1.65^2}{2} = 281.2\ \text{kN}\cdot\text{m}$$ Trying 7–20M across the 1.3 m width ($A_s = 2\,100$ mm$^2$), $a = 41.6$ mm and $$M_r = 0.85(2\,100)(400)(455 - 20.8)/10^6 = \boxed{310.0\ \text{kN}\cdot\text{m}} \ \ge\ 281.2\ \text{kN}\cdot\text{m}$$ This exceeds $A_{s,\min} = 0.002bh = 1\,430$ mm$^2$. Bars at about 180 mm centres.
Design the flexural steel, short direction. The cantilever is only $(1.3 - 0.6)/2 = 0.35$ m, giving $M_f = 38.0$ kN$\cdot$m over the full 3.9 m — trivial. Minimum steel governs: $A_{s,\min} = 0.002(3\,900)(550) = 4\,290$ mm$^2$, so provide 15–20M ($4\,500$ mm$^2$) at about 260 mm centres. Cl. 15.4.4 requires a fraction $2/(\beta+1) = 0.5$ of this steel to be banded within a central strip equal to the short side, so 8 of the 15 bars are concentrated in the middle 1.3 m.
Proportion the pier and check bearing. Place an 800 mm square pier, 900 mm high, between the 600 mm column and the pad. Cl. 10.8 bearing on the pier, with the load spread allowed by the larger area, gives $$B_r = 0.85\phi_cf'_cA_1\sqrt{A_2/A_1} = 0.85(0.65)(25)(360\,000)(1.333)/10^3 = 6\,628\ \text{kN} \ \gg\ 806\ \text{kN}$$
Detail the connection as a pin. The frame analysis assumes zero moment at A, so the joint must be detailed to deliver that rather than to develop the column bars:
- 4–25M dowels bundled at the centre of the column, satisfying the Cl. 15.8.2.1 minimum of $0.005A_g = 1\,800$ mm$^2$, projecting 700 mm into the column and turned down into the pier with 90° standard hooks. Keeping the dowels in a central cluster gives them almost no lever arm, so the joint has negligible flexural stiffness.
- No dowels near the column faces, and a 20 mm compressible joint filler around the perimeter of the column base so that the edges can lift as the joint rotates.
- Shear transfer by shear friction across the intentionally roughened (6 mm amplitude) construction joint. With $c = 0.25$ MPa and $\mu = 1.0$, $V_r = \phi_c(cA_{cv} + \mu N_f) = 0.65[0.25(360\,000) + 806\,200]/10^3 = 582.5$ kN against the 18.6 kN applied — the compression alone is more than sufficient and no shear key is required.
- Confinement: 10M ties at 100 mm through the top 400 mm of the pier and the bottom 400 mm of the column, to contain the bearing stresses at the reduced contact area.

Question 6 — results
Quantity	Value
Footing plan	1.3 m × 3.9 m (3:1), 550 mm thick, $f'_c = 25$ MPa
Service bearing pressure (mean / peak)	130.9 / 133.0 kPa (allowable 150)
Eccentricity at base / kern limit	10.3 mm / 650 mm
Net factored pressure	158.9 kPa (ultimate 225)
One-way shear $V_f$ / $V_c$	246.8 / 282.3 kN
Punching $V_f$ / $V_r$	629.1 / 2 372 kN
Flexure, long direction $M_f$ / $M_r$	281.2 / 310.0 kN$\cdot$m — 7–20M
Flexure, short direction	15–20M (minimum steel; 8 bars banded centrally)
Pier	800 mm square × 900 mm; $B_r = 6\,628$ kN
Connection	4–25M central dowels, hooked; shear friction $V_r = 582.5$ kN

Topic. Rectangular spread footing on a prescribed aspect ratio, with a pinned column-to-pier joint.

Key relations. Service sizing $q = (N_{ser}+W_{ftg})/A \le q_a$; kern limit $e \le L/6$; one-way $V_c = \phi_c\lambda\beta\sqrt{f'_c}\,bd_v$ with $\beta = 230/(1\,000+d_v)$; two-way $v_c = \min[0.38,\ (1+2/\beta_c)0.19,\ \alpha_sd/b_o+0.19]\phi_c\sqrt{f'_c}$; bearing $B_r = 0.85\phi_cf'_cA_1\sqrt{A_2/A_1}$; shear friction $V_r = \phi_c(cA_{cv}+\mu N)$.

Why this works. The 3:1 aspect ratio is imposed, not derived — it is what you get on a site with a property line or a services corridor — and it changes the design character completely: the long cantilever of 1.65 m is what sets both the thickness (through one-way shear) and the reinforcement, while punching, which usually governs a square pad, is here beaten by a factor of nearly four. Because the base is pinned, no moment reaches the footing except the small couple from the base shear acting through the pad thickness, so the eccentricity is negligible and bearing stays almost uniform. The pier detail then has to be honest about that assumption: dowels spread to the column faces would build a partially fixed base, redistributing moment into the footing that the analysis never accounted for.

Common pitfall. Including the footing self weight in the factored pressure used for bending and shear. Soil reaction under the pad's own weight is self-equilibrating and produces no flexure; adding it inflates $q_f$ by about 13% and can push a correct design into an unnecessary thickness increase. The second is sizing the pad on the ultimate bearing capacity of 225 kPa — that value is a factored check, and the service check against 150 kPa is the one that governs the plan area here.

Source. CSA A23.3:19 Cl. 10.8, 11.3.6, 13.3.4, 15.3–15.4, 15.8.2 and 11.5 (shear friction); MacGregor & Bartlett, Reinforced Concrete, Canadian ed., Ch. 15.

← Question 5 Next: Question 7 →