16-Civ-B2 Advanced Structural Design · December 2014

Question 5 of 7: Footing at E and the beam-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 5: Footing at E and the beam-column CE (12 + 8 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
Structure	Figure 4 frame, section W610 × 125 from Question 4
Allowable soil bearing pressure	300 kPa (service)
Lateral support	at joints C and E and at the load point D only
Concrete for the footing	$f_c' = 30$ MPa, unit weight 24 kN/m³

Find. A footing plan size and thickness that keeps the service bearing pressure below 300 kPa with the whole base in compression, and a verdict on the adequacy of W610 × 125 as the beam-column CE.

Approach. Soil bearing is a serviceability question, so the frame must be re-analysed elastically under unfactored loads — the plastic redundant of Question 4 is not valid at service. The same elastic analysis, scaled by 1.5, then supplies the factored actions for the Clause 13.8 beam-column checks.

Part (a) — analyse the frame elastically. With one redundant, the unit-load method on the primary structure (roller released at A) gives $$\delta_{11} = \frac{16^3}{3} + 16^2(10) = 3925, \qquad \delta_{10} = -724\,267 \quad\Rightarrow\quad V_A = \frac{724\,267}{3925} = 184.5\ \text{kN}$$ Both members are the same section, so a uniform $EI$ is exact here.
Extract the service base reactions. $$V_E = 800 - 184.5 = 615.5\ \text{kN}, \qquad H_E = 80\ \text{kN}, \qquad M_E = 16(184.5) - 3600 = \boxed{-648\ \text{kN}\cdot\text{m}}$$ The base moment is large relative to the axial load — an eccentricity of more than a metre — which is what drives the footing size.
Size the footing so the resultant stays in the middle third. Try a pad 5.0 m (in the plane of the frame) × 3.0 m × 1.0 m thick, weighing $5(3)(1)(24) = 360\ \text{kN}$. Then $$P = 615.5 + 360 = 975.5\ \text{kN}, \qquad e = \frac{648}{975.5} = 0.664\ \text{m} \;<\; \frac{L}{6} = 0.833\ \text{m}$$ The footing's own weight is doing real work here: it is what pulls the eccentricity back inside the kern and keeps the whole base bearing.
Check the bearing pressures. $$q_{max} = \frac{P}{A}\left(1 + \frac{6e}{L}\right) = \frac{975.5}{15}(1.797) = \boxed{117\ \text{kPa}} \;<\; 300\ \text{kPa}$$ $$q_{min} = \frac{975.5}{15}(0.203) = 13.2\ \text{kPa} \;>\; 0 \quad\text{(no uplift)}$$
Check sliding. With a base friction coefficient of 0.45 on a granular subgrade, $$FS = \frac{0.45(975.5)}{80} = 5.5 \;>\; 1.5 \quad\checkmark$$ No shear key is needed.
Confirm the thickness structurally. Under factored loads the elastic redundant scales to $V_A = 276.8\ \text{kN}$, giving $V_E = 923.2\ \text{kN}$ and $M_E = 972\ \text{kN}\cdot\text{m}$, and a factored bearing distribution of 169 kPa to 14 kPa. With $d_v \approx 905\ \text{mm}$ the one-way shear critical section falls 1295 mm from the toe; the shear there is well inside $V_c = \phi_c\lambda\beta\sqrt{f_c'}b\,d_v$ for the 3.0 m width, and punching around a 600 mm base plate is not close. Flexural steel is governed by the minimum $0.002A_g$: provide 20M at 250 mm each way, top and bottom, with the heavier mat in the 5.0 m direction.
Part (b) — assemble the beam-column actions. From the same factored elastic analysis, $$C_f = 923\ \text{kN}, \qquad M_{f,E} = 972\ \text{kN}\cdot\text{m}, \qquad M_{f,C} = 372\ \text{kN}\cdot\text{m}$$ The column bends in single curvature between C and E, with the moment constant from C down to D and rising linearly to the base.
Compute the compressive resistances. The frame sways in plane (no horizontal restraint at A), so take $K_x = 1.5$; out of plane the member is braced at C, D and E, so $K_yL = 5000\ \text{mm}$. With $r_x = 248.4$ and $r_y = 49.9\ \text{mm}$ and $n = 1.34$, $$C_{r,x} = 3574\ \text{kN}, \qquad C_{r,y} = 2105\ \text{kN}, \qquad C_{r0} = \phi AF_y = 4975\ \text{kN}$$ and $M_{rx} = \phi Z_xF_y = 1145\ \text{kN}\cdot\text{m}$.
Apply the first two Clause 13.8.2 checks. $$\text{(a) cross-section: } \frac{923}{4975} + 0.85\frac{972}{1145} = 0.186 + 0.721 = 0.91 \le 1.0 \quad\checkmark$$ $$\text{(b) in-plane member: } \frac{923}{3574} + 0.85\frac{972}{1145} = 0.258 + 0.721 = 0.98 \le 1.0 \quad\checkmark$$ Both pass, the second only just. The axial-moment interaction of plastic design is also satisfied: with $C_f/C_y = 0.167$, the reduced plastic moment is $M_{pc} = 1.18M_p(1 - C_f/C_y) = 1251 > 1080\ \text{kN}\cdot\text{m}$.
Apply the lateral-torsional buckling check — and find it fails. The unbraced segment D to E is 5.0 m with $\kappa = 372/972 = 0.383$ in single curvature, so $\omega_2 = 1.39$ and $$M_u = \frac{\omega_2\pi}{L}\sqrt{EI_yGJ + \left(\frac{\pi E}{L}\right)^2I_yC_w} = 1523\ \text{kN}\cdot\text{m}$$ $$M_{r,LTB} = 1.15\phi M_p\left(1 - \frac{0.28M_p}{M_u}\right) = 1009\ \text{kN}\cdot\text{m}$$ $$\text{(c) } \frac{923}{2105} + 0.85\frac{972}{1009} = 0.439 + 0.819 = \boxed{1.26 \;>\; 1.0} \quad\text{FAILS}$$
Give the remedy and re-check. Add one lateral brace at mid-height of the segment D–E, i.e. 2.5 m above the base. The unbraced length halves, the moment at the brace is 672 kN·m so $\omega_2 = 1.17$, and $$M_u = 4511\ \text{kN}\cdot\text{m} \;\Rightarrow\; M_{r,LTB} = \phi M_p = 1145\ \text{kN}\cdot\text{m}, \qquad C_{r,y} = 4003\ \text{kN}$$ $$\frac{923}{4003} + 0.85\frac{972}{1145} = 0.231 + 0.721 = \boxed{0.95 \le 1.0} \quad\checkmark$$ Verdict: W610 × 125 is not adequate as the beam-column CE with lateral support only at C, D and E, because lateral-torsional buckling of the 5 m base segment overstresses it by 26 per cent. It becomes adequate, with a five per cent reserve, once a single additional brace is added at mid-height of D–E. Upsizing instead of bracing does not help much: W610 × 140 unbraced over 5 m still returns 1.10, because the deficiency is slenderness, not area.

Result	Value
Service redundant / base reactions	$V_A = 184.5$ kN; $V_E = 615.5$ kN, $H_E = 80$ kN, $M_E = 648$ kN·m
Footing size	5.0 m × 3.0 m × 1.0 m thick
Eccentricity / kern	0.664 m < 0.833 m (whole base bearing)
Bearing pressures	117 kPa max, 13.2 kPa min (allowable 300 kPa)
Sliding factor of safety	5.5
Footing reinforcement	20M at 250 mm each way, top and bottom
Factored column actions	$C_f = 923$ kN, $M_f = 972$ kN·m at E
Clause 13.8.2(a) cross-section	0.91 — passes
Clause 13.8.2(b) in-plane	0.98 — passes
Clause 13.8.2(c) lateral-torsional	1.26 — FAILS as braced
With extra brace at mid-height of DE	0.95 — adequate

Topic. Service-load footing design under a large base moment, and the three-part beam-column check of CSA S16 Clause 13.8.

Key relations. $q = \frac{P}{A}\left(1 \pm \frac{6e}{L}\right)$ valid while $e \le L/6$; $C_r = \phi AF_y(1+\lambda^{2n})^{-1/n}$ with $n = 1.34$; $M_u = \frac{\omega_2\pi}{L}\sqrt{EI_yGJ + (\pi E/L)^2I_yC_w}$; $M_r = 1.15\phi M_p(1 - 0.28M_p/M_u)$ when $M_u > 0.67M_p$; Clause 13.8.2 checks (a) cross-section, (b) in-plane member, (c) lateral-torsional.

Why this works. The three Clause 13.8 checks are not redundant restatements of one another; each fails for a different physical reason and each uses a different resistance pair. Check (a) asks whether the section can take the actions at all, check (b) whether the member buckles in the plane of bending, and check (c) whether it buckles out of plane while bending. A beam-column with a slender weak axis and a stocky strong axis — which describes almost every deep W-shape used as a column — will sail through (a) and (b) and fail (c), so the third check is the one to run first when the answer is in doubt. On the footing side, the middle-third rule is not a convention but the condition under which the linear pressure formula is valid at all; once $e$ exceeds $L/6$ part of the base lifts and the pressure must be recomputed on the reduced contact area.

Common pitfall. Using the plastic-analysis redundant to compute service bearing pressures. The plastic redundant belongs to the collapse state, and at service the frame is still elastic: here $V_A$ is 184.5 kN elastically against 270 kN at collapse, which changes the base moment by nearly half. The second pitfall is omitting the footing's own weight from the eccentricity calculation — without it this pad lifts off.

Source. CSA S16:19 Clauses 13.3, 13.6 and 13.8; CFEM Canadian Foundation Engineering Manual, 4th ed., Ch. 10 for shallow-foundation bearing; Kulak and Grondin, Limit States Design in Structural Steel, Ch. 8.

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