16-Civ-B2 Advanced Structural Design · May 2018

Question 7 of 7: Long-Term Deflection of AB and the Footing at C

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

Notes on this paper

Paper format. 16-Civ-B2 Advanced Structural Design, May 2018 — a three-hour examination. The cover page allows design handbooks and textbooks but no notes, states that any five questions constitute a complete paper, that all seven questions carry equal value, and that all loads shown on the figures are unfactored. The mark split printed on page 1 is Q1 (12+6+2), Q2 (12+8), Q3 (12+4+4), Q4 (14+6), Q5 (10+5+5), Q6 (10+4+6) and Q7 (12+8). All seven questions are worked below, because the set is a study resource rather than an answer book.

Design data (page 1, reproduced verbatim). These values are used throughout and are not repeated in every question.

Design data supplied with the paper
Material	Property	Value
Concrete	Cylinder strength $f'_{c}$	30 MPa
Structural steel	$F_{y}$	350 MPa
Reinforcing steel	$f_{y}$	400 MPa
Prestressed concrete	$f_{ci}$ at transfer	35 MPa
	$f'_{c}$	50 MPa
	Modular ratio $n$	6
	$f_{ult}$	1750 MPa
	$f_{y}$ (strand)	1450 MPa
	$f_{initial}$	1200 MPa
	Loss of prestress	240 MPa

Reference texts. The answers are written to the Canadian limit-states codes that Engineers Canada lists for this examination:

CSA S16:19, Design of Steel Structures (with the CISC Handbook of Steel Construction, 11th ed.).
CSA A23.3:19, Design of Concrete Structures (with the Cement Association of Canada Concrete Design Handbook, 4th ed.).
National Building Code of Canada 2020, Part 4, for load combinations.
CSA S6:19, Canadian Highway Bridge Design Code, for the bridge floor of Question 4.
Kulak and Grondin, Limit States Design in Structural Steel, 9th ed.; MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design (Canadian edition); Collins and Mitchell, Prestressed Concrete Structures.

Check — load factors. Page 1 says only that the loads shown are unfactored; it gives no dead-to-live split. Every question below therefore applies a single $\alpha = 1.5$ to the loads printed on the figures (the NBCC live-load factor, which is the conservative reading of an unidentified load) and $\alpha_{D} = 1.25$ to any self weight the solution itself introduces. Mechanisms, section classifications and all interaction equations are independent of that choice; only the magnitudes move with it. State this assumption on the answer paper, as the cover page invites.

Question 7: Long-Term Deflection of AB and the Footing at C (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. The frame and sections of Questions 5 and 6, now under service loads — 400 kN unfactored at each load point plus 14.4 kN/m of beam self weight — and an allowable bearing pressure of 400 kPa at C.

Service actions from the frame analysis
Quantity	Value
Reaction at A / D	133.8 kN / 329.7 kN
Reaction at C (vertical)	909.3 kN
Moment at mid-span of AB	+336.6 kN·m
Moment at B, AB side	−656.4 kN·m
Column moment (constant)	24.4 kN·m
Beam section	500 × 1200 mm, 6-25M top, 4-25M bottom
Allowable bearing pressure	400 kPa

Find. (i) the long-term deflection at mid-span of AB; (ii) a plan size for the spread footing under C.

Approach. For the deflection, compare the service moments with the cracking moment section by section, form Branson effective inertias, average them as Clause 9.8.2.4 directs, recover the immediate deflection from the frame model and multiply by the creep factor. For the footing, size the base on service loads and the allowable pressure, then check the thickness for punching and one-way shear at factored load.

Part (i) — is the mid-span section cracked? The modulus of rupture is $f_{r} = 0.6\lambda\sqrt{f'_{c}} = 3.286$ MPa, so with $I_{g} = 7.2 \times 10^{10}\ \text{mm}^{4}$ and $y_{t} = 600$ mm, $$M_{cr} = \frac{f_{r}I_{g}}{y_{t}} = 394\ \text{kN}\cdot\text{m}$$ At mid-span $M_{a} = 337 \lt 394$ kN·m, so that section is still uncracked and $I_{e} = I_{g}$. At B, $M_{a} = 656$ kN·m exceeds $M_{cr}$ and the section is cracked.
Cracked inertia at B. With $n = E_{s}/E_{c} = 8.11$, 6-25M in tension at $d = 1130$ mm and 4-25M in compression at 70 mm, the transformed neutral axis follows from $b c^{2}/2 + (n-1)A_{s}^{\prime}(c-d^{\prime}) = nA_{s}(d-c)$, giving $c = 269$ mm and $$I_{cr} = \frac{bc^{3}}{3} + (n-1)A_{s}^{\prime}(c-d^{\prime})^{2} + nA_{s}(d-c)^{2} = 2.185 \times 10^{10}\ \text{mm}^{4} = 0.30 I_{g}$$ Branson then gives $I_{e} = I_{cr} + (I_{g}-I_{cr})(M_{cr}/M_{a})^{3} = 3.27 \times 10^{10}\ \text{mm}^{4}$ at B.
Averaged inertia and immediate deflection. AB is continuous at one end only, so Clause 9.8.2.4 weights the two values as $$I_{e,avg} = 0.85I_{e,mid} + 0.15I_{e,cont} = 0.85(7.20) + 0.15(3.27) = 6.61 \times 10^{10}\ \text{mm}^{4}$$ Re-running the frame with that stiffness under service load gives an immediate mid-span deflection of $$\Delta_{i} = 0.59\ \text{mm} = \frac{L}{10\,200}$$ The hogging restraint at B is what makes this so small: the same span treated as simply supported would deflect about 1.3 mm.
Creep and shrinkage. Clause 9.8.2.5 multiplies the sustained part of the immediate deflection by $\zeta = s/(1+50\rho^{\prime})$, with $s = 2.0$ beyond five years. Taking $\rho^{\prime} = 0$ at the critical section (the compression face at mid-span carries no design steel) is the conservative reading, so $$\Delta_{LT} = \Delta_{i}(1 + \zeta) = 0.59(3.0) = \boxed{1.8\ \text{mm}} = \frac{L}{3400}$$

Check — sustained fraction. The paper gives no dead-to-live split, so the whole 400 kN has been treated as sustained, which is the upper bound. If the point loads were entirely transient, only the 14.4 kN/m self weight would creep and the long-term value would fall to about 0.5 mm. Either way the answer is the same engineering statement: at 1200 mm deep over a 6 m span the member is far stiffer than serviceability requires — the limits are $L/240 = 25$ mm and $L/360 = 17$ mm, and the beam uses seven per cent of the tighter one.

Part (ii) — plan area of the footing. The service load at C is 909.3 kN from the frame plus 29.4 kN of column self weight. Trying a 1.8 m square pad 600 mm thick adds $1.8^{2}(0.6)(24) = 46.7$ kN, so $N = 985.3$ kN with $M = 24.4$ kN·m. The eccentricity $$e = \frac{M}{N} = \frac{24.4}{985.3} = 0.025\ \text{m} \ll \frac{B}{6} = 0.30\ \text{m}$$ keeps the whole base in contact, so $$q_{\max,\min} = \frac{N}{B^{2}}\left(1 \pm \frac{6e}{B}\right) = 304.1(1 \pm 0.083) \;\Rightarrow\; \boxed{q_{\max} = 329\ \text{kPa} \lt 400\ \text{kPa}}$$ A 1.7 m square would also work at 369 kPa; 1.6 m fails at 417 kPa. Adopt 1.8 m square for a round dimension and a sensible margin.
Thickness: two-way (punching) shear. At factored load $N_{f} = 1374$ kN the net pressure is $1374/1.8^{2} = 424$ kPa. With 75 mm cover and 15M bars, $d = 505$ mm, and the critical perimeter at $d/2$ from the column face is $b_{o} = 4(600+505) = 4420$ mm. Deducting the pressure inside that perimeter, $V_{f} = 1374 - 424(1.105)^{2} = 856$ kN, while $$v_{c} = 0.38\phi_{c}\lambda\sqrt{f'_{c}} = 1.353\ \text{MPa} \;\Rightarrow\; V_{r} = v_{c}b_{o}d = 3020\ \text{kN}$$ a ratio of 0.28. One-way shear at $d$ from the face is trivial at 73 kN against 583 kN.
Flexural steel in the footing. The cantilever from the column face is $(1.8-0.6)/2 = 0.6$ m, so $M_{f} = 424(1.8)(0.6)^{2}/2 = 137$ kN·m across the full width, needing only $842\ \text{mm}^{2}$. Shrinkage and temperature steel governs instead: $A_{s,\min} = 0.002A_{g} = 0.002(1800)(600) = 2160\ \text{mm}^{2}$. Provide 11-15M each way ($2200\ \text{mm}^{2}$, about 160 mm centres) in the bottom mat.

Spread footing at C sized on the 400 kPa allowable bearing pressure.

Question 7 — final results
Quantity	Value
$M_{cr}$ for the beam	394 kN·m
Mid-span service moment	337 kN·m — uncracked
$I_{cr}$ at B / $I_{e}$ at B	$2.19 \times 10^{10}$ / $3.27 \times 10^{10}$ mm$^{4}$
$I_{e,avg}$ (Cl 9.8.2.4)	$6.61 \times 10^{10}$ mm$^{4}$
Immediate deflection	0.59 mm ($L/10\,200$)
Long-term deflection	1.8 mm ($L/3400$), all load sustained
Deflection limits	$L/240 = 25$ mm, $L/360 = 17$ mm — both easily met
Footing size	1.8 m square × 600 mm thick
Service bearing pressure	279 to 329 kPa against 400 kPa allowed
Punching shear	856 kN against 3020 kN, ratio 0.28
Footing reinforcement	11-15M each way, bottom

Concept & Theory

Topic. Serviceability of a continuous concrete beam, and proportioning a spread footing on an allowable bearing pressure.

Key relations. $M_{cr} = f_{r}I_{g}/y_{t}$ with $f_{r} = 0.6\lambda\sqrt{f'_{c}}$; Branson $I_{e} = I_{cr} + (I_{g}-I_{cr})(M_{cr}/M_{a})^{3}$; $I_{e,avg} = 0.85I_{e,mid} + 0.15I_{e,cont}$ for a member continuous at one end; $\Delta_{LT} = \Delta_{i}[1 + s/(1+50\rho^{\prime})]$; $q = (N/A)(1 \pm 6e/B)$ valid while $e \le B/6$; $v_{c} = 0.38\phi_{c}\lambda\sqrt{f'_{c}}$ on a perimeter at $d/2$.

Why this works. A reinforced concrete member is neither fully uncracked nor fully cracked — cracks form only where the moment has exceeded $M_{cr}$, and the concrete between them still carries tension. Branson interpolates between the two limits with the cube of the moment ratio, which reproduces test data well and collapses to $I_{g}$ the moment the section stops cracking, as it does at mid-span here. Creep then roughly triples whatever part of that deflection is permanent. On the foundation side, the kern rule is the whole story: while the resultant stays within the middle third the pressure diagram is a trapezoid and the linear formula applies; outside it, part of the base lifts and the arithmetic changes entirely.

Common pitfall. Using $I_{g}$ everywhere for deflection and $0.35I_{g}$ everywhere for forces. Neither is right: the analysis stiffnesses of Clause 10.14.1.2 are calibrated for force distribution in a frame, whereas deflections need the section-by-section Branson value. Using $0.35I_{g}$ here would have doubled the computed deflection to 1.1 mm.

Source. CSA A23.3:19, Clauses 9.8.2 and 15.3–15.4; CSA A23.3 Table 9.3 for the deflection limits; Das, Principles of Geotechnical Engineering, Chapter 16, for allowable bearing pressure.

← Question 6 Back to the paper →