07-Str-A5 · May 2014

Question 6 of 7: Member CDE as a beam-column; long-term deflection of AC

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, May 2014 — 3 hours, closed book (handbooks and textbooks permitted, no notes on them; Casio or Sharp approved calculator). Seven questions of equal value; any five constitute a complete paper and only the first five presented are marked. 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, $f_{pi} = 1200$ MPa, losses $= 240$ MPa, hence $f_{pe} = 1200 - 240 = 960$ MPa.

Reference texts. CSA S16 Design of Steel Structures with the CISC Handbook of Steel Construction — plastic design (Cl. 8.6, 13.7), lateral–torsional buckling (Cl. 13.6), beam-columns (Cl. 13.8), plate girders (Cl. 14), composite beams (Cl. 17), welded connections (Cl. 13.13, 21); CSA A23.3 Design of Concrete Structures — flexure and shear (Cl. 10, 11), slenderness (Cl. 10.13–10.16), deflection (Cl. 9.8), 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 taken from the trial section and factored at $1.25$. A candidate assuming a different split obtains proportionally different sizes; the method is what is examined.

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

Check: readings taken from the examination drawing. Three details are shown on the drawing but not stated in the question text. In Figure 1 the bases at A and D carry the fixed-support hatching and the question text says "the frame is fixed at the bases A and D", so both bases are fixed and the frame is three times statically indeterminate. In Figure 4 the 100 kN arrow at D points towards the column, i.e. to the left; A and E both carry pin symbols, making that frame indeterminate to the first degree. In Figure 5 the deck is 3 + 3 = 6 m wide over three beams spaced at 2.5 m, leaving 0.5 m overhangs — the slab-thickness annotation is not legible on the printed figure, so the deck thickness is designed in Question 7 rather than read off.

[Figure not reproduced: Figure 1 as printed on the examination paper: rectangular steel portal, fixed at A and D, beam capacity 1.5 M₊ and column capacity M₊. All loads shown are unfactored. See the official exam paper.]

Question 6: Member CDE as a beam-column; long-term deflection of AC (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. From the Question 5 analysis, member CDE carries $N_f = 1452$ kN with $M_f = 721$ kN·m at C, falling to $M_f \approx 23$ kN·m at D and zero at the pin at E; the axial load rises to 1560 kN at E. Height 9 m, frame braced against sway. Member AC is $450 \times 1000$ mm with 6–30M bottom and 2–30M top continuing through mid-span; service moments 764 kN·m sagging at B and 493 kN·m hogging at C.

Find. A column section and reinforcement for CDE including slenderness, and the long-term mid-span deflection of AC.

Approach. Test whether slenderness must be considered; if so, magnify the moment; then design the section from strain compatibility, recognising that the eccentricity places it well below the balanced point. For the deflection, compute Branson's effective moment of inertia, obtain the immediate deflection from the moment diagram, and apply the A23.3 sustained-load multiplier.

Question 6: member CDE, 500 × 800 mm with symmetric reinforcement. The factored axial load is well below the balanced load, so the section behaves as an under-reinforced beam with axial help.

Part (a) — trial section and slenderness. Try $500 \times 800$ mm bending about the 800 mm dimension, so $r \approx 0.3h = 240$ mm and $$\frac{k\ell_u}{r} = \frac{1.0(9000)}{240} = 37.5$$ For a braced member A23.3 Cl. 10.15.2 permits slenderness to be ignored only if $k\ell_u/r < 25 - 10M_1/M_2$; with $M_1 = 0$ that limit is 25, so slenderness must be considered.
Moment magnification. With $I_g = 21\,333\times10^6$ mm⁴, $E_c = 4500\sqrt{30} = 24\,648$ MPa and a sustained-to-total factored axial ratio $\beta_d = 0.053$ (only self-weight is dead here), $$EI = \frac{0.4E_cI_g}{1 + \beta_d} = 1.998\times10^{14}\ \text{N}\cdot\text{mm}^2, \qquad P_c = \frac{\pi^2EI}{(k\ell_u)^2} = 24\,347\ \text{kN}$$ $$\delta_b = \frac{C_m}{1 - P_f/(\phi_mP_c)} = \frac{0.60}{1 - 1452/(0.75 \times 24\,347)} = 0.65 \;\Rightarrow\; \text{use } \delta_b = 1.0$$ so the design moment stays at $M_f = 721$ kN·m. The minimum-eccentricity moment, $P_f(15 + 0.03h) = 57$ kN·m, is far smaller and does not control.
Where the section sits on its interaction diagram. With $d = 740$ mm and $d' = 60$ mm the balanced neutral axis is $c_b = 700d/(700 + f_y) = 471$ mm, at which $$P_b = \alpha_1\phi_cf'_cb\beta_1c_b = 0.805(0.65)(30)(500)(0.895)(471) = 3308\ \text{kN}$$ Since $N_f = 1452$ kN is well below $P_b$, the section is tension-controlled: the tension steel yields and the section behaves as an under-reinforced beam with the axial load helping.
Neutral-axis depth at the design axial load. For symmetric reinforcement in which both faces yield, the steel forces cancel in the axial equation, so $$c = \frac{N_f}{\alpha_1\phi_cf'_cb\beta_1} = \frac{1452\times10^3}{7025} = 207\ \text{mm}, \qquad a = \beta_1c = 185\ \text{mm}$$ and the compression steel strain is $0.0035(207 - 60)/207 = 0.00248 > f_y/E_s = 0.0020$, confirming that it too yields.
Reinforcement. Taking moments about the plastic centroid, $$M_r = \alpha_1\phi_cf'_cba\left(\frac{h}{2} - \frac{a}{2}\right) + \phi_sA'_sf_y\left(\frac{h}{2} - d'\right) + \phi_sA_sf_y\left(d - \frac{h}{2}\right)$$ The concrete term alone supplies 447 kN·m, so the steel must supply 274 kN·m, needing only 1189 mm² per face. The 1 % minimum for columns governs instead: $$\boxed{A_{st} = 0.01(500 \times 800) = 4000\ \text{mm}^2 \Rightarrow \text{8--25M, four per face}}$$ $$\boxed{M_r = 447 + 462 = 909\ \text{kN}\cdot\text{m} \ \ge\ M_f = 721\ \text{kN}\cdot\text{m} \ \text{at}\ N_f = 1452\ \text{kN}}$$
Remaining column checks. At the base, where the moment is zero, $P_{r,\max} = 0.80[\alpha_1\phi_cf'_c(A_g - A_{st}) + \phi_sA_{st}f_y] = 6061$ kN, far above 1560 kN. For shear, the 155 kN in CD is below the concrete contribution $V_c = \phi_c(0.18)\sqrt{f'_c}bd_v = 213$ kN with $d_v = 666$ mm, so the ties are not asked to carry shear. Tie spacing is the least of $16d_b = 400$ mm, $48d_{\text{tie}} = 480$ mm and the least column dimension 500 mm. $$\boxed{500 \times 800\ \text{column, 8--25M, 10M ties at 400 mm}}$$
Part (b) — cracked-section properties of AC. $I_g = 37\,500\times10^6$ mm⁴, $f_r = 0.6\lambda\sqrt{f'_c} = 3.29$ MPa and $M_{cr} = f_rI_g/y_t = 246$ kN·m. With $n = E_s/E_c = 8.11$ and $A_s = 4200$ mm², the transformed neutral axis at mid-span solves $bx^2/2 = nA_s(d - x)$, giving $x = 302$ mm and $$I_{cr} = \frac{bx^3}{3} + nA_s(d - x)^2 = 16\,523\times10^6\ \text{mm}^4$$
Effective moment of inertia. Branson's expression with the service sagging moment $M_a = 764$ kN·m gives $(M_{cr}/M_a)^3 = 0.034$, so $$I_e = 0.034I_g + 0.966I_{cr} = 17\,228\times10^6\ \text{mm}^4$$ barely above $I_{cr}$ — at this load level the member is thoroughly cracked and the gross section would overstate its stiffness by more than a factor of two.
Immediate deflection at mid-span. AC is pinned at A and carries the end moment $M_C = 493$ kN·m at the far end, so superposing the three standard cases, $$\Delta_i = \frac{PL^3}{48E_cI_e} + \frac{5wL^4}{384E_cI_e} - \frac{M_CL^2}{16E_cI_e} = 17.17 + 3.31 - 7.25 = 13.2\ \text{mm}$$
Long-term multiplier and result. Two 30M bars run through mid-span, so $\rho' = 1400/(450 \times 905) = 0.00344$ and, for loads sustained five years or more, $$\zeta = \frac{s}{1 + 50\rho'} = \frac{2.0}{1.172} = 1.71$$ Taking the whole specified load as sustained — the conservative reading, since the paper does not separate dead from live — $$\boxed{\Delta_{\text{total}} = \Delta_i(1 + \zeta) = 13.2(2.71) = 35.8\ \text{mm} = \frac{L}{279}}$$ This satisfies the A23.3 Table 9.3 limit of $L/240 = 41.7$ mm for members not supporting damage-sensitive construction, but not the $L/480 = 20.8$ mm limit that applies if brittle partitions are attached — in that case the beam would need to be deepened or given compression steel through mid-span.

Quantity	Result
(a) Member CDE	$500 \times 800$ mm, 8–25M, 10M ties at 400 mm
Slenderness	$k\ell_u/r = 37.5 > 25$, but $\delta_b = 1.0$ ($P_c = 24\,347$ kN)
Design actions at C	$N_f = 1452$ kN with $M_f = 721$ kN·m
Section resistance	$M_r = 909$ kN·m at that axial load ($P_b = 3308$ kN)
(b) $I_e$ of AC	$17\,228 \times 10^6$ mm⁴ (vs $I_g = 37\,500 \times 10^6$)
(b) Immediate deflection	13.2 mm
(b) Long-term deflection	35.8 mm $= L/279$ ($\zeta = 1.71$)

Topic. Slender reinforced concrete columns under combined axial load and moment, and time-dependent deflection of cracked flexural members.

Key relations. Slenderness threshold $k\ell_u/r < 25 - 10M_1/M_2$ (braced); magnifier $\delta_b = C_m/[1 - P_f/(\phi_mP_c)] \ge 1.0$ with $P_c = \pi^2EI/(k\ell_u)^2$ and $EI = 0.4E_cI_g/(1 + \beta_d)$; balanced depth $c_b = 700d/(700 + f_y)$; Branson $I_e = (M_{cr}/M_a)^3I_g + [1 - (M_{cr}/M_a)^3]I_{cr} \le I_g$; sustained-load factor $\zeta = s/(1 + 50\rho')$.

Why this works. A column's capacity is a curve, not a number: above the balanced point adding axial load reduces the moment it can carry, below it adding axial load increases the moment capacity because the compression zone the concrete must supply is partly paid for by $N_f$. Establishing which side of $P_b$ the design point lies on is therefore the first move, and it also tells you the failure will be ductile. On the deflection side, the cube in Branson's formula is what makes $I_e$ collapse onto $I_{cr}$ so quickly: once the service moment is about three times the cracking moment, as here, the uncracked regions contribute almost nothing. The $\zeta$ multiplier then represents creep and shrinkage, and compression steel reduces it because the concrete in the compression zone can shed sustained stress into steel that does not creep.

Common pitfall. Computing deflections on the gross section. Here $I_g$ would give 6.1 mm immediate and about 16 mm long-term, roughly 55 % of the true value, and would wrongly certify the member against the $L/480$ limit. The second trap is skipping the slenderness test because $\delta_b$ turns out to be 1.0 — the test itself is what proves the magnification is unnecessary, and with a larger $M_1/M_2$ ratio the same column would need magnifying.

Source. CSA A23.3 Cl. 10.9–10.16 (columns and slenderness) and Cl. 9.8 with Table 9.3 (deflection); MacGregor & Wight, Reinforced Concrete, Ch. 9, 11 and 12.

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