07-Str-A5 · May 2014

Question 5 of 7: Reinforced concrete frame — member AC in flexure and shear

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 5: Reinforced concrete frame — member AC in flexure and shear (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. Frame of Figure 4: horizontal member A–B–C, 10 m long (B at mid-span), pinned at A; vertical member C–D–E, 9 m tall, pinned at E, rigidly joined to AC at C. Specified loads 350 kN down at B, 700 kN down at C, 100 kN horizontal at D (mid-height). Materials $f'_c = 30$ MPa, $f_y = 400$ MPa. Trial member AC: $450 \times 1000$ mm, self-weight 10.8 kN/m.

Find. The longitudinal reinforcement for the sagging and hogging regions of AC, and its transverse reinforcement.

Figure 4 with the factored bending-moment diagram of member AC superimposed. Pinned at A and E, rigid at C — one degree of static indeterminacy, resolved with equal EI as the paper directs.

Approach. The frame has four reaction components and three equations of statics, so it is indeterminate to the first degree; with equal $EI$ throughout, as the paper directs, a stiffness analysis gives the moment distribution. Design the critical sections for flexure by the rectangular stress block, then check shear by the CSA A23.3 simplified method.

Factor the loads. $P_B = 1.5(350) = 525$ kN, $P_C = 1.5(700) = 1050$ kN, $H_D = 1.5(100) = 150$ kN acting towards the column, plus $1.25 \times 10.8 = 13.5$ kN/m on AC and $1.25 \times 9.6 = 12.0$ kN/m on CE from self-weight.
Analysis. Taking equal $EI$ in both members, the stiffness solution gives

Action	Factored	Service
Sagging moment at B	1120 kN·m	764 kN·m
Hogging moment at C	721 kN·m	493 kN·m
Shear at A	258 kN	180 kN
Shear at C	402 kN	277 kN
Axial compression in AC	155 kN	105 kN
Axial in column at C / at E	1452 / 1560 kN	978 / 1065 kN

Part (a) — sagging steel at B. With two layers of bars, $d = 905$ mm, $b = 450$ mm. For $f'_c = 30$ MPa, $\alpha_1 = 0.85 - 0.0015f'_c = 0.805$ and $\beta_1 = 0.97 - 0.0025f'_c = 0.895$. From $$\rho = \frac{\alpha_1\phi_cf'_c}{\phi_sf_y}\left[1 - \sqrt{1 - \frac{2M_f/bd^2}{\alpha_1\phi_cf'_c}}\right]$$ with $M_f/bd^2 = 3.04$ MPa the result is $\rho = 0.01003$ and $A_s = 4085$ mm². $$\boxed{\text{Provide 6--30M} = 4200\ \text{mm}^2\ \text{in two layers}}$$
Confirm the resistance and the ductility. $$a = \frac{\phi_sA_sf_y}{\alpha_1\phi_cf'_cb} = \frac{0.85(4200)(400)}{0.805(0.65)(30)(450)} = 202\ \text{mm}, \qquad c = \frac{a}{\beta_1} = 226\ \text{mm}$$ $$M_r = \phi_sA_sf_y\left(d - \frac{a}{2}\right) = 0.85(4200)(400)(804) = 1148\ \text{kN}\cdot\text{m} \ge 1120\ \text{kN}\cdot\text{m}$$ with $c/d = 0.250$, well below the balanced value $700/(700 + f_y) = 0.636$, so the steel yields long before the concrete crushes.
Hogging steel at C. Top steel in one layer gives $d = 930$ mm and $M_f/bd^2 = 1.85$ MPa, so $A_s = 2435$ mm². $$\boxed{\text{Provide 4--30M} = 2800\ \text{mm}^2\ \text{top},\ M_r = 821\ \text{kN}\cdot\text{m} \ge 721\ \text{kN}\cdot\text{m}}$$ Minimum steel, $A_{s,\min} = 0.2\sqrt{f'_c}\,b_th/f_y = 1232$ mm², is satisfied in both regions. Two of the four top bars run through to mid-span as hangers, which the deflection calculation of Question 6 then uses as compression steel.
The 155 kN of axial compression. It is only $155\times10^3/(0.805 \times 0.65 \times 30 \times 450 \times 1000) = 1.3\ \%$ of the section's axial capacity, well under the $0.1f'_cA_g$ threshold at which A23.3 requires the member to be treated as a compression member, so AC is designed as a beam. Its small favourable effect on cracking is neglected.
Part (b) — concrete contribution to shear. The critical section is at the face of C, where $V_f = 402$ kN. With $d_v = \max(0.9d,\,0.72h) = 814.5$ mm and, because at least minimum stirrups are provided, $\beta = 0.18$ and $\theta = 35^{\circ}$: $$V_c = \phi_c\lambda\beta\sqrt{f'_c}\,b_wd_v = 0.65(1.0)(0.18)(5.477)(450)(814.5) = 235\ \text{kN}$$
Stirrups. Ten-millimetre double-leg stirrups give $A_v = 200$ mm². At $s = 300$ mm, $$V_s = \frac{\phi_sA_vf_yd_v\cot\theta}{s} = \frac{0.85(200)(400)(814.5)(1.428)}{300} = 264\ \text{kN}$$ $$\boxed{V_r = V_c + V_s = 235 + 264 = 499\ \text{kN} \ \ge\ V_f = 402\ \text{kN}}$$ Checks: $A_{v,\min} = 0.06\sqrt{f'_c}b_ws/f_y = 111 < 200$ mm²; $V_f = 402$ kN is below $0.125\phi_cf'_cb_wd_v = 893$ kN, so the spacing limit is $\min(0.7d_v,\,600) = 570$ mm and 300 mm satisfies it; and $V_f$ is far below the crushing limit $0.25\phi_cf'_cb_wd_v = 1787$ kN. $$\boxed{\text{10M double-leg stirrups at 300 mm throughout member AC}}$$

Quantity	Result
Member AC	$450 \times 1000$ mm reinforced concrete
Factored moments	$+1120$ kN·m at B, $-721$ kN·m at C
(a) Bottom steel at B	6–30M ($A_s = 4200$ mm²), $M_r = 1148$ kN·m
(a) Top steel at C	4–30M ($A_s = 2800$ mm²), $M_r = 821$ kN·m
Neutral-axis ratio at B	$c/d = 0.250$ (ductile)
(b) Design shear	$V_f = 402$ kN at C
(b) Transverse steel	10M double-leg stirrups at 300 mm, $V_r = 499$ kN

Topic. Limit-states design of a beam in a statically indeterminate reinforced concrete frame.

Key relations. $M_r = \phi_sA_sf_y(d - a/2)$ with $a = \phi_sA_sf_y/(\alpha_1\phi_cf'_cb)$, $\alpha_1 = 0.85 - 0.0015f'_c$, $\beta_1 = 0.97 - 0.0025f'_c$; ductility limit $c/d \le 700/(700 + f_y)$; $A_{s,\min} = 0.2\sqrt{f'_c}b_th/f_y$; CSA A23.3 simplified shear $V_c = \phi_c\lambda\beta\sqrt{f'_c}b_wd_v$ with $\beta = 0.18$, $\theta = 35^{\circ}$ when $A_v \ge A_{v,\min}$, and $V_s = \phi_sA_vf_yd_v\cot\theta/s$.

Why this works. The instruction "assume the same stiffness for all members" is not a throwaway: in an indeterminate frame the moments depend only on the relative stiffnesses, so fixing them equal makes the analysis independent of the sections that have not yet been chosen, and breaks what would otherwise be a circular design loop. Once the moments are known, the rectangular stress block converts each into a required steel area in a single closed-form step, and the $c/d$ check confirms the assumption behind that step — that the steel is yielding when the concrete reaches $0.0035$. On the shear side, the simplified method is a truss analogy: the concrete carries a share $V_c$ through aggregate interlock and the compression zone, and the stirrups carry the rest as the vertical ties of a truss whose diagonals are inclined at $\theta$.

Common pitfall. Designing AC as a simply supported 10 m beam because A is a pin. The rigid joint at C attracts 721 kN·m of hogging moment, which both relieves the span (the sagging moment is 1120 kN·m rather than the $PL/4 + wL^2/8 = 1481$ kN·m of a simple span) and demands top steel that a simple-span design would omit entirely — the more dangerous half of the error.

Source. CSA A23.3 Cl. 10.1–10.5 (flexure) and Cl. 11.3 (shear); MacGregor & Wight, Reinforced Concrete: Mechanics and Design, Ch. 4, 6 and 10.

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