Question 6 of 7: Limit states design of the reinforced concrete member ABC

Question 6: Limit states design of the reinforced concrete member ABC (14 + 6 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 Figure 5: an L-frame whose horizontal member A–B–C is built into a wall at A and spans $6 + 6 = 12$ m to joint C, from which a 10 m column runs down to a pinned base at D. Point loads of 400 kN act at B and 600 kN at C. Materials are $f'_c = 30$ MPa and $f_y = 400$ MPa.

Find. Flexural and shear reinforcement for member ABC, with a layout drawing.

Approach. Analyse the twice-redundant frame by the force method using the Cl 10.14.1.2 cracked-section stiffnesses, take the envelope of hogging and sagging moments and the peak shear, then proportion the section with the rectangular stress block and detail the stirrups.

1. Factor the loads. $$\begin{gathered}P_B = 1.5(400) = 600\ \text{kN}, \\ P_C = 1.5(600) = 900\ \text{kN}, \\ w = 1.25(14.4) = 18.0\ \text{kN/m}.\end{gathered}$$ The 900 kN at C acts directly over the column and does no work on the beam except through the joint.
2. Set up the analysis. A is fully fixed (3 restraints) and D is pinned (2), so with three equations of equilibrium the frame is twice redundant. Cl 10.14.1.2 gives the stiffnesses to use in a frame analysis: $0.35I_g$ for the beam and $0.70I_g$ for the column, i.e. $$I_b = 0.35\frac{500(1200)^3}{12} = 2.52\times10^{10},\qquad I_c = 0.70\frac{600(800)^3}{12} = 1.792\times10^{10}\ \text{mm}^4 .$$
3. Solve for the redundants. Releasing the horizontal and vertical reactions at D and applying the unit-load method over both members gives $$\begin{gathered}V_D = 1222.9\ \text{kN}, \\ H_D = 43.6\ \text{kN}, \\ V_A = 493.1\ \text{kN}.\end{gathered}$$ Because the column base is pinned, the horizontal reaction is simply the column shear, $H = |M_C|/h$.
4. Design moments and shear. The resulting beam diagram is $$\begin{gathered}M_A = -1456\ \text{kN}\cdot\text{m}\ \text{(hogging)}, \\ M_B = +1178\ \text{kN}\cdot\text{m}\ \text{(sagging)}, \\ M_C = -436\ \text{kN}\cdot\text{m},\end{gathered}$$ with the maximum shear $V_f = 493.1$ kN at A. The value at A sits between the propped-cantilever bound (1674 kN·m, a pin at C) and the fixed-fixed bound (1116 kN·m, a rigid support at C), as it must — the column supplies partial rotational restraint.
5. Effective depth. With 40 mm cover, 10M stirrups and 30M bars, $$d = 1200 - 40 - 11.3 - \tfrac{29.9}{2} = 1134\ \text{mm}.$$
6. Flexural steel at A. Solving $\phi_s A_s f_y(d - a/2) = M_f$ with $a = \phi_s A_s f_y/(\alpha_1\phi_c f'_c b)$ and $\alpha_1 = 0.805$: $$A_s = 4099\ \text{mm}^2 \;\Rightarrow\; \boxed{\text{6 – 30M top}} \ (4200\ \text{mm}^2).$$ Then $a = 181.9$ mm, $c/d = 0.179 \ll 0.5$, so the section is comfortably tension-controlled, and $M_r = 1489$ kN·m against $M_f = 1456$ kN·m. Six 30M bars fit in one layer: they need $6(29.9) + 5(42) = 389$ mm against the 400 mm clear width.
7. Flexural steel at B and at C. The same solve gives $A_s = 3259$ mm$^2$ at B, so provide 5 – 30M bottom ($3500$ mm$^2$, $M_r = 1259$ kN·m). At C the demand is only $A_s = 1155$ mm$^2$, below the minimum $$A_{s,min} = \frac{0.2\sqrt{f'_c}}{f_y}b_t h = \frac{0.2\sqrt{30}}{400}(500)(1200) = 1643\ \text{mm}^2,$$ so minimum steel governs: provide 3 – 30M top ($2100$ mm$^2$) across the joint and anchor it into the column.
8. Shear. With $d_v = \max(0.9d, 0.72h) = 1020$ mm and the simplified method ($\beta = 0.18$, $\theta = 35^\circ$), $$V_c = \phi_c\lambda\beta\sqrt{f'_c}\,b_w d_v = 0.65(0.18)\sqrt{30}(500)(1020)/10^3 = 327\ \text{kN},$$ so the stirrups must carry $V_s = 493.1 - 327 = 166$ kN. Ten-millimetre double-leg stirrups ($A_v = 200$ mm$^2$) would satisfy strength at $$s = \frac{\phi_s A_v f_y d_v \cot\theta}{V_s} = 597\ \text{mm}.$$
9. Detailing limits govern the stirrup spacing. Because $V_f = 493$ kN is far below $0.125\phi_c f'_c b_w d_v = 1244$ kN, Cl 11.3.8.1 caps the spacing at $\min(0.7d_v, 600) = 600$ mm; but the minimum-shear- reinforcement rule $A_v \ge 0.06\sqrt{f'_c}\,b_w s/f_y$ caps it at 487 mm. Adopt $$\boxed{\text{10M double-leg stirrups at 450 mm throughout}},$$ tightened to 250 mm over the first 1.5 m from A where the shear is largest.