Question 7 of 7: Member ABC as a beam-column, and the long-term deflection at B

Question 7: Member ABC as a beam-column, and the long-term deflection at B (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. The 500 × 1200 section of Question 6 with 6 – 30M top and 5 – 30M bottom, $d = 1134$ mm and $d' = 66.3$ mm. From the frame analysis it carries a factored axial compression of 43.6 kN together with $M_f = 1456$ kN·m at A. Service loads are 400 kN at B, 600 kN at C and 14.4 kN/m self weight.

Find. (a) whether the point $(N_f, M_f)$ lies inside the factored interaction diagram of the section; (b) the long-term vertical deflection at B.

Approach. Locate the axial demand relative to the balanced load to decide whether slenderness matters, build the interaction diagram by strain compatibility, read off $M_r$ at $N_f$, then compute the immediate deflection with an effective moment of inertia and multiply by the long-term creep factor.

1. Part (a) — the axial force is real but small. The beam carries the horizontal thrust of the frame; with the column pinned at D that thrust is $H = |M_C|/h = 436/10 = 43.6$ kN. As a fraction of the section capacity, $$\frac{N_f}{0.1f'_c A_g} = \frac{43.6\times10^3}{0.1(30)(600\,000)} = 0.024,$$ so A23.3 would allow the member to be designed as a pure flexural member. The question asks for the beam-column check, so it is carried out properly below.
2. Slenderness first. For a rectangular section $r = h/\sqrt{12} = 346.4$ mm, so $$\frac{k l_u}{r} = \frac{12\,000}{346.4} = 34.6 .$$ The joint at C moves vertically only by the column's axial shortening, so the member is braced against sidesway in the plane of bending. The Cl 10.15.2 limit for a braced member, $25 - 10(M_1/M_2)/\sqrt{N_f/(f'_c A_g)}$, is enormous here because $N_f/(f'_c A_g) = 0.0024$; the slenderness ratio is therefore far inside it and $\delta_b = 1.00$. No moment magnification applies.
3. Build the interaction diagram. For a chosen neutral-axis depth $c$, with $\alpha_1 = 0.805$, $\beta_1 = 0.895$ and $\varepsilon_{cu} = 0.0035$, $$N_r = \alpha_1\phi_c f'_c b(\beta_1 c) + \phi_s A'_s f'_s - \phi_s A_s f_s,\qquad M_r = \sum N_i\!\left(\frac{h}{2} - y_i\right),$$ with $f_s$ and $f'_s$ from strain compatibility and capped at $f_y$. The pure axial capacity is $N_0 = 11\,916$ kN and the balanced point is $c_b = 721$ mm, $(M_b, N_b) = (2773\ \text{kN}\cdot\text{m}, 4775\ \text{kN})$.
4. Read the diagram at the design axial load. Solving $N_r(c) = 43.6$ kN gives $c = 106.0$ mm, and at that neutral-axis depth $$M_r = \boxed{1562\ \text{kN}\cdot\text{m}} \;\gt\; M_f = 1456\ \text{kN} \cdot\text{m},$$ a utilisation of $0.93$. The section is satisfactory as a beam-column. The small axial load helps: it closes part of the tension in the top steel, so $M_r$ at $N_f = 43.6$ kN is 5 per cent above the pure-flexure value of 1489 kN·m computed in Question 6.
5. Tie requirements. Because $N_f$ is under $0.1f'_c A_g$, Cl 10.9.1 does not impose the 1 per cent minimum longitudinal steel of a column, and the beam stirrups already satisfy the tie spacing. No change to the Question 6 detailing is needed.

Part (b) is a serviceability calculation, so it uses unfactored loads and an effective, not a cracked or gross, moment of inertia.

1. Part (b) — service moments. Repeating the frame analysis with 400 kN at B, 600 kN at C and 14.4 kN/m gives $$M_A = -1008\ \text{kN}\cdot\text{m},\qquad M_B = +804\ \text{kN}\cdot \text{m}.$$
2. Cracking moment. With $f_r = 0.6\lambda\sqrt{f'_c} = 3.29$ MPa and $I_g = 7.20\times10^{10}$ mm$^4$, $$M_{cr} = \frac{f_r I_g}{y_t} = \frac{3.29(7.20\times10^{10})}{600} = 394\ \text{kN}\cdot\text{m}.$$ Both critical sections are well past it, so the member is cracked throughout and the gross inertia would badly overstate the stiffness.
3. Cracked and effective inertia. With $n = E_s/E_c = 8.11$, the transformed cracked section at A ($A_s = 4200$ mm$^2$) has $kd = 331$ mm and $I_{cr} = 2.80\times10^{10}$ mm$^4$; at mid-span ($A_s = 3500$ mm$^2$), $I_{cr} = 2.42\times10^{10}$ mm$^4$. Branson's expression $$I_e = \left(\frac{M_{cr}}{M_a}\right)^{\!3}I_g + \left[1 - \left(\frac{M_{cr}}{M_a}\right)^{\!3}\right]I_{cr}$$ gives $3.06\times10^{10}$ at A and $2.99\times10^{10}$ mm$^4$ at mid-span, and Cl 9.8.2.4 weights them $0.85I_{e,mid} + 0.15I_{e,end} = 3.00\times10^{10}$ mm$^4$ for a member continuous at one end.
4. Immediate deflection. Applying a unit vertical load at B to the released structure and integrating $\int Mm\,\mathrm{d}x/EI$ over the beam and the column, $$\Delta_i = 8.98\ \text{mm} \;\ \text{downward} = \frac{L}{1336}.$$
5. Long-term multiplier. Cl 9.8.2.5 multiplies the sustained part of the immediate deflection by $$\frac{\zeta}{1 + 50\rho'},\qquad \rho' = \frac{A'_s}{bd} = \frac{3500}{500(1134)} = 0.00617,$$ so with $\zeta = 2.0$ the factor is $2.0/1.309 = 1.53$ and the total multiplier is $2.53$. The compression steel is doing very little here: without it the multiplier would be 3.0.
6. Long-term deflection. Treating the whole load as sustained — the conservative reading, since the paper gives no dead/live split — $$\Delta_{LT} = 2.53(8.98) = \boxed{22.7\ \text{mm}} = \frac{L}{528}.$$ This is comfortably inside the $L/240 = 50$ mm total-deflection limit of Table 9.3. If instead the 400 kN and 600 kN loads were entirely transient, only the 14.4 kN/m self weight would creep and the long-term value would fall to about 11 mm.