Question 3 of 7: Long-term deflection of BCD and the footing at F

Question 3: Long-term deflection of BCD and the footing at F

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 section of Questions 1 and 2: beam 400 × 900 mm with 4-30M bottom and 4-25M top, $d = 830$ mm, $f'_c = 30$ MPa. Service (unfactored) moments from the same analysis are $M_B = 328.4$, $M_C = 447.1$ and $M_D = 288.4$ kN·m, and the service vertical reaction at F is 448.5 kN with a base moment of 35.4 kN·m and a horizontal thrust of 15.5 kN. Allowable bearing pressure 500 kPa.

Find. (i) the long-term deflection at C; (ii) a footing size at F.

Approach. Compute the effective second moment of area by Branson's expression at the three critical sections, scale the elastic deflection, then apply the A23.3 creep multiplier. Size the footing on service pressure and check it for punching, one-way shear and flexure at factored load.

1. Part (i) — find the cracking moment. With $f_r = 0.6 \lambda \sqrt{f'_c} = 3.29$ MPa and $I_g = 2.43 \times 10^{10}$ mm$^4$, $$M_{cr} = \frac{f_r I_g}{y_t} = 177.5 \text{ kN}\cdot\text{m}$$ All three critical sections carry more than this under service load, so the whole span is cracked and the gross section badly overstates the stiffness.
2. Compute the cracked and effective stiffnesses. With $n = E_s/E_c = 8.11$, the cracked transformed second moment at mid-span (4-30M tension, 4-25M compression) is $I_{cr} = 1.0161 \times 10^{10}$ mm$^4$ with the neutral axis at $kd = 236.0$ mm. Branson's expression $$I_e = I_{cr} + (I_g - I_{cr})\left(\frac{M_{cr}}{M_a}\right)^{3}$$ gives $1.1045 \times 10^{10}$ mm$^4$ at C, $1.0441 \times 10^{10}$ at B and $1.1677 \times 10^{10}$ at D. The Cl 9.8.2.4 weighting for a member continuous at both ends, $I_e = 0.70 I_{e,mid} + 0.15(I_{e1} + I_{e2})$, gives $1.1049 \times 10^{10}$ mm$^4$, i.e. $I_g/I_e = 2.199$.
3. Scale the immediate deflection. The elastic frame analysis on gross sections gives 7.056 mm at C under service load. Deflection is inversely proportional to $EI$, so $$\Delta_i = 7.056 \times 2.199 = 15.52 \text{ mm}$$
4. Apply the creep and shrinkage multiplier. Cl 9.8.2.5 gives the additional long-term deflection as $[s/(1 + 50\rho')]$ times the sustained immediate deflection, with $s = 2.0$ beyond five years. The 4-25M bars lapped through mid-span give $\rho' = 0.602$ %, so the multiplier is $2.0/(1 + 50 \times 0.00602) = 1.537$ and $$\boxed{\Delta_{total} = \Delta_i (1 + 1.537) = 39.4 \text{ mm} = L/305}$$ if the whole service load is treated as sustained. The A23.3 Table 9.3 limit for a member not supporting brittle finishes is $L/240 = 50$ mm, so the beam passes with about 20 % to spare.
5. Part (ii) — size the footing on bearing pressure. The service load at F is the 448.5 kN reaction plus 86.4 kN of column self weight; with a 1.6 m square pad 500 mm thick the total is 565.6 kN, and the base moment is 35.4 kN·m plus the thrust times the pad depth, $43.2$ kN·m in all. The eccentricity $e = 76.3$ mm is well inside the middle third ($e/B = 0.048$), so $$q = \frac{P}{B^{2}}\left(1 \pm \frac{6e}{B}\right) = 284.2 \text{ and } 157.7 \text{ kPa} \; \lt \; 500 \text{ kPa}$$ The pad is not governed by the ground — 500 kPa is a stiff-till or weak-rock value — but by the need to project far enough beyond a 500 × 900 mm column to develop the reinforcement.
6. Check punching and one-way shear. At factored load $C_f = 767.8$ kN and $q_f = 300$ kPa, with $d = 405$ mm. The critical punching perimeter at $d/2$ from the column face is $b_o = 4420$ mm, and $$V_f = 414 \text{ kN}, \qquad v_r = 0.38 \phi_c \lambda \sqrt{f'_c} \;\Rightarrow\; V_r = 2422 \text{ kN}$$ a utilisation of only 0.171. One-way shear at $d$ from the face gives 69.6 kN against 374 kN. Both are trivial because the pad is thick relative to its plan size.
7. Design the footing steel. The cantilever moment at the column face is 72.6 kN·m over the 1.6 m width, which needs far less steel than the Cl 7.8.1 shrinkage and temperature minimum of $0.002 A_g = 1600$ mm$^2$ in each direction. Provide 8-15M each way in the bottom mat, and dowel the column bars into the pad with a compression lap.