[ W_conc = 7\times7\times1.5\times25 = 1837.5 , \textkN ] [ N_total = 850 + 1837.5 = 2687.5 , \textkN ] [ e = 4200 / 2687.5 = 1.563 , \textm ] [ L/6 = 7/6 = 1.167 , \textm; \quad e > L/6 \rightarrow \textstill partial uplift ] [ L' = 3\times(3.5 - 1.563) = 5.811 , \textm ] [ q_max = \frac2\times2687.57 \times 5.811 = \frac537540.677 \approx 132.2 , \textkPa < 150 , \textkPa \quad \text✓ OK ]
Overturning moment includes wind, eccentric lifting, and dynamic effects. 4. Foundation Sizing – Bearing Pressure Check (SLS) 4.1 Self-weight of foundation [ W_conc = L \times B \times t \times \gamma_conc = 6.0 \times 6.0 \times 1.2 \times 25 = 1080 , \textkN ] Soil above base (ignore – removed during excavation and not replaced for simplicity – conservative). 4.2 Total vertical load (SLS) [ N_total = V_k + W_conc = 850 + 1080 = 1930 , \textkN ] 4.3 Eccentricity [ e = \fracM_kN_total = \frac42001930 = 2.176 , \textm ]
(M_Ed, per m = 4473 / 7 = 639 , \textkNm/m) [ A_s,req = \frac639\times10^60.87\times500\times0.9\times1409 = \frac639\times10^6551,000 \approx 1160 , \textmm^2/\textm ] Tower Crane Foundation Design Calculation Example
7.0 m × 7.0 m × 1.5 m thick. 5. Stability Checks 5.1 Overturning (ULS) [ M_overturning,ULS = M_d = 6300 , \textkNm ] Restoring moment (about edge): [ M_restoring = N_total,ULS \times \fracL2 = (1148 + 1837.5) \times 3.5 = 2985.5 \times 3.5 = 10449 , \textkNm ] Factor of safety: [ FOS = \frac104496300 = 1.66 > 1.5 \quad \text✓ OK ] 5.2 Sliding (ULS) Sliding force (H_d = 97.5 , \textkN) Friction resistance: (\mu = 0.45) (concrete on stiff clay) [ R_friction = N_total,ULS \times \mu = 2985.5 \times 0.45 = 1343.5 , \textkN ] [ FOS_sliding = 1343.5 / 97.5 = 13.8 \gg 1.5 \quad \text✓ OK ] 6. Structural Design of Pad (ULS) 6.1 Bending moment at column base interface Ultimate bearing pressure distribution (simplified for ULS) – Use factored loads and effective area.
Moment per meter width: [ M_Ed = q_max \times \fracc^22 \times \texteffective width factor ] Better to compute total moment = integral of pressure × distance. Given complexity, use standard formula from crane manuals: [ W_conc = 7\times7\times1
Effective width (L') (ULS) with (e = M_d / N_total,ULS = 6300 / 2985.5 = 2.11 , \textm) [ L' = 3\times(3.5 - 2.11) = 4.17 , \textm ] [ q_max,ULS = \frac2 \times 2985.57 \times 4.17 = \frac597129.19 \approx 204.5 , \textkPa ]
Maximum moment at crane column face (assume column base plate 2 m × 2 m): Structural Design of Pad (ULS) 6
For 7 m square, 2.5 m projection, (M_Ed \approx 0.5 \times q_max \times B \times c^2 = 0.5 \times 204.5 \times 7 \times 6.25 = 4473 , \textkNm) – that’s total moment.
Moment about column edge = pressure resultant × lever arm. Use trapezoidal distribution? For simplicity, take average pressure = (204.5 + 0)/2? No, partial uplift. Actually, use effective width method:
Net bearing pressure at SLS = (q_max \approx 132.2 , \textkPa) Influence factor (I_s) for square footing ≈ 0.88 [ \delta = q_max \times B \times \frac1-\nu^2E_s \times I_s = 132.2 \times 7 \times \frac1-0.122530000 \times 0.88 ] [ \delta \approx 132.2\times7\times0.8775/30000\times0.88 = 0.0239 , \textm = 23.9 , \textmm ]
For a 6 m square foundation, (L/6 = 1.0 , \textm). Since (e > L/6) (2.176 > 1.0), the resultant lies outside the middle third → partial uplift. Effective width (L' = 3 \times (L/2 - e) = 3 \times (3.0 - 2.176) = 2.472 , \textm). [ q_max = \frac2 \times N_totalB \times L' = \frac2 \times 19306.0 \times 2.472 = \frac386014.832 \approx 260.3 , \textkPa ]