Maximum moment at crane column face (assume column base plate 2 m × 2 m):
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 ] Tower Crane Foundation Design Calculation Example
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: Maximum moment at crane column face (assume column
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. No, partial uplift
Cantilever projection from column edge to foundation edge: [ c = (7.0 - 2.0)/2 = 2.5 , \textm ] Average pressure under cantilever (triangular variation) – Use integration: Equivalent linear pressure distribution – conservative approach: [ M_Ed = q_max,ULS \times B \times \fracc^22 \times \text(shape factor) ] Simplified: (M_Ed \approx 204.5 \times 7.0 \times \frac2.5^22 = 204.5 \times 7.0 \times 3.125 = 4473 , \textkNm/m width?) – Wait, that’s too high – correct 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 ]