This exceeds (q_allow = 150 , \textkPa) → or must be deepened or widened. 4.5 Revised foundation size Try (L = B = 7.0 , \textm, t = 1.5 , \textm):
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. Tower Crane Foundation Design Calculation Example
Provide T20 @ 200 mm c/c (both directions top and bottom) → (A_s = 1570 , \textmm^2/m) ✓. Maximum tension per bolt from overturning (ULS): [ T_bolt = \fracM_dn \times r - \fracV_dn ] where (n=12) bolts, (r) = bolt circle radius ≈ 1.5 m. Approximate: [ T = \frac630012 \times 1.5 = 350 , \textkN \quad\text(ignoring vertical load compression) ] Check bolt capacity (M36, 8.8): (A_s = 817 , \textmm^2), (f_yb = 640 , \textMPa) [ N_Rd = 0.9 \times A_s \times f_yb / \gamma_M2 = 0.9\times817\times640 / 1.25 = 376 , \textkN > 350 , \textkN \quad \text✓ OK ] 8. Settlement Analysis Using elastic settlement for stiff clay ((E_s \approx 30 , \textMPa), (\nu=0.35)): This exceeds (q_allow = 150 , \textkPa) →
Maximum moment at crane column face (assume column base plate 2 m × 2 m): Stability Checks 5
