We need a heavier or wider foundation. Let's increase the width to 5.5 m and keep the depth at 1.2 m.
Revised Properties:
Revised Check: $$e = \frac1,2001,457.5 = 0.823 \text m$$ $$e_limit = \frac5.56 = 0.917 \text m$$
Result: $0.823 \text m < 0.917 \text m$. PASS. The foundation is stable against overturning.
Crane model: Potain MD 265 (typical for 6–10 story buildings)
Max working load: 12 t at 15 m radius
Max free-standing height: 45 m tower crane foundation design calculation example link
Manufacturer-provided loads at foundation level (serviceability):
| Load type | Value | |-----------|-------| | Vertical load ( V_k ) | 950 kN | | Horizontal load ( H_k ) | 75 kN (wind + slewing) | | Overturning moment ( M_k ) | 2,600 kNm |
Soil data (assumed):
Foundation geometry (trial size):
Tower cranes are the backbone of high-rise construction. However, a tower crane is only as reliable as the ground it stands on. A catastrophic foundation failure can lead to loss of equipment, project delays, injuries, or fatalities. Unlike standard building foundations, tower crane foundations are subjected to extreme overturning moments, torque, and horizontal forces.
Despite its critical nature, finding a clear, step-by-step tower crane foundation design calculation example link can be frustrating. Many resources offer theory but lack a practical numerical example.
In this article, we provide a detailed design calculation example for a typical tower crane foundation (pad footing + anchor bolts) and, crucially, direct you to authoritative, downloadable calculation links at the end.
Sliding resistance:
( R_slide = (V_d) \times \tan(2/3 \phi') + c' \cdot A_base )
( V_d ) (ULS, with wind) = ( 1.35 \times 950 + 1.0 \times 1134 = 2,416 , kN )
( \tan(2/3 \times 30°) = \tan(20°) = 0.364 )
Friction term = ( 2416 \times 0.364 = 879 , kN )
Cohesion term = ( 5 \times 30.25 = 151 , kN )
Total resistance = ( 879 + 151 = 1,030 , kN ) We need a heavier or wider foundation
Driving force ( H_d = 1.5 \times 75 = 112.5 , kN )
( FOS_slide = 1030 / 112.5 = 9.2 \gg 1.5 ) → OK.
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What this looks like practically: