. Tower Crane Foundation Design Calculation Example Link File

Tower Crane Foundation Design Calculation Example Link File

We must check if the pressure exerted on the soil exceeds the soil's capacity (3.0 ksf).

Eccentricity ($e$): This measures how far off-center the load is applied. $$e = \fracMP_total$$ $$e = \frac1,500 \text ft-kips393 \text kips = \mathbf3.82 \text ft$$

The "Middle Third" Rule: For the entire base of the foundation to remain in contact with the ground (no lifting/tilting), the eccentricity must be less than $B/6$. $$B/6 = 18 / 6 = 3.0 \text ft$$

Result: $3.82 \text ft > 3.0 \text ft$. Failure! The foundation is unstable; the load is outside the kern. The crane would tip backward under this specific load case if the foundation is not anchored or enlarged.

While the math above covers the theory, real-world execution requires the following:

  • Uplift (Out-of-Service Wind): The storm condition ($M_out$) is almost always the governing case for stability. Never rely solely on in-service loads.
  • Proximity to Excavations: If the tower crane is located near a basement excavation, the soil bearing capacity is reduced significantly due to the loss of lateral confinement. A slip circle analysis may be required.
  • Ground Water: If the water table is high, the submerged weight of the concrete reduces. You must subtract the buoyancy force from the stabilizing weight ($W_f$).

    • This example provides a complete workflow from load definition to stability checks and reinforcement detailing.

    Now we calculate the maximum pressure the foundation exerts on the soil and compare it to the soil's bearing capacity.

    For eccentric loading where $e < B/6$: The pressure distribution is trapezoidal.

    $$q_max = \fracNA \left( 1 + \frac6eB \right)$$ tower crane foundation design calculation example link

    Using Factored Loads (Conservative approach):

    Factored Vertical Load ($N_Ed$): $N_Ed = 1.35 \times (907.5 + 150) + 1.35 \times 400 = 1,428 + 540 = 1,968 \text kN$.

    Factored Moment ($M_Ed$): $M_Ed = 1.50 \times 1,200 = 1,800 \text kNm$.

    Factored Eccentricity ($e$): $e = \frac1,8001,968 = 0.914 \text m$. Check Kern: $B/6 = 0.917 \text m$. (Still just inside).

    Maximum Bearing Pressure: $$q_max = \frac1,96830.25 \left( 1 + \frac6 \times 0.9145.5 \right)$$ $$q_max = 65.0 \times (1 + 0.997)$$ $$q_max = 65.0 \times 1.997 = 129.8 \text kN/m^2$$

    Compare to Allowable Bearing Capacity: Allowable $q_all = 200 \text kN/m^2$.

    Result: $129.8 \text kN/m^2 < 200 \text kN/m^2$. PASS. The soil can easily support the crane.

    (Note: In some codes, the allowable stress is compared directly to unfactored loads. In Eurocode, we compare $q_max$ to the Design Bearing Resistance $R_d$, which is usually $q_all \times$ safety factors. Since our calculated pressure is significantly lower than the allowable, this design is safe.) We must check if the pressure exerted on

    Some cities (e.g., Dubai Municipality, London Building Control) publish approved tower crane foundation calculation sheets as part of temporary works guidance.

    Overturning moment M = 3,200 kNm
    Horizontal force H = 180 kN → effective moment at base:
    M_eff = M + H × h = 3,200 + 180 × 1.2 = 3,416 kNm

    Eccentricity e = M_eff / V = 3,416 / 1,600 = 2.135 m

    Check: e > B/6 = 5.0/6 = 0.833 m → partial uplift occurs.
    Max bearing pressure under trapezoidal distribution:

    q_max = (2 × V) / [3 × L × (B/2 – e)] = (2 × 1,600) / [3 × 5 × (2.5 – 2.135)]
    = 3,200 / [15 × 0.365] = 3,200 / 5.475 ≈ 584 kN/m²

    This exceeds allowable (150 kN/m²) → increase foundation size.

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    Design moment at column face (using factored loads: γ = 1.5 typical for crane loads in ULS):

    Factored V = 1.5 × 850 + 1.35 × 1,837.5 ≈ 1,275 + 2,480 = 3,755 kN
    Factored M_eff = 1.5 × 3,470 = 5,205 kNm

    Critical cantilever projection from column (assume column base plate 1.0×1.0 m) → projection = (7.0 – 1.0)/2 = 3.0 m

    Soil pressure under ULS (triangular with uplift) → average pressure on cantilever approx 0.7×q_max,ULS
    q_max,ULS ≈ 1.5×115.8 ≈ 174 kN/m²

    Moment at column face per meter width = 174 × 3.0²/2 = 783 kNm/m

    Required steel area (d = 1.5 m – cover 0.075 m – 0.025 m = 1.4 m)
    As = M / (0.87 fy z) ≈ 783×10⁶ / (0.87×500×0.9×1,400) ≈ 1,430 mm²/m

    Provide T20 @ 200 mm c/c (As,prov ≈ 1,570 mm²/m) top and bottom both ways.

    They offer PDF examples for MDT and MR series cranes.
    📍 Link: potain.com → “Documentation” → “Foundation Design Examples”