### Design Example: Cantilever Balanced Foundation.

**Fig. 12.10**. The design is required to be adjusted accordingly.

Before redesigning the foundation, the designer should explore the possibilities, and relative costs, of either persuading the services engineer to relocate these services, or setting back the two columns on grid line 1, and cantilevering the building out to the site boundary at each ﬂoor level. Either solution may well prove more economic than changing the foundation.

If these options fail to bear fruit, the designer will need to design the combined base to cantilever over the service zone without loading it. As in the previous example, the base will be designed as a balanced foundation.

Fig. 12.10 Cantilever balanced foundation design example. |

**Size of base**

The column loads and positions are unchanged, and therefore the centroid of the superstructure loads remains in the same place as in the last example. Again a balanced

foundation will be achieved by making the centre of gravity of the effective base (i.e. the centroid of the uniform stress block below the base) coincide with that of the applied loads.

The service zone does not affect the centre of gravity of the base in the Y direction, and the overall dimension in this direction for a balanced foundation therefore remains at 9 m. In the x direction, the 1.5 m width of the service zone is discounted in considering the effective base area.

The weight of the cantilever section of the slab acts as a net applied load in this direction and must be taken into account in calculating the centroid of all applied loads.

It will therefore be included as part of the superstructure load, P.

The weight of this strip of foundation is

With reference to

**Fig. 12.10**, the distance from the centroid to the effective left-hand edge of the base is 5.0 − 2.25 = 2.75 m. Thus, in order to align the centre of gravity with the centroid of applied loads, the right-hand edge of the base must also be located at 2.75 m from the centroid of the applied loads. This gives an effective horizontal base width of 2 × 2.75 m = 5.5 m, and a total horizontal base width of 5.5 + 1.5 = 7.0 m. The effective area of the base is given by

**Bearing pressure**

The actual bearing pressure will be

**Ultimate design pressure**

From design example 2, the imposed load Q is 55% of the 4500 kN column loads, i.e.

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