### Design Example: Trapezoidal balanced foundation.

This example deals with the same building considered in the previous two examples, and designs the foundations for the perimeter columns where these occur along a site boundary, as shown in Fig. 12.11. As in the previous examples, the close proximity of the perimeter columns to the site boundary means that isolated pad foundations are not suitable, and that the foundations of the perimeter columns must be combined with those of the adjacent  internal columns.

Since the ratio of internal to perimeter loads is 2 : 1, i.e. the same as in Design Example 2 (section 12.3.4), the centroid  of loads will again be 2.0 m from grid line B. A 9.0 m long base, as in Design Example 2, would therefore again be required to achieve a balanced rectangular foundation.

This relatively long base would however be associated with comparatively large bending moments and reinforcement areas. A more economic foundation is likely to be achieved using a shorter trapezoidal balanced foundation.

 Fig. 12.11 Trapezoidal balanced foundation designexample.

Condition for a balanced trapezoidal foundation
Again the condition for a balanced foundation is for the centre of gravity of the base to coincide with the centroid of the applied loads.

With reference to Fig. 12.11, and taking moments of area about x–x, the location of the centre of gravity of the base of area A is given by

Area of base
The values of B1 and B2 would normally be chosen to minimize the size of the base. This would result in a bearing  pressure equal to the allowable bearing pressure, na, giving a base area

Dimensions of base
The end of the base furthest from the site boundary will,  in this instance, be chosen to extend beyond grid line B  by the same amount as a standard 3.65 m × 3.65 m internal pad foundation (see section 12.3.4), i.e. extending by 3.65/2 = 1.825 m.

Thus, from Fig. 12.11,

These values will give a balanced trapezoidal foundation, with a bearing pressure of p = 150 kN/m2.

Ultimate design pressure
The combined dead and imposed partial load factor is  γP = 1.51, as in the previous examples. The ultimate design pressure for reinforcement design, pu, is given by