## Friday, January 11, 2013

### Design Example 5: Pad base – axial load plus bending moment (small eccentricity).

A column pad base is subject to an axial load of 200 kN (dead) plus 300 kN (imposed), and a bending moment of  40 kNm. To suit site constraints, the base is limited to a length of L = 1.8 m.

When moments act on a foundation, it is normal to replace them by positioning the vertical load at an equivalent eccentricity. The resultant vertical superstructure load is

P = G + Q
= 200 + 300
= 500 kN

Q as a percentage of P is 100Q/P = (100 × 300)/500 = 60%.
From  "Fig. 11.22 Combined partial safety factor for dead + imposed loads",  the combined partial factor for superstructure loads is γP = 1.52.
The resultant eccentricity is given by

Bearing pressure check – design chart approach
A suitable base size can be checked or calculated using design chart H.1 in Appendix H. For the purpose of this example this is reproduced in Fig. 11.26 below. Assuming a superstructure bearing pressure of p = na = 300 kN/m2,

Assuming a base length of L = 1.8 m,

From Fig. 11.26, this gives a required base area of

A = BL = 2.1 m2

Thus

A width of B = 1.2 m will be adopted.

Bearing pressure check – calculation approach
The eccentricity eP = 0.08 m is less than L/6 = 1.8/6 = 0.3 m, and thus the formation is loaded in compression over the full plan area of the base. Assume a width of B = 1.2 m.

Thus pmax = 293 kN/m2 and pmin = 169 kN/m2. These are less than the allowable bearing pressure of na = 300 kN/m2 ; the width of B = 1.2 m is therefore satisfactory.

Resultant ultimate design pressures
Since the base is fully in compression, ultimate design pressures, pu, are obtained by simply factoring up these pressures using the combined partial safety factor γP.

This is shown in Fig. 11.27.

Effect of offsetting the base
Where the moment always acts in one direction, economies in the base size can be achieved by positioning the base eccentric to the vertical load. Thus if the centroid of the base is offset by eP = 0.08 m, the pressure becomes uniform, and is simply given by p = P/A. This would give

Compared to A = 1.8 × 1.2 = 2.16 m2, this would be a reduction of 23%. This approach is used in Design Example 8.

Fig. 11.26 Pad base (small eccentricity) design example – design chart H1 (Appendix H) for base size. Fig. 11.27 Pad base (small eccentricity) design example – loads and bearing pressures.

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