Moist clay in inclined strata is liable to slide when loaded. Clay mixed with sand or gravel will bear more load than pure clay. Sand will bear more load than ordinary clay, and when in beds of sufficient thickness and extent to prevent running, will bear heavy loads with little settlement. Sand sufficiently fluid to run, as quicksand, cannot be easily employed to carry foundations. Grillage Foundations. The simple grillage foundation is illustrated by Fig. 151. The method of calculating the beams composing the grillage involves assumptions as to the conditions of distribution of loading and stresses. One method is given in Cambria, Page 263. This method involves the assumption that the beams can deflect from the line of axis of column. Such a condition, however, would lead to the cast-iron base bearing at its outer edges only; this would involve strains for which these bases are rarely designed. Another assumption and one more in harmony with the assumption of the ordinary beam theory, is that the beams of the upper tier are fixed for the portion under the column base. Under this assumption the load is distributed uniformly by the upper tier and the stress in the free portion is calculated by the formula for a beam fixed at one end and free at the other.

DIAGRAM OF DISTRIBUTION OF COLUMN LOAD ON GRILLAGE FOOTING. Fig. 153.

DIAGRAM OF DISTRIBUTION OF COLUMN LOAD ON GRILLAGE FOOTING. Fig. 153.

Referring to Fig. 153; suppose the column load is P, and by the principles already given the extreme dimensions of footing are L and L' in feet. The length of the beams in the two tiers can be taken as L and L' also. Then if b and b' are the dimensions in feet of the column base, and the beams in the upper tier are placed the same width out to out of flanges as the column base, L-b/2 = projection of the upper tier, and L'-b'/2 = projection of the lower tier. The load per square foot on the upper tier is

P/b'L and on the lower tier is .P/LL' The moment in inch pounds, therefore, is M = 1/2 x b'P/b'L x (L-b)2/4 x 12

= 3/2 x P/L (L-b)2 for the upper tier and M' = 1/2 x PL/LL' x (L'-B')2/4 x 12

=3/2 x P/L' (L'-b')2 for the lower tier.

These formulas give the total moment borne by all the beams in the tier. The number of beams is generally determined by the dimensions of the footing, the beams of the upper tier being placed with their flanges generally not much more than 6 inches apart in the clear, and those of the lower tier from 6 inches to 12 inches. The number of beams being determined, the moment each bears is obtained by dividing the total moment by the number of beams; and by dividing this individual moment by the allowable fibre stress the required moment of resistance and hence the size of beams is obtained. Since the concrete and steel act together, a higher fibre strain can be safely allowed; this should in general be not more than 20,000 pounds per square inch, however.

Some trial and reproportioning of dimensions may sometimes be necessary to keep within the limits of depth and number of beams desired. Grillage beams in foundations should have the concrete thoroughly tamped around them, and it is preferable that the steel should be coated with neat cement instead of a coat of paint.

Diagram of Grillage Footing. Fig. 154.

Diagram of Grillage Footing. Fig. 154.

The following problem will illustrate the method of procedure in the case of combined footings.

Suppose two columns loaded and spaced as shown by Fig. 154, and let the allowable bearing on soil be 5,000 pounds per square foot. Let the dimensions of the footing be 20' - O" x ll' - O"=220 square feet. The determination of the size of base is largely a matter of judgment and depends upon the amount of load and the degree of spreading necessary to keep the size of grillage beams, or masonry offsets, within the limits which are economical. Suppose in this case the base is 3' - 6" X 3' - 6." The load per square foot in the upper tier is therefore1,100,000 =

20x 3.5

15,714 pounds. The moment on this tier will be a maximum either at one of the columns or at some point between them. The center of gravity of load is 600,000 x 11 = 6, or 6 feet from the

1,100,000 lighter load. This fixes the projection of the footing beyond the loads as 4 feet from the light load and 5 feet from the heavy load. The beams between the column loads are in the condition of a beam fixed at the ends and loaded with a uniformly distributed load. The moment may therefore be taken as approximately 2/3 of that for a beam simply supported. The moment between the columns will be a maximum where the shear is zero. To determine this start from one end, say the left-hand end, and determine the distance to the point of no shear by dividing the concentrated load by the load per linear foot; 600,000/55,000 = 10.9.

If the load is assumed uniformly distributed over the upper tier the greatest moment outside of the column load will be at the end having the greatest free length. The maximum moment therefore in this case will be at the edge of the base plate of the column at the left-hand end or 10.9 feet from this end. Call these moments M and M' respectively.

M=1/2 x 55,000 X 3.25 X 3.25 X 12 = 3,487,000 inch pounds and M' = 2/3 X [55,000 X 10.9x 5.45 - 600,000 X 5.9] X 12 = 2,181,800 inch pounds.

If the allowable fibre strain is taken at 18,000 pounds per square inch, the required moment of resistance = 3,487,000/18,000 =

194.

The offsets in masonry footings can be determined by the formula for a beam fixed at one end and loaded uniformly. A general practice and one in fairly close accord with the results of the above formula is to draw lines at 60 degrees with the horizontal from the edges of the column bases and where these cross the joint lines (the thickness of the courses having been assumed) will be the vertical face of the course. When the structure is of such a character that wind load affects the foundations, this must be considered in addition to the other live loads. Such cases would be narrow and very high buildings, chimneys, monuments, etc.