## 54. A. Under a Wall

As the duty of the beams is to distribute the load coming from the foundation wall or base plate evenly over the ground, so that the pressure on each square foot of the soil will be the same, it is obvious that the beams must have sufficient transverse strength to keep them from bending, so that they will settle as much at the outer ends as under the centre. The effect on the beams shown in Fig. 15, when resting on a compressible soil and heavily loaded from above, is to cause the ends of the beams to bend upward, thus straining the beams most at the centre; the stress in the beams being the same as if they were supported at the centre and loaded with a distributed load. The maximum bending moment is also the same as for a beam fixed at one end and uniformly loaded, so that the beams are usually calculated by the formula for a beam fixed and loaded in that way.

CROSS SECTION.

SIDE VIEW.

Fig. 15.

The readiest method of determining the size of the beams is by computing the required coefficient of strength and finding in the tables of the manufacturers the size of beam which has a coefficient equal to or next above the value obtained by the formula. The coefficient of strength, generally represented by the letter C, is given in the catalogues of the companies that roll beams, and may also be found in the tables of beams in the Architects' and Builders' Pocket Book.

The formula for the coefficient of strength for beams under a wall, as in Fig. 15, and also for the lower tier of beams under a pier, is

C = 4 X w X p2 X s.............................. (1) in which w represents the assumed bearing power in pounds per square foot; p, the projection of the beam in feet, and s, the spacing or distance between centres of beams, also in feet.

Owing to the tendency of the beams in bending, to concentrate the load on the outer edges of the masonry footing, and thus crush them, which action would have the same effect on the beam as lengthening the arm or projection (see article in Architecture and Building previously referred to), the author recommends that when the course above the beams is of stone, brick or concrete, at least one-third the width of the masonry footing be added to the actual projection. The calculations above indicated will be more clearly shown by the following example :

Example 1. - A building is to be erected on a soil of which the safe bearing power is but 2 tons, and the pressure on each lineal foot of wall is 20 tons. It is decided to build the footings as shown in Fig. 15. What should be the dimensions and weight of the beams?

Answer. - As the total pressure under each lineal foot of wall is 20 tons, and the safe bearing power of the soil 2 tons, the footings must be 20÷2, or 10 feet wide. We will use 4-foot granite blocks for the bottom course of the wall, which will give an actual projection (P) of 3 feet for the beams. For making the calculations we will add to the actual projection one-third of 4 feet, or 16 inches, making the value of p 4 1/3 feet. We will assume 1 foot for the spacing of the beams, so that s will equal 1. The beams must then have a coefficient of strength = 4XWXp2 Xs = 4X4000X (4 1/3)2 X 1=304,000 lbs. Examining the table giving the properties of Carnegie steel beams, we find that a 10-inch 33-pound steel beam has a coefficient of 344,-000 pounds, and a 25-pound beam 261,000 pounds; therefore, we must use 33-pound steel beams 10 feet long. If we spaced the beams 10 inches on centres, s would equal 5/6 and C would equal 4X4000X(4 1/3)2X5/6, or 253,500 pounds, which would enable us to use 25-pound beams, thereby effecting a saving of 30 pounds to the lineal foot of wall.

55. To facilitate making the above calculations, the Carnegie Steel Company publishes the following table giving the safe projection of Carnegie steel beams, spaced 1 foot on centres, and for bearing values ranging from 1 to 5 tons :

### Table IV. - Safe Projections In Feet Of Steel Beams In Foundations

 DEPTH OF BEAM. WEIGHT PER FOOT. BEARING POWER IN TONS PER SQUARE FOOT. 1 1¼ 1½ 2 2¼ 2½ 3 3½ 4 4½ 5 In. Lbs. 20 80. 14.0 12.5 11.5 10.0 9.0 9.0 8.0 7.5 7.0 6.5 6.0 20 64. 12.5 11.0 10.0 8.5 8.0 8.0 7.0 6.5 6.0 6.0 5.5 15 75. 11.5 10.5 9.5 8.0 7.5 7.5 6.5 6.0 6.0 5.5 5.0 15 60. 10.5 9.5 8.5 7.5 7.0 6.5 6.0 5.5 5.5 5.0 5.0 15 50. 9.5 8.5 8.0 7.0 6.5 6.0 5.5 5.0 5.0 4.5 4.5 15 41. 8.5 8.0 7.0 6.0 6.0 5.5 5.0 4.5 4.5 4.0 4.0 12 40. 8.0 7.0 6.5 5.5 5.5 5.0 4.5 4.0 4.0 3.5 3.5 12 32. 7.0 6.5 5.5 5.0 4.5 4.5 4.0 4.0 3.5 3.5 3.0 10 33. 6.5 6.0 5.5 4.5 4.5 4.0 4.0 3.5 3.5 3.0 3.0 10 25.5 5.5 5.0 4.5 4.0 4.0 3.5 3.5 3.0 3.0 2.5 2.5 9 27. 5.5 5.0 4.5 4.0 4.0 3.5 3.5 3.0 3.0 2.5 2.5 9 21. 5.0 4.5 4.0 3.5 3.5 3.0 3.0 2.5 2.5 2.5 2.0 8 22. 5.0 4.5 4.0 3.5 3.5 3.0 3.0 2.5 2.5 2.5 2.0 8 18. 4.5 4.0 3.5 3.0 3.0 3.0 2.5 2.5 2.0 2.0 2.0 7 20. 4.5 4.0 3.5 3.0 3.0 3.0 2.5 2.5 2.0 2.0 2.0 7 15.5 4.0 3.5 3.0 2.5 2.5 2.5 2.0 2.0 2.0 2.0 1.5 6 16. 3.5 3.0 3.0 2.5 2.5 2.0 2.0 2.0 1.5 1.5 1.5 6 13. 3.0 3.0 2.5 2.5 2.0 2.0 2.0 1.5 1.5 1.5 1.5 5 13. 3.0 2.5 2.5 2.0 2.0 2.0 1.5 1.5 1.5 1.5 1.5 5 10. 2.5 2.5 2.0 2.0 1.5 1.5 1.5 1.5 1.5 ... ... 4 10. 2.5 2.0 2.0 1.5 1.5 1.5 1.5 ... ... ... ... 4 7.5 2.0 2.0 1.5 1.5 1.5 1.5 ...

Values given based on extreme fibre strain of 16,000 pounds per square inch.

By the use of this table no calculations are necessary except to determine the length and projection of the beams. If the beams are to be spaced more or less than 1 foot from centres, the bearing power must be increased or decreased in the same ratio in using the table. The results obtained by this table should agree with the result obtained from formula 1.

Thus, in the above example, to use the table, we simply look down the column headed 2 until we find the projection nearest to (above)

4 1/3 feet, which in this case is 4.5, and opposite it we find a 10-inch, 33-pound beam.

To use the table for a spacing of 10 inches we must take five-sixths of the bearing power, or 1 2/3 tons. There is no column headed 1 2/3, but it would come between 1½ and 2. For 1½ tons the projection of a 10-inch, 25-pound beam is 4.5, and for 2 tons, 4 feet. At the same ratio the projection for 1 2/3 tons would be about 4.3 feet.

When there is no column corresponding with the bearing power it will be safer to use formula 1.