TO DETERMINE DEPTH OF BEAM TO SUPPORT A GIVEN LOAD. There has just been given the method of finding the breadth of a beam when the depth was already fixed, the length (span) and load being also known. The case worked out is only one of numerous instances in which the depth of a beam is limited. A very common case is that of a floor where a beam has to be placed to carry a partition above, but is flush with the lower edges of the joist below. The last formula given will apply to that and all similar cases.
Very often, however, it happens that it is the breadth of a beam that is fixed; and the problem then is to find what depth the timber should be to carry the load safely. To find this requires, of course, merely a slight turning about of the formulas used in the cases dealt with previously, and presents no difficulty to anyone at all ac quainted with mathematics. It will be worked out, however, in the same manner as the other problems, and divested as much as possible of all complicated and repellant algebraic signs and symbols. The writer's attempts to interest and instruct his two craftsmen friends showed that it is usually the look of the formulas in books that scares the seeker after knowledge who is some years away from his schooldays. Also, that if these mystifying formulas are properly and simply explained, any man with a knowl edge of the elementary rules of arithmetic can make calculations for the size of practically all the timbers of the average building.
As before, certain conditions will be laid down, being as follows: A beam is to carry an 8-inch wall over an opening. The beam is of Southern hard (or pitch) pine; the opening is 10 feet wide; the height of the wall above the beam is 13 feet 6 inches, giving 90 cubic feet (10 feet by 13 feet 6 inches by 8 inches)—say, 90 cwt. as weight to be carried on beam.
The method of putting down these particu lars is shown in Fig. 168, which will be seen to resemble the earlier formulas, although the sev eral factors are differently arranged. Fig. 169, A, shows the calculations for our present case, which works out in exactly the same manner as earlier ones; that is, all the values above the line are multiplied together, and divided by the product of all the values below the line.
The only difficulty is in working out the final answer, for the result obtained at first will be the square of the required depth—that is, the depth multiplied by itself. To arrive at the exact depth, it is necessary to extract the square root of the answer, which means to find what number multiplied by itself will give the answer. For all ordinary practical purposes of wooden beam calculations, however, the exact result is not absolutely necessary, and a sufficiently ac curate one can be obtained by inspection of the first answer. For instance, in the foregoing problem the answer is Now, the nearest square of a whole number to this is 49, the square of 7 (7 X7=49) ; therefore the depth of the beam is more than 7. The next square of a whole number is 64, the square of 8 (8 x 8=64) ; and, as that is greater than our answer, evi dently our beam should be somewhere between 7 and 8 inches deep. As a matter of fact, it proves in this case to be exactly inches ; but if the sum had not worked out so exactly as that, a result quite good enough for practical purposes could have been arrived at by the method indicated above— namely, by finding the nearest squares of the whole numbers above and below the answer, and allowing a sufficient amount over and above the root number of the square below. As in the pre vious articles, the nature and position of the load must be considered in any calculations made. The factor of safety used is 5, the usual for a dead load. That is, one-fifth of the break ing weight _ is considered to be the amount a beam can safely carry when the load is a sta tionary one.
The position of the load is important, for, as shown in the earlier lessons, a beam will carry twice as much if the load is evenly distributed along it, as it would if the load were in the cen ter only. The figure 2 is therefore placed below the line in the present case. The effect of this is shown by comparing the result at A with that at B, Fig. 169.