TO DETERMINE WIDTH OF BEAM TO SUPPORT A GIVEN LOAD. "Well, how are you getting along with your calculations of the strength of tim ber'?" was the writer's greeting as his two craftsmen friends and pupils came into the office again one day recently.
"Oh! we are doing pretty well, and we can now remember the first rule or formula which you gave us without having to turn to refresh our memories. But we struck a small snag yes terday. We have a sort of solution; but we felt that there was a proper way to work it out, and that is the reason of our visit to-day." The writer having expressed his readiness to help, the spokesman went on to explain the little problem which was worrying him, somewhat as follows: "Up to now we have been finding the strength of some particular piece of timber when fixed and loaded in various ways. But suppose we knew the load that was to be carried by a beam, and wanted to calculate the size of the timber—is there not some way of doing that just as easily as finding the strength of a piece whose size we know?" Having been assured that the calculations for such a problem were only a trifle more diffi cult than for the one now familiar to them, the speaker went on to state his problem more particularly.
"There is an 8-foot driveway to be made through a new brick store, and we want to find the size of the beam that will safely carry the weight of the brickwork in the story above the beam. We want to have the beam of such a depth that it will be equal to so many courses of the brickwork, and thus save a lot of cutting for the bricklayer and a poor appearance after wards. The wall is one brick thick (8 inches), and is to be carried up nine feet above the beam. The bricks are inches thick, and four courses measure just 10 inches." Having these particulars, we proceeded as follows: First, we found the weight of the brickwork to be carried by the beam. Nine feet high, 8 feet wide, and 8 inches thick, gave us 48 cubic feet. A cubic foot of brickwork weighs about 1 cwt. (112 lbs.); so the weight to be car ried was, of course, 48 cwt.
But the roof had also to be considered, and we next proceeded to calculate the weight of that portion of the roof carried on the wall over the beam. From ridge to eaves, the roof slope measured 20 feet; 20 feet times 8 feet gave us 160 square feet. Slates were to be used as cov erings; and as the weight of a slate covered roof, allowing for wind pressure, is usually taken at 1/2 cwt. per square foot, we obtained as the weight to be considered, 80 cwt. (160 times The problem then resolved itself into this: One hundred and twenty-eight cwt. was to be carried as a distributed load on a beam 10 inches deep over an opening of 8 feet span. The load was to be stationary, or "dead." How broad should our beam be to carry the weight safely'? It seemed best to follow the plan adopted in the earlier lessons, and to give another rule or formula which could be easily referred to and in time remembered. This was written down
as in Fig. 166, which is merely a simple way of stating a reversal of our first rule for finding the strength of a given beam.
With this rule before us, we put down our particulars as in Fig. 167, Calculation No. 1— namely, 128 cwts. for our load; 8 as our length; 5 as our factor of safety (dead load); 10 times 10 as our depth (squared), equal to four courses of brickwork; 5 as our figure for oak (see table); and 2 because our load was to be evenly dis tributed along the beam.
The result, inches, being less than the thickness of the wall to be carried, led us to try what breadth our beam would have to be if we made it only inches deep—that is, equal to three courses of brickwork—Calculation No. 2 shows this, which gave as a result rather over nine inches (see Calculation No. 2, Fig. 167). As this was more than the thickness of the wall to be carried, it was decided to make the beam 10 inches deep, and, for the better convenience of the bricklayer in starting his courses, to make the beam the full thickness of the wall. This, of course, was a stronger beam than was actually required, but was an error in the right direction.
In point of fact, it is always wise to err on the side of strength in deciding on the sizes of tim ber to be used in construction, for no two pieces give the same results if tested. As already re marked, the strengths of the various kinds of woods as given in most textbooks were obtained by averaging the results of hundreds oLactual tests; and it was found that different pieces, even from the same tree, varied considerably. So serious is this variation, that many architects require that beams or girders carrying heavy weights shall be ripped in half and the ends re versed. It is very common to find in specifica tions for beams in positions similar to that which is the subject of this present calculation, a requirement that (in addition to the usual clauses as to the quality, etc., of the lumber) the piece be halved, reversed, and bolted together again. By doing this, any possible weakness at one end of the stick would be obviated, and a piece of even strength obtained.
Of course, in many parts of the world, iron beams have largely superseded wooden ones; but for a long time to come, wood will hold its own in many districts for a variety of reasons.
To sum up this lesson, then, the first thing is the new rule or formula for putting down the known particulars. In doing this the things to be remembered are: (1) the factor of safety (5 for a dead load, 8 or 10 for a live load); (2) the figure for the particular timber used; and (3) the fact that a beam carries a distributed load just double of what it would carry when loaded in the center only.