BEAMS SUPPORTED AT BOTH ENDS, IRREGULARLY LOADED.
We have now covered all the cases of loading for which a simple rule can be given. To find the size of a beam to support several loads applied at different places, without determining the bending moment, the beam should be considered as made up of as many thicknesses as there are loads, and the thickness necessary to support each load calculated, using the same depth for each thickness; the sum of the thicknesses will be the required breadth of the beam.
Example VI.—To find the size of beam necessary to safely support the loads shown in Fig. 5, the wood being Oregon pine.
Answer.—It will be necessary to consider this as three beams, placed side by side, and each loaded as in Fig. 3. We will assume 12 inches for the depth of the beam. Then by Rule 7 the thickness required to support the load A will equal (4 × 2500 × 4 × 12) ÷ (A × 144 × 16) = 480,000 ÷ 207,360 = 2.31 inches.
Thickness for load B = (4 × 3000 × 7 × 9) ÷ (90 × 144 × 16) = 756,000 ÷ 207,360 = 3.64 inches.
Thickness for load C = (4 × 2500 × 10 × 6) ÷ (90 × 144 × 16) = 600,000 ÷ 207,360 = 2.89 inches.
Total breadth of beam = 2.31 + 3.64 + 2.89 = 8.84 inches, or it will require a 9 × 12 inch beam to support the three loads.*
Example VII.—The girder shown in Fig. 6 supports a partition over its entire length and heavy floor beams, placed at equal distances of 4 feet from the supports and from each other. The partition supports a flat roof; its own weight and the weight of the roof supported by it is 16,000 pounds. The weight coming on the girder from each of the beams is 6000 pounds. The wood is long leaf Georgia pine. What should be the size of the beam?
Fig. 5.—Beam Supported at Both Ends and Irregularly Loaded.
Fig. 6.—Girder Supporting Partition and Heavy Floor Beams.
Answer.—As the girder is symmetrically loaded, its size can be determined by two operations. The load at the center is equivalent to a distributed load of 12,000 pounds† We have then to determine the size of beam to support a distributed load of 28,000 pounds, and the size of beam to support loads of 6000 pounds 4 feet from each support. The first should be determined by Rule 3* and the latter by Rule 9. We will assume 14 inches for the depth of the beam. The thickness of beam required to support a distributed load of 28,000 pounds is, by Rule 3, equal to 28,000 × 16 divided by 2 × 14 × 14 × 100 = 448,000 ÷ 39,200 = 11 1/2 inches.
The thickness required for the loads at A and C is, by Rule 9, equal to 4 × 6000 × 4 ÷ 14 × 14 × 100 = 5 inches. The thickness required for all of the loads will be 11 1/2 + 5 = 16 1/2 inches, or the beam must be 16 1/2 × 14 inches. If it were practicable to obtain a beam 15 or 16 inches deep, it would be better to use the deeper beam. Thus if we had assumed 15 inches for the depth of our beam we would have obtained 10 inches for the thickness required for the distributed and center loads and 4 1/4 inches for the loads at A and C, or 14 1/4 inches for all of the loads.
When it is necessary to support as great loads as in the above example by wooden beams, it may be necessary to make the breadth greater than the depth, and in such a case the beam may be built up of two or more pieces of the same depth, placed side by side and bolted together. There is no objection to building up girders in this way, and in fact such a girder is often better than one made of a solid stick; but wooden girders should not be built up by spiking or bolting two or more timbers one on top of the other, if it is possible to obtain planks of the full depth. Compound girders made by placing one timber above another may be built so as to obtain about three-fourths of the strength of a solid timber, but it requires a particular system of keying and bolting, which is expensive, and requires careful calculation. The best method of building compound girders, and rules for computing their strength, are given in Part II of the author’s work on Building Construction and Superintendence.
EQUIVALENT DISTRIBUTED LOAD, FOR A GIVEN CONCENTRATED LOAD.
Rule 10.—When a beam supported at both ends supports one or more concentrated loads applied at even fractions of the span from one support, the size of beam required to support the given loads may be most easily computed by first finding the equivalent distributed load and then finding the size of beam required to support this load by Rule 3. The equivalent distributed load for concentrated loads applied at different proportions of the span from either support may be found by multiplying the concentrated load by the corresponding factor given below:
For concentrated load at center of span, multiply by 2
For concentrated load applied at 1-3 the span, multiply by 1.78
For concentrated load applied at 1-4 the span, multiply by 1.5
For concentrated load applied at 1-5 the span, multiply by 1.28
For concentrated load applied at 1-6 the span, multiply by 11-9
For concentrated load applied at 1-7 the span, multiply by .98
For concentrated load applied at 1-8 the span, multiply by 7-8
For concentrated load applied at 1-9 the span, multiply by .79
For concentrated load applied at 1-10 the span, multiply by .72
For concentrated load applied at 1-12 the span, multiply by .61
For two equal loads applied one-third of the span from each support multiply one load by 2 2-3.
For two equal loads applied one-fourth the span from each end multiply one load by 2.
Application.—To show the application of Rule 10, we will apply it to Example VII. Here we have a distributed load of 16,000 pounds. A concentrated load at center of 6000 pounds and two concentrated loads of 6000 pounds each applied at one-fourth of the span from each support. Then by the above rule, the concentrated load at center is equal to a distributed load of 2 × 6000, or 12,000 pounds.
The two loads, 4 feet from each end (one-fourth the span), are together equal to a distributed load of 2 × 6000, or 12,000 pounds, and the total equivalent distributed load is 16,000 + 12,000 + 12,000, or 40,000 pounds.
Assuming 14 inches for the depth of the beam, the width of the beam should be, by Rule 3,
[NOTE.—The above rule (Rule 10) for finding the equivalent distributed load, and also the method used in Example VI, while absolutely correct for single loads, and also for a symmetrical application of loads, as in Fig.
