Читать книгу Strength Of Beams, Floor And Roofs - Including Directions For Designing And Detailing Roof Trusses, With Criticism Of Various Forms Of Timber Construction - Frank E. Kidder - Страница 10

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, = 16 1/3 inches, agreeing, practically, with the value found in Example VII.

[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. 6, will give an excess of strength when several concentrated loads are applied unsymmetrically as regards the span, as for instance in Figs. 5 and 7.

For a beam loaded as in Fig. 7, the equivalent distributed load found by Rule 10 would be:

For load A, 1000 × 1 1-9 = 1111 lbs.

For load B, 1000 × 1.78 = 1780 lbs.

For load C, 1000 × 2 = 2000 lbs.

Equivalent distributed load, for all three loads, 4891 pounds.

But by the correct method of bending moments, the equivalent distributed load would be but 4,000 pounds, so that Rule 10 will give an excess of strength of a little more than one-fifth. For Example VI, Fig. 5, the error is about 18 per cent. This error, however, is always on the safe side. The exact method is to first find the greatest bending moment produced by the load, and to then proportion the beam to the bending moment. The method of finding the bending moment for any system of loading is explained in Chapter IX of The Architects’ and Builders’ Pocket Book.]

Fig. 7.