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.
CANTILEVER BEAMS.
To determine the maximum safe load for a beam of known dimensions, loaded and supported as in Fig. 8 or Fig. 9.*
Rule 11.—To find safe load W, multiply the breadth of the beam by the square of the depth, both in inches, and this product by the value of A, and divide by 4 × L, in feet.
Example VIII.—What is the safe concentrated load for a spruce beam, 6 × 8 inches, fixed at one end, the point of application of the load being 6 feet from the support?
Answer.—Safe load equals 6 × 8 × 8 × 70 divided by 4 × 6 = 26,880 ÷ 24 = 1120 pounds.
Fig. 8.—Cantilever Beam Supporting Load at Its Outer End.
Fig. 9.—Another Form of Cantilever.
The beam will have the same strength whether loaded and supported as in Fig. 8 or as in Fig. 9.
To determine the SIZE OF BEAM to support a given load applied at a fixed point from the support, as in Fig. 8 or Fig. 9.
Rule 12.—First assume the depth. To find the breadth, multiply four times the load, in pounds, by the distance L, in feet, and divide by the square of the depth multiplied by the value for A.
Example IX.—What size of spruce beam will be required to support a load of 1120 pounds, applied 6 feet from the support?
Answer.—Assume 8 inches for the depth of the beam. Then the breadth will be equal to 4 × 1120 × 6 divided by 8 × 8 × 70 = 26,880 ÷ 4480 = 6 inches.
CANTILEVER BEAM WITH DISTRIBUTED LOAD.
To determine the maximum safe distributed load for a cantilever beam of known dimensions.
Fig. 10.—Load Extending from the Support.
Fig. 11.—Beam Supported at the Center.
Let W = the amount of the load, in pounds, and L the distance in feet that the load extends from the support, as in Fig. 10. If the beam is supported at the center, as in Fig. 11, W should equal the load on each side of the support.
Rule 13.—To find the safe load W, multiply the breadth by the square of the depth and the product by the value for A, and divide by two times L (in feet).
To determine the SIZE OF BEAM required to support a known load, uniformly distributed on the beam, as in Figs. 10 and 11.*
Rule 14.—Assume the depth. To find the breadth, multiply twice the load by the distance L (in feet), and divide by the square of the depth multiplied by the value for A.
STRENGTH OF CYLINDRICAL BEAMS.
Rule 15.—To find the safe load for a cylindrical beam, as a log, first find the strength of a square beam (loaded in the same way) whose sides are equal to the diameter of the round beam, and divide the answer by 1.7. If the beam tapers slightly, as in the case of the trunk of a tree, measure the diameter at the center of the span.
Example X.—What is the safe center load for a spruce pole 12 inches in diameter at the center and with a span of 16 feet?
Answer.—By Rule 2 we find that the strength of a spruce beam, 12 inches square and 16 feet span, equals 12 × 12 × 12 × 70 divided by 16 = 7560 pounds. Dividing this by 1.7 we have 4447 pounds for the answer.
To determine the diameter of a cylindrical beam to support a given load at the center.
Rule 16.—Multiply the span by the load and the product by 1.7 and divide by the value of A. The cube root of the result will be the answer.
Example XI.—Find the diameter of a round spruce pole of 16 feet span to support a center load of 4447 pounds.
Answer.—4447 × 16 × 1.7 = 120,958.4. Dividing this by 70, the value of A, we have 1728. The cube root of 1728 is 12, the required diameter of the pole.
If the load is distributed, divide it by 2 and then proceed by the above rule.
STIFFNESS OF BEAMS.
When
