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 the span of a floor or ceiling joist measured in feet exceeds the depth of the joist in inches, the beam or joist, if loaded to its full safe load, will bend more than is desirable, and often enough to crack a plastered ceiling supported by it. For this reason the size of floor joists that support plastered ceilings should be calculated by the rule for stiffness. This rule is based upon the principle of the deflection of beams, and involves a quantity known as the modulus of elasticity, which varies for different woods, and is determined by experiments upon the flexure or bending of beams under known loads. Simple rules for the stiffness of beams can only be given for the two cases of beams uniformly loaded over the entire span and of beams loaded with a concentrated load applied at the center of the span. The rules for these cases are as follows:
To determine the maximum uniformly distributed load for a rectangular beam supported at both ends that will not produce a deflection exceeding 1-30 inch per foot of span.
Rule 17.—Multiply eight times the breadth by the cube of the depth, and the product by the value for E (Table II), and divide by five times the square of the span.
To determine the maximum center load for a rectangular beam supported at both ends that will not produce undue deflection.
Rule 18.—Multiply the breadth by the cube of the depth, and the product by the value for E, and divide by the square of the span.
To determine the SIZE OF BEAM to support a given distributed load without producing undue deflection, the beam being supported at both ends.
Rule 19.—Assume the depth. Multiply five times the load by the square of the span, and divide by eight times the cube of depth times E. The answer will be the breadth in inches.
To determine the SIZE OF BEAM to support a given center load without producing undue deflection, the beam being supported at both ends.
Rule 20.—Assume the depth. Multiply the load by the square of the span, and divide by the cube of the depth multiplied by E. The answer will be the breadth in inches.
Table II.—Value of E, to be Used in Rules 17-20.
| Kind of Wood. | E, in Pounds. |
| Chestnut | 72 |
| Hemlock | 80 |
| Oak, white | 95 |
| Pine, Georgia yellow | 137 |
| Pine, Norway | 100 |
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