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The strength of a beam subject to almost any of the different variations of loading may be determined with about the same degree of accuracy as if simply loaded at the center, but the calculations require a considerable knowledge of mathematics, so that only a few of the more common cases can be covered by simple rules. These we will now consider.

When considering the strength of beams we usually have either one of two problems to solve—namely, to find the strength of a given beam or to determine the necessary size of beam to support a given load. The same algebraic formula really answers for both, but for the benefit of those not proficient in algebraic equations we will give a simple rule for each question, and also for each of the common conditions of support and loading. When we have to determine the strength of a given beam all of the conditions are known, but when we wish to determine the size of beam to carry a given load we must guess at or assume one dimension of the beam and solve for the other. If our first guess gives a badly proportioned beam we must guess again, and do the problem over again a second time. The quantity which represents the strength of the wood or the resistance of the fibers to breaking is now commonly designated as “fiber stress.” In text books written previous to the year 1885 the same quantity is called “modulus of rupture.” This quantity, of course, varies with different woods, and has been determined by numerous experiments on beams of the different kinds of woods. For convenience in making calculations one-eighteenth of the modulus of rupture is generally used for determining the breaking strength of wooden beams, and one-third of this latter value for determining the safe strength.

In the following rules this quantity will be represented by the letter A, the values of this letter for the different woods used in construction being given in Table I:

Table I.—Values of A, Used in Determining the Safe Strength of Beams.

Kind of Wood.	A, in Pounds.
Chestnut	60
Hemlock	55
Oak, white	75
Pine, Georgia yellow	100
Pine, Norway	70
Pine, Oregon	90
Pine, Texas yellow	90
Pine, common white	60
Redwood	60
Spruce	70
Whitewood (poplar)	65

To find the strength of a rectangular beam, supported at both ends and uniformly loaded over its entire length.

Rule 1.—Multiply twice the breadth of the beam by the square of the depth and by the value of A in Table I, and divide by the span in feet.*

To find the strength of a rectangular beam, supported at both ends and loaded at the center.

Rule 2.—Multiply the breadth of the beam by the square of the depth and by the value of A, Table I. Divide the product by the span in feet.

To determine the SIZE of a rectangular beam, supported at both ends and uniformly loaded over its entire length.

Rule 3.—Assume the depth of the beam. Multiply the span by the load, and divide by twice the square of the depth multiplied by the value of A. The answer will be the breadth of the beam.

Example I.—A rectangular spruce beam having a span of 16 feet is required to support a uniformly distributed load of 3780 pounds; what should be the size of the beam?

Answer.—We will assume 12 inches for the depth of the beam. The span multiplied by the load = 60,480. Twice the square of the depth multiplied by A, for spruce = 2 × 12 × 12 × 70 = 20,160. Divide 60,480 by 20,160 and we have 3 inches for the breadth of the beam, or a beam 3 × 12 inches will just support the load. If we assume 8 inches for the depth of the beam we shall obtain 6 3/4 inches for the breadth. One beam would have the same strength as the other, but the deeper beam would contain the least material and bend less.

To determine the size of a rectangular beam, supported at both ends and loaded at the center.

Rule 4.—Multiply the load by 2, and then proceed by Rule 3. That is, a load of 1000 pounds at the center will require the same size beam as a load of 2000 pounds distributed.

To determine the size of a rectangular beam, supported at both ends and carrying both a distributed load and a concentrated load at the center.

Rule 5.—Multiply the concentrated or center load by 2, and add the product to the distributed load, then proceed by Rule 3.

Example II.—A hard pine girder of 12-foot span supports a distributed load of 18,000 pounds, and also a post at the center, which sustains a load of 9600 pounds; what should be the dimensions of the girder?

Answer.—Twice the center load = 19,200 pounds. This added to the distributed load = 37,200 pounds. Assume 14 inches for the depth, and proceed by Rule 3. The product of the span by the load = 12 × 37,200 = 446,400. Twice the square of the depth multiplied by A, for hard pine = 2 × 14 × 14 × 100 = 39,200. Now 446,400 divided by 39,200 = 11 3/8 inches, or it will require an 11 3/8 × 14 inch girder to support both loads.

To determine what amount of concentrated load a given beam supported at both ends will safely carry at a given distance, N, from the left support (see Fig. 3).

Rule 6.—Multiply together the breadth, the square of the depth, the span and A, and divide the final product by four times the product of N multiplied by M, both in feet. The result will be the maximum safe load in pounds.

Example III.—A 10 × 12 inch hard pine girder, having a span of 14 feet, supports a post 4 feet from the left support; what is the maximum load that should be put on the post?

Answer.—The product of the breadth, the square of the depth, the span, and A = 10 × 144 × 14 × 100 = 2,016,000. If the span is 14 feet and N is 4 feet, M will be 10 feet. Four times the product of N by M = 160, and 2,016,000 divided by 160 = 12,600 pounds, the maximum safe load.

When the load is at the center this rule will give the same result as Rule 4.

Fig. 3.—Diagram Illustrating Rule 6.

Fig. 4.—Showing Equal Loads Concentrated at Equal Distances from Supports.

To determine the SIZE OF BEAM, supported at both ends, required to support a concentrated load applied at a given distance, from the left support.

Rule 7.—Multiply four times the load by the product of M by N, and divide the final product by the product of A times the square of the depth times the span. The result will be the breadth in inches.

Example IV.—What size of hard pine beam will be required to support a load of 12,600 pounds 4 feet from the left support, the span being 14 feet?

Answer.—Four times the load multiplied by the product of M by N = 2,016,000. Assume 12 inches for the depth; then A multiplied by the square of the depth, and the product by the span = 100 × 144 × 14 = 201,600, and 2,016,000 divided by 201,600 equals 10 inches, the required breadth. If we had taken 14 for the depth we would have obtained a breadth of 7 34-100 inches.

To determine the strength of a rectangular beam, loaded as in Fig. 4, M being equal to M¹ and W equal to W¹.

Rule 8.—Multiply the breadth by the square of the depth and their product by A, and divide by four times M (in feet). The result will be the safe load at each point. It should be noted that in this case the strength is not affected by the span, if we neglect the weight of the beam itself.

Example V.—What are the greatest safe loads a 10 × 12 inch hard pine beam of 12 feet span will support at a distance of 4 feet from each end?

Answer.—10 × 144 × A = 144,000, which divided by four times M = 144,000 ÷ 16 = 9000 pounds at each point.

To determine the SIZE OF BEAM required to support equal loads concentrated at equal distances from the supports, as in Fig. 4.

Rule 9.—Assume the depth: Multiply four times the load at one point by M (in feet), and divide by the square of the depth multiplied by A. The answer will be the breadth of the beam in inches.