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
Example XII.—What is the maximum distributed load that a 2 × 12 inch spruce beam, 16 feet span, will support without undue deflection?
Answer.—Apply Rule 17. Eight times the breadth times the cube of the depth times E = 8 × 2 × 1728 × 100 = 2,764,800. This divided by five times the square of the span or 1280 = 2160 pounds, the answer.
Example XIII.—A white pine floor joist of 18-foot span has to support a uniformly distributed load of 1440 pounds; what should be the size of the beam that the deflection may not be excessive?
Answer.—We will try 10 inches for the depth of the beam, and use Rule 19. Five times the load multiplied by the square of the span = 5 × 1440 × 324 = 2,332,800. Eight times the cube of the depth × E = 8 × 1000 × 82 = 656,000, and 2,332,800 ÷ 656,000 = 3 1/2 inches, the breadth.
If we use 12 inches for the depth we will have 2,332,800 ÷ 1,133,568, which gives 2 inches for the breadth, showing that a 2 × 12 joist of 18-foot span has the same stiffness as one 3 1/2 × 10 inches, although the latter beam contains nearly 50 per cent. more lumber than the former.
CONTINUOUS BEAMS.
A continuous beam is one which extends over three or more supports.
The formulas for determining the strength and stiffness of such beams are too elaborate to be reduced into simple rules, but it is worth while to know how the strength and stiffness of such beams compare with the strength and stiffness of a single span.
The strength of a continuous girder of two spans, Fig. 12, is the same as if the girder were cut over the support, when the load is distributed over the full length of the beam, but when the load is applied at the center of each span the strength of the beam is increased one-third by making it continuous over the center support, provided that the spans are equal. If the spans are unequal the increase in the strength will be less.
Fig. 12.—Continuous Girder of Two Spans.
Fig. 13.—Continuous Beam of Three Equal Spans.
The stiffness of a continuous beam of two equal spans is more than doubled by having the beam continuous over the center support, whether the load is distributed or concentrated.
In the case of a continuous beam of three equal spans, Fig. 13, the strength is increased one-fourth when the load is distributed, and two-thirds when equal loads are applied at the center of each span, and the stiffness is increased about 90 per cent. in both cases. It is therefore desirable to make beams continuous whenever practicable.
When beams are made continuous over three or more supports the points of greatest strain are those which come over the center supports or support, hence the beam should not be cut into at those points.
BEARING OF BEAMS ON THE WALL OR SUPPORT.
The transverse strength of a beam is not affected by the distance that the end of the beam extends onto the support, but the bearing must be sufficient that the beam will not pull off from the support when it is loaded, or that the bottom fibers of the beam will not be crushed by the load. This latter consideration is one which should be considered in the case of short beams loaded to their full capacity. Every wooden beam should have a bearing area—that is, the product of the breadth of the beam by the bearing, Fig. 1—equal to the load divided by 1000 for hard pine or oak, 500 for spruce and 400 for soft pine.
Thus a 10 × 12 inch white pine beam, of 8-foot span, might be safely loaded with 21,600 pounds if the load were uniformly distributed. Then the beam should have a bearing area = 21,600 ÷ 400, or 54 square inches. As the breadth of the beam is 10 inches, the bearing should equal 54 ÷ 10, or 5.4 inches. For floor joists a bearing of 4 inches is usually ample, and for girders a bearing of six inches is usually sufficient; 4 inches, however, should be considered as the minimum bearing, unless the beams are securely tied in place.
TABLES FOR THE STRENGTH AND STIFFNESS OF WOODEN BEAMS.
Tables III and IV will be found very convenient in figuring the safe loads for beams supported at both ends, and loaded either with a distributed load or a concentrated load applied at the center.
By following the directions given, the results obtained by the use of the tables should be the same as by using the corresponding rule, and with less figuring.
When the bending or deflection of the beam is of no importance use Table III, and when excessive bending must be avoided use Table IV.
TABLE III.—STRENGTH OF HARD-PINE BEAMS.
Table of safe quiescent loads for horizontal rectangular beams of Georgia yellow pine, one inch broad, supported at both ends, load uniformly distributed. For concentrated load at center divide by two. For permanent loads (such as masonry) reduce by 10 per cent.
For beams of any width greater than 1 inch, multiply the load in table by the width of the beam in inches.
For beams of Oregon pine, use 9-10 of tabular loads; for spruce beams, 7-10; for common white pine, 3-5, and for white oak, 3-4.
To use Table III for beams that run less than the nominal dimensions. In many localities floor joists as carried in stock are more or less scant of the nominal dimensions, and for such joists a reduction in the safe load must be made to correspond to the reduction in size. For beams having the full depth multiply the load in table by the actual breadth, as 1 5/8, 1 3/4, 2 7/8, or whatever it may be. For beams 1/4 inch scant in both dimensions the safe load may be obtained by multiplying the safe load as given in the table by the following factors:
For beams 1 3/4″ × 5 3/4″ by 1.6 | For beams 1 3/4″ × 11 3/4″ by 1.67 |
2 3/4″ × 5 3/4″ by 2.52 | 2 3/4″ × 11 3/4″ by 2.63 |
1 3/4″ × 6 3/4″ by 1 5/8 | 1 3/4″ × 13 3/4″ by 1.68 |
2 3/4″ × 6 3/4″ by 2.55 | 2 3/4″ × 13 3/4″ by 2.65 |
1 3/4″ × 7 3/4″ by 1.64 | 1 3/44″ × 14 3/4″ by 1.69 |
2 3/4″ × 7 3/4″ by 2.58
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