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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
Pine, Oregon	110
Pine, Texas yellow	120
Pine, common white	82
Redwood	60
Spruce	100
Whitewood (poplar)	95

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.