117,000 In

lb. The modulus of rupture of yellow pine, 7.000 lb., divided by the factor of safety 4, gives a unit fiber stress of 1,750 lb. Then as the resisting moment of any beam section,

M1 = Q S, by transposition, Q =M1/S, or 117,000/1,750 =67, the required section modulus.

By Table XII, for any rectangular section, Q =bd2/6 . By trial, the value of Q for a 3" X 10" beam is 50; as this is evidently too small, try a 3" X 12" beam; for this Q which is ample, and this beam is selected, being the nearest stock size.

Rolled-Steel Beams

In calculating the size of steel beam sections, the greatest bending moment is first calculated; then the section modulus required is equal to the bending moment in inch-pounds divided by the safe unit fiber stress, obtained by dividing the modulus of rupture of the material by the factor of safety. This may be expressed in formula as

Q =M/S, in which, as before, Q = section modulus, M = bending moment in inch-pounds, and S = safe or allowable unit fiber stress in pounds per square inch. Having found the section modulus, the proper size channel or I beam may be selected from Tables XIII and XIX. the one being chosen which has a section modulus (see column headed Q) nearest to that required, the deepest beam having the required section modulus and lightest weight being preferred.

In selecting rolled structural-steel beams the depth of the beam in inches should never be less than one-half the span in feet, in order to avoid excessive deflection, which causes cracks in plastered ceilings. For instance, if the span-is 20 ft., a beam should be not less than 10 in. deep.

Example

The brick-and-concrete floor of an office building weighs 110 lb. per sq. ft., and is designed for a live load of 40 lb. per sq. ft. The span of the beams is 20 ft., and they are spaced 5 ft. on centers. Using a factor of safety of 4, what size steel I beams are required?

Solution

Total load is 110 lb. + 40 lb. = 1501b. per sq. ft.: floor surface supported upon one beam is 20 ft. X 5 ft. = 100 sq.ft.; total load on one beam is 100 sq. ft. X 1501b. = 15.0001b.,

=W. The load being uniformly distributed, the bending moment, if, by Table XXVI, is WL/8=(15,000x20) / 8=37,500 ft.-lb. or 450,000 in.-lb. The modulus of rupture for structural steel being about 60,000 lb. (see Table VIII), and the required factor of safety, 4, the allowable unit fiber stress is 60,000 / 4 = 15,000 lb. The section modulus, Q = M/S ; substituting the values of M and S, Q =450,000/15,000 = 30. From

Table XIX, page 90, it is found that a 10" beam weighing 33 lb. per ft. has a section modulus of 32.3, and would meet the requirements. It is, however, seen that for a 12" beam, weighing 31 1/2 lb. per ft., Q = 36.7; and on account of its greater strength and less weight, this beam would be by far the most economical.

Stone Beams

The strength of lintels, flagstones, etc. may be calculated as rectangular beams, except that it is usual to use a factor of safety at least of 10. It is, however, more convenient to use the formula,

W =(bd2 / l) x c, in which W = safe uniformly distributed load in tons of 2,000 pounds; b = breadth of beam in inches; d = depth of beam in inches; I = span of beam in inches; c = coefficient taken from the following table:

Kind of Stone.

Coefficient.

Bluestone............................................

.18

Granite...................................................

.12

Limestone..............................................

.10

Sandstone...........................................

.08

Slate ...........................................

.36

Example

A limestone lintel 20 in. wide X 14 in. thick spans a 42" opening. What is the safe distributed load?

Solution

Substituting values in the formula, (bd2 / l) x c,

W =[(20x14x14) / 42 ] x .10 = 9.33 tons = 18,660 lb.

If the load is concentrated at the center of the span, the safe load will be one-half the safe uniform load.