## 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.