Spruce.

Georgia pine.

White pine.

White oak.

Hemlock

Calculate (x) for transverse strength only if d is greater than

1 1/6L

L

1 1/19L

1 1/5L

14/15L

Calculate (x1) for deflection only if d is less than

1 1/6L

L

1 1/19L

1 1/5L

14/15L

Where L = the length of span, in feet.

Where d = the depth of beam, in inches.

Where x or x1 = is found according to Table IX.

To find the safe load (x) or (x1) per running foot of span, which a beam supported at both ends, and 1" thick will carry, use the following table. (Beams two inches thick will safely carry twice as much per running foot, as found pur table, beams three inches thick three times as much, four inches thick four times as much, etc)

Table IX

Spruce.

Georgia pine.

White pine.

White oak.

Hemlock.

If calculating for transverse

Strength only use

x=111(d/L)2

x=133(d/L)2

x= 100 (d/L)2

x = 122(d/L)2

x=83(d/L)2

If calculating for a deflection

(not to crack plaster) use

x1=95(d/L)3

x1=133(d/L)3

x, = 95(d/L)3

x1 =100 (d/L)3

x1 =89 (d/L)3

Where x = the safe load in lbs., per running foot of span, which a beam one-inch thick will carry regardless of deflection, if supported at both ends, and x1 = the same, but without deflecting the beam enough to crack plaster; for thicker beams multiply x or x1 by breadth, in inches.

Calculate for either x or x1 as indicated in Table VIII.

Where d= the depth of beam, in inches.

Where L = the length of span in feet.

If a beam is differently supported, or not uniformly loaded, also for cantilevers, add or deduct from above result, as directed in cases 1 to 8, page 57.

Example.

A floor of 19' clear span is to be built with spruce beams, to carry, 100 lbs. per square foot; what size beams would be the most economical?

According to Table VIII, if d= 1 1/6. L = 1 1/6. 19 = 22 1/6" we can calculate for either deflection or rupture and the result would be the same. If we make the beam deeper it will be so stiff that it will break before deflecting enough to crack plastering underneath; while if we make the beam more shallow it will deflect enough to crack plaster before it carries its total safe load. The former would be more economical of material, but, of course, in practice we should certainly not make a wooden beam as deep as 22". Whatever depth we select, therefore, less than 22", we need calculate for deflection only. We have, then, according to Table IX, second column, x1 = 95(d/L)3

If we use a beam 12" deep, we should have x1 = 95. 123/193 = 24 or a beam l"x 12" would carry 24 lbs. per foot; as the load is 100 lbs. per foot we should need a beam 100/24 = 4 1/6 wide, or say a beam

4" x 12", and of course 12" from centres.

If we use a beam 14" deep we should have x1 = 95. 143/193 = 38 or a beam 1" x 14" would carry 38 lbs. per foot, we need, therefore, a beam of width b= 100/38 = 2 12"/19 or we must use a beam say 3" x 14" and 12" from centre, or a beam 4" xl4" and 16" from centre. For if the beams are 16" from centres each beam will carry per running foot 1 1/3.100 lbs. = 133 lbs. and a 4" x 14" will carry per foot

4.x, = 4.38 =152.

We could even spread the beams farther apart, except for the difficulty of keeping the cross-furring strips sufficiently stiff for lathing.

Of course the 14" beam is the most economical, for in the 12" beam we use 4" x 12" = 48 square inches (cross-section) of material, and our beam is a trifle weak. While with the 14" beam we use only 3" x 14" = 42 square inches of material, and our beam has strength to spare. The 4" x 14" beam 16" from centres would be just as strong and use just as much material as the 3" x 14" beam 12" from centres. If we wished to be still more economical of material, we might use a still deeper beam, but in that case it would be less than 3" thick and might twist and warp. If the beam is not cross-bridged or supported sideways it might be necessary to calculate its strength for lateral flexure. That it will not shear off transversely we can see readily, as the load is so light, nor is there much danger of longitudinal shearing, still for absolute safety it would be better to calculate each strain.

The comparative strength of columns of same cross-section is approximately inversely as the square of their lengths. Thus, if x be the strength of a column, whose length

Strength of columns different lengths.