The distances of the load from each support in this case are 2 feet and 4 feet.
Therefore (2 × 4 × 4)/6 = 51⁄3 feet, effective length of beam.
Posts and Struts subject to Compression.[5]—Posts and struts, when above 30 diameters in length, tend to fail by bending and subsequent cross breaking. When below 30 and above 5 diameters in length, their weakness may partly be in their bending and partly by crushing, and when below 5 diameters in length by crushing alone; but, as these latter are rarely met with in scaffolding, they need not be considered.
On the resistance of long posts.—To find the greatest weight, W, that a square post of 30 diameters and upwards will carry:
Multiply the fourth power of the side of the post in inches by the value of D (Table II.), and divide by the square of the length in feet. The quotient will be the weight required in pounds.
The formula is as follows:
W = (d4 × D)/L2 = weight in lbs. for square posts to resist bending,
where d = depth in inches,
D = constant in pounds, for flexure,
L = length in feet.
Example.—Find the weight that may be placed upon a post of elm 12 feet long and 4 inches square.
The value of D (Table II.) being 1,620,
(44 × 1,620)/122 = (256 × 1,620)/144
= 2,880 lbs.