When A Beam Or Bar Is Fixed At One End, And Loaded At The Other. Rule

Multiply the value of the material in the preceding tables, or as may be ascertained, by the breadth and square of the depth in inches, and divide the product by the length in feet.

Note - When the beam is loaded uniformly throughout its length, the result must be doubled.

Example

What are the weights each that a cast and wrought iron bar, 2 inches square and projecting 80 inches in length, will bear without permanent injury?

The values for cast and wrought iron in this and the following calculations are assumed to be 225 and 180.

Hence 225x2x22 = 1800, which, + 2.5 =720 lbs.; and 180X2X22 = 1440, which, +2 5=576 lbs.

If The Dimensions Of A Beam Or Bar Are Required To Support A Given Weight At Its End. Rule

Divide the product of the weight and the length in feet by the talus of the material, and the quotient will give the product of the breadth and the square of the depth.

* An inch-square batten from the same plank as this specimen broke at 139 lbs.

Example

What is the depth of a wrought-iron beam, 2 inches broad, necessary to support 576 lbs. suspended at 30 inches from the fixed end?

576X2.5 .

---------=8, whlch, /2 ins. for the breadth=4, and √4=2 ins., the 180 breadth.