When a piece of timber is employed as a column or support, its tendency to yielding by compression is different according to the proportion between its length and area of its cross section; and supposing the form that of a cylinder whose length is less than seven or eight times its diameter, it is impossible to bend it by any force applied longitudinally, as it will be destroyed by splitting before that bending can lake place; but when the length exceeds this, the column will bend under a certain load, and be ultimately destroyed by a similar kind of action to that which has place in the transverse strain. Columns of cast iron and of other bodies are also similarly circumstanced.

When the length of a cast iron column with flat ends equals about thirty limes its diameter, fracture will be produced wholly by bending of the material. When of less length, fracture takes place partly by crushing and partly by bending. But, when the column is enlarged in the middle of its length from one and a half to twice its diameter at the ends, by being cast hollow, the strength is greater by one-seventh than in a solid column containing the same quantity of material. To determine the dimensions of a support or column to bear, without Sensible curvature, a given pressure in the direction of its axis.

Rule. - Multiply the pressure to be supported in lbs. by the square of the column's length in feet, and divide the product by twenty times the tabular value of E; and the quotient will be equal to the breadth multiplied by the cube of the. least thickness, both being expressed in inches.

Note 1. - When the pillar or support is a square, its side will be the fourth root of the quotient.

Note 2.-If the pillar or column be a cylinder, multiply the tabular value of E by 12, and the fourth root of the quotient equal the diameter.

Ex. 1. What should be the least dimensions of an oak support, to bear a weight of 2240 lbs, without sensible flexure, its breadth being 3 inches, and its length 5 feet?

Tabular value of E = 105, and 2240 X 52 / 20 X 105 X 3 = Resistance Of Bodies To Flexure By Vertical Pressu 48 =2.05 inches.

Ex. 2 Required the side of a square piece of Riga fir, 9 feet in length, to bear a permanent weight of 6000 lbs.

Tabular value of E =96, and 6000 X 92 / 20 X 96 = Resistance Of Bodies To Flexure By Vertical Pressu 49 = 4 inches nearly.