127. To find the weight that would break a rectangular beam when applied at the middle of its length, the beam being supported at the ends.

Rule XI. Multiply the breadth in inches by the square of the depth in inches, divide this product by the length in feet, and the quotient multiplied by the value of c in Table XIV. corresponding to the kind of wood; the product will be the weight in pounds.

Example. - The length of a girder of Riga fir between the supports is 21 feet, its depth is 14 inches, and breadth 12 inches; to find the weight that would break it when applied in the middle. Opposite Riga fir in the Table we find c = 530; and 12x14x14x530/21 = 59,360 pounds, or above 26 tons.

If a beam of the same scantling and length had been supported at one end only, one-fourth of the weight would have broken it.

128. To find the weight that would break a solid cylinder when applied at the middle of its length, the cylinder being supported at the ends.

Rule XII. - Find the value of c for the kind of wood in Table XIV., and divide it by 1.7; multiply the quotient by the cube of the diameter in inches, and divide the product by the length in feet; the quotient will be the weight in pounds that would break the cylinder.

Example. - What weight would break a solid cylinder of ash, 12 feet long and 8 inches diameter? For ash the value of c is 635 in the Table, therefore

635x8x8x8/1.7x12=15,937pounds 129. If the weight be uniformly diffused over the length of a beam, it will require to break it twice the weight that would break it when applied at the middle of its length.

130. If a beam be fixed at both ends and loaded in the middle, it has to break in three places instead of one; it will in that case carry 1 1/2 times more weight than if merely supported at both ends.