The strength of a square or rectangular beam to resist lateral pressure, octing in a perpendicular direction to its Length, is as the breadth and square of the depth, and inversely us the length;-thus, a beam twice the breadth of another, all other circumstances being alike, equal twice the strength of the other; or twice the depth, equal four times the strength, and twice the length, equal only half the strength, etc, according to the rule.

Table of Data, containing the Results of Experiments on the Elasticity and Strength of various Species of Timber, by Mr. Barlow.

To find the dimensions of a beam capable of sustaining a given weight, with a given degree of defection, when supported at both ends.

Rule.- .Multiply the weight to be supported in lbs. by the cube of the length in feet; divide the product by 32 times the tabular value of E, multiplied into The given deflection in inches; and the quotient is the breadth multiplied by the cube of the depth in inches.

Note 1 .-When the beam is intended to be square, then the fourth root of the quotient is the breadth and depth required.

Note 2.-If the beam is to be cylindrical, multiply the quotient by 1.7, and the fourth root of the product is the diameter.

Ex. The distance between the supports of a beam of Riga fir is J6 feet, and the weight it must be capable of sustaining in the middle of its length is 8000 lbs, with a deflection of not more than ¾ of an inch; what must be the depth of the beam,supposing the breadth 8 inches?

16 X 8000 / 90x32x.75 = 15171 ÷ 8 = 3√1897 = 12.35in.,the depth

To determine the absolute strength of a rectangular beam of timber, when supported at both ends, and loaded in the middle of its length, as beams in general ought to be calculated to, so that they may be rendered capable of withstanding all accidental cases of emergency.

Rule.-Multiply the tabular value of S by four times the depth of the beam in inches, and by the area of the cross section in inches; divide the product by the distance between the supports in inches, and the quotient will be the absolute strength of the beam in lbs.

Note ].-If the beam be not laid horizontally, the distance between the supports,for Calculation, must be the horizontal distance.

Note 2.-One fourth of the weight obtained by the rule, is the greatest weight that ought to be applied in practice as permanent load.

Note 3.-If the load is to be applied at any other point than the middle, then the strength will be as the product of the two distances is to the square of half the length of the beam between the supports;-or, twice the distance from one end, multiplied by twice from the other, and divided by the whole length, equal the effective length of the beam.

Ex. In a building 18 feet in width, an engine boiler of 5½ tons (2240 lbs. to a ton) is to be fixed, the center of which to be 7 feet from the wall, and having two pieces of red pine, 10 inches by 6, which I can lay across the two walls for the purpose of slinging it at each end,-may I with sufficient confidence apply them, so as to effect this object ?

2240X5.5 ÷ 2 = 6160 lbs. to carry at each end.

And 18 feet - 7 = 11, double each, or 14 and 22, then 14X22 ÷18 = 17 feet, or 204 inches, effective length of beam.

Tabular value of S, red pine, =1341X4X10X60 ÷204 = 15776 lbs. the abso-lute strength of each piece of timber at that point.

To determine the dimensions of a rectangular beam capable of supporting a required weight, with a given degree of deflection, when fixed at one end.

Rule.-Divide the weight to be supported, in lbs., by the tabular value of E, multiplied by the breadth and deflection, both in inches; and the cube root of the quotient, multiplied by the length in feet, equal the depth required in inches.

Ex. A beam of ash is intended to bear a load of 700 lbs. at its extremity; its length being 5 feet, breadth 4 inches, and the deflection not to exceed ½ au inch.

Tabular value of E = 119X4X.5 = 238 the divisor; then 700 ÷ 238 = X 5 = 7.25 inches, depth of the beam.

To find the absolute strength of a rectangular beam, when fixed at one end, and loaded at the other

Rule-Multiply the value of S by the depth of the beam, and by the area of its section, both in inches; divide the product by the leverage in inches, and the quotient equal the absolute strength of the beam in lbs.

Ex. A beam of Riga fir. 12 inches by 4½, and projecting 6½feet from the wall; what is the greatest weight it will support" at the extremity of its length ?

Tabular value of S = 1100. 12X4.5 = 54 sectional area. Then, 1100X12X54 ÷ 78 = 913S.4 lbs.

When fracture of a beam is produced by vertical pressure, the fibres of the lower section of fracture are separated by extension, whilst at the same time those of the upper portion are destroyed by compression; hence exists a point in section where neither the one nor the other takes place, and which is distinguished as the point of neutral axis. Therefore, by the law of fracture thus established, and proper data of tenacity and compression given, as in the preceding table, we are enabled to form metal beams of strongest section with the least possible material. Thus, in cast iron, the resistance to compression is nearly as 0} to 1 of tenacity, consequently a beam of cast iron, to be of strongest section, must be of the following form, and a parabola in the direction of its length, the quantity of material in the bottom flange bing about 6 ½| times that of the upper. But such is not the case with beams of timber; for although the tenacity of timber be on an average twice that of its resistance to compression, its flexibility is so great, that any considerable length of beam, where columns cannot be situated to its support, requires to be strengthened or trussed by iron rods, as in the following manner.