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 17, and the fourth root of the product is the diameter.

Ex. The distance between the supports of a beam of Riga fir is 16 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 & of an inch; what must be the depth of the beam, supposing the breadth 8 inches?

16 × 8000

90 × 32 × 75

= 15175÷8 = 31897

=

12:35 in. 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 1.-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 51⁄2 tons is to be fixed, the centre 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?

2240 × 5.5

2

=

6160 lbs. to carry at each end.

And 18 feet-7=11, double each, or 14 and 22, then 14 × 22

18

= 17 feet, or 204 inches, effective length of beam.

1341 x 4 x 10 × 60

204

Tabular value of S, red pine: - 15776 lbs. the absolute 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, its breadth 4 inches, and the deflection not to exceed of an inch.

Tabular value of E

=

119 x 4 x 5=238 the divisor; then 700÷2383 √2·94 × 57.25 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 42, and projecting 6 feet from the wall; what is the greatest weight it will support at the extremity of its length?

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 Table (p. 87), 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 6 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 being about 6 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.

And these applications of principle not only tend to diminish deflection, but the required purpose is also more effectively attained, and that by lighter pieces of timber.

To ascertain the absolute strength of a cast iron beam of the preceding form, or that of strongest

section.

Rule.-Multiply the sectional area of the bottom flange in inches by the depth of the beam in inches, and divide the product by the distance between the supports also in inches; and 514 times the quotient equal the absolute strength of the beam in cwts.

The strongest form in which any given quantity of matter can be disposed is that of a hollow cylinder; and it has been demonstrated that the maximum of strength is obtained in cast iron, when the thickness of the annulus or ring amounts to 4th of the cylinder's external diameter; the relative strength of a solid to that of a hollow cylinder being as the diameters of their sections.

A Table showing the Weight or Pressure a Beam of Cast Iron, 1 inch in breadth, will sustain without destroying its elastic force, when it is supported at each end, and loaded in the middle of its length, and also the deflection in the middle which that weight will produce. By Mr. Hodgkinson, Manchester.

10

11

12

13

14

12100 098 10600 128 9500 162 85002

12826 117 11495.15 10285182 15264 107 13680 135 12240 17

16100 125 14400154 18600 115 16700143|

Note.-This Table shows the greatest weight that ever ought to be laid upon a beam for permanent load, and if there be any liability to jerks, &c., ample allowance must be made; also the weight of the beam itself must be included.