 | Peter Barlow - 1826 - 372 pages
...3360 For Riga Fir, E = 3075. f Example I. — A beam of Riga fir is intended to bear a load of 665 Ibs. at its extremity, its length being 5 feet, its...breadth 4 inches, and the deflection not to exceed 5 of an inch. In this case the tabular value of E is 3075; hence, ,=6'92 ; the cube root of which is... | |
 | William Templeton (engineer.) - 1845 - 210 pages
...feet, equal the depth required in inches. Ex. A beam of ash is intended to bear a load of 700 fts. 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 = 1 1 9 x 4 x -5 = 238 the divisor ; then 700 -:- 238 = 3 1/2-94 x... | |
 | 1847 - 190 pages
...the cube root of the quotient, multiplied by the length in feet, equal the depth required in inchesEx A beam of ash is intended to bear a load of 700 Ibs-...breadth 4 inches, and the deflection not to exceed 4 of an inchTabular value ofE_= 1 19 X 4 X -5 = 238 the divisor ; then ft0-:- 238 = a V2'9* X 5 = 7-25... | |
 | Thomas Kelt - 1849 - 424 pages
...the cube root of the quotient, multipliedJg the length in feet, equal the depth required in iffs. Ex. A beam of ash is intended to bear a load of 700 Ibs....breadth 4 inches, and the deflection not to exceed 4 of an inch. Tabular value of.E = 1 19 X 4 X -5 = 238 the divisor; then 700-:- 238 = 3 Va-94 X 5 =... | |
 | Peter Barlow - 1851 - 608 pages
...the depth required in inches.16 Example i. — A beam of Riga fir is intended to bear a load of 665 ibs. at its extremity, its length being 5 feet, its...breadth 4 inches, and the deflection not to exceed J of an inch. In this case the tabular value of E is 192 ; hence, 665 — = 3-44 ; the cube root of... | |
 | Oliver Byrne - 1852 - 604 pages
...the cube root of the quotient, multiplied by the length in feet, equal the depth required in inches. A beam of ash is intended to bear a load of 700 Ibs....exceed |- an inch. Tabular value of E = 119 X 4 X -5 = 238, the divisor ; then 700 ~ 238 = ,s/2T94 x 5 = 7•25 inches, depth of the beam. RULE. — To... | |
 | Oliver Byrne - 1852 - 600 pages
...the cube root of the quotient, multiplied by the length in feet, equal the depth required in inches. A beam of ash is intended to bear a load of 700 Ibs....deflection not to exceed \ an inch. Tabular value ofJE = 119 x 4 x -5 = 238, the divisor ; then 700 -r- 238 = ^2-94 x 5 = 7-25 inches, depth of the beam,... | |
 | William Templeton (engineer.) - 1855 - 326 pages
...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 Ibs....deflection not to exceed } an inch. Tabular value of E=119 x 4 x -5=238, the divisor; then 700rf-238=3 -v/2-94 x 5=7-25 inches, depth of the beam. 6zam,... | |
 | Charles Haslett - 1855 - 482 pages
...length in feet, equal the depth required in inches.. ., EXAMPLE. A beam of ash isJuteaded tO-i«ar a load of 700 Ibs. at its extremity, its .length being....breadth 4 inches, and the deflection not to exceed ^ of an inch. . . .' .1 i . , Tabular value of E = 119 x 4 x -5 = 238 the divisor; then 700 -f- 238... | |
 | Charles W. Hackley - 1856 - 530 pages
...deflection, both in inches; and the cube root of the quotient, multiplied by the length in feet, equal the depth required in inches. EXAMPLE. A beam of ash...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 -7- 238 = \/2'94 x... | |
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Divide the weight to be supported, in Ibs., 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. 
