 | Peter Barlow - 1845 - 526 pages
...loaded in the middle of its length. Mule. — Multiply the value of S, in the Table of Data, by four times the depth in inches, and by the area of the...will be the greatest weight the beam will bear in fts. Note 1. — If the beam be not horizontal, the distance between the supports must be the horizontal... | |
 | William Templeton (engineer.) - 1845 - 210 pages
...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... | |
 | 1847 - 190 pages
...strongest sectionRule 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 cwtsThe strongest... | |
 | Thomas Kelt - 1849 - 424 pages
...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... | |
 | Oliver Byrne - 1852 - 598 pages
...strongest section. — 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... | |
 | Charles Haslett - 1855 - 482 pages
.... .... y. 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 ewts. The strongest... | |
 | Charles Haslett - 1855 - 544 pages
...of S by four times the deptli 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 Ibs. ' i \off I.— If the lieitm be not laid horizontally, the distance... | |
 | Charles W. Hackley - 1856 - 530 pages
...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 cAvts. The strongest... | |
 | William Ezra Worthen - 1857 - 632 pages
...is : — Multiply the sectional area of the bottom flange in square inches by the depth of the beam in inches, and divide the product by the distance between the supports in feet, and 2.16 times the quotient will be the breaking weight in tons (2240 Ibs.) As has already been... | |
 | W.E. WORTHEN - 1857 - 600 pages
...222. is:—Multiply the sectional area of the bottom flange in square inches by the depth of the beam in inches, and divide the product by the distance between the supports in feet, and 2.16 times the quotient will be the breaking weight in tons (2240 Ibs.) As has already been... | |
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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. 
