 | Peter Barlow - 1826 - 372 pages
...is given by the above rule. PROBLEM II. To find the ultimate Transverse Strength of any Rectangular Beam of Timber, fixed at one End and loaded at the other. Rule. — 1. Compute the ultimate deflection by note 3 of the last problem, and divide the deflection by... | |
 | William Galbraith - 1827 - 412 pages
...one End and loaded at the other. Hule I. — Multiply the value of S, in the preceding table of data, by the breadth and square of the depth, both in inches,...divide that product by the length, also in inches, and the quotient will be the weight in Ibs. This is approximative. Rule II. — 1. Take the ultimate... | |
 | Charles Hutton - 1831 - 662 pages
...at one End and haded at the other. Rule i. Multiply the value of a, in the preceding table of data, by the breadth and square of the depth, both in inches,...divide that product by the length, also in inches, and the quotient will be the weight in Ibs. This is approximative. Rule. ii. 1. Take the ultimate deflection... | |
 | James Hann, Isaac Dodds - 1833 - 234 pages
...timber, foced at one end, and loaded at the other. Rule 1. Multiply the value given in the table of data by the breadth and square of the depth, both in inches,...divide that product by the length, also in inches, and the quotient will be the weight in pounds. Example 1. What weight will it require to break a piece... | |
 | William Galbraith - 1834 - 454 pages
...one End and loaded at the other. Rule I .— Multiply the value of S, in the preceding table of data, by the breadth and square of the depth, both in inches,...divide that product by the length, also in inches, and the quotient will be the weight in Ibs. This is approximative. Rule II. — 1. Take the ultimate... | |
 | Andrew Bell - 1842 - 402 pages
...fixed at one end, and loaded at the other. 1 Find the continued product of the tabular value of S, the breadth and square of the depth, both in inches, and divide this product by the length in inches, and the quotient will be the weight in pounds.1 Or, W „ . ld*S... | |
 | Charles Haynes Haswell - 1844 - 298 pages
...suspended from the middle to break it. TO FIND THE TRANSVERSE STRENGTH. When a Rectangular Bar or Beam is Fixed at one End, and Loaded at the other. RULE. — Multiply the Value in the preceding table by the breadth, and square of the depth, in inches, and divide the product... | |
 | William Templeton (engineer.) - 1845 - 210 pages
...-5 = 238 the divisor ; then 700 -:- 238 = 3 1/2-94 x 5 = 7'25 inches, depth of the beam. 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... | |
 | William Templeton - 1846 - 518 pages
...and loaded at both ends ; also, when the weight is between the middle and the end ; likewise, when fixed at one end and loaded at the other. RULE. — Multiply the strength of an inch square bar, 1 foot long, (as in the table,) by the breadth, and square of the depth... | |
 | James Hann - 1848 - 352 pages
...— Multiply the value given in the table of data by four times the breadth and square of the depth in inches, and divide that product by the length, also in inches, for the weight. Ex. 1. — What weight will be necessary to break a beam of Canadian oak, the length... | |
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S, in the preceding table of data, by the breadth and square of the depth, both in inches, and divide that product by the length, also in inches, and the quotient will be the weight in Ibs. 
