 | American School (Lansing, Ill.) - Strength of materials - 1907 - 80 pages
...= 2,500 tf - 0.6 X 264', or 2,500 d* = 400,000 + 0.6 x 264" = 441,817.6. Hence (P = 176.73, or d = 13.3 inches. 2. What size of cast-iron column is needed...safety of 10, the length of the column being 14 feet ? 100,000 X 10 == 1,000,000 pounds. The length is 14 feet or 168 inches; hence the formula oecomes... | |
 | American School (Lansing, Ill.) - Architecture - 1907 - 424 pages
...2,500 f/2 - 0.6 X 264°, or 2,500 (P = 400,000 + 0.6 X 264* = 441,817.6. Hence d2 = 176.73, or d = 13.3 inches. 2. What size of cast-iron column is needed to sustain a load of 100.000 pounds with a factor of safety of 10, the length of the column being 14 feet ? We shall suppose... | |
 | Frederick Eugene Turneaure - Civil engineering - 1909 - 472 pages
...= 2,500 d2 - 0.6 X 264', or 2,500 rf2 = 400,000 + 0.6 X 264' = 441,817.6. Hence d* = 176.73, or d = 13.3 inches. 2. What size of cast-iron column is needed...(equation 10'). The breaking load for the column would be 115 100,000 X 10 = 1,000,000 pounds. The length is 14 feet or 168 inches; hence the formula oecomes... | |
 | American School (Lansing, Ill.) - Civil engineering - 1909 - 476 pages
...176.73, or <l = 13.3 inches. 2. \Vhat sixc of cast-iron column is needed to sustain a load of 100.000 pounds with a factor of safety of 10, the length of...(equation 10'). The breaking load for the column would be 100,000 X 10 = 1,000,000 pounds. The length is 14 feet or 108 inches; hence the formula Becomes... | |
 | Civil engineering - 1909 - 472 pages
...2,500 d 1 - 0.6 X 264', or 2,500 J 1 = 400,000 + 0.6 x 264 ' = 441,817.6. Hence d 1 = 176.73, or d = 13.3 inches. 2. What size of cast-iron column is needed to sustain a load of 100.000 pounds with a factor of safety of 10, the length of the column being 14 feet ? We shall suppose... | |
 | 1912 - 514 pages
...2,500 <Z, - 0.6 X 264,, or 2,500 d» = 400,000 + 0.6 X 264' = 441,817.6. Hence d1 = 176.73, or d = 13.3 inches. 2. What size of cast-iron column is needed...(equation 10'). The breaking load for the column would be 100,000 X 10 = 1,000,000 pounds. The length is 14 feet or 168 inches; hence the formula oecomea... | |
 | Edward Rose Maurer - Strength of materials - 1917 - 144 pages
...= 2,500 d2 - 0.6 X 264", or 2,500 d? = 400,000 + 0.6 X 264' = 441,817.6. Hence d2 = 176.73, or d = 13.3 inches. 2. What size of cast-iron column is needed...safety of 10, the length of the column being 14 feet ? 100,000 X 10 = 1,000,000 pounds. The length is 14 feet or 168 inches; hence the formula oecomes 1,000,000... | |
 | Edward Rose Maurer - Strength of materials - 1919 - 144 pages
...2,500 d2 - 0.6 X 264", or • 2,500 & = 400,000 + 0.6 X 264" = 441,817.6. Hence d2 = 176.73, or d = 13.3 inches. 2. What size of cast-iron column is needed...make the crosssection circular, and shall compute by Eankine's formula modified for cast-iron columns (equation 10'). The breaking load for the column would... | |
 | Civil engineering - 1920 - 440 pages
...= 2,500 d* - 0.6 X 2642, or 2,500 d* = 400,000 + 0.6 X 264' = 441,817.6. Hence d2 = 176.73, or d = 13.3 inches. 2. What size of cast-iron column is needed...(equation 10'). The breaking load for the column would be 100,000 X 10 = 1,000,000 pounds. The length is 14 feet or 168 inches; hence the formula oecomes... | |
 | Alfred Peter Poorman - Strength of materials - 1925 - 338 pages
...Problem 4. A hollow rectangular concrete pier is 2 by 3 feet outside dimensions. If it is to carry a load of 100,000 pounds with a factor of safety of 10, what must be the thickness of the walls? Ans. 5 inches. 21. Effect of Stresses above the Elastic Limit.... | |
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X 264' = 441,817.6. Hence d2 = 176.73, or d = 13.3 inches. 2. What size of cast-iron column is needed to sustain a load of 100,000 pounds with a factor of safety of 10, the length of the column being 14 feet ? We shall suppose that it has been decided... 
