 | American School (Lansing, Ill.) - Strength of materials - 1907 - 80 pages
...X 12,000 = 31,500 pounds; Ps= 6 X 0.7854 X (-j-)1 X 7,500 = 19,880 pounds ; • Pc = 6 X 4- X -|- X 15,000 = 33,750 pounds. Since P, is the least of these...safe tension which the joint can bear ? Here nl = 3, nt = 12, and «3 = 6; hence, as in the preceding example, Pt = 31,500; and P0 = 33,750 pounds; but... | |
 | American School (Lansing, Ill.) - Architecture - 1907 - 424 pages
...pounds; Pg= 6 X 0.7854 X (-|-)' X 7,500 = 19,880 pounds ; PC= ° XTXTX 15,000 = 33,750 Pound9Since PB is the least of these three values, the strength of...is the safe tension which the joint can bear ? Here «., = 3, n, = 12, and n2 = 6; hence, as in the preceding example, Pt = 31,500; and Pc = 33,750 pounds;... | |
 | Frederick Eugene Turneaure - Civil engineering - 1909 - 472 pages
...-|- X 12,000 = 31,500 pounds; P,= 6 X 3.7854 X (-J-)' X 7,500 = 19,880 pounds ; Pc = 6 X \ X -J- X 15,000 = 33,750 pounds. Since P, is the least of these...the preceding example are joined by means of a butt pint (two cover-plates), and 12 rivets are used, being spaced as before. What is the safe tension which... | |
 | American School (Lansing, Ill.) - Civil engineering - 1909 - 476 pages
...-g- X 12,000 = 31,500 pounds; PB-^= 6 X 0.7854 X (~|-)2 X 7,500 = 19,880 pounds ; Pe= 6 X -|- X -|- X 15,000 = 33,750 pounds. Since P, is the least of these...is the safe tension which the joint can bear ? Here nt = 3, n2 = 12, and ns — 6; hence, as in the preceding example, P, = 31,500; and Pc = 33,750 pounds;... | |
 | Civil engineering - 1909 - 472 pages
...X 12,000 = 31,500 pounds; P s = 6 X 3.7854 X (-j-) 1 X 7,500 = 19,880 pounds ; P c== 6 X \ x -|- X 15,000 = 33,750 pounds. Since P. is the least of these...is the safe tension which the joint can bear ? Here n t — 3, », = 12, and « 2 = G; hence, as in the preced ing example, . P t = 31,500; and P c = 33,750... | |
 | 1912 - 514 pages
...-|' X 12,000 = 31,500 pounds; P,= 6 X ).7854 X (-j-)1 X 7,500 = 19,880 pounds ; P0= 6 X J- X -|- X 15,000 = 33,750 pounds. Since P. is the least of these...are used, being spaced as before. What is the safe tensio:. which the joint can bear ? Here nt = 3, nt = 12, and n, = 6; hence, as in the preceding example,... | |
 | Edward Rose Maurer - Strength of materials - 1917 - 144 pages
...X ~ X 12,000 = 31,500 pounds; Ps= 6 X 3.7854 X (^-)a X 7,500 = 19,880 pounds ; po= 6 X -|- X -|- X 15,000 = 33,750 pounds. Since P. is the least of these...is the safe tension which the joint can bear ? Here nt = 3, n2 = 12, and n2 — 6; hence, as in the preceding example, Pt = 31,500; and Pc = 33,750 pounds;... | |
 | Edward Rose Maurer - Strength of materials - 1919 - 144 pages
...-|-) X -|- X 12,000 = 31,500 pounds; Pg= 6 X 0.7854 X (~|-)" X 7,500 = 19,880 pounds ; pc=6X-^-X-|-X 15,000 = 33,750 pounds. Since P, is the least of these...is the safe tension which the joint can bear ? Here nt = 3, n2 = 12, and n2 = 6; hence, as in the preceding example, Pt = 31,500; and P0 = 33,750 pounds;... | |
 | Ernest L. Wallace - Machine design - 1919 - 186 pages
...33,750 pounds Since PB is the least of these three values, it determines the strength of the joint, viz, 19,880 pounds. 2. Suppose that the plates described...preceding example are joined by means of a butt joint (two cover plates), and 12 rivets are used, being spaced as before. What is the safe tension which the joint... | |
 | Ernest L. Wallace - Machine design - 1919 - 186 pages
...the plates described in the preceding example are joined by means of a butt joint (two cover plates), and 12 rivets are used, being spaced as before. What is the safe tension which the joint can bear? Solution. Here «, = 3, n2 = 12, and n3 = 6; hence, as in the preceding example, Pt = 31,500; and P0... | |
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X 15,000 = 33,750 pounds. Since P. is the least of these three values, the strength of the joint depends on the shearing value of its rivets, and it equals 19,880 pounds. 2. Suppose that the plates described in the preceding example are joined by means... 
