This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1882 Excerpt: ...I---;6, we can assume, that a parallelopipedical element A B C D, Fig. 460, of the body, is acted upon by the normal forces A B. S, and---C' D. S, upon A B and CD and by the couple (A B. b'.,, 0). 8,) along A B and (YD and by the opposite couple (B-'C. Z,-ABC. Z) along B (' and A D. If the diagonal plane U forms an angle 1,0 with the axis of the body or with the direction of the strain iS',, the components of the forces S,, S, and Z upon one side of A C' are A B. S, sin. 1)), A B. S, cos. 1/, and B C'. Z sin. 1/», and consequently the total normal force upon A C is or, since the moment of (B C. Z,---A D. Z) is equal to the ino A U is in the direction of the ax and the allowable force F 'P a c 2)Q--FT_'T( W) In order to find the dimensions of the cross-section corresponding to the forces P and Q, we put F' _ P 'Z 8 1 Q T' 1/ T-7. roduciiig torsior is the greater, and, on the when the force p contrary, 9 A 1. T W is is the greater. when that in the.direction of the ax. ' '. ' at 0OZ"Zt77?.7'2-, whose dimensions are b and ii,. b.. If we know the ratio v: 7 of the dimensions, we can calculate, the dimensions themselves by means of this formula. For a pillar with a square base b: /z, and therefore we I1-at For a cylindrical pillar or shaft we have 'Z Tr 7,4 F: 7T ¢, W:-5-, and e: r, whence If the force Q in the direction of the axis is a compressive one, the formulas found above still hold good; for F1G-461-not only the direction of the force 15', (Fig. A S, B 461) is opposite, but also the forces S, and Z P can be assumed to act in the opposite direc-Z tion, when we wish to obtain the maximum-Z resultant Sm. EXAMPLE.----If a vertical wooden shaft weigh-D 39 C ing 10000 pounds is subjected to a...
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