Notes and Examples in Mechanics; With an Appendix on the Graphical Statics of Mechanism

Notes and Examples in Mechanics; With an Appendix on the Graphical Statics of Mechanism

By (author) 

List price: US$14.14

Currently unavailable

Add to wishlist

AbeBooks may have this title (opens in new window).

Try AbeBooks

Description

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. 1893 edition. Excerpt: ...effect of the component Tis such that if the rate of retardation remained constant for one second, at the close of that second the velocity in the path would be 700--37.06 = 662.94 ft. per sec. 71. Example 19. Steam Working Expansively and Raising a HW-G. Weight.--In Prob. 4 of p. 61, M. of E., supposing the boiler-gauge to read 80 lbs. per sq. in. (above one atmosphere) and the total length of stroke, K sn--OJV, to be 16 inches, with cutoff at one third stroke (so that s, = i of 16 in.), the diameter of piston L being 10 inches; how great a weight Flo-87-G can be raised if the (circular) pis ton is to come to rest at the end of the stroke, having started from rest at the beginning of the stroke? Required also the time occupied from O to B, and the position of the piston when its velocity is a maximum. From p. 62 we have the equation.S, l + log.() = + Gsn, ... (2) now to be solved for G. sn =-ft.; s, = $ ft., and hence the ratio = 3. The area of piston = -' = if. x (5)3 = 78.57 sq. in..-. Air-pressure above piston, = A, = constant = 78.57 X (15 lbs. per sq. in.) = 1178 lbs.; while the steam-pressure under piston while it is passing from O to B, = Sl, = 78.57 X 95 = 7464.15 lbs. Noting that log, = common log X 2.302, we have from eq. (2) (using the foot, pound, and second) 7464 X 1 + 2.302 X.47712 = 1178 X f + G X f Solving, G = 4044 lbs. (so that M = 4044 x.031 = 125.36). The acceleration from O to B is constant and = pl = (#, -A-G)-h M= 2241-=-125.36 = 17.88 ft. per sec. per sec.; and eq. (2), p. 54, M. of E. sl = ipj,"; hence Above B, the steam-pressure S diminishes, and when at some point m it has become = A-4-G, i.e., to 5222 lbs., the resultant or accelerating force, S--(A--G), is zero; above this point m that force is negative, i.e., ...show more

Product details

  • Paperback | 46 pages
  • 189 x 246 x 3mm | 100g
  • Rarebooksclub.com
  • Miami Fl, United States
  • English
  • black & white illustrations
  • 1236654684
  • 9781236654687