Question

A jack similar to the one shown in the Figure uses a M24 single square coarse thread power screw. The jack is used to compress a spring to half of its free length. The spring at its free length is hosted between the movable jaw of the jack and a fixed support. The following data is known about the spring free length is 100 mm, wire diameter is 5.4 mm, outside diameter is 60 mm, and the spring ends are square and grounded. The static friction coefficients are estimated to be 0.15 for the screw and 0.12 for the collar. The spring is made out of steel that has a shear modulus of rigidity G = 79 GPa. If a force F which is perpendicular to the lever is used to rotate the nut. Estimate the force Frequired at the end of the 300 mm lever to compress the spring to half of its free length. Note: all dimensions are in millimeters,

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Answer 1

Ane) for Spring let arrume force exerled by spring on dellection is pthon for free length of Spring (lo) dos (i+n)d + yta ,' i - no of active coil as clearance Uw coul n= o for free end = 0.254 1 for lingle square end y= deflection -2 for both square end - somm d = wire dia (mm) 100 = (i+2) 5.4 + 1.25 (50) i = 4.94 = 5 coil Deflection of spring ý = 8P 33 Gdt D= mean coil diameter = Outside dia - wire dig = GO - 5.4 = 54.6 mm G = 79 GPa = 79 8103 Nimm2 50 = 8xP X 54.63X5 79 x103 x 5.44 p=515. BGN = 516 N

NEXU Curdol a thread of So then for lifting load 1- lead of screw UN = friction force due to thread " dm meon dia of thread = dy piten Tendm - de = major dia for figure F - UNCOSK -N Sindro Ptunsind-NCosa =0 Mcosa tu Sina N = F/cosd-sing' CP PCM Cold+ sond) Coed- ju sind . from tig Tronsmit Torque TR = F dm tonal L I Tudm TR = P. dan (W20844 je Sinan) TR = p de operation for collar friction pdm / Midmth .. - 2 T dm-42 / 2 de a collar dia for friction Mc = co-effic of friction due to conar To -) + Me P dc

( Single Stant) for M24 thread de = 24 P = L = 3 mm dm = 24 - 3 = 22.5 mm de: 100 mm (from fig) M=0.15 Mc=0.12 R = 516 22.5 (0.15% TT X 225+3 i 0:12X516X100 ( 22.5x7. - 0.1583) 2 1124.28 + 3096 4220.28 N-mm TR = FX l de lever length . 4220.28 = F x 300 F = 14.06 N.