Question

1. A 10-in circular saw runs idle at 7200 rev/min. The saw blade is 14 gauge (0.0747 in) steely = 0.282 lbf in and y = 0.292) and is used on a 3/4-in arbor. The thickness of the blade is uniform a. Find the maximum radial stress in the blade *12* *8* b. Find the maximum tangential stress in the blade *5* c. Find maximum shear stress in the blade, use MSS theory

Answer 1

Since the circular saw blade is rotating idle, it is equivalent to a rotating ring. Given, poisson's ratio, V = 0, 292 density of the ring material, s = 0,282 lbf /in. 3 angular velocity of the ring, w: 26 (7200) rad's 60 W75 3,9822 vad/s. internal radius of the ring, ris ( 311 in= 0.375 in. outer radius of the ring, 10 = 1/10) in. = 5 in. The tangen tial stress in rotating ring is given by Of = $w? ( 34*) (****+ **** - (**)2) where r is the boint from the center of the ring where stress is to be determined.

is maximum at 8= ria 0-375 in. + Coemar = $w? (***)( *** *** vero ***) = sco(st) (*** ***- (1988) **) = (037) (753-982)* (31 292) (0-360°+ 2(0)2 386 1+30-292) 3+ (0-292) (0-375) $) = (170.9046) (50-06049) 8555.56 76 psi tmax = 855 5. 5676 psi Now, Radial stress is given by G = fw( 23) (ri* ***-) to be maximum,

as ( ) ( - 20) = 0 .9 ;: (r Y Yo i ra160,375in.) (Sin.) = 1.3693 in. 2 max = sw² / ritro2-8,2 ro2 - Viro - Ply So, (6) max = $w(9") (uitzol - *** - 1070) = $w (270) (x*+*o* - 24270) = $w?( 3+ v) (46-70) - (0-282) (52-3623° ( 2191292) (5- 01375) ? 386 3 3655,757 psi. (6) max = 3655.757 psi .

a) Maximum radial stress in the blade, (Or max = 3655.757 psi b) maximum tangential Stress in the blade (1) max = 8555.5676 Psi c) Since, ot and so are both positive ;, Tmax = ( Pumax [: @t)max> (6) mar] = (8555-5676) psi 2 4277-7838 PSL