Gear
Train Application (continued)
Step 5: Calculate φA/B: Here: φA/B = T1
L / J G = TA L / J G |
Step 6: The total angle of
rotation of the shaft at A = φA = φA/B + φgear 1 φA = TA
L / J G + [TA (R2 / R1)2
L] / J G = TA
L [ 1 + (R2/R1)2 ] / J G (result) |
For the Given Data: Shearing modulus of
elasticity, G, of 10 x 106 psi.
Shaft length, L, is 100 inches. Each shaft diameter is 5 inches. T = 5,000 ft lb. R1 = 12 in, R2 = 4 in. φA = (5,000)(12)
lb in (100 in) [1 + 9 ] / [π(5)4/32 in4][107
lb/in2 ] φA = 0.0978
rad = 5.6 o (result) |
Maximum shear stress in shaft 1: τ1max = T1(d/2)/J = TA(d/2)/J = 16
TA/ (π d3 )
(result) τ1max = (16)(5,000)(12)
/ (π 53 ) = 2445 psi (result) |
Maximum shear stress in shaft 2: τ2max = T2(d/2)/J = TA(R2/R1)(d/2)/J = 16
TA(R2/R1)/ (π d3 ) (result) Thus τ2max = (R2/R1) τ1max τ2max = 3
τ1max = 7335 psi (result) |
Return to Notes on Solid Mechanics |
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