Question

The built-up shaft is designed to rotate at 540 rpm. The radius of the fillet weld connecting the shafts is = 7.20 mm, and the allowable shear stress for the material is Tallow 59 MPa. Part A Determine the maximum power the shaft can transmit (Figure 1) Express your answer with the appropriate units. P= 101 kW Submit Previous Answers Request Answer * Incorrect; Try Again; 5 attempts remaining Figure Provide Feedback 75 mm 60 mm

Answer 1

Figure A-15-8 Round shaft with shoulder fillet in torsion. To = Tc/), where c=d/2 and J = ndº/32. CIT? AT Did = 2 1.20 1.09 0 0.05 0.100.150.200.250.30 rld
fig for Kt
Given data : The speed of the shaft = N=540rpm The radius of fillet = r= 7:20mm The allowable strear stress Callow = 59 mpa The larger diameter of shaft = D = 75mm The smaller diameter of shafta daboum The maximum power transmission by the smaller diameter would be considered - The polar moment of inestia of shaft - 32 J = a (sommit - 32 - î ( 60 x1073m) - 32 = 1.2723 x 10-6 mt The shear centre of the shaft is, ..ca diq = 60 = 30mm ( 30x10 0.030 The ratio of the larger and the smaller diameter is 94 - = 1.25 The ratio of the radius of filled and smaller diameter is I = 7:20 = 0.12

Refer the graph between the stress concentration a factor kt and the rid) at given ratio of Old) kt = 1:35 (see in the attached graph) The expression for the allowable shean stress is - Collow = ke To Here, the maximum torque is T. subsitute the known valuer 59 M1 = 1.35 x T x 103 m ) (1.2723 X 16-6mt) 59 X10 (Mm2) = 1.35 XT X (0.03 m) (1.2723 10-6 m4) » T = 1853.474 N.m angular speed of shaft is, wa ZIN the maximum power that shaft can The and transmit Rmax = TW 7 Pmax = T (2AN) = 1853.474 (2*** 54 0 rpm) 1853.474 x (2** * 540*) Penax = 104811.405 to Watt [Pmax= 104.811 kW/ Ano - Answer

Given data : The speed of the shaft = N=540rpm The radius of fillet = r= 7:20mm The allowable strear stress Callow = 59 mpa The larger diameter of shaft = D = 75mm The smaller diameter of shafta daboum The maximum power transmission by the smaller diameter would be considered - The polar moment of inestia of shaft - 32 J = a (sommit - 32 - î ( 60 x1073m) - 32 = 1.2723 x 10-6 mt The shear centre of the shaft is, ..ca diq = 60 = 30mm ( 30x10 0.030 The ratio of the larger and the smaller diameter is 94 - = 1.25 The ratio of the radius of filled and smaller diameter is I = 7:20 = 0.12

Refer the graph between the stress concentration a factor kt and the rid) at given ratio of Old) kt = 1:35 (see in the attached graph) The expression for the allowable shean stress is - Collow = ke To Here, the maximum torque is T. subsitute the known valuer 59 M1 = 1.35 x T x 103 m ) (1.2723 X 16-6mt) 59 X10 (Mm2) = 1.35 XT X (0.03 m) (1.2723 10-6 m4) » T = 1853.474 N.m angular speed of shaft is, wa ZIN the maximum power that shaft can The and transmit Rmax = TW 7 Pmax = T (2AN) = 1853.474 (2*** 54 0 rpm) 1853.474 x (2** * 540*) Penax = 104811.405 to Watt [Pmax= 104.811 kW/ Ano - Answer