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EXAMPLE 5.4 A solid steel shaft AB, shown in Fig. 5-11, is to be used to transmit 5 hp from the motor M to which it is attached. If the shaft rotates at w- 175 rpm and the steel has an allowable shear stress of 7allow - 100 MPa determine the required diameter of the shaft to the nearest mm. M Fig. 5-11

EXAMPLE 5.4 A solid steel shaft AB, shown in Fig. 5-11, is to be used to transmit 5 hp from the motor M to which it is attached. If the shaft rotates at w- 175 rpm and the steel has an allowable shear stress of 7allow - 100 MPa determine the required diameter of the shaft to the nearest mm. M Fig. 5-11

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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41,9 15% O 4G 10:21 KB/dtk Zoom v Leave Is in SI Units-Pearson (2017).pdf - Adobe Reader 213 (216 of 899) 100% EXAMPLE 5.4 A solid steel shaft AB,shown in Fig. 5-11, is to be used to transmit 5 hp from the motor M to which it is attached. If the shaft rotates at e = 175 rpm and the steel has an allowable shear stress of 7,llow = 100 MPa determine the required diameter of the shaft to the nearest mm. M Fig. 5-11 SOLUTION The torque on the shaft is determined from Eq. 5-10, that is, P = To. Expressing Pin watts and o in rad/s, we have (746 W P = (5 hp)(he) = 3730 W 2m rad 1 rev 175 rev I min 60 s = 18.33 rad/s min Thus, P = Tw: 3730 W/s = T18.33 rad/s) 8,00 x 9,25 in •.. Unmute Start Video Share Participants More
Transcribed Image Text:41,9 15% O 4G 10:21 KB/dtk Zoom v Leave Is in SI Units-Pearson (2017).pdf - Adobe Reader 213 (216 of 899) 100% EXAMPLE 5.4 A solid steel shaft AB,shown in Fig. 5-11, is to be used to transmit 5 hp from the motor M to which it is attached. If the shaft rotates at e = 175 rpm and the steel has an allowable shear stress of 7,llow = 100 MPa determine the required diameter of the shaft to the nearest mm. M Fig. 5-11 SOLUTION The torque on the shaft is determined from Eq. 5-10, that is, P = To. Expressing Pin watts and o in rad/s, we have (746 W P = (5 hp)(he) = 3730 W 2m rad 1 rev 175 rev I min 60 s = 18.33 rad/s min Thus, P = Tw: 3730 W/s = T18.33 rad/s) 8,00 x 9,25 in •.. Unmute Start Video Share Participants More
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