Shaft ABC comprises a brass section AB and a steel section BC with different diameters shown in the diagram below. Alever is located at B with a force F applied at its end. What is the required diameter of the steel section in order tosupport 70% of the total applied torque to the shaft given the following parameters:d₁=90mmGs = 75MPaGb===40MPaGive your answer in mm.T2mB0.45mACdpds A2MB0.45Steelbrass701 Torque supponed by steelTS = 0.7TT Total:. Tb = 0·35+TorqueOAC = 0 BC + OABOAC = 0· OBC = OABTs LsGsJs"077-(0.45)Tb LbGbJb75×10 6 x πTds²0.431532275X106X Tids²320315 x 32GS75MPaGb = 40MP:db =• 90mmds = ?0.37 + ( 2 )T40x106x 7d2b=0-61175x106x πds 210.083240×106x πdb²321175 X 10.6 x πids 20-6x3240 x 106 x 71 x0.09219.2400106 XA CO09)10.08 x 40×106 XπT(0.09)² = 19.2 (75×106) x 77052ds = 47.62mmX65.47mm

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Answered: Shaft ABC comprises a brass section AB…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Question

Shaft ABC comprises a brass section AB and a steel section BC with different diameters shown in the diagram below. A
lever is located at B with a force F applied at its end. What is the required diameter of the steel section in order to
support 70% of the total applied torque to the shaft given the following parameters:
d₁
=
90mm
Gs = 75MPa
Gb
===
40MPa
Give your answer in mm.
T
2m
B
0.45m
A
C
dp
ds

A
2M
B
0.45
Steel
brass
701 Torque supponed by steel
TS = 0.7TT Total
:. Tb = 0·35+
Torque
OAC = 0 BC + OAB
OAC = 0
· OBC = OAB
Ts Ls
GsJs
"
077-(0.45)
Tb Lb
GbJb
75×10 6 x πTds²
0.4315
32
2
75X106X Tids²
32
0315 x 32
GS
75MPa
Gb = 40MP
:
db =
• 90mm
ds = ?
0.37 + ( 2 )
T
40x106x 7d2b
=
0-6
11
75x106x πds 2
10.08
32
40×106x πdb²
32
11
75 X 10.6 x πids 2
0-6x32
40 x 106 x 71 x0.092
19.2
400106 XA CO09)
10.08 x 40×106 XπT(0.09)² = 19.2 (75×106) x 77052
ds = 47.62mm
X
65.47mm

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