4. The figure below shows a countershaft with helical gear (B), bevel gear (D), and two supporting bearings (A and C). All shoulder fillets (at points where diameter changes) have a radius of 5 mm. Note that the shaft is designed so that only bearing A takes thrust (There is a resultant force in the x-direction). F= 0.3675Fy Fx =0.2625Fy (1) Forces act at 500-mm dia. с B 550- 400- D = 1.37 kN 400 450 F = 5.33 KN 120 dia. Keyway 80 dia. F = 1.37 kN (K 1.6 for bend and torsion; 1.0 for axial load all at the keyway. Use Cs = 1 with these values.) Forces act at 375-mm dia. (2) The shaft is made of hardened steel, with ultimate strength = 1069 MPa and yield strength = MPa. All surfaces are finished by grinding, and a 99% reliability is desired. a) Find F on the 500 mm diameter bearing. (Hint: Use equilibrium of torques) F, y b) Draw VMNT diagrams for the shaft in the xy and xz planes. c) At point B, calculate the equivalent stresses. 0 d) Estimate the safety factor for infinite life if ea = constant σ em $896

Elements Of Electromagnetics
7th Edition
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Author:Sadiku, Matthew N. O.
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4. The figure below shows a countershaft with helical gear (B), bevel gear (D), and two supporting
bearings (A and C). All shoulder fillets (at points where diameter changes) have a radius of 5 mm.
Note that the shaft is designed so that only bearing A takes thrust (There is a resultant force in the
x-direction).
F= 0.3675Fy Fx =0.2625Fy
(1)
Forces act at 500-mm dia.
с
B
550-
400-
D
= 1.37 kN
400
450
F = 5.33 KN
120 dia.
Keyway
80 dia.
F = 1.37 kN
(K 1.6 for bend and torsion; 1.0 for axial
load all at the keyway. Use Cs = 1 with these values.)
Forces act at 375-mm dia.
(2)
The shaft is made of hardened steel, with ultimate strength = 1069 MPa and yield strength =
MPa. All surfaces are finished by grinding, and a 99% reliability is desired.
a) Find F on the 500 mm diameter bearing. (Hint: Use equilibrium of torques)
F,
y
b) Draw VMNT diagrams for the shaft in the xy and xz planes.
c) At point B, calculate the equivalent stresses.
0
d) Estimate the safety factor for infinite life if
ea
= constant
σ
em
$896
Transcribed Image Text:4. The figure below shows a countershaft with helical gear (B), bevel gear (D), and two supporting bearings (A and C). All shoulder fillets (at points where diameter changes) have a radius of 5 mm. Note that the shaft is designed so that only bearing A takes thrust (There is a resultant force in the x-direction). F= 0.3675Fy Fx =0.2625Fy (1) Forces act at 500-mm dia. с B 550- 400- D = 1.37 kN 400 450 F = 5.33 KN 120 dia. Keyway 80 dia. F = 1.37 kN (K 1.6 for bend and torsion; 1.0 for axial load all at the keyway. Use Cs = 1 with these values.) Forces act at 375-mm dia. (2) The shaft is made of hardened steel, with ultimate strength = 1069 MPa and yield strength = MPa. All surfaces are finished by grinding, and a 99% reliability is desired. a) Find F on the 500 mm diameter bearing. (Hint: Use equilibrium of torques) F, y b) Draw VMNT diagrams for the shaft in the xy and xz planes. c) At point B, calculate the equivalent stresses. 0 d) Estimate the safety factor for infinite life if ea = constant σ em $896
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