5. 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). F₂ = 0.3675Fy F₁ = 0.2625F (1) Forces act at 500-mm dia. D B 550 400 Fy = 1.37 kN D 400 450 F=5.33 kN 120 dia. E 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 = 896 MPa. All surfaces are finished by grinding, and a 99% reliability is desired. a) Find Fy on the 500 mm diameter bearing. (Hint: Use equilibrium of torques) b) Draw VMNT diagrams for the shaft in the xy and xz planes.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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Please answer part a and b
5. 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.
C
B
550
400
= 1.37 kN
400
450-
F = 5.33 KN
120 dia.
E
Keyway
80 dia.
Fx = 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 = 896
MPa. All surfaces are finished by grinding, and a 99% reliability is desired.
a) Find Fy on the 500 mm diameter bearing. (Hint: Use equilibrium of torques)
b) Draw VMNT diagrams for the shaft in the xy and xz planes.
c) At point B, calculate the equivalent stresses.
σea
σem
d) Estimate the safety factor for infinite life if = constant
Transcribed Image Text:5. 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. C B 550 400 = 1.37 kN 400 450- F = 5.33 KN 120 dia. E Keyway 80 dia. Fx = 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 = 896 MPa. All surfaces are finished by grinding, and a 99% reliability is desired. a) Find Fy on the 500 mm diameter bearing. (Hint: Use equilibrium of torques) b) Draw VMNT diagrams for the shaft in the xy and xz planes. c) At point B, calculate the equivalent stresses. σea σem d) Estimate the safety factor for infinite life if = constant
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