(a) The reaction forces at the two ends of the shaft A and D. (b) The bending moments at shoulder fillets B and C. (c) The nominal stresses at shoulder fillets B and C.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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Consider the loaded shaft illustrated in Figure Q4 (i). The
shaft is simply supported at A and D. The shoulder fillet
radius is 0.3 cm at B and C.
If the shaft is subjected to a force of F-800 N, determine the
following:
(a)
The reaction forces at the two ends of the shaft A and D.
(b)
The bending moments at shoulder fillets B andC.
(c)
The nominal stresses at shoulder fillets B and C.
(d)
The stress concentration factor at shoulder fillet B.
(e)
The maximum stress in the shaft taking into account stress
concentrations.
800 N
T-0.3 can
A
d, - 3 cm
d, = 5 cm
d, - 3 cm
5 cm
8 cm
8 cm
12 cm
Figure Q4 (i)
The reference parameters are shown in Figure Q4 (ii) in the
page that follows. For 0.002 srlds 0.3 and 1.01 s Dlds 6.0,
the stress concentration factor for bar bending is given as:
kn
-4.4
= 0.632 + 0.377
-0.14 – 0.363(P/a)" + 0.503(P/a)
1- 2.39(P/a) + 3.368(2la)*
-0.5
The nominal stress is:
32M
%3D
nd3
Transcribed Image Text:Consider the loaded shaft illustrated in Figure Q4 (i). The shaft is simply supported at A and D. The shoulder fillet radius is 0.3 cm at B and C. If the shaft is subjected to a force of F-800 N, determine the following: (a) The reaction forces at the two ends of the shaft A and D. (b) The bending moments at shoulder fillets B andC. (c) The nominal stresses at shoulder fillets B and C. (d) The stress concentration factor at shoulder fillet B. (e) The maximum stress in the shaft taking into account stress concentrations. 800 N T-0.3 can A d, - 3 cm d, = 5 cm d, - 3 cm 5 cm 8 cm 8 cm 12 cm Figure Q4 (i) The reference parameters are shown in Figure Q4 (ii) in the page that follows. For 0.002 srlds 0.3 and 1.01 s Dlds 6.0, the stress concentration factor for bar bending is given as: kn -4.4 = 0.632 + 0.377 -0.14 – 0.363(P/a)" + 0.503(P/a) 1- 2.39(P/a) + 3.368(2la)* -0.5 The nominal stress is: 32M %3D nd3
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