sm9 25

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PROBLEM (9.25) A circular bar is fixed at one and loaded as illustrated in Fig. P9.23. Find the
required diameter d of the shaft.
Given: P = 0, R = 50
π
N, T = 120
π
N
⋅
m, L = 2 m, all
σ
= 120 MPa, all
τ
= 70 MPa
SOLUTION
50 N
π
D
E
120TNm
π
=
⋅
2 m
33
32(2 50 ) 3200
xdd
π
σπ
×
==
33
16(120 ) 1920
tdd
π
τπ
−
==−
3
16Td
3
32
x
M
d
σ
π
=
D
τ
π =
Equation (9.1):
62
33 3
1600 1600 1920 2
10 ( ) ( )
dd d
−
×= + +
120
or
63
1
120 10 (1600 2499.2798), 0.0325
dm
d
×= + =
Equation (9.2):
mdd 0329.0,2798.24991070
36 ==×
Similarly,
From Example 9.4:
4
3
dVQ V
I
bA
τ ==−
E
0
x
σ
=
22
4 50 267
34
dd
π
π
−
=−
3
1920 d
d=
τ
τ =+
Equation (9.1):
62
32
1920 276
120 10 0 ( )
dd
×=+− − 3
1(1920 267 )
d
d
=+
Solve by trial & error : .0252.0 md
=
Equation (9.2):
62
3
1
70 10 (0) ( 1920 267 )
d
d
×= +− − 2
Solve by trial & error : .0302.0 md
=
Thus, 25.2
all
dm=m
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