sm9 84

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PROBLEM (9.84) A 6-m-long steel shaft of allowable strength all
σ
= 120 MPa carries a torque of
T = 400 N⋅m and its own weight. Calculate the required shaft diameter d in accordance with the von
Mises theory of failure.
Assumptions: Use
ρ
= 7.86 Mg/m3 as the mass per unit volume for steel (see Table B.4).
The shaft is supported by frictionless bearings that act as simple supports at its ends.
SOLUTION
22
7.86(9.81)( 4) 60.6wddkNm π ==
2236
33
32( 8) 32(60.6 )(36)10 2.78 10
8
xwL d
ddd
σππ
×
== =
3
33
16(400) 2.04 10
dd
τπ
×
=− =−
Equation (9.19),
222
3
x
all
σ
τσ +=
66
2262
3
2.78 10 2.04 10
( ) 3( ) (120 10 )
dd
×−×
+=×
or 68
26
7.7284 10 12.4848 144 10
dd
×+=×
Solving by trial & error: 34.2 .dmm
=
Use a
35 mm− diameter shaft.
C
3
16Td
τ
π =
3
32
x
M
d
σ
π
=
T
B
A
w=60.6d2
C
T=400 N⋅m 3
m
3 m