HW 8

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PROBLEM 3.70
The hollow steel shaft shown (G = 77.2GPa. r*, = 50MPa)
rotates at 240 rpm. Determine (a) the maximum power that can
be transmitted, (D) the corresponding angle of twist of the shaft.
25 nrm
SOLUTION
' 2'
".=!d,=3omm
c,t2r=
,r =
=
I
j-d,
= 12.5 mm
tki-,1)
= lxtolo -
1.234 x 106mma
t* = 50x 106 Pa
*Tc -_t*J
-,,,=j
, =-;
Angular
(q)
speed.
(12.s)41
= 1.234 x 10-6ma
_(50xlo6)0.234x[0-6i
#=2056.7N'm
.f = 240 fpfil = 4 rev/sec = 4 Hz
P =2n.fT=2tt(4)(2056.7)= 51.7x103W
ry.
P=51.7 kW <
(b) A{rgleoftwist.
a=L=
' GJ
(20-56'7X5)
(77 .2
'-
x r oe)(r .234 x t0-6)
=0.1078rad
i
PROPRIETARY IIIATERIAL, O 2012 The McGraw-Hill Companies, Inc.
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PROBLEM 3.72
The design of a machine element calls for a 40-mur-outer-diameter
shaft to transmit 45 kW. (a) If the speecl of rotation is 720 rpm,
detennine the maximum shearing stress in shaft a. (b) lf the speed of
rotation can be increased 50% to 1080 rpm, detennine the largest inner
diameter of shaft 6 for which the maximum shearing stress will be the
same in each shaft.
$OLUTION
(.t)
f =720 =l2Hz
"60
P=45kW=45x103W
(1))
.r
P
=-2rf
c,
=
= 20 mrn = 0.020 m
-rl
2
r
=
L - 27',.- (2X596.81) = 47.194x t0opa
J /Tc' z(0.020)r
45x103
=596.83N.rn
2n(12)
I
lo8o
/=
"60
2r(18)
4
I
{
2Tc.
J
C'I
dz = 2cr = 30.4 mm
=397.8eN.m
Tc.
t*:
"l
{
= ISHz
r=gr-lo1
-,
r*"" = 47.5 MPa
o('!
2Tc'.
-'l)
'
"fiT
= 0.020a
=
15.20
-
(2X397'89X0'0?0)
/r(47 .49.1 x
x l0-3m =
l0')
15.20 mm
PROPRIETARY \IATERIAL, O 2012'I'he
= 5i.333 x t0-e
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or distrihuled iti any ftrrnr or by any nreans, rvithout the prior rvritteu perntission of the publisher. or used heygn6 the limited
tlistribution to teachers and educators pernritted by McGrarv-tlill lirr tlreir individual coilrse preparation. A student using tlris manual is usi.g it
repr<iduced,
rvithout permission.
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