Problem 3.9
3.9 Knowing that each of the shafts AB, BC, and CD consist of solid circular rods,
determine (~ the shaft in which the maximum shearing stress occurs. (b) the magnitude
of that stress.
T
ShA.-tt AS ':
c. ,. tel
800III' in,
~
'(,
:IT
-=
J
7rc.S
7f (P.&.l):!o
T,: - 800 + 2&fOO
t. :'-tJ
j",
0.4 ;".
-= ('Z.)(Soo) -= 7QS7'
".
ShA.ff B~:
':
T£.
~
~
Boo lb.
-=
psi
.
1'00
':'
11:.-~
o.s i,...
~
(2.\( I~o() ')
'l'
ShA.ft CD:
T:-800
+ 2400 +1000
(OS)3
g ILfq fS"
':'
R~OO Rh.in
-::
(~)(~'oo)
'rift-..,,&.':
If
':
TI {O_b )3
7"
=
~=icl :: 0.' in.
g pSI
(a.) Sh4ft
Hole: :
c, = toll :- t(O.4b)
Sh4f+
AB:
T:-
T
II
\J
= ~
L ,t.- -J
Shc.J't BC:
T =-- goo + 2.'/00 ::
J"-: ¥(cz"- c,") = f(o.sY-
$hc..fr
J
':' Tc,,-::.
CD:
~
(I boo)(
J
T
0.$)
D_OftSb{.'
= -800
',"00
::-
~J
':'
'WI.
0-~o
I ()
(.
i","
-
.
O.o37'lfQ
CL1<1
0,
0g
.
ps, ( KA~S rt)
'1)
C:t= t~1.. = 0.5' ;r).
o.:lO") -: O.oCfS"661iM"
=
83(,3
+ 2'-100 ... 1000
(~bOO)(O.")
;WI.
'I
(8o0)«().~)-
)1. IW"I
rsi
:'
~(C1.'t- C. 'f) :- ~(0.' y - O.~OCf)::
t:"'-,&> =
o. ~O
1ft"
"
( -c, ) ::. ~\O.'t
- 0.2.0 )
'I
C~
=0.031'"
,.., - Tcz
~
gOD .P~:.~
C?,. ': ~cI~ = 0.4
800 Ib . in.
,..
.48
(6) 8.. IS ksi
BC
3.10 Knowing that a O.40-in.-diameter hole has been drilled through each of the
sbaftsAB, BC, and CD, determine (0) the shaft in which the maximum shearing stress
occurs, (b) the magnitude of that stress.
Problem 3.10
1:
)
.
AnslUe.r::
,y
(Jecrjes!
::;
77SCf
A.,s..,~1"S~ La) sh~
~'OO
0.;(0'1°'
11.iVl
c~= ~Jt = 0-";"&-
j,. If
fS/
AB
(6) '8.~~
ks;
~
3.15 The solid shaft shown is formed of a brass for which the allowable shearing
stress 55 MPa. Neglecting the effect of stress concentrations, determine smallest
diameters dABand doc for which the allowable shearing stress is not exceeded..
Problem 3.15
~'t,..,,* ;:: Ss- 1'1P4.
1:'"","", -=
T~J
# 55"
:
ID" p~
c-= O/?..
~T'
,;(T
"nc'1
..,.'
sh#t Aa ~
S
T,.,s ':
\ '200 -'too
= 'V (2.)(800)
1iCSS>t'Ic::F- -=
~, .00'>"10-1-M
W'I~VI"tIo)V,-, dAB -= ~<:..
Sh ~.f't Be. ~ T Be ::
c- --~ ~(Z-)(LfbC:»
(ss'~,cJ)
Y-oO
=
N.V\'I
-s ~
,,- "7 '>I/o
-
)G. br
-: 800 ~~
-=
~
~/.o
4-2. b
YO
MW'\ ~
~ M
'M\tI~'W\I.IW"\&I3c.=:<e.
-::: .33..3"",W\
~
Problem 3.29
3.29
(a)
For a given allowable
stress, determine
the ratio T/w of the maximum
allowable torque T and the weight per unit length w for the hollow shaft shown. (b)
Denoting by (T/w)o the value of this ratio computed for a solid shaft ofthe same radius
c2, express the ratio T/w for the hollow shaft in terms of (T/w)o and C/C2°
w = we-t"1-~t p"eJ..~d- j)~+h
J f' ~ ~ s..pec.;;"G ~e.:~-t
iIV 1Il I-,,~j
we."t~+,)
L ':: J~~~'"
-W
IN
T,.,' :: li~
C2.
(I)~
Wi'"
CI
~
~
T
= 1k G;4t~
~'
-:. r-r:-u
I
':: f~A
C I .,. ~~
::
Cz.
1f (C2-"'-+ C, ~ )(4~'-C.t.)
~
C2,.
(C2.7.+c,2..)'l'.,Ji
2.f>~Cz.
D
(7
+0 If"' soJi.J
Iw ) h
(T /w')o
-
a.
1
= jJ'fflT' ~'Cz, - C,
)
1:'..1
(hoiJ;,J s ~Jt)
5h&tft
t:"
+ --L
c~~
I
( w )0
-= c~~
'2,
(Solid
( ~ \ = (~ )0 ( 1+ ~~)
.......
s~c.#)
......
Problem 3.41
3.41 A coder F, used to record in digital form the rotation of shaft A, is connected
to the shaft by means of the gear train shown, which consists of four gears and three
solid steel shafts each of diameter d. Two of the gears have a radius r and the other
two a radius fIT.If the rotation of the coder F is prevented, determine in terms of T.
/, G. J, and n the angle through which end A rotates.
TAB = T~
~o:'
-ra. TAB::'
f'e
TAs -= TA
V)
n
TI!iF- fL.,..Ico-:: ~I>
- --
fo
n
rf) ':
't'C
(f)
'f'EF
~
!A.
qJ 11
tD
:
=- ~FJ
&J
6F:,
&J"
~
(()
-
'j't:. -
to
A)
- rC.r()
't'8 -
+,()
'i't)
('8 'f"C
'rCf).
=
TA.R
GJ
-
L.L
y\3GJ'""
Li
<:Pc.
Y)
t').3
O'
~
-
J
Y\z&T
<PE
rf1 = Te.D;~J) 't'CD
T",.
n --
:
-T..
V'J'Z..
':
GU
+
JAL
'(\&f
TAP
GJ
( L.,3 + Ln )
( nilI + .Lna. )
<PAa
': ~~JAa :: ~
CPA~ ~
qI.4B :
~;
...L+..1. '" \ )
(,oJ" ( Y)\ ~"
....
Problem 3.49
3.49 The solid cylindrical rod BC is attached to the rigid lever AB and to the fixed
support at C. The vertical force P applied at A causes a small displacement 11at point
A. Show that the corresponding maximum shearing stress in the rod is
Gd
.=-/1
2La.
where d is the diameter of the rod and G its modulus of rigidity.
A
A
Q..
B
Leve." AB 4-u>I""5 ~V"O ~ '"
dh~je
c.f>
to f,°$ iti 0,"" A'13 ~S
-=shONII\ IV! the
Vel/'+iGJ J"SfjAc:.~,..J
~D'"
N J..H'<:!.~
C(.I.1)1t'ij q...,
is
b. == a. SI"'P
1>
The M~I'\NI Ulofl .shetcf"~r\~ 5tN!.~S
--P - /"'"\/1 - G ~
e...""
- ~,
~
1=0i' smJ)
/Y
L"",~
-
-
",...,c
J
C:;U"Gs,,'A
GJ A
2L 0..
+,,~
,
(...
V'
~
= aV'c.S", a
i~ rod
8 C.
- G~
2.L
L
A~.A
Q..
'0$
11
. I.l
-- Q..:r
Z.L etV"C.SI '"
~.
Gt
-!!SI
Problem 3.69
L=
4D
l->--.
C1.. := --kdo::- 7.00
Jo::
""IV)""
::-
T
Tc~
-J
)(;0
::- -Go
o. ~oo
=
-=- '7
A. G(,,(."7
S'L'
C_-
c, ~
""
t
01;. -= 100
""'"
2.3SC:.2.>i/O-3
:::- o.
JDO
"..,
mil
(~.3S";()o'/O'~)(bO><'IOq
O. ~oo
~
70C:,.g'><lo~
N-I'>-t
i-h.
'Po::
(Q.) M"...)(t'"".J"'" powe'v"
p::-
~
*(O.~()o't_O.I()O417
?f(c.:z:'1-CI'/)""
T
+'
3.69 One of two hollow drive shafts of a cruise ship is 40 m long, and its outer
and inner diameters are 400 rom and 200 rom, respectively. The shaft is made of a
steel for which Tan= 60 MPa and G =77.2 GPa. Knowing that the maximum speed
of rotation of the shaft is 160 rpm, determine (a) the maximum power that can be
transmitted by one shaft to its propeller, (b) the corresponding angle of twist of the
shaft.
Z TIt T
l1T (-z.(;(,.(.7)(70f,.Z(,x/O3)
-= 1/.~44>(/OG.
W
p=
1/.'B4Iv1W'"
- TL
(b) A~~y'e of' +1<11'.$
t
-
'1 - G-J""
(70 G.. '3b >:(03 ) ( Lf0 )
p - (77.;.
y
loQ)(?.
3,562;.< 10-3
=- /55.4'-1
></0--" "vi
cp ~
8.ctl°
~
Problem 3.76
3.76 The two solid shafts and gears shown are used to transmit 16 hp !Tom the
motor at A operating at a speed of 1260 rpm, to a machine tool at D. Knowing that each
shaft has a diameter of! in., determine the maximum shearing stress (a) in shaftAB, (b)
in shaft CD.
(ci) Sh~-P+ A9:
?=- \b hp"
~ -= ~~O -= ~I j..{z.
1;.s=
'P
I05.b
-:lllf
-= '21f(~J)
c -= ~ol
~
~ -= Te--
I:
><[03
= 800_3Z
)
'
b'/l')
=- 2T
--
( 0.5):&
Ii D o(.) ><0
I.
I
:!.
r
.
~1
~a
CD
~
.s
l
c.o=- Y'e, 1M =- '6 (800.3Z)~
1.'33387
Y{O1
2.T - ('2.)(I.'3~3~7)(IO~)
T
~
.
iV/.
J
1fc3
(Z)(800.3;Z)
11
(6) sLc!1
0.5
(/£ )U;'DO)= IOSJ>lu'?> lb.;",/se.c
TIC:!.
-
\\
to_5)~
,+.08 lbi
~
..u-.\V\
:l
::. 6.7Q
~
X 1o psi
'ko'"
b. 7Ct J<ws~.....
Problem 3.86
3.86lU1owing that the stepped shaft shown must transmit 45 kW speed of21 00 rpm,
determine the minimum radius r of the fillet if an allowable shearing stress of 50 MPa is
not to be exceeded.
60mm
f'
-=:
p
=-
~bO
'1-5)1 103
-
f
T '= ~;Ttf Fo\l"
SMA.)l-e\l"
K:.-
%
SiJ~
T'1:'C."
7.\
-:
If' =
;~
Co::"
-=
~
o. /l J
:(
tJ
W
3
4-5')( J0
'2.1\ (~5)
,.
F"j~
().6L5"'rvI
=-
(2J(~O't.'3)
~ ( o. n )(30)"::"
-= :<'0'1.'?J
N. WI
\<T Co - Z K \
"'" 1S'w\hfI
11(50)<'10&)(0.015)3
Fr~
Hz
35
=-
3.3Z
S - I Iv!WI
1:'::"...r
-
1\ C. S
~_Z.::;S
Y..d -
0. It
V'::-5. I """",,'"
Problem 3.90
3.90 In the stepped shaft shown, which has a full quarter -circular fillet, the allowable
shearing stress is 80 MPa. Knowing that D = 30 nun, detennine the largest allowable
torque that can be applied to the shaft is (0) d= 26 nun, (b) d= 24 nun.
1::::
= t(D -
r
d)
Ca.)
8D )lID'" Pa...
D
30
"d=
~,
V'-=-i{o.c:ol)"
-:: 1./54
2............
-J- '" ;2~ ::" O. Dl!l;g
F~o~
E'~,~.
SMAjjetl'
Full quarter-circular
~l., -
fillet
.
3:Z
siJe.
K ':" 1_..3~
C
:
-iJ '"
13"",10'\"O.oi3 ~
k'lc2KT
J"
- lfc 3
extends to edge of larger shaft
T-
-
1TC.31:'- li(O.OI3)~(80)(IOC)
;l K (2 )(/.gg)
-= 2.03
(b)
~ ~
~
~
/. ~S
Frow. h'~,
-\ -
3. '32
V'" tCD-of)
=- :3..."""
T~ ;lO3 N- ~ ..
N-~
L_~
-=-O.I~5
d - 2'1
c = * ch. 11"""",,'"0.012........
k'". I. 21
TrC.'!:.1:'
- 1i(O.ol~)'!.(80)(/o');( K
-
(). ")( /. 31")
-I ,5". 8 N. 'M
~"I -= 16oS:11N. M 4
Problem 3.103
3.103 A steel rod is machined to the shape shown to form a tapered solid shaft
to which torques of magnitude T = 75 kip . in. are applied. Assuming the steel to be
elastoplastic
with
t"y = 21 ksi and G
= I 1.2
x 106 psi, determine
(0) the radius
of the
elastic core in portionAB of the shaft, (b) the length of portion CD which remains
fully elastic.
(0-)
po \-{o", ABc::
In
T'(=
~~1:'Y'
kel
=- ~C!T'(
""
I. 'ZS ;n.
-= ~(1.25Y'(ZI)tLd)=btf.I.f~7>C/O!l
Jb.i\1
If
n~ !.
T= -""
( /- LL)
3 I'{
c."
3 in.
Cl>
E:.:t.
t:. =-
tT'
~~ - ~
1Y
" .8 '4 - (3GLf.l.fn>c
)(7$)C /03
10"I
P'("="D.7"1775C
:(0.71775")(./.2.5')::
-
= 0 71775
O_C;cr719> iVl.
f'(
(6)
J
1='011"'
Y iePJj"'a
T= Jc~
C«.
=
foin1
1:. C;'Ty
I.~
- ~
1.31'1'1~
} -r-
~T
I
1
I_.
I
C
1- 5f> --~"Lf'
1: =
"Lr.)
c
C\t, -=3fJ:T
11(;:'(
USin'd
I
5,
':: c;..>
pNr°-t..OV\~
- '.:l..5
=:
'......
7.5y.IO~.It.;~
~ :3 (2.)(1Sx/O I = LgILf'tLf.in.
If (,ZI )( IcY)
-tf'o
\_So - \.3 \lt~¥ =: X1.50
T
-=O-Cfi1
5
+ke
ske.+c..-h
- '7
~ -
.:::>_7Dao.