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Chapter 4
41
For a torsion bar,
k
T
=
T/
=
Fl/
, and so
=
Fl/k
T
.
For a cantilever,
k
l
=
F/
,
=
F/k
l
.
For
the assembly,
k
=
F/y
, or,
y
=
F/k
=
l
+
Thus
2
Tl
FF
l
F
y
kk
k
Solving for
k
2
2
1
.
1
lT
kA
l
kl
k
n
s
______________________________________________________________________________
42
For a torsion bar,
k
T
=
T/
=
Fl/
, and so
=
Fl/k
T
.
For each cantilever,
k
l
=
F/
l
,
l
=
F/k
l
, and
,
L
=
F/k
L
.
For the assembly,
k
=
F/y
, or,
y
=
F/k
=
l
+
l
+
L
.
Thus
2
l
y
k
k
L
Solving for
k
2
2
1
.
11
LlT
lL
TL
TlL
kkk
l
kkl
k k
k
n
s
______________________________________________________________________________
43
(a
) For a torsion bar,
k =T/
=GJ/l
.
Two springs in parallel, with
J =
d
i
4
/32,
and
d
1
=
d
1
=
d
,
44
12
1
2
4
32
.(
1
)
32
JG JG
d
d
kG
x
lx
x lx
Gd
Ans
xlx
Deflection equation,
2
1
2
1
results in
(2)
Tl x
Tx
JG
JG
T
x
From statics,
T
1
+
T
2
=
T
= 1500. Substitute Eq. (2)
Chapter 4  Rev B, Page 1/81
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2
1500
1500
.
(3)
lx
x
TT
T
A
n
s
xl
Substitute into Eq. (2) resulting in
1
1500
.
(4)
TA
n
l
s
(
b
) From Eq. (1),
46
3
11
0.5 11.5 10
28.2 10
lbf in/rad
.
32
5
10 5
kA
n
s
From Eq. (4),
1
10 5
1500
750 lbf in
.
10
T
Ans
From Eq. (3),
2
5
1500
750 lbf in
.
10
T
Ans
From either section,
3
3
3
16 1500
16
30.6 10
psi
30.6 kpsi
.
0.5
i
i
T
Ans
d
______________________________________________________________________________
44
Deflection to be the same as Prob. 43 where
T
1
= 750 lbf
in,
l
1
=
l
/ 2 = 5 in, and
d
1
= 0.5
in
1
=
2
=
12
3
44
4
4
6
750 5
60 10
(1)
0.5
32
32
32
dd
dG
G
Or,
34
15 10
(2)
Td
10 10
(3)
Equal stress,
1
2
33
3
3
1
2
16
16
(4)
T
T
d
d
Divide Eq. (4) by the first two equations of Eq.(1) results in
21
1.5
(5)
Statics,
T
1
+
T
2
= 1500
(6)
Substitute in Eqs. (2) and (3), with Eq. (5) gives
4
3
15 10
10 10
1.5
1500
Solving for
d
1
and substituting it back into Eq. (5) gives
d
1
= 0.388 8 in,
d
2
= 0.583 2 in
Ans.
Chapter 4  Rev B, Page 2/81
From Eqs. (2) and (3),
T
1
= 15(10
3
)(0.388 8)
4
= 343 lbf
in
Ans.
T
2
= 10(10
3
)(0.583 2)
4
= 1 157 lbf
in
Ans.
Deflection of
T
is
11
1
46
1
343 4
0.053 18 rad
/ 32 0.388 8 11.5 10
Tl
JG
Spring constant is
3
1
1500
28.2 10
lbf in
.
0.053 18
T
k
Ans
The stress in
d
1
is
3
1
1
3
3
1
16 343
16
29.7 10
psi
29.7 kpsi
.
0.388 8
T
Ans
d
The stress in
d
1
is
3
2
2
3
3
2
16 1157
16
29.7 10
psi
29.7 kpsi
.
0.583 2
T
Ans
d
______________________________________________________________________________
45
(
a
) Let the radii of the straight sections be
r
1
=
d
1
/2 and
r
2
=
d
2
/2. Let the angle of the
taper be
where tan
= (
r
2
r
1
)/2. Thus, the radius in the taper as a function of
x
is
r
=
r
1
+
x
tan
, and the area is
A
=
(
r
1
+
x
tan
)
2
. The deflection of the tapered portion
is
2
1
00
1
0
1
21
12
1
tan
tan
tan
1
tan
tan
tan
tan
tan
tan
tan
4
.
l
ll
FF
d
x
F
dx
AE
E
E r
x
rx
Er
r l
E
r r
rr
l
F
l
E
E
rrE
Fl
Ans
ddE
2
1
(
b
) For section 1,
4
1
22
6
1
4
4(1000)(2)
3.40(10 ) in
.
(0.5 )(30)(10 )
Fl
Fl
Ans
AE
d E
For the tapered section,
4
6
4
4
1000(2)
2.26(10 ) in
.
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This note was uploaded on 11/20/2011 for the course MCE 325 taught by Professor Ahm during the Spring '11 term at American Dubai.
 Spring '11
 Ahm

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