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Chapter 4
4-1
(a)
F
k 1
k 2
k 3
y
k
=
F
y ;
y
=
F
k 1 +
F
k 2 +
F
k 3
so
k
=
1
Ans.
(1
/ k 1 )
+
(1
/ k 2 )
+
(1
/ k 3 )
(b)
k 1
F
F
=
k 1 y
+
k 2 y
+
k 3 y
k 2
y
k
=
F
/
y
=
k 1 +
k 2 +
k 3 Ans.
k 3
(c)
1
k
1
k 1 +
1
k 2 +
1
k 1 +
1
k 2 +
− 1
k 2
=
k
=
k 3
k 3
k 1
k 3
4-2 For a torsion bar, k T
= T /θ = Fl /θ
, and so
θ = Fl / k T .
For a cantilever, k C = F /δ,
δ = F / k C .
For the assembly, k = F / y , y = F / k = l θ + δ
So
y =
F
k =
Fl 2
k T +
F
k C
Or
k =
1
Ans.
( l 2
/
k T )
+
(1
/
k C )
4-3 For a torsion bar, k = T /θ = GJ / l where J = π d 4
/
32 . So k = π d 4 G /
(32 l )
= Kd 4
/ l . The
springs, 1 and 2, are in parallel so
k = k 1 + k 2 = K d 4
l 1 + K d 4
l 2
= Kd 4 1
x +
1
l
−
x
And
θ =
T
k =
Kd 4 1
x
T
1
+
l
−
x
Then
T
=
k
θ =
Kd 4
x θ +
Kd 4
θ
l
−
x
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Chapter 4
71
Thus
T 1 =
Kd 4
x θ ;
T 2 =
Kd 4
θ
l
−
x
If x = l /
2 , then T 1 = T 2 .
If x < l /
2 , then T 1 > T 2
Using
τ =
16 T /π d 3
and
θ =
32 Tl /
( G π d 4 ) gives
T
= π
d 3
τ
16
and so
d 3
32 l
G
d 4 · π
τ
16 =
2 l
τ all
Gd
θ all =
π
Thus, if x < l /
2 , the allowable twist is
θ all =
2 x
τ all
Gd
Ans.
k = Kd 4 1
Since
x +
1
l
−
x
1
x +
= π
Gd 4
32
1
Ans.
l
−
x
Then the maximum torque is found to be
T max = π
d 3 x
τ all
1
x +
1
Ans.
16
l
−
x
4-4 Both legs have the same twist angle. From Prob. 4-3, for equal shear, d is linear in x . Thus,
d 1 =
0
.
2 d 2 Ans.
1
258 d 2
k = π
G
32
(
0
.
2 d 2 )
4
d 2
0
= π
G
32 l
2 l +
.
Ans.
0
.
.
8 l
θ all =
2(0
.
8 l )
τ all
Ans.
Gd 2
T max = k θ all =
0
.
198 d 2 τ all Ans.
4-5
F
A = π r 2
= π( r 1 + x tan
α)
2
r 1
d δ =
Fdx
AE =
Fdx
E
π
( r 1 +
x tan
α
) 2
x
F
π
l
dx
δ =
E
( r 1 +
x tan
α
) 2
l
0
l
dx
F
π
1
=
−
E
tan
α
( r 1 +
x tan
α
)
0
=
F
π
E
1
r 1 ( r 1 +
l tan
α
)
F
672657233.005.png
72
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Then
F
δ
= π
Er 1 ( r 1 +
l tan
α
)
k
=
l
1
=
EA 1
l
+
2 l
d 1
tan
α
Ans.
4-6
F =
( T + dT )
+ w dx − T
=
0
x
Enlarged free
body of length dx
dT
dx =− w
T
Solution is T
=− w x + c
l
dx
w dx
T | x = 0 = P + w l = c
T
=− w x + P + w l
T
dT
T
= P + w
( l − x )
w is cable’s weight
per foot
P
The infinitesmal stretch of the free body of original length dx is
d δ =
Tdx
AE
=
P
+ w
( l
−
x )
dx
AE
Integrating,
l
[ P
+ w
( l
−
x )] dx
δ =
AE
0
δ =
Pl
AE +
l 2
2 AE
w
Ans.
4-7
M = w lx − w
2 − w
l 2
x 2
2
EI dy
dx = w
2 − w
l 2
2
x − w
x 3
6 + C 1 ,
dy
dx =
0 at x =
0,
C 1 =
0
EIy = w
6 − w
l 2 x 2
4 − w
x 4
24 + C 2 ,
y =
0 at x =
0,
C 2 =
0
y =
x 2
24 EI (4 lx −
6 l 2
− x 2 ) Ans.
lx 2
lx 3
w
672657233.006.png 672657233.007.png 672657233.008.png 672657233.009.png 672657233.010.png
Chapter 4
73
4-8
M
=
M 1 =
M B
EI dy
dx =
M B x
+
C 1 ,
dy
dx =
0 at x
=
0,
C 1 =
0
EIy
=
M B x 2
2
+
C 2 ,
y
=
0 at x
=
0,
C 2 =
0
y
=
M B x 2
2 EI
Ans.
4-9
y
dx 2
dy
dx
2
ds
dy
dx
ds
=
+
dy 2
=
dx
1
+
Expand right-hand term by Binomial theorem
1
dy
dx
2 1 / 2
1
2
dy
dx
2
+
=
1
+
+ ···
Since dy
/
dx is small compared to 1, use only the first two terms,
d
λ =
dx 1
−
dx
2
dy
dx
1
2
=
+
−
dx
1
2
dy
dx
2
=
dx
dy
dx
2
1
2
l
λ =
dx Ans.
0
This contraction becomes important in a nonlinear, non-breaking extension spring.
4-10 y
=
Cx 2 (4 lx
−
x 2
−
6 l 2 )
where C
=
w
24 EI
dy
dx =
Cx (12 lx
−
4 x 2
−
12 l 2 )
=
4 Cx (3 lx
−
x 2
−
3 l 2 )
dy
dx
2
=
16 C 2 (15 l 2 x 4
−
6 lx 5
−
18 x 3 l 3
+
x 6
+
9 l 4 x 2 )
1
2
l
dy
dx
2
l
λ =
dx
=
8 C 2
(15 l 2 x 4
−
6 lx 5
−
18 x 3 l 3
+
x 6
+
9 l 4 x 2 ) dx
8 C 2 9
0
14 l 7
24 EI 2 9
0
14 l 7
8
112 EI 2
w
1
=
=
=
l 7
Ans.
ds
672657233.011.png 672657233.012.png
74
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
4-11 y
=
Cx (2 lx 2
−
x 3
−
l 3 )
where C
=
w
24 EI
dy
dx =
C (6 lx 2
−
4 x 3
−
l 3 )
dy
dx
2
=
C 2 (36 l 2 x 4
−
48 lx 5
−
12 l 4 x 2
+
16 x 6
+
8 x 3 l 3
+
l 6 )
l
dy
dx
2
l
1
2
1
λ =
dx =
2 C 2
(36 l 2 x 4
− 48 lx 5
− 12 l 4 x 2
+ 16 x 6
+ 8 x 3 l 3
+ l 6 ) dx
0
0
C 2 17
70 l 7
2 17
70 l 7
w
24 EI
EI
2
=
=
=
17
40 320
l 7
Ans.
4-12
I
=
2(5
.
56)
=
11
.
12 in 4
y max =
y 1 +
y 2 =− w
l 4
8 EI +
Fa 2
6 EI ( a
−
3 l )
Here
w =
50
/
12
=
4
.
167 lbf/in, and a
=
7(12)
=
84 in, and l
=
10(12)
=
120 in.
y 1 =−
167(120) 4
8(30)(10 6 )(11
4
.
12) =−
0
.
324 in
.
y 2 =−
600(84) 2 [3(120)
12) =−
84]
0
.
584 in
6(30)(10 6 )(11
.
So
y max =−
0
.
324
−
0
.
584
=−
0
.
908 in Ans .
M 0 =− Fa −
(
w l 2
/
2)
=−
600(84)
−
[4
.
167(120) 2
/
2]
=−
80 400 lbf
·
in
c =
4
−
1
.
18
=
2
.
82 in
σ max = −
My
I =−
(
−
80 400)(
−
2
.
82)
(10 − 3 )
11
.
12
=− 20 . 4 kpsi Ans.
σ max is at the bottom of the section.
−
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