6
Answers to the exercises of chapter 6
Exercises
15
16
Answers to the exercises of chapter 6
6.4
Exercises
17
6.6
The distributed load
q
is neglected and
A
=
1
4
π
(
D
2

d
2
).
(a) Integrating once leads to:
EA
du
dx
=
c
1
=

P
Note the negative sign in front of P (compressive force)!
σ
=

P
A
=

P
1
4
π
(
D
2

d
2
)
(b)
u
(
x
) =

P
EA
x
+
c
2
u
(
‘
) = 0
→
c
2
=
P‘
EA
u
(
x
) =
P
(
‘

x
)
1
4
πE
(
D
2

d
2
)
Supplementary, the strain can be determined:
ε
=
du
dx
=

P
1
4
πE
(
D
2

d
2
)
Note that the strain is negative!
6.7
(a) Everywhere in the muscle tendon complex the internal force
is
F
.
(b) Between A and B:
σ
1
=
F/A
1
. Between B and C:
σ
2
=
F/A
2
.
(c) Forces and stresses remain unchanged.
(d)
u
B
=
F‘
1
E
1
A
1
u
C
=
F‘
1
E
1
A
1
+
F‘
2
E
2
A
2
7
Answers to the exercises of chapter 7
Exercises
18
Exercises
19
7.5
L
=

α
2
‘
2
(
~e
x
~e
y
+
~e
y
~e
x
)
8
Answers to the exercises of chapter 8
Exercises
8.1
σ
xy
=

2
axy
+ 2
a
8.2
(a)
σ
= 10
~e
x
~e
x
+ 20
~e
y
~e
y
+ 5(
~e
x
~e
y
+
~e
y
~e
x
)
(b)
~s
=
1
2
√
2(15
~e
x
+ 25
~e
y
)
(c)
~s
n
=
10
√
2 (
~e
x
+
~e
y
)
~s
t
=
5
2
√
2(

~e
x
+
~e
y
)
8.3
(a)
σ
a
=
10
0
0

20
;
σ
b
=
0
5
5
0
(b)
~s
a
=
10 cos(
α
)
~e
x

20 sin(
α
)
~e
y
~s
b
=
5 sin(
α
)
~e
x
+ 5 cos(
α
)
~e
y
(c)

~s
a

=
10
q
cos
2
(
α
) + 4 sin
2
(
α
)

~s
b

=
5
20
Exercises
21
9
Answers to the exercises of chapter 9
Exercises
9.1
v
∼
(
x
∼
) =
a
+
b
(
y

at
)
a
0
9.2
a
∼
(
x
∼
, t
) =
1
(1 +
αt
)
2
bcx

αay

αbz
0

αcx
+
acy
+
bcz
9.3
t
= 3
/
2 [s]
9.4
~a
=

ω
2
x~e
x

ω
2
y~e
y
9.5
~a
=
c
2
x~e
x
+
c
2
y~e
y
9.6
x
∼
P
=
1
2
‘
+
1
6
‘
sin
(
2
π
t
T
)
1
3
‘
0
22
10
Answers to the exercises of chapter 10
Exercises
10.1
x
∼
0
Q
=

1
9
6
10.2
~a
0
·
~a
=

~a
0

~a

cos(
φ
)
~a
0
·
F
·
~a
0
=
p
~a
0
·
~a
0
q
(
~a
0
·
F
T
)
·
(
F
·
~a
0
) cos(
φ
)
cos(
φ
) =
~a
0
·
F
·
~a
0
q
(
~a
0
·
~a
0
)(
~a
0
·
F
T
·
F
·
~a
0
)
10.3
~x
Q
= 11
~e
x
+ 6
~e
y
+ 3
~e
z
10.4
J
= 2
10.5
‘
= 3 e
0
.
02
t
[cm]
23
24
Answers to the exercises of chapter 10
10.6
˙
F
=
1
2
˙
F
·
F
T
+
F
·
˙
F
T
=
1
2
˙
F
·
F

1
·
F
·
F
T
+
F
·
F
T
·
F

T
·
˙
F
T
=
1
2
L
·
B
+
B
·
L
T
11
Answers to the exercises of chapter 11
Exercises
11.1
ρ
A
V
A
=
ρ
B
V
B
11.2
ρ
=
1500
1 + 0
.
02
t
expressed in [kg/m
3
]
25
12
Answers to the exercises of chapter 12
Exercises
12.1
ε
v
= 1
/
8
12.2
F
v
=

65
24
G‘
2
12.3
E
=
5
h K
v
6
π δ R
2
12.4
ν
= 0
.
2
12.5
(a)
ε
yy
= 0
.
1
(b)
ε
xx
=

0
.
067
(c)
σ
yy
= 0
.
8 [MPa]
12.6
ε
=
1
2
αR
L
(
~e
x
~e
z
+
~e
z
~e
x
)
σ
=

E
2(1 +
ν
)
αR
L
(
~e
x
~e
z
+
~e
z
~e
x
)
σ
M
=
E
2(1 +
ν
)
αR
L
√
3
26
13
Answers to the exercises of chapter 13
Exercises
13.1
The surfaces with a normal in
y
direction are stress free. So at
these surfaces
σ
yz
=
σ
zy
= 0. This contradicts the assumptions
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