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Answer:
L
−
1
s
4
=
3
6
(a)
e
4
()
=
L
−
1
=
1
1
a
3
(b)
e
4
()
=
(
sin
a
−
a
)
(
)
s
2
s
2
−
a
2
L
−
1
=
1
1
a
3
(c)
e
4
()
=
(
a
−
sin
a
)
(
)
s
2
s
2
+
a
2
1
1
2
a
3
L
−
1
(d)
e
4
()
=
=
(
sinh
a
−
sin
a
)
s
4
−
a
4
1
1
4
a
4
L
−
1
(e)
e
4
()
=
=
(
cosh
a
sin
a
−
sinh
a
cos
a
)
s
4
+
a
4
4.20 Find the initial parameters for a transfer matrix solution of a beam that is simply
supported on both ends.
Answer:
0
b
LM
b
/(
M
b
/
w
=
M
0
=
0
,
θ
=
w
/
L
+
6
EI
)
,
0
=−
L
0
0
=
(
+
/).
4.21 Suppose a beam has a variable moment of inertia
I
Place the governing
differential equations in first order form. Integrate these equations to find an expres-
sion for the state of variables at position
x
. That is, find the transfer matrix for this
beam.
I
0
1
x
Hint:
A
is not constant.
b
,
V
b
, and
M
b
for a transfer matrix of a beam element
with a parabolically distributed applied loading.
4.23 The Runge-Kutta method is a numerical integration technique for the solution of
initial value problems. For the system of differential equations
z
1
=
0
0
4.22 Find the loading functions
w
b
,
θ
f
1
(
x, z
1
,z
2
)
z
2
=
f
2
(
x, z
1
,z
2
)
with the initial values
z
1
(
x
0
)
=
z
1
,
0
and
z
2
(
x
0
)
=
z
2
,
0
, one Runge-Kutta approximation
for moving one step (0 to
h
)is
1
6
(
z
1
,
1
=
z
1
,
0
+
k
1
+
2
k
2
+
2
k
3
+
k
4
)
1
6
(
=
+
+
+
3
+
4
)
z
2
,
1
z
2
,
0
2
2
1
2
with
k
1
=
(
)
1
=
(
)
hf
1
x
0
,z
1
,
0
,z
2
,
0
hf
2
x
0
,z
1
,
0
,z
2
,
0
hf
1
x
0
hf
2
x
0
h
2
,z
1
,
0
k
1
2
+
h
2
,z
1
,
0
k
1
2
+
1
2
1
2
=
+
+
2
=
+
+
k
2
,z
2
,
0
,z
2
,
0
hf
1
x
0
hf
2
x
0
h
2
,z
1
,
0
k
2
2
+
h
2
,z
1
,
0
k
2
2
+
2
2
2
2
=
+
+
3
=
+
+
k
3
,z
2
,
0
,z
2
,
0
=
(
+
+
+
)
4
=
(
+
+
+
)
k
4
hf
1
x
0
h, z
1
,
0
k
3
,z
2
,
0
hf
2
x
0
h, z
1
,
0
k
3
,z
2
,
0
3
3
As an example, integrate the beam equations
w
w
0
0
10 0
000
1
EI
000 0
001 0
−
θ
θ
0
d
dx
=
+
V
M
V
M
−
p
z
(
x
)
0
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