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In transfer matrix form
4
24
EI
3
6
EI
2
2
EI
−
−
−
−
w
1
w
0
3
6
EI
V
2
2
EI
EI
V
01
=
+
p
0
00
1
0
M
M
2
2
00
1
4.24 Find the deflection, slope, moment, and shear along a uniform beam fixed on both
ends.
U
i
z
a
+
z
i
, with
U
i
Answer:
z
b
=
taken from Eq. (4.8c) and
w
0
=
θ
0
=
0
,
12
L
3
o
b
6
L
2
0
b
2
0
b
6
L
2
0
b
V
0
=−
EI
(
w
+
θ
)
,
and
M
0
=
EI
(
L
θ
+
w
)
4.25 Use the Laplace transform to derive the transfer matrix for a bar in extension.
4.26 The governing differential equations for a bar in extension on an elastic foundation
(modulus
k
x
)are
du
EA
,
d
2
u
dx
2
/
dx
=
N
/
EA, dN
/
dx
=
k
x
u
or
du
/
dx
=
N
/
/
=
k
x
u
EA
. Do not include the effects of applied loading. Derive the element transfer
matrix, using
(a) An exponential series expansion
(b) The Cayley-Hamilton theorem
(c) The Laplace transform
Use this transfer matrix to derive the element stiffness matrix.
/
Answer:
cosh
β
sinh
β/(
EA
β)
U
i
2
=
β
=
k
x
/
EA
EA
β
sinh
β
cosh
β
cosh
β
−
1
EA
β
k
i
=
sinh
β
−
1
cosh
β
Flexibility Matrices
4.27 Show that the flexibility matrix for a beam element that is hinged at the left end (
x
=
a
)
and guided at the right end (
x
=
b
) is given by
w
b
θ
a
2
V
b
M
a
6
EI
2
−
3
=
−
3
6
4.28 Derive the complete (not reduced) element stiffness matrix of a beam beginning with
any of the flexibility matrices discussed in this chapter.
4.29 For an extension bar on an elastic foundation (modulus
k
x
) with the left end (
x
a
)
fixed, find the nonsingular stiffness matrix and then find the flexibility matrix for an
element of length
=
.
Hint:
Remove a dependent DOF from the stiffness matrix of Problem 4.16.
Answer:
Reduced Stiffness Matrix
Flexibility Matrix
β
β
cosh
l
sinh
l
N
b
=
EA
β
l
u
u
=
l
N
b
β
β
β
sinh
EA
cosh
2
u
=
u
b
−
u
a
=
u
b
β
=
k
x
/
EA
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