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8.2 Solve the problem
d
2
u
dx
2
−
u
=
0
,
u
(
0
)
=
0
,u
(
1
)
=
1
Answer:
Exact solution gives
u
x
=
1
/
3
=
0
.
29
,u
x
=
2
/
3
=
0
.
61
8.3 Solve the problem
d
2
u
dx
2
−
0
,u
(
u
=
0
,
u
(
0
)
=
1
)
=
1
Answer:
Exact solution gives
u
x
=
1
/
3
=
0
.
22
,u
x
=
2
/
3
=
0
.
46
,u
x
=
1
=
0
.
76
8.4 Use a Taylor series to derive the difference expressions
1
12
h
(
u
0
=
u
−
8
u
−
+
8
u
1
−
u
2
)
−
2
1
and
1
12
h
(
−
u
0
=
3
u
−
10
u
0
+
18
u
1
−
6
u
2
+
u
3
)
−
1
8.5 Use the multiple position difference method to solve the differential equation
u
−
2
u
−
0
,u
(
3
u
=
3
x
+
1
,
subject to
u
(
0
)
=
0
)
=
1
1
1
6
e
3
x
2
e
−
x
Answer:
You can compare your result with the exact solution
u
=
−
−
1
3
8.6 The governing equation for the extension of a straight uniform bar is
u
+
x
+
p
x
/
EA
=
0
.
The central difference quotient about point
i
leads to
p
xi
h
Show that this relationship resembles that obtained using a structural analysis of a
straight bar formed of two elements with nodes at
i
(
EA
/
h
)(
−
u
i
−
1
+
2
u
i
−
u
i
+
1
)
=
−
1
,i,
and
i
+
1
.
Hint:
For
h
=
,
the global stiffness relation for a two-element, three-node bar
would be
=
u
i
−
1
u
i
u
i
+
1
p
x,i
−
1
p
x,i
p
x,i
+
1
1
−
1
EA
−
11
+
1
−
1
−
1
1
Note that the finite difference relation is the same as the central equation of the
global structural equations.
8.7 Suppose a fixed-fixed beam of length
L
has a moment of inertia
I
(
x
)
=
2
I
0
[1
−
2
(
x
/
L
)(
1
−
x
/
L
)
]
m
2
,
The applied loading of magnitude
p
0
is uniformly distributed. Let
EI
0
=
30 000 kN
·
4 using a finite difference
solution. Verify your answer by using a variationally based finite difference solution.
L
=
12 m
,p
0
=
10 kN/m. Find the deflection at
x
=
L
/
Answer:
w(
L
/
4
)
≈
0
.
01 m
8.8 Consider a beam of variable cross-section with the applied loading
p
z
=
p
0
(
1
+
ξ)
,
2
where
ξ
=
x
/
L
. The moment of inertia varies as
I
(
x
)
=
I
0
(
9
ξ
−
6
ξ
+
1
).
Find the
deflection at
x
4 using (a) a simple finite difference mesh and (b) an improved
finite difference discretization
=
L
/
w(
/
)
≈
.
9
aL
2
,a
=
p
0
L
2
/(
)
Answer:
L
4
0
6
EI
0
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