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makes no physicalsense to prescribe both the slope
y
and the bendingmoment
E I y
at the same point.
EXAMPLE 8.8
P
x
L
v
The uniformbeamof length
L
andbending rigidity
E I
is attached to rigid supports
at both ends. The beam carries a concentrated load
P
at its mid-span. If we utilize
symmetryandmodel only the left half of the beam, the displacement
v
canbeobtained
by solving the boundary value problem
E I
d
4
v
dx
4
=
0
x
=
0
=
x
=
L
/
2
=
dx
3
x
=
L
/
2
=−
dv
dx
dv
dx
E I
d
3
v
v
|
x
=
0
=
0
0
0
P
/
2
Use the finite difference method to determine the displacement and the bending
moment
M
E I
(
d
2
v
dx
2
) at the mid-span (the exact values are
v
PL
3
=−
/
=
/
(192
E I
)
=
/
and
M
PL
8).
Solution
By introducing the dimensionless variables
x
L
E I
PL
3
v
ξ
=
y
=
the problembecomes
d
4
y
d
=
0
4
ξ
ξ
=
0
=
ξ
=
1
/
2
=
ξ
=
1
/
2
=−
d
3
y
d
dy
d
dy
d
1
2
y
|
ξ
=
0
=
0
0
0
ξ
ξ
3
ξ
We nowproceed to writing Eqs. (8.13)taking into account the boundary condi-
tions.Referring to Table 8.1, weobtain the finite difference expressions of the bound-
ary conditions at the left end as
y
1
=
0 and
y
0
=
y
2
. Hence Eqs. (8.13a) and (8.13b)
become
y
1
=
0
(a)
−
4
y
1
+
7
y
2
−
4
y
3
+
y
4
=
0
(b)
Equation (8.13c) is
y
1
−
4
y
2
+
6
y
3
−
4
y
4
+
y
5
=
0
(c)
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