Civil Engineering Reference
In-Depth Information
These become the following:
1
b
≅∴≅
≅∴=
()
=
m
m
br
2
r
1
1
1
1
2
1
2
2
m
br r b r
m
2
1
1
2
2
1
∴=
b
b
m
2
mm
m
brr br
≅
=
r
3
3
1
3
2 3
3
2
After the multiple roots have been passed, the rest of the terms have the
regular solution relationship and will appear as follows:
1
≅∴=
b
b
b
b
m
m
n
n
r
r
n
n
n
−
1
n
−
1
If the second term was ½ the square of the term in the previous cycle, then
the solution would appear as follows, assuming
r
2
= r
3
and considering
only the dominant terms in each expression:
b
=++=∴≅
=
r
mmmm
r
r
r br
m
1
1
2
3
1
1
1
mm mm mm mm
mmm
mm
brr
+
r r
+
r r
=
r r
+
r
r brr
∴≅
2
2
1
2
1
3
2
3
1
2
1
2
2 12
b rrr rrr brr
=
mmm mmm
= ∴≅
mm
2
3
1 23 122
3 12
These become the following:
1
≅∴≅
()
m
br r b
m
1
1
1
1
1
= ∴=
b
b
m
mm
m
br r
≅
2
2
b r
r
2
=
r
2
1
2
1 2
2
3
2
1
1
2
= ∴=
b
b
m
mmm m
2
2
3
br
≅
r r
r
=
r
3
1
2
12
2
3
1
Similar to the previous case where
r
1
= r
2
, after the multiple roots have
been passed, the rest of the terms have the regular solution relationship
and will appear as follows:
1
≅∴=
b
b
b
b
m
m
n
n
r
r
n
n
n
−
1
n
−
1
