Civil Engineering Reference
In-Depth Information
by taking two derivatives from previously used deflection and moment
differential equations.
P
EI
cr
y
′′ +
y
=
0
P
EI
cr
y
′′′ ′ +
y
′′ =
0
′′ ==
−
y
4
y
+
6
y
−
4
y
+
y
y
−+
2
y
y
i
+
2
i
+
1
i
i
−
1
i
−
2
i
+
1
i
i
−
1
y y
′′′ ′ +
0
+
Q
4
2
h
h
1
Q
h
(
)
+
(
)
0
=
y
− +− +
4
y
6
y
4
y
y
y
−
2
yyy
i
+
i
+
2
i
+
1
i
i
−
1
i
−
2
i
+
1
i
−1
4
2
h
The solution process is similar to Example 3.10 and 3.11 except that the
two operators are placed on the model at the location of the unknown
deflections. The value at point 0 of
y
0
is zero and can be eliminated from
the solutions. This becomes a homogeneous linear algebraic solution set.
1296
741
47 4
286
−
−
21 0
121
022
y
y
y
0
0
0
1
Q
L
−
−
+
36
−
=
2
4
2
L
−
−
3
252
144
36
−
2
Q
−
+
Q
2
2
2
L
L
L
y
y
y
0
0
0
1
144
252
144
−+ −
Q
2
Q
−
+
Q
=
2
2
2
2
L
L
L
72
−+
16
288
2
3
2
Q
−
2
Q
2
2
L
2
L
L
A non-trivial solution to a homogeneous linear algebraic set exists if the
determinant of the coefficient matrix is zero. Therefore, we can find
Q
by
setting the determinant of the coefficient matrix equal to zero. This will be
done using the basket weave method for a 3×3 matrix.
252
252
216
144
144
72
+− +
0
=
−
2
Q
−
2
Q
−
2
Q
Q
−+
Q
L
2
L
2
L
2
L
2
L
2
L
2
+
36
−+
144
288
−
36
252
72
−+
Q
2
Q
LL
Q
−
2
2
2
2
2
2
2
L
L
L
L
