Civil Engineering Reference
In-Depth Information
EI
x
dx
EI
PS MS
PS M
∫
2
S
=
3
y
∆
q
=
+
iz
iz
3
iy
2
=
+
S
1
iy
iz
2
iy
Written in matrix form this is the flexibility matrix:
∆
iz
P
M
SS
SS
f F
=
3
2
iz
q
iy
iy
2
1
[]
=
[[]
d
i
ii
i
We can solve this matrix equation for the forces P
iz
and M
iy
by the cofactor
method. Substituting the determinant of the flexibility matrix,
D = S
3
S
1
−
S
2
, to simplify the equation. This is the stiffness form of the equation.
P
M
∆
q
=
S S
SS
−
1
iz
1
2
iz
D
−
iy
2
3
iy
[]
=
[[]
FK
d
i
ii
i
We can use the transmission matrix Equation 4.6 to find the forces at
the
j
-end of the member, where the values of
x
are cause minus effect or
x
=
x
i
−
x
j
= 0−
L
= −
L
.
10
1
10
1
=
[]
=
T
−
x
L
=−
[][]
K
T K
ji
ii
−
−−
10
1
S S
SSD
−
S S
SSLSLSD
−
−
1
1
=
1
2
1
2
K
=
ji
L
−
−
2
3
2
1
2
3
P
M
∆
θ
S S
SSLSLS
−
−
1
D
jz
iz
1
2
=
−
jy
iy
2
1
2
3
d
[]
FK
=
j
ji
The carry-over factor (COF) used in the moment distribution method can
be found by observing the ratio of the moment at the
j-
end to the moment
at the
i-
end.
