Information Technology Reference
In-Depth Information
Also,
M Ya
= M a
=
M a
(5.89c)
In matrix notation, for end a of the i th element,
p i a =
T i aa p i a
(5.90)
where
i
cos
α
sin
α
0
T i aa =
sin
α
cos
α
0
(5.91)
0
0
1
The same transformation applies also to displacements. The transformation of Eq. (5.23) is
appropriate when both ends ( a and b ) of element i are to be treated together. Then:
Global to Local
Local to Global
v i
T i v i
v i
T iT
v i
=
=
(5.92)
p i
=
T i p i
p i
=
T iT
p i
where
T i aa
0
T i
=
T i bb
0
with T i aa
T i aa .
The stiffness matrix transformation from local to global coordinates is given by [Eq. (5.28)]
given by Eq. (5.91) and T i bb =
k i jj
k i jk
k i T i
k i
T iT
=
=
(5.93)
k i kj
k i kk
3 matrices, and k i
where k i jj , k i jk , k i kj , and k i kk are 3
×
is the element stiffness matrix referred
to local coordinates. From Chapter 4, Eq. (4.80),
EA
EA
0
0
0
0
12 EI
6 EI
12 EI
6 EI
0
0
3
2
3
2
6 EI
4 EI
6 EI
2 EI
0
0
k i
2
2
=
(5.94)
EA
EA
0
0
0
0
12 EI
6 EI
12 EI
6 EI
0
0
3
2
3
2
6 EI
2 EI
6 EI
4 EI
0
0
2
2
Now that the coordinate transformations have been established for an element, the dis-
placement method analysis of a frame proceeds in the same fashion as for a truss.
EXAMPLE 5.5 Displacement Method for a Frame
The three-element frame of Fig. 5.21a is modeled and labeled as shown in Fig. 5.21b. Suppose
it is subjected to the concentrated loadings shown in Fig. 5.22. Find the response if both
legs are treated as being fixed. The element properties are E
m 2
=
200 GN
/
and
2356 cm 4 ,
32 cm 2
Elements 1 and 2: I
=
A
=
5245 cm 4 ,
66 cm 2
Elements 3: I
=
A
=
The displacement method analysis of a frame is basically the same as that used for a
truss; only the stiffness matrices differ. Fundamental to the method is the identification
Search WWH ::




Custom Search