Advanced Methods of Structural Analysis

11.4 Resolving Equations 389

coordinates for uniform beam with fixed ends becomes

kff D

EI

l



42

24

(11.7a)

Let us form the stiffness matrix in local coordinates for frame shown in
Fig.11.22. Assume that bending stiffness equalsEIfor horizontal members 2
and 4, and for vertical members 3EI.

Fig. 11.22 Design diagram
of the frame 4m 7.5m

1 6m

2

3

4

According to primary system of the displacement method, we have the fixed-
fixed members 1 and 2, and fixed-pinned members 3 and 4. Stiffness matrices for
each member in local coordinates are

k 1 D

EI 1

l 1



42

24

D

3 EI

6



42

24

DEI



21

12

;

k 2 D

EI 2

l 2



42

24

D

EI

4



42

24

DEI



10:5

0:5 1

;

k 3 D

EI 3

l 3

Œ3D

3 EI

6

Œ3DEIŒ1:5 ;

k 4 D

EI 4

l 4

Œ3D

EI

7:5

Œ3DEIŒ0:4 :

Each of these stiffness matrices is presented in the form that contains general mul-
tiplierEI. The internal stiffness matrix of the frame becomes

kQDEI

2 6 6 6 6 6 6 6 6 6 6 4

21

::

:0 0 0 0

12

::

:0 0 0 0



00

::

:1 0:5

::

:0 0

00

::

:0:5 1

::

:0 0



000 0

::

:1:5

::

:0



000 0 0

::

:0:4

3 7 7 7 7 7 7 7 7 7 7 5

Internal stiffness matrix for combinedstructure (e.g., the frame with tie) can be
constructed by the similar way.