Advanced Methods of Structural Analysis

408 11 Matrix Stiffness Method

The equivalent moment at joint 1 (Fig.11.27e) isM D 15:364:1667 D
11:1993kNm (clockwise); the equivalent force for cross bar (Fig.11.27f) is
PD 5 kN. CorrespondingJ-Ldiagram is shown in Fig.11.27g.
The joint-load andZ-Pdiagrams allows us to construct the vector of external
equivalent joint loads
PEDb11:1993 5:0cT:

The entries of this vector present the free terms, which are written on the right side
of canonical equations of the Displacement method (compare with Example 8.2).
The vector of fixed-end moments (vector of internal forces of the first state) at
sections 1–4 on the basis of theMP^0 andS-ediagrams becomes

ES 1 Db4:1667 4:1667 15:36 0cT:

This vector corresponds to the last term in expressionMDM 1 Z 1 CCMP^0 for
final bending moment.

Static matrixThis matrix is constructed on the basis of theZ-PandS-ediagrams.
Figure11.27h shows free body diagram for joint 1 subjected to three unknown in-
ternal forces in vicinity of joint 1 (bending moments) and momentP 1. Equilibrium
condition for joint 1 isP 1 DS 2 CS 3 CS 4.
It is not difficult to show that

P 2 D

S 1

h 1



S 2

h 1

C

S 4

h 2

Indeed, the positive momentsS 1 andS 2 at the ends of the memberA-1 may be
equilibrated by two forcesS 1 =h 1 CS 2 =h 1 (Fig.11.27i). Then this force should be
transmitted on the cross bar; similar procedure should be done for member 1-B.
Equilibrium equation

P
XD 0 for cross bar leads to the above expression forP 2 ,
so the static matrix becomes

AD



0111

1= h 1 1= h 1 0 1h 2

For given parametersh 1 andh 2 we get

AD



0111

0:2 0:2 0 0:333

Stiffness matrices of the elements in the local coordinates The finite elements
are shown in Fig. refch11:fig11.27j; stiffness matrix for each member in local coor-
dinates is

k 1 D

EI 1

l 1



42

24

D

EI

5



42

24

;