Advanced Methods of Structural Analysis

11.4 Resolving Equations 387

The second equation shows that if introducedconstraint 1 has angular displacement
Z 1 then corresponding deformation (angle of rotation) at the section 2 will be same,
i.e.,e 2 D 1 Z 1. If introduced constraint2 has angular displacementZ 2 then corre-
sponding deformation at the section 2 willbe zero, because introduced constraint 1.
Note that deformation matrixBand static matrixAare connected as follows:
BDAT. This is a general rule, so for calculation of matrixBwe can apply two
approaches.

11.4.3 Physical Equations and Stiffness Matrix

in Local Coordinates

These equations present the relationships between unknown internal forcesSand
deformationseof the elements. The required relationships is

ESDkQEe; (11.3)

whereSE.m1/D
S 1 S 2 Sm

̆T

is a vector of unknown internal forces;Ee.m1/D

e 1 e 2 em

̆T

is a vector of deformation;kQis a stiffness matrix of the system.

In general formkQis diagonal matrix

kQD

2

6

6

4

k 1 0  0

0k 2  0



00 km

3

7

7

5 (11.4)

The diagonal entrykiis a stiffness matrix ofi-th finite element of a structure. The
each diagonal entrykiis called theinternal stiffness matrixor stiffness matrix in
local coordinates for specified memberi;matrix(11.4) in whole is internal stiffness
matrix or stiffness matrix in local coordinates for all structure.
For truss element (bar with hinged at the ends), a deformation is

eD

Sl

EA

;soSD

EA

l

e:

Thus internal stiffness matrix for truss element contains only one entry and pre-
sented as

kD

EA

l

Œ1 : (11.5)

This expression allows to determine the axial forceSif axial deformation of element
eD 1. The symbolŒ1means a matrix with the sole entry equals 1.
Let us form the stiffness matrix for truss shown in Fig.11.21; this figure con-
tains the numeration of the members. Assume that for all membersEAis constant.