Advanced Methods of Structural Analysis

396 11 Matrix Stiffness Method

2.Computation of shear may be performed on the basis of theMPdiagram. Reac-
tions of supports can be calculated on the basis of the shear diagram. Reaction of
intermediate support isR 1 DQright 1 Q 1 left.
Example 11.4.Design diagram of a nonuniform continuous beam is presented in
Fig.11.24a. The angle of rotation of the clamped support is' 0 D'D0:01rad
and the vertical displacement of the support 2 is 2 DD0:04m. Construct the
bending moment diagram.

Solution.Degree of kinematical indeterminacy equals two. Both joints at supports

BandChave angular possible displacements. Now we need to present the effect of

the settlements of supports in form of moments at the jointsBandC. The fixed end

moments are

MABD

4 EI

lAB

'D

4 EI 0

6

0:01D0:00667EI 0

MBCDMCBD

6 EI

lBC^2

D

6  2 EI 0

42

0:04D0:03EI 0

MBAD

2 EI

lAB

'D

2 EI 0

6

0:01D0:00333EI 0

MCDD

3 EI

lCD^2

D

3  2 EI 0

42

0:04D0:015EI 0 :

Corresponding bending moment diagramM^0 in the primary system of the dis-

placement method and the joint-load diagram are shown in Fig.11.24b, c.

The positive possible angular displacements of the joints and corresponding

possible external forces are presented on theZ-Pdiagram (Fig.11.24d). TheS-e

diagram and positive bending moments diagram are shown in Fig.11.24e. The vec-

tor of external joint loads is

PEDEI 0

0:02667 0:015

̆T

:

Vector of fixed end moments in the first state is

ES 1 DEI 0 0:00667 0:00333 0:030:03 0:015 ̆T

Static matrix Considering theZ-PandS-ediagrams we get

P 1 DS 2 CS 3

P 2 DS 4 CS 5

so the static matrix becomes

A.25/D



01100

00011