11.6 Analysis of Continuous Beams 395
Stiffness matrices for left and right spans are
k 1 D
3 EI 1
l 1
Œ1D
3 EI 1
8
Œ1D
3 EI 1
40
Œ5 ;
k 2 D
3 EI 2
l 2
Œ1D
3 EI 1
10
Œ1D
3 EI 1
40
Œ4 :
Internal stiffness matrix for all beam in local coordinates is
kQD
k 1 0
0k 2
D
3 EI
40
50
04
Stiffness matrix for all structure in global coordinates and its inverse are
KDAkAQ
T
D
11
̆
3 EI
40
50
04
1
1
D
27
40 EI
!K^1 D
40 EI
27
Displacement of the joint 1
ZEDK^1 EPD^40 EI
27
4:16D
6:163
EI
Vector of internal forces of the second state
SE 2 DkAQ TZED^3 EI
40
50
04
1
1
6:163
EI
D
2:3111
1:8489
Final internal forces
ESfinDES 1 CSE 2 D
16
20:16
C
2:3111
1:8489
D
18:31
18:31
The negative sign means that momentS 2 , according toS-ediagram, is directed in
opposite direction, so external fibers right at the joint 1 are located above the neutral
line. Final bending moment diagram is shown in Fig.11.23g.
Equilibrium condition
P
M 1 D 0 for joint 1 is satisfied. The bending moment
at pointkmay be calculated considering second span as simply supported beam
subjected to forcePand moment 18.31 kNm counterclockwise.
Notes:
1.Analysis of this beam has been performed early by the displacement method
(Chap. 8 ) so the reader has an opportunity to compare analysis of the same beam
by different methods.