11.6 Analysis of Continuous Beams 395Stiffness matrices for left and right spans arek 1 D3 EI 1
l 1Œ1D3 EI 1
8Œ1D3 EI 1
40Œ5 ;k 2 D3 EI 2
l 2Œ1D3 EI 1
10Œ1D3 EI 1
40Œ4 :Internal stiffness matrix for all beam in local coordinates iskQD
k 1 0
0k 2D3 EI
40
50
04Stiffness matrix for all structure in global coordinates and its inverse areKDAkAQT
D
11̆
3 EI
40
50
04
1
1
D27
40 EI!K^1 D40 EI
27Displacement of the joint 1ZEDK^1 EPD^40 EI
274:16D6:163
EIVector of internal forces of the second stateSE 2 DkAQ TZED^3 EI
40
50
04
1
1
6:163
EID
2:3111
1:8489Final internal forcesESfinDES 1 CSE 2 D
16
20:16
C
2:3111
1:8489
D
18:31
18:31The 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 conditionP
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.