Advanced Methods of Structural Analysis

394 11 Matrix Stiffness Method

q=2kN/m

l 1 =8 m l 2 =10 m

ul 2 =6 m

P=12 kN

ul 2 =4 m

AB^1

EI

a

d 4.16 Joint load diagram

q P

M^0

1 A M 10 B

MP^0

M^0 k

k

b

16

1

20.16

c

f

Positive bending moments

S-e diagram

S 2

S 2

S 1

S 1

e

Z-P diagram

1

S 1 S 2

P 1

g

q P

k MP

M^0

K

18.31

Fig. 11.23 (a–d) Fixed load. Continuous beam and corresponding joint-load diagram; (e,f) Con-
tinuous beam and correspondingZ-PandS-ediagrams; (g) Design diagram and final bending
moment diagram

of external joint moments; in this simplest case, the vectorPEDŒ4:16, so this vector
have only one entry. Positive sign means that moment at joint load andZ-Pdiagrams
act at one direction.
Unknown internal forces (momentsS 1 andS 2 /and their positive directions are
shown onS-e diagram (Fig.11.23f). To construct the vectorSof internal forces in
the state 1, we need to take into account bending moment diagramMPo(Fig.11.23b).
This vector is

SE 1 D

16

20:16



:

The signs of the entries correspond toS-e diagram.
A static matrixAcan be constructed on the basis of theZ-PandS-ediagrams.
Figure11.23e shows free body diagram for joint 1 subjected to unknown internal
“forces” S 1 and S 2 in vicinity of joint 1 and loadP 1 , which corresponds to possible
displacement of the joint 1. Since this displacement is angle of rotation then this
load is a moment. Equilibriumcondition leads to the equationP 1 DS 1 CS 2 .So
the static matrix becomesAD
11

̆

.