Advanced Methods of Structural Analysis

11.4 Resolving Equations 383

a F

AB

c 12345

M 1 M^2

M 3

M 4

M 5

S-e

b

(^12) Z-P
d
Joint A
P 1
M 2 M 3 P 1 = M 2 + M 3 Joint B
P 2
M 4 M 5 P 2 = M 4 + M 5
Fig. 11.17 (a–c) Continuous beam and correspondingZ-PandS-ediagrams; (d) Construction of
static matrixAfor beam
In matrix form, these equations can be rewritten as follows

P 1
P 2
D

01100
00011

2 6 6 6 6 6 6 6 6
M 1
M 2
M 3
M 4
M 5
3 7 7 7 7 7 7 7 7
;where static matrixA.25/D

01100
00011
Each row of matrixApresents the corresponding equilibrium equation; the entries
of the matrix are coefficients atMi. Note again that equilibrium equations are for-
mulated forpossible loads P, which correspondsto possible displacements, but not
for given loadF.
The truss in Fig.11.18a contains fourpossiblelinear displacements and corre-
sponding external loadsP, labeled as1–4(Fig.11.18b). SupportAhas no linear
displacements; therefore,Z-Pdiagram does not contain the vectors of displacement
at supportA. The possible displacements 1 and 4 describe the rolled supportsB
andDas well as their orientation.
Equilibrium equations should be formulated for each joint withpossibleload
Joint B
X
YD 0 W P 1 DS 1
Joint C
X
XD 0 W P 2 DS 2 S 3 cos ̨CS 5 cos ̨
P 2 DS 2 0:6S 3 C0:6S 5
X
YD 0 W P 3 DS 3 sin ̨CS 5 sin ̨
P 3 D0:8S 3 C0:8S 5
Joint D
X
XD 0 W P 4 DS 4 CS 3 cos ̨
P 4 D0:6S 3 CS 4