Advanced Methods of Structural Analysis

6.6 Elastic Loads Method 187

are zero, the members with nonzero internal forces are located within only two
panels of the truss, and as a result, multiplication of both axial force diagrams related
to only for members which belong to two adjusted panels.

Actual state 2 EA

EA

2 EA

d = 4m

3m

PP

152 3 4

abc de

0

a

PP

NP

–8P/3

• 5P/3

+4P/3 +8P/3 +4P/3

+5P/3 –P –P

b

c

N 1

1/4 1/4 1/4 1/4

–5/12

+1/2

+1/3

–5/12

0 1

ab

2

d

a

b

c

–1/4 +5/12 –1/4 N 2

+5/12

–1/3

1/4 1/4 1/4 1/4

2

a

b

1

c

3

–1/4 +5/12 –1/4

+5/12

–1/3

Fig. 6.25 (a,b) Design diagram of the truss, and internal forces due to given loads. (c) Calculation
of elastic loadW 1 .(d) Calculation of elastic loadW 2

Elastic loadW 2 Two unit couples are applied to members 1-2 and 2-3. Correspond-
ing distribution of internal forces is presented in diagramNN 2 (Fig.6.25d).
Elastic loadW 2 is obtained by “multiplication” of two axial force diagramsNN 2
andNp. Multiplication of diagrams within only three members (ab,bc,and
a 2 ) has nonzero result.

W 2 D

XNN 2 Npl

EA

D

1

2 EA

8P

3



1

3

 4  2 C

1

EA

5P

3



5

12

 5 D

253

36

P

EA

: (c)