Advanced Methods of Structural Analysis

8.2 Canonical Equations of Displacement Method 287

abc

q

4m 6m

5m

3m

1

2

1

P

i6-8

i4-5

P

1

q i1-3

2 8

q

P

1

5

2

7

6

3

4

d

0.333EI

r 21

3 i 4 − 5

l 4 − 5 = 0.333EI

0.24EI 0.24EI

6 i 1 − 3

l 1 − 3 = 0.24EI

3 i4-5

4 i1-3

2 i1-3

r 21

*

M 1

r 11

3 i4-5=1.0EI

4 i1-3=0.8EI 3 i6-8=0.6EI

2 i1-3=0.4EI

State 1: The unit angular displacement Z 1 =1

Z 1 =1

M 2

Z 2 =1

*

r 12

r 22

3 i 4 − 5

l 4 − 5

3 i 4 − 5

l 4 − 5

= 0.333EI

6 i 1 − 3

12 i 1 − 3

l 1 − 3

l^21 − 3

= 0.24EI

State 2: The unit linear displacementZ 2 =1

0.111EI

0.096EI

0.096EI

= 0.096EI

3 i 4 − 5

l^24 − 5

= 0.111EI

6 i 1 − 3

l 1 − 3

6 i 1 − 3

l 1 − 3

r 22

e

f Loaded state

5.0 R 2 P

5 q/2 =5

5.0

q

R 2 P

R 1 P

MP^0

M 3 =4.1667

M 7 = 9.984

M 2 = 2.0833

M 5 =15.36 ul ul

*

*

*

g

MkNm

13.084

1.814 11.349

11.27

1.324

8.037

Fig. 8.8 (a) Design diagram; (b) Primary system; (c) Specified sections; (d) State 1. Bending mo-
ment diagram due to unit angular displacementZ 1 D 1 and free-body diagram for the calculation
ofr 21 .(e) State 2. Bending moment diagram due to unit linear displacementZ 2 D^1 and free-body
diagram for the calculation ofr 22 .(f) Bending moment diagram in the loaded state and free-body
diagram for the calculation of load reactionR2P.(g) Final bending moment diagram M (kN m)