Nonlinear Static Analysis - Theory of Nonlinear Structural Analysis - page 55

To analyze the structural response, the first step is to assume that the response is linear,
i.e. θ 0 1 = θ 0 2 = θ 0 3 = θ 0 4 = θ 0 5 = θ 0 6 = 0. As a result, the first matrix equation of Eq. (2.150) becomes

84 EI
5 L 3 x 1 =2 F o

ð2 : 154Þ

Solving for the displacement x 1 in Eq. (2.154) gives

x 1 =5 F o L 3

= 42 EI ð2 : 155Þ

Substituting Eq. (2.155) back into the second matrix equation of Eq. (2.150) and solving for the
moments gives

< = < =

2
4

3
5

m 1
m 2
m 3
m 4
m 5
m 6

24 EI = 5 L 2
18 EI = 5 L 2
24 EI = 5 L 2
18 EI = 5 L 2
−18 EI = 5 L 2
−18 EI = 5 L 2

4 F o L = 7
3 F o L = 7
4 F o L = 7
3 F o L = 7
−3 F o L = 7
−3 F o L = 7

5 F o L 3
42 EI

= = ð2 : 156Þ

: ; : ;

Comparing the moment demands in Eq. (2.156) with the corresponding yield moments in
Eq. (2.151) show that m 1 , m 3 , m 5 , and m 6 have all exceeded their corresponding yield moment.
This means the original assumption that the structure is linear is incorrect. Therefore, the second
step is to assume that plastic hinges are formed at these locations, and the resulting moment
versus plastic rotation relationships become

m 1 = F o L = 2+ EI = ð Þ θ 0 1 ð2 : 157aÞ

m 3 = F o L = 2+ EI = ð Þ θ 0 3

ð2 : 157bÞ

m 5 = m 6 = − F o L = 3 ð2 : 157cÞ

θ 2 = θ 4 =0

ð2 : 157dÞ

where the minus sign in front of F o L /3 in Eq. (2.157c) denotes that the negative yield moment
has been exceeded. Substituting Eq. (2.157) into Eq. (2.150) gives

3 2 2 2 2

x 2 F

84 EI 5 L 24 EI 5 L 24 EI 5 L − 18 EI 5 L − 18 EI 5 L o

1

2

−θ″ F L 2

24 EI 5 L 67 EI 15 L 2 EI 15 L − 14 EI 15 L − 4 EI 15 L o

1
−θ 3
−θ 5
−θ 6

2

= F L 2

24 EI 5 L 2 EI 15 L 67 EI 15 L − 4 EI 15 L − 14 EI 15 L o

2

− F L 3

− 18 EI 5 L − 14 EI 15 L − 4 EI 15 L 28 EI 15 L 8 EI 15 L o

2

− F L 3

− 18 EI 5 L − 4 EI 15 L − 14 EI 15 L 8 EI 15 L 28 EI 15 L o

ð2 : 158Þ