Advanced Methods of Structural Analysis

428 12 Plastic Behavior of Structures

P

0.4P 0.3P 0.2P 0.1P

P

d ddl

1 2 3 4

K

a b

Ny 0.75Ny 0.5Ny 0.25Ny

P=2.5Ny

c

Ny

DP 2

N 2 N 3 N 4

d

Ny Ny 0.6Ny 0.20Ny

P=P+DP 2 =2.8Ny

e

Ny Ny

DP 3

N 3 N 4

f

Ny Ny

3.0Ny

Ny N 4 =0

g

3.0

2.8

2.5

0.75 DK

P

1.0 2.0

h

Fig. 12.3 (a, b) Design diagram and distribution of internal forces according to elastic analysis;
(c)Step1–Plasticstateinthemember1andinternalforcesintherestmembers;(d)Step2–
Internal forces in the members 2–4 due to loadP 2 ;(e)Step2–Plasticstateinthemembers1
and 2; (f) Step 3 – Internal forces in the members 3–4 due to loadP 3 ;(g)Step4–Plasticstate
in the members 1, 2, and 3; (h)PKdiagram in plastic analysis

The limit load for the second hanger isN 2 D Ny. Thus equationN 2 D
0:75NyC0:833P 2 D Nyleads to the following value for increment of load

P 2 D0:25N0:833yD0:3Ny.
Thus, the valueP 2 D0:3Nyrepresents additional load, which is required so
that the second hanger reaches its yielding state. Therefore, if load

PD2:5NyC0:3NyD2:8Ny;