Advanced Methods of Structural Analysis

12.2 Direct Method of Plastic Analysis 429

then both members 1 and 2 reach their limit state. As this takes place, the internal
forces in hangers 3 and 4 are (Fig.12.3e) are following

N 3 D0:5NyC0:3330:3NyD0:6NyI N 4 D0:25Ny0:1670:3NyD0:20Ny

Step 3: Since internal forces in hangers 1 and 2 reached the limit values, then the
following increase of the load by valueP 3 (Fig.12.3f) affects the members 3 and
4 only. Elastic analysis of this statically determinate structure leads to the following
internal forces in members 3 and 4:N 3 D2P 3 andN 4 DP 3.
Step 4: Similarly as above, the limit state for this case occurs if internal force in
hanger 3 reaches its limit value

N 3 D0:6NyC2P 3 DNy:

This equation leads to the following value for increment of the load

P 3 D

0:4Ny

2

D0:2Ny

The total value of external force (Fig.12.3g)

PD2:5NyCP 2 CP 3 D2:5NyC0:3NyC0:2NyD3:0Ny

The first term in this formula corresponds to limit load in the first member; in-
crement of the force by0:3Nyleads to the limit state in the second member. The
following increment of the force by0:2Nyleads to the limit state in the third mem-
ber. After that the load carrying capacity of the structure is exhausted. From the
equilibrium equation for the entire structure, we can see that on this stageN 4 D 0
(Fig.12.3g).
All forces satisfy to equilibrium condition.Plastic behavior analysis leads to the
increment of the limit load by^3 2:52:5^100 %D^20 %.

Plastic displacements If some of the elements reached its limiting value and the
load continues to increase, then we cannotdetermine displacements of the system
using only elastic analysis. However, plastic analysis allows calculating displace-
ments of a structure on the each stage of loading. Let us show the graph of
displacement of the point application of forceP(pointK).
If loadP D 2:5Ny, then internal force in second element equals0:75Ny

(see Fig.12.3c) and vertical displacement of pointKisKD0:75NEAyl.
If loadP D 2:8Ny, then internal force in second element equalsNy

(see Fig.12.3e) and vertical displacement of pointKisKDNEAyl.
If loadPD3:0Ny, then internal force in third element equalsNy(see Fig.12.3g)

and deflection of this element equalsNEAyl. Since internal force in fourth element

equals zero, its deflection is zero and required displacementKD 2 NEAyl. Corre-
spondingPKdiagram is shown in Fig.12.3h; the factorsl=EAandNyfor
horizontal and vertical axis, respectively.This diagram shows that plastic analysis
is nonlinear analysis.