12.3 Fundamental Methods of Plastic Analysis 431P=∞2.P=6Ny1.Ny NyNy NyNy Ny NyNy Ny NyNyP=4Ny3.NyP= 3 Ny4.
DaP1.NyNy Ny NyNy Ny NyNy Ny NyNy Ny
N 4 =0P3.N 2 =∝P4.N 1 =3NyP2.N 3 =−NybFig. 12.4 (a) Kinematical method. Different forms of failure of the structure; (b) Static method.
Different stress conditions of the structureLet us consider variousdistributions of internal forcesin the structure. For all
these schemes, the following two conditions should be met:1.Since the structure contains four hangers, then forces in any three members have
to be equal to the limit loadNy, which means that these three members corre-
spond to horizontal part of Prandtl diagram
2.The force in the remaining forth member should be less thanNy.
There is possible the following cases.1.N 1 DN 2 DN 3 DNy
2.N 1 DN 2 DN 4 DNy
3.N 1 DN 3 DN 4 DNy
4.N 2 DN 3 DN 4 DNy
The force in the remaining forth member should be defined from equilibrium con-
dition. The sum of the moments with respect to point of application of the forceP
leads to the following results:1: N 4 D0; 2: N 3 DNy;3:N 4 D1;4:N 4 D3NyThese internal forces are shown in Fig.12.4b.