Advanced Methods of Structural Analysis

12.5 Limit Plastic Analysis of Frames 441

For partCD

qliml 32

2

DMy!qlimD

2My

l 32

D

2 60 .kNm/

22 .m^2 /

D 30 kN=m;

which leads to
PlimD2:0 30  6 D 360 kN>70kN: (g)

Thus, limit distributed loadqin both cases leads to the limit forcePwhich can not
be accepted.
The final limit load is governed by minimum value given by formulae (e), (f),
and (g), so the limit loadPlimD 70 kN, and correspondingqlimD P2llim 2 D 270  6 D
5:83kN=m.

Discussion:

On the basis of obtained numerical results, we can explain the order of appearance
of the plastic hinges. The limit loadPD 70 kN and corresponding loadqare de-
termined from the conditions of appearance of plastic hinges at the supportsA, B
and at pointKof application forceP. Thus, the failure of the structure in whole
is defined by a failure of the spanABbecause this simply supported beam is be-
ing transformed in mechanism. In thiscase, with further increase of the loadq,the
relationshipPD2ql 2 is not held anymore, since loadPcannot reach the greater
value thanPlim. It is obvious that spanBCstill can resist to increased loadq,how-
ever, the structure in whole is differ from the original one.
The problem of determination of limit load for continuous beam with given bear-
ing capacity has unique solution.

12.5 Limit Plastic Analysis of Frames......................................

A frame can be failed by different ways. The different schemes of failure are pre-
sented in Fig.12.9. They are following: beam mechanism of failure.B 1 ;B 2 ;B 3 /,
mechanism of sidesway failure.S /, joint failure.J /, framed.F /and different com-
bined mechanisms.
The type of mechanism of failure, which will occur, is not known in advance.
This is the principal difficulty for plastic analysis of frames. Therefore, for each
mechanism of failure and their different combinations, the equilibrium conditions
should be considered and then that mechanism of failure should be adopted, which
occurs at the minimum load.
Let us consider the limit load determination for different types of failure. Design
diagram of the portal frame is presented in Fig.12.10a. The loading of the frame
is simple. Assume thatQD2Pand the limit moments for vertical and horizontal
members satisfy to conditionMyhorD2Myvert.
Now let us consider the different mechanisms of failure.