Advanced Methods of Structural Analysis

12.4 Limit Plastic Analysis of Continuous Beams 437

hinge. It is shown by solid circle; corresponding plastic moment isMy(Fig.12.6b).
Now we have simply supported beam, i.e., the appearance of the plastic hinge do
not destroy the beam but led to the changing of design diagram. Therefore, we can
increase the loadPuntil the second plastic hinge appears at the point of application
of the load. Thus, we have three hinges (Fig.12.6c), which are located on the one
line, so this structure becomes instantaneously changeable.

Using superposition principle, the moment at the point of application of force is

M.Plim/CM



My



D

Pliml

4



My

2

:

In the limit condition, the moment at the point of application of force equals to
plastic moment:
Pliml
4



My

2

DMy: (a)

This equation leads to the limit loadP

PlimD

6My

l

:

The procedure (a) may be presented graphically as shown in Fig.12.6d.

Example 12.2.Design diagram of a pinned-clamped beam is presented in Fig.12.7.
Calculate the limit loadqand find the location of a plastic hinges.

Fig. 12.7 Plastic analysis
of pinned-clamped beam

RA

A B

l

x

q

MB

RB

A B

l

x 0

q

MB=My

RA RB

Solution.It is obvious that the first plastic hinge appears at the clamped support
B; corresponding plastic moment isMy. Now we have simply supported beamAB
(this design diagram is not shown), so we can increase the loadquntil the second
plastic hinge appears. Location of this plastic hinge will coincide with position of
maximum bending moment of simply supported beam subjected to plastic moment
at the supportBand given loadq.