Advanced Methods of Structural Analysis

440 12 Plastic Behavior of Structures

Bending moment at the middle point of the second span caused by distributed load
qand plastic moments at supportsBandCequals

MD

ql 22

8



My

2



My

2

D

ql 22

8

My: (b)

The following increase of the load leads to the appearance of the plastic moments
within the first and second spans. In the limit state, the bending moments of the first
and second spans must be equal to plastic moments; therefore expressions (a) and
(b) should be rewritten for both spans as follows:

For first span

Pab

l 1

MyDMy or

Pab

l 1

D2My: (c)

For second span

ql^22

8

MyDMy or

ql 22

8

D2My: (d)

Equations (c) and (d) show that in limit state, the maximum bending moment caused
by external load equals twice plastic moment. This can be presented geometrically
as shown in Fig.12.8c. Limit plastic moments (LPM) are shown by two horizontal
dotted lines. These lines show that limit moments for supports and for any section
of the beam are equal, may be negative or positive; however, they cannot be more
thanMy. Now we can fit a space between twoLPMlines by bending moment
diagrams caused by external load for eachsimply supported beam. This procedure
is called equalizing of bending moments and can be effectively applied for any con-
tinuous beam.
For the first span, (c) allows calculating the limit forceP

Plimab

l 1

D2My!PlimD

2Myl 1

ab

D

2 60 .kNm/7.m/

3.m/4.m/

D 70 kN: (e)

For the second span, (d) allows calculating the limit distributed loadq

qliml^22

8

D2My!qlimD

16My

l 22

D

16 60 .kNm/

36 .m^2 /

D26:67kN=m:

Now we need to take into account the condition of simple loading as well as limiting
forcePD 70 kN and loadqD26:67kN=m.
Knowing the distributed load wecan calculate, according conditionPD2ql 2 ,
corresponding limit forceP

PlimD2:026:67 .kN=m/6.m/D 320 kN>70kN: (f)