Advanced Methods of Structural Analysis

438 12 Plastic Behavior of Structures

For this beam, the general expressions for shear and bending moment at any
sectionxare

Q.x/DRAqxD



ql

2



MB

l



qx;

M.x/DRAx

qx^2

2

D



ql

2



MB

l



x

qx^2

2

:

In these expressions, the momentMBat the supportBequals to plastic momentMy.
The maximum moment occurs at the point whereQ.x/D 0. This condition
leads tox 0 D 2 lMqlB.
Corresponding bending moment equals

Mmax.x 0 /D



ql

2



MB

l







l

2



MB

ql





q

2



l

2



MB

ql

 2

:

The limit condition becomes when this moment and the moment at the supportB
will be equal toMy,i.e.MmaxDMy. This condition leads to the following equation

My^2 3Myql^2 C

q^2 l^4

4

D0: (a)

If we consider this equation as quadratic with respect toMy, then solution of this

equation isMy D ql

2

2



3 ̇

p

8



. Minimum root isMy D ql

2

2



3  2

p

2



D

0:08578ql^2. The limit load becomes

qlimD

2My



3  2

p

2



l^2

D11:657

My

l^2

: (b)

The plastic hinge occurs atx 0 Dl

p

2  1



D0:4142l.

If we consider (a) as quadratic with respect toq, then solution of this equation is

qmaxD

2My

l^2



3 C 2

p

2



:

This result coincides with (b).

Example 12.3.Two-span beam with overhang is subjected to forcePand uni-
formly distributed loadqasshowninFig.12.8a. The loading of the beam is simple;
assume that relationship between loads is alwaysP D2ql 2. Determine the limit
load, ifaD 3 m;bD 4 m;l 2 D 6 m;l 3 D 2 m, and bearing capacity of all cross
sections within the beam isMyD 60 kNm.