Plastic Analysis in framed structures

Dr.-Ing. Girma Z. and Adil Z. 1

PLASTIC ANALYSIS IN FRAMED STRUCTURES

Dr.-Ing. Girma ZerayohannesDr.-Ing. Adil Zekaria


Chapter 5- Plastic Hinge Theory in Framed Structures

• 5.1 Introduction• All codes for concrete, steel and steel-composite

structures (EBCS-2, EBCS-3, EBCS-4) allow the plastic method of analysis for framed structures

• The requirement is that, sufficient rotation capacity is available at the plastic hinges



• In this chapter we will introduce the plastic method of analysis for line elements. It is called the “plastic hinge theory”

• The method is known as the “yield line theory” for 2D elements (e.g. slabs)

• Both are based on the upper bound theorem of the theory of plasticity

• Recall that the strip method is also a plastic method of analysis based on the lower bound theorem



• Therefore the capacity of the line elements are greater or at best equal to the actual capacity of the member. a concern for the designer,



• 5.2 Design Plastic Moment Resistances of Cross-Sections

• 5.2.1 RC Sections• Such plastic section capacities are essential in the

plastic hinge theory, because they exist at plastic hinges



• Determine using the Design Aid (EBCS-2: Part 2), the plastic moment resistance (the design moment resistance) of the RC section shown in following slide, if the concrete class and steel grade are C-25 and S-400 respectively.



Fig. Reinforced Concrete Section



• Steps:– Assume that the reinforcement has yielded– Determine Cc c

– Determine MR,ds

– Check assumption of steel yielding



• Assume Reinforcement has yielded • Ts = Asfyd = 2 314 (400/1.15) = 218435 N

•

• from General design chart No.1 Sd.s= 0.195

• Check the assumption that the reinforcement has yielded

22.035025033.11

218435

bdfC

cd

cc

kNm

bdfM cdsSdsSd

66.6735025033.11195.0 2

2,,

NbdfCT cdccs 218435



• yd = fyd/Es = 347.8/200000 = 1.739(0/00)

• s = 9.4(0/00) 1.739(0/00) reinforcement has yielded

• Exercise for section with compression reinforcement





• 5.2.2 Structural Steel Sections• Consider the solid rectangular section

shown in the next slide• The plastic section capacity, Mpl is:

• Mpl = y(bd2/4); (bd2/4) is called the plastic section modulus and designated as Wpl

• The elastic section modulus Wel = bd2/6



Fig. Rectangular section :– Stress Distribution ranging from elastic, partially plastic, to fully plastic




y

y

Fig. Elasto-plastic behavior



• From the stress distribution in the previous figure

• Total bending moment M about the neutral axisdbFandddbF yy

21 2

elpl

y

y

MMMWhen

M

bd

dFddFM

5.10

223

223

6

32

242

2

22

21




• The ratio between Mpl and Mel which is equal to the ratio between Wpl and Wel is called shape factor pl.

• For the solid rectangular section,

• It is different for different sections• For I-sections pl 1.14

5.1

64

2

2

el

pl

y

y

el

plpl W

Wbdbd

MM


• Shape factors for common cross sections (check as a home work)

Shape Shape factor, pl Rectangle 1.5

Circular solid 1.7 (16/3π)

Circular hollow

1.27 (4/π)

Triangle 2.34

I-sections (major axis)

1.1-1.2

Diamond 2




• For simply and doubly symmetric sections, the plastic neutral axis (PNA) coincides with the horizontal axis that divides the section in to 2 equal areas



• 5.3 Plastic Hinge Theory• It is based on the hypothesis of a localized

(concentrated) plastic hinge.



• The load carrying capacity of a structure is reached when sufficient numbers of plastic hinges have formed to turn the structure into a mechanism.

• The load under which the mechanism forms is called the ultimate load.

• As an example, let us consider a typical interior span of a continuous beam (see next slide)





• The ultimate state is reached when 3 plastic hinges form (2 over the supports plus 1 in the span)

• The ultimate load Ppl corresponding to the ultimate state is:

•

2

2

16

28

lM

P

MlP

From

plpl

plpl



• Compare with the elastic strength of the continuous beam, Pel

• Here section capacities are determined on the basis of linear elastic stress distribution where only the extreme fibers have plasticized

• From structural analysis,• From

12

2lPM elel

22

2

121212 l

MlWP

WlP

WM elel

yele

el

el

ely



• So that

where pl = (16/12) = 1.3333

• Summary- in continuous beams or frames (statically indeterminate) there exist:

a) plastic cross-section reserve pl

b) plastic system reserve pl

plplel

elpl

el

pl

el

pl

el

pl

MM

MM

lMlM

PP

1216

1216

1216

2

2



• In the above example with an I-section (pl = 1.14)

• plpl = 1.52 52% increase



• 5.4 Method of Analysis• As in the linearly elastic method,

– either the equilibrium method or – the principle of virtual work is applicable for the

plastic method of analysis.• Examples for different types of framed

structures follow



• 5.4.1 Single span and continuous beams• (a) single span-fixed end beam• System and loading see next slide• Goal is to determine Fpl

• First we solve using the equilibrium method and then repeat with the virtual method



• (i) Equilibrium method• From FBD of element 1

• From FBD of element 2

•

FaMQMaQFM A )/2(02)( 2323

bMQMbQM B /202 2323

ablMFbMFaM pl 2/2)/2(





• (ii) Principle of virtual work• External virtual work = internal virtual work

•

ablMF

baMF pl 222



• (b) Propped cantilevers under UDL• System and loading see next slide• NB- position of the plastic hinge in the span is not

known. Must be determined from the condition of zero shear at location of Mmax

• (i) Equilibrium method

PlMlx

lMxlP

dxxdM

lMxxlPxPxAxxM

lMPlB

lMPlA

o

20)2(

2)(

2)(

2)(

2;

22





• Substituting xo in the expression for M(x) and equating the maximum moment to Mpl (MplM) results, after simplification in a quadratic equation in P.

•

• • • (ii) Principle of virtual work• Knowledge of the location of the plastic hinge in the span

is a requirement for VWM

24

2

22 65.110412

lMP

lMP

lMP pl

lM

ll

Mlxo 414.065.112

2

Substituting for x



• Of course, the correct location of the plastic hinge can be determined by trial and error, i.e., keep trying new locations until the minimum Ppl is found

• For the present example, check the result using the PVW

•

265.11

586.0414.02586.0

2414.0

lMP

llMlPlP



• (c) Continuous beams• System and loading see next slide• The ultimate capacity of a continuous beam is

reached when a mechanism forms in one of the spans. The ultimate load is determined as the minimum of the different mechanisms in all the spans





• (i) Equilibrium method• Locations of plastic hinges are simple to

determine. They are at 1, 2, 3, 4, and 5.• The two mechanisms I and II are to be

investigated. It is not immediately obvious which one governs

• Mechanism I•

lMFMl

lMFMlA

lMFA

8083

34

20

83:21

432:31



• Mechanism II

• Mechanism II governs and Fpl=6M/l lMFMMFl

Mll

MFMMlQ

lMF

lM

lMFC

andl

MFl

Ml

MFQ

Br

Br

602

569

4

02

5323

402

33

:43

232

232

31

234

232

32:53



• (ii) Principle of virtual work• Mechanism I • Mechanism II • Mechanism II governs with Fpl=6M/l• PVW is much simpler in this case• Figure at the bottom shows the moment diagram at the

ultimate capacity. Observe that the moment at all sections is less than or equal to the respective plastic section capacities

lMF

llMF 8

832

83

lMF

llM

lMF 6

322

323

32



• 5.5.1 Frames• One of the important application areas of the

method of plastic hinge theory, which has been proved by experiments are frames

• The procedure is one of trial and error as in continuous beams using the basic or combined modes



• The combination procedure, based on selective combination of the elementary mechanisms leads to result more quickly

• Three elementary(basic) mechanisms (basic modes of failure) are to be distinguished

• They are the beam mechanism, frame mechanism, and joint mechanism (see next slide)





• The beam and frame mechanisms represent independent failure mechanisms.

• Joint mechanism can occur only in combination with another elementary failure mechanism. It does not represent a failure mechanism alone

• Number of elementary (basic) mechanisms k is determined from:



• k = m-n; where m=possible no of plastic hinges depending on system and loading, and n=degree of statical indeterminacy

• The no of possible combination including the basic modes (elementary mechanisms) is given by:

• q=2k-1• The combination method will be explained by

means of the portal frame





• k = m-n = 5-3 = 2• The no of possible combination q, which includes

the basic mode I and II is:• q = 2k – 1 = 22 – 1 = 3• See the three mechanisms in the next slide with

the plastic moments. When a plastic hinge forms at a joint, it must be on the columns and the hinge must be shown on the column side of the joint





• All member rotation angles are equal in this example. In more complicated structures, the relationships b/n the various rotations must be determined.

• The virtual work equations are:• Mechanism I:

• Mechanism II: F(h)=(M+M+M+M) Fh=4MF=4M/h

MlFMFlMMMlF 462322

23



• Mechanism III:

3F(l/2)+F(h) =(M+22M+M+M+M)

(3/2)Fl+Fh=8M• Substituting the values for l and h• Mechanism I: F=0.666M• Mechanism II: F=1.000M• Mechanism III: F=0.615M• Therefore Mechanism III governs with Fpl=0.615Mpl

hlMF

238



• Two-bay frames• System and loading- See next slide• The frame is statically indeterminate to the 6th

degree n=6• The no of hinges m are 10 so that the no of basic

mechanisms (modes) are:• k=m-n=10-6=4 (I to IV)and the no of possible

combinations including the basic ones are:• q=24-1=15 (too many!)





• Basic mode IV is the joint mode and is not an independent mode. Virtual work equations for the 3 other basic modes are:

• Mechanism I: • 1.5F(3.0)=(299+21172+1172)

F=848kN• Mechanism II: • F(2.0)=(1172+21172+299) F=1908kN



• Mechanism III: • F(4.0)=(2299+2863+ 2299)

F=730.5kN• Now the basic modes will be combined in

search of a governing mechanism• (i) Combination: I+III, the plastic hinge 4 will

be eliminated• See the resulting mechanism on next slide





• I+III: 1.5F(3.0)+F(4.0)=(299+863+299+21172+1172+863+299) F=722.2kN

• (ii) Combination: II+III+IV, the plastic hinges 5 and 10 will be eliminated

• See resulting mechanism on next slide• II+III+IV: F(2.0)

+F(4.0)=(299+863+299+299+1172+21172+2299) F=974.5kN





• (iii) Combination: I+II+III+IV, the plastic hinges4, 5 and 10 will be eliminated

• See resulting mechanism on next slide• I+II+III+IV: • 1.5F(3.0)+F(2.0)

+F(4.0)=(299+863+299+21172+21172+21172+2299) F=865.8kN

• Other combinations involve more hinges resulting in higher values for internal virtual work w/o increased external virtual work and therefore in higher values of Fpl not governing

• The plastic limit load is thus Fpl = 722 kN



Plastic Analysis in framed structures

Documents

plastic analysis

plastic hingesdr

plastic section modulus

plastic method of analysis

plastic hinge theorythe

elastoplastic behaviorch

framed structures5

theory of plasticityrecall