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Dr.-Ing. Girma Z. and Adil Z. 1
PLASTIC ANALYSIS IN FRAMED STRUCTURES
Dr.-Ing. Girma ZerayohannesDr.-Ing. Adil Zekaria
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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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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,
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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.
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Chapter 5- Plastic Hinge Theory in Framed Structures
Fig. Reinforced Concrete Section
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Chapter 5- Plastic Hinge Theory in Framed Structures
• Steps:– Assume that the reinforcement has yielded– Determine Cc c
– Determine MR,ds
– Check assumption of steel yielding
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
Fig. Rectangular section :– Stress Distribution ranging from elastic, partially plastic, to fully plastic
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Chapter 5- Plastic Hinge Theory in Framed Structures
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y
y
Fig. Elasto-plastic behavior
Chapter 5- Plastic Hinge Theory in Framed Structures
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• 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
Chapter 5- Plastic Hinge Theory in Framed Structures
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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• 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
Chapter 5- Plastic Hinge Theory in Framed Structures
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 5.3 Plastic Hinge Theory• It is based on the hypothesis of a localized
(concentrated) plastic hinge.
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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)
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Chapter 5- Plastic Hinge Theory in Framed Structures
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• In the above example with an I-section (pl = 1.14)
• plpl = 1.52 52% increase
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• (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(
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Chapter 5- Plastic Hinge Theory in Framed Structures
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Chapter 5- Plastic Hinge Theory in Framed Structures
• (ii) Principle of virtual work• External virtual work = internal virtual work
•
ablMF
baMF pl 222
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Chapter 5- Plastic Hinge Theory in Framed Structures
• (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
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Chapter 5- Plastic Hinge Theory in Framed Structures
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• (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
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Chapter 5- Plastic Hinge Theory in Framed Structures
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Chapter 5- Plastic Hinge Theory in Framed Structures
• (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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• (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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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)
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Chapter 5- Plastic Hinge Theory in Framed Structures
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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:
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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!)
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Chapter 5- Plastic Hinge Theory in Framed Structures
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
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Chapter 5- Plastic Hinge Theory in Framed Structures
• 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
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Chapter 5- Plastic Hinge Theory in Framed Structures
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Chapter 5- Plastic Hinge Theory in Framed Structures
• (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
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Chapter 5- Plastic Hinge Theory in Framed Structures