PLASTIC ANALYSIS IN FRAMED
STRUCTURES
Dr.-Ing. Girma Zerayohannes
Dr.-Ing. Adil Zekaria
Dr.-Ing. Girma Z. and Adil Z.
1
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
Dr.-Ing. Girma Z. and Adil Z.
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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
Dr.-Ing. Girma Z. and Adil Z.
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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,
Dr.-Ing. Girma Z. and Adil Z.
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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.
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in Framed
Structures
Fig. Reinforced Concrete Section
Dr.-Ing. Girma Z. and Adil Z.
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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
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in
Framed Structures
• Assume Reinforcement has yielded
•  Ts = Asfyd = 2 314 (400/1.15) = 218435 N
• 
Ts  Cc   c f cd bd  218435 N
Cc
218435
c 

 0.22
f cd bd 11.33  250  350
•  from General design chart No.1  Sd.s= 0.195 
M Sd , s   Sd , s  f cd  bd 2
 0.195 11.33  250  350 2  67.66kNm
• Check the assumption that the reinforcement has
yielded
Dr.-Ing. Girma Z. and Adil Z.
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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
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in
Framed Structures
Dr.-Ing. Girma Z. and Adil Z.
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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
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in Framed
Structures
y
y
Fig. Elasto-plastic behavior
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Chapter 5- Plastic Hinge Theory in Framed
Structures
• From the stress distribution in the previous figure
d

F1   y b  d  and
2

F2   y bd
• Total bending moment M about the neutral axis
 d  d 
2

M  2 F1  
  F2   d 
2 
3

 4
bd 2  3
2
y

2



6 2

3

 M y   2 2 
2

When   0  M  M pl  1.5M el
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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,
 y bd 2 4 Wpl
 pl 


 1.5
2
M el  y bd 6 Wel
M pl
• It is different for different sections
• For I-sections pl  1.14
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in Framed
Structures
• 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
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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
Dr.-Ing. Girma Z. and Adil Z.
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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
Dr.-Ing. Girma Z. and Adil Z.
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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:
Ppl l 2
•
From
 2M pl
 Ppl 
8
16M pl
l2
Dr.-Ing. Girma Z. and Adil Z.
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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
Pel l 2
• From structural analysis, M el 
12
• From
2
M el Pel l
y 

Wel 12We
12Wel
M el
 Pel   y 2  12 2
l
l
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in Framed
Structures
• So that
Ppl
Pel

16  M pl  16   pl M el
  
 
2
l
12  M el  12  M el
16M pl l 2
12M el
where pl = (16/12) = 1.3333

   pl   pl

• Summary- in continuous beams or frames
(statically indeterminate) there exist:
a) plastic cross-section reserve pl
b) plastic system reserve pl
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in Framed
Structures
• In the above example with an I-section
(pl = 1.14) 
• plpl = 1.52  52% increase
Dr.-Ing. Girma Z. and Adil Z.
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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
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in Framed
Structures
• (i) Equilibrium method
• From FBD of element 1
M
A
 ( F  Q23 )a  2M  0  Q23  (2M / a)  F
• From FBD of element 2
M
• 
B
 Q23b  2M  0  Q23  2M / b
 l 
(2 M / a )  F  2 M / b  Fpl  2 M  
 ab 
Dr.-Ing. Girma Z. and Adil Z.
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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
• 
 2 2 
 l 
F  M 


F

2
M

 
pl
a
b


 ab 
Dr.-Ing. Girma Z. and Adil Z.
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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
Pl M
Pl M

; B

2
l
2
l
Px 2 Px(l  x) Mx
M ( x)  Ax 


2
2
l
dM ( x) P
M
l M
 (l  2 x) 
 0  xo  
dx
2
l
2 Pl
A
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in
Framed Structures
Dr.-Ing. Girma Z. and Adil Z.
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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 (MplM) results, after
simplification in a quadratic equation in P.
•  2  12M  4M 2
M
P   2 P  4
l
 l 
 0  Ppl11.65
l2
•
l M
l2
xo  
 0.414l
Substituting for x
2 l 11.65M
• 
• (ii) Principle of virtual work
• Knowledge of the location of the plastic hinge in the
span is a requirement for VWM
Dr.-Ing. Girma Z. and Adil Z.
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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


 
 


•  P  0.414l  P  0.586l  M 
2
2
 0.414l
0.586l 
M
 P  11.65 2
l
Dr.-Ing. Girma Z. and Adil Z.
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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
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in
Framed Structures
Dr.-Ing. Girma Z. and Adil Z.
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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
• 1 3 : A  F  M
2
1 2 : A
3 4l
3l
8M
 F 4 M  3l
M 0 
 M 0 F 
8
3l  8
l
2
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in Framed
Structures
• Mechanism II
M 3M 4 F M
2
3  5 : QBr   2 F 



l
2l
3
2l
3
M 3M 2 F M
1
C   2 F 



l
2l
3
2l
3
and
3M
l
 4 F M  l 5M
3  4 : QBr    M 
0
  
0
2
2l  3
2
 3
 3
4
M 5M
6M
 Fl 

0  F 
9
6
2
l
•  Mechanism II governs and Fpl=6M/l
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in Framed
Structures
• (ii) Principle of virtual work
• Mechanism I
 
2 
8M
  F 
F  M 

•
l
 3l 8 3l 8 
Mechanism II
 3 
2 
6M
  F 
2 F  M
 M 

l 3 2  l 3 2l 3 
l
• 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
Dr.-Ing. Girma Z. and Adil Z.
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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
Dr.-Ing. Girma Z. and Adil Z.
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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:
Dr.-Ing. Girma Z. and Adil Z.
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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 = 2 k – 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
Dr.-Ing. Girma Z. and Adil Z.
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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:
l
3
3F   M  2  2M  M  
Fl  6M  F  4Ml
2
2
• Mechanism II:
F(h)=(M+M+M+M) Fh=4MF=4M/h
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Chapter 5- Plastic Hinge Theory in
Framed Structures
• Mechanism III:
3F(l/2)+F(h) =(M+22M+M+M+M)
F
•
•
•
•
•
8M
3 2l  h
(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
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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+21172+1172)
F=848kN
• Mechanism II:
• F(2.0)=(1172+21172+299) F=1908kN
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Chapter 5- Plastic Hinge Theory in
Framed Structures
• Mechanism III:
• F(4.0)=(2299+2863+ 2299)
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
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in
Framed Structures
Dr.-Ing. Girma Z. and Adil Z.
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Chapter 5- Plastic Hinge Theory in
Framed Structures
• I+III:
1.5F(3.0)+F(4.0)=(299+863+299+21172
+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
+21172+2299) F=974.5kN
Dr.-Ing. Girma Z. and Adil Z.
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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+21172
+21172+21172+2299) 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
Dr.-Ing. Girma Z. and Adil Z.
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