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.
2
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.
3
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.
4
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
Dr.

Ing. Girma Z. and Adil Z.
5
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.
6
Chapter 5

Plastic Hinge Theory in Framed
Structures
Dr.

Ing. Girma Z. and Adil Z.
7
Fig. Reinforced Concrete Section
Chapter 5

Plastic Hinge Theory in Framed
Structures
•
Steps:
–
Assume that the reinforcement has yielded
–
Determine
C
c
c
–
Determine
M
R,ds
–
Check assumption of steel yielding
Dr.

Ing. Girma Z. and Adil Z.
8
Chapter 5

Plastic Hinge Theory in
Framed Structures
•
Assume Reinforcement has yielded
•
T
s
=
A
s
f
yd
= 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
Dr.

Ing. Girma Z. and Adil Z.
9
22
.
0
350
250
33
.
11
218435
bd
f
C
cd
c
c
kNm
bd
f
M
cd
s
Sd
s
Sd
66
.
67
350
250
33
.
11
195
.
0
2
2
,
,
N
bd
f
C
T
cd
c
c
s
218435
Chapter 5

Plastic Hinge Theory in
Framed Structures
•
yd
=
f
yd
/E
s
= 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.
10
Chapter 5

Plastic Hinge Theory in
Framed Structures
Dr.

Ing. Girma Z. and Adil Z.
11
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,
M
pl
is:
•
M
pl
=
y
(
bd
2
/4
); (
bd
2
/
4
) is called the
plastic
section modulus
and designated as
W
pl
•
The
elastic section modulus
W
el
=
bd
2
/6
Dr.

Ing. Girma Z. and Adil Z.
12
Chapter 5

Plastic Hinge Theory in Framed
Structures
Dr.

Ing. Girma Z. and Adil Z.
13
Fig. Rectangular section :
–
Stress Distribution ranging from elastic,
partially plastic, to fully plastic
Dr.

Ing. Girma Z. and Adil Z.
14
Chapter 5

Plastic Hinge Theory in Framed
Structures
Dr.

Ing. Girma Z. and Adil Z.
15
y
y
Fig.
Elasto

plastic behavior
Chapter 5

Plastic Hinge Theory in Framed
Structures
•
From the stress distribution in the previous figure
•
Total bending moment M about the neutral axis
Dr.

Ing. Girma Z. and Adil Z.
16
d
b
F
and
d
d
b
F
y
y
2
1
2
el
pl
y
y
M
M
M
When
M
bd
d
F
d
d
F
M
5
.
1
0
2
2
3
2
2
3
6
3
2
2
4
2
2
2
2
2
1
Chapter 5

Plastic Hinge Theory in Framed
Structures
Chapter 5

Plastic Hinge Theory in Framed
Structures
•
The ratio between
M
pl
and
M
el
which is equal
to the ratio between
W
pl
and
W
el
is called
shape factor
pl
.
•
For the solid rectangular section,
•
It is different for different sections
•
For
I

sections
pl
1.14
Dr.

Ing. Girma Z. and Adil Z.
17
5
.
1
6
4
2
2
el
pl
y
y
el
pl
pl
W
W
bd
bd
M
M
•
Shape factors for common cross sections (check as a
home work)
Dr.

Ing. Girma Z. and Adil Z.
18
Shape
Shape factor,
pl
Rectangle
1.5
Circular solid
1.7 (16/3
π
)
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
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.
19
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
.
Dr.

Ing. Girma Z. and Adil Z.
20
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)
Dr.

Ing. Girma Z. and Adil Z.
21
Chapter 5

Plastic Hinge Theory in Framed
Structures
Dr.

Ing. Girma Z. and Adil Z.
22
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
P
pl
corresponding to the
ultimate state is:
•
Dr.

Ing. Girma Z. and Adil Z.
23
2
2
16
2
8
l
M
P
M
l
P
From
pl
pl
pl
pl
Chapter 5

Plastic Hinge Theory in Framed
Structures
•
Compare with the elastic strength of the
continuous beam,
P
el
•
Here section capacities are determined on the
basis of linear elastic stress distribution where
only the extreme fibers have plasticized
•
From structural analysis,
•
From
Dr.

Ing. Girma Z. and Adil Z.
24
12
2
l
P
M
el
el
2
2
2
12
12
12
l
M
l
W
P
W
l
P
W
M
el
el
y
el
e
el
el
el
y
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
Dr.

Ing. Girma Z. and Adil Z.
25
pl
pl
el
el
pl
el
pl
el
pl
el
pl
M
M
M
M
l
M
l
M
P
P
12
16
12
16
12
16
2
2
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.
26
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
Dr.

Ing. Girma Z. and Adil Z.
27
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
F
pl
•
First we solve using the equilibrium method
and then repeat with the virtual method
Dr.

Ing. Girma Z. and Adil Z.
28
Chapter 5

Plastic Hinge Theory in Framed
Structures
•
(
i
) Equilibrium method
•
From FBD of element
1
•
From FBD of element 2
•
Dr.

Ing. Girma Z. and Adil Z.
29
F
a
M
Q
M
a
Q
F
M
A
)
/
2
(
0
2
)
(
23
23
b
M
Q
M
b
Q
M
B
/
2
0
2
23
23
ab
l
M
F
b
M
F
a
M
pl
2
/
2
)
/
2
(
Chapter 5

Plastic Hinge Theory in Framed
Structures
Dr.

Ing. Girma Z. and Adil Z.
30
Chapter 5

Plastic Hinge Theory in Framed
Structures
•
(ii) Principle of virtual work
•
External virtual work = internal virtual work
•
Dr.

Ing. Girma Z. and Adil Z.
31
ab
l
M
F
b
a
M
F
pl
2
2
2
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
M
max
•
(
i
) Equilibrium method
Dr.

Ing. Girma Z. and Adil Z.
32
Pl
M
l
x
l
M
x
l
P
dx
x
dM
l
Mx
x
l
Px
Px
Ax
x
M
l
M
Pl
B
l
M
Pl
A
o
2
0
)
2
(
2
)
(
2
)
(
2
)
(
2
;
2
2
Chapter 5

Plastic Hinge Theory in
Framed Structures
Dr.

Ing. Girma Z. and Adil Z.
33
Chapter 5

Plastic Hinge Theory in Framed
Structures
•
Substituting
x
o
in the expression for
M
(
x
) and equating
the maximum moment to
M
pl
(
M
pl
M
)
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
Dr.

Ing. Girma Z. and Adil Z.
34
2
4
2
2
2
65
.
11
0
4
12
l
M
P
l
M
P
l
M
P
pl
l
M
l
l
M
l
x
o
414
.
0
65
.
11
2
2
Substituting for
x
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
P
pl
is found
•
For the present example, check the result
using the PVW
•
Dr.

Ing. Girma Z. and Adil Z.
35
2
65
.
11
586
.
0
414
.
0
2
586
.
0
2
414
.
0
l
M
P
l
l
M
l
P
l
P
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.
36
Chapter 5

Plastic Hinge Theory in
Framed Structures
Dr.

Ing. Girma Z. and Adil Z.
37
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
•
Dr.

Ing. Girma Z. and Adil Z.
38
l
M
F
M
l
l
M
F
M
l
A
l
M
F
A
8
0
8
3
3
4
2
0
8
3
:
2
1
4
3
2
:
3
1
Chapter 5

Plastic Hinge Theory in Framed
Structures
•
Mechanism II
•
Mechanism II governs and
F
pl
=
6M/
l
Dr.

Ing. Girma Z. and Adil Z.
39
l
M
F
M
M
Fl
M
l
l
M
F
M
M
l
Q
l
M
F
l
M
l
M
F
C
and
l
M
F
l
M
l
M
F
Q
Br
Br
6
0
2
5
6
9
4
0
2
5
3
2
3
4
0
2
3
3
:
4
3
2
3
2
2
3
2
3
1
2
3
4
2
3
2
3
2
:
5
3
Chapter 5

Plastic Hinge Theory in Framed
Structures
•
(ii) Principle of virtual work
•
Mechanism I
•
Mechanism II
•
Mechanism II governs with
F
pl
=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.
40
l
M
F
l
l
M
F
8
8
3
2
8
3
l
M
F
l
l
M
l
M
F
6
3
2
2
3
2
3
3
2
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
Dr.

Ing. Girma Z. and Adil Z.
41
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)
Dr.

Ing. Girma Z. and Adil Z.
42
Chapter 5

Plastic Hinge Theory in Framed
Structures
Dr.

Ing. Girma Z. and Adil Z.
43
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.
44
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=2
k

1
•
The
combination method
will be explained by
means of the
portal frame
Dr.

Ing. Girma Z. and Adil Z.
45
Chapter 5

Plastic Hinge Theory in
Framed Structures
Dr.

Ing. Girma Z. and Adil Z.
46
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 = 2
2
–
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
Dr.

Ing. Girma Z. and Adil Z.
47
Chapter 5

Plastic Hinge Theory in
Framed Structures
Dr.

Ing. Girma Z. and Adil Z.
48
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
=4M
F=4M/h
Dr.

Ing. Girma Z. and Adil Z.
49
Ml
F
M
Fl
M
M
M
l
F
4
6
2
3
2
2
2
3
Chapter 5

Plastic Hinge Theory in
Framed Structures
•
Mechanism III:
3F
(
l/2
)
+
F
(
h
) =(
M+2
2M+M+M+M
)
(
3/2
)
F
l
+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
F
pl
=0.615M
pl
Dr.

Ing. Girma Z. and Adil Z.
50
h
l
M
F
2
3
8
Chapter 5

Plastic Hinge Theory in
Framed Structures
•
Two

bay frames
•
System and loading

See next slide
•
The frame is statically indeterminate to the 6
th
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=2
4

1=15
(too many!)
Dr.

Ing. Girma Z. and Adil Z.
51
Chapter 5

Plastic Hinge Theory in Framed
Structures
Dr.

Ing. Girma Z. and Adil Z.
52
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
Dr.

Ing. Girma Z. and Adil Z.
53
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.
54
Chapter 5

Plastic Hinge Theory in
Framed Structures
Dr.

Ing. Girma Z. and Adil Z.
55
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.
56
Chapter 5

Plastic Hinge Theory in
Framed Structures
Dr.

Ing. Girma Z. and Adil Z.
57
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
F
pl
not governing
•
The plastic limit load is thus
F
pl
=
722 kN
Dr.

Ing. Girma Z. and Adil Z.
58
Chapter 5

Plastic Hinge Theory in
Framed Structures
Dr.

Ing. Girma Z. and Adil Z.
59
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