# PLASTIC ANALYSIS IN FRAMED STRUCTURES

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15 Νοε 2013 (πριν από 4 χρόνια και 7 μήνες)

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PLASTIC ANALYSIS IN FRAMED
STRUCTURES

Dr.
-
Ing
.
Girma

Zerayohannes

Dr.
-
Ing
.

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
-
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
.

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)

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

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

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

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
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

-

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