Theory of Plasticity – lecture 4, Adam Paul Zaborski
Limit capacity of the cross-section
We assume the cross-section has one (vertical) symmetry axis. Moreover:
E , dla  < Re
 uniaxial state of stress   
 Re
 Bernoulli’s hypothesis of plane cross-sections.
We use the conditions of equilibrium:
 dA  N ,
 zdA  M
The stages of the cross-section work are:
a)
b)
c)
Re
d)
Re
e)
Re
Re
-Re
-Re
 elastic range, the principle of superposition is valid due to linearity of equations
 yielding of the first extreme fibers, it is elastic limit of the cross-section bearing capacity,
 one-sided yielding in elastic-plastic range,
 two-sided yielding, the elastic-plastic range,
 full yielding with the plastic hinge arising, it is plastic limit of the cross-section bearing
capacity.
The elastic-plastic range calculations are the most complicated. The problem consists on
determination of two process parameters from the equations of equilibrium. There are two
oś centralna
z0
Re
E


0
Re
E
z0


-Re
regions in the cross-section: elastic and plastic, divided by the plasticity front. The position of
the front, the integral limits, is unknown.
From Bernoulli’s hypothesis, we have:
   0  z .
Two parameters determine the strain distribution: the bar axis strain and its curvature.
Sometimes, it is easier way to apply other parameters, like:
 position of neutral axis
 ( z0 )  0  z0  
-
0
,

position of the plastic front
1
Theory of Plasticity – lecture 4, Adam Paul Zaborski
 0  z p   Re E  z p  
-
Re  0
 ,
E 
range of the elastic zone
 E   Re   
Re
E
Only two parameters are independent.
Elastic and plastic limit capacity of the cross-section
Elastic limit capacity of the cross-section is the set of cross-section forces causing yielding
of the (first) extreme fibers.
For simple bending, the value of the bending moment causing the first yielding can be easily
calculated from the equation of elastic range (Hooke’s equations).
Plastic limit capacity of the cross-section is the set of cross-section forces causing full
yielding of the whole cross-section.
For simple bending the plastic limit value can be calculated from rectangular distribution of
the normal stress. From the first condition of zero normal force we have:
 dF  0,   RedF   ( Re )dF  0,  A1  A2
A1
A2
so, the neutral axis halved cross-section (divides cross-section into two equal areas).
From the second condition we get that the plastic limit value of the bending moment is equal
to the sum of static moments of the cross-section halves.
M   zdF

 R zdF   R zdF  R (S
e
A1
e
e
A1
 S A 2 )  Re  S A1  S A 2  .
A2
The condition of equilibrium is valid in every coordinate set. If one axis is the principal
central axis the static moments differ only by their sign, so the formula may be rewritten:
M  2 Re S 0 A1, A 2 .
Similarly to elastic cross-section factor we introduce the plastic cross-section factor as:
W  2 S 0 A1, A 2 .
Examples
triangular cross-section b·h
Wel 
bh 3 3 bh 2

36 2h
24
the neutral axis: A1  12 A  14 bh  h1 
h
b
, b1 
, so:
2
2
2 2 2
bh
6
2
27 bh
 q  64 2 Re
l
W  2S A1  2 12 b1h1  23 h  23 h1  
elastic limit M  M max
plastic limit M  M max
bh 2 8 2
 Re
 ql
24 81
2 2 2
 Re
bh 
6
8
81
ql 2
 q
27
16
2  2 bhl
2
2
R,
2
Theory of Plasticity – lecture 4, Adam Paul Zaborski
proportion:
q
q
 2.34
I section
7a
a
10a
a
a
5a
position of the central and neutral axis:
zc  6.5a,
z0  7a, W  90a 3 R
Rectangular cross-section and the equation of the plastic front
There are two symmetry axes. The neutral axis agrees with the central axis:  0  z0  0 .
bh 2
bh 2
Re , and plastic limit: M 
Re , it means an increase of
4
6
the cross-section capacity of 50%. For the bending moment less than plastic limit and greater
Elastic limit is: M  ReWspr 
than elastic limit, M  M  M , we have:
h

 2

 h2 M 
 h2

2

M  b  Ez dz  2b  Re zdz     bRe   13  2  , so:    3 
 

 4

 4 bRe 

The function  = (M) is the plastic front equation.
3
Theory of Plasticity – lecture 4, Adam Paul Zaborski
Theorems of limit analysis
Introduction
Analyzing the structure, we seek:
 limit value of the load that begins the plastic mechanism
 stress field corresponding to the equilibrium state and the static boundary conditions
 appropriate displacement field or the rate of this field which fulfills kinematic boundary
conditions.
The exact solution fulfills the principle of virtual work.
The work (the power) of stress on the displacements (or their rates) ie equal to the work
(the power) of external forces on the displacements (or their rates).
 p u dA   F u dV   
j
j
j
AT
j
v
ij
ij dV
v
Neglecting mass forces, we can write the equation by a coefficient of external forces. The
coefficient has an exact value for true forces.
m  p j 0 u j dA    ij ij dV
AT
v
Lemma
If the limit plastic state is reached and displacements increase under constant load, the
stress remains constant and only plastic (not elastic) strain increases.
Proof: („rate” form of principle of virtual work)
l l
l l
l l
l l
A q i u i dA  Au q i u i dA  V Fi u i dV V  ij  ij dV
(index l means limit state)
For the limit load, left side of the equation vanishes, from the definition:
 Fil  0 in volume V,
 q il  0 at Aσ,
 u il  0 at Au.
We decompose the strain rate into elastic and plastic parts:


V  ij  ij dV  V  ij  ijl   ijl dV  0
From the associated flow rule for perfectly plastic material follows that the vector  ij is
tangent to the flow surface, if the plastic strain appears. So:
l e
V  ij  ijl dV  0 ,
and as consequence, the stress is constant and the rate of elastic strain is zero. For the limit
load only the plastic strain exists. The elastic properties of material are not important.
In this way, the model of perfectly elastic-plastic material is equivalent t the model of
perfectly plastic material.
l
l
l
e
p
Statically admissible solutions
The stress field is statically admissible, if the following conditions are fulfilled:
 the equations of internal equilibrium
 static boundary conditions
4
Theory of Plasticity – lecture 4, Adam Paul Zaborski
 the yield criterion in the form of weak inequality (and in particular, the stress does not
exceed the plasticity limit).
In such case, the multiplier will be different:
m s  p j 0 u j dA    *ij ij dV
AT
v
 ij
 *ij
 ij
Statically admissible stress field and the sign of the integral
Subtracting the above equations, we get:
m  ms  p j 0 u j dA    ij   *ij ij dV .
AT
V
From the Drucker’s stability postulate follows that the sign of the integral on the right side is
nonnegative and:
ms  m .
Lower bound theorem:
The structure does not undergo destruction, or, at the most is in the state of limit equilibrium,
if the statically admissible state of stress balances the actual loading.
In other words, the structure does not collapse if the external loading can be balanced by the
statically admissible state of stress. The real bearing capacity is at least as the balanced load
and it is the lower bound estimation.
Kinematically admissible solutions
The field of displacement rates is kinematically admissible if the following conditions are
fulfilled:
 kinematic boundary conditions and compatibility equations
 the condition of nonnegative work (power) of external forces:
D z  m  p j 0 u kj  0
A
Applying the principle of virtual work to an arbitrary kinematically admissible field of
displacements, we have:
mk  p j 0 u kj dA    ijk ijk dV .
AT
v
Subtracting the above equation from this of “real” state, we get:
mk  m  p j 0 u kj dA    ijk   ij ijk dV .
AT
V
From the Drucker’s stability postulate follows that the integral on the right side is
nonnegative. The integral on the left is positive (positive work of internal forces) and the
multiplier of kinematically admissible displacement field is not less than real (exact) value:
mk  m .
5
Theory of Plasticity – lecture 4, Adam Paul Zaborski
Upper bound theorem:
The structure collapses (becomes a mechanism) or, at least is in limit equilibrium state if for
kinematically admissible field of displacements the total work (power) of external forces is
not less than the work (power) of internal forces.
In other words, if the structure collapses under external load, its bearing capacity is less or
equal to the applied load (upper bound estimation).
Approximate and exact solutions
Comparing the theorems, we have:
ms  m  mk .
If the statically admissible state of stress is associated at the same time with kinematically
admissible field of displacements, the solution is exact to the real value of limit capacity. The
theorems are very attractive because, in most cases, the upper and lower estimations can be
found very easy. Many satisfactory engineering solutions were found for bar structures, plates
and soil.
Example
Find the limit bearing capacity of the truss, where A1  3 cm2, A2  2 cm2, A3  5 cm2,
 0  400 MPa.
3
1
1
3
2
3
5
P
N1
N2
N3
P
Statically indeterminate truss and the node equilibrium
Static approach
We calculate the bearing capacity of the bars:
N 1  120 kN, N 2  80 kN, N 3  200 kN
The structure becomes a mechanism if two bars reach yield point. There are three
possibilities:
a) N1  N 1 , N 2  N 2
from the sum of projections, we have:
3
1
3
 X  0 : 120 34  80 26  N 3 34  N 3  150.5 kN
5
5
5
 Y  0 : P  120 34  80 26  150.5 34  310.4 kN
N
 301 MPa   0 (statically admissible scheme)
we verify  3  3
A3
b) N1  N 1 , N 3  N 3
from the projections sum on x axis, we have:
6
Theory of Plasticity – lecture 4, Adam Paul Zaborski
X  0:
 120
3
1
 N2
 200
34
26
so, the scheme is not statically admissible
3
34
 0  N 2  209.9 kN  N 2
c) N 2  N 2 , N 3  N 3
from the projection on horizontal axis we have:
3
1
3
 X  0 : N1 34  80 26  200 34  0  N1  169.5 kN  N 1
and the scheme is not statically admissible.
Actually, we have only one scheme statically admissible and the lower bound estimation is
310.4 kN.
Kinematic approach
As before, there are three possibilities of changing the structure into mechanism.We verify
only one of them that corresponds to yielding of bars 1 and 2. The system has an instant
centre of rotation at the end of the bar no 3.
N1
N2
N3
α
α
β
Δ
Δ2
γ
Δ1
Kinematic scheme of the mechanism
Comparing the work of external and internal forces, we have:
3
P  Δ  N1  Δ1  N 2  Δ 2  P
 1201  80 2
34
and from geometrical relations follows:
3
5
5
  arccos
 59.04 0 ,   arccos
 30.96 0 ,   arccos
 11.310
34
34
26
1   cos(   )  0.8823,  2   cos(   )  0.6726
so:
P  310.4 kN,
and the result is identical with the static approach result. We have exact solution for the
model.
Example
Find limit bearing capacity of the beam.
P
A
2
B
2P
1
C
D
1
Beam scheme
Static approach
We apply the method of consecutive plastic hinges. The plastic hinges carry on the limit
plastic bending moments, M , which direction correspond with the stretched fibers. To
7
Theory of Plasticity – lecture 4, Adam Paul Zaborski
determine the section of the first hinge we have to have the diagram of bending moments. Due
to its form of linear segments we consider three possible sections only.
a) For the hinge in the section A, we get the statically
determined beam
M
P
A
B
2
2P
1
RA  P 
D
C
1
M
M
, RD  2 P 
4
4
M
M
, M C  2P 
2
4
Because of MB < MC, we assume the next plastic hinge in section C. The limit load and the
bending moment in the section B will be:
M B  2P 
P  85 M , M B  34 M  M .
The scheme is statically admissible because the bending moment MB is less than plastic limit
moment.
b) For the first plastic hinge at the section B, we have:
M
A
P
RD  P  M 2 , RB  P  M 2 , M A  2M  4P ,
2P
B
D
C
2
1
1
MC  P  M 2 .
For two possibilities for second hinge we get, at the section A:
P  14 M , M C  34 M  M (admissible)
and at the section C:
P
A
M
B
2
P  M 2 , and M A  2M  M (not admissible).
c) For the first hinge at the section C, we have:
2P
D
C
1
1
RD  M , RC  M , M B  2M  2P , M A  4M  8P .
The next hinge at the section A, we have: P  85 M ,
M B  34 M  M ,
and for the next hinge at the section B, we get: P  12 M , M A  0  M .
From these lower bound estimations we chose the biggest value. So, the limit plastic bearing
capacity of the beam is equal to the maximum of lower bound estimations.
P  max( 85 M , 14 M , 12 M )  85 M .
Kinematical approach
The most probable sections of plastic hinges are the sections at the intervals ends. To obtain
the kinematic mechanism with one degree of freedom (DOF) we need two plastic hinges.
Only three sections are involved and, consequently, three kinematic schemes.


P
2P
P
2
2P
3
P

4
2P


2
8
Theory of Plasticity – lecture 4, Adam Paul Zaborski
Comparing the external and internal work, we get:
2P  2 P  3M   P1  0.75M
2P  2 P3  5M   P 2  0.625M
2 P  3M   P 3  1.5M
The kinematic approach is the upper bound estimation, so we chose the smallest value of
estimation (the beam collapses under the force P1 , P 2 as well as P 3 ), so the best value is
P  0.625M . The same result we got from the static approach, so, the result is exact.
Example
Find the limit bearing capacity of the statically undetermined beam.
q

1
l
b
+1
l-b
Beam and the kinematic scheme of collapse
Similarly as before, two plastic hinges will be necessary to create the kinematical mechanism
with one DOF. One hinge will be at the fixed end but the position of the second hinge is
unknown. We assume hypothetically the second hinge in the middle of the span,
b  12 l , 1   . We get:
0.5l
2  qxdx  3 M
 q  12
0
M
.
l2
From the static approach the hypothesis give us:
q
RB
l
RC
Static scheme of the beam
The reaction in the upper beam is:
ql
RB   M
4
and the bending moment at the fixed end of the lower beam is
ql 2
Mu 
 2M
4
and with the second hinge at the fixed and, we have:
M
M u  M  q  12 2 ,
l
so, the same value as from the kinematic approach.. Seemingly, the solution looks to be exact,
but it is not the case. When we calculate the reaction for the limit value found, we get:
M
M M
2

 0,
l
l
l
and this signifies that in the on the right of the hinge the shear force change the sign and has a
zero-value point:
RB  3
9
Theory of Plasticity – lecture 4, Adam Paul Zaborski
ql M
M
M

6
 5 .
2
l
l
l
It means that there is extreme bending moment in the right span of the beam. This extreme
value must be greater than the value at the hinge due to convexity of the bending moment
diagram. The value exceeds the limit bending moment and the scheme is statically not
admissible.
Let’s change the sequence of the hinges. For the plastic hinge at the fixe end, we have:
RB 
ql M

2
l
and the shear force in the span:
Q( x)  RA  qx .
From the condition of zero-value of the shear force we get the position of the second hinge:
R
Q ( x)  0  xextr  A
q
and extreme value of the moment will be:
2
qx
R2
M ( xextr )  R A xextr  M  extr  A  M
q
2q
Assuming the second hinge created we find the limit load of the beam:
RB 
2
2
q l2
M
M ( xextr )  M 
 3q M  2  0
4
l
We change the equation introducing a new variable:

ql 2

1
4
 2  3  1  0  1  0.343,  2  11.66 .
M
Because the static approach gives the lower bound estimation we take the second core and the
corresponding limit load value:
M
.
l2
Similarly, we find from the kinematical scheme the exact value of load capacity. The position
of the hinge at the span we find from the principle of virtual work:
q  11.66
b
l b
0
0
 qxdx 
 q x dx
1 1
1
 M 2  1 
and, from the figure is:
b
1 
,
l b
we have:
2M 2l  b
.
bl l  b
We seek the extreme (minimal) value of load capacity:
q
min q

 0  b 2  4bl  2l 2  0  b  l 2  2  0.59l ,
b
q


M
M
 11.66 2
2
l
64 2 l
Both solutions are identical.
and finally: q 
4
10
Theory of Plasticity – lecture 4, Adam Paul Zaborski
Example
Find the limit load of the frame below.
q
P=ql
l
2l
Portal frame
1. Kinematically admissible schemes of collapse
We verify 3 schemes of collapse: beam type, frame type and mixed:
θ
θ
θ
2θ
θ
θ
θ
θ
2θ
θ
Kinematical schemes of collapse
 beam scheme
l
2  xqdx  4M
 q4
0
M
l2
 frame scheme
ql l  4M
 q4
 mixed scheme
M
l2
l
M
l2
o
We get upper bound estimation for the smallest value from the mixed scheme.:
ql l  2   xqdx  M  2 M  2M  M
 q3
M
l2
2. We check is the mixed scheme statically admissible?
We calculate:
q  3
HA  
M
l
ql M

2
l
the shear force at spandrel beam from the left:
RA 
QR  R A  ql 
M ql
M


l
2
2l
11
Theory of Plasticity – lecture 4, Adam Paul Zaborski
q
M
ql
B
M
HA
VA
Calculation scheme
The shear force changes the sign, the extreme value of the bending moment exceeds
admissible limit value. The scheme is not admissible.
We look for the hinged section at the spandrel beam.
q
M
ql
M
M
HA
VA
Calculation scheme
we calculate:
M ql
2M
 , HA 
 ql
l
2
l
and the shear force in the spandrel beam is:
RA 
Q( x) 
M ql
  qx  0
l
2
so:
x 
M
ql

l
2
and:
M ( x)  R A x  H A l  M  12 qx 2
and in the same time
M ( x)  M
so, after the transformations, we have:
2
9 2
M M 
q  7q 2   2   0
l 
4
l
 
and finally:
M
.
l2
3. We verify the solution by kinematic approach, assuming the kinematic scheme of collapse
with the hinge at the spandrel beam is located at a, to the left from the middle:
q  2.96
12
Theory of Plasticity – lecture 4, Adam Paul Zaborski
l a
l a
0
0
q l    qxdx 
2
l a
l a
  l  a qxdx  4M  2M l  a
after simple transformations, we have:
2M
3l  a
l 2l  a l  a 
We calculate the extreme:
q
 0  2l  a l  a   3l  a l  2a   0
a
we get the equation:
a 2  6al  l 2  0
with the core:
a  0.162l
and finally:
q
2M
3  0.162
M
 2.96 2 .
2
l 2  0.1621  0.162
l
The result is the same as from the static approach..
q
Example
Find the limit load capacity of the beam with variable cross-section capacity: 2M from the
left and M from the right.
P
1
θ
0.4l
0.2l
0.4l
2
θ
Beam with variable stiffness and collapse schemes
For the kinematical schemes we have:
M
l
1) 1  23  ,
2M (  53 )  P0.4l  P  13.33
2) 1  1,5 ,
2M   2.5M   P0.4l  P  11.25
M
l
The lower value is the solution from the kinematic approach.
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