CE 579_Lecture3_based on the w chin give more explanation Jan24.ppt

About structure stability

1. CE 579: STRUCTRAL STABILITY AND DESIGN
Amit H. Varma
Assistant Professor
School of Civil Engineering
Purdue University
Ph. No. [phone]
Email: [email protected]
Office hours: M-T-Th 10:30-11:30 a.m.

2. Chapter 1. Introduction to Structural Stability
OUTLINE

Definition of stability

Types of instability

Methods of stability analyses

Bifurcation analysis examples – small deflection analyses

Energy method

Examples – small deflection analyses

Examples – large deflection analyses

Examples – imperfect systems

Design of steel structures

3. ENERGY METHOD

We will currently look at the use of the energy method for an
elastic system subjected to conservative forces.

Total potential energy of the system –  – depends on the work
done by the external forces (W
e) and the strain energy stored in
the system (U).
=U - W
e.

For the system to be in equilibrium, its total potential energy 
must be stationary. That is, the first derivative of  must be
equal to zero.

Investigate higher order derivatives of the total potential energy
to examine the stability of the equilibrium state, i.e., whether the
equilibrium is stable or unstable

4. ENERGY METHD

The energy method is the best for establishing the equilibrium
equation and examining its stability

The deformations can be small or large.

The system can have imperfections.

It provides information regarding the post-buckling path if large
deformations are assumed

The major limitation is that it requires the assumption of the
deformation state, and it should include all possible degrees of
freedom.

5. ENERGY METHOD

Example 1 – Rigid bar supported by rotational spring

Assume small deflection theory
Step 1 - Assume a deformed shape that activates all possible d.o.f.
Rigid bar subjected to axial force P
Rotationally restrained at end
P
k
L

L P
L cos
L (1-cos)
k

6. ENERGY METHOD – SMALL DEFLECTIONS

Write the equation representing the total potential energy of system
L (1-cos)

L P
L cos
k
L sin
L
k
PTherefore
LPksdeflectionsmallFor
LPkTherefore
d
d
mequilibriuFor
LPk
d
d
LPk
LPW
kU
WU
cr
e
e











,
0;
0sin,
0;
sin
)cos1(
2
1
)cos1(
2
1
2
2









7. ENERGY METHOD – SMALL DEFLECTIONS

The energy method predicts that buckling will occur at the same load
P
cr
as the bifurcation analysis method.
At P
cr, the system will be in equilibrium in the deformed.

Examine the stability by considering further derivatives of the total
potential energy

This is a small deflection analysis. Hence  will be  zero.

In this type of analysis, the further derivatives of  examine the stability of
the initial state-1 (when  =0)
PLk
d
d
LPkLPk
d
d
LPk





2
2
2
sin
)cos1(
2
1




sureNot
d
d
PPWhen
mequilibriuUnstable
d
d
PPWhen
mequilibriuStable
d
d
PPWhen
cr
cr
cr









0
0
0
2
2
2
2
2
2




8. ENERGY METHOD – SMALL DEFLECTIONS
In state-1, stable when P<P
cr
, unstable when P>P
cr

No idea about state during buckling.

No idea about post-buckling equilibrium path or its stability.
P
cr

P
Stable
Unstable
Indeterminate

9. ENERGY METHOD – LARGE DEFLECTIONS

Example 1 – Large deflection analysis (rigid bar with rotational spring)
L (1-cos)

L P
L cos
k
L sin
abovegivenisiprelationshPbucklingpostThe
mequilibriufor
L
k
PTherefore
LPkTherefore
d
d
mequilibriuFor
LPk
d
d
LPk
LPW
kU
WU
e
e





















sin
,
0sin,
0;
sin
)cos1(
2
1
)cos1(
2
1
2
2

10. ENERGY METHOD – LARGE DEFLECTIONS

Large deflection analysis

See the post-buckling load-displacement path shown below
The load carrying capacity increases after buckling at P
cr
P
cr is where   0
Rigid bar with rotational spring
0
0.2
0.4
0.6
0.8
1
1.2
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
End rotation
q
L
o
a
d

P
/
P
c
r

q
0=0




sin
sin


crP
P
mequilibriufor
L
k
P

11. ENERGY METHOD – LARGE DEFLECTIONS

Large deflection analysis – Examine the stability of equilibrium using
higher order derivatives of 
00,
).,.(0
)
tan
1(
cos
sin
sin
,
cos
sin
)cos1(
2
1
2
2
2
2
2
2
2
2
2
2
2




































for
d
d
But
STABLEAlways
ofvaluesalleiAlways
d
d
k
d
d
L
L
k
k
d
d
L
k
PBut
LPk
d
d
LPk
d
d
LPk

12. ENERGY METHOD – LARGE DEFLECTIONS

At  =0, the second derivative of =0. Therefore, inconclusive.

Consider the Taylor series expansion of  at =0

Determine the first non-zero term of ,
Since the first non-zero term is > 0, the state is stable at P=P
cr and =0
n
n
n
d
d
nd
d
d
d
d
d
d
d












0
4
0
4
4
3
0
3
3
2
0
2
2
0
0
!
1
.....
!4
1
!3
1
!2
1





















cos
sin
cos
sin
)cos1(
2
1
4
4
3
3
2
2
2
LP
d
d
LP
d
d
LPk
d
d
LPk
d
d
LPk









kPLLP
d
d
LP
d
d
d
d
d
d

























cos
0sin
0
0
0
0
4
4
0
3
3
0
2
2
0
0
0
24
1
!4
1
44
0
4
4







k
d
d

13. ENERGY METHOD – LARGE DEFLECTIONS
Rigid bar with rotational spring
0
0.2
0.4
0.6
0.8
1
1.2
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
End rotation
q
L
o
a
d

P
/
P
c
r

q
0=0
STABLE
STABLE
STABLE

14. ENERGY METHOD – IMPERFECT SYSTEMS

Consider example 1 – but as a system with imperfections
The initial imperfection given by the angle 
0 as shown below

The free body diagram of the deformed system is shown below
P
k L
0
L cos(
0
)
L (cos
0
-cos)
L P
L cos
k(

 L sin


0

15. ENERGY METHOD – IMPERFECT SYSTEMS
abovegivenisiprelationshPmequilibriuThe
mequilibriufor
L
k
PTherefore
LPkTherefore
d
d
mequilibriuFor
LPk
d
d
LPk
LPW
kU
WU
e
e






















sin
)(
,
0sin)(,
0;
sin)(
)cos(cos)(
2
1
)cos(cos
)(
2
1
0
0
0
0
2
0
0
2
0
L (cos
0
-cos)
L P
L cos
k(

 L sin


0

16. Rigid bar with rotational spring
0
0.2
0.4
0.6
0.8
1
1.2
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
End rotation
q
L
o
a
d

P
/
P
c
r

q
0=0
q
0=0.05
q
0=0.1
q
0=0.2
q
0=0.3
ENERGY METHOD – IMPERFECT SYSTEMS
:
sinsin
)(
0
00
belowshownofvaluesdifferentforipsrelationshP
P
P
L
k
P
cr











17. ENERGY METHODS – IMPERFECT SYSTEMS

As shown in the figure, deflection starts as soon as loads are
applied. There is no bifurcation of load-deformation path for
imperfect systems. The load-deformation path remains in the
same state through-out.

The smaller the imperfection magnitude, the close the load-
deformation paths to the perfect system load –deformation path

The magnitude of load, is influenced significantly by the
imperfection magnitude.

All real systems have imperfections. They may be very small but
will be there

The magnitude of imperfection is not easy to know or guess.
Hence if a perfect system analysis is done, the results will be
close for an imperfect system with small imperfections

18. ENERGY METHODS – IMPERFECT SYSTEMS

Examine the stability of the imperfect system using higher order
derivatives of 

Which is always true, hence always in STABLE EQUILIBRIUM











tan.,.
cossin
)(
.,.
cos
.,.
0cos.,.
0
cos
sin)(
)cos(cos)(
2
1
0
0
2
2
2
2
0
0
2
0













ei
L
k
L
k
ifei
L
k
Pifei
LPkifei
d
d
if
stablebewillpathmEquilibriu
LPk
d
d
LPk
d
d
LPk

19. ENERGY METHOD – SMALL DEFLECTIONS
P
k
L
P
L

L (1-cos)
L cos
L sin
k L sin
O
Example 2 - Rigid bar supported by translational spring at end
Assume deformed state that activates all possible d.o.f.
Draw FBD in the deformed state

20. ENERGY METHOD – SMALL DEFLECTIONS
Write the equation representing the total potential energy of system
P
L

L (1-cos)
L cos
L sin
k L sin
O
LkPTherefore
LPLksdeflectionsmallFor
LPLkTherefore
d
d
mequilibriuFor
LPLk
d
d
LPLk
LPW
LkLkU
WU
cr
e
e











,
0;
0sin,
0;
sin
)cos1(
2
1
)cos1(
2
1
)sin(
2
1
2
2
2
22
222









21. ENERGY METHOD – SMALL DEFLECTIONS

The energy method predicts that buckling will occur at the same
load P
cr
as the bifurcation analysis method.
At P
cr, the system will be in equilibrium in the deformed.
Examine the stability by considering further derivatives of the
total potential energy

This is a small deflection analysis. Hence  will be  zero.

In this type of analysis, the further derivatives of  examine the
stability of the initial state-1 (when  =0)
LPLk
d
d
andsdeflectionsmallFor
LPLk
d
d
LPLk
d
d
LPLk








2
2
2
2
2
2
2
22
0
cos
sin
)cos1(
2
1







ATEINDETERMIN
d
d
kLPWhen
UNSTABLE
d
d
LkPWhen
STABLE
d
d
LkPWhen









0
0,
0,
2
2
2
2
2
2




22. ENERGY METHOD – LARGE DEFLECTIONS
abovegivenisiprelationshPbucklingpostThe
mequilibriuforLkPTherefore
LPLkTherefore
d
d
mequilibriuFor
LPLk
d
d
LPLk
LPW
LkU
WU
e
e




















cos,
0sincossin,
0;
sincossin
)cos1(sin
2
1
)cos1(
)sin(
2
1
2
2
22
2
P
L

L (1-cos)
L cos
L sin
O
Write the equation representing the total potential energy of system

23. ENERGY METHOD – LARGE DEFLECTIONS

Large deflection analysis

See the post-buckling load-displacement path shown below
The load carrying capacity decreases after buckling at P
cr
P
cr
is where   0
Rigid bar with translational spring
0
0.2
0.4
0.6
0.8
1
1.2
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
End rotation
q
L
o
a
d

P
/
P
c
r



cos
cos


cr
P
P
mequilibriuforLkP

24. ENERGY METHOD – LARGE DEFLECTIONS

Large deflection analysis – Examine the stability of equilibrium using
higher order derivatives of 
UNSTABLEHENCEALWAYS
d
d
Lk
d
d
LkLk
d
d
LkLk
d
d
LkPmequilibriuFor
LPLk
d
d
LPLk
d
d
LPLk
.0
sin
cos)sin(cos
cos2cos
cos
cos2cos
sincossin
)cos1(sin
2
1
2
2
22
2
2
22222
2
2
222
2
2
2
2
2
2
22
































25. ENERGY METHOD – LARGE DEFLECTIONS

At  =0, the second derivative of =0. Therefore, inconclusive.

Consider the Taylor series expansion of  at =0

Determine the first non-zero term of ,
Since the first non-zero term is < 0, the state is unstable at P=P
cr and =0
n
n
n
d
d
nd
d
d
d
d
d
d
d












0
4
0
4
4
3
0
3
3
2
0
2
2
0
0
!
1
.....
!4
1
!3
1
!2
1












0sin2sin2
0cos2cos
0sin2sin
2
1
0)cos1(sin
2
1
2
3
3
2
2
2
2
22














LPLk
d
d
LPLk
d
d
LPLk
d
d
LPLk
occursbucklingwhenatUNSTABLE
d
d
LkLkLk
d
d
LPLk
d
d
0
0
34
cos2cos4
4
4
222
4
4
2
4
4















26. ENERGY METHOD – LARGE DEFLECTIONS
Rigid bar with translational spring
0
0.2
0.4
0.6
0.8
1
1.2
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
End rotation
q
L
o
a
d

P
/
P
c
r

UNSTABLE
UNSTABLE
UNSTABLE

27. ENERGY METHOD - IMPERFECTIONS

Consider example 2 – but as a system with imperfections
The initial imperfection given by the angle 
0 as shown below

The free body diagram of the deformed system is shown below
P
k
L cos(
0
)
L
0
P
L

L (cos
0
-cos)
L cos
L sin
O

0
L sin


28. ENERGY METHOD - IMPERFECTIONS
abovegivenisiprelationshPmequilibriuThe
mequilibriuforLkPTherefore
LPLkTherefore
d
d
mequilibriuFor
LPLk
d
d
LPLk
LPW
LkU
WU
e
e






















)
sin
sin
1(cos,
0sincos)sin(sin,
0;
sincos)sin(sin
)cos(cos)sin(sin
2
1
)cos(cos
)sin(sin
2
1
0
0
2
0
2
0
2
0
2
0
2
0
2
P
L

L (cos
0
-cos)
L cos
L sin
O

0
L sin


29. 











3
max
3
2
0
max
00
cos
sinsin0)
sin
sin
sin(0
)
sin
sin
1(cos)
sin
sin
1(cos
LkP
Lk
d
dP
P
P
P
LkP
cr



ENERGY METHOD - IMPERFECTIONS
0
0.2
0.4
0.6
0.8
1
1.2
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
End rotation
q
L
o
a
d

P
/
P
c
r

q
0=0
q
0=0.05
q
0=0.1
q
0=0.2
q
0=0.3
Envelope of peak
loads P
max

30. ENERGY METHOD - IMPERFECTIONS
As shown in the figure, deflection starts as soon as loads are
applied. There is no bifurcation of load-deformation path for
imperfect systems. The load-deformation path remains in the
same state through-out.
The smaller the imperfection magnitude, the close the load-
deformation paths to the perfect system load –deformation path.
The magnitude of load, is influenced significantly by the
imperfection magnitude.
All real systems have imperfections. They may be very small but
will be there
The magnitude of imperfection is not easy to know or guess.
Hence if a perfect system analysis is done, the results will be
close for an imperfect system with small imperfections.
However, for an unstable system – the effects of imperfections
may be too large.