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Buckling and stability analysis
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CE 579: STRUCTRAL STABILITY AND DESIGN
Amit H. Varma
Assistant Professor
School of Civil Engineering
Purdue University
Ph. No. (765) 496 3419
Email: [email protected]
Office hours: M-W-F 9:00-11:30 a.m.
Chapter 1. Introduction to Structural Stability
OUTLINE
Definition of stability
Types of instability
Methods of stability analyses
Examples – small deflection analyses
Examples – large deflection analyses
Examples – imperfect systems
Design of steel structures
STABILITY DEFINITION
Change in geometry of a structure or structural component under compression – resulting in loss of ability to resist loading is defined as instability in the book.
Instability can lead to catastrophic failure must be accounted in design. Instability is a strength-related limit state.
Why did we define instability instead of stability? Seem strange!
Stability is not easy to define. Every structure is in equilibrium – static or dynamic. If it is not in
equilibrium, the body will be in motion or a mechanism. A mechanism cannot resist loads and is of no use to the civil
engineer. Stability qualifies the state of equilibrium of a structure. Whether it
is in stable or unstable equilibrium.
STABILITY DEFINITION
Structure is in stable equilibrium when small perturbations do not cause large movements like a mechanism. Structure vibrates about it equilibrium position.
Structure is in unstable equilibrium when small perturbations produce large movements – and the structure never returns to its original equilibrium position.
Structure is in neutral equilibrium when we cant decide whether it is in stable or unstable equilibrium. Small perturbation cause large movements – but the structure can be brought back to its original equilibrium position with no work.
Thus, stability talks about the equilibrium state of the structure.
The definition of stability had nothing to do with a change in the geometry of the structure under compression – seems strange!
STABILITY DEFINITION
BUCKLING Vs. STABILITY
Change in geometry of structure under compression – that results in its ability to resist loads – called instability.
Not true – this is called buckling.
Buckling is a phenomenon that can occur for structures under compressive loads.
The structure deforms and is in stable equilibrium in state-1. As the load increases, the structure suddenly changes to
deformation state-2 at some critical load Pcr. The structure buckles from state-1 to state-2, where state-2 is
orthogonal (has nothing to do, or independent) with state-1.
What has buckling to do with stability? The question is - Is the equilibrium in state-2 stable or unstable? Usually, state-2 after buckling is either neutral or unstable
equilibrium
BUCKLING
P=Pcr
P
P<Pcr
P P
P>Pcr
P
BUCKLING Vs. STABILITY
Thus, there are two topics we will be interested in this course Buckling – Sudden change in deformation from state-1 to state-2 Stability of equilibrium – As the loads acting on the structure are
increased, when does the equilibrium state become unstable? The equilibrium state becomes unstable due to:
Large deformations of the structure Inelasticity of the structural materials
We will look at both of these topics for Columns Beams Beam-Columns Structural Frames
TYPES OF INSTABILITY
Structure subjected to compressive forces can undergo:
1. Buckling – bifurcation of equilibrium from deformation state-1 to state-2. Bifurcation buckling occurs for columns, beams, and symmetric
frames under gravity loads only
2. Failure due to instability of equilibrium state-1 due to large deformations or material inelasticity Elastic instability occurs for beam-columns, and frames subjected
to gravity and lateral loads. Inelastic instability can occur for all members and the frame.
We will study all of this in this course because we don’t want our designed structure to buckle or fail by instability – both of which are strength limit states.
TYPES OF INSTABILITY
BIFURCATION BUCKLING
Member or structure subjected to loads. As the load is increased, it reaches a critical value where:
The deformation changes suddenly from state-1 to state-2. And, the equilibrium load-deformation path bifurcates.
Critical buckling load when the load-deformation path bifurcates Primary load-deformation path before buckling Secondary load-deformation path post buckling Is the post-buckling path stable or unstable?
SYMMETRIC BIFURCATION
Post-buckling load-deform. paths are symmetric about load axis. If the load capacity increases after buckling then stable symmetric
bifurcation. If the load capacity decreases after buckling then unstable
symmetric bifurcation.
ASYMMETRIC BIFURCATION
Post-buckling behavior that is asymmetric about load axis.
INSTABILITY FAILURE
There is no bifurcation of the load-deformation path. The deformation stays in state-1 throughout
The structure stiffness decreases as the loads are increased. The change is stiffness is due to large deformations and / or material inelasticity.
The structure stiffness decreases to zero and becomes negative. The load capacity is reached when the stiffness becomes zero. Neutral equilibrium when stiffness becomes zero and unstable
equilibrium when stiffness is negative. Structural stability failure – when stiffness becomes negative.
INSTABILITY FAILURE
FAILURE OF BEAM-COLUMNSP
P
M
M
K
K=0
K<0
M
No bifurcation.
Instability due to material and geometric nonlinearity
INSTABILITY FAILURE
Snap-through buckling
P
Snap-through
INSTABILITY FAILURE
Shell Buckling failure – very sensitive to imperfections
Chapter 1. Introduction to Structural Stability
OUTLINE
Definition of stability
Types of instability
Methods of stability analyses
Examples – small deflection analyses
Examples – large deflection analyses
Examples – imperfect systems
Design of steel structures
METHODS OF STABILITY ANALYSES
Bifurcation approach – consists of writing the equation of equilibrium and solving it to determine the onset of buckling.
Energy approach – consists of writing the equation expressing the complete potential energy of the system. Analyzing this total potential energy to establish equilibrium and examine stability of the equilibrium state.
Dynamic approach – consists of writing the equation of dynamic equilibrium of the system. Solving the equation to determine the natural frequency () of the system. Instability corresponds to the reduction of to zero.
STABILITY ANALYSES Each method has its advantages and disadvantages. In fact,
you can use different methods to answer different questions
The bifurcation approach is appropriate for determining the critical buckling load for a (perfect) system subjected to loads.
The deformations are usually assumed to be small. The system must not have any imperfections. It cannot provide any information regarding the post-buckling load-
deformation path.
The energy approach is the best when 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.
STABILITY ANALYSIS
The dynamic method is very powerful, but we will not use it in this class at all.
Remember, it though when you take the course in dynamics or earthquake engineering
In this class, you will learn that the loads acting on a structure change its stiffness. This is significant – you have not seen it before.
What happens when an axial load is acting on the beam. The stiffness will no longer remain 4EI/L and 2EI/L. Instead, it will decrease. The reduced stiffness will reduce the
natural frequency and period elongation. You will see these in your dynamics and earthquake engineering
class.
a
Ma
Mb
bbaa L
IEM
L
IEM 24
P
STABILITY ANALYSIS
FOR ANY KIND OF BUCKLING OR STABILITY ANALYSIS – NEED TO DRAW THE FREE BODY DIAGRAM OF THE DEFORMED STRUCTURE.
WRITE THE EQUATION OF STATIC EQUILIBRIUM IN THE DEFORMED STATE
WRITE THE ENERGY EQUATION IN THE DEFORMED STATE TOO.
THIS IS CENTRAL TO THE TOPIC OF STABILITY ANALYSIS
NO STABILITY ANALYSIS CAN BE PERFORMED IF THE FREE BODY DIAGRAM IS IN THE UNDEFORMED STATE
BIFURCATION ANALYSIS
Always a small deflection analysis
To determine Pcr buckling load
Need to assume buckled shape (state 2) to calculate
Example 1 – Rigid bar supported by rotational spring
Step 1 - Assume a deformed shape that activates all possible d.o.f.
Rigid bar subjected to axial force P
Rotationally restrained at end
Pk
L
L P
L cosL (1-cos)
k
BIFURCATION ANALYSIS
Write the equation of static equilibrium in the deformed state
Thus, the structure will be in static equilibrium in the deformed state when P = Pcr = k/L
When P<Pcr, the structure will not be in the deformed state. The structure will buckle into the deformed state when P=Pcr
L (1-cos)
L P
L cos
k L sin
L
k
L
kP
nsdeformatiosmallFor
L
kP
LPkM
cr
o
sin
sin
0sin0
BIFURCATION ANALYSIS
P
k
L
PL
L (1-cos)
L cos
L sin
k L sinO
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
BIFURCATION ANALYSIS
Write equations of static equilibrium in deformed stateP
L
L (1-cos)
L cos
L sin
k L sinO
LkL
LkP
nsdeformatiosmallFor
L
LkP
LPLLkM
cr
o
2
2
sin
sin
sin
0sin)sin(0
• Thus, the structure will be in static equilibrium in the deformed state when P = Pcr = k L. When P<Pcr, the structure will not be in the deformed state. The structure will buckle into the deformed state when P=Pcr
BIFURCATION ANALYSIS
Example 3 – Three rigid bar system with two rotational springs
PP
A B CD
kk
L L L
Assume deformed state that activates all possible d.o.f.Draw FBD in the deformed state
PP
A D
kk
LL
12
L sin 1
L sin 2
BC1 – 2)
1 – 2)
Assume small deformations. Therefore, sin=
BIFURCATION ANALYSIS
Write equations of static equilibrium in deformed statePP
A D
kk
LL
12
L sin 1
L sin 2
BC1 – 2)
1 – 2)
P
A
L
1
B
L sin 1
1+(1-2)
0)2(0sin)2(0 121121 LPkLPkM B
k(21-2)
P
D
k
L
2L sin 2
C
1 – 2)
k(22-1)
0)2(0sin)2(0 212212 LPkLPkMC
BIFURCATION ANALYSIS
Equations of Static Equilibrium
Therefore either 1 and 2 are equal to zero or the determinant of the coefficient matrix is equal to zero.
When 1 and 2 are not equal to zero – that is when buckling occurs – the coefficient matrix determinant has to be equal to zero for equil.
Take a look at the matrix equation. It is of the form [A] {x}={0}. It can also be rewritten as [K]-[I]){x}={0}
0
0
2
2
2
1
PLkk
kPLk0)2( 121 LPk
0)2( 212 LPk
0
0
10
012
2
2
1
P
L
k
L
kL
k
L
k
BIFURCATION ANALYSIS
This is the classical eigenvalue problem. [K]-[I]){x}={0}.
We are searching for the eigenvalues () of the stiffness matrix [K]. These eigenvalues cause the stiffness matrix to become singular
Singular stiffness matrix means that it has a zero value, which means that the determinant of the matrix is equal to zero.
Smallest value of Pcr will govern. Therefore, Pcr=k/L
L
kor
L
kP
PLkPLk
kPLkkPLk
kPLk
PLkk
kPLk
cr
3
0)()3(
0)2()2(
0)2(
02
2
22
BIFURCATION ANALYSIS
Each eigenvalue or critical buckling load (Pcr) corresponds to a buckling shape that can be determined as follows
Pcr=k/L. Therefore substitute in the equations to determine 1 and 2
All we could find is the relationship between 1 and 2. Not their specific values. Remember that this is a small deflection analysis. So, the values are negligible. What we have found is the buckling shape – not its magnitude.
The buckling mode is such that 1=2 Symmetric buckling mode
21
21
121
121
0
0)2(
0)2(
kk
kkL
kPPLet
LPk
cr
PP
A D
kk
LL
1 2=1
B C
21
21
212
212
0
0)2(
0)2(
kk
kkL
kPPLet
LPk
cr
BIFURCATION ANALYSIS
Second eigenvalue was Pcr=3k/L. Therefore substitute in the equations to determine 1 and 2
All we could find is the relationship between 1 and 2. Not their specific values. Remember that this is a small deflection analysis. So, the values are negligible. What we have found is the buckling shape – not its magnitude.
The buckling mode is such that 1=-2 Antisymmetric buckling mode
21
21
121
121
0
03)2(
3
0)2(
kk
kkL
kPPLet
LPk
cr
PP
A D
kk
L
L
1
2=-1
B
C
21
21
212
212
0
03)2(
3
0)2(
kk
kkL
kPPLet
LPk
cr
BIFURCATION ANALYSIS
Homework No. 1 Problem 1.1 Problem 1.3 Problem 1.4 All problems from the textbook on Stability by W.F. Chen
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
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 (We) and the strain energy stored in the system (U).
=U - We.
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
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.
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
Pk
L
L P
L cosL (1-cos)
k
ENERGY METHOD – SMALL DEFLECTIONS
Write the equation representing the total potential energy of systemL (1-cos)
L P
L cos
k L sin
L
kPTherefore
LPksdeflectionsmallForLPkTherefore
d
dmequilibriuFor
LPkd
d
LPk
LPW
kU
WU
cr
e
e
,
0;0sin,
0;
sin
)cos1(2
1)cos1(
2
1
2
2
ENERGY METHOD – SMALL DEFLECTIONS
The energy method predicts that buckling will occur at the same load Pcr as the bifurcation analysis method.
At Pcr, 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)
PLkd
d
LPkLPkd
d
LPk
2
2
2
sin
)cos1(2
1
sureNotd
dPPWhen
mequilibriuUnstabled
dPPWhen
mequilibriuStabled
dPPWhen
cr
cr
cr
0
0
0
2
2
2
2
2
2
ENERGY METHOD – SMALL DEFLECTIONS
In state-1, stable when P<Pcr, unstable when P>Pcr
No idea about state during buckling.
No idea about post-buckling equilibrium path or its stability.
Pcr
P
Stable
Unstable
Indeterminate
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
mequilibriuforL
kPTherefore
LPkTherefored
dmequilibriuFor
LPkd
d
LPk
LPW
kU
WU
e
e
sin,
0sin,
0;
sin
)cos1(2
1)cos1(
2
1
2
2
ENERGY METHOD – LARGE DEFLECTIONS
Large deflection analysis See the post-buckling load-displacement path shown below The load carrying capacity increases after buckling at Pcr
Pcr is where 0Rigid 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
Lo
ad P
/Pcr
00
sin
sin
crP
P
mequilibriuforL
kP
ENERGY METHOD – LARGE DEFLECTIONS
Large deflection analysis – Examine the stability of equilibrium using higher order derivatives of
00,
).,.(0
)tan
1(
cossin
sin,
cos
sin
)cos1(2
1
2
2
2
2
2
2
2
2
2
2
2
ford
dBut
STABLEAlways
ofvaluesalleiAlwaysd
d
kd
d
LL
kk
d
d
L
kPBut
LPkd
d
LPkd
d
LPk
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=Pcr and =0
nn
n
d
d
nd
d
d
d
d
d
d
d
0
4
04
43
03
32
02
2
00 !
1.....
!4
1
!3
1
!2
1
cos
sin
cos
sin
)cos1(2
1
4
4
3
3
2
2
2
LPd
d
LPd
d
LPkd
d
LPkd
d
LPk
kPLLPd
d
LPd
d
d
d
d
d
cos
0sin
0
0
0
04
40
3
30
2
20
00
24
1
!4
1 44
04
4
kd
d
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
Lo
ad P
/Pcr
00
STABLE
STABLESTABL
E
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
Pk L0
L cos(0)
L (cos0-cos)
L P
L cos
k(0 L sin
0
ENERGY METHOD – IMPERFECT SYSTEMS
abovegivenisiprelationshPmequilibriuThe
mequilibriuforL
kPTherefore
LPkTherefored
dmequilibriuFor
LPkd
d
LPk
LPW
kU
WU
e
e
sin
)(,
0sin)(,
0;
sin)(
)cos(cos)(2
1
)cos(cos
)(2
1
0
0
0
02
0
0
20
L (cos0-cos)
L P
L cos
k(0 L sin
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
Lo
ad P
/Pcr
00 00.05 00.1 00. 00.3
ENERGY METHOD – IMPERFECT SYSTEMS
:
sinsin
)(
0
00
belowshownofvaluesdifferentforipsrelationshP
P
P
L
kP
cr
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
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
02
0
eiL
k
L
kifei
L
kPifei
LPkifeid
dif
stablebewillpathmEquilibriu
LPkd
d
LPkd
d
LPk
ENERGY METHOD – SMALL DEFLECTIONS
P
k
L
PL
L (1-cos)
L cos
L sin
k L sinO
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
ENERGY METHOD – SMALL DEFLECTIONS
Write the equation representing the total potential energy of system
PL
L (1-cos)
L cos
L sin
k L sinO
LkPTherefore
LPLksdeflectionsmallFor
LPLkTherefore
d
dmequilibriuFor
LPLkd
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
ENERGY METHOD – SMALL DEFLECTIONS
The energy method predicts that buckling will occur at the same load Pcr as the bifurcation analysis method.
At Pcr, 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)
LPLkd
d
andsdeflectionsmallFor
LPLkd
d
LPLkd
d
LPLk
22
2
22
2
2
22
0
cos
sin
)cos1(2
1
ATEINDETERMINd
dkLPWhen
UNSTABLEd
dLkPWhen
STABLEd
dLkPWhen
0
0,
0,
2
2
2
2
2
2
ENERGY METHOD – LARGE DEFLECTIONS
abovegivenisiprelationshPbucklingpostThe
mequilibriuforLkPTherefore
LPLkTherefore
d
dmequilibriuFor
LPLkd
d
LPLk
LPW
LkU
WU
e
e
cos,
0sincossin,
0;
sincossin
)cos1(sin2
1
)cos1(
)sin(2
1
2
2
22
2
PL
L (1-cos)
L cos
L sin
O
Write the equation representing the total potential energy of system
ENERGY METHOD – LARGE DEFLECTIONS
Large deflection analysis See the post-buckling load-displacement path shown below The load carrying capacity decreases after buckling at Pcr
Pcr is where 0Rigid 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
Lo
ad P
/Pcr
cos
cos
crP
P
mequilibriuforLkP
ENERGY METHOD – LARGE DEFLECTIONS
Large deflection analysis – Examine the stability of equilibrium using higher order derivatives of
UNSTABLEHENCEALWAYSd
d
Lkd
d
LkLkd
d
LkLkd
d
LkPmequilibriuFor
LPLkd
d
LPLkd
d
LPLk
.0
sin
cos)sin(cos
cos2cos
cos
cos2cos
sincossin
)cos1(sin2
1
2
2
222
2
222222
2
2222
2
22
2
2
22
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=Pcr and =0
nn
n
d
d
nd
d
d
d
d
d
d
d
0
4
04
43
03
32
02
2
00 !
1.....
!4
1
!3
1
!2
1
0sin2sin2
0cos2cos
0sin2sin2
1
0)cos1(sin2
1
23
3
22
2
2
22
LPLkd
d
LPLkd
d
LPLkd
d
LPLk
occursbucklingwhenatUNSTABLEd
d
LkLkLkd
d
LPLkd
d
0
0
34
cos2cos4
4
4
2224
4
24
4
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
Lo
ad P
/Pcr
UNSTABLE
UNSTABLE
UNSTABLE
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)
L0
PL
L (cos0-cos)
L cos
L sin
O0
L sin0
ENERGY METHOD - IMPERFECTIONS
abovegivenisiprelationshPmequilibriuThe
mequilibriuforLkPTherefore
LPLkTherefore
d
dmequilibriuFor
LPLkd
d
LPLk
LPW
LkU
WU
e
e
)sin
sin1(cos,
0sincos)sin(sin,
0;
sincos)sin(sin
)cos(cos)sin(sin2
1
)cos(cos
)sin(sin2
1
0
02
02
02
02
0
20
2
PL
L (cos0-cos)
L cos
L sin
O0
L sin0
3max
302
0max
00
cos
sinsin0)sin
sinsin(0
)sin
sin1(cos)
sin
sin1(cos
LkP
Lkd
dPP
P
PLkP
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
Lo
ad P
/Pc
r
00 00.05 00.1 00. 00.3
Envelope of peak loads Pmax
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.
ENERGY METHODS – IMPERFECT SYSTEMS
Examine the stability of the imperfect system using higher order derivatives of
sin
sin1
cos)sinsin2(cos
sincos)sin(sin
)cos(cos)sin(sin2
1
0
02
2
2
02
02
02
LkPmequilibriuFor
LPLkd
d
LPLkd
d
LPLk
sin
sinsin
sin
)cos(sinsinsin
sin
cossinsinsinsin
sin
cossincossinsinsincos
cossin
sin1)sinsin2(cos
03
22
2
220
32
2
2
20
022
2
2
202
0222
2
2
2020
22
2
Lkd
d
Lkd
d
Lkd
d
Lkd
d
LkLkd
d
0sin
sinsinsinsin
sin1sin
sin1
cossin
sin1
cos)sin
sin1(cos
cos)sin
sin1(cos
302
2
23
0
20
20
30
max
3max
0
Lkd
dand
LkLk
PPWhen
LkPandLkP
ENERGY METHOD – IMPERFECT SYSTEMS
UnstablePPwhend
d
StablePPwhend
d
Lkd
d
max2
2
max2
2
03
22
2
0
0
sin
sinsin
0sin
sinsinsinsin
sin1sin
sin1
cossin
sin1
cos)sin
sin1(cos
302
2
23
0
20
20
30
max
Lkd
dand
LkLk
PPWhen
Chapter 2. – Second-Order Differential Equations
This chapter focuses on deriving second-order differential equations governing the behavior of elastic members
2.1 – First order differential equations
2.2 – Second-order differential equations
2.1 First-Order Differential Equations
Governing the behavior of structural members Elastic, Homogenous, and Isotropic Strains and deformations are really small – small deflection theory Equations of equilibrium in undeformed state
Consider the behavior of a beam subjected to bending and axial forces
2.1 First-Order Differential Equations
Assume tensile forces are positive and moments are positive according to the right-hand rule
Longitudinal stress due to bending
This is true when the x-y axis system is
a centroidal and principal axis system.
xI
My
I
M
A
P
y
y
x
x x
I
My
I
M
A
P
y
y
x
x
inertiaofmomentprincipalareIandI
IdAyIdAxAdA
axisCentroidaldAyxdAxdAy
yx
x
A
y
AA
AAA
22 ;;
0
2.1 First-Order Differential Equations
The corresponding strain is
If P=My=0, then
Plane-sections remain plane and perpendicular to centroidal axis before and after bending
The measure of bending is curvature which denotes the change in the slope of the centroidal axis between two point dz apart
xIE
My
IE
M
EA
P
y
y
x
x
yIE
M
x
x
xyyyxx
x
xy
y
yy
y
IEMsimilarlyandIEM
IE
M
y
nsdeformatiosmallFor
y
tan
tan
2.1 First-Order Differential Equations
Shear Stresses due to bending
s
Oy
x
s
Ox
y
dstxI
Vt
dstyI
Vt
2.1 First-Order Differential Equations
Differential equations of bending
Assume principle of superposition Treat forces and deformations in y-z and x-z
plane seperately Both the end shears and qy act in a plane
parallel to the y-z plane through the shear center S
yyx
yyx
yx
yx
yy
qIE
qdz
IEd
qdz
Md
Vdz
dM
qdz
dV
2
2
2
2
)(
2.1 First-Order Differential Equations
Differential equations of bending
Fourth-order differential equations using first-order force-deformation theory
directionypositiveindeflectionv
directionxpositiveindeflectionu
quIESimilarly
qvIE
v
sdeflectionsmallFor
v
v
qIE
xiv
y
yiv
x
y
y
yyx
2/32)(1
Torsion behavior – Pure and Warping Torsion
Torsion behavior – uncoupled from bending behavior
Thin walled open cross-section subjected to torsional moment This moment will cause twisting and warping of the cross-section. The cross-section will undergo pure and warping torsion behavior. Pure torsion will produce only shear stresses in the section Warping torsion will produce both longitudinal and shear stresses The internal moment produced by the pure torsion response will be
equal to Msv and the internal moment produced by the warping torsion response will be equal to Mw.
The external moment will be equilibriated by the produced internal moments
MZ=MSV + MW
Pure and Warping Torsion
MZ=MSV + MW
Where,
MSV = G KT ′ and MW = - E Iw "‘
MSV = Pure or Saint Venant’s torsion moment
KT = J = Torsional constant =
is the angle of twist of the cross-section. It is a function of z.
IW is the warping moment of inertia of the cross-section. This is a new cross-sectional property you may not have seen before.
MZ = G KT ′ - E Iw "‘ ……… (3), differential equation of torsion
Lets look closely at pure or Saint Venant’s torsion. This occurs when the warping of the cross-section is unrestrained or absent
For a circular cross-section – warping is absent. For thin-walled open cross-sections, warping will occur.
The out of plane warping deformation w can be calculated using an equation I will not show.
Pure Torsion Differential Equation
A
T
TSV
AA
SV
dArJKwhere
KGM
dArGdArM
rG
rdz
dr
drdz
2
2
,
The torsional shear stresses vary linearly about the center of the thin plate
Pure Torsion Stresses
tG
rG
SV
SV
max
sv
Warping deformations
The warping produced by pure torsion can be restrained by the: (a) end conditions, or (b) variation in the applied torsional moment (non-uniform moment)
The restraint to out-of-plane warping deformations will produce longitudinal stresses (w) , and their variation along the length will produce warping shear stresses (w) .
Warping Torsion Differential Equation
propertytionnewinertiaofmomentwarpingisIwhere
IEM
hIEM
huIEM
hVM
VuIE
bendingofedborrowingMuIE
flangetheinforceShearV
flangetheinmomentM
ntdisplacemelateralflangeuwhere
hu
W
WW
fW
ffW
fW
fff
fff
f
f
f
f
sec
2
..
2
2
Lets take a look at an approximate derivation of the warping torsion differential equation.
This is valid only for I and C shaped sections.
Torsion Differential Equation Solution
Torsion differential equation MZ=MSV+MW = G KT ’- E IW ’’’
This differential equation is for the case of concentrated torque
Torsion differential equation for the case of distributed torque
The coefficients C1 .... C6 can be obtained using end conditions
W
Z
W
Z
W
T
ZwT
IE
M
IE
M
IE
KG
MIEKG
2
W
Ziv
W
Z
W
Tiv
Ziv
wT
ZZ
IE
m
IE
m
IE
KG
mIEKG
dz
dMm
2
W
z
IE
zMzCzCC
2321 sinhcosh
T
z
KG
zmzCzCzCC
2sinhcosh
2
7654
Torsion Differential Equation Solution
Torsionally fixed end conditions are given by
These imply that twisting and warping at the fixed end are fully restrained. Therefore, equal to zero.
Torsionally pinned or simply-supported end conditions given by:
These imply that at the pinned end twisting is fully restrained (=0) and warping is unrestrained or free. Therefore, W=0 ’’=0
Torsionally free end conditions given by ’=’’ = ’’’= 0
These imply that at the free end, the section is free to warp and there are no warping normal or shear stresses.
Results for various torsional loading conditions given in the AISC Design Guide 9 – can be obtained from my private site
0
0
Warping Torsion Stresses
Restraint to warping produces longitudinal and shear stresses
The variation of these stresses over the section is defined by the section property Wn and Sw
The variation of these stresses along the length of the beam is defined by the derivatives of .
Note that a major difference between bending and torsional behavior is
The stress variation along length for torsion is defined by derivatives of , which cannot be obtained using force equilibrium.
The stress variation along length for bending is defined by derivatives of v, which can be obtained using force equilibrium (M, V diagrams).
opertySectionMomentStaticalWarpingS
opertySectionWarpingUnitNormalizedW
where
SEt
WE
W
n
WW
nW
Pr
Pr
,
Torsional Stresses
Torsional Stresses
Torsional Section Properties for I and C Shapes
and derivatives for concentrated torque at midspan
Summary of first order differential equations
)3(
)2(
)1(
zWT
yy
xx
MIEKG
MuIE
MvIE
NOTES:
(1) Three uncoupled differential equations
(2) Elastic material – first order force-deformation theory
(3) Small deflections only
(4) Assumes – no influence of one force on other deformations
(5) Equations of equilibrium in the undeformed state.
HOMEWORK # 3
Consider the 22 ft. long simply-supported W18x65 wide flange beam shown in Figure 1 below. It is subjected to a uniformly distributed load of 1k/ft that is placed with an eccentricity of 3 in. with respect to the centroid (and shear center).
At the mid-span and the end support cross-sections, calculate the magnitude and distribution of:
Normal and shear stresses due to bending Shear stresses due to pure torsion Warping normal and shear stresses over the cross-section.
Provide sketches and tables of the individual normal and shear stress distributions for each case.
Superimpose the bending and torsional stress-states to determine the magnitude and location of maximum stresses.
HOMEWORK # 2
22 ft.
W18x65
3in.
Cross-section
Span
Chapter 2. – Second-Order Differential Equations
This chapter focuses on deriving second-order differential equations governing the behavior of elastic members
2.1 – First order differential equations
2.2 – Second-order differential equations
2.2 Second-Order Differential Equations
Governing the behavior of structural members Elastic, Homogenous, and Isotropic Strains and deformations are really small – small deflection theory Equations of equilibrium in deformed state The deformations and internal forces are no longer independent.
They must be combined to consider effects.
Consider the behavior of a member subjected to combined axial forces and bending moments at the ends. No torsional forces are applied explicitly – because that is very rare for CE structures.
Member model and loading conditions
Member is initially straight and prismatic. It has a thin-walled open cross-section
Member ends are pinned and prevented from translation.
The forces are applied only at the member ends
These consist only of axial and bending moment forces P, MTX, MTY, MBX, MBY
Assume elastic behavior with small deflections
Right-hand rule for positive moments and reactions and P assumed positive.
Member displacements (cross-sectional)
Consider the middle line of thin-walled cross-section
x and y are principal coordinates through centroid C
Q is any point on the middle line. It has coordinates (x, y).
Shear center S coordinates are (xo, y0)
Shear center S displacements are u, v, and
Member displacements (cross-sectional)
Displacements of Q are: uQ = u + a sin
vQ = v – a cos
where a is the distance from Q to S
But, sin = (y0-y) / a
cos = (x0-x) / a
Therefore, displacements of Q are: uQ = u + (y0-y)
vQ = v – (x0 – x)
Displacements of centroid C are:uc = u + (y0)
vc = v - (x0)
Internal forces – second-order effects
Consider the free body diagrams of the member in the deformed state.
Look at the deformed state in the x-z and y-z planes in this Figure.
The internal resisting moment at a distance z from the lower end are:
Mx = - MBX + Ry z + P vc
My = - MBY + Rx z - P uc
The end reactions Rx and Ry are:
Rx = (MTY + MBY) / L
Ry = (MTX + MBX) / L
Therefore,
Internal forces – second-order effects
0BYTYBYy
0BXTXBXx
yuPMML
zMM
xvPMML
zMM
Internal forces in the deformed state
x
y
z
uc
vc
In the deformed state, the cross-section is such that the principal coordinate systems are changed from x-y-z to the system
x
y
z
uc
vc
P
RxRy
MBx
MBY
P
Mx
My
z
Ry
Rx
Mξ
Mη
Mς σ+dσ
σ
a
Internal forces in the deformed state
The internal forces Mx and My must be transformed to these new axes
Since the angle is small
MMx + My
M = My – Mx
BXTXBX0BYTYBY
BYTYBY0BXTXBX
0BYTYBYy
0BXTXBXx
MML
zMyPuPMM
L
zMM
MML
zMxPvPMM
L
zMM
yuPMML
zMM
xvPMML
zMM
Twisting component of internal forces
Twisting moments M are produced by the internal and external forces
There are four components contributing to the total M
(1) Contribution from Mx and My – M1
(2) Contribution from axial force P – M
(3) Contribution from normal stress – M3
(4) Contribution from end reactions Rx and Ry – M
The total twisting moment M = M1 + M + M3 + M
Twisting component – 1 of 4
Twisting moment due to Mx & My
M1 = Mx sin (du/dz) + Mysin (dv/dz)
Therefore, due to small angles, M1 = Mx du/dz + My dv/dz
M1 = Mx u’ + My v’
uv
Twisting component – 2 of 4
The axial load P acts along the original vertical direction
In the deformed state of the member, the longitudinal axis is not vertical. Hence P will have components producing shears.
These components will act at the centroid where P acts and will have values as shown above – assuming small angles
u v
Twisting component – 2 of 4
These shears will act at the centroid C, which is eccentric with respect to the shear center S. Therefore, they will produce secondary twisting.
M = P (y0 du/dz – x0 dv/dz)
Therefore, M = P (y0 u’ – x0 v’)
Twisting component – 3 of 4
The end reactions (shears) Rx and Ry act at the shear center S at the ends. But, along the member ends, the shear center will move by u, v, and .
Hence, these reactions will also have a twisting effect produced by their eccentricity with respect to the shear center S.
M + Ry u + Rx v = 0
Therefore,
M = – (MTY + MBY) v/L – (MTX + MBX) u/L
Twisting component – 4 of 4
Wagner’s effect or contribution – complicated.
Two cross-sections that are d apart will warp with respect to each other.
The stress element dA will become inclined by angle (a d/d with respect to d axis.
Twist produced by each stress element about S is equal to
dAad
dM
d
dadAadM
A
23
3
Twisting component – 4 of 4
anglessmallfordz
dKM
d
dKM
KdAa,Let
3
3
A
2
Twisting component – 4 of 4
anglessmallfordz
dKM
d
dKM
KdAa,Let
3
3
A
2
x y
yx
Total Twisting Component
M = M1 + M + M3 + M
M1 = Mx u’ + My v’
M = P (y0 u’ – x0 v’)
M = – (MTY + MBY) v/L – (MTX + MBX) u/L
M3= -K ’
Therefore,
MMx u’ + My v’+ P (y0 u’ – x0 v’) – (MTY + MBY) v/L – (MTX + MBX) u/L-K ’
While,
BXTXBX0BYTYBY
BYTYBY0BXTXBX
MML
zMyPuPMM
L
zMM
MML
zMxPvPMM
L
zMM
Total Twisting Component
M = M1 + M + M3 + M
M1 = Mx u’ + My v’ M = P (y0 u’ – x0 v’) M3= -K ’
M = – (MTY + MBY) v/L – (MTX + MBX) u/L
Therefore, 0 0
0 0
0
0
0
( ) ( ) ( ) ( )
, ( ) ( )
, ( ) ( )
( ( ) ) (
x y TY BY TX BX
x y TY BY TX BX
x BX BX TX
y BY BY TY
BX BX TX BY
v uM M u M v P y u x v M M M M K
L Lv u
M M P y u M P x v M M M M KL L
zBut M M M M P v x
Lz
and M M M M P u yL
zM M M M P y u M
L
0( ) )
( ) ( )
BY TY
TY BY TX BX
zM M P x v
Lv u
M M M M KL L
Thus, now we have the internal moments about the axes for the deformed member cross-section.
Internal moments about the axes
0
0
0 0( ( ) ) ( ( ) )
( ) ( )
BX TX BX BY TY BY
BY TX BX BX TY BY
BX BX TX BY BY TY
TY BY TX BX
z zM M M M P v P x M M M
L L
z zM M M M P u P y M M M
L L
z zM M M M P y u M M M P x v
L Lv u
M M M M KL L
x
y
z
MTX+MBXMTY+MBY
Internal Moment – Deformation Relations
The internal moments M, M, and M will still produce flexural bending
about the centroidal principal axis and twisting about the shear center.
The flexural bending about the principal axes will produce linearly varying longitudinal stresses.
The torsional moment will produce longitudinal and shear stresses due to warping and pure torsion.
The differential equations relating moments to deformations are still valid. Therefore,
M = - E I v” …………………..(I = Ix)
M = E I u” …………………..(I= Iy)
M = G KT ’ – E Iw ’”
Internal Moment – Deformation Relations
0
0
0
0
,
( ( ) )
( ( ) ) ( ) (
x BX TX BX BY TY BY
y BY TX BX BX TY BY
T w BX BX TX
BY BY TY TY BY TX
Therefore
z zM E I v M M M P v P x M M M
L L
z zM E I u M M M P u P y M M M
L L
zM G K E I M M M P y u
Lz v
M M M P x v M M ML L
)BX
uM K
L
MTX+MBXMTY+MBY
Second-Order Differential Equations
0
0
0
0
,
( ) ( ( ) )
( ( ) ) ( ) ( ) 0
x BY TY BY BX TX BX
y BX TY BY BY TX BX
w T BX BX TX
BY BY TY TY BY TX BX
Therefore
z zE I v P v P x M M M M M M
L L
z zE I u P u P y M M M M M M
L L
zE I G K K u M M M P y
Lz v u
v M M M P x M M M ML L L
1
2
3
You end up with three coupled differential equations that relate the applied forces and moments to the deformations u, v, and .
These differential equations can be used to investigate the elastic behavior and buckling of beams, columns, beam-columns and also complete frames – that will form a major part of this course.
MTX+MB
X
MTX+MB
X
MTY+MBY
Chapter 3. Structural Columns
3.1 Elastic Buckling of Columns
3.2 Elastic Buckling of Column Systems – Frames
3.3 Inelastic Buckling of Columns
3.4 Column Design Provisions (U.S. and Abroad)
3.1 Elastic Buckling of Columns
Start out with the second-order differential equations derived in Chapter 2. Substitute P=P and MTY = MBY = MTX = MBX = 0
Therefore, the second-order differential equations simplify to:
This is all great, but before we proceed any further we need to deal with Wagner’s effect – which is a little complicated.
0
0
0 0
0
0
( ) ( ) ( ) 0
x
y
w T
E I v P v P x
E I u P u P y
E I G K K u P y v P x
1
2
3
Wagner’s effect for columns
2
0
0
20 0
20 0
,
( )
( )
( ) ( )
( ) ( )
A
nx y
nx yA
nx y A
K a dA
where
M y M xPE W
A I I
M P v x
M P u y
P v x y P u y xPK E W a dA
A I I
P v x y P u y xPK E W a dA
A I I
Neglecting higher order t
2;A
Perms K a dA
A
Wagner’s effect for columns2 2 2
0 0
2 2 20 0
2 2 2 2 20 0 0 0
2 2 2 2 20 0 0 0
2 2 20 0
2 20 0
, ( ) ( )
( ) ( )
2 2
2 2
( )
,
( )
A A
A A
A A A A A A
x y
A
x y
But a x x y y
a dA x x y y dA
a dA x y x y x x y y dA
a dA x y dA x dA y dA x x dA y y dA
a dA x y A I I
Finally
PK x y A I I
A
2 20 0
2 2 20 0 0
20
( )
( )
x y
x y
I IK P x y
A
I ILet r x y
A
K P r
Second-order differential equations for columns
Simplify to:
Where
0
0
20 0 0
0
0
( ) ( ) ( ) 0
x
y
w T
E I v P v P x
E I u P u P y
E I P r G K u P y v P x
1
2
3
2 2 20 0 0
x yI Ir x y
A
Column buckling – doubly symmetric section
For a doubly symmetric section, the shear center is located at the centroid xo= y0 = 0. Therefore, the three equations become uncoupled
Take two derivatives of the first two equations and one more derivative of the third equation.
20
0
0
( ) 0
x
y
w T
E I v P v
E I u P u
E I P r G K
1
2
3
20
0
0
( ) 0
ivx
ivy
ivw T
E I v P v
E I u P u
E I P r G K
1
2
3
22 2 2 0, T
v ux y w
P r G KP PLet F F F
E I E I E I
Column buckling – doubly symmetric section
All three equations are similar and of the fourth order. The solution will be of the form C1 sin z + C2 cos z + C3 z + C4
Need four boundary conditions to evaluate the constant C1..C4
For the simply supported case, the boundary conditions are:
u= u”=0; v= v”=0; = ”=0
Lets solve one differential equation – the solution will be valid for all three.
2
2
2
0
0
0
ivv
ivu
iv
v F v
u F u
F
1
2
3
Column buckling – doubly symmetric section
2
1 2 3 4
2 21 2
2 4
2
1 2 3 4
2 21 2
0
sin cos
sin cos
:
(0) (0) ( ) ( ) 0
0 (0) 0
0 (0) 0
sin cos ( ) 0
sin
ivv
v v
v v v v
v v
v v v
v F v
Solution is
v C F z C F z C z C
v C F F z C F F z
Boundary conditions
v v v L v L
C C v
C v
C F L C F L C L C v L
C F F L C F
1
2
32 2
4
cos ( ) 0
0 1 0 1 00 1 0 0 0
sin cos 1 00sin cos 0 0
v
v v
v v v v
F L v L
CC
F L F L L CCF F L F F L
2
2 2
2
2
2
0
sin 0
sin 0
1:
v v
v
v
vx
x x
xx
The coefficient matrix
F F L
F L
F L n
P nF
E I L
nP E I
LSmallest value of n
E IP
L
Column buckling – doubly symmetric section
2 2
2
2
2
,
sin 0
1:
u
u
uy
y y
yy
Similarly
F L
F L n
P nF
E I L
nP E I
L
E ISmallest value of n P
L
20
2 2
22
0
2 2
22
0
,
sin 0
1
1:
1
T
w
w T
w T
Similarly
F L
F L n
P r G K nF
E I L
nP E I G K
L r
Smallest value of n
nP E I G K
L r
2
2
2
2
2
22
0
1
xx
yy
wT
E IP
L
E IP
L
E IP G K
L r
1
2
3
Summary
Column buckling – doubly symmetric section
Thus, for a doubly symmetric cross-section, there are three distinct buckling loads Px, Py, and Pz.
The corresponding buckling modes are:
v = C1 sin(z/L), u =C2 sin(z/L), and = C3 sin(z/L).
These are, flexural buckling about the x and y axes and torsional buckling about the z axis.
As you can see, the three buckling modes are uncoupled. You must compute all three buckling load values.
The smallest of three buckling loads will govern the buckling of the column.
Column buckling – boundary conditions
2
1 2 3 4
1 2 3
2 4
1 3
1 2 3 4
1 2
0
sin cos
cos sin
:
(0) (0) ( ) ( ) 0
0 (0) 0
0 (0) 0
sin cos ( ) 0
cos sin
ivv
v v
v v v v
v
v v
v v v
v F v
Solution is
v C F z C F z C z C
v C F F z C F F z C
Boundary conditions
v v v L v L
C C v
C F C v
C F L C F L C L C v L
C F F L C F F
3
1
2
3
4
( ) 0
0 1 0 1 00 1 0 0
sin cos 1 00cos sin 1 0
v
v
v v
v v v v
L C v L
CF CF L F L L C
CF F L F F L
Consider the case of fix-fix boundary conditions:
2 2
2
2 2
2 2
0
sin 2cos 2 0
2 sin cos 2sin 02 2 2
22
4
1:
0.5
v v v
v v vv
v
v
x x
x xx
The coefficient matrix
F L F L F L
F L F L F LF L
F Ln
nF
L
nP E I
LSmallest value of n
E I E IP
L K L
Column Boundary Conditions
The critical buckling loads for columns with different boundary conditions can be expressed as:
Where, Kx, Ky, and Kz are functions of the boundary conditions:
K=1 for simply supported boundary conditions
K=0.5 for fix-fix boundary conditions
K=0.7 for fix-simple boundary conditions
2
2
2
2
2
2 2
0
1
xx
x
yy
y
wT
z
E IP
K L
E IP
K L
E IP G K
K L r
1
2
3
Column buckling – example.
Consider a wide flange column W27 x 84. The boundary conditions are:
v=v”=u=u’==’=0 at z=0, and v=v”=u=u’==”=0 at z=L
For flexural buckling about the x-axis – simply supported – Kx=1.0
For flexural buckling about the y-axis – fixed at both ends – Ky = 0.5
For torsional buckling about the z-axis – pin-fix at two ends - Kz=0.7
2 2 2 2
2 2 2
22 2 2 2
2 2 2
2 22
2 2 2 2
0
1
x xx
x xx
x
y y yy
xy yy
x
w wT T x
x x yzz
x
E I E A r E AP
K L K L LK
r
E I E A r rE AP
rK L K L LK
r
E I E I AP G K G K r
r I IK L r LK
r
Column buckling – example.
2 2
2 2 2
2 2 22
2 2 2
22
2 2
1 5823.066
( / ) ( / ) 791.02
1
x
Y Y
x Y xx x x
y y x y x
Y Y
y Y yx x x
wT x
Y Yx x y
zx
P E A E
P AL L LK K
r r r
P r r E r rE A
P AL L LK K
r r r
P E I AG K r
P Ar I ILK
r
P
2
22 2
2
1
578.260.2333
wT x
Y x x y Y
zx
Y
x
E IG K r
P r I ILK
r
P
P Lr
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0 10 20 30 40 50 60 70 80 90 100
L-rx (Slenderness Ratio)
Cri
tica
l b
uck
lin
g l
oad
/ y
ield
lo
ad (
Pcr
/PY)
Px - flexural buckling Py - flexural buckling Pz - torsional buckling
Flexural buckling about y-axis governs
Torsional buckling about z-axis governs
Yield load PY
Cannot be exceeded
Flexural buckling about x-axis
Flexural buckling about y-axis
Torsional buckling about z-axis
Column buckling – example.
Column buckling – example.
When L is such that L/rx < 31; torsional buckling will govern
rx = 10.69 in. Therefore, L/rx = 31 L=338 in.=28 ft.
Typical column length =10 – 15 ft. Therefore, typical L/rx= 11.2 – 16.8
Therefore elastic torsional buckling will govern.
But, the predicted load is much greater than PY. Therefore, inelastic buckling will govern.
Summary – Typically must calculate all three buckling load values to determine which one governs. However, for common steel buildings made using wide flange sections – the minor (y-axis) flexural buckling usually governs.
In this problem, the torsional buckling governed because the end conditions for minor axis flexural buckling were fixed. This is very rarely achieved in common building construction.
Column Buckling – Singly Symmetric Columns
Well, what if the column has only one axis of symmetry. Like the x-axis or the y-axis or so.
y
xC
S
As shown in this figure, the y – axis is the axis of symmetry.
The shear center S will be located on this axis.
Therefore x0= 0.
The differential equations will simplify to:
0
20 0
0
0
( ) ( ) 0
x
y
w T
E I v P v
E I u P u P y
E I P r G K u P y
1
2
3
Column Buckling – Singly Symmetric Columns
The first equation for flexural buckling about the x-axis (axis of non-symmetry) becomes uncoupled.
2
2
1 2 3 4
2
2
1
0 (1)
0
0
,
sin cos
sin 0
( )
mod sin
x
ivx
ivv
vx
v v
v
xx
x x
v
E I v P v
E I v P v
v F v
Pwhere F
E I
v C F z C F z C z C
Boundary conditions
F L
E IP
K L
Buckling v C F z
Equations (2) and (3) are still coupled in terms of u and .
These equations will be satisfied by the solutions of the form
u=C2 sin (z/L) and =C3 sin (z/L) 0
20 0
0
( ) ( ) 0
y
w T
E I u P u P y
E I P r G K u P y
2
3
Column Buckling – Singly Symmetric Columns
0
20 0
0
20 0
2 3
4
2
0 (2)
( ) ( ) 0 (3)
0
( ) ( ) 0
, sin ; sin
, 2 3
sin
y
w T
ivy
ivw T
y
E I u P u P y
E I P r G K u P y
E I u P u P y
E I P r G K u P y
z zLet u C C
L LTherefore substituting these in equations and
zE I C P
L L
2 2
2 0 3
4 2 22
3 0 3 0 2
sin sin 0
sin ( ) sin sin 0w T
z zC P y C
L L L L
z z zE I C P r G K C P y C
L L L L L L
Column Buckling – Singly Symmetric Columns
2
2 0 3
22
0 3 0 2
2 2
22 2
0
2 0 3
20 3 0 2
0 22
30 0
0
( ) 0
1,
0
0
0( )
y
w T
y wy T
y
y
E I P C P y CL
and E I P r G K C P y CL
E I E ILet P and P G K
L L r
P P C P y C
P P r C P y C
P P P y CCP y P P r
P
0
20 0
0( )
y P P y
P y P P r
Column Buckling – Singly Symmetric Columns
2 2 20 0
2 2 2 20 0
22 0
20
22 0
20
202
0
202
2 02
202
0
( )( ) 0
( ) 0
(1 ) ( ) 0
( ) ( ) 4 (1 )
2 (1 )
4 (1 )
( ) ( ) 1( )
2 (1 )
y
y y
y y
y y y
y
y yy
P P P P r P y
P P P P P P r P y
yP P P P P P
r
yP P P P P P
rP
y
r
yP P
rP P P P
P P
Py
r
202
02 202
0
202
02 202
0
4 (1 )( )
1 1( )
2 (1 )
,
4 (1 )( )
1 1( )
2 (1 )
yy
y
yy
y
yP P
P P rP
y P Pr
Thus there are two roots for P
Smaller value will govern
yP P
P P rP P
y P Pr
Column Buckling – Singly Symmetric Columns
The critical buckling load will the lowest of Px and the two roots shown on the previous slide.
If the flexural torsional buckling load govern, then the buckling mode will be C2 sin (z/L) x C3 sin (z/L)
This buckling mode will include both flexural and torsional deformations – hence flexural-torsional buckling mode.
Column Buckling – Asymmetric Section
No axes of symmetry: Therefore, shear center S (xo, yo) is such that neither xo not yo are zero.
For simply supported boundary conditions: (u, u”, v, v”, , ”=0), the solutions to the differential equations can be assumed to be:
u = C1sin (z/L)
v = C2 sin (z/L)
= C3 sin (z/L)
These solutions will satisfy the boundary conditions noted above
0
0
20 0 0
0 (1)
0 (2)
( ) ( ) ( ) 0 (3)
x
y
w T
E I v P v P x
E I u P u P y
E I P r G K u P y v P x
Column Buckling – Asymmetric Section
Substitute the solutions into the d.e. and assume that it satisfied too:
2
1 1 0 3
2
2 2 0 3
3
3
sin sin sin 0
sin sin sin 0
cos
x
y
w
z z zE I C P C P x C
L L L L
z z zE I C P C P y C
L L L L
zE I C
L L
20 3 0 1 0 2( ) cos cos cos 0
T
z z zP r G K C P y C P x C
L L L L L L
2
0 1
2
0 2
22
30 0 0
0 sin
00 sin 0
0
cos( )
x
y
w T
zE I P P x CL L
zE I P P y C
L Lz
CP x P y E I P r G KL LL
Column Buckling – Asymmetric Section
Either C1, C2, C3 = 0 (no buckling), or the determinant of the coefficient matrix =0 at buckling.
Therefore, determinant of the coefficient matrix is:
1
0
0 2
20 0 0
3
2 2 2
22
0
sin
0 00 sin 0
0
cos
,
1
x
y
wx x y y T
zC
LP P P x
zP P P y C
LP x P y P P r z
CL L
where
E IP EI P EI P G K
L L L r
2 22 2o o
x y x y2 2
o o
y xP P P P P P P P P P P P 0
r r
Column Buckling – Asymmetric Section
This is the equation for predicting buckling of a column with an asymmetric section.
The equation is cubic in P. Hence, it can be solved to obtain three roots Pcr1, Pcr2, Pcr3.
The smallest of the three roots will govern the buckling of the column.
The critical buckling load will always be smaller than Px, Py, and P
The buckling mode will always include all three deformations u, v, and . Hence, it will be a flexural-torsional buckling mode.
For boundary conditions other than simply-supported, the corresponding Px, Py, and P can be modified to include end condition effects Kx, Ky, and K
2 22 2o o
x y x y2 2
o o
y xP P P P P P P P P P P P 0
r r
Homework No. 4 See word file Problem No. 1
Consider a column with doubly symmetric cross-section. The boundary conditions for flexural buckling are simply supported at one end and fixed at the other end.
Solve the differential equation for flexural buckling for these boundary conditions and determine the eigenvalue (buckling load) and the eigenmode (buckling shape). Plot the eigenmode.
How the eigenvalue compare with the effective length approach for predicting buckling?
What is the relationship between the eigenmode and the effective length of the column (Refer textbook).
Problem No. 2 Consider an A992 steel W14 x 68 column cross-section. Develop the normalized
buckling load (Pcr/PY) vs. slenderness ratio (L/rx) curves for the column cross-section. Assume that the boundary conditions are simply supported for buckling about the x, y, and z axes.
Which buckling mode dominates for different column lengths? Is torsional buckling a possibility for practical columns of this length? Will elastic buckling occur for most practical lengths of this column?
Problem No. 3 Consider a C10 x 30 column section. The length of the column is 15 ft. What is the
buckling capacity of the column if it is simply supported for buckling about the y-axis (of non-symmetry), pin-fix for flexure about the x-axis (of symmetry) and simply supported in torsion about the z-axis. Which buckling mode dominates?
Column Buckling - Inelastic
A long topic
Effects of geometric imperfection
EIx v Pv 0
EIy u Pu 0Leads to bifurcation buckling of perfect doubly-symmetric columns
P
vo o sinz
L
vv
vo
P
Mx
Mx P(v vo ) 0
EIx v P(v vo ) 0
v Fv2(v vo ) 0
v Fv2v Fv
2vo
v Fv2v Fv
2(o sinz
L)
Solution vc v p
vc A sin(Fvz) Bcos(Fvz)
v p C sinz
L Dcos
z
L
Effects of Geometric Imperfection
Solve for C and D first
v p Fv2v p Fv
2o sinz
L
L
2
C sinz
LDcos
z
L
Fv
2 C sinz
LDcos
z
L
Fv
2o sinz
L0
sinz
L C
L
2
Fv2C Fv
2o
cos
z
L
L
2
D Fv2D
0
CL
2
Fv2C Fv
2o 0 and L
2
D Fv2D
0
C Fv
2o
L
2
Fv2
and D 0
Solution becomes
v A sin(Fvz) Bcos(Fvz) Fv
2o
L
2
Fv2
sinz
L
Geometric Imperfection
Solve for A and B
Boundary conditions v(0) v(L) 0
v(0) B 0
v(L) A sin FvL 0
A 0
Solution becomes
v Fv2o
L
2
Fv2
sinz
L
v
Fv2
L
2 o
1 Fv2
L
2
sinz
L
P
PE
o
1 P
PE
sinz
L
v
P
PE
1 PPE
o sinz
L
Total Deflection
v vo
P
PE
1 PPE
o sinz
Lo sin
z
L
P
PE
1 PPE
1
o sinz
L
1
1 PPE
o sinz
L
AFo sinz
L
AF = amplification factor
Geometric Imperfection
AF 1
1 PPE
amplification factor
Mx P(v vo )
Mx AF (Po sinz
L)
i.e., Mx AF (moment due to initial crooked)
0
2
4
6
8
10
12
0 0.2 0.4 0.6 0.8 1
P/P E
Am
pli
ficati
on
Facto
r A
F
Increases exponentiallyLimit AF for designLimit P/PE for design
Value used in the code is 0.877This will give AF = 8.13Have to live with it.
Residual Stress Effects
Residual Stress Effects
History of column inelastic buckling
Euler developed column elastic buckling equations (buried in the million other things he did).
Take a look at: http://en.wikipedia.org/wiki/EuleR An amazing mathematician
In the 1750s, I could not find the exact year.
The elastica problem of column buckling indicates elastic buckling occurs with no increase in load.
dP/dv=0
History of Column Inelastic Buckling
Engesser extended the elastic column buckling theory in 1889.
He assumed that inelastic
buckling occurs with no
increase in load, and the
relation between stress
and strain is defined by
tangent modulus Et
Engesser’s tangent modulus theory is easy to apply. It compares reasonably with experimental results.
PT=ETI / (KL)2
History of Column Inelastic Buckling
In 1895, Jasinsky pointed out the problem with Engesser’s theory.
If dP/dv=0, then the 2nd order moment (Pv) will produce incremental strains that will vary linearly and have a zero value at the centroid (neutral axis).
The linear strain variation will have compressive and tensile values. The tangent modulus for the incremental compressive strain is equal to Et and that for the tensile strain is E.
History of Column Inelastic Buckling
In 1898, Engesser corrected his original theory by accounting for the different tangent modulus of the tensile increment.
This is known as the reduced modulus or double modulus The assumptions are the same as before. That is, there is no
increase in load as buckling occurs.
The corrected theory is shown in the following slide
History of Column Inelastic Buckling
The buckling load PR produces critical stress R=Pr/A
During buckling, a small curvature d is introduced
The strain distribution is shown.
The loaded side has dL and dL
The unloaded side has dU and dU
dL (y y1 y) ddU (y y y1) dd L E t ( y y1 y) ddU E(y y y1) d
History of Column Inelastic Buckling
d v
d L E t (y y1 y) v
dU E(y y y1) v
But, the assumption is dP 0
dU dA y y1
y
d L dA ( d y )
y y1
0
E( y y y1) dA y y1
y
E t ( y y1 y) dA ( d y )
y y1
0
ES1 E tS2 0
where, S1 ( y y y1) dAy y1
y
and S2 ( y y1 y) dA ( d y )
y y1
History of Column Inelastic Buckling
S1 and S2 are the statical moments of the areas to the left and right of the neutral axis.
Note that the neutral axis does not coincide with the centroid any more.
The location of the neutral axis is calculated using the equation derived ES1 - EtS2 = 0
M Pv
M dU ( y y y1) dA y y1
y
d L (y y1 y) dA ( d y )
y y1
M Pv v ( EI1 E tI2 )
where, I1 ( y y y1)2 dAy y1
y
and I2 ( y y1 y)2 dA ( d y )
y y1
History of Column Inelastic Buckling
M Pv v (EI1 E tI2 )
Pv (EI1 E tI2 ) v 0
v P
EI1 E tI2
v 0
v Fv2v 0
where, Fv2 P
EI1 E tI2
PE Ix
and E EI1
Ix
E t
I2
Ix
PR 2E Ix
(KL)2
E is the reduced or double modulus
PR is the reduced modulus buckling load
History of Column Inelastic Buckling
For 50 years, engineers were faced with the dilemma that the reduced modulus theory is correct, but the experimental data was closer to the tangent modulus theory. How to resolve?
Shanley eventually resolved this dilemma in 1947. He conducted very careful experiments on small aluminum columns.
He found that lateral deflection started very near the theoretical tangent modulus load and the load capacity increased with increasing lateral deflections.
The column axial load capacity never reached the calculated reduced or double modulus load.
Shanley developed a column model to explain the observed phenomenon
History of Column Inelastic Buckling
History of Column Inelastic Buckling
History of Column Inelastic Buckling
History of Column Inelastic Buckling
Column Inelastic Buckling
Three different theories Tangent modulus Reduced modulus Shanley model
Tangent modulus theory assumes
Perfectly straight column Ends are pinned Small deformations No strain reversal during
buckling
P
dP/dv=0
vElastic buckling analysis
Slope is zero at bucklingP=0 with increasing v
PT
Tangent modulus theory
Assumes that the column buckles at the tangent modulus load such that there is an increase in P (axial force) and M (moment).
The axial strain increases everywhere and there is no strain reversal.
PT
v
PT
Mx
v
T=PT/A
Strain and stress state just before buckling
Strain and stress state just after buckling
T
T
T
T
T=ETT
Mx - Pv = 0
Curvature = = slope of strain diagram
T
h
T h
2 y
where y dis tance from centroid
T h
2 y
ET
Tangent modulus theory
Deriving the equation of equilibrium
The equation Mx- PTv=0 becomes -ETIxv” - PTv=0 Solution is PT= 2ETIx/L2
Mx yA dA
T T
T ( y h / 2) ET
Mx T (y h / 2)ET yA dA
Mx T y dA ET y 2dA A h / 2)ET y dA
A
A
Mx 0 ET Ix 0
Mx ET Ix v
Example - Aluminum columns
Consider an aluminum column with Ramberg-Osgood stress-strain curve
E
0.002
0.2
n
1
E 0.002
0.2n n n 1
1
0.002
0.2n nE n 1
E
1 0.002
0.2
nE
0.2
n 1
E
E
1 0.002
0.2
nE
0.2
n 1 ET
E 10100 ksi
0.2 40.15 ksin 18.55
ET ET (KL/r) cr
0.000E+00 0 differences equation1.980E-04 2 10100.0 10100.0 223.25210463.960E-04 4 10100.0 10100.0 157.86307715.941E-04 6 10100.0 10100.0 128.89466277.921E-04 8 10100.0 10100.0 111.62605239.901E-04 10 10100.0 10100.0 99.841376411.188E-03 12 10100.0 10100.0 91.14228981.386E-03 14 10100.0 10100.0 84.38136041.584E-03 16 10100.0 10100.0 78.931502751.782E-03 18 10100.0 10099.9 74.417101531.980E-03 20 10099.8 10099.5 70.596906792.178E-03 22 10098.8 10097.6 67.30487952.376E-03 24 10094.2 10088.7 64.41136912.575E-03 26 10075.1 10054.2 61.778574342.775E-03 28 10005.7 9934.0 59.174309522.979E-03 30 9779.8 9563.7 56.092082863.198E-03 32 9142.0 8602.6 51.50976563.458E-03 34 7697.4 6713.6 44.145664153.829E-03 36 5394.2 4251.9 34.14196854.483E-03 38 3056.9 2218.6 24.004640135.826E-03 40 1488.8 1037.0 15.99612018.771E-03 42 679.2 468.1 10.488274751.529E-02 44 306.9 212.4 6.9025161442.949E-02 46 140.8 98.5 4.5966334065.967E-02 48 66.3 46.9 3.1054403611.221E-01 50 32.1 23.0 2.129145204
Tangent Modulus Buckling
Ramberg-Osgood Stress-Strain
0
10
20
30
40
50
60
0.000 0.010 0.020 0.030 0.040 0.050
Strain (in./in.)
Str
ess
(ksi
)
Stress-tangent modulus relationship
0
2000
4000
6000
8000
10000
12000
0 10 20 30 40 50
Stress (ksi)Ta
ngen
t Mod
ulus
(ks
i)
ET differences ET equation
Tangent Modulus Buckling
Column Inelastic Buckling Curve
0
10
20
30
40
50
60
0 30 60 90 120 150
KL/r
Tang
ent
Mod
ulus
Buc
klin
g S
tres
s
(KL/r) cr
02 223.25210464 157.86307716 128.89466278 111.6260523
10 99.8413764112 91.142289814 84.381360416 78.9315027518 74.4171015320 70.5969067922 67.304879524 64.411369126 61.7785743428 59.1743095230 56.0920828632 51.509765634 44.1456641536 34.141968538 24.0046401340 15.996120142 10.4882747544 6.90251614446 4.59663340648 3.10544036150 2.129145204
PT 2ET Ix
L2
PT
AT
2ET Ix
AL2 2ET
KL / r 2
KL / r cr
2ET
T
Residual Stress Effects
Consider a rectangular section with a simple residual stress distribution
Assume that the steel material has elastic-plastic stress-strain curve.
Assume simply supported end conditions
Assume triangular distribution for residual stresses
x
y
b
d
rc rc
rt
E
y
0.5y 0.5y
0.5y
y/b
Residual Stress Effects
One major constrain on residual stresses is that they must be such that
Residual stresses are produced by uneven cooling but no load is present
rdA 0
0.5 y 2 y
bx
d dx
b / 2
0
0.5 y 2 y
bx
d dx
0
b / 2
0.5 yd b 2 0.5 yd b 2 2d y
b
b2
8
2d y
b
b2
8
0
Residual Stress Effects
Response will be such that - elastic behavior when
0.5 y
Px 2EIx
L2and Py
2EIy
L2
Yielding occurs when
0.5 y i.e., P 0.5PY
Inelastic buckling will occur after 0.5 y
x
y
b
d
x
y
b b
Y Y
Y 2Y
bb
Y (1 2 )
Y/b
Residual Stress Effects
Total axial force corresponding to the yielded sec tion
Y b 2b d Y Y (1 2 )
2
bd 2
Y 1 2 bd Y (2 2 )bd
Ybd 2bdY 2Ybd 2 2bdY
Ybd(1 2 2 ) PY (1 2 2 )
If inelastic buckling were to occur at this load
Pcr PY (1 2 2 )
1
21
Pcr
PY
If inelastic buckling occurs about x axis
Pcr PTx 2E
L2 (2b)d3
12
PTx 2EIx
L2 2
PTx Px 2 1
21 Pcr
PY
PTx Px 2 1
21 PTx
PY
Pcr PTx
PTx
PY
Px
PY
2 1
21 PTx
PY
Let,
Px
PY
1
x2 2 E
Y
rx
KxLx
2
PTx
PY
1
x2 2 1
21 PTx
PY
x2
2 1 PTx
PY
PTx
PY
x
y
b b
If inelastic buckling occurs about y axis
Pcr PTy 2E
L2 (2b)3 d
12
PTy 2EIy
L2 2 3
PTy Py 21
21
Pcr
PY
3
PTy Py 2 1PTy
PY
3
Pcr PTy
PTy
PY
Py
PY
2 1PTy
PY
3
Let,Py
PY
1
y2 2 E
Y
ry
KyLy
2
PTy
PY
1
y2 2 1
PTy
PY
3
y2
2 1PTy
PY
3
PTy
PY
x
y
b b
Residual Stress Effects
P/P Y x y
0.200 2.236 2.2360.250 2.000 2.0000.300 1.826 1.8260.350 1.690 1.6900.400 1.581 1.5810.450 1.491 1.4910.500 1.414 1.4140.550 1.313 1.2460.600 1.221 1.0920.650 1.135 0.9490.700 1.052 0.8150.750 0.971 0.6870.800 0.889 0.5620.850 0.803 0.4400.900 0.705 0.3150.950 0.577 0.1820.995 0.317 0.032
Column I nelastic Buckling
0.000
0.200
0.400
0.600
0.800
1.000
1.200
0.0 0.5 1.0 1.5 2.0
LambdaN
orm
alized c
olu
mn
capacit
y
0.000
0.200
0.400
0.600
0.800
1.000
1.200
Tangent modulus buckling - Numerical
Centroidal axis
Afib
yfib
Discretize the cross-section into fibersThink about the discretization. Do you need the flange
To be discretized along the length and width?
For each fiber, save the area of fiber (Afib), the distances from the centroid yfib and xfib,
Ix-fib and Iy-fib the fiber number in the matrix.
Discretize residual stress distribution
Calculate residual stress (r-fib)each fiber
Check that sum(r-fib Afib)for Section = zero
1
2
3
4
5
Tangent Modulus Buckling - Numerical
Calculate effective residual strain (r) for each fiber
r=r/E
Assume centroidal strain
Calculate total strain for each fibertot=+r
Assume a material stress-strain curve for each fiber
Calculate stress in each fiber fib
Calculate Axial Force = P Sum (fibAfib)
Calculate average stress = = P/A
Calculate the tangent (EI)TX and (EI)TY for the (EI)TX = sum(ET-fib{yfib
2 Afib+Ix-fib})(EI)Ty = sum(ET-fib{xfib
2 Afib+ Iy-fib})
Calculate the critical (KL)X and (KL)Y for the (KL)X-cr = sqrt [(EI)Tx/P](KL)y-cr = sqrt [(EI)Ty/P]6
7
8
910
11
12
13
14
Tangent modulus buckling - numerical
Section Dimensionb 12 fiber no. Afib xfib yfib r-fib r-fib Ixfib Iyfib
d 4 1 2.4 -5.7 0 -22.5 -7.759E-04 3.2 78.05y 50 2 2.4 -5.1 0 -17.5 -6.034E-04 3.2 62.50
3 2.4 -4.5 0 -12.5 -4.310E-04 3.2 48.67No. of fibers 20 4 2.4 -3.9 0 -7.5 -2.586E-04 3.2 36.58
5 2.4 -3.3 0 -2.5 -8.621E-05 3.2 26.216 2.4 -2.7 0 2.5 8.621E-05 3.2 17.57
A 48 7 2.4 -2.1 0 7.5 2.586E-04 3.2 10.66Ix 64 8 2.4 -1.5 0 12.5 4.310E-04 3.2 5.47Iy 576.00 9 2.4 -0.9 0 17.5 6.034E-04 3.2 2.02
10 2.4 -0.3 0 22.5 7.759E-04 3.2 0.2911 2.4 0.3 0 22.5 7.759E-04 3.2 0.2912 2.4 0.9 0 17.5 6.034E-04 3.2 2.0213 2.4 1.5 0 12.5 4.310E-04 3.2 5.4714 2.4 2.1 0 7.5 2.586E-04 3.2 10.6615 2.4 2.7 0 2.5 8.621E-05 3.2 17.5716 2.4 3.3 0 -2.5 -8.621E-05 3.2 26.2117 2.4 3.9 0 -7.5 -2.586E-04 3.2 36.5818 2.4 4.5 0 -12.5 -4.310E-04 3.2 48.6719 2.4 5.1 0 -17.5 -6.034E-04 3.2 62.5020 2.4 5.7 0 -22.5 -7.759E-04 3.2 78.05
Tangent Modulus Buckling - numerical
Strain Increment Fiber no. tot fib Efib Tx-fib Ty-fib Pfib
-0.0003 1 -1.076E-03 -31.2 29000 92800 2.26E+06 -74.882 -9.034E-04 -26.2 29000 92800 1.81E+06 -62.883 -7.310E-04 -21.2 29000 92800 1.41E+06 -50.884 -5.586E-04 -16.2 29000 92800 1.06E+06 -38.885 -3.862E-04 -11.2 29000 92800 7.60E+05 -26.886 -2.138E-04 -6.2 29000 92800 5.09E+05 -14.887 -4.138E-05 -1.2 29000 92800 3.09E+05 -2.888 1.310E-04 3.8 29000 92800 1.59E+05 9.129 3.034E-04 8.8 29000 92800 5.85E+04 21.12
10 4.759E-04 13.8 29000 92800 8.35E+03 33.1211 4.759E-04 13.8 29000 92800 8.35E+03 33.1212 3.034E-04 8.8 29000 92800 5.85E+04 21.1213 1.310E-04 3.8 29000 92800 1.59E+05 9.1214 -4.138E-05 -1.2 29000 92800 3.09E+05 -2.8815 -2.138E-04 -6.2 29000 92800 5.09E+05 -14.8816 -3.862E-04 -11.2 29000 92800 7.60E+05 -26.8817 -5.586E-04 -16.2 29000 92800 1.06E+06 -38.8818 -7.310E-04 -21.2 29000 92800 1.41E+06 -50.8819 -9.034E-04 -26.2 29000 92800 1.81E+06 -62.8820 -1.076E-03 -31.2 29000 92800 2.26E+06 -74.88
Tangent Modulus Buckling - Numerical
P Tx Ty KLx-cr KLy-cr T/Y (KL/r)x (KL/r)y
0.0003 -417.6 15000 10000 209.4395102 628.3185307 0.174 181.3799364 181.37993640.000 -556.8 15000 10000 181.3799364 544.1398093 0.232 157.0796327 157.0796327
-0.0005 -696 1856000 16704000 162.231147 486.6934411 0.29 140.4962946 140.4962946-0.0006 -835.2 1856000 16704000 148.0960979 444.2882938 0.348 128.254983 128.254983-0.0007 -974.4 1856000 16704000 137.1103442 411.3310325 0.406 118.7410412 118.7410412-0.0008 -1113.6 1856000 16704000 128.254983 384.764949 0.464 111.0720735 111.0720735-0.0009 -1252.8 1856000 16704000 120.9199576 362.7598728 0.522 104.7197551 104.7197551
-0.001 -1384.8 1670400 12177216 109.11051 294.5983771 0.577 94.49247352 85.04322617-0.0011 -1510.08 1670400 12177216 104.4864889 282.1135199 0.6292 90.48795371 81.43915834-0.0012 -1624.32 1484800 8552448 94.98347542 227.960341 0.6768 82.25810265 65.80648212-0.0013 -1734.72 1299200 5729472 85.97519823 180.5479163 0.7228 74.45670576 52.11969403-0.0014 -1832.16 1299200 5729472 83.65775001 175.681275 0.7634 72.44973673 50.71481571-0.0015 -1924.8 1113600 3608064 75.56517263 136.0173107 0.802 65.44135914 39.26481548-0.0016 -2008.32 1113600 3608064 73.97722346 133.1590022 0.8368 64.06615482 38.43969289-0.0017 -2083.2 928000 2088000 66.30684706 99.46027059 0.868 57.423414 28.711707-0.0018 -2152.8 928000 2088000 65.22619108 97.83928663 0.897 56.48753847 28.24376924-0.0019 -2209.92 742400 1069056 57.58118233 69.0974188 0.9208 49.86676668 19.94670667
-0.002 -2263.2 556800 451008 49.27629185 44.34866267 0.943 42.67452055 12.80235616-0.0021 -2304.96 556800 451008 48.8278711 43.94508399 0.9604 42.28617679 12.68585304-0.0022 -2340.48 371200 133632 39.56410897 23.73846538 0.9752 34.26352344 6.852704688-0.0023 -2368.32 371200 133632 39.33088015 23.59852809 0.9868 34.06154136 6.812308273-0.0024 -2386.08 185600 16704 27.70743725 8.312231176 0.9942 23.99534453 2.399534453
-0.00249 -2398.608 185600 16704 27.63498414 8.290495243 0.99942 23.9325983 2.39325983
Tangent Modulus Buckling - Numerical
Inelastic Column Buckling
0
0.2
0.4
0.6
0.8
1
1.2
0 20 40 60 80 100 120 140 160 180 200
KL/r ratio
Norm
alize
d c
riti
cal str
ess
( T/
Y)
(KL/r)x (KL/r)y
Column Inelastic Buckling
0
0.2
0.4
0.6
0.8
1
1.2
0.0 0.5 1.0 1.5 2.0
Lambda
Norm
ali
zed
colu
mn
cap
acit
y
0.0
0.2
0.4
0.6
0.8
1.0
1.2
Num-x Num-y Analytical-x
Elastic AISC-Design Analytical-y
ELASTIC BUCKLING OF BEAMS
Going back to the original three second-order differential equations:
0
0
0
0
,
( ) ( ( ) )
( ( ) ) ( ) ( ) 0
x BY TY BY BX TX BX
y BX TY BY BY TX BX
w T BX BX TX
BY BY TY TY BY TX BX
Therefore
z zE I v P v P x M M M M M M
L L
z zE I u P u P y M M M M M M
L L
zE I G K K u M M M P y
Lz v u
v M M M P x M M M ML L L
1
2
3
(MTX+MBX) (MTY+MBY)
ELASTIC BUCKLING OF BEAMS
Consider the case of a beam subjected to uniaxial bending only: because most steel structures have beams in uniaxial bending Beams under biaxial bending do not undergo elastic buckling
P=0; MTY=MBY=0
The three equations simplify to:
Equation (1) is an uncoupled differential equation describing in-plane bending behavior caused by MTX and MBX
( ) ( ) ( ) 0
x BX TX BX
y BX TX BX
w T BX BX TX TX BX
zE I v M M M
Lz
E I u M M ML
z uE I G K K u M M M M M
L L
1
2
3
ELASTIC BUCKLING OF BEAMS
Equations (2) and (3) are coupled equations in u and – that describe the lateral bending and torsional behavior of the beam. In fact they define the lateral torsional buckling of the beam.
The beam must satisfy all three equations (1, 2, and 3). Hence, beam in-plane bending will occur UNTIL the lateral torsional buckling moment is reached, when it will take over.
Consider the case of uniform moment (Mo) causing compression in the top flange. This will mean that
-MBX = MTX = Mo
ELASTIC BUCKLING OF BEAMS For this case, the differential equations (2 and 3) will become:
2
2 2
2 2 2 20 0
0
( ) 0
:
'
,
( ) ( )
2 2
y o
w T o
A
o
x
oo o
xA
oo o
x
E I u M
E I G K K u M
where
K Wagner s effect due to warping caused by torsion
K a dA
MBut y neglecting higher order terms
I
MK y x x y y dA
I
MK y x x xx y y yy dA
I
2 2 2 2 20 2 2
A
oo o o
x A A A A A
MK x y dA y x y dA x xy dA y y dA y y dA
I
ELASTIC BUCKLING OF BEAMS
2 2
2 2
2 2
2
2
, 2
sec
oo x
x A
Ao o
x
Ao x x o
x
x
MK y x y dA y I
I
y x y dA
K M yI
y x y dA
K M where yI
is a new tional property
:
(2) 0
(3) ( ) 0
y o
w T o x o
The beam buckling differential equations become
E I u M
E I G K M u M
ELASTIC BUCKLING OF BEAMS
2
2
2
2
1 2 2
1 2
(2)
(2) (3) :
( ) 0
sec : 0
0
,
0
o
y
iv ow T o x
y
x
iv oT
w y w
oT
w y w
iv
MEquation gives u
E I
Substituting u from Equation in gives
ME I G K M
E I
For doubly symmetric tion
MG K
E I E I I
MG KLet and
E I E I I
. .becomes the combined d e of LTB
ELASTIC BUCKLING OF BEAMS
1 1 2 2
4 21 2
4 21 2
2 21 1 2 1 2 12
2 21 1 2 1 1 2
1 2
1 2 3 4
0
0
4 4,
2 2
4 4,
2 2
, ,
z
z
z z i z i z
Assume solution is of the form e
e
i
Let and i
Above are the four roots for
C e C e C e C e
collect
1 1 2 1 3 2 4 2cosh( ) sinh( ) sin( ) cos( )
ing real and imaginary terms
G z G z G z G z
ELASTIC BUCKLING OF BEAMS
Assume simply supported boundary conditions for the beam:
12 21 2 2
1 1 2 2 32 2 2 2
41 1 1 1 2 2 2 2
(0) (0) ( ) ( ) 0
. .
1 0 0 10 0
0cosh( ) sinh( ) sin( ) cos( )
cosh( ) sinh( ) sin( ) cos( )
L L
Solution for must satisfy all four b c
GG
L L L L GGL L L L
For buckl
2 21 2 1 2
2
2
sin :
det min 0
sinh( ) sinh( ) 0
:
sinh( ) 0
ing coefficient matrix must be gular
er ant of matrix
L L
Of these
only L
L n
ELASTIC BUCKLING OF BEAMS
2
21 2 1
22
1 2 1 2
22 2 22
1 1 12 2 2
2
2 2
2 12 2
2 2 2
2 2 2 2
2 22
2 2
4
2
24
2 2 22
4 4
o T
y w w
To y w
w
n
L
L
L
L L L
L L
M G K
E I I L E I L
G KM E I I
L E I L
2 2
2 2
y wo T
E I E IM G K
L L