Structural Stability & Design by Varma

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.


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!


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


OUTLINE




Examples – small deflection analyses

Examples – large deflection analyses

Examples – imperfect systems


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


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


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


Write equations of static equilibrium in deformed stateP

L

L (1-cos)

L cos

L sin

k L sinO

LkL

LkP


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


Example 3 – Three rigid bar system with two rotational springs

PP

A B CD

kk

L L L


PP

A D

kk

LL

12

L sin 1

L sin 2

BC1 – 2)

1 – 2)

Assume small deformations. Therefore, sin=


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


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


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


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


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


Homework No. 1 Problem 1.1 Problem 1.3 Problem 1.4 All problems from the textbook on Stability by W.F. Chen


OUTLINE




Bifurcation analysis examples – small deflection analyses

Energy method Examples – small deflection analyses Examples – large deflection analyses Examples – imperfect systems


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


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


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


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


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


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


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


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


:

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


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


P

k

L

PL

L (1-cos)

L cos

L sin

k L sinO

Example 2 - Rigid bar supported by translational spring at end



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


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


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


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


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


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


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


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


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


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.


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


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


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


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


y

tan

tan


Shear Stresses due to bending

s

Oy

x

s

Ox

y

dstxI

Vt

dstyI

Vt


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

)(


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


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


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’)


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


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


anglessmallfordz

dKM

d

dKM

KdAa,Let

3

3

A

2


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’)


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 ’


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


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


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


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


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


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



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


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


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


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


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


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


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


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


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


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


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.


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


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


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


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


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


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





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


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


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


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


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


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


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


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)


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


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


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


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


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


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


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

Date post:	29-Jan-2016
Category:	Documents
Upload:	customerx
View:	39 times
Download:	4 times

Structural Stability & Design by Varma

Documents