Elastic Instability

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Chapter 7

ELASTIC INSTABILITY

Dr Rendy Thamrin; Zalipah Jamellodin 7.1 INTRODUCTION TO ELASTIC INSTABILITY OF COLUMN AND FRAME

In structural analysis problem, the aim is to determine a configuration of loaded system, which satisfies the condition of: • Equilibrium • Compatibility, and • Force-displacement relations of material

For a structure to be satisfactory → whether the equilibrium configuration so determined is stable.

The stability loss under compressive load is usually termed structural or geometrical instability commonly known as “buckling”.

In the case of a column loaded by a compressive load, the ability of the column

to resist axial load terminates as soon as buckling occurs and, consequently, the critical load of the column is its failure load.

It is to be emphasized that the load at which instability occurs depends upon the

stiffness of structure (or parts of the structure), rather than on the strength of material.

7.1.1 Concept of Stability

Initial state

Position after displacing force is removed

Position after displacing force is applied

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7.2 TYPES OF INSTABILITY

Instability is a condition wherein a compression member loses the ability to resist increasing loads and exhibits instead a decrease in load carrying capacity.

Generally, the types of instability of a structure (or a member of structure) can

be classified as follows : • Buckling with respect to principal minor axis, • Buckling with respect to principal major axis, • Pure torsional instability.

Classification of instability can be also provided as follows : • Flexural buckling, • Torsional buckling, • Torsional – flexural buckling, and • Snap-through buckling

Figure 7.1: Mode of Failures

(a) Stable equilibrium

(b) Neutral equilibrium

(c) Unstable equilibrium

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7.3 METHODS OF ANALYSIS

Stability analysis consists in determining the mode of loss of structural stability and corresponding load called critical load.

Four definitely different classical methods available for the solution of buckling

problems are: • Non-trivial equilibrium state approach, • Work approach, • Energy approach, and • Kinetic or dynamical approach

7.4 CRITICAL LOAD ANALYSIS USING STABILITY FUNCTIONS

The slope deflection equation of a member related moments and slopes can be written as follows:

Where; MAB = moment at A

EI = flexural stiffness L = member length; θA = slope at A θB = slope at B Δ = member deformation

The equation of a member subjected to axial load can be written as follows:

Where; s = member stiffness c = carry-over factor

⎟⎠⎞

⎜⎝⎛ Δ

−θ+θ=L

624LEIM BAAB

( ) ⎟⎠⎞

⎜⎝⎛ Δ

+−θ+θ=L

c1sscsLEIM BAAB

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The parameters s and c are called ”stability functions”. The values of parameter s and c are functions of the ratio of axial load, P, and

Euler’s load, Pe. This ratio can be expressed and notate as:

EIPL

PP

2

2

e π==ρ

The selected values of these functions (s and c) are tabulated and will be shown in the next following pages.

7.4.1 Stability Functions Derivation (Non-Sway)

Consider a member AB with length L subjected to flexural moment and compression load. In linear elastic analysis, if the axial load is neglected (only flexure):

kM i.e. LEI4M A1A1 θ∝θ=

LEIk : where kM i.e.

LEI2M A2A2 =θ∝θ=

LMMV 21

1+

=

For a member with axial load P , the stiffness coefficient of moment M1 and M2

should be changed and can write:

cskM and skM A2A1 θ=θ= where: s = stiffness factor and c = carry-over factor

hence: ( ) L

kc1sVor L

cskskV A1AA

1 θ+

=θ+θ

=

The parameters s and c are known as ”stability functions”.

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From the figure above, for each location x from A: xVPyMM 11x −+= ………. (1)

From pure bending:

2

2

x dxydEIM −=

substitute to Eq. (1): 112

2

MxVPydx

ydEI −=+

EIM

EIxVy

EIP

dxyd 112

2

−=+

replace: EIP2 =μ →

EIM

EIxVy

dxyd 1122

2

−=μ+ ………. (2)

The homogeneous differential equation (DE) has the form:

0ydx

yd 22

2

=μ+ or 0x"y 2 =μ+

The solution of homogeneous DE, y , consist of the part: kc yyy +=

(i) The complementary solution is: xcosBxsinAyc μ+μ= ………. (3)

(ii) The particular solution may be assumed to be the form: DCxyk +=

( )EIM

EIxV)DCx 112 −=+μ∴

Take the coefficient of x: PV

EIVC

EIxVCx 1

2112 =

μ=∴→=μ

Coefficient of constant: P

MEI

MD EIMD 1

2112 =

μ−=∴→−=μ

Substitute to particular solution: P

MPxVy 11

k −=∴ ………. (4)

Thus, the complete solution is:

PM

PxVxcosBxsinAy 11 −+μ+μ=

The constants A and B may be determined by applying the end boundary

conditions:

(i) 0y ; 0x == P

MB 1=∴

(ii) 0y ; Lx == ; 0P

MPLVLcos

PMLsinA 111 =−+μ+μ

L

MMV 211

+=

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0P

MP

MP

MLcosP

MLsinA 1211 =−++μ+μ∴

LsinPM

LtanPMA 21

μ−

μ−=∴

( )P

MxPL

MMxcosP

MxsinLsinP

MLtanP

My 121121 −+

+μ+μ⎟⎟⎠

⎞⎜⎜⎝

⎛μ

−μ

−=∴ ………. (6)

(iii) 0dydx ; Lx ==

substitute cskM and skM A2A1 θ=θ=

0Lsin

LcosLLsincLsin

LLsinPL

sk A =⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛μ

μμ−μ+⎟⎟⎠

⎞⎜⎜⎝

⎛μμ−μθ

0PL

sk A ≠θ

LsinLcosLLLsincμ−μμ

μ−μ= ………. (7)

replace 2 assume and L α=ρπρπ=μ

we find: α−αα

α−α=

2sin2cos222sinc ………. (8)

(iv) Adydx ; 0x θ==

PLM

PLMLcosec

PMLcot

PM 2121

A ++⎟⎠⎞

⎜⎝⎛ μ−μμ−=θ∴

substitute cskM and skM A2A1 θ=θ=

⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛μμ−μ

+⎟⎟⎠

⎞⎜⎜⎝

⎛μ

μμ−μθ=θ

LsinLLsinc

LsinLcosLLsin

PLsk A

A

replace Lk

PL 22μ=

( )

2L

2Ltan

LcotL12L

sμ

−μ

μμ−μ

= ………. (9)

replace ρπ=α=μ 2L we obtain: ( )

α−ααα−α

=tan

2cot21s ………. (10)

The values of parameter s and c are functions of the strength ratio, ρ, which can be expressed as:

EIPL

PP

2

2

e π==ρ

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7.4.2 Stability Functions for the Compression and Tension Member (Non-Sway)

For the compression member, the stability functions parameter s and c as:

α−ααα−α

=2sin2cos2

22sinc ………. (11)

( )

α−ααα−α

=tan

2cot21s .......... (12)

Where, ρπ=α 2

The selected values of the stability functions for axial compression are listed in Table 1 and for intermediate values interpolation may be adopted.

For the tension member, the stability functions parameter s and c can be expressed

as:

( )( )S 2C22

SC 2 2sα+−−αα

= .......... (13)

( )( )S 2C22

2S 2scα+−α−α

= .......... (14)

Where ρπ=αα=α= 2 and, , 2cosh C , 2sinhS

The selected values of the stability functions for axial tension are listed in Table 2 and for intermediate values interpolation may also be adopted.

7.4.3 Stability Functions Curve

The curve below shows the plot of Equation 11 ~ 14 against the strength ratio, ρ , note that : if s and c (+) → stable, and if s and c (-) → instable.

-10

-8

-6

-4

-2

0

2

4

6

8

10

-4 -3 -2 -1 0 1 2 3 4

s c

Equation 14/s

Equation 12

Equation 11

Equation 13

-10

-8

-6

-4

-2

0

2

4

6

8

10

-4 -3 -2 -1 0 1 2 3 4

s c

Equation 14/s

Equation 12

Equation 11

Equation 13

ρ

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Table 1: Tabulated Selected Values of Stability Functions (Compression)

ρ s c s(1 – c 2) (sc)2

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Table 1: Tabulated Selected Values of Stability Functions (Compression)

Table2: Tabulated Selected Values of Stability Functions (Tension)

ρ s c s(1 – c 2) (sc)2

ρ s c s(1 – c 2) (sc)2

Elastic Instability

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BCABBC

AB 36.1 36.1 ρ=ρ→=ρρ

⎪⎭

⎪⎬

⎫

848.047174000

kk

BC

AB ==⎪⎭

⎪⎬

⎫

EXAMPLE 1 Considering a frame structure shown in figure below, the internal forces in members AB and BC can be calculated as follows:

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

21

1

1

AB

hLvh

Lv

WP

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

12

2

2

BC

hLvh

Lv

WP ;

Summing moment at point B :

( ) BBCBCABABB skskM θ+=∑ Note that : ∑ += )stable( MB

∑ −= )unstable( MB

∑ = )critical( 0MB Instability condition requires:

( ) 0skskM BCBCABABB =+=∑ Given: m 5.1h , m 6.2h , m 12.2L , m 0.3L 2121 ====

m 5.1v , cm 0.1000I , cm 0.1200I 4BC

4AB ===

1. Calculate the internal forces, assuming that the frame is pin-jointed :

367.1W

5.1 x 0.35.1 x 6.2

0.35.1

WPAB =⎟⎠⎞

⎜⎝⎛ +

= ; 116.1W

6.2 x 12.25.1 x 5.1

12.25.1

WPBC =⎟⎠⎞

⎜⎝⎛ +

=

2. Calculate the Euler’s load and the strength ratio:

kN 272510x 3

10 x 1200 x 207 x LEIP 62

42

2

2

EAB =π

=π

=

kN 454710x 12.2

10 x 1000 x 207 x LEIP 62

42

2

2

EBC =π

=π

=

3725W

27251 x

367.1W

PP

EAB

ABAB ===ρ

5074W

45471 x

116.1W

PP

EBC

BCBC ===ρ

33

4

1

ABAB mm 4000

10 x 310 x 1200

LIk ===

33

4

2

BCBC mm 4717

10 x 12.210 x 1000

LIk ===

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3. Instability criteria of the structure: ( ) 0skskM BCBCABABB =+=∑ → 0ss848.0 BCAB =+

4. Calculate the critical load, Wcr :

kN 87203725 x 34.2W

3725W

crAB ==→=ρ

kN 87300745 x 72.1W

5074W

crBC ==→=ρ

EXAMPLE 2 A rigid jointed steel frame ABC carry a vertical load W at B as shown in figure beside. Formulate the instability equation and find the critical load (Wcr) for the frame. Given :

m 0.4 v, m 0.4h , m 0.4h 21 === 4

BC4

AB cm 0.1000I , cm 0.1200I == 1. Calculate the internal forces, assuming that the frame is pin-jointed:

404.1W

0.4 x 7.50.4 x .04

7.50.4

W

hLvh

Lv

WP

21

1

1

AB =⎟⎠⎞

⎜⎝⎛ +

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

404.1W

0.4 x 7.50.4 x .04

7.50.4

W

hLvh

Lv

WP

12

2

2

BC =⎟⎠⎞

⎜⎝⎛ +

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

1st Trial 2nd Trial 3rd Trial 4th Trial 5th Trial 6th Trial ρBC 0.00 1.00 2.00 1.50 1.80 1.72 ρAB 0.00 1.36 2.72 2.04 2.45 2.34 sBC 4.00 2.47 0.14 1.46 0.72 0.93 sAB 4.00 1.76 -2.93 0.02 -1.52 -1.05

0.848sAB 3.39 1.49 -2.48 0.02 -1.29 -0.89 0.848sAB + sBC 7.39 3.96 -2.34 1.48 -0.57 -0.04

Different due to rounded of calculation

⎪⎭

⎪⎬

⎫

Elastic Instability

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BCABBC

AB 83.0 83.0 ρ=ρ→=ρρ

⎪⎭

⎪⎬

⎫

2.18.17673.2121

kk

BC

AB ==⎪⎭

⎪⎬

⎫

2. Calculate the Euler’s load and the strength ratio:

( ) kN 2.74010 x 24

10 x 1200 x 200 x LEIP

62

42

2

2

EAB =π

=π

=

( ) kN 8.61610 x 24

10 x 1000 x 200 x LEIP

62

42

2

2

EBC =π

=π

=

2.1039W

740.21 x

404.1W

PP

EAB

ABAB ===ρ

0.866W

616.81 x

404.1W

PP

EBC

BCBC ===ρ

3

3

4

1

ABAB mm 3.2121

10 x 2410 x 1200

LIk ===

33

4

2

BCBC mm 8.1767

10 x 2410 x 1000

LIk ===

3. Instability criteria of the structure:

( ) 0skskM BCBCABABB =+=∑ → 0ss2.1 BCAB =+

Instability occur when: 26.2 and 876.1 BCAB =ρ=ρ

4. Calculate the critical load, Wcr:

kN 5.1949039.21 x 876.1W 2.1039

WcrAB ==→=ρ

kN 2.195766.08 x 26.2W

0.866W

crBC ==→=ρ

1st Trial 2nd Trial 3rd Trial 4th Trial 5th Trial 6th Trial ρBC 0.00 1.00 2.00 2.10 2.20 2.26 ρAB 0.00 0.83 1.660 1.743 1.826 1.876 sBC 4.00 2.467 0.143 -0.176 -0.519 -0.739 sAB 4.00 2.765 1.078 0.875 0.663 0.496

1.2sAB 4.80 3.318 1.294 1.050 0.760 0.595 1.2sAB + sBC 8.80 5.785 1.437 0.874 0.241 -0.144

⎪⎭

⎪⎬

⎫Different due to rounded of calculation

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BCABBC

AB 83.0 83.0 ρ=ρ→=ρρ

⎪⎭

⎪⎬

⎫

2.10.20000.2400

kk

BC

AB ==⎪⎭

⎪⎬

⎫

EXAMPLE 3

A rigid jointed steel frame ABC carry a vertical load W at B as shown in below. Formulate the instability equation and find the critical load (Wcr) for the frame. Given: h1 = h2 = 3 m, v = 4 m, IAB = 1200 cm4, IBC = 1000 cm4

1. Calculate the internal forces, assuming that the frame is pin-jointed:

6.1W

0.3 x 0.50.4 x .03

0.50.4

W

hLvh

Lv

WP

21

1

1

AB =⎟⎠⎞

⎜⎝⎛ +

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

6.1W

0.3 x 0.50.4 x .03

0.50.4

W

hLvh

Lv

WP

12

2

2

BC =⎟⎠⎞

⎜⎝⎛ +

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

2. Calculate the Euler’s load and the strength ratio:

kN 5.94710x 5

10 x 1200 x 200 x LEIP 62

42

2

2

EAB =π

=π

=

kN 6.78910x 5

10 x 1000 x 200 x LEIP 62

42

2

2

EBC =π

=π

=

0.1516W

947.51 x

6.1W

PP

EAB

ABAB ===ρ

4.1263W

789.61 x

6.1W

PP

EBC

BCBC ===ρ

3

3

4

1

ABAB mm 0.2400

10 x 510 x 1200

LIk ===

33

4

2

BCBC mm 0.2000

10 x 510 x 1000

LIk ===

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3. Instability criteria of the structure:

( ) 0skskM BCBCABABB =+=∑ → 0ss2.1 BCAB =+

Instability occur when: 26.2 and 876.1 BCAB =ρ=ρ

4. Calculate the critical load, Wcr:

kN 0.2844.01651 x 876.1W

0.1516W

crAB ==→=ρ

kN 3.2855263.41 x 26.2W

4.1263W

crBC ==→=ρ

1st Trial 2nd Trial 3rd Trial 4th Trial 5th Trial 6th Trial ρBC 0.00 1.00 2.00 2.10 2.20 2.26 ρAB 0.00 0.83 1.660 1.743 1.826 1.876 sBC 4.00 2.467 0.143 -0.176 -0.519 -0.739 sAB 4.00 2.765 1.078 0.875 0.663 0.496

1.2sAB 4.80 3.318 1.294 1.050 0.760 0.595 1.2sAB + sBC 8.80 5.785 1.437 0.874 0.241 -0.144

⎪⎭

⎪⎬

⎫Different due to rounded of calculation