Buckling of Column With Two Intermediate Elastic Restraints Thesis Presentation 15.11.2007...

Buckling of Column With Two Intermediate Elastic Restraints

Thesis Presentation 15.11.2007

Author: Md. Rayhan ChowdhuryMohammad Misbah UddinMd. Abu Zaed KhanMd. Monirul Islam Masud

Supervisor: Dr. Mohammad Nazmul Islam

PRESIDENCY UNIVERSITY

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Slide title 40 pt

Slide subtitle 24 pt

Text 24 pt

Bullets level 2-520 pt

Department of Civil Engineering Undergraduate Thesis Presentation 2007-11-152 PRESIDENCY UNIVERSITY

Introduction Background

– In a laterally loaded cross-bracing system, one bracing member will be under compression while the other member subjected to tension. The tension brace may be modeled as a discrete, lateral elastic spring attached to the compression member. Thus, the prediction of the elastic buckling loads of columns with two intermediate elastic restraints is therefore of practical interest.

Objectives– The main objective of this theoretical research is to find a set of

stability criteria for Euler columns with two intermediate elastic restraints.

Scope– The scope of this thesis is the derivation of Euler column buckling

theory.


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Text 24 pt



Model

Fig. Column with two intermediate elastic restraints

P

z, w

x

c1

c2L

a2L

a1L

Spring 1

Spring 2

Here,

Column Length, L Flexural Rigidity, EI Spring 1

– Stiffness, c1

– Located at a1L (a1 < L) Spring 2

– Stiffness, c2

– Located at a2L (a1 ≤ a2 ≤ L) The entire column can be divided into

three segments as – Segment-1: 0 ≤ x ≤ a1L– Segment-1: a1L ≤ x ≤ a2L– Segment-1: a2L ≤ x ≤ L


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Governing equation for column buckling

4 22

4 20i id w d w

dx dx

Where i = 1, 2, 3 denote the quantity belonging to segment 1, segment 2 and segment 3.

and/ ,x x L / ,w w L 2 2 /( )PL EI

1 1 2 3 4sin cosw A x A x A x A 10 x a

2 1 2 3 4sin cosw B x B x B x B 1 2a x a

3 1 2 3 4sin cosw C x C x C x C 2 1a x

(1)for

(2)for

(3)for

General Solution:


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Continuity condition at a1

31 1 /( )c L EI

(4)

(5)

Where,

(6)

(For deflection, slope, bending moment and shear force )

11

1 2 0x a x aw w

11

1 2 0x ax a

dw dw

dx dx

1

1

2 21 2

2 20x a

x a

d w d w

dx dx

1

1 1

3 22 21 1 2 2

1 13 30x a

x a x a

d w dw d w dww

dx dxdx dx

(7)


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Text 24 pt



Continuity condition at a2

32 2 /( )c L EI

(8)

(9)

Where,

(10)

(For deflection, slope, bending moment and shear force )

(11)

22

2 3 0x a x aw w

22

32 0x ax a

dwdw

dx dx

2

2

2232

2 20x a

x a

d wd w

dx dx

2

2 2

232 23 32 2

2 23 30x a

x a x a

d w dwd w dww

dx dxdx dx


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Continuation…

(12)

(13)

(14)

(15)

11 1 1 1 2 1 3 1 4 13

cos( )[ sin( ) cos( ) ]B a A a A a A a A A

12 1 1 1 2 1 3 1 4 23

sin( )[ sin( ) cos( ) ]B a A a A a A a A A

21

3 1 1 2 1 3 1 422

[ sin( ) cos( ) ( ) ]B A a A a A a A

21 1

4 1 1 2 1 3 1 421 1

[ sin( ) cos( ) (1 )a

B A a A a A a Aa

Substituting Eqs. (1) and (2) into Eqs. (4)-(7), a set of homogeneous equations is obtained which may be expressed in forms of Bi in terms of Ai, i.e.,


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Continuation…

(16)

(17)

(18)

(19)

Substituting Eqs. (2) and (3) into Eqs. (8)-(11), another set of homogeneous equations is obtained which may be expressed in forms of Ci in terms of Bi, i.e.,

21 2 1 2 2 2 3 2 4 13

cos( )[ sin( ) cos( ) ]C a B a B a B a B B

22 2 1 2 2 2 3 2 4 23

sin( )[ sin( ) cos( ) ]C a B a B a B a B B

22

3 1 2 2 2 3 2 422

[ sin( ) cos( ) ( ) ]C B a B a B a B

22 2

4 1 2 2 2 3 2 422 2

[ sin( ) cos( ) (1 )a

C B a B a B a Ba


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Continuation…

(20)

(21)

(22)

(23)

Hence,

Substituting Eqs. (12) - (15) into Eqs. (16)-(19), we get Ci in terms of Ai

1 1 2 3 4( , , , )C A A A A

1 1 2 3 4( , , , )C A A A A

1 1 2 3 4( , , , )C A A A A

1 1 2 3 4( , , , )C A A A A


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Boundary Condition (fixed – free)

(24)

(25)

(26)

(27)

At the fixed end:

Fig. Boundary condition, fixed - free

P

z, w

x

c1

c2L

a2L

a1L

Spring 1

Spring 2

1 00

xw

1

0

0x

dw

dx

23

21

0x

d w

dx

223 3

31

0x

d w dw

dxdx

At the free end:


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B. C. (fixed – free) continuation

(24)

(25)

(26)

(27)

Differentiating the boundary equation:

2 4 0A A

1 3 0A A

21 2( sin( ) cos( )) 0C C

23 0C

If we substitute the value of Ci into Eqs. (26) and (27), the buckling problem involves only four constants

Ai (i = 1,2,3,4).

[ ]{ } {0}M A

a.

Hence, we can develop the Eigen value equation from the boundary equation in form

b.

Where {A} = (A1, A2, A3, A4) and [M] is the coefficient matrix of {A}.

Finally the determinant of matrix[M] yields the stability criteria.


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B. C. (fixed – free) continuation

ξξ11=20, ξ2=20: (ξ=20, ξ2=20: (ξ11=c=c11LL33 and ξ and ξ22=c=c22LL33) )

λλ22= PL= PL22/(EI)/(EI)

a2a2 α = 0α = 0 α = 0.2α = 0.2 α = 0.4α = 0.4 α = 0.6α = 0.6 α = 0.8α = 0.8 α = 1α = 1

0.10.1 4.71014.7101 4.71014.7101 4.714.71 4.70984.7098 4.70914.7091 4.70784.7078

0.30.3 4.56674.5667 4.56634.5663 4.56094.5609 4.53954.5395 4.48884.4888 4.40044.4004

0.50.5 4.14764.1476 4.14464.1446 4.1074.107 3.98553.9855 3.7943.794 3.6423.642

0.70.7 4.10444.1044 4.09624.0962 4.00594.0059 3.8073.807 3.69133.6913 3.78163.7816

0.90.9 4.46024.4602 4.43964.4396 4.24324.2432 3.99093.9909 4.07484.0748 4.35744.3574

The following tables present the buckling load parameter for different locations and stiffness of the intermediate restraints:


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Conclusions

Exact stability criteria for columns with two intermediate elastic restraints at arbitrary location along the column length are derived.

This stability criteria can be used to determine the buckling capacity of compressive member in a cross-bracing system.


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?


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Thank you all.

PRESIDENCY UNIVERSITY

Buckling of Column With Two Intermediate Elastic Restraints Thesis Presentation 15.11.2007...

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