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PAPER PRESENTATION ON BUCKLING OF COLUMNS BY ARJUN JAYADEV(19ME63D08) 1.

Aim

To study the effect buckling of columns

2.

Preview (a)

Introduction

(b)

Basic concepts

(c)

Assumptions and sign condition for Euler’s equation

(c)

Euler’s equation for different end support

(d)

Limitations of Euler’s equation

(e)

Rankine formula and its importance

(f)

Conclusion

Introduction 3. The purpose of this paper presentation is to study the effects of buckling on a simple column under load. Buckling is one of the most common modes of failure in structural engineering. The Buckling majorly depends upon the slenderness ratio (l/d ratio), axial load and eccentric load. By varying the end supports, a column will experience different modes of buckling. Buckling of Columns is a form of deformation as a result of axial- compression forces. This leads to bending of the column, due to the instability of the column. This mode of failure is quick, and hence dangerous. If the load on a column is applied through the center of gravity of its cross section, it is called an axial load. A load at any other point in the cross section is known as an eccentric load. A short column under the action of an axial load will fail by direct compression before it buckles, but a long column loaded in the same manner will fail by buckling (bending), the buckling effect being so large that the effect of the direct load may be neglected. The intermediate-length column will fail by a combination of direct compressive stress and bending. .

Basic Concept: Buckling of Columns 2. A long column of uniform cross-sectional area A and of length l, subjected to an axial compressive load P. A column is known as long column if the length of the column in comparison to its lateral dimensions is very large. Such columns do not fail by crushing alone, but also by bending (also known buckling) The load, at which the column just buckles, is known as buckling load and it is less than the crushing load for a long column. Buckling load is also known as critical just (or) crippling load. The value of buckling load for long columns are long columns is low whereas for short columns the value of buckling load is high. Euler’s formula can be used to determine the buckling load since the stress in the column remains elastic. The critical load at which buckling occurs is a functions of slenderness ratio. The details of which is as appended below:(a) (L/r) - slenderness ratio. Buckling will occur about the axis when the ratio gives the greatest value. The graph below is used to identify the (L/r) for the column made of a structural steel. (b) For a steel column if (L/r) ≥ 89 , Euler’s formula can be used to determine the buckling load since the stress in the column remains elastic. But if the (L/r) s < 89, the column’s stress will exceed the yield point and the Euler formula is not valid.

fig (i)- Critical Stress Vs L/r

3. In order to understand the mathematical connotation. A numerical solution is derived as shown:Let l p

= =

A e

= =

length of the long column Load (compressive) at which the column has jus buckled. Cross-sectional area of he column Maximum bending of the column at the centre.

=

P A

σ0

=

Stress due to direct load

σb

=

Stress due to bending at the centre of the column

=

P×e Z

Where Z = Section modulus about the axis of bending. The extreme stresses on the mid-section are given by

Maximum stress =

σ0

+

σb

Minimum stress =

σ0 -

σb

σ σ 4. The column will fail when maximum stress (i.e) 0 + b is more the crushing stress fc. In case of long column, the direct compressive stresses are negligible as compared to buckling stresses. Hence very long columns are subjected to buckling stresses.

Equilibrium Conditions. 5. Analysing simplified column consisting of two rigid beams of length l/2 each with simple supports is joined together with a torsion . The torsion spring has a spring constant of K which relates the torsion (or moment) at the angle of rotation given by M = 2Kθ (when the spring is loaded equally from both ends). The free body diagram in figure 1 allows us to write the bending moment balance as:∑M = M - Pd = M – P*θ*L/2 =0

-------- eqn (i)

Therefore , 2Kθ – P*θ*L/2=0 P(critical)=4*K/L which gives a value for the load which will cause an angular rotation other than 0. So, there are three possibilities: (a)

0 < P < Pcr: Stable equilibrium. i.e. no buckling at all.

(b)

P = Pcr: Neutral equilibrium.

(c)

P > Pcr: Unstable equilibrium. i.e. will buckle if perturbed.

Fig (ii)- Analysis of Equilibrium Condition on Column with Torsion Spring

6.

ASSUMPTIONS OF EULER’S COLUMN THEORY: (a) The column is initially perfectly straight and the load is applied axially (b) The cross-section of the column is uniform throughout its length. (c) The column material is perfectly elastic, homogeneous and isotropic and obeys Hooke’s law. (d) The length of the column is very large as compared to its lateral dimensions (e) The direct stress is very small as compared to the bending stress (f) The column will fail by buckling alone.

(g)

The self-weight of column is negligible.

EULERS EQN FOR DIFFERENT END SUPPORT 7. The boundary conditions of a column may, however, be idealized in one the following ways . All the cases have been derived subsequently P P

P

P

A A

A

A

B

Le=L

Le=0.7L

Le= 0.5L

B

B

B

Both ends hinged

Both ends fixed.

Le=2L

One end fixed other end hinged

One end fixed other end free

Fig (iii)- Different End Support 8.

Condition I : Both Ends Hinged (a) Consider a column AB of length L hinged at both its ends A and B carries an axial crippling load at A. Consider any section X-X at a distance of x from B. Let the deflection at X-X is y.  The bending moment at X-X due to the load P, M = −P. y 2

d y −Py = =−k 2 y 2 EI dx

k2= (Where

2

d y 2 +k y=0 dx 2

` Solution of this differential equation is

y= A cos kx+ B sin kx

p EI )

y= A cos x

(√ )

p +B sin x EI

 By using Boundary conditions, At B, (x = 0, y = 0  A = 0) At A, (x = l, y = 0)

0=B sin l



l×

l

√

√

√

p EI

Sinl

√

(√ ) p EI

p =0 EI

p =0 , π , 2 π , 3 π . . .. .. EI (Now taking the least significant value (i.e) )

p =π EI

l2 ;

( EIp )=π

2

π 2 EI p= 2 l

`The Euler’s crippling load for long column with both ends hinged.

9.

Condition II :Both Ends Fixed. (a) Consider a column AB of length l fixed at both the ends A and B and caries an axial crippling load P at A due to which buckling occurs. Under the action of the load P the column will deflect as shown in fig. (b) Consider any section X-X at a distance x from B.Let the deflection at X-X is y.Due to fixity at the ends, let the moment at A or B is M.  Total moment at XX = M – P.y Differential equation of the elastic curve is 2

d y =M−Py dx 2 2 d y py M + = dx 2 EI IE EI

2

d y py M p + = × dx 2 EI IE p 2 d y py P M + = × dx 2 EI EI P

The general solution of the above differential equation is

y= A cos x ( √ P /EI ) +B sin x ( √ P/ EI ) +

M P

Where A and B are the integration constant At, N. x = 0 and y = 0  From (i)

0 =A×1 +B× 0 +

A=−

M p

M p

Differentiating the equation (i) with respect to x,

√

√ (√)

dy P P P +0 Sin ( x . √ P/ EI )+B Cos x . =−A EI EI EI dx At the fixed end B, x = 0 and



B

√

P =0 EI

Either B = 0 (or) Since

√

P ≠0 EI

√

P =0 EI

as p ¿ 0 B=0

dy =0 dx

(i)

A=− Subs

M p

and B = 0 in equation (i)

y=−

( √ )

M cos x . P

M P + EI P

[ ( √ )]

P M 1−cos x .. EI P

y=

Again at the fixed end A, x = l, y = 0

0=

M [ 1−Cos (l . √ P /EI ) ] P

l.√ P/ EI=0,2π ,4 π ,6π ........ Now take the least significant value 2

√

P =2 π EI P l.2 × =4 π 2 EI l.

2

P= 

4 π EI l2

The crippling load for long column when both the ends of the column are fixed

2

P=

10.

4 π EI L2

Condition III : One End Fixed Other End Hinged. (a) Consider a column AB of length l fixed at B and hinged at A. It carries an axial crippling load P at A for which the column just buckles. As here the column AB is fixed at B, there will be some fixed end moment at B. Let it be M. To balance this fixing moment M, a horizontal push H will be exerted at A.

Consider any section X-X at a distance x from the fixed end B. Let the deflection at xx is y. Bending moment at xx = H (l-x) - Py

Differential equation of the elastic curve is, 2

EI

d y =H ( l−x) −Py dx 2

2 14 (l−x ) d y P + y= 2 EI dx EI 2 d y P y = H (l−x ) p + × P EI dx 2 EI 2 H (l−x ) p d y P + y= × 2 EI EI dx EI

The general solution of the above different equation is

) ( √ EIp )+B sin( x . √ EIp )+ H (l−x P

y= A cos x.

Where A and B are the constants of integration.

(i)

At B, x = 0, y = 0

From (i)

B

√

A=

−Hl P

P H = EI P

√

H EI B= × P p Again at the end A, x = l, y=0.  substitute these values of x, y, A and B in equation (i)

0=−

H P

Hl H Cos ( l . √ P / EI )+ P P

√

EI Sin ( l . √ P / EI ) P

(√ EIp Sin . (l . √ P / EI ) )= HlP Cos ( l .√ P/ EI)

tan (l . √ P/ EI . l )=√ P/ EI . l (b) The value of tan ( P/ EI . l ) in radians has to be such that its tangent is equal to itself. The only angle whose tangent is equal to itself, is about 4.49 radians.

√

√P /EI .l=4 . 49 P 2 l =( 4 . 49 )2 EI P 2 l =2 π 2 EI

(approx)

2 π 2 EI P= l2 The crippling load (or) buckling load for the column with one end fixed and one end hinged. 2

2 π EI P= l2

11. Condition IV : One End Fixed Other End Free. Consider a column AB of length l, fixed at B and free at A, carrying an axial rippling load P at D de to which it just buckles. The deflected form of the column AB is shown in fig. Let the new position of A is A1. Let a be the deflection at the free end. Consider any section X-X at a distance x from B. Let the deflection at xx is y. Bending moment due to critical load P at xx,

d2 y M=EI 2 =P ( a− y ) dx 2 d y EI 2 =Pa − py dx d 2 y py pq + = dx 2 EI EI The solution of the above differential equation is,

( √ EIP )+ B sin ( x. √ EIP )+a

y= A cos x.

Where A and B are constants of

integration. At B, x = 0, y = 0  From (i), A = 0 Differentiating the equation (I w.r. to x

√ (√ ) √ (√ )

dy P P P P +B Sin x. Cos x. =−A EI EI EI EI dx dy 0 At the fixed end B, x = 0 and dx

0=B

As

√

√

P EI

P ≠0 EI

( ∴ p≠0)

Substitute A = -a and B = 0 in equation (i) we get,

( √ EIP )+a

y=−a cos x .

[ ( √ )]

y=a 1−cos x..

P EI

(ii)

At the free end A, x = l, y = a, substitute these values in equation (ii)

[ ( √ )] ( √ )

a=a 1− cos 1 . .

cos 1 . .

1

P EI

P =0 EI

√

P π 3 π 5π = , , EI 2 2 2

√

P π = EI 2

Now taking the least significant value,

1

2 P π = 1 EI 4 2

2

π EI P= 2 4l The crippling load for the columns with one end fixed and other end free. 2

π EI P= 2 4l

12.

Rankine’s Formula and its Importance. (a) Euler’s formula gives correct results only for long columns, which fail mainly due to buckling. Whereas Rankine’s devised an empirical formula based on practical experiments for determining the crippling or critical load which is

applicable to all columns irrespective of whether they a short or long.The derivation of Rankine formula is as under:If P is the crippling load by Rankine’s formula. Pc is the crushing load of the column material PE is the crippling load by Euler’s formula.

Then the Empirical formula devised by Rankine known as Rankine’s formula stand as:

1 1 1 = + P Pe P E (b)

For a short column, if the effective length is small, the value of P E will be

very high and the value of is negligible. For the short column, (i.e)

1 PE

will be very small as compared to

1 1 = Pc P

1 PC and

P = PC

(c) Thus for the short column, value of crippling load by Rankine is more or less equal to the value of crushing load. For long column having higher effective

length, the value of P E is small and

to

1 PC

. So

1 PC

1 PE

will be large enough in comparison

is ignored.

1 1 PC  P E  For the long column,

(i.e) p  PE (d) Thus for the long column the value of crippling load by Rankine is more or less equal to the value of crippling load by Euler.

1 1 1 =P + P P E c P × Pc 1 = E Pc × PE P p=

p=

Pc ×P E P E × Pc

;

Substitute the value of Pc = fc A and equation,

p= 1+

Pc P 1+ c PE

π 2 EI PE = 2 L

in the above

f c× A f c×A

π 2 EI / L2 Where, fc = Ultimate crushing stress of the column material. A

=

Cross-sectional are of the column

L

=

Effective length of the column

I

=

Ak2

Where k = Least radius of gyration.

f c× A

p= 1+

p=

f c×A π 2 EI / L2

f c× A

= 1+

f c × A × L2 π 2 EAk 2

f c× A

( )

L 1+α K

2

= where  = Rankine’s constant

fc π2 E

CrushingLoad 1+α ( L/k ) 2 P= CONCLUSION 14. If buckling deflections become too large then the structure fails - this is a geometric consideration, completely divorced from any material strength consideration. If a component or part is prone to buckling then its design must satisfy both strength and buckling safety .Buckling has become more of a problem in recent years since the use of high strength material requires less material for load support - structures and components have become generally more slender and buckle- prone 15.

BIBLIOGRAPHY (a)

(b)

BOOKS: (i) Prof SS Ratan: Strength of Materials, Tata Macgraw hill, second edition (ii) Prof RK Bansal: Strength of Materials,2001 WEB SOURCE (i) https://web.mit.edu/16.unified (II) https://wp.optics.arizona.edu/optomech/wp-content/uploads/sites...