Stability

of

columns

Columns and sturts: Structural members subjected to compression and which are relatively long compared to their lateral dimensions are called columns or Struts. Generally, the term column is used to denote vertical members and the term strut denotes inclined members

Examples: strut in a truss, Piston rods, side links in forging machines, connecting rods etc.

Stable, Neutral and unstable Equilibrium

Stable equilibrium: A stable equilibrium is one in which a body in static equilibrium on being displaced slightly, returns to its original position and continues to remain in equilibrium.

Neutral equilibrium: A neutral equilibrium is one in which a body in equilibrium, on being displaced does not returns to its original position, but its motion stops and resumes its equilibrium state in its new position.

Unstable equilibrium: An unstable equilibrium is one in which a body in equilibrium on being slightly disturbed, moves away from its equilibrium position and loses its state of equilibrium.

Buckling Load: The maximum load which a column can support before becoming unstable is known as buckling load or crippling load or critical load

The buckling takes place about the axis having minimum radius of gyration or least moment of inertia.

At this stage, the maximum stress in the column will be less than the yield stress (crushing stress) of the material.

Safe load: It is the load to which a column is subjected to and is well below the buckling load. It is obtained by dividing the buckling load by a suitable factor of safety.

safe load = buckling load / factor of safety

Stability factor: The ratio of critical load to the allowable load on a column is called stability Factor

MODES OF FAILURE OF THE COLUMNS

P

P

The load carrying capacity of a short column depends only on its cross sectional area(A) and the crushing stress of the material(σcu). The crushing load Pu for axially

loaded short column is given by Pcu =σcu × A .

The safe load on the column is obtained by dividing the crushing load by suitable factor of safety. i.e., Psafe =Pcu/ FS

The mode of failure of columns depends upon their lengths and depending on the mode of failure columns are classified as

a. Short columns b. Long columns

Short Columns: A short column buckles under compression as shown in figure and fails by crushing. The load causing failure is called crushing load.

Long columns: Long columns, which are also called slender columns, when subjected to compression, deflects or bends in a lateral direction as shown in the figure. The lateral deflection of the long column is called buckling.

The load carrying capacity of long column depends upon several factors like the length of the column, M.I of its cross–section, Modulus of elasticity of the material, nature of its support, in addition to area of cross section and the crushing strength of the material.

Critical load denotes the maximum load carrying capacity of the long column.

The long column fails when there is excessive buckling .ie when the load on the column exceeds critical load.

SC – 10

Short columns fails by crushing or yielding of the material under the load P1

Long column fails by buckling at a substantially smaller load P2 (less than P1).

P

1

P2

The buckling load is less than the crushing load for a long column

The value of buckling load for long column is low whereas for short column the value of buckling load is relatively high.

sc-11Failure of long columns(contd)

Stress due to buckling

σb = ( M.ymax)/ I

= {(P.e). ymax}/ I

= ( P.e) / ZWhere e = maximum bending of the column

at the centre

Consider a long column of uniform cross sectional area A throughout its length L subjected to an axial compressive load P. The load at which the column just buckles is known as buckling load or crippling load.

Stress due to axial load σc = P/AP

L e

 

Sc-12Failure of long columns(contd)

σmax = σc + σb

Extreme stress at centre of column will be the sum of direct compressive stress and buckling stress

In case of long columns, the direct compressive stresses are negligible when compared to buckling stress. So always long columns fail due to buckling.

Modes of failures (contd.)

Intermediate Columns: These are columns which have moderate length, length lesser than that of long columns and greater than that of short columns.

sc-13

In these columns both bulging and buckling effects are predominant. They show the behavior of both long columns and short columns when loaded.

Euler’s Theory (For long columns)

Assumptions:

1. The column is initially straight and of uniform lateral dimension

2. The material of the column is homogeneous, isotropic, obeys Hookes law

3. The stresses are within elastic limit

4. The compressive load is axial and passes through the centroid of the section

5. The self weight of the column itself is neglected.

6. The column fails by buckling alone

Euler’s Theory (For long columns)

A Bending moment which bends the column as to present convexity towards the initial centre line of the member will be regarded as positive

Bending moment which bends the column as to present concavity towards the initial centre line of the member will be regarded as negative

Sign convention for Bending Moments

Euler’s Formula for Pin-Ended Beams (both ends hinged)

Consider an axially loaded long column AB of length L. Its both ends A and B are hinged. Due to axial compressive load P, let the deflection at distance x from A be y.

d2ydx2

d2y dx2 +

PyEI = 0

L

y x

P

P A

B The bending moment at the section is given by

EI

= - P y

-ve sign on right hand side, since as x increases curvature decreases

At x =0, y =0,we get c1=0 (from eq.1)

Also at x=L, y =0 we get

c2 .sin [L√P/(EI)] =0

If c2 = 0, then y at any section is zero, which means there is no lateral deflection which is not true

Therefore sin [L√P/(EI)] =0

This is the linear differential equation, whose solution is

Y = c1.cos [x√P/(EI)] + c2.sin[x √P/(EI)] …(1)

Where c1 and c2 are the constants of integration. They can be found using the boundary conditions.

sin [L√P/(EI)] =0

=> [L√P/(EI)] = 0, π, 2 π ,……n π

Taking least non zero value we get

[L√P/(EI)] = π

Squaring both sides and simplifying

PE =π2E I

L2

This load is called critical or buckling load or crippling load

case End condition Equivalent

length(Le) Euler’s Buckling load

1 Both ends hinged Le=L PE= (π 2E I) / Le2

2 One end fixed, other end free

Le=2L PE= (π 2EI) / 4L2

3 One end fixed, other end pin jointed

Le=L / √2 PE= 2(π 2EI) / L2

4 Both ends fixed Le=L/2 PE= 4(π 2 EI) / L2

Note: L is the actual length of respective column and Le is to be considered in calculating Euler's buckling load

Extension of Euler’s formula

Slenderness ratio: It is the Ratio of the effective length of the column to the least radius of gyration of the cross sectional ends of the column.

The Effective length: of a column with given end conditions is the length of an equivalent column with both ends hinged, made up of same material having same cross section, subjected to same crippling load (buckling load) as that of given column.

Slenderness ratio, λ =Le /k

Least radius of gyration, k= √ Imin/A

Imin is the least of I xx and I yy

L =actual length of the column

Le=effective length of the column

A= area of cross section of the column

Based on slenderness ratio ,columns are classified as short ,long and intermediate columns.

Generally the slenderness ratio of short column is less than 32 ,and that of long column is greater than 120, Intermediate columns have slenderness ratio greater than 32 and less than 120.

Limitation of Euler's theory

Pcr= (π 2 EI) / Le

2

But I =Ak2

∴ Pcr/A= π 2E/(Le/K)2

σcr = π2E/(Le/K)2

Where σcr is crippling stress or critical stress or stress at failure

The validity of Euler’s theory is subjected to condition that failure is due to buckling. The Euler’s formula for crippling is

The term Le/K is called slenderness ratio. As slenderness ratio increases critical load/stress reduces. The variation of critical stress with respect to slenderness ratio is shown in figure 1. As Le/K approaches to zero the critical stress tends to infinity. But this cannot happen. Before this stage the material will get crushed.

∴ σc = π 2E/(Le/K)2

Le/K= √ (π2E / σc)

For steel σc = 320N/mm2

and E =2 x 105 N/mm2

Limiting value (Le/K) is given by

(Le/K)lim =√ (π2E / σc) = √ π2 × 2 × 105/320) = 78.54

Hence, the limiting value of crippling stress is the crushing stress. The corresponding slenderness ratio may be found by the relation

σcr = σc

Hence if Le /k < (Le /k)lim Euler's formula will not be valid.

Empirical formula or Rankine - Gordon formula

PR = crippling load by Rankine’s formula

Pc = crushing load = σc .A

PE = buckling load= PE= (π 2 EI) / Le 2

We know that, Euler’s formula for calculating crippling load is valid only for long columns.

But the real problem arises for intermediate columns which fails due to the combination of buckling and direct stress.

The Rankine suggested an empirical formula which is valid for all types of columns. The Rankine’s formula is given by,

1PR

1PE

1PC

= +

For short columns: The effective length will be small and hence the value of PE =(π

2 EI) / Le

2 will be very large.

Hence 1/ PE is very small and can be neglected.

therefore 1/ PR= 1/ Pc or PR =Pc

For long column: we neglect the effect direct compression or crushing and hence the term 1/ Pc can be neglected.

therefore 1/ PR= 1/ PE or PR =PE

Hence Rankines formula,

1/ PR= 1/ Pc + 1/ PE is satisfactory for all types of columns

Eawhere

KLea

AP

E

KLe

A

KE

Le

A

LeAKE

AA

LeEI

AA

P

Le

EIPandAPngsubstituti

PP

P

PP

PPP

PP

PP

P

PPP

c

cR

c

c

c

c

c

c

c

cR

EcC

E

C

C

EC

ECR

EC

EC

R

ECR

2

2222

2

2

2

2

2

2

2

2

2

)/(1

)/(11

)(11

1

1

111

(I=AK2)

where a = Rankine’s constant =σc / π 2EIand λ = slenderness ratio = Le/ k

PR = σcA / (1+a.λ2 )

ILLUSTRATIVE NUMERICAL EXAMPLES

Euler’s crippling load =PE= (π2EI) / Le 2

= [π 2 × 200 × 109 × π × (0.06)4 /64] / (1.7682)

= 401.7 ×103 N =401.4 kN Safe compressive load = PE /3 =133.9kN

1. A solid round bar 60mm in diameter and 2.5m long is used as a strut. One end of the strut is fixed, while its other end is hinged. Find the safe compressive load, for this strut, using Euler’s formula. Assume E=200GN/m2 and factor of safety =3.

end condition: one end hinged, other end fixed

effective length Le = L /(√ 2)= 2.5/ (√ 2)= 1.768m

Solution:

outside diameter of the column =D =50mm =0.05m; E=70 × 10 9N/m2

Inside diameter = ?

2.A slender pin ended aluminium column 1.8m long and of circular cross-section is to have an outside diameter of 50mm. Calculate the necessary internal diameter to prevent failure by buckling if the actual load applied is 13.6kN and the critical load applied is twice the actual load. Take Ea = 70GN/m2.

Solution:

End condition: pin-ended ( hinged)

Le =L =1.8m

Euler’s crippling load =PE= π 2 (EI) / Le 2

d = 0.0437m = 43.7mm

2

44

3

8.164

)05.0(1070

102.27

92d

Critical load =PE = 2 × safe load (given condition) = 2 × 13.6=27.2kN

I= π (D4-d4) /64 = π (0.054-d4) /64

3. A built up beam shown in the figure is simply supported at its ends. Compute its length, given that when it subjected to a load of 40kN per metre length. It deflects by 1cm. Find the safe load, if this beam is used as a column with both ends fixed. Assume a factor of safety of 4. use Euler’s formula. Take E = 210GN/m2.

300 mm 50 mm

1000 mm

20 mm

L = 14.15m

Moment of inertia of section about X-X axis,

Load =40kN/m , length of the beam =?

= 994166 × 104 mm4.= 99.41×10-4m

Using the relation, δ = 5 wL4 384EI

0.01 = 5 × 40 × 10 3 × L 4

( 384 × 210 × 109 × 99.41 × 10-4

12

100020525)50300(

12

503002

332

xxI

Safe load, the beam can carry as column:

End condition: Both ends fixed

PE = π 2 (EIyy) / Le 2

= (π 2 × 210 × 109 × 2.25 × 10 -4) / (7.07)2

= 9.33 × 10 6 N = 9.33 × 10 3 kN

Safe load = Pe /F.S = 9.33 × 10 3 / 4 = 2.333 × 10 3 kN

Le = L/2 = 14.15/2 = 7.07m

Iyy = 2[ (50 × 300 3) /12] + (1000 × 20 3 ) /12

= 22567 × 10 4 mm4 = 2.25 × 10 -4 m4

4.From the test on steel struts with ends fixed in position and fixed in direction the following results are obtained.

Assuming the values in agreement with Rankine’s formula ,find the two constants

Rankine’s critical load = PR = σcA / (1+a.λ2 )

Rankine’s critical stress = PR / A

200= σc / [1+a.(70 2 ) ] …. (1)

69= σc / [1+a.(170 2 ) ] …(2)

(1) / (2) gives

constant a = 1.29 × 10 -4

substituting ‘a ‘ in (1) or (2)

we get σc = 326.4 N/ mm

22

2

2

)170(1)70(18986.2

)70(1

)170(1

69

200

aa

a

a

5.Find the Euler’s crushing load for a hollow cylindrical cast iron column, 15cm external diameter and 2cm thick, if it is 6m long and hinged at both ends. E = 80GPa. Compare this load with the crushing load as given by the Rankine’s formula, using σc= 550MPa and a =1/600. For what length of the strut of this

cross-section does the Euler’s formula ceases to apply ?

Solution: Internal diameter = 15 – 2 × 2 = 11 cm

A = π/4[ 0.152- 0.112 ] = 81.7 × 10-4 m2.

I = π [ 0.154 - 0.114 ] /64

=17.66 × 10-6 m4 .

A

IK min = 0.0465 m

Euler’s critical load is given by

PE = π 2 (EI) / Le 2

= (π 2 × 80 × 10 9 × 17.66 × 10 -6 ) / 62

= 387327.14 N (higher)

Rankine’s critical load, PR = (σc A) / [1+ a (Le / K)2]

= (550 × 10 6 × 81.7 × 10 -4)/ [1+1/600 (6/0.0465)2]

= 156301.78

To calculate limiting length : σc = 550 MPa =550N/mm2 = PE / A

550 = π 2 (EI) / A Le 2

therefore Le = 1.761m

6. The built up column shown in the figure consisting of 150mm × 100mm RSJ with 120mm wide plate riveted to each flange. Calculate the safe load, the column can carry , if it is 4m long having one end fixed and other end hinged with a factor of safety 3.5. Take the properties of the joist as

A = 21.67 × 102 mm2 ; Ixx = 839.1 × 104 mm4 ; Iyy = 94.8 × 104 mm4.

Assume Rankine’s constant as 315N/mm2 and a =1/7500

120mm

100mm

150mm

12m

m

Solution:

Ixx = 839.1 × 104 + 2[ (120 × 123)/12 + 120 × 12 ×( 75 +6)2]

= 2732.1 × 104 mm4

= 2732.1 × 10-8 m4

Similarly, Iyy = 94.8 × 10 4 + 2[ (12 × 1203 )/12 ] mm4

= 440.4 × 104 mm4

= 440.4 × 10-8 m4

(Iyy is the lower value, column will tend to buckle in YY direction, Iyy has to be considered)

A = 21.67×102 + 2 (120×12) = 5047 mm2=50.47×10-4 m2

Exercise problems

1. Calculate the safe compressive load on a hollow cast iron column one end fixed and other end hinged of 150mm external diameter,100mm internal diameter and 10m length. Use Euler's formula with a factor of safety of 5 and E=95GN/m2

Ans: 74.8kN

2. Bar of length 4m when used as a simply supported beam and subjected to a u.d.l of 30kN/m over the whole span., deflects 15mm at the centre. Determine the crippling loads when it is used as a column with the following end conditions:

(i) Both ends pin jointed (ii) one end fixed and other end hinged (iii) Both ends fixed

Ans: (i) 4108 kN (ii) 8207kN (iii) 16432 kN

sc - 42Exercise problems (contd)

3.Determine the ratio of the buckling strengths of two columns of circular cross-section one hollow and other solid when both are made of the same material, have the same length, cross sectional area and end conditions. The internal diameter of the hollow column is half of its external diameter

Ans: 1.66

4. Calculate the critical load of a strut 5m long which is made of a bar circular in section and pin jointed at both ends. The same bar when freely supported gives mid span deflection of 10mm with a load of 80N at the centre.

Ans: 8.22kN

sc - 43Exercise problems (contd)

5. A hollow C.I column whose outside diameter is 200mm has a thickness of 20mm. It is 4.5m long and is fixed at both ends. Calculate the safe load by Rankine’s formula using a factor of safety of 4. Take σc =550MN/m2 , a=1/1600

Ans: 0.877 MN

6. A hollow cylindrical cast iron column is 4m long with both ends fixed. Determine the minimum diameter of the column, if it has to carry a safe load of 250kN with a factor of safety of 5. Take the internal diameter as 0.8 times the external diameter.

σC =550MN /m 2 a= 1/1600

Ans: D= 136mm d= 108.8mm