11. column stability supports

7/23/2019 11. Column Stability Supports

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COLUMN & SUPPORT

COLUMN STABILITYAND

SUPPORT

Ahmad Shufni Othman

July 2012


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LEARNING OUTCOMES

Explaining the following :

Column and support

Type of column failure

Type of column ends and support

Understanding the Euler Theorem and buckling load

(crippling load) formulae.

Determining the effective length and buckling load for a

slender column with different types of supports.


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COLUMNS (or struts)

Columns are vertical member of a structure, which issubjected to axial compressive loads. Struts is a member of

a structure which is not vertical with one or both ends

hinged or pinned jointed.

The cross-sectional dimensions are relatively small

compared with their length in the direction of thecompressive force.

They are also referred to as pillars, posts, and stanchions.


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Columns fail in compression. In civil

engineering they are often made of brittlematerial which is strong in compression such

as cast iron, stone and concrete. These

materials are weak in tension so it is important

to ensure that bending does not produce

tensile stresses in them. If the compressivestress is too big, they fail by crumbling and

cracking.

COLUMNS (or struts)


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TYPE OF COLUMNS FAILURE

(crushing)

All short columnsfail due to crushing

whereas long

columns fail due to

buckling and

crushing.
http://www.world-housing.net/uploads/101123_111_25.jpg


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FORMATIONS OF SHORT COLUMNS


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Euler's buckling equation

What is buckling?

In engineering, buckling is a

failure mode characterized by a

sudden failure of a structuralmember subjected to high

compressive stresses, where

the actual compressive stress at

the point of failure is less than

the ultimate compressivestresses that the material is

capable of withstanding.A column under a concentric

axial load exhibiting the

characteristic deformation of

buckling
http://en.wikipedia.org/wiki/File:Buckled_column.svg


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Buckling in columns

If the load on a column is applied through thecenter 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 itbuckles, 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 acombination of direct compressive stress and

bending.

eccentric load

axial load

The load at which the column just buckles is

known as buckling load or critical load or

crippling load.


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1757, mathematician Leonhard Euler (pronounced Oiler) derived

a formula that gives the maximum axial load that a long, slender,

ideal column can carry without buckling.

An ideal column is one that is perfectly straight, homogeneous,

and free from initial stress.The maximum load, sometimes called the critical load, causes

the column to be in a state of unstable equilibrium; that is, any

increase in the load, or the introduction of the slightest lateral

force, will cause the column to fail by buckling.

The formula derived by Euler for columns with no considerationfor lateral forces is given below. However, if lateral forces are

taken into consideration the value of critical load remains

approximately same.

Euler's Buckling Equation


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The following assumptions are made in the Eulers column theory;

1. Initially the column is perfectly straight, and the load applied is

truly axial.

2. The cross-section of the column is uniform throughout its length.

3. The column material is perfectly elastic, homogeneous and

isotropic, and thus obeys Hookes law.

4. The length of column is very large as compare to its cross-

sectional dimensions.

5. The shortening of column, due to direct compression (being

very small) is neglected.

6. The failure of column occurs due to buckling alone.


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Types of end cond i t ions o f columns


In actual practice, there are number of and conditions, for

column. For the Eulers column theory, there are four types of

end condition.

Both end

FixedBoth end

Pinned

(Hinged)

One end Fixed,

one end Pinned

One end Fixed,

one end Free


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http://localhost/var/www/apps/conversion/tmp/scratch_1//upload.wikimedia.org/wikipedia/commons/6/69/Buckledmodel.JPG


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Sign convent ions


1. A moment, which tends to

bend the column with

convexity towards its initial

central line, as shown in figure(a), is taken as positive.

2. A moment, which tends to

bend the column with itsconcavity towards its initial

central line, as shown in figure

(b), is taken as negative.

Figure (a)

Figure (b)


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Derivation of Eulers Equation

Eulers Buckling Load for Columns or Struts

In 1757, Leonhard Euler developed a

relationship for the critical column

load which would produce buckling.

A very brief derivation of Euler's

equation goes as follows:

M

x

y

A loaded pinned-pinned column is

shown in the diagram. A top section of

the diagram is shown with the bending

moment indicated. In terms of theload P, and the lateral deflection y, we

can write an expression for the

bending moment M:

)1.......()( EqxPyM


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We can also state that for beams and columns, the bending moment isproportional to the curvature of the beam, which, for small deflection

can be expressed as:

Where E = Young's modulus and I = moment of Inertia. Substituting

Equation 1 into Equation 2, we obtain the following differential

equation:

or

This is a second order differential equation which has the general

solution of:

)2.......(2

2

Eqdx

yd

EI

M

yEI

P

dx

yd

2

2

)3.......(02

2

Eqy

EI

P

dx

yd

)4.......(cossin

)sin()cos(

EqxEI

PBx

EI

PAy

cxBcxAy


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We next apply boundary conditions: at x = 0, y = 0 and at x = L, y = 0.That is, the deflection of the column must be zero at each end since it

is pinned.

Applying the first boundary condition, it is noted that B must be zero

since cos (0) = 1. The second boundary condition implies that either A

must be zero (which leaves us with no equation at all) or that:

Noting that sin ( ) = 0, we can solve for P:

where Pcrstands for the critical load (or buckling load) at which the

column is predicted to buckle.

)5.......(0sin EqLEI

P

LEI

P

)6......(2

2

EqL

EIPcr


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Therefore, Eulers Buckling Load is;

2

2

Le

EIPe

Where; E = Modulus of Elasticity of Column Material.

I = Minimum Moment of Inertia of Column Section.

Le = Equivalent length of Column or Effective Length


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Le= Equivalent length of Column or effective length(is defined as the distance between successive inflection

points or points of zero moment)



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Moment of Inertia

OR

2nd. Moment Area


axis


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Let us consider a column with both end hinged.

It is subjected to axial compressive load P.


CASE 1 - Both end Hinged (pinned).

Let, d= diameter of the column

L = length of the column


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CASE 1 - Both end Pinned (Hinged).

Let, d= diameter of the column

L = length of the column

Since Both Ends are Pinned, Le = L

Column material has Modulus of Elasticity = E

Moment of Inertia of Circular Section,

Due to axial compressive load P, the column will buckle

Now, at a distance of x from Point B.

Let us take a section x-x

Let the lateral deflection at section x-x be y.

4

64dI

Now, Eulers Formula will be,2

2

L

EIPe


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CASE 2 With both end Fixed.

Since Both Ends are Fixed,

42

2

2

2

2

2

2

L

EI

L

EI

Le

EIPe

2

LLe

2

24

L

EIPe

&


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CASE 3 With one end Fixed & one end Hinged (pinned).

When one end Fixed and one end Hinged,2

LLe

22

2

2

2

2

2

2

L

EI

L

EI

Le

EIPe

2

22

L

EIPe

CO & S O


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CASE 4 With one end Fixed & one end Free.

When one end Fixed and one end Free, LLe 2

2

2

2

2

2

2

42 L

EI

L

EI

Le

EIPe

2

2

4L

EIPe

A column with one fixed and one free

end (a), will behave as the upper-half of

a pin-connected (hinged) column (b).

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Summary


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The Buckling Stress, cr, is the Euler Buckling Load divided bythe columns cross-sectional area:

2

2

ALe

EI

A

Pecr

Buckling Stress (Crippling Stress)


Radius of Gyration

A

Ir

If all of the cross-sectional area of a column were massed a distance r away

from the neutral axis, the lumped-area cross-section would have the same

Moment of Inertia as the real cross-section. ris termed the RADIUS OF

GYRATION and is given by:

The Moment of Inertia I, can be expressed in terms of Radius of Gyration, r

as:2ArI

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2

ArI

Buckling or Crippling Stress in terms of Effective Length and Radius ofGyration.


2

2

ALe

EI

A

Pecr

2

2

2

22

2

22 )(

rL

E

L

Er

AL

ArE

)( ratiosslendernessr

Lwhere

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Slenderness Ratio

r

Ls

SLENDERNESS RATIO is a measure of how long the column is

compared to its cross-section's effective width (resistance to bending

or buckling). The Slenderness Ratio, s, is simply the column's Length

divided by the Radius of Gyration.

2

22

2

2

2

s

E

Le

rE

Le

EIPe


Applying the Slenderness Ratio and the

Radius of Gyration reduces the Euler BucklingFormula to:

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EXAMPLE 1

A steel rod 5 m long and 4 cm d iameter is us ed as a colum n, wi th one end

fixed and the other free. Determine the crippling load by Eulers formula.Take E as 2.0 x 106 kg/cm2

Solution

Given. Length, L = 5 m = 500 cm

Dia. Of column, d= 4 cm

Moment of Inertia,

444

446464 cmdI

Crippling load (critical load or buckling load),

Since the column is fixed at one end and free at the other, therefore

equivalent length of the column, Le= 2L = 2 x 500 = 1000 cm

kgLe

EIPe 1.248

1000

4100.22

62

2

2

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EXAMPLE 2

A hol low al loy tube 5 m long w ith diameters 4 cm and 2.5 cm respect ive ly

was fo und to extend 6.4 mm under a tens i le load of 60 kN. Find th ebuc kl ing load for the tube, when used as a strut wi th both ends pinn ed.

Als o f ind th e safe load on the tube, taking factor o f safety as 4.

Solution

Given. Length, L = 5 m

Outer diameter, D= 4 cm = 0.04 m

Moment of Inertia,474444

1006.1)025.004.0(64

)(64

mdDI

Therefore the Area,

Inner diameter, d= 2.5 cm = 0.025 m

242222

1066.7)025.004.0(4)(4 mdDA

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EXAMPLE 2 : Solution (cont.)

Deflection, = 6.4 mm = 6.4 X 10-3

m

Buckling Load (Crippling Load or Critical Load),

Since the column is pinned at its both end, therefore equivalent length of

the column, Le= L = 5 m

kNLe

EIPe 56.2

5

)1006.1()1012.6(2

772

2

2

Tensile load, P = 60 kN

And Youngs Modulus,

l

27

34 /1012.6)104.6()1066.7(

560mkN

lA

PL

Ll

APE

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We know that the safe load on the tube,

kN64.04

56.2

SafetyofFactor

LoadBucklingLoadSafe

Safe Load

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EXAMPLE 3

An I sect ion jo ist 40 cm x 20 cm x 2 cm and 6 m long is u sed as a strut

with both ends fixed. What is Eulers crippling load for the column?. TakeYoungs modulus for the joist as 2.0 x 106 N/cm2

Solution

Given; Length, L = 6 m

Outer width, b= 20 cm

4333

11

3c67.36682)36184020(

12

1)(

12

1mdbbdI

XX

Therefore, Moment of Inertia about X-X;

Inner depth, d1= 40 2x2 = 36 cm

Inner width, b1= 20 2 = 18 cm

Outer depth, d= 40 cm 40 cm

20 cm

2 cm

2 cm

2 cm

xx

y

y

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433c67.26902467.2666)236(

12

1)202(

12

12 mI

YY

And ;

Since IYY is less than IXX, therefore the joist will tend to buckle along Y-Y

axis. Thus we shall take the value of Ias IYY= 2690.67 cm4

Since the column is fixed at its both ends, therefore equivalent length of

the column,cm300

2

600

2

LLe

Crippling Load,

kNNLe

EIPe 5909.590129

300

)67.2690()102(2

62

2

2

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EXAMPLE 4

A T-sect ion 150 mm x 120 mm x 20 mm is used as a strut of 4 m lon g w ith

hinged at its both ends. Calculate the crippling load, if Youngs modulusfor t he material be 2.0 x 106 N/cm2

Solution

Given; Length, L = 4 m

First, determine the c.g. of the section; 20 mm

20 mm

xx

y

y

120 mm

150 mm

Flange; a1= 150 X 20 = 3000 mm2

y1= 20 / 2 = 10 mm

Web; a2= 100 X 20 = 2000 mm2

y2

= 20 + 100 / 2 = 70 mm

y26/102 cmNE

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2sec

23

1sec

23

1212 tiontionXX ay

bday

bdI


And,

We know that the moment of inertia the section about X-X axis,

mmaa

yaya

y 3420003000

)70(2000)10(3000

21

2211

442

3

2

3

1060936200012

10020243000

12

20150mmI

XX

44

33

1056912

20100

12

15020mmI

YY

Since IYYis less than IXX, therefore the column will buckle along Y-Y axis,

Thus we shall take the value ofIasIYY

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The column is pinned at its both ends, therefore the equivalent length ofthe column, Le = L = 4 m = 4000 mm

24

2

6

26/102

1010

1020/102 mmN

mm

NcmNE

Youngs Modulus,

Crippling Load (or Buckling Load),

kNNLe

EIPe 2.700197.67

4000

10569)102(2

442

2

2

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EXAMPLE 5

If an al lowable strain for a column are 0.00035 and the min imum radius o f

gyrat ion are 50 mm before the co lumn beg in to buck le . Determine themaximum length for that colum n, that are f ixed at one end and free on th e

other.

Since the column is fixed at one end and free at the other, therefore

equivalent length of the column, Le= 2L

Solution

Buckling Load (or Crippling Load),2

2

2

2

2

2

4)2( L

EI

L

EI

Le

EIPe

We know that,

2

22

2

22

2

2

44.

)(

4.

L

Er

LA

ArE

LA

EI

A

Pecr

2

where,

ArI

A

Ir

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And we know that, Youngs Modulus,

cr

crE

2

22

4,previously

L

Ercr

2

22

2

22

4

4.

L

r

LE

Er

E

cr

cr

mmL

L

L

4196

00035.04

50

45000035.0

22

2

2

22


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EXAMPLE 6

A squ are steel sect ion , 38 mm X 50 mm, f ixed at both end and loaded

axially. If the buc klin g streng th = 23.9 N/mm2and E = 21.7 kN/mm2 .Determine the minim um length for that sect ion.

45

33

1029.212

3850

12 mm

bd

I

Solution

Secon d m oment area,

Column is fixed at both ends, therefore,2

LLe

And, the Bu ckl ing Lo ad,

22

2

2

2

2 4

2 L

EI

L

EI

Le

EIPe

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Buc kl ing Strength,

2

532

)3850(

)1029.2)(107.21(49.23

L


2

24

AL

EI

A

Pecr

mmL

L

2077

9.23)3850(

)1029.2)(107.21(4 5322

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CLASS ASSIGNMENT

QUESTION 1

A colum n of t imb er sect ion, 15 cm X 20 cm is 6 m lo ng w ith both ends

being fixed. If the Youngs modulus for timber = 17.5 kN/mm2, determ ine;

i . Buck ling load .

i i . Safe load for the co lumn i f fac tor of safety = 3.

A sol id ro und bar 3 m lon g and 5 cm in diameter is u sed as a strut . If the

Youngs modulus for the strut, 2.0 x 105 N/mm2 , determine the cr ippl in g

(or bu ckl ing) load, when the given s trut is used w ith the fo l low ing

cond i t ions ;

i . B oth of th e en ds h in ged .

i i . One end of the stru t is f ixed and the other end is f ree.i i i. Both the ends of stru t are f ixed.

iv . One end is f ixed and the other end is h inged.

QUESTION 2

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CLASS ASSIGNMENT (cont.)

QUESTION 3a) A ho l low cy l indr ical cast i ron co lumn wi th both ends f ixed has to

carry a safe load of 250 kN with a factor of safety of 5. Determine the

m inim um diameter of the co lumn. Take the internal diameter as 0.8

t imes the external diameter and the maximum crus hin g stress as 550

N/mm2.

QUESTION 4

A simply sup por ted beam of length 4 m is sub jected to a uni form lydistr ib uted load of 30 kN/m over the wh ole span and def lects 15 mm at the

centre. Determine the buc kl ing loads w hen th is b eam is us ed as a column

wi th the fo l lowing cond i t ions;

i . One end i s fi xed and the o ther end i s h inged .

i i. Bo th the ends pin join ted .

b) If an al lowable stra in are 0.00055. Determine the maximum length for

the co lumn .

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QUESTION 5Determine the rat io of buck l ing strengths of two c olum ns, one hol low and

the other so l id. Both are made of th e same mater ial and have the same

length, cros s-sect ional area and end con dit ion s. The internal diameter of

ho l low c olumn is half of i ts external diameter.


A sol id round bar 4 m long and 5 cm in diameter was found to extend

4.6 mm under a tensi le load of 50 kN. This bar is used as a strut wi th

both ends pinned. Determine the buc kl ing load for the bar and also the

safe load taking a factor of safety of 4.

QUESTION 6

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QUESTION 7Determine Eulers crippling load for an I-

sect ion jo ist 40 cm X 20 cm X 1 cm and 5

m long wh ich is used as a st ru t wi th both

ends fixed. Take Youngs modulus for the

jo is t as 2.1 X 105N/mm2.

20 cm

1 cm

1 cm

1 cm

40 cm

Calculate the Eulers critical load for a

st rut of T-sect ion , the f lange width

being 10 cm , overall depth 8 cm and

both f lange and stem 1 cm thick. The

strut is 3 m lo ng and is bui l t -in at both

ends. Take Youngs modulus for the

strut as 2 x 105 N/mm2 .

QUESTION 8

1 cm

1 cm

8 cm

10 cm

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It is the resu lt of,

preparat ion,

learn ing from fai lure.

and
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11. column stability supports

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