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PDT 201 – Strength of Materials

Chapter 6

Buckling of Columns

Dr. Khor Chu Yee

Office: S4-L2-64, UniCITI

Email: [email protected]

H/P: 019-5637283

1

• Analyze the critical load in supports.

• Analyze the column with various types of supports.

• Discuss the design of Column – Centric Load

Learning objectives

2

Introduction • Load-carrying structures may fails in the

variety ways, depending upon the type of structure, conditions of support, the kinds of loads, and the materials used.

• These kinds of failure are prevented by designing structures so that the maximum stresses and maximum displacement remain within the tolerance limits.

• Thus, strength and stiffness are important factors in design.

3

Introduction • A long and slender structural member

(column) loaded axially compression are considered in this chapter.

• Buckling is a mode of failure generally resulting from structural instability due to compressive action on the structural member or element involved.

4

Buckling of a slender column due to an axial compressive load, P

Compressive load, P

Slender column

5

Introduction • The phenomenon of buckling is not limited to

columns, which can occur in many kinds of structures.

• For example, when you step on the top of empty aluminum can, the thin cylindrical walls buckle under your weight and the can collapses.

6

Examples of buckling

Overloaded metal building columns.

Buckling of storage tank

7

Buckling of concrete columns

Before and after applied compressive load to the aluminum can.

8

9

• In the design of columns, cross-sectional area is

selected such that

- allowable stress (σall) is not exceeded

allA

P

• After these design calculations, may discover

that the column is unstable under loading and

that it suddenly becomes sharply curved or

buckles.

- deformation falls within specifications

specAE

PL

Stability of Structures

10


• If the two rods and the two forces P and P’ are perfectly aligned, the system will remain in the position of equilibrium. This system is said to be stable, Figure (a).

• When the system move further away from the original position. This system is said to be unstable, Figure (b).

• The maximum axial load when a column can support on the verge of buckling is called the critical load, Pcr.

• The column is buckled when the compressive load exceeds the critical load, P>Pcr

11


12

• Consider model with two rods and

torsional spring. After a small

perturbation,

moment ingdestabiliz 2

sin2

moment restoring 2

LP

LP

K


• Column is stable (tends to return to

aligned orientation) if

22

KL

PSpring

L

KP

stablePP

cr

cr

4

)(

13

sin4

2sin2

crP

P

K

PL

KL

P


• Assume that a load P is applied.

After a perturbation, the system

settles to a new equilibrium

configuration at a finite deflection

angle.

• Noting that sin < , the assumed

configuration is only possible if P >

Pcr (unstable).

14


P < Pcr , the system is stable P > Pcr , the system is unstable P = Pcr , the system is neutral equilibrium

L

KP

4

L

KP

4

L

KP

4

Critical load for Pin-Ended Beams

15 Column Buckled column

P > Pcr


16

• Consider an axially loaded beam.

After a small perturbation, the system

reaches an equilibrium configuration

such that

02

2

2

2

yEI

P

dx

yd

yEI

P

EI

M

dx

yd

• Euler’s formula for critical load and

critical stress:

2

2

2

22

2

2

rL

E

AL

ArE

L

EIP

cr

cr


17

s ratioslendernesr

L

tresscritical srL

E

AL

ArE

A

P

A

P

L

EIP

cr

crcr

cr

2

2

2

22

2

2

• The value of stress corresponding to the critical load,

Pcr = critical or maximum axial load on column just before it begins to buckle.

E = modulus of elasticity of material I = moment of inertia for column’s x-sectional

area. L = unsupported length of pinned-end

columns. cr = critical stress, an average stress in column

just before the column buckles. r = smallest radius of gyration of column,

determined from r = √(I/A), where I is least moment of inertia of column’s x-sectional area A.

18

A 7-m long steel tube having the x-section shown is to be used a pin-ended column. Determine the maximum allowable axial load of the column can support so that it does not buckle. Given Est = 200 GPa.

Example 1

19

Solution:

Use equation to obtain critical load with

Est = 200 GPa.

Est = 200 GPa = 200 × 106 kN/m2.

20

This force creates an average compressive stress in the column

Since cr < Y = 250 MPa. cr is not exceeded the yield strength of steel (Y).

The maximum allowable axial load the column can support is 241.4 kN

21

Plot of critical stress for structural steel.

Yield strength (Y ) or yield point is the stress at which a material begins to deform plastically. Once the yield point is passed, some fraction of the deformation will be permanent and non-reversible.

Yield strength (Y ) or yield point or yield stress

23

The A-36 steel W20046 member shown is to be used as a pin-connected column. Determine the largest axial load it can support before it either begins to buckle or the steel yields. Given the

column’s x-sectional area

and moments of inertia

are A = 5890 mm2, Ix = 45.5106 mm4,

and Iy = 15.3106 mm4.

Example 2

24

Solution:

Column’s x-sectional area and moments of inertia are A = 5890 mm2, Ix = 45.5106 mm4,

and Iy = 15.3106 mm4.

25

When fully loaded, average compressive stress in column is

Since this stress exceeds yield stress (250 MPa), the load Pallowable is determined from simple compression:

kN5.1472

m105890250

26

allowable

all

allY

P

PMPa

A

P

26

Effective length

• If a column is not supported by pinned-ends, then Euler’s formula can also be used to determine the critical load.

• “L” must then represent the distance between the zero-moment points.

• This distance is called the columns’ effective length, Le.

Columns having various types of supports

Columns having various types of supports

27

• A column with one fixed and one free

end, will behave as the upper-half of a

pin-connected column.

• The critical loading is calculated from

Euler’s formula,

length effective 2

2

2

2

2

LL

rL

E

L

EIP

e

e

cr

e

cr

28

Effective length, Le

22

2

2

2

2

rL

E

L

EIP

cr

cr

22

2

2

rL

E

L

EIP

cr

cr

Free end

Fixed end

Pinned end

Pinned end

29

Effective length, Le

22

2

2

7.0

7.0

rL

E

L

EIP

cr

cr

22

2

2

5.0

5.0

rL

E

L

EIP

cr

cr

Pinned end

Fixed end Fixed end

Fixed end

30

A W15024 steel column is 8 m long and is fixed at its ends as shown. Its load-carrying capacity is increased by bracing it about the y-y axis using struts that are assumed to be pin-connected to its mid-height. Determine the load it can support that the column does not buckle nor material exceed the yield stress.

Take Est = 200 GPa; Y = 410 MPa;

Ix = 13.4106 mm4 and Iy = 1.83106 mm4

Radius of gyration: rx = 66.2mm; ry = 24.5mm

Area of cross-section is 3060 mm2

Example 3

31

Buckling behavior is different about the x and y axes due to bracing.

Buckled shape for each case is shown.

The effective length for buckling about the x-x axis is 0.5 L= 0.5(8 m) = 4 m.

For buckling about the y-y axis, 0.7 L= 0.7(8 m/2) = 2.8 m.

Example 3

Fixed end

Fixed end

Pinned

32

Applying equations, calculate the critical load for x-x and y-y axis:

kN8.460

m8.2

m1083.1kN/m10200

7.0

kN2.1653

m4

m104.13kN/m10200

5.0

2

46262

2

2

2

46262

2

2

ycr

y

y

ycr

xcr

x

x

xcr

P

L

EIP

P

L

EIP

Example 3

33

Calculate the slenderness ratio of both axes:

Note: Buckling always occur about the column axis having the largest slenderness ratio.

By comparison, buckling will occur about the y-y axis.

4.60m0662.0

m45.0

xr

L

Example 3

3.114m0245.0

m8.27.0

yr

L

x-x axis

y-y axis

34

Area of cross-section is 3060 mm2, so average compressive stress in column will be

Since cr < Y = 410 MPa. Thus, buckling will occur before the material yields.

2

2

3

N/mm6.150

mm3060

N108.460

cr

cr

crcr

A

P

Example 3

Note: N/mm2 = MPa

Design of Columns Under Centric Load

35

• Previous analyses assumed stresses below the proportional limit and initially straight, homogeneous columns

• Experimental data demonstrate

- for large Le/r, cr follows Euler’s formula and depends upon E but not Y.

- for intermediate Le/r, cr depends on both Y and E.

- for small Le/r, cr is determined by the yield strength Y and not E.

36


Structural Steel

American Inst. of Steel Construction

• For Le/r > Cc

92.1

/2

2

FS

FSrL

E crall

e

cr

• For Le/r < Cc

3

2

2

/

8

1/

8

3

3

5

2

/1

c

e

c

e

crall

c

eYcr

C

rL

C

rLFS

FSC

rL

• At Le/r = Cc

YcYcr

EC

22

21 2

Note: FS = Factor of safety

37


• Alloy 6061-T6

Le/r < 66:

MPa /868.0139

ksi /126.02.20

rL

rL

e

eall

Le/r > 66:

2

3

2/

MPa 10513

/

ksi 51000

rLrL ee

all

• Alloy 2014-T6

Le/r < 55:

MPa /585.1212

ksi /23.07.30

rL

rL

e

eall

Le/r > 55:

2

3

2/

MPa 10273

/

ksi 54000

rLrL ee

all

Aluminum

Aluminum Association, Inc.

38

Example 4

Using the aluminum alloy2014-T6, determine the smallest diameter rod which can be used to support the centric load P = 60 kN if a) L = 750 mm, b) L = 300 mm

SOLUTION:

• With the diameter unknown, the

slenderness ratio can not be evaluated.

Must make an assumption on which

slenderness ratio regime to utilize.

• Calculate required diameter for

assumed slenderness ratio regime.

• Evaluate slenderness ratio and verify

initial assumption. Repeat if necessary.

39

Example 4

2

4

gyration of radius

radiuscylinder

2

4 c

c

c

A

I

r

c

• For L = 750 mm, assume L/r > 55

• Determine cylinder radius:

mm44.18

c/2

m 0.750

MPa 103721060

rL

MPa 10372

2

3

2

3

2

3

cc

N

A

Pall

• Check slenderness ratio assumption:

553.81

mm 18.44

mm750

2/

c

L

r

L

assumption was correct

mm 9.362 cd

40

• For L = 300 mm, assume L/r < 55

• Determine cylinder radius:

mm00.12

Pa102/

m 3.0585.1212

1060

MPa 585.1212

62

3

c

cc

N

r

L

A

Pall

• Check slenderness ratio assumption:

5550

mm 12.00

mm 003

2/

c

L

r

L

assumption was correct

mm 0.242 cd

Example 4

Chapter review

41

• Buckling is a mode of failure generally resulting from structural instability due to compressive action on the structural member or element involved.

• The maximum axial load when a column can support on the verge of buckling is called the critical load, Pcr.

• The column is buckled when the compressive

load exceeds the critical load, P>Pcr

42

Chapter review • Columns having various types of supports:

Chapter review

43

P

PFS

FS

cr

all

cr

• Factor of safety:

44

THE END

pdt 201 strength of materials - universiti malaysia...

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