Chapter 5 Buckling

8/13/2019 Chapter 5 Buckling

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Euler's Buckling Formula

Empirical Formula


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DEFINITIONA column is a bar subjected to a longitudinal compressive

loadBucklingor lateral deflectionis a common failure modeto react the bending moment generated by a compressiveload.


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Short column carries direct compressive stress

Long column subjected primarily to flexural stress


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Maximum axial load a column can support when on theverge of buckling is called the cr i t ical load, Pcr.

The easiest concept to grasp is that the design load Pdesmust be less than the critical buckling load Pcr.

Euler's Buckling Formula


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Ideal Column with Pin Supports

A column will buckle about the principal

axis of the cross section having the least

moment o f iner t ia (the weakest axis).

Pcris given by a formula;

22

2

2

/

rL

E

L

EIP

cr

cr

Pcr= maximum axial load

cr= critical stress,

E = modulus of elasticity for the material

I = least moment of inertia for the columnscross-sectional area

L = unsupported length of the column

r = smallest radius of gyration of the column

( )

L/r = slenderness ratioA

Ir

A

PCR

CR


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Example 5.1

The A-36 steel W200 X 46 member is to beused as a pin-connected column.

Determine the largest axial load it can

support before it either begins to buckle or

the steel yields.


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Solution:

From Appendix B,

2N/mm5.3205890

10006.1887

A

Pcrcr

46462 mm103.15,mm105.45,mm5890 yx IIA

kN.

/.

L

EIPcr 61887

4

100011031510200

2

4692

2

2

By inspection, buckling will occur about the yy axis because of

least moment of inertia.

When fully loaded, the average compressive stress in

the column is

Since this stress exceeds the yield stress,

(Ans)kN5.14725890

250 PP


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Columns Having Various Types of Supports

Euler is used to determine the critical loadprovided L represents the distance between the

zero-moment points.

It is called the columns effect iv e length, L e.

A dimensionless coefficient K, effect ive-length

factor, is used to calculate Le.

Thus we have,

KLLe

22

2

2

/

rKL

E

KL

EIP

crcr

KL/r =effective-slenderness ratio


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Example 5.2

The aluminium column is fixed at

its bottom and is braced at its top

by cables so as to prevent

movement at the top along thex

axis. If it is assumed to be fixed atits base, determine the largest

allowable load P that can be

applied. Use a factor of safety for

buckling of FS = 3.0. Take Eal=70GPa, Y= 215MPa, A = 7.5(10

-

3)m2,

Ix= 61.3(10-6)m4, Iy= 23.2(10

-6)m4.


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Solution:

MN31.1,kN424

2

2

2

2

y

ycr

x

xcrKLEIP

KLEIP

MPa215MPa.

.A

P

kN.FS

PP

cr

cr

cr

allow

556

1057

424

14103

424

3

m1052 xKLForxx axis buckling, K = 2,

For yy axis buckling, K = 0.7, m..KL y 53570

The critical loads for each case are

The allowable load and critical stress are


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Example 5.3

Knowing that a factor of safety of 2.6 is required,determine the largest load P which may be

applied to the structure shown. Use E = 200

GPa and consider only buckling in the plane of

the structure.

15 mm

diameter

P

20 mm diameter

0.5 m

A

B

C

0.5 m 1.0 m

P


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Solution:

51

01510

51

50510

.

PAP.A;M

.

P.C00.5P-.C;M

yyC

yyA

15 mm

diameter

P

20 mm diameter

0.5 m

A

B

C

0.5 m 1.0 m

AyCy

CPP

FFP

FP

TPPP

F

PF

CPP

FFP

F

P

ABAB

AB

AC

AC

BCBC

BC

943.075.0

5.0

5.0

5.0

5.1

0sin5.1

;0F

667.05.1

1

25.1

0.1

5.1

25.1

0cos5.1

25.1;0F

745.05.1

25.10

25.1

5.0

5.1

5.0

0sin5.1

5.0

;0F

y

x

y

51

50

.

P.

FAC

FBC

251.50.

51.

P

FAB

51.

P


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kN.

...

..P

L.

EI

.

P

.

P.F

kN.

...

..P

L.EI

.P.

L.

EI

.

P

.

P.F

Cr

AB

Cr

BC

04

4506250

0075010200750

6262750

50

46

425162251

0101020051

6251251

626251

251

492

2

2

492

2

2

2

2

Hence, the largest load P = 4.0 kN

Chapter 5 Buckling

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