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MECHANICS OF

MATERIALS

Third Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

CHAPTER

10Columns

.

Lecture Notes:

J. Walt Oler

Texas Tech University

© 2002 The McGraw-Hill Companies, Inc. All rights reserved.



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Beer • Johnston • DeWolf

Columns

Stability of Structures

Euler’s Formula for Pin-Ended Beams

Extension of Euler’s Formula

Sample Problem 10.1

© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 2

Sample Problem 10.2

Design of Columns Under Centric Load

Sample Problem 10.4

Design of Columns Under an Eccentric Load



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

selected such that

- allowable stress is not exceeded

all A

P σ σ ≤=

- deformation falls within specifications


spec AE δ δ ≤=

• After these design calculations, may discover

that the column is unstable under loading and

that it suddenly becomes sharply curved orbuckles.



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• Consider model with two rods and torsional

spring. After a small perturbation,

( )

momentingdestabiliz2

sin2

momentrestoring2

=∆=∆

=∆

θ θ

θ

LP

LP

K

•


orientation) if

( )

L

K PP

K L

P

cr 4

22

=<

∆<∆ θ θ



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• Consider an axially loaded beam.

After a small perturbation, the system

reaches an equilibrium configuration

such that

2

2

2

−==

P yd

y EI

P

EI

M

dx

yd


2

=

EI dx

• Solution with assumed configuration

can only be obtained if

( )( )2

2

2

22

2

2

r L

E

A L

Ar E

A

P

L EI PP

cr

cr

π π σ σ

π

==>=

=>



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A

P

A

P

L

EI PP

cr cr

cr 2

2

=>=

=>

σ σ

π

• The value of stress corresponding to

the critical load,


( )

s ratioslendernesr

L

tresscritical sr L

E

A L

r cr

2

2

2

=

==

=

π

π

σ

• Preceding analysis is limited to

centric loadings.



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• 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,

2


( )

length equivalent2

2

2

2

==

=

=

L L

r L

E

LP

e

e

cr

ecr

π σ



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Sample Problem 10.1

An aluminum column of length L and

rectangular cross-section has a fixed end at B

and supports a centric load at A. Two smooth

and rounded fixed plates restrain end A from

moving in one of the vertical planes ofsymmetry but allow it to move in the other

plane.


a) Determine the ratio a/b of the two sides ofthe cross-section corresponding to the most

efficient design against buckling.

b) Design the most efficient cross-section for

the column.

L = 20 in.

E = 10.1 x 106 psi

P = 5 kips

FS = 2.5



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Sample Problem 10.1

• Buckling in xy Plane:

23

121

2 aaba I z

SOLUTION:

The most efficient design occurs when the

resistance to buckling is equal in both planes of

symmetry. This occurs when the slenderness

ratios are equal.


12

7.01212

,

a

L

r

Lab A

z

ze

z z

=

• Buckling in xz Plane:

12 /

2

1212

,

23121

2

b

L

r

L

br

b

ab

ab

A

I r

y

ye

y y

y

=

====

• Most efficient design:

2

7.0

12 /

2

12

7.0

,,

=

=

=

b

a

b

L

a

L

r

L

r

L

y

ye

z

ze

35.0=

b

a



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Sample Problem 10.1

• Design:

( )

( ) ( )( )

( )cr0.35

lbs12500

kips5.12kips55.2

6.138

12

in202

12

2

bb A

P

PFS P

bbb

L

r

L

cr

cr

y

e

==

===

===

σ


L = 20 in.

E = 10.1 x 106 psi

P = 5 kips

FS = 2.5

a/b = 0.35

( ) ( )

( )

( )( )2

62

22cr

6.138

psi101.10

0.35

lbs12500

6.138

psi101.10

bbb

br L

E

e

×=

×

==

π

π π

σ

in.567.035.0

in.620.1

==

=

ba

b



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Eccentric Loading; The Secant Formula

• Eccentric loading is equivalent to a centricload and a couple.

• Bending occurs for any nonzero eccentricity.

Question of buckling becomes whether the

resulting deflection is excessive.

2PePy yd −−

=

• The deflection become infinite when P = Pcr


2

2

max 12

sec

e

cr cr L

EI P

P

Pe y

dx

π π =

−

=

• Maximum stress

( )

+=

++=

r

L

EA

P

r

ec

A

P

r

ce y

A

P

e

2

1sec1

1

2

2max

maxσ



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Eccentric Loading; The Secant Formula


+==

r

L

EA

P

r

ec

A

P eY

2

1sec1

2max σ σ



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Sample Problem 10.2

The uniform column consists of an 8-ft section

of structural tubing having the cross-section

shown.

a) Using Euler’s formula and a factor of safety

of two, determine the allowable centric load

for the column and the corresponding

normal stress.


b) Assuming that the allowable load, found in

part a, is applied at a point 0.75 in. from the

geometric axis of the column, determine the

horizontal deflection of the top of the

column and the maximum normal stress inthe column.

.psi1029 6×= E



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Sample Problem 10.2

SOLUTION:

• Maximum allowable centric load:

( ) in.192ft16ft82 ===e L

- Effective length,

- Critical load,


( )

kips1.62

in192in0.8psi1029 22

=

×==

π π

e

cr L

EI P

2in3.54

kips1.31

2

kips1.62

==

==

A

P

FS

PP

all

cr all

σ

kips1.31=allP

ksi79.8=σ

- Allowable load,



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Sample Problem 10.2

• Eccentric load:

( )

−

=

−

=

122

secin075.0

1

2

sec

π

π

cr

m

P

Pe y

- End deflection,


in.939.0=m

y

( )( )

( )

+=

+=

22sec

in1.50

in2in75.01

in3.54

kips31.1

2sec1

22

2

π

π σ

cr

mP

P

r

ec

A

P

ksi0.22=m

- Maximum normal stress,



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• Previous analyses assumed

stresses below the proportional

limit and initially straight,

homogeneous columns

• Experimental data demonstrate

- for large Le /r, σ cr follows


Euler’s formula and depends

upon E but not σ Y .

- for intermediate Le /r, σ cr

depends on both σ Y and E .

- for small Le /r, σ cr is

determined by the yield

strength σ Y and not E .



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Structural Steel

American Inst. of Steel Construction

• For Le /r > C c

( )

92.1

/ 2

2

=

==

FS

FS r L

E cr all

e

cr σ

σ π

σ

• For Le /r > C c


3

2

/

8

1 /

8

3

3

5

2

/ 1

−+=

=

−=

c

e

c

e

cr allc

eY cr

C

r L

C

r LFS

FS C

r L σ

σ σ σ

• At Le /r = C c

Y

cY cr E

C σ

π σ σ

22

21 2

==



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Aluminum

Aluminum Association, Inc.

• Alloy 6061-T6 Le /r < 66:

( )[ ]

( )[ ]MPa / 868.0139

ksi / 126.02.20

r L

r L

e

eall

−=

−=σ

Le /r > 66:

( ) ( )2

3

2 /

MPa10513

/

ksi51000

r Lr L ee

all×

==σ


• Alloy 2014-T6

Le /r < 55:

( )[ ]

( )[ ]MPa / 585.1212

ksi / 23.07.30

r L

r L

e

eall

−=

−=σ

Le /r > 66:

( ) ( )2

3

2 /

MPa10273

/

ksi54000

r Lr L ee

all×

==σ

T E



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Sample Problem 10.4

SOLUTION:

• With the diameter unknown, the

slenderness ration can not be evaluated.

Must make an assumption on which

slenderness ratio regime to utilize.

•


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

assumed slenderness ratio regime.

• Evaluate slenderness ratio and verify

initial assumption. Repeat if necessary.

T E



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Sample Problem 10.4

• For L = 750 mm, assume L/r > 55

• Determine cylinder radius:

( )

mm44.18

MPa103721060

rL

MPa10372

2

3

2

3

2

3

=×

=×

×

==

c N

A

Pallσ


2

4

gyrationof radius

radiuscylinder

2

4c

c

c

A

I

r

c

===

=

=

π

π

c/2

.

• Check slenderness ratio assumption:

( ) 553.81mm18.44

mm750

2 / >===

c

L

r

L

assumption was correct

mm9.362 == cd

T E



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Sample Problem 10.4

• For L = 300 mm, assume L/r < 55

• Determine cylinder radius:

Pa102 /

m3.0585.1212

1060

MPa585.1212

6

2

3

×

−=

×

−==

cc

N

r

L

A

Pall

π

σ


mm00.12=c

• Check slenderness ratio assumption:

( )5550

mm12.00

mm003

2 / <===

c

L

r

L

assumption was correct

mm0.242 == cd

T E



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Design of Columns Under an Eccentric Load

• An eccentric load P can be replaced by a

centric load P and a couple M = Pe.

• Normal stresses can be found from

superposing the stresses due to the centricload and couple,

bendingcentric += σ


• Allowable stress method:

all

I

Mc

A

Pσ ≤+

• Interaction method:

( ) ( )1≤+

bendingallcentricall

I Mc AP

σ σ

I

c

A+=

maxσ

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