Third Edition MECHANICS OF 10 MATERIALS - site. OF MATERIALS Third Edition Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University CHAPTER

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• MECHANICS OF

MATERIALS

Third Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

John T. DeWolf

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

10 Columns

Presented by:

Dr. Mohammed Arafa

MECHANICS OF MATERIALS

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10 - 2

Columns

Stability of Structures

Euler’s Formula for Pin-Ended Beams

Extension of Euler’s Formula

Sample Problem 10.1

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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MECHANICS OF MATERIALS

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10 - 3

Stability of Structures

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

selected such that

- allowable stress is not exceeded

all A

P  

- deformation falls within specifications

spec AE

PL  

• After these design calculations, may discover

that it suddenly becomes sharply curved or

buckles.

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10 - 4

Stability of Structures

• Consider model with two rods and torsional

spring. After a small perturbation,

 

moment ingdestabiliz 2

sin 2

moment restoring 2







L P

L P

K

orientation) if

 

L

K PP

K L

P

cr 4

2 2



 

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10 - 5

Stability of Structures

• Assume that a load P is applied. After a perturbation, the system settles to a new

equilibrium configuration at a finite

deflection angle.

 



sin4

2sin 2



crP

P

K

PL

K L

P

• Noting that sin <  , the assumed configuration is only possible if P > Pcr.

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10 - 6

Euler’s Formula for Pin-Ended Beams

• Consider an axially loaded beam.

After a small perturbation, the system

reaches an equilibrium configuration

such that

0 2

2

2

2





y EI

P

dx

yd

y EI

P

EI

M

dx

yd

• Solution with assumed configuration

can only be obtained if

   2

2

2

22

2

2

rL

E

AL

ArE

A

P

L

EI PP

cr

cr

 





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10 - 7

Euler’s Formula for Pin-Ended Beams

 

 

s ratioslendernes r

L

tresscritical s rL

E

AL

ArE

A

P

A

P

L

EI PP

cr

cr cr

cr

2

2

2

22

2

2







 



• The value of stress corresponding to

• Preceding analysis is limited to

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10 - 8

Extension of Euler’s Formula

• A column with one fixed and one free

end, will behave as the upper-half of a

pin-connected column.

Euler’s formula,

 

length equivalent 2

2

2

2

2



LL

rL

E

L

EI P

e

e

cr

e cr

 

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10 - 9

Extension of Euler’s Formula

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10 - 10

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 of

symmetry but allow it to move in the other

plane.

a) Determine the ratio a/b of the two sides of

the 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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10 - 11

Sample Problem 10.1

• Buckling in xy Plane:

12

7.0

1212

,

23 12 1

2

a

L

r

L

a r

a

ab

ba

A

I r

z

ze

z z

z



• Buckling in xz Plane:

12/

2

1212

,

23 12 1

2

b

L

r

L

b r

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

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.

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10 - 12

Sample Problem 10.1

L = 20 in.

E = 10.1 x 106 psi

P = 5 kips

FS = 2.5

a/b = 0.35

• Design:

 

    

 

 

   

     2

62

2

62

2

2

cr

cr

6.138

psi101.10

0.35

lbs 12500

6.138

psi101.10

0.35

lbs 12500

kips 5.12kips 55.2

6.138

12

in 202

12

2

bbb

brL

E

bbA

P

PFSP

bbb

L

r

L

e

cr

cr

y

e

 

 







 

in. 567.035.0

in. 620.1



ba

b

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