Third Edition MECHANICS OF 10 MATERIALS 212 (SUMMER... · 2015. 7. 13. · MECHANICS OF MATERIALS Edition Beer • Johnston • DeWolf 10 - 20 Design of Columns Under Centric Load

MECHANICS OFMATERIALS

Third Edition

Ferdinand P. BeerE. Russell Johnston, Jr.John T. DeWolf

Lecture Notes:J. Walt OlerTexas Tech University

CHAPTER

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

10Columns


MECHANICS OF MATERIALS

ThirdEdition

Beer • Johnston • DeWolf

10 - 2

Columns

Stability of Structures

Euler’s Formula for Pin-Ended Beams

Extension of Euler’s Formula

Sample Problem 10.1

Eccentric Loading; The Secant Formula

Sample Problem 10.2

Design of Columns Under Centric Load

Sample Problem 10.4

Design of Columns Under an Eccentric Load



ThirdEdition


10 - 3

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

selected such that

- allowable stress is not exceeded

allA

P

- deformation falls within specifications

specAE

PL

• After these design calculations, may discoverthat the column is unstable under loading andthat it suddenly becomes sharply curved orbuckles.



ThirdEdition


10 - 4


• Consider model with two rods and torsionalspring. After a small perturbation,

momentingdestabiliz2

sin2

momentrestoring2

LP

LP

K

• Column is stable (tends to return to alignedorientation) if

L

KPP

KL

P

cr4

22



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10 - 5


• Assume that a load P is applied. After aperturbation, the system settles to a newequilibrium configuration at a finitedeflection angle.

sin4

2sin2

crP

P

K

PL

KL

P

• Noting that sinq<q, the assumedconfiguration is only possible if P > Pcr.



ThirdEdition


10 - 6


• Consider an axially loaded beam.After a small perturbation, the systemreaches an equilibrium configurationsuch that

02

2

2

2

yEI

P

dx

yd

yEI

P

EI

M

dx

yd

• Solution with assumed configurationcan only be obtained if

2

2

2

22

2

2

rL

E

AL

ArE

A

P

L

EIPP

cr

cr



ThirdEdition


10 - 7


s ratioslendernesr

L

tresscritical srL

E

AL

ArE

A

P

A

P

L

EIPP

cr

crcr

cr

2

2

2

22

2

2

• The value of stress corresponding tothe critical load,

• Preceding analysis is limited tocentric loadings.



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10 - 8


• A column with one fixed and one freeend, will behave as the upper-half of apin-connected column.

• The critical loading is calculated fromEuler’s formula,

lengthequivalent2

2

2

2

2

LL

rL

E

L

EIP

e

ecr

ecr



ThirdEdition


10 - 9




ThirdEdition


10 - 10

Sample Problem 10.1An aluminum column of length L andrectangular cross-section has a fixed end at Band supports a centric load at A. Two smoothand rounded fixed plates restrain end A frommoving in one of the vertical planes ofsymmetry but allow it to move in the otherplane.

a) Determine the ratio a/b of the two sides ofthe cross-section corresponding to the mostefficient design against buckling.

b) Design the most efficient cross-section forthe column.

L = 20 in.

E = 10.1 x 106 psi

P = 5 kips

FS = 2.5



ThirdEdition


10 - 11

Sample Problem 10.1

• Buckling in xy Plane:

127.0

1212

,

23121

2

a

L

r

L

ar

a

ab

ba

A

Ir

z

ze

zz

z

• Buckling in xz Plane:

12/2

1212

,

23121

2

b

L

r

L

br

b

ab

ab

A

Ir

y

ye

yy

y

• Most efficient design:

27.0

12/2

127.0

,,

b

a

b

L

a

L

r

L

r

L

y

ye

z

ze

35.0b

a

SOLUTION:

The most efficient design occurs when theresistance to buckling is equal in both planes ofsymmetry. This occurs when the slendernessratios are equal.



ThirdEdition


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

psi101.100.35

lbs12500

6.138

psi101.10

0.35lbs12500

kips5.12kips55.2

6.13812in202

122

bbb

brL

E

bbA

P

PFSP

bbb

L

r

L

e

cr

cr

y

e

in.567.035.0

in.620.1

ba

b



ThirdEdition


10 - 13

Eccentric Loading; The Secant Formula• Eccentric loading is equivalent to a centric

load and a couple.

• Bending occurs for any nonzero eccentricity.Question of buckling becomes whether theresulting deflection is excessive.

2

2

max

2

2

12

sece

crcr L

EIP

P

Pey

EI

PePy

dx

yd

• The deflection become infinite when P = Pcr

• Maximum stress

r

L

EA

P

r

ec

A

P

r

cey

A

P

e21

sec1

1

2

2max

max



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10 - 14

Eccentric Loading; The Secant Formula

r

L

EA

P

r

ec

A

P eY 2

1sec1

2max



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10 - 15

Sample Problem 10.2

The uniform column consists of an 8-ft sectionof structural tubing having the cross-sectionshown.

a) Using Euler’s formula and a factor of safetyof two, determine the allowable centric loadfor the column and the correspondingnormal stress.

b) Assuming that the allowable load, found inpart a, is applied at a point 0.75 in. from thegeometric axis of the column, determine thehorizontal deflection of the top of thecolumn and the maximum normal stress inthe column.

.psi1029 6E



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10 - 16

Sample Problem 10.2SOLUTION:

• Maximum allowable centric load:

in.192ft16ft82 eL

- Effective length,

kips1.62

in192

in0.8psi10292

462

2

2

ecr

L

EIP

- Critical load,

2in3.54

kips1.31

2kips1.62

A

P

FS

PP

all

crall

kips1.31allP

ksi79.8

- Allowable load,



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10 - 17

Sample Problem 10.2• Eccentric load:

in.939.0my

122

secin075.0

12

sec

crm P

Pey

- End deflection,

22sec

in1.50

in2in75.01

in3.54

kips31.1

2sec1

22

2

crm P

P

r

ec

A

P

ksi0.22m

- Maximum normal stress,



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10 - 18

Design of Columns Under Centric Load• Previous analyses assumed

stresses below the proportionallimit and initially straight,homogeneous columns

• Experimental data demonstrate

- for large Le/r,scr followsEuler’s formula and dependsupon E but notsY.

- for intermediate Le/r,scr

depends on bothsY and E.

- for small Le/r,scr isdetermined by the yieldstrengthsY and not E.



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10 - 19


Structural Steel

American Inst. of Steel Construction

• For Le/r > Cc

92.1

/ 2

2

FS

FSrL

E crall

ecr

• For Le/r > Cc

3

2

2

/81/

83

35

2

/1

c

e

c

e

crall

c

eYcr

C

rL

C

rLFS

FSC

rL

• At Le/r = Cc

YcYcr

EC

2

221 2



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10 - 20


Aluminum

Aluminum Association, Inc.

• Alloy 6061-T6Le/r < 66:

MPa/868.0139

ksi/126.02.20

rL

rL

e

eall

Le/r > 66:

23

2 /

MPa10513

/

ksi51000

rLrL eeall

• Alloy 2014-T6Le/r < 55:

MPa/585.1212

ksi/23.07.30

rL

rL

e

eall

Le/r > 66:

23

2 /

MPa10273

/

ksi54000

rLrL eeall



ThirdEdition


10 - 21

Sample Problem 10.4

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

SOLUTION:

• With the diameter unknown, theslenderness ration can not be evaluated.Must make an assumption on whichslenderness ratio regime to utilize.

• Calculate required diameter forassumed slenderness ratio regime.

• Evaluate slenderness ratio and verifyinitial assumption. Repeat if necessary.



ThirdEdition


10 - 22

Sample Problem 10.4

24

gyrationofradius

radiuscylinder

2

4 c

c

c

A

I

r

c

• For L = 750 mm, assume L/r > 55

• Determine cylinder radius:

mm44.18

c/2m0.750

MPa103721060

rL

MPa10372

2

3

2

3

2

3

cc

N

A

Pall

• Check slenderness ratio assumption:

553.81mm18.44

mm7502/

c

L

r

L

assumption was correct

mm9.362 cd



ThirdEdition


10 - 23

Sample Problem 10.4• For L = 300 mm, assume L/r < 55

• Determine cylinder radius:

mm00.12

Pa102/m3.0

585.12121060

MPa585.1212

62

3

c

cc

N

r

L

A

Pall

• Check slenderness ratio assumption:

5550mm12.00

mm0032/

c

L

r

L

assumption was correct

mm0.242 cd



ThirdEdition


10 - 24

Design of Columns Under an Eccentric Load

• Allowable stress method:

allI

Mc

A

P

• Interaction method:

1bendingallcentricall

IMcAP

• An eccentric load P can be replaced by acentric load P and a couple M = Pe.

• Normal stresses can be found fromsuperposing the stresses due to the centricload and couple,

I

Mc

A

P

bendingcentric

max

Documents

Third Edition MECHANICS OF 10 MATERIALS 212 (SUMMER... · 2015. 7. 13. · MECHANICS OF MATERIALS Edition Beer • Johnston • DeWolf 10 - 20 Design of Columns Under Centric Load