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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
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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
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
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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.
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Stability of Structures
• 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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Stability of Structures
• 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.
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Euler’s Formula for Pin-Ended Beams
• 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
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Euler’s Formula for Pin-Ended Beams
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
Extension of Euler’s Formula
• 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
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10 - 9
Extension of Euler’s Formula
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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
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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.
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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
psi101.100.35
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
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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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Eccentric Loading; The Secant Formula
r
L
EA
P
r
ec
A
P eY 2
1sec1
2max
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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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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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Design of Columns Under Centric Load
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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Design of Columns Under Centric Load
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
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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.
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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
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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
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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