mecanica solidos beer-10
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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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T h i r d
E d i t i on
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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T h i r d
E d i t i on
Beer • Johnston • DeWolf
Stability of Structures
• 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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 3
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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Beer • Johnston • DeWolf
Stability of Structures
• Consider model with two rods and torsional
spring. After a small perturbation,
( )
momentingdestabiliz2
sin2
momentrestoring2
=∆=∆
=∆
θ θ
θ
LP
LP
K
•
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 4
orientation) if
( )
L
K PP
K L
P
cr 4
22
=<
∆<∆ θ θ
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Beer • Johnston • DeWolf
Euler’s Formula for Pin-Ended Beams
• 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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 6
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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Beer • Johnston • DeWolf
Euler’s Formula for Pin-Ended Beams
A
P
A
P
L
EI PP
cr cr
cr 2
2
=>=
=>
σ σ
π
• The value of stress corresponding to
the critical load,
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 7
( )
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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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.
• The critical loading is calculated from
Euler’s formula,
2
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 8
( )
length equivalent2
2
2
2
==
=
=
L L
r L
E
LP
e
e
cr
ecr
π σ
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T h i r d
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Beer • Johnston • DeWolf
Extension of Euler’s Formula
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 9
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Beer • Johnston • DeWolf
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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 10
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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Beer • Johnston • DeWolf
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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 11
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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T h i r d
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Beer • Johnston • DeWolf
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 12
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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Beer • Johnston • DeWolf
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 13
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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Beer • Johnston • DeWolf
Eccentric Loading; The Secant Formula
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 14
+==
r
L
EA
P
r
ec
A
P eY
2
1sec1
2max σ σ
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Beer • Johnston • DeWolf
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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 15
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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Beer • Johnston • DeWolf
Sample Problem 10.2
SOLUTION:
• Maximum allowable centric load:
( ) in.192ft16ft82 ===e L
- Effective length,
- Critical load,
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 16
( )
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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Beer • Johnston • DeWolf
Sample Problem 10.2
• Eccentric load:
( )
−
=
−
=
122
secin075.0
1
2
sec
π
π
cr
m
P
Pe y
- End deflection,
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 17
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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T h i r d
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Beer • Johnston • DeWolf
Design of Columns Under Centric Load
• Previous analyses assumed
stresses below the proportional
limit and initially straight,
homogeneous columns
• Experimental data demonstrate
- for large Le /r, σ cr follows
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 18
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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Beer • Johnston • DeWolf
Design of Columns Under Centric Load
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 19
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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Beer • Johnston • DeWolf
Design of Columns Under Centric Load
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 20
• 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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Beer • Johnston • DeWolf
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.
•
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 21
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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Beer • Johnston • DeWolf
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 22
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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Beer • Johnston • DeWolf
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 23
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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Beer • Johnston • DeWolf
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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 24
• 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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