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PDT 201 – Strength of Materials
Chapter 6
Buckling of Columns
Dr. Khor Chu Yee
Office: S4-L2-64, UniCITI
Email: [email protected]
H/P: 019-5637283
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• Analyze the critical load in supports.
• Analyze the column with various types of supports.
• Discuss the design of Column – Centric Load
Learning objectives
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Introduction • Load-carrying structures may fails in the
variety ways, depending upon the type of structure, conditions of support, the kinds of loads, and the materials used.
• These kinds of failure are prevented by designing structures so that the maximum stresses and maximum displacement remain within the tolerance limits.
• Thus, strength and stiffness are important factors in design.
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Introduction • A long and slender structural member
(column) loaded axially compression are considered in this chapter.
• Buckling is a mode of failure generally resulting from structural instability due to compressive action on the structural member or element involved.
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Buckling of a slender column due to an axial compressive load, P
Compressive load, P
Slender column
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Introduction • The phenomenon of buckling is not limited to
columns, which can occur in many kinds of structures.
• For example, when you step on the top of empty aluminum can, the thin cylindrical walls buckle under your weight and the can collapses.
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Examples of buckling
Overloaded metal building columns.
Buckling of storage tank
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Buckling of concrete columns
Before and after applied compressive load to the aluminum can.
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• In the design of columns, cross-sectional area is
selected such that
- allowable stress (σall) is not exceeded
allA
P
• After these design calculations, may discover
that the column is unstable under loading and
that it suddenly becomes sharply curved or
buckles.
- deformation falls within specifications
specAE
PL
Stability of Structures
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Stability of Structures
• If the two rods and the two forces P and P’ are perfectly aligned, the system will remain in the position of equilibrium. This system is said to be stable, Figure (a).
• When the system move further away from the original position. This system is said to be unstable, Figure (b).
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• The maximum axial load when a column can support on the verge of buckling is called the critical load, Pcr.
• The column is buckled when the compressive load exceeds the critical load, P>Pcr
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Stability of Structures
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• Consider model with two rods and
torsional spring. After a small
perturbation,
moment ingdestabiliz 2
sin2
moment restoring 2
LP
LP
K
Stability of Structures
• Column is stable (tends to return to
aligned orientation) if
22
KL
PSpring
L
KP
stablePP
cr
cr
4
)(
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sin4
2sin2
crP
P
K
PL
KL
P
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.
• Noting that sin < , the assumed
configuration is only possible if P >
Pcr (unstable).
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Stability of Structures
P < Pcr , the system is stable P > Pcr , the system is unstable P = Pcr , the system is neutral equilibrium
L
KP
4
L
KP
4
L
KP
4
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Critical load for Pin-Ended Beams
15 Column Buckled column
P > Pcr
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Critical load for Pin-Ended Beams
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• Consider an axially loaded beam.
After a small perturbation, the system
reaches an equilibrium configuration
such that
02
2
2
2
yEI
P
dx
yd
yEI
P
EI
M
dx
yd
• Euler’s formula for critical load and
critical stress:
2
2
2
22
2
2
rL
E
AL
ArE
L
EIP
cr
cr
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Critical load for Pin-Ended Beams
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s ratioslendernesr
L
tresscritical srL
E
AL
ArE
A
P
A
P
L
EIP
cr
crcr
cr
2
2
2
22
2
2
• The value of stress corresponding to the critical load,
Pcr = critical or maximum axial load on column just before it begins to buckle.
E = modulus of elasticity of material I = moment of inertia for column’s x-sectional
area. L = unsupported length of pinned-end
columns. cr = critical stress, an average stress in column
just before the column buckles. r = smallest radius of gyration of column,
determined from r = √(I/A), where I is least moment of inertia of column’s x-sectional area A.
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A 7-m long steel tube having the x-section shown is to be used a pin-ended column. Determine the maximum allowable axial load of the column can support so that it does not buckle. Given Est = 200 GPa.
Example 1
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Solution:
Use equation to obtain critical load with
Est = 200 GPa.
Est = 200 GPa = 200 × 106 kN/m2.
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This force creates an average compressive stress in the column
Since cr < Y = 250 MPa. cr is not exceeded the yield strength of steel (Y).
The maximum allowable axial load the column can support is 241.4 kN
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Plot of critical stress for structural steel.
Yield strength (Y ) or yield point is the stress at which a material begins to deform plastically. Once the yield point is passed, some fraction of the deformation will be permanent and non-reversible.
Yield strength (Y ) or yield point or yield stress
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The A-36 steel W20046 member shown is to be used as a pin-connected column. Determine the largest axial load it can support before it either begins to buckle or the steel yields. Given the
column’s x-sectional area
and moments of inertia
are A = 5890 mm2, Ix = 45.5106 mm4,
and Iy = 15.3106 mm4.
Example 2
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Solution:
Column’s x-sectional area and moments of inertia are A = 5890 mm2, Ix = 45.5106 mm4,
and Iy = 15.3106 mm4.
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When fully loaded, average compressive stress in column is
Since this stress exceeds yield stress (250 MPa), the load Pallowable is determined from simple compression:
kN5.1472
m105890250
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allowable
all
allY
P
PMPa
A
P
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Effective length
• If a column is not supported by pinned-ends, then Euler’s formula can also be used to determine the critical load.
• “L” must then represent the distance between the zero-moment points.
• This distance is called the columns’ effective length, Le.
Columns having various types of supports
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Columns having various types of supports
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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,
length effective 2
2
2
2
2
LL
rL
E
L
EIP
e
e
cr
e
cr
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Effective length, Le
22
2
2
2
2
rL
E
L
EIP
cr
cr
22
2
2
rL
E
L
EIP
cr
cr
Free end
Fixed end
Pinned end
Pinned end
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Effective length, Le
22
2
2
7.0
7.0
rL
E
L
EIP
cr
cr
22
2
2
5.0
5.0
rL
E
L
EIP
cr
cr
Pinned end
Fixed end Fixed end
Fixed end
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A W15024 steel column is 8 m long and is fixed at its ends as shown. Its load-carrying capacity is increased by bracing it about the y-y axis using struts that are assumed to be pin-connected to its mid-height. Determine the load it can support that the column does not buckle nor material exceed the yield stress.
Take Est = 200 GPa; Y = 410 MPa;
Ix = 13.4106 mm4 and Iy = 1.83106 mm4
Radius of gyration: rx = 66.2mm; ry = 24.5mm
Area of cross-section is 3060 mm2
Example 3
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Buckling behavior is different about the x and y axes due to bracing.
Buckled shape for each case is shown.
The effective length for buckling about the x-x axis is 0.5 L= 0.5(8 m) = 4 m.
For buckling about the y-y axis, 0.7 L= 0.7(8 m/2) = 2.8 m.
Example 3
Fixed end
Fixed end
Pinned
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Applying equations, calculate the critical load for x-x and y-y axis:
kN8.460
m8.2
m1083.1kN/m10200
7.0
kN2.1653
m4
m104.13kN/m10200
5.0
2
46262
2
2
2
46262
2
2
ycr
y
y
ycr
xcr
x
x
xcr
P
L
EIP
P
L
EIP
Example 3
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Calculate the slenderness ratio of both axes:
Note: Buckling always occur about the column axis having the largest slenderness ratio.
By comparison, buckling will occur about the y-y axis.
4.60m0662.0
m45.0
xr
L
Example 3
3.114m0245.0
m8.27.0
yr
L
x-x axis
y-y axis
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Area of cross-section is 3060 mm2, so average compressive stress in column will be
Since cr < Y = 410 MPa. Thus, buckling will occur before the material yields.
2
2
3
N/mm6.150
mm3060
N108.460
cr
cr
crcr
A
P
Example 3
Note: N/mm2 = MPa
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Design of Columns Under Centric Load
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• Previous analyses assumed stresses below the proportional limit and initially straight, homogeneous columns
• Experimental data demonstrate
- for large Le/r, cr follows Euler’s formula and depends upon E but not Y.
- for intermediate Le/r, cr depends on both Y and E.
- for small Le/r, cr is determined by the yield strength Y 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
e
cr
• For Le/r < Cc
3
2
2
/
8
1/
8
3
3
5
2
/1
c
e
c
e
crall
c
eYcr
C
rL
C
rLFS
FSC
rL
• At Le/r = Cc
YcYcr
EC
22
21 2
Note: FS = Factor of safety
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Design of Columns Under Centric Load
• Alloy 6061-T6
Le/r < 66:
MPa /868.0139
ksi /126.02.20
rL
rL
e
eall
Le/r > 66:
2
3
2/
MPa 10513
/
ksi 51000
rLrL ee
all
• Alloy 2014-T6
Le/r < 55:
MPa /585.1212
ksi /23.07.30
rL
rL
e
eall
Le/r > 55:
2
3
2/
MPa 10273
/
ksi 54000
rLrL ee
all
Aluminum
Aluminum Association, Inc.
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Example 4
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
SOLUTION:
• With the diameter unknown, the
slenderness ratio can not be evaluated.
Must make an assumption on which
slenderness ratio regime to utilize.
• Calculate required diameter for
assumed slenderness ratio regime.
• Evaluate slenderness ratio and verify
initial assumption. Repeat if necessary.
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Example 4
2
4
gyration of radius
radiuscylinder
2
4 c
c
c
A
I
r
c
• For L = 750 mm, assume L/r > 55
• Determine cylinder radius:
mm44.18
c/2
m 0.750
MPa 103721060
rL
MPa 10372
2
3
2
3
2
3
cc
N
A
Pall
• Check slenderness ratio assumption:
553.81
mm 18.44
mm750
2/
c
L
r
L
assumption was correct
mm 9.362 cd
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40
• For L = 300 mm, assume L/r < 55
• Determine cylinder radius:
mm00.12
Pa102/
m 3.0585.1212
1060
MPa 585.1212
62
3
c
cc
N
r
L
A
Pall
• Check slenderness ratio assumption:
5550
mm 12.00
mm 003
2/
c
L
r
L
assumption was correct
mm 0.242 cd
Example 4
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Chapter review
41
• Buckling is a mode of failure generally resulting from structural instability due to compressive action on the structural member or element involved.
• The maximum axial load when a column can support on the verge of buckling is called the critical load, Pcr.
• The column is buckled when the compressive
load exceeds the critical load, P>Pcr
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42
Chapter review • Columns having various types of supports:
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Chapter review
43
P
PFS
FS
cr
all
cr
• Factor of safety:
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44
THE END