Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Buckling of ColumnsENGR0145 - Statics and Mechanics of Materials 2
William S. Slaughter
Mechanical Engineering DepartmentUniversity of Pittsburgh
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Student Learning Objectives
Students should be able to:I Determine the buckling load for long, slender
columns in axial compression.I Account for different types of end conditions.I Apply empirical formulas for the design of
columns.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Buckling of Columns
Consider a simply-supported column in axialcompression.
LP
I Boundary conditions:I Both ends of the column are free to rotate, i.e.
there is zero bending moment at both ends.I At the left end, both axial and lateral deflection
are prevented.I At the right end, there is unconstrained axial
motion while lateral deflection is prevented.
I According to our axial loading theory, the column“fails” when the axial compressive stress P/Areaches the material’s yield stress.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Buckling of Columns
However, for long, slender beams, buckling may occurat a lower load.
LP
x
y
I y is the lateral deflection at a distance x from theleft side of the column.
I We want to determine the load at which bucklingoccurs.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Buckling of Columns
Consider the free body diagram of a portion of abuckled column.
LP
PP
x
x
y
y
M
o
∑Mo = Py + M = 0 =⇒ M = −Py
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Buckling of Columns
Using the beam deflection equation y′′ = M/EI andM = −Py gives
d2y
dx2= − P
EIy
ord2y
dx2+ p2y = 0 where p2 =
P
EI
I The general solution of this differential equation is
y = A sin px + B cos px
where A and B are constants.
dy
dx= pA cos px− pB sin px
d2y
dx2= −p2A sin px− p2B cos px = −p2y
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Buckling of Columns
LP
x
y
y = A sin px + B cos px , p2 =P
EI
I y|x=0 = B = 0 =⇒ y = A sin px
I y|x=L = A sin pL = 0
=⇒
A = 0 (the column is straight)orp = nπ
L where n is an integer
I The A = 0 solution is unstable and not what weare looking for.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Euler Buckling Load
n = 3 n = 2
n = 1x
y
p =nπ
L=⇒ y = A sin
nπx
L
I Different integer values of n correspond todifference modes of buckling.
I For each buckling mode:
p2 =P
EI
p =nπ
L
=⇒ P =n2π2EI
L2
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Euler Buckling Load
n = 3 n = 2
n = 1x
y
y = A sinnπx
L, P =
n2π2EI
L2
I n = 1 gives the smallest nonzero load, so, undernormal circumstances, buckling will occur via thefirst mode:
y = A sinπx
L, Pcr =
π2EI
L2
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Euler Buckling Load
LP
x
y
y = A sinπx
L, Pcr =
π2EI
L2
I Pcr is the Euler buckling load, the compressiveaxial load at which a column with these supportswill buckle.
I The deflection coefficient A is undetermined in thelinear theory.
I Important: when calculating Pcr, use theminimum second moment I for the cross section.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example - Stainless Steel Ruler
E = 28× 106 psi
σy = 36× 103 psiL = 13 in
I =112
(1 in)(0.04 in)3 = 5.33× 10−6 in4
A = (1 in)(0.04 in) = 0.04 in2
I In the absence of buckling, the axial stress reachesthe yield stress when the compressive axial load is
Py = σyA = 1440 lb
I In comparison, the Euler buckling load is
Pcr =π2EI
L2= 8.77 lb
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example - Air-Dried, Douglas Fir 2× 4
E = 1.9× 106 psi
σy = 6.4× 103 psi
A = 5.89 in2
I 6= 6.45 in4
I =112
(3.625 in)(1.625 in)3 = 1.2962 in4
I In the absence of buckling, the axial stress reachesthe yield stress when the compressive axial load is
Py = σyA = 37.7 kip
I When L = 1 ft, the Euler buckling load is
Pcr =π2EI
L2= 169 kip
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example - Air-Dried, Douglas Fir 2× 4
E = 1.9× 106 psi
σy = 6.4× 103 psi
A = 5.89 in2
I 6= 6.45 in4
I =112
(3.625 in)(1.625 in)3 = 1.2962 in4
I In the absence of buckling, the axial stress reachesthe yield stress when the compressive axial load is
Py = σyA = 37.7 kip
I When L = 10 ft, the Euler buckling load is
Pcr =π2EI
L2= 1.69 kip
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example - Tacoma Narrows Bridge Disaster
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example - Tacoma Narrows Bridge Disaster
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example - Tacoma Narrows Bridge Disaster
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Critical Stress
The compressive axial stress at buckling (i.e. thecritical stress) is
σcr =Pcr
A=
π2EI
AL2
where A is the cross sectional area of the column.I Recall that the radius of gyration r is defined such
that I = Ar2.
=⇒ σcr =π2E
(L/r)2
I L/r is the slenderness ratio.I Doubling the slenderness ratio decreases the
buckling load by a factor of four.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Critical Stress
Comparison of plastic yielding versus buckling forstructural steel (E = 200 GPa, σy = 250 MPa).
L/r
s (MPa)
250
89
scr
sy
I A column with L/r < 89 will deform plasticallywithout buckling.
I A column with L/r > 89 will buckle first.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Different End Conditions
L
L = 2L
P
e
y = 0y' = 0
"effective length"
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Different End Conditions
Pcr =π2EI
L2e
, σcr =π2E
(Le/r)2
I Effective length Le = 2L yields a 4-fold decreasein Pcr when compared to the simply supportedcolumn.
I The effective lengths for a number of different endconditions have been tabulated (see Figure 11-4,page 661). Use the above equations with theappropriate effective length for all cases.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Different End Conditions
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Comparing Experiments and Theory
L/r
s (MPa)scr
sy
compressionblock range intermediate
range
slenderrange
I Experimental data differ from theoreticalpredictions, particularly in the “intermediaterange” between plastic deformation (compressionblock range) and Euler buckling (slender range).
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Empirical Column Formulas
I Design codes for columns based on experimentaldata.
I For example, the allowable axial compressivestress for 2014-T6 aluminum alloy columns(Specifications for Aluminum Structures,Aluminum Association, Inc, Washington, D.C.,1986):
Le
r≤ 12 =⇒ σall = 28 ksi
12 ≤ Le
r≤ 55 =⇒ σall =
[30.7− 0.23
(Le
r
)]ksi
Le
r≥ 55 =⇒ σall =
54, 000(Le/r)2
ksi
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Empirical Column Formulas
I The allowable axial compressive stress forstructural steel with a yield point σy (Manual ofSteel Construction, 9th ed., American Institute ofSteel Construction, New York, 1959):
0 ≤ Le
r≤ Cc =⇒ σall =
σy
FS
[1− 1
2
(Le/r
Cc
)2]
C2c =
2π2E
σy
FS =53
+38
(Le/r
Cc
)− 1
8
(Le/r
Cc
)3
Le
r≥ Cc =⇒ σall =
π2E
1.92(Le/r)2
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Eccentrically Loaded Columns
x
y
ce
P
When a column is subjected to an eccentrical load, theresult is combined axial loading and flexure.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Eccentrically Loaded Columns
x
y
ce
P
M
In this case, the statically equivalent centric axial loadand bending moment are P and M = Pe.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Combined Axial Loading and Flexure ofColumns
x
y
c
P
M
What is an allowable combination of axial load P andbending moment M?
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Allowable Stress Method
I The axial loading compressive stress is P/A,where A is the cross sectional area.
I The maximum compressive stress due to bendingis Mc/I, where I is the second moment withrespect to the neutral axis and c is the maximumdistance from the neutral axis (depends on theaxis about which bending occurs).
I The sum shall not exceed the allowable bucklingstress σall from Table 11-1:
P
A+
Mc
I≤ σall
I Usually yields a conservative design.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Interaction Method
I Starting first from the allowable stress method:
P
A+
Mc
I≤ σall =⇒ P/A
σall+
Mc/I
σall≤ 1
I However, for the interaction method, use
P/A
σa+
Mc/I
σb≤ 1
whereI σa is the allowable buckling stress from
Table 11-1.I σb is the allowable flexural stress as would be used
for a beam bending problem.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example
x
y
P125 mm
I A W457× 144 wide-flange section is used for thecolumn shown.
I Made of steel (E = 200 GPa, σy = 290 MPa,σb = 190 MPa).
I Effective length of 6 m.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example
x
y
P125 mm
I An eccentric load P is applied on the centerline asshown.
I Determine the maximum safe load according to(a) The allowable stress method.(b) The interaction method.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Relevant Section Properties
x
y
P125 mm
For a W457× 144 wide-flange section (Table A-2):I A = 18.365× 10−3 m2
I c = 0.472/2 = 0.236 mI rmin = 0.0673 m, rmax = 0.199 mI Imin = 83.7× 10−6 m4, Imax = 728× 10−6 m4
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Allowable Buckling Stress - Empirical
Use Code 1 for steel:The allowable axial compressive stress for structuralsteel with a yield point σy:
0 ≤ Le
r≤ Cc =⇒ σall =
σy
FS
[1− 1
2
(Le/r
Cc
)2]
C2c =
2π2E
σy
FS =53
+38
(Le/r
Cc
)− 1
8
(Le/r
Cc
)3
Le
r≥ Cc =⇒ σall =
π2E
1.92(Le/r)2
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Allowable Buckling Stress - Empirical
I First determine Cc (recall that E = 200 GPa andσy = 290 MPa):
Cc =
√2π2E
σy= 116.68
I Always use the minimum radius of gyration whendetermining the effective slenderness ratio (recallthat Le = 6 m and rmin = 0.0673 m):
Le
r= 89.15
I Since Le/r < Cc, use the intermediate formulawith a factor of safety of
FS =53
+38
(Le/r
Cc
)− 1
8
(Le/r
Cc
)3
= 1.90
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Allowable Buckling Stress - Empirical
I Thus, using the intermediate formula (recall thatσy = 290 MPa, Le/r = 89.15, Cc = 116.68, andFS = 1.90, the allowable stress for buckling is
σall =σy
FS
[1− 1
2
(Le/r
Cc
)2]
= 108.08 MPa
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
(a) Allowable Stress Method
According to the allowable stress method, a combinedaxial load P and bending moment M is safe if
P
A+
Mc
I≤ σall ,
where I and c are with respect to the axis about whichbending will occur.
I Recall that A = 18.365× 10−3 m2, c = 0.236 m,I = Imax = 728× 10−6 m4, and σall = 108.08 MPa.
I Note that M = (0.125 m)P .I It follows that the maximum safe load is
Pmax = 1138 kN
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
(b) Interaction Method
According to the interaction method, a combined axialload P and bending moment M is safe if
P/A
σa+
Mc/I
σb≤ 1 ,
where σa is the allowable axial stress for buckling, σb isthe allowable flexural stress, and I and c are withrespect to the axis about which bending will occur.
I Recall that A = 18.365× 10−3 m2, c = 0.236 m,I = 728× 10−6 m4, σa = σall = 108.08 MPa, andσb = 190 MPa.
I Note that M = (0.125 m)P .I It follows that the maximum safe load is
Pmax = 1395 kN
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Post-buckling Behavior
P
P
maxy
maxy
stablebuckling
unstablebuckling
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Post-buckling Behavior
I Buckling is said to be stable if, after buckling, onemust increase the load in order to cause anincrease in the lateral deflection.
I Buckling is said to be unstable (or catastrophic) ifthe lateral deflection keeps increasing withouthaving to increase the load.
I Whether or not buckling will be stable depends onthe cross sectional geometry of the column andthe type of supports at the ends.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Other Buckling Mechanisms
I Columns are not the only structure wherebuckling is an issue.
I The flanges and/or web of I-type beams inbending can buckle.
I Buckling can occur in plates and shells (e.g.,thin-walled pressure vessels subject to externalpressure; ICBM nose cones during reentry).
I The stability of buckling in these case is often anissue.
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example - Buckling of a Soda Can
I Ro = 1.25 inI Ri = 1.24 inI L = 4.75 inI E = 10, 000 ksiI σy = 40 ksi
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example - Buckling of a Soda Can
Euler buckling loadI I = π
4 (R4o −R4
i ) = 0.0606 in4
I Le = 2L = 9.5 in (flag pole-like support)
Pcr =π2EI
L2e
= 265 kip
Compressive yield loadI A = π(R2
o −R2i ) = 7.823× 10−2 in2
Py = σyA = 3.13 kip
Compare with demonstration
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example - Vitreous Silo
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example - Diamond Pattern
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example - “Elephant Foot”
Buckling ofColumns
ENGR0145
Learning Objectives
Euler BucklingExamples
Critical Stress
End Conditions
Empirical Formulas
Eccentric LoadingExample
Post-bucking
Other Mechanisms
Example - “Squidged” Column