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
1
• Analyze the critical load in supports.
• Analyze the column with various types of supports.
• Discuss the design of Column – Centric Load
Learning objectives
2
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.
3
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.
4
Buckling of a slender column due to an axial compressive load, P
Compressive load, P
Slender column
5
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.
6
Examples of buckling
Overloaded metal building columns.
Buckling of storage tank
7
Buckling of concrete columns
Before and after applied compressive load to the aluminum can.
8
9
• 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
10
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).
• 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
11
Stability of Structures
12
• 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
)(
13
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).
14
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
Critical load for Pin-Ended Beams
15 Column Buckled column
P > Pcr
Critical load for Pin-Ended Beams
16
• 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
Critical load for Pin-Ended Beams
17
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.
18
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
19
Solution:
Use equation to obtain critical load with
Est = 200 GPa.
Est = 200 GPa = 200 × 106 kN/m2.
20
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
21
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
22
23
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
24
Solution:
Column’s x-sectional area and moments of inertia are A = 5890 mm2, Ix = 45.5106 mm4,
and Iy = 15.3106 mm4.
25
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
26
allowable
all
allY
P
PMPa
A
P
26
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
Columns having various types of supports
27
• 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
28
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
29
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
30
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
31
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
32
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
33
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
34
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
Design of Columns Under Centric Load
35
• 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.
36
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
37
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.
38
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.
39
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
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
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
42
Chapter review • Columns having various types of supports:
Chapter review
43
P
PFS
FS
cr
all
cr
• Factor of safety:
44
THE END