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8/15/2019 Structural stability of column
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August 2014 11-1
P r i n t e d f r o m : h
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Stability of Columns
• Bending due to a compressive axial load is called Buckling.
• Structural members that support compressive axial loads are called Columns.
• Buckling is the study of stability of a structure’s equilibrium.
Learning objectives
• Develop an appreciation of the phenomena of buckling and the various types ofstructure instabilities.
• Understand the development and use of buckling formulas in analysis anddesign of structures.
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Buckling Phenomenon
Energy Approach
Bifurcation/Eigenvalue Problem
Stable Equilibrium
Neutral Equilibrium
Unstable Equilibrium
Potential Energy is Minimum
P P
K θθ
θ
L
Lsinθ
(c)
O O
R
StableStable
UnstableUnstable
PL K
θ ⁄ θ θsin ⁄ =
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Snap Buckling Problem
L
P
P
(45−θ)
P
45oFs
Fs
L cos(45-θ)
L s i n ( 4 5 - θ )
0 θ 45o
< <
θ 450
>
θ 0=
Fs
P
L cos(θ−45)
Fs
P
(θ−45)
s
n
−
P
K L
L---------- 45 θ – ( )cos 45cos – ( ) 45 θ – ( )tan= 0 θ 45
0< <
P
K L L---------- θ 45 – ( )cos 45cos – ( ) θ 45 – ( )tan= θ 45
0>
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Local Buckling
crinkling
Axial Loads
Compressive
Torsional Loads
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C11.1 Linear deflection springs and torsional springs are attached to
rigid bars as shown.The springs can act in tension or compression and
resist rotation in either direction. Determine Pcr , the critical load value.
Fig. C11.1
30 in
O
P
k = 8 lbs/in
30 in
k = 8 lbs/in
K = 2000 in-lbs/rads
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Euler Buckling
Boundary Value Problem
Differential Equation:
Boundary conditions:
Solution
Trivial Solution:
Non-Trivial Solution:
where:
Characteristic Equation:
Euler Buckling Load:
• Buckling occurs about an axis that has a minimum value of I.
v(x)
M z
P
N = P
A
y
x
L
PA
EI
x2
2
d
d v P v+ 0=
v 0( ) 0= v L( ) 0=
v 0=
v x( ) A λ xcos B λ xsin+=
λ P EI ------=
λ Lsin 0=
P n
n2
π2 EI
L2
------------------= n 1 2 3. ⋅ ⋅, ,=
P cr
π2 EI
L2
------------=
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Buckling Mode: v B nπ x
L---
⎝ ⎠⎛ ⎞sin=
Mode shape 2
Mode shape 3
Pcr Pcr
Pcr Pcr
L/2
Pcr
Pcr
L/2
L/3 L/3 L/3
P cr
4π2 EI
L2
----------------=
P cr
9π2 EI
L2
----------------=
I
I I
Mode shape 1
PcrPcr
Pcr
L P cr
π2 EI
L2
------------=
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Effects of End Conditions
Case
1.
Pinned at
both Ends
2.
One end fixed,
other end free
3.
One end fixed,
other end pinned
4.
Fixed at both ends.
Differen-
tial
Equation
Boundary
Conditions
Character-
istic
Equation
Critical
Load
Pcr
[
Effective
Length—
Leff
L 2L 0.7L 0.5L
y
x
L
P
A
B
x
L
P
B
A
y
x
L
P
B
Ay
x
L
P
B
A
EI x
2
2
d
d v P v+
0=
EI x
2
2
d
d v P v+
P v L( )=
EI x
2
2
d
d v P v+
R B L x – ( )=
EI x
2
2
d
d v P v+
R B L x – ( ) M B+=
v 0( ) 0=
v L( ) 0=
v 0( ) 0=
xd
dv0( ) 0=
v 0( ) 0=
xd
dv0( ) 0=
v L( ) 0=
v 0( ) 0=
xd
dv0( ) 0=
v L( ) 0=
xd
dv L( ) 0=
λ P EI ------=
λ Lsin 0= λ Lcos 0= λ Ltan λ L= 2 1 λ Lcos – ( )
λ L λ Lsin – 0=
π2 EI
L2
------------ π2 EI
4 L2
------------ π2 EI
2 L( )2--------------=
20.13 EI
L2
------------------- π2 EI
0.7 L( )2------------------=
4π2 EI
L2
--------------- π2 EI
0.5 L( )2------------------=
P cr π
2
EI L
eff
2------------=
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Axial Stress:
Radius of gyration:
Slenderness ratio: (Leff /r).
Failure Envelopes
• Short columns: Designed to prevent material elastic failure.
• Long columns: Designed to prevent buckling failure.
σcr
P cr
A--------=
r I
A---=
σcr
P cr
A--------
π2 E
Leff
r ⁄ ( )2
-----------------------= =
A A A
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C11.2 Columns made from an alloy will be used in a construction of
a frame. The cross-section of the columns is a hollow-cylinder of thick-
ness 10 mm and outer diameter of ‘d’ mm. The modulus of elasticity is
E = 200 GPa and the yield stress is σyield = 300 MPa. Table below showsa list of the lengths ‘L’ and outer diameters ‘d’. Identify the long and the
short columns. Assume the ends of the column are built in.
L
(m)
d
(mm)
1 60
2 80
3 100
4 150
5 200
6 225
7 250
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Class Problem 1
Identify the members in the structures that you would check for buckling. Circle the
correct answers.
Structure 1 AP BP Both None
Structure 2 AP BP Both None
Structure 3 AP BP Both None
Structure 4 AP BP Both None
110o
FA
B
P
Structure 1 110o
F
A
B
P 40o
Structure 2
75o
F
A
B
P60
o
F
AB
P
30o
Structure 3
Structure 4
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C11.4 A hoist is constructed using two wooden bars to lift a weight
of 5 kips. The modulus of elasticity for wood is E = 1,800 ksi and the
allowable normal stress 3.0 ksi. Determine the maximum value of L to
the nearest inch that can be used in constructing the hoist.
Fig. C11.4
W
B
L ft. CA
A
A
A2 in
4 in
Cross-section AA
30o
A