Chapter 7: Buckling

Transcription

Chapter 7: Buckling

9. Buckling of Columns
CHAPTER OBJECTIVES
• Discuss the behavior of
columns.
• Discuss the buckling of
columns.
• Determine the axial load
needed to buckle an ideal
column.
Chapter 7: Buckling
©2005 Pearson Education South Asia Pte Ltd
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CHAPTER OUTLINE
Introduction
1. Critical Load
2. Ideal Column with Pin Supports
3. Columns Having Various Types of Supports
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• Buckling is a failure mode characterized by a sudden
failure of a structural member subjected to high
compressive stresses, where the actual compressive
stresses at failure are smaller than the ultimate
compressive stresses that the material is capable of
withstanding.
• This mode of failure is also described as failure due to
elastic instability.
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7.1 CRITICAL LOAD
7.1 CRITICAL LOAD
• Long slender members subjected to axial
compressive force are called columns.
• The lateral deflection that occurs is
called buckling.
• The maximum axial load a column can
support when it is on the verge of
buckling is called the critical load, Pcr.
• Spring develops restoring force F = k∆, while
applied load P develops two horizontal
components, Px = P tan θ, which tends to push the
pin further out of equilibrium.
• Since θ is small,
∆ = θ(L/2) and tan θ ≈ θ.
• Thus, restoring spring
force becomes
F = kθL/2, and
disturbing force is
2Px = 2Pθ.
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7.1 CRITICAL LOAD
7.2 IDEAL COLUMN WITH PIN SUPPORTS
• For kθL/2 > 2Pθ,
• An ideal column is perfectly straight before loading,
made of homogeneous material, and upon which
the load is applied through the centroid of the xsection.
• We also assume that the material behaves in a
linear-elastic manner and the column buckles or
bends in a single plane.
kL
stable equilibrium
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• For kθL/2 < 2Pθ,
P<
P>
kL
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unstable equilibrium
• For kθL/2 = 2Pθ,
kL
Pcr =
neutral equilibrium
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8
• Summing moments, M = −Pν, Eqn 13-1
becomes
d 2υ ⎛ P ⎞
(13 - 2)
+
⎟υ = 0
2 ⎜
• In order to determine the critical load and buckled
shape of column, we apply Eqn 12-10,
d 2υ
(13 - 1)
EI 2 = M
dx
• Recall that this eqn assume
the slope of the elastic
curve is small and
deflections occur only in
bending. We assume that
the material behaves in a
linear-elastic manner and
the column buckles or
bends in a single plane.
dx
• General solution is
⎛ P ⎞
⎛ P ⎞
υ = C1 sin ⎜
x ⎟ + C2 cos⎜
x⎟
⎝ EI ⎠
⎝ EI ⎠
(13 - 3)
• Since ν = 0 at x = 0, then C2 = 0.
Since ν = 0 at x = L, then
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⎝ EI ⎠
⎛ P ⎞
C1 sin ⎜
L⎟ = 0
⎝ EI ⎠
• Smallest value of P is obtained for n = 1, so critical
load for column is
π 2 EI
Pcr =
4L2
• This load is also referred to
as the Euler load. The
corresponding buckled
shape is defined by
πx
υ = C1 sin
L
• Disregarding trivial soln for C1 = 0, we get
⎛ P ⎞
sin ⎜
L⎟ = 0
⎝ EI ⎠
• Which is satisfied if
P
L = nπ
EI
• or
P=
n 2π 2 EI
L2
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n = 1,2,3,...
• C1 represents maximum
deflection, νmax, which occurs
at midpoint of the column.
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• A column will buckle about the principal axis of the
x-section having the least moment of inertia
(weakest axis).
• For example, the meter stick shown will
buckle about the a-a axis and not
the b-b axis.
• Thus, circular tubes made excellent
columns, and square tube or those
shapes having Ix ≈ Iy are selected
for columns.
• Buckling eqn for a pin-supported long slender
column,
π 2 EI
(13 - 5)
Pcr = 2
L
Pcr = critical or maximum axial load on column just
before it begins to buckle. This load must not cause
the stress in column to exceed proportional limit.
E = modulus of elasticity of material
I = Least modulus of inertia for column’s x-sectional
area.
L = unsupported length of pinned-end columns.
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• Expressing I = Ar2 where A is x-sectional area of
column and r is the radius of gyration of x-sectional
π 2E
area.
(13 - 6)
σ cr =
2
(L r )
σcr = critical stress, an average stress in column just
before the column buckles. This stress is an elastic
stress and therefore σcr ≤ σY
E = modulus of elasticity of material
L = unsupported length of pinned-end columns.
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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• The geometric ratio L/r in Eqn 13-6 is known as the
slenderness ratio.
• It is a measure of the column’s flexibility and will be
used to classify columns as long, intermediate or
short.
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IMPORTANT
• Columns are long slender members that are
subjected to axial loads.
• Critical load is the maximum axial load that a
column can support when it is on the verge of
buckling.
• This loading represents a case of neutral
equilibrium.
IMPORTANT
• An ideal column is initially perfectly straight, made
of homogeneous material, and the load is applied
through the centroid of the x-section.
• A pin-connected column will buckle about the
principal axis of the x-section having the least
moment of intertia.
• The slenderness ratio L/r, where r is the smallest
radius of gyration of x-section. Buckling will occur
about the axis where this ratio gives the greatest
value.
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EXAMPLE 13.1
EXAMPLE 13.1 (SOLN)
A 7.2-m long A-36 steel tube
having the x-section shown is to
be used a pin-ended column.
Determine the maximum
allowable axial load the column
can support so that it does not
buckle.
Use Eqn 13-5 to obtain critical load with
Est = 200 GPa.
Pcr =
π 2EI
L2
=
π 2 [200(106 ) kN/m2 ]⎢ π (75)4 − π (70)4 ⎥(1 m/1000mm)4
⎡1
⎣4
1
4
(7.2 m)2
⎤
⎦
= 228.2 kN
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EXAMPLE 13.1 (SOLN)
EXAMPLE 13.2
The A-36 steel W200×46 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.
This force creates an average compressive stress in
the column of
σ cr =
Pcr 228.2 kN(1000 N/kN )
=
A
π (75)2 − π (70)2 mm2
[
]
= 100.2 N/mm2 = 100 MPa
Since σcr < σY = 250 MPa, application of Euler’s eqn
is appropriate.
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EXAMPLE 13.2 (SOLN)
EXAMPLE 13.2 (SOLN)
From table in Appendix B, column’s x-sectional area
and moments of inertia are A = 5890 mm2,
Ix = 45.5×106 mm4,and Iy = 15.3×106 mm4.
By inspection, buckling will occur about the y-y axis.
Applying Eqn 13-5, we have
Pcr =
When fully loaded, average compressive stress in
column is
P
1887.6 kN(1000 N/kN )
σ cr = cr =
A
5890 mm2
= 320.5 N/mm2
Since this stress exceeds yield stress (250 N/mm2),
the load P is determined from simple compression:
P
250 N/mm2 =
5890 mm2
P = 1472.5 kN
π 2 EI
L2
=
π 2 [200(106 ) kN/m 2 ](15.3(10 4 ) mm 4 )(1 m / 1000 mm )4
(4 m )2
= 1887.6 kN
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7.3 COLUMNS HAVING VARIOUS TYPES OF SUPPORTS
• From free-body diagram, M = P(δ − ν).
• Differential eqn for the deflection curve is
d 2υ
• Thus, smallest critical load occurs when n = 1, so
that
π 2EI
(13-9)
Pcr = 2
L
• By comparing with Eqn 13-5, a column fixedsupported at its base will carry only one-fourth the
critical load applied to a pin-supported column.
P
P
(13 - 7 )
=
υ
δ
EI
dx 2 EI
• Solving by using boundary conditions
and integration, we get
+
⎡
P ⎞⎤
x ⎟⎥
⎝ EI ⎠⎦
⎛
υ = δ ⎢1 − cos⎜
⎣
(13 - 8)
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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.
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Effective length
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Effective length
• Many design codes provide column formulae that
use a dimensionless coefficient K, known as the
effective-length factor.
(13 - 10)
Le = KL
Effective length
• Here (KL/r) is the column’s effective-slenderness
ratio.
Factor of safety ;
• Thus, Euler’s formula can be expressed as
Pcr =
σ cr =
π 2 EI
( KL )2
π 2E
(KL r )2
(13 - 11)
• Pcr = critical or maximum axial load on column just
before it begins to buckle.
• P = largest allowable axial load on column
(13 - 12)
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EXAMPLE 13.3
EXAMPLE 13.3 (SOLN)
A W150×24 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 or
material exceed the yield stress.
Take Est = 200 GPa and σY = 410 MPa.
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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 (KL)x = 0.5(8 m) = 4 m.
For buckling about the y-y
axis, (KL)y = 0.7(8 m/2) = 2.8 m.
We get Ix = 13.4×106 mm4 and Iy = 1.83×106 mm4
from Appendix B.
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EXAMPLE 13.3 (SOLN)
EXAMPLE 13.3 (SOLN)
Applying Eqn 13-11,
(Pcr )x =
π 2 EI x
( KL )2x
=
(Pcr ) y =
( KL )2y
=
] ( )
π 2 200 106 kN/m 2 13.4 10−6 m 4
(Pcr )x = 1653.2 kN
π 2 EI y
[ ( )
Area of x-section is 3060 mm2, so average
compressive stress in column will be
[ ( )
(4 m )2
σ cr =
] ( )
π 2 200 106 kN/m 2 1.83 10−6 m 4
( )
Pcr 460.8 103 N
=
= 150.6 N/mm2
2
A
3060 m
Since σcr < σY = 410 MPa, buckling will occur before
the material yields.
(2.8 m )
2
(Pcr ) y = 460.8 kN
By comparison, buckling will occur about the y-y axis
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EXAMPLE 13.3 (SOLN)
CHAPTER REVIEW
• Buckling is the sudden instability that occurs in
columns or members that support an axial load.
• The maximum axial load that a member can
support just before buckling occurs is called the
critical load Pcr.
• The critical load for an ideal column is determined
from the Euler eqn, Pcr = π2EI/(KL)2, where
K = 1 for pin supports, K = 0.5 for fixed supports,
K = 0.7 for a pin and a fixed support, and K = 2 for
a fixed support and a free end.
NOTE: From Eqn 13-11, we see that buckling always
occur about the column axis having the largest
slenderness ratio. Thus using data for the radius of
gyration from table in Appendix B,
⎛ KL ⎞ = 4 m(1000 mm/m) = 60.4
⎜
⎟
66.2 mm
⎝ r ⎠x
⎛ KL ⎞ = 2.8 m(1000 mm/m ) = 114.3
⎜
⎟
24.5 mm
⎝ r ⎠y
Hence, y-y axis buckling will occur, which is the same
conclusion reached by comparing Eqns 13-11 for
both axes.
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Chapter 7: Buckling

Transcription

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