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6 The Bucking of plates
6.1 Introduction
one dimensional members
beams and columns,
ordinary differential equations;
critical load=the failure load
two dimensional members
plates (local buckling)
partial differential equations;
failure load>the critical load;
(post-buckling)
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Middle-surface
Located on the plane of half thickness, paralleled with the surface
of plate (Figure 6-1) .
the coordinate surface XOY on the middle-surface.The axis Z obeys the corkscrew rule of right hand.
x
y
z
t/2t/2
Middle-surface
x
y
z
Figure 6-1
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x
y
z
Nx
Nx
Nxy
Nxy
Ny
Ny
Nyx
Nyx
forces distribute on the unit width
Figure 6-2
⎭⎬
⎫
==
==
t N t N
t N t N
yx yx xy xy
y y x x
τ τ
σ σ
(
6.2)
σ x
σ x
σ y
σ y
τ xyτ yx
middle-surface forces
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Fig 6.4 bending moments, twisting moments andshears
x
y
z
Mx
Mx
My
My
MxyMxy
Myx
Myx
Qx
Qx
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dx
dy
t
σ x
σ yτ
xyτ yx
τ xz
τ yz
Fig. the bending stress in thin plates
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The relationship between the bending forces and the
bending stresses is:
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⋅=⋅=
⋅=⋅=
⋅=⋅=
∫∫
∫∫
∫∫
−−
−−
−−
2
2
2
2
2
2
2
2
2
2
2
2
11
11
11
t
t yz y
t
t xz x
t
t yx yx
t
t xy xy
t
t y y
t
t x x
dzQdzQ
dz z M dz z M
dz z M dz z M
τ τ
τ τ
σ σ
(6.1)
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The first denotes the plane on which it acts and thesecond the direction of the stress.
For stress, if the normal line of the plane is the sameas the coordinate, the positive when the second is thesame as the coordinate. versa.
For strain, tension is positive for normal stress andchanging to smaller angle is positive for shear strain.
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Thick plates / 1/5t a > Shear stresses aren’t negligible.
Thin plates / 1/5t a ≤ Shear stresses are negligible and
some assumptions made.
rigid plates / 1/5w t ≤ small deflection problem,
membrane strain are negligible.
no tension or compression or
shear and deformation.
Flexural plates 1/5 / 5w a≤ <large deflection problem,
membrane strain aren’t
negligible
membranes / 5w a ≥ membrane strain
bending strain> >
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following assumptions:
• The shear strains are negligible and lines normal to the middle
surface prior to bending remain straight and normal to the middle
surface during bending.
• The normal stress and the corresponding strain are negligible,
and therefore the transverse deflection at any point is equal to that ofthe corresponding point(x,y,0) along the middle surface.
• The transverse deflections of the plates are small compared to the
thickness of the plate. Thus middle surface stretching caused by bending can be neglected (membrane action is negligible).
Rigid plates.
• Homogeneous, isotropic, and obeys Hooke’s law.
Small deflection of uniform thickness thin plates
with rigid and flexure
z
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6.2 Differential equation of plate bucking
To determine the critical loading by neutral equilibrium, it is
necessary to have the equation of equilibrium in a slight bent
configuration.
constant biaxial compression forces
Loading constant in-plane shears;
in-plane forces=externally applied loads
Two sets of forces moments and shears=transverse bending
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Equilibrium of in-plane forces
y
dx
dyx
y
a
b c
d
w
a′
b′
d′
c′
wdy
y
w
∂
∂
w dx xw
∂
∂
x
w
∂
∂
y
w
∂
∂
Nx
x
w
∂
∂
Nx
dx xw
x xw )(
∂∂
∂∂+
∂∂
Ny
y
w
∂
∂
Ny
dy
y
w
y y
w)(
∂
∂
∂
∂+
∂
∂
Nxy
Nxy
Nyx
Nyx dx
y
w
x y
w)(
∂
∂
∂
∂+
∂
∂
dy
x
w
y x
w)(
∂
∂
∂
∂+
∂
∂
x
y
z
NxNx
Nxy
Nxy
Ny
Ny
Nyx
Nyx
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the resultant of the middle-surface forces in the Z
direction :
dxd y
w N y x
w N x
w N y xy x ⎟⎟
⎞
⎜⎜⎝
⎛
∂
∂+∂∂
∂+∂
∂2
22
2
2
2 (6.3)
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Q x
x
w
∂
∂
Q y
y
w
∂
∂
Q x+(∂Q x / ∂x)dx
dx x
w
x
w2
2
∂
∂+
∂
∂
dy y
w
y
w2
2
∂
∂+
∂
∂
Q y+(∂Q y / ∂y)dy
x
y
The sum of components of Qx and Qy in the Z direction is:
dxdy y
Q
x
Q y x ⎟⎟ ⎠
⎞⎜⎜⎝
⎛
∂
∂
+∂
∂(6.4)
Equilibrium of bending moments, twisting moments and
shears
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Addition of these terms to the Z components of the middle-
surface force, given by equation 6.3 to 6.4, leads to the
equation of equilibrium in Z direction:
022
22
2
2
=∂
∂+
∂∂
∂+
∂
∂+
∂
∂+
∂
∂
y
w N
y x
w N
x
w N
y
Q
x
Q y xy x
y x (6.5)
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If the higher-order terms are neglected, this relation
reduces to:
Similarly, moment equilibrium about the Y axis leads to:
0=−∂
∂+∂
∂ y xy y Q
x M
y M
(6.6)
0=−∂
∂+
∂
∂ x
yx x Q y
M
x
M (6.7)
Equation(
6.5),
(
6.6)
and(
6.7)
are equilibrium equations
of plate bucking.
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Moment-displacement relations
The relationships between the force moments of Mx, My, Mxy and
stresses are given in the equation 6.1.
(1) The stress-strain relations are:
( )
( )
( ) ⎪
⎪⎪
⎭
⎪
⎪⎪
⎬
⎫
+
=
+−
=
+−
=
xy xy
x y y
y x x
E
E
E
γ
µ
τ
µε ε µ
σ
µε ε µ
σ
12
1
1
2
2
(
6.10)
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(2)
The bending strains-deflection relations
··m′
A′
w
xw
∂
∂
u
xw
∂∂
z
x
z
o ··
m
A
According to the hypothesis 1 and hypothesis 3, gives;
x
w zu
∂
∂−= (6.11)
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2
2
x
w z x
∂
∂−=ε (6.13)
2
2
y
w z y
∂
∂−=ε (6.14)
y xw z xy∂∂∂−=
2
2γ (6.15)
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(3) The bending stress-deflection relations
The equation (6.13)-(6.15) express the relationship betweenthe bending strains and the deflections. Substituting them into
equation (6.10), gives:
⎪⎪
⎪⎪
⎭
⎪
⎪⎪⎪
⎬
⎫
∂∂
∂
+
−=
⎟⎟ ⎠
⎞⎜⎜⎝
⎛
∂
∂+∂
∂
−−=
⎟⎟ ⎠ ⎞⎜⎜
⎝ ⎛
∂∂+
∂∂
−−=
y x
w Ez
xw
yw Ez
yw
xw Ez
xy
y
x
2
2
2
2
2
2
2
2
2
2
2
1
1
1
µ τ
µ µ
σ
µ µ
σ
(6.16)
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⎟⎟ ⎞
⎜⎜⎝
⎛
∂
∂+
∂
∂−=
2
2
2
2
y
w
x
w D M x µ (6.17)
⎟⎟ ⎠ ⎞⎜⎜
⎝ ⎛
∂∂+
∂∂−=
2
2
2
2
x
w
y
w D M y µ (6.18)
( ) y x
w D M xy
∂∂
∂−−=
2
1 µ (6.19)
(6.20)( )2
3
112 µ −=
Et D
in which
(4)
The relationship between the moments and deflections
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The quantity D is the flexural rigidity per unit width
of plate, and it called the flexural rigidity of plate as well.
It corresponds to the bending stiffness EI of a beam.
Comparison of the beam rigidity with that of the plate,
indicates that a strip of plate is stiffer than a beam of
similar width and depth by a factor.
The difference in stiffness exists because the beam is free
to deform laterally, whereas the plate strip is constrained
from deforming in this manner by the adjacent material.
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Differential equation of plate buckling
Substitution of (6.17)-(6.19) into equation (6.8), gives:
⎟⎟ ⎞
⎜⎜⎝
⎛
∂
∂+
∂∂
∂+
∂
∂4
4
22
4
4
4
2 y
w
y x
w
x
w D
y x
w
N y
w
N x
w
N xy y x ∂∂
∂+
∂
∂+
∂
∂=
2
2
2
2
2
2 (6.21)
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The critical load for uniaxial compression, pure shear or due
to a combination of compression and shear can be determined.
A primary difference between plates and columns is the
existence of two independent variables in the former as
opposed to a single independent variable in the latter.
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6.3 Critical load of a rectangular plate uniformly
compressed in one directiona simply supported rectangular plate with sides a and b units
long.
coordinates system is shown as Figure 6-8.
The plate is acted on by a compression force per unit length,
distributed uniformly along the edges X=0 and X=a.
a
bσ xσ x
x
y
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1 The differential equation of plate bucking
0122
2
4
4
22
4
4
4
=∂∂+
∂∂+
∂∂∂+
∂∂
xwt
D yw
y xw
xw
xσ (6.22)
t N
N N
x x
xy y
σ −=
== 0
The differential equation of plate bending reduces to:
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2 The boundary conditions
simply supported, the boundary conditions are:
when x=0 a,w=0 Mx=0
when y=0
b,
w=0
My=0 a
b
x
y
when y=0 b ,w=0, 02
2
2
2
=∂
∂+
∂
∂
x
w
y
wµ
02
2
2
2
=
∂
∂+
∂
∂
y
w
x
wµ
when x=0
a ,w=0,
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3 To determine the critical stresses by solving
the equationLet us assume that the solution to equation ( 6.22) is of the form
∑∑
∞
=
∞
== 1 1 sinsinm nmn b
yn
a
xm
Aw
π π
(a)
The coefficient that Amn is undetermined in the type , m and n
are the number of half-waves that plate buckles into in the x
and y directions, respectively.
m=1 2 3…
n=1
2
3…
The assumed solution already satisfies the boundary conditions
already.
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Substitution (a ) into equation (6.22 ), gives:
∑∑∞
=
∞
= ⎢⎢⎣
⎡
⎟⎟ ⎠
⎞⎜⎜⎝
⎛ +
1 1
2
2
2
2
24
m n
mnb
n
a
m A π 0sinsin
2
22
=⎥⎦
⎤−
b
yn
a
xm
a
m
D
t x π π π σ ( b)
The left-hand side of equation (b) consists of the sum of an infinite
number of independent functions, The only way such a sum can
vanish is if the coefficient of every one of the terms is equal to zero.
02
222
2
2
2
24 =
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−⎟⎟
⎠
⎞⎜⎜
⎝
⎛ +
a
m
D
t a
b
n
a
m A x
mn
π π (c)
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(
1)
Amn
=0
Substituting (1) into equation (a), we get:
w=0
solve is zero .
(
2)
The term in the square brackets vanishes
02
222
2
2
2
24 =−⎟⎟
⎞⎜⎜
⎝
⎛ +
a
m
D
t
b
n
a
m x π σ π
2
2
2
2
2
2
22
⎟⎟ ⎞
⎜⎜⎝
⎛ +=
b
n
a
m
t m
Da x
π σ
or22
2
2
⎟⎟
⎠
⎞⎜⎜
⎝
⎛ +=
mb
an
a
mb
t b
D x
π σ (d )
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The critical value ofσ x is the smallest value.
The balancecurved state
the dimension,
physics properties of plate,the number of half-waves that
the plate buckles into.
n=1 results in a minimum value of x that is the plate buckles
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2
2
2
⎟ ⎞
⎜⎝
⎛ +=
mb
a
a
mb
t b
D x
π σ (6.25)
n=1 results in a minimum value ofσ
x , that is, the plate buckles
in a single half-wave in the y direction
0=dm
d xσ
02 22
2
=⎟ ⎞⎜⎝ ⎛ −⎟ ⎞⎜⎝ ⎛ + bma
ab
mba
amb
t b Aπ
from which (6.26)
Substitution (6.26)
into equation(
6.25)
b
am =
( )t b
D
cr x 2
24 π σ = (6.27)
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m must be an integer. The critical load given by(
6.27)
is
thus valid only when a/b is a whole number.For plates in this category, the bucking pattern consists of a
single half-wave in y direction and a/b half-waves in x direction.
But a/b is seldom an integer.
The more general case, where a/b is not an integer:
t b
Dk cr 2
2π σ = (6.28)
2
⎟ ⎞
⎜⎝
⎛ +=
mb
a
a
mbk (6.29)
The factor k depends on the aspect ratio a/b and on m and n,
the number of half-waves that the plate buckles into.
The minimal value of k is called bulking stress coefficient.
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0
2
4
6
8
k
1 2 3 4a/b
2 6
m=1
m=2 m=3
m=4
Fig 6-9 k varies with a/b
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The critical stress of the column thus depends on its length,
whereas that of the plate depends on the width of the plate and itsindependent of the length.
2
2 212(1 )( / )cr
k E
b t
π
σ
µ
=−
(6.51)
2
2
( / )
cr
C E
l r
π
σ = (6.52)
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6.4 Strain energy of bending in a plate
1
Strain energy of the plate in tiny bending state
In small bending state, the plate has two kinds of force:
middle-surface force and bending force. Because a tiny bendingdeformation can not generate middle surface strain, so the work
done by middle surface force are negligible, the corresponding
strain energy is zero. Only bending force need to be considered.
x
y
z
Nx
Nx
Nxy
Nxy
Ny
Ny
Nyx
Nyxσ x
τ xy
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work done by bending force or strain energy of the plate
gives:
( )∫∫∫ ++ dxdydz xy xy y y x x γ τ ε σ ε σ 2
1
[∫ ∫ ∫−−+= 2
20 0
22 22
1 l
l
b a
y x y x E
U σ µσ σ σ
( ) dxdydz xy
2
12 τ µ ++ (6.32)
substituting the relationship between bending stress and
curvature given by ( 6.16) :
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2. The potential energy of external load
The load acted on the plate is the middle surface
load. Nx , Ny , Nxy . Base on tiny deformationhypothesis, the changes of middle surface load
produced by bending deformation are neglected.
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when it buckles, the slab bends and the end A has a
translation along the X axis, it has been assumed that the
middle surface can not deform, so the length of AB is always
the same.
a
A A′ BN
x
Nx
dx dx
ds
w
dx x
w
∂
∂
B A B A ′−′=∆ (a)
2dx
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2
2⎟ ⎠
⎞⎜⎝
⎛
∂
∂+= dx
x
wdxds
dx x
wds
⎥⎦
⎤
⎢⎣
⎡
⎟ ⎠
⎞
⎜⎝
⎛
∂
∂+=
2
2
11 (b)
∫ ∫∫ ′ ′′ ⎟
⎞⎜⎝
⎛ ∂
∂+==
B
A
B
A
B
Adx
x
wdxds
2
2
1A′B
(c)∫ ′ ⎟
⎞⎜⎝
⎛ ∂
∂+′= B
Adx
x
w B A
2
2
1
Substitution(
c)
into(
a),
gives:
∫ ⎟ ⎞
⎜
⎝
⎛
∂
∂=∆ dx
x
w2
2
1(6.35)
ds
w
dx x
w
∂∂
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Take out a right-angle AOB (Fig 6-12) from middle-
surface of the plate along x and y direction, after
bending AOB turn into A1O1B1, and the shear
strain γ xy is:
1112
BO A xy ∠−= π
γ o1 w
dx x
w
∂
∂
dy y
w
∂
∂
o A
B
dx
dy
A1
B1
dx
dy
x
yz
is a minim,
so)2
( 111 BO A∠−π
)
2
sin( 111 BO A xy ∠−= π
γ
Fig 6—12
111cos BOA∠=
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111cos BO A∠
And l 1 m1 n1—cosine in three axes of lineO1A1
l 2
m2
n2
—cosine in three axes of lineO1
BI
from Fig6-12,
212121 nnmmll xy ++=
γ (
a)
⎪
⎪⎪⎪⎪⎪
⎪
⎭
⎪⎪⎪⎪
⎪⎪⎪
⎬
⎫
∂
∂=
⎟ ⎠
⎞⎜⎝
⎛
∂
∂−≈
=
∂
∂=∂
∂
=
=
⎟ ⎠
⎞⎜⎝
⎛ ∂
∂−≈
⎟ ⎠
⎞⎜⎝
⎛ ∂
∂−
=
y
wn
y
wm
l
x
w
dx
dx x
w
n
m
x
w
dx
dx x
wdx
l
2
2
2
1
1
22
1
2
11
0
0
2
11
( b)
substitution b into a gives
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substitution(b)into(a)gives
During bending,the work done by in-plane force in
strip AB is:
yw
xw xy
∂∂⋅∂∂=γ (6.36)
∫ ⎟ ⎞
⎜⎝
⎛ ∂
∂=∆ dxdy
x
wdy N dy N x x
2
2
1
For the entire plate, the work done by in-plane force Nx
is
∫∫ ⎟ ⎞
⎜⎝
⎛ ∂
∂= dxdy
x
w N T x
2
12
1
Similarly, for the entire plate, the work done by in-plane force Ny is
∫∫ ⎟ ⎞
⎜
⎝
⎛
∂
∂= dxdy
y
w N T y
2
2
2
1
F th ti l t th k d b i l f N i
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For the entire plate, the work done by in-plane force Nxy is
∫∫ ∂∂
∂∂= dxdy
y
w
x
w N T xy3
∫∫⎢⎢⎣
⎡⎟ ⎠
⎞⎜⎝
⎛
∂
∂+⎟
⎠
⎞⎜⎝
⎛ ∂
∂=
22
2
1
y
w N
x
w N V y x dxdy
y
w
x
w N xy ⎥
⎦
⎤
∂
∂
∂
∂+ 2 (6.37)
The total potential energy for the bending plate under the
action of Nx , Ny , Nxy are:
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6.5 Critical load of uniaxially compressed plate, fixed
along all edges, by the energy method
1
boundary conditionwhen x=0 a,
when y=0 a,
0=w 0=∂
∂
x
w
0=w 0=∂
∂
y
w
In accordance with these conditions the plate is prevented from
moving in the z direction or rotating at the boundaries. The edges
of the plate are however, free to move in the xy plane.
a
a
x
y z
2 Boundary conditions are satisfied if w is assumed
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2 Boundary conditions are satisfied if w is assumed
to be of the form:
⎟ ⎞
⎜⎝
⎛ −⎟
⎞⎜⎝
⎛ −=
a
y
a
x Aw
π π 2cos1
2cos1
(
6.39)
It is easy to prove,
formula(
6.39)
can satisfy all theboundary conditions .
3
Calculate the strain energy due to bending and thepotential energy of the external loads
∫ ∫ ⎢⎢⎣
⎡
∂∂
∂∂+⎟⎟
⎠ ⎞⎜⎜
⎝ ⎛
∂∂+⎟⎟
⎠ ⎞⎜⎜
⎝ ⎛
∂∂= a a
y xw
yw
xw DU
0 0 2
2
2
22
2
22
2
2
22
µ ( ) dxd y xw
⎥⎥⎦
⎤
⎟⎟ ⎠
⎞⎜⎜⎝
⎛
∂∂
∂−+
22
12 µ (6.40)
To evaluate this expression, the following derivatives of w are
needed:
⎫⎞⎛⎞⎛∂ A 222
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⎪
⎪⎪⎪⎪
⎪
⎭
⎪⎪
⎪⎪⎪⎪
⎬
⎫
⎟ ⎠
⎞⎜⎝
⎛ ⎟ ⎠
⎞⎜⎝
⎛ =∂∂
∂
⎟ ⎠ ⎞⎜
⎝ ⎛ −⎟
⎠ ⎞⎜
⎝ ⎛ =
∂∂
⎟ ⎠
⎞⎜⎝
⎛ −⎟
⎠
⎞⎜⎝
⎛ =
∂
∂
⎟ ⎠
⎞⎜⎝
⎛ −⎟
⎠
⎞⎜⎝
⎛ =
∂
∂
⎟ ⎠
⎞⎜⎝
⎛ −⎟
⎠
⎞⎜⎝
⎛ =
∂
∂
a
y
a
x
a
A
y x
w
a x
a y
a A
yw
a
y
a
x
a
A
x
w
a
x
a
y
a
A
y
w
a
y
a
x
a
A
x
w
π π π
π π π
π π π
π π π
π π π
2sin
2sin
4
2cos12cos4
2cos1
2cos
4
2cos1
2sin
2
2cos1
2sin
2
2
22
2
2
2
2
2
2
2
2
(a)
Substitution(a)into(6.40),gives:
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∫ ∫ ⎢⎣
⎡
⎜⎝
⎛
−⎟ ⎠
⎞
⎜⎝
⎛
=
a a
a
y
a
x
a
A D
U 0 0
2
4
242
cos21
2
cos
16
2
π π π
( )
dxdya
y
a
x
a
y
a
y
a
x
a
x
a
x
a
x
a
y
a
y
⎥⎦
⎤⎟
⎠
⎞⎜⎝
⎛ ⎟
⎠
⎞⎜⎝
⎛ ⋅
−+⎟ ⎠
⎞⎜⎝
⎛ −⋅
⎟ ⎠
⎞⎜⎝
⎛ −+⎟
⎠
⎞+
⎜⎝
⎛ −⎟
⎠
⎞⎜⎝
⎛ +⎟ ⎠
⎞+
π π
µ π π
π π µ
π
π π π
2sin
2sin
122
cos2
cos
2cos
2cos2
2cos
2cos21
2cos
2cos
22
2
22
22
( b)
Making use of the following definite integrals:
⎫a ax2π
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⎪⎪⎪
⎭
⎪⎪⎪
⎬
⎫
=
=
=
∫
∫∫
a
a
a
dxa
x
adx
a
x
adx
a
x
0
0
2
0
2
02
cos
2
2cos
2
2sin
π
π
π
(c)
Equation(
b)
reduces into:
⎢⎣
⎡⎟ ⎠
⎞⎜⎝
⎛ ++⎟
⎠
⎞⎜⎝
⎛ +=
2222
84
24a
aaa
aa
a
A DU
π
( ) ⎥⎦
⎤−+⎟⎟
⎠
⎞⎜⎜⎝
⎛ +
412
42
22aa
µ µ
The strain energy of the bending plate is equals
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substitution into(
a),
==>
gy g p q
to:2
24
16a
A DU π = (6.41)
The potential energy of the external loads derive
from(
6.37):
∫ ∫ ⎟ ⎞
⎜⎝
⎛ ∂∂−=
a a x dxdy
x
w N V
0 0
2
2(a)
x
w
∂
∂
∫ ∫ ⎜⎝
⎛ −−=
a a x
a
y
a
x
a
N AV
0 0
2
2
22 2cos21
2sin
2
4 π π π dxdy
a
y⎟
⎞+
π 2cos2
⎥⎦
⎤⎢⎣
⎡⎟ ⎠
⎞⎜⎝
⎛ +−=
22
22
22a
aa
a
N AV xπ
(6.42)
or 3 22
x N AV
π −=
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2V −=
The total potential energy of the system is obtained by adding
the expressions in(6.41) and (6.42):
2316
22
2
24
x N Aa
A DV U π π −=+ (6.43)
( )
( ) 0332
0
2
2
4
=−
=+
cr x N A
a
A D
dA
V U d
π π
The critical loading and critical stress can be determined by
finding the slightly bent configuration for which the total
potential energy has a stationary value.
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( )2
2
2
2
67.103
32
a
D
a
D N
cr x
π π == (6.44
)
Critical stress:
( ) ( )
t a
D
t
N cr x
cr x 2
2
67.10 π
σ == (6.45)
Exact solution :
( ) 2
2
07.10a
D N cr x
π
=
Critical load:
6.7 CRITICAL STRESSS OF A RECTANGULAR
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6.7 CRITICAL STRESSS OF A RECTANGULAR
PLATE IN SHEARThis shows for bucking to take place, it is not necessary
that a member be loaded in axial compression. All that is
necessary is that compression stresses exist in some part ofthe member.
x
y
a
a xy
We solve this problem with Galerkin method introducing of 2 4
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We solve this problem with Galerkin method introducing of 2.4.
1, the brief introduction of Galerkin equation
Consulting 2.4, general form of Galerkin equation is:
( ) ( )∫ ∫ =a a
i dxdy xgwQ0 0
0 (6.59)
Q (w ) is the right type of the differential equation about flexion;
gi(x) is nember i function of flexion modal
( ) y x
w N
y
w
y x
w
x
wwQ xy
∂∂
∂+
∂
∂+
∂∂
∂+
∂
∂=
2
4
2
22
4
4
4
22 (6.60)
To determine critical loading and critical stress by solving theGalerkin equation
2 Calculate critical shearing stress
Choosing an expression for the deflected shape of the member
that satisfies the boundary conditions
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t at sat s es t e bou da y co d t o s:
in Galerkin equation:
i=1,2
a y
a x A
a y
a x Aw π π π π 2sin2sinsinsin 21 += (6.58)
( )a
y
a
x xg
π π sinsin1 = (6.61)
( )a y
a x xg π π 2sin2sin2 = (6.62)
Substitution(6.58)into(6.60)and according to i=1
and i=2, using(6.61)and(6.62)separately,get two
Galerkin equations:
A44
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∫ ∫ +
a a
a
y
a
x
a
A
0 0
22
4
4
1
sinsin
4 π π π
a
y
a
x
a
x
a
A
a
y
a
y
a
x
a
x
a
A
D
N a
y
a
y
a
x
a
x
a
A
xy
π π π π π
π π π π
π π π π π
2cossin
2cos
4sin
cossincos2
sin2
sinsin2
sin64
2
2
2
2
2
1
4
4
2
+⋅
⎜⎜
⎝
⎛ +
0sin =⎥⎦
⎤⎟ ⎠
⎞⋅ dxdya
yπ
(6.63)
⎡ A 4 224
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∫ ∫ ⎢⎣
⎡a a
a
y
a
y
a
x
a
x
a
A
0 0 4
41 2
sinsin
2
sinsin
4 π π π π π
a
x
a
x
a
A
a
y
a
y
A
x
a
x
a
A
D
N a
x
a
x
a
A
xy
π π π π π
π π π
π π π
2sin
2cos
42sincos
2sincos
2
2sin
2sin
64
2
22
2
21
22
4
42
+⋅
⎜⎜⎝
⎛ +
+
02sin2cos =⎥⎦
⎤⎟ ⎠ ⎞⋅ dxdy
a
y
a
y π π
(6.64)
The definite integrals appearing in(
6.64)
and(
6.64)
have
the following value:
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∫ =
a a
dxa
x
0
2
2sin
π
∫
∫
=
=
a
a
dxa x
a x
dxa
x
a
x
0
0
0sincos
0sin2
sin
π π
π π
∫
∫
=
−=
a
a
adxa x
a x
adx
a
x
a
x
0
0
34cos2sin
3
2sin
2cos
π π π
π
π π
Hence the equation (6.63) and (6.64) can be reduced to
03
42
2
64
03242
24
2
2
21
2
4
42
2
2
2
2
2
4
4
1
=⎟ ⎞
⎜⎝
⎛ +⎟
⎞⎜⎝
⎛
=⎟ ⎠ ⎞⎜
⎝ ⎛ −+⎟
⎠ ⎞⎜
⎝ ⎛
π
π π
π π π
a
a
A
D
N a
a
A
aa
A D N a
a A
xy
xy
After simplifying ,gives
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⎪⎪⎭
⎪⎪⎬
⎫
=+
=+
016
9
32
09
32
22
4
1
212
4
Aa
A D
N
A D
N Aa
xy
xy
π
π
(
6.65)
The indifferent equilibrium can become possible , only
and isn’t zero completely 。 So, the coefficient
determinant of the equation (6.65 ) must be zero.
016
9
329
32
2
4
2
4
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
a D
N D
N
a
xy
xy
π
π
(6.66)
1 A
2 A
( )2
2
1.11a
D N
cr xy
π =
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a
Critical shear stress:
( )t a
Dcr xy 2
2
1.11 π
τ =
Bending bucking coefficient: k =11.1
using double trigonometric series with energy law to
have former 18 terms , obtained an exacter solution:
k =9.34
Standardize formulae in Germany's steel construction:4
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(
6.67)
2
4
5.34k = +1
a
b = ≥
( 6.68)2
5.344k
= +
result in fig
6-3Germany
standardizes
formulae
α
1
k
5.34
9.34
1.0
Fig. bending bucking coefficients varies with
1a
b = <
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6.10 Inelastic bucking of plates
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A theory of the inelastic behavior is extremely complex and
beyond the scope of this book.
However, the conclusions that have been drawn from inelastic
plate-buckling studies are simple and straightforward and will
be briefly considered here.
Young’ modulus is replaced by a reduced modulus.
( )
2
2
2
2
2
112⎟ ⎠
⎞⎜⎝
⎛
−==
b
t E k
t b
Dk cr
µ
η π η π σ (6.92)
Where is a plasticity reduction factor. .η 1<η
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No generally applicable expression, like the tangent modulus in the
columns, therefore exists for the reduced modulus of a plate.
η
the shape of the stress-strain curve,
the type of loading,
the length-to-width ratio of the plate,
the boundary conditions
A relation for considerably more suited for design purposes has
been derived by Bleich. Using an approximate theory, he obtained
η
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E
E t =η
The advantage of this relation is that it leads to conservative results
for any long plate, regardless of the boundary conditions, and that it
can be used for shear as well as axial compression.
But remember: k should be decided by
4 η
α α =
y g pp y,
the simple expression:
6.11 finite deflection theory of plates
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Plates, unlike columns, do not collapse when the critical load isreach.it is thus obvious that the postbuckling behavior should be
considered in formulating design criteria for plates. However, if one
desires to study the behavior of the member, it is necessary to
consider deformations of finite magnitude.
If a plate is bent into a nondevelopable surface or if its edges are
restrained from approaching one another during bending, membrane
strains will be induced in the middle surface of the plate.once the
transverse deflections become of the order of magnitude of the plate
thickness, stretching of the middle surface is no longer negligible.
The main difference between that in this article and theinfinitesimal deformation theory is that middle-surface strains due
to bending that were neglected previously will now be considered.
Derivation of equations
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Equilibrium
No change occurs. On the other hand there are in-plane forces due to
membrane action in addition to the forces applied along the edges of the
plate when large deflections are considered.Forces in the x direction of in-plane to zero:
For the y direction of in-plane
In the z direction::
0=∂
∂+
∂
∂
y
N
x
N xy x
0=∂
∂+
∂
∂
x
N
y
N xy y
⎟⎟ ⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂∂
∂+
∂
∂4
4
22
4
4
4
2 y
w
y x
w
x
w D
y x
w N
y
w N
x
w N xy y x
∂∂
∂+
∂
∂+
∂
∂=
2
2
2
2
2
2 (6.113)
Compatibility
The displacements and of a point in the plate consist of t o parts:
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The displacements u and v of a point in the plate consist of two parts:
(1)The displacement in the middle plane
(2)the bending displacements
2)(2
1
x
w
x
u o xo
∂
∂+
∂
∂=ε
y
w
x
w
x
v
y
u oo xyo
∂
∂
∂
∂
+∂
∂
+∂
∂
=γ
2)(21
yw
yv o
yo∂∂+
∂∂=ε
Relationship
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)(1
y x xo N N
Eh
µ ε −=
)(1
x y yo N N Eh
µ ε −=
xy xyo N Eh
)1(2 µ γ
+=
To reduce the number of equations that must be solved
simultaneously a stress function is introduced
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simultaneously, a stress function is introduced.
2
2
y
F h N x
∂
∂=
2
2
x
F h N y
∂
∂=
y x
F h N xy
∂∂
∂−=
2
Von Karman large deflection plate equations:
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])[(
2
2
2
2
22
2
4
4
22
4
4
4
y
w
x
w
y x
w E
y
F
y x
F
x
F
∂
∂
∂
∂−
∂∂
∂
=∂
∂+∂∂
∂+∂
∂
)2(
2
22
2
2
2
2
2
2
2
2
4
4
22
4
4
4
y x
w
y x
F
y
w
x
F
x
w
y
F
D
h
yw
y xw
xw
∂∂
∂
∂∂
∂−
∂
∂
∂
∂+
∂
∂
∂
∂
=∂∂+∂∂ ∂+∂∂
6.12 postbuckling behavior of axially compressed plates
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Simply supported square plate subjected to a uniaxial compressionforce:
2)
2cos
2(cos
32
22 y
a
y
a
x Ef F xaσ π π
−+=
2
22
8a
f E cr xa
π σ σ +=
a
ycr xa xa x
π σ σ σ σ 2cos)( −+=
a x
cr xa yπ σ σ σ 2cos)( −=
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Conclusions:
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1.plates can continue to carry increasing load
subsequent to reaching the critical stress; this is, they
exhibit postbuckling strength.
2.transverse tensile stresses that arise subsequent to thestart of buckling are primarily responsible for the
presence of postbuckling strength in plates.
3. The material near the longitudinal edges of the plateresists most of the increase in load that occurs in the
postbuckling range.
6.13 ultimate strength of axially compressed plates
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Where is the average longitudinal stress at failure.
)1/1(
21 +=
ycr cr
fa
σ σ σ σ
fa
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yeu t bP σ =
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6.14 design provisions for local buckling
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In designing, it is common to proportion the member so thatoverall failure occurs prior to local buckling.
ycr F F >
For I section axial column, web:
flange: y
f
t
b/235)1.010( λ +≤
y
w
o f t h /235)5.025( λ +≤
For I section eccentric column, web:
flange: y f
t
b/235151 ≤
yo
w
o f t
h/235)255.016( ++≤ λ α
yo
w
o f t
h/235)2.265.048( −+≤ λ α
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For beam, it is the most complex in this class.
Disadvantages:
1. Elastic analysis;
2. No postbuckling is calculated but considered by using
high buckling stress coefficient k, no initial
imperfections.
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