Buckling of Stepped Thickness Plates
in the Theory of Plasticity
Momčilo Dunjić
Assistant Professor
University of Belgrade
Faculty of Mechanical Engineering
The problem of buckling has been treated for a long time as a single
rectangular field with different boundary conditions loaded with inplate
loads and usualy applying Levy method, when two opposite edges were
always simply supported. This paper is concerned with the plastic buckling
of rectangular plates in the region of plasticity. In this paper, it has been
treated the elasto-plastic deformations for two coupled plates with
different thicknesses, loaded with inplane constant forces Nx = const.
Keywords: buckling, plates, plasticity, deformations.
1. INTRODUCTION
The problem of buckling has been treated for many
years as one rectangular field with different boundary
conditions loaded with in-plate loads Nx, and Ny and
usually applying Levy method, when two opposite
edges were always simply supported.
In plastic region A. A. Ilyushin [1] and E. Z. Stowell
[2], using their differential equations solved many
buckling problems in elasto-plastic region, but always
with plates of constant thickness. Now, in this paper, the
author has solved the problem of buckling when two
rectangular plates form the unity, but of different
thicknesses. All sides are simply supported and pressed
along two parallel opposite sides with Nx = const.
It is known, Fig. 1, that in the plastic region the angle
α0 corresponds to modulus of elasticity (E), elastic
region, and direction ON0, determined by angle α0 gives
“cutting modulus” of elasticity ( Ec0 ), where the point N0
gives the initial appearance of plasticity [3]. The value of
angle αk gives “tangential modulus of plasticity”.
E 0k =
dσ
.
dε
(1)
The direction ON determines the angle αc which
determines primary modulus of elasticity
Ec0 =
σ
.
ε
(2)
According to the hypothesis that the radius of the
Mises circle may be written, as it is known, as
generalized stress, effective stress or equivalent
stress.
In the case of complete stresses the intensity of
stress σi and intensity of deformation εi are introduced
in the forms
σi =
(
⋅ σ x −σ y
) + (σ y −σ z )
2
2
1
2
⋅
(
2
)
2
2
+ (σ z −σ x ) + 6 τ xy
+τ 2yz +τ zx
(3)
and
εi =
(
⋅ εx −ε y
) + (ε y −ε z )
2
2
2
⋅
3
+ (εz −ε x ) +
2
(
)
3 2
2
γ xy + γ 2yz + γ zx
. (4)
2
2. FORMULATION OF THE PROBLEM
In our case we are solving the problem of stability of
plates, when we have small elasto-plastic deformations,
using partial differential equations of A. A. Ilyushin
[1,2]
⎛ 3 1 − s σ xσ y + 2τ 2 ⎞
⎛ 3 1 − s σ x2 ⎞ ∂ 4 w
⎟⋅
+ 2 ⎜1 −
⎜1 −
⎟
2
⎜ 4 1 − r σ 2 ⎟ ∂x 4
⎜
⎟
4
1
r
−
σ
i ⎠
i
⎝
⎝
⎠
⎛ 3 1 − s σ 2y ⎞ ∂ 4 w 1 − s τ
⎟
+ ⎜1 −
−3
⋅
1 − r σ i2
∂x 2 ∂y 2 ⎜⎝ 4 1 − r σ i2 ⎟⎠ ∂y 4
⎛
∂4w
∂4 w ⎞
1
⋅⎜σ x
+σ y
∏ (σ , w ) = 0 . (5)
⎟+
3 ⎟ (1 − r ) D ′
⎜ ∂x3∂y
∂x∂y ⎠
⎝
⋅
Figure 1. Relation between (α, αc, αk) and (E, Ec, Ek)
Received: November 2010, Accepted: February 2011
Correspondence to: Dr Momčilo Dunjić
Faculty of Mechanical Engineering,
Kraljice Marije 16, 11120 Belgrade 35, Serbia
E-mail: [email protected]
© Faculty of Mechanical Engineering, Belgrade. All rights reserved
∂4w
In (5) is introduced the sign
∂2w
∂2w
.
+σ y
∂x∂y
∂y 2
(6)
FME Transactions (2011) 39, 41-47
41
∏ (σ , w ) = σ x
∂2w
∂x 2
+ 2τ
In our case we consider two plates (1) and (2) (Fig.
2), it is plate of dimensions a × b which has contact
along the edge y = η, (0 < η < b).
functions, while the functions f1 (y) and f2 (y) are giving
behaviours of plates (1) and (2) in y direction.
The conditions on edges x = 0 and x = a must be
satisfied:
⎛ ∂ 2 w1
∂ 2 w1 ⎞
w1 = 0 x =0 and D1 ⎜
+ν
⎟= 0
⎜ ∂x 2
∂y 2 ⎟⎠
⎝
x =a
⎛ ∂ 2 w2
∂ 2 w2
w2 = 0 x = a and D2 ⎜
+ν
⎜ ∂x 2
∂y 2
⎝
⎞
⎟=0
⎟
⎠
(11)
(12)
x =a
which means that plates (1) and (2) are simply
supported on x = 0 and x = a, or the deflections and
moments are equal to zero.
Supposed solutions must also satisfy the differential
equations for plates (1) and (2), respectively:
Figure 2. Two coupled plates with load NX = const.
Now, we have
• plate (1): 0 ≤ x ≤ a; 0 ≤ y ≤ η
• plate (2): 0 ≤ x ≤ a; η ≤ y ≤ b.
In (5) are given: r = 1 – ϕc, s = 1 – (ϕc – ϕk).
Along the edges y = 0 and y = b we could have any
kind of boundary conditions. Along the plates (1) and
(2), for x = 0 and x = a are pressed with forces Nx =
const. and they have thicknesses h1 and h2. The other
forces are Ny = 0 and Nxy = 0.
In actual case is used modified differential equation
given by E. Z. Stowell
⎡ 3⎛ ϕ
⎡ 3 ⎛ ϕk ⎞ σ x2 ⎤ ∂ 4 w
⎢1 − ⎜1 − ⎟ 2 ⎥ 4 + 2 ⎢1 − ⎜1 − k
⎢ 4 ⎝ ϕc
⎢⎣ 4 ⎝ ϕc ⎠ σ i ⎥⎦ ∂x
⎣
⋅
2
⎞ σ xσ y + 2τ ⎤
⎥⋅
⎟
⎥
σ i2
⎠
⎦
⎡ 3 ⎛ ϕ ⎞ σ 2y ⎤ ∂ 4 w ⎛ ϕ
⎥
+ ⎢1 − ⎜ 1 − k ⎟
− 3 ⎜1 − k
2
2 ⎢
2⎥
4
ϕ
4
∂x ∂y
c ⎠ σ i ⎦ ∂y
⎝
⎝ ϕc
⎣
⎛
∂4w
∂4w ⎞ h
⋅⎜σ x
+σ y
∏ (σ , w ) = 0
⎟+
⎜ ∂x3∂y
∂x∂y 3 ⎟⎠ Dc′
⎝
∂4w
⎞
⎟⋅
⎠
(7)
⎛ 1 3 ϕk ⎞ ∂ 4 w
∂4 w
∂4w σ xh ∂2w
= 0 (8)
⎜ +
⎟ 4 +2 2 2 + 4 +
Dc′ ∂x 2
∂x ∂y
∂y
⎝ 4 4 ϕc ⎠ ∂x
where
E
ϕk = k ,
E
Dc′ = Dc′1 = D1
or
Dc′ = Dc′2 = D2 .
According to Morris-Lavy method, for the plate (1)
and (2) are supposed functions w1 and w2:
w1 = f1 ( y ) sin
mπx
a
(9)
w2 = f 2 ( y ) sin
mπx
.
a
(10)
Using (9) and (10), under the load Nx = σx · h, the
plates are going to deflect in x direction in the form of sin
42 ▪ VOL. 39, No 1, 2011
⎛ 1 3 ϕk
⎜ +
⎝ 4 4 ϕc
⎞ ∂ 4 w2
∂ 4 w2
∂ 4 w2 N x ∂w2
2
+
+
+
= 0 . (14)
⎟ 4
∂x 2 ∂y 2 ∂y 4
D2 ∂x 2
⎠ ∂x
According to obtained solutions for (13) and (14),
we have to find the derivations of (9) and (10):
w1 = f1 ( y ) f0 ( x) = f1 ( y ) sin
mπx
a
∂w1
mπ
mπx ∂ 2 w1 ∂3 w1 ∂ 4 w1
= f1 ( y )
;
cos
;
;
∂x
a
a
∂x 2
∂x3
∂x 4
⎛ m 2 π2 ⎞
∂ 4 w1
mπx
= f1′′( y ) ⎜ −
⎟ sin
2 2
2 ⎟
⎜
a
∂x ∂y
a ⎠
⎝
∂w1
mπx ∂ 2 w1 ∂3 w1 ∂ 4 w1
= f1′( y ) sin
;
;
;
∂y
a
∂y 2
∂y3
∂y 4
where values for aluminium and for steel are given in
[5].
For our case, given with (Fig. 2), (7) is reduced to:
E
ϕc = c ,
E
⎛ 1 3 ϕk ⎞ ∂ 4 w1
∂ 4 w1
∂ 4 w1 N x ∂w1
+2
+
+
= 0 (13)
⎜ +
⎟
4
∂x 2 ∂y 2 ∂y 4
D1∂x 2
⎝ 4 4 ϕc ⎠ ∂x
w2 = f 2 ( y ) f0 ( x) = f 2 ( y ) sin
(15)
mπx
a
∂w2
mπ
mπx ∂ 2 w2 ∂3 w2 ∂ 4 w2
= f2 ( y)
cos
;
;
;
a
a
∂x
∂x 2
∂x3
∂x 4
⎛ m 2 π2 ⎞
∂ 4 w2
mπx
= f 2′′ ( y ) ⎜ −
⎟ sin
2 2
2 ⎟
⎜
a
a ⎠
∂x ∂y
⎝
∂w2
mπx ∂ 2 w2 ∂ 3 w2 ∂ 4 w2
;
;
;
. (16)
= f 2′ ( y ) sin
a
∂y
∂y 2
∂y 3
∂y 4
Using (15) and (16) we obtain:
⎛ 1 3 ϕk ⎞
m 4 π4
mπx
+
⎜ +
⎟ f1 ( y ) 4 sin
a
a
⎝ 4 4 ϕc ⎠
⎛ m 2 π2 ⎞
mπx
mπx
+2 f1′′( y ) ⎜ −
+ f1IV ( y ) sin
+
⎟ sin
2 ⎟
⎜
a
a
a ⎠
⎝
⎛ m 2 π2 ⎞
mπx
+ f1 ( y ) ⎜ −
= 0.
(17)
⎟ sin
2 ⎟
⎜
a
a ⎠
⎝
At last from (17) we get:
FME Transactions
f1IV ( y ) − 2 f1′′( y )
⎡ m 4 π4 ⎛ 1 3 ϕk
+ f1 ( y ) ⎢
+
4 ⎜
⎣⎢ a ⎝ 4 4 ϕc
m 2 π2
a2
The roots of (25) are
+
Partial differential equation of forth order is reduced
to ordinal differential equation with constant
coefficients of the same order (18).
The solution for (18) is supposed to be:
f1 ( y ) = eλ y .
(19)
From (18) we obtain characteristic equation
λ4 − 2
m 2 π2
a
2
⎡ m 4 π4
λ2 + ⎢
⎣⎢ a
4
κ −k
π4 m 2 ⎤
⎥
a 2b 2 ⎦⎥
(20)
⎛1
⎝4
3 ϕk
4 ϕc
⎞
N x b2
,
and
.
k
=
⎟
1
D1π2
⎠
(21)
⎛
a2 ⎞
⎜ 1 + k1 2 2 ⎟ − κ 0 .
⎜
b m ⎟⎠
⎝
mπ
1±
a
or
λ1,′ 2 = ±
λ3,′ 4 = ± i
mπ
1±
a
The solution for the plate (2) is now:
mπx
= ( C5 chα 2 y + C6 shα 2 y +
a
mπx
+C7 cosβ 2 y + C8 sinβ 2 y ) sin
.
(30)
a
w1 = 0
⎛
a2
⎜ 1 + k1 2 2
⎜
b m
⎝
⎛
mπ
a2
λ3, 4 = ± i
−1 + ⎜1 + k1
⎜
a
b2 m2
⎝
⎞
⎟ − κ 0 = ±α1
⎟
⎠
∂ 2 w1
∂y
a
2
For
2 4
FME Transactions
mπx
= f1 (0) = 0
a
mπx
a
∂w1
= (α1C1 shα1 y + α1C2 chα1 y −
∂y
∂ 2 w1
∂y
2
1
(26)
(
mπx
a
= α 2C1 chα1 y + α 2C2 shα1 y −
1
1
1
)
mπx
.
a
(33)
Including (33) into (32) we obtain:
(C1α12 ch0 + C2α12 sh0 − C3β12 cos0 − C4 β12 ⋅ 0) ⋅
introducing the coefficient ψ:
D1
.
D2
y =0
−C3 β 2 sinβ1 y + C4 β 2 cosβ1 y sin
2 4
Nx m π
N b m π D1
D m π
= x
= k1 1
2
2 D
2
D2 a
D
D1π
b
2
2 b
ψ =
(32)
−C3 β1 sinβ1 y + C4 β1 cosβ1 y ) sin
⎛ N x m 2 π2 m 4 π4 ⎞
k1 ⎟ = 0 . (25)
−
⎜ D2 a 2
⎟
a4
⎝
⎠
2
=0
II: w1 = ( C1 chα1 y + C2 shα1 y +
(24)
λ2 − ⎜
2 2
∂x 2
+C3 cosβ1 y + C4 sinβ1 y ) sin
For the plate (2) we have:
m 2 π2
∂w1
C1 + C3 = 0
w1 = ( C1 chα1 y + C2 shα1 y +
λ4 − 2
+ν
I: ( C1 ch0 + C2 sh0 + C3 cos0 + C4 sin0) sin
In this case we obtain our function of deflection:
mπx
.
a
2
means that the moment is equal to zero.
Using (31) and (32) we get:
⎞
⎟ − κ 0 = β1 ⋅ i . (23)
⎟
⎠
+C3 cosβ1 y + C4 sinβ1 y ) sin
(31)
y =0
means that deflection for y = 0 is zero.
II condition:
is:
mπ
1+
a
⎛
⎞
a2
⎜ 1 + k1 2 2 ψ ⎟ − κ 0 = ±α 2
⎜
⎟
b m
⎝
⎠
⎛
⎞
a2
⎜1 + k1 2 2 ψ ⎟ − κ 0 − 1 = ± iβ 2 . (29)
⎜
⎟
b m
⎝
⎠
mπ
a
(22)
⎛
k a2 ⎞
As it is κ0 < 1, then ⎜1 + 1 ⎟ − κ 0 > 0 .
⎜ b2 m2 ⎟
⎝
⎠
One possible combination for resolving our problem
λ1, 2 = ±
(28)
Now we must use the boundary conditions (I, II, ...,
VII, VIII), in order to get unknown coefficients of (24)
and (28):
I condition:
From (20), we get four solutions for λ:
λ1, 2,3, 4 = ±
⎛
⎞
a2
⎜ 1 + k1 2 2 ψ ⎟ − κ 0
⎜
⎟
b m
⎝
⎠
w2 = f 2 ( y ) sin
where κ0 and k1 are:
κ0 = ⎜ +
mπ
1±
a
λ1,′ 2,3, 4 = ±
⎞ N x m 2 π2 ⎤
⎥ = 0 . (18)
⎟−
2
⎠ D1 a ⎦⎥
(27)
⋅ sin
⎛ m 2 π2
mπx
mπx ⎞
+ f1 (0) ⎜ −
ν
sin
⎟=0.
⎜
a
a ⎟⎠
a2
⎝
(34)
VOL. 39, No 1, 2011 ▪ 43
Taking into account (I), f1 (0) = 0, condition (II) is
reduced to:
C1α12 − C3 β12 = 0 .
∂2w
∂y 2
(35)
−C7 β 2 sinβ 2 y + C8 β 2 cosβ 2 y ) sin
C1 = C3 = 0 .
(37)
w1 = ( C2 shα1 y + C4 sinβ 2 y ) sin
mπx
a
w2 = 0 y =b .
(39)
IV condition:
=0.
(
= C5α 22 chα 2 y + C6α 22 shα 2 y − C7 β 22 cosβ 2 y −
)
−C8 β 22 sinβ 2 y sin
⋅ sin
⎛ m 2 π2
mπx
+ν ⎜
⎜ a2
a
⎝
mπx
a
−C7 β 2 sinβ 2 y + C8 β 2 cosβ 2 y ) sin
C5α 22 chα 2 b + C6α 22 shα 2 b −
−C7 β 22 cosβ 2 y − C8 β 22 sinβ 2 y
)
When condition (III) is multiplied first by α 22 and
)
(41)
)
(46)
(
)
(
= ( C5 chα 2b + C6 shα 2b +
)
−C7 α 22 + β 22 cosβ 2b − C8 α 22 + β 22 sinβ 2b = 0
C8 = − C7
mπx
+C7 cosβ 2b + C8 sinβ 2 b ) sin
a
f 2 (b) = 0 = C5 chα 2 b + C6 shα 2b +
44 ▪ VOL. 39, No 1, 2011
(
– III⋅ α 22 + IV:
From (III) we obtain:
+C7 cosβ 2b + C8 sinβ 2b = 0
(45)
α 22 + β 22 ≠ 0 ⇒ C5 chα 2b + C6 shα 2b = 0
chα 2b
.
C6 = − C5
shα 2b
mπx
a
mπx
sin
.
a
−C7 β 22 cosβ 2 b − C8 β 22 sinβ 2b = 0 .
(
(
and from IV:
(44)
C5 α 22 + β 22 chα 2b + α 22 + β 22 shα 2 b = 0
= C5α 22 chα 2 y + C6α 22 shα 2 y −
y =b
⎞
mπx
=0.
⎟ f 2 (b) sin
⎟
a
⎠
then by β 22 , and in combination with condition (IV), by
addition and subtraction we get:
∂w2
= ( C5α 2 shα 2 y + C6α 2 chα 2 y −
∂y
mπx
a
(43)
(C5α 22 chα 2b + C6α 22 shα 2b −
−C7 β 22 cosβ 2b − C8 β 22 sinβ 2b ) ⋅
y =b
+C7 cosβ 2 y + C8 sinβ 2 y ) sin
w2 = f 2 (b) sin
mπx
.
a
Condition IV gives us:
(40)
w2 = ( C5 chα2 y + C6 shα2 y +
∂y
mπx
a
Taking into account condition (III): f2 (b) = 0,
condition (IV) is reduced to:
mπx
, we have:
From the function w2 = f 2 ( y ) sin
a
2
2
(38)
where α1 and β1 are given with (23).
The conditions (III) and (IV) at the end of plate (2)
are for y = b, the deflection and the moment are also
equal to zero.
III condition for plate (2):
∂ 2 w2
∂ 2 w2
∂y
The function w1 is now
⎞
⎟
⎟
⎠
mπx
a
∂w2
= ( C5α 2 shα 2 y + C6α 2 chα 2 y −
∂y
(36)
C1 = −C3 = 0
⎛ ∂ w2
∂ w2
D2 ⎜
+ν
⎜ ∂y 2
∂x 2
⎝
=0
y =b
+C7 cosβ 2 y + C8 sinβ 2 y ) sin
Then from (I):
2
∂x 2
w2 = ( C5 chα 2 y + C6 shα 2 y +
−C1α12 − C2α12 + C1α12 + C3 β12 = 0
2
∂2w
f 2′′(b) f 0 ( x) + ν f 2 (b) f0′′( x) = 0
Multiplying by α12 , and subtracting (I) and (II):
(α12 + β12 ) C3 = 0 , α12 + β12 ≠ 0 ⇒ C3 = 0 .
+ν
cosβ 2b
.
sinβ 2b
(47)
When introducing (46) and (47), from (41) we get
⎡
chα 2b
shα 2 y + C7 cosβ 2 y −
w2 = ⎢C5 shα 2 y − C5
shα 2b
⎣
−C7
(42)
w2 = C5
⎤
cosβ 2 b
mπx
sinβ 2 y ⎥ sin
sinβ 2 b
a
⎦
shα 2 ( b − y )
shα 2 b
+ C7
sinβ 2 ( b − y )
sinα 2b
.
(48)
FME Transactions
Now, having functions w1 (x,y) and w2 (x,y):
w1 = f1 ( y ) f0 ( x) = ( C2 shα1 y + C4 sinβ 2 y ) f 0 ( x)
w2 = f 2 ( y ) f0 ( x) =
⎛ chα 2 ( b − y )
sinβ 2 ( b − y ) ⎞
= ⎜ C5
+ C7
⎟ f0 ( x) . (49)
shα 2b
sinα 2 b ⎠
⎝
Condition VII: Along the line of discontinuity, the
moment of plate (1) is equal to the moment of plate (2),
for y = η:
⎛ ∂2 w1
⎛ ∂2 w2
∂2 w1 ⎞
∂2 w2 ⎞
. (56)
D1 ⎜
+ν
= D2 ⎜
+ν
⎟
⎟
⎜ ∂y 2
⎜ ∂y 2
∂x2 ⎟⎠
∂x2 ⎟⎠
⎝
⎝
y =η
y =η
Now, we use the condition at y = η. Along the edge
y = η, deflections between (1) and (2) are equal
w1 ( x, y )
= w2 ( x, y )
y =η
y =η
.
(50)
Using (53), and notation
derivatives we get:
⎡
⎤
mπx
=
f1 ( y ) ⎥ sin
2
a
a
⎢⎣
⎥⎦
⎡
⎤
m 2 π2
mπx
= ⎢ f 2′′( y ) −ν
f 2 ( y ) ⎥ sin
2
a
a
⎣⎢
⎦⎥
ψ ⎢ f1′′( y ) −ν
The condition (50) is reduced to:
f1 ( y )
y =η
= f2 ( y)
y =η
C2 shα1η + C4 sinβ1η =
shα 2 ( b − η )
C5
shα 2 b
+ C7
sinβ 2 ( b − η )
sinβ 2b
,
(51)
⎡
−C7
shα 2 ( b − η )
sinβ 2 ( b − η )
sinβ 2b
shα 2b
⎢⎣
−
−ν
=0.
(52)
∂w1
mπx
= ( C2α1 chα1 y + C4 β1 cosβ1 y ) sin
∂y
a
∂y
2
3
∂ w1
∂y
3
(
)
mπx
a
(
)
mπx
a
= C2α12 shα1 y − C4 β12 sinβ1 y sin
= C2α13 chα1 y − C4 β13 cosβ1 y sin
∂y
2
(
)
⎛
shα2 ( b − y )
sinβ2 ( b − y ) ⎞
mπx
= ⎜ C5α22
− C7 β22
⎟ sin
shα2b
sinβ2b ⎠
a
⎝
⎛
chα 2 ( b − y )
= ⎜ −C5α 23
+
shα 2b
∂y3
⎝
cosβ 2 ( b − y ) ⎞
mπx
.
+C7 β 23
⎟ sin
sinβ 2 b ⎠
a
m 2 π2
2
⎤
( C2 shα1η + C4 sinβ1η )⎥ =
⎥⎦
⎡
shα 2 ( b − η )
sinβ 2 ( b − η ) ⎤
= ⎢C5α 22
− C7 β 22
⎥−
shα 2 b
sinβ 2b ⎦
⎣
sinβ 2 ( b − η ) ⎤
m 2 π2 ⎡ shα 2 ( b − η )
−ν
C
− C7
⎥ . (58)
2 ⎢ 5
sh
α
sinβ 2b ⎦
b
a ⎣
2
a
The final form is:
ch ( b − y )
cosβ2 ( b − y ) ⎞
∂w2 ⎛
mπx
= ⎜ −C5α 2 2
− C7 β2
⎟ sin
∂y ⎝
shα 2b
sinβ2b ⎠
a
∂2 w2
∂ 3 w2
⎛
⎛
⎞
m2 π2 ⎞
m2 π2
shα1η − C4ψ ⎜ β12 +ν
sinβ1η ⎟ −
C2ψ ⎜ α12 −ν
⎟
2 ⎟
2
⎜
⎜
⎟
a ⎠
a
⎝
⎝
⎠
2 2 ⎞ shα b − η
⎛ 2
)
m π
2(
−C5 ⎜ α 2 −ν
+
⎟
2 ⎟
⎜
a ⎠ shα 2b
⎝
⎛
m 2 π2 ⎞ sinβ 2 ( b − η )
+C7 ⎜ β 22 + ν
.
(59)
⎟
⎜
a 2 ⎟⎠ sinβ 2b
⎝
Condition VIII: Sharing forces along the contact of
plate (1) and (2), for y = η must be equal:
(53)
⎡ ∂ 3 w1
∂3 w1 ⎤
+ ( 2 −ν )
=
D1 ⎢
⎥
3
∂x 2 ∂y ⎦⎥
⎣⎢ ∂y
y =η
Condition VI is: the slope of plate (1) is equal to the
slope of plate (2) for y = η:
⎡ ∂ 3 w2
∂ 3 w3 ⎤
= D2 ⎢
+ ( 2 −ν )
⎥
3
∂x 2 ∂y ⎦⎥
⎣⎢ ∂y
y =η
∂w1 ( x, y )
∂y
=
y =η
∂w2 ( x, y )
∂y
,
(54)
y =η
this means that using (53) we come to f1′(η ) = f 2′ (η ) , or
C2α1 chα1η + C4 β1 cosβ1η + C5α 2
+C7 β 2
FME Transactions
cosβ 2 ( b − η )
sinβ 2b
chα 2 ( b − η )
=0.
(57)
ψ ⎢ α12C2 shα1η − C4 β12 sinβ1η −
Condition VI: We need the derivatives of (49):
∂ 2 w1
m 2 π2
or
or
C2 shα1η + C4 sinβ1η − C5
D1
= ψ , and having
D2
shα 2b
+
(55)
(60)
which can be written, introducing ψ, in the form
⎡
m 2 π2
⎣⎢
a2
ψ ⎢ f1′′′( y ) − ( 2 −ν )
⎤
mπx
sin
=
f1′( y ) ⎥
a
⎦⎥ y =η
⎡
⎤
m 2 π2
mπx
f 2′ ( y ) ⎥ sin
= ⎢ f 2′′′( y ) − ( 2 −ν )
2
a
a
⎣⎢
⎦⎥
(61)
or, using necessary derivatives (16), for (60), we obtain:
VOL. 39, No 1, 2011 ▪ 45
⎪⎧
ψ ⎨ ⎡C2α13 chα1η − C4 β13 cosβ1η ⎤ − ( 2 −ν )
⎪⎩ ⎣
⎦
m 2 π2
a2
⋅
or
⎡
ψ C2 ⎢α13 − ( 2 −ν ) α1
⎣⎢
⎡
m 2 π2 ⎤
−ψ C4 ⎢ β13 + ( 2 −ν ) β1
⎥ cosβ1η +
a 2 ⎦⎥
⎣⎢
⎫⎪
mπx
⋅ [α1C2 chα1η + β1C4 cosβ1η ]⎬ sin
=
a
⎪⎭
⎧⎪⎡
chα 2 ( b − η )
cosβ2 ( b − η ) ⎤
= ⎨⎢ −C5α 23
+ C7 β 23
⎥ − ( 2 −ν ) ⋅
shα 2b
sinβ2b ⎦
⎩⎪⎣
⋅
chα 2 ( b − η )
cosβ 2 ( b − η ) ⎤ ⎫⎪
m 2 π2 ⎡
−C5α 2
− C7 β 2
⎥⎬ ⋅
2 ⎢
shα 2η
sinβ 2 b ⎦ ⎭⎪
a ⎣
mπx
⋅ sin
.
(62)
a
Form (62) can be written
⎪⎧
⎡
⎩⎪
⎣⎢
ψ ⎨C2 ⎢α13 − ( 2 −ν ) α1
m 2 π2 ⎤
⎥ chα1η −
a 2 ⎦⎥
⎡
m 2 π2 ⎤ chα 2 ( b − η )
+C5 ⎢α 23 − ( 2 −ν ) α 2
−
⎥
a 2 ⎥⎦ shα 2b
⎢⎣
⎡
m2 π2 ⎤ cosβ 2 ( b − η )
−C7 ⎢ β 23 + ( 2 −ν ) β 2
= 0 . (64)
⎥
a 2 ⎦⎥ sinβ 2b
⎣⎢
We have obtained 4 homogeneous algebraic equations
with unknown constants C2, C4, C5 and C7. As those
systems have trivial solution when all constants are equal
to zero, the solution could be obtained only in the case
when the determinant of the system (65) is equal to zero.
3. CONCLUSION
⎫⎪
⎡
m 2 π2 ⎤
−C4 ⎢ β13 + ( 2 −ν ) β1
⎥ cosβ1η ⎬ =
2
a ⎦⎥
⎣⎢
⎭⎪
The determinant (65) has given solutions for critical
coefficient k for different ratios of a/b (Fig. 3), for 3
cases: m = 1, 2 and 3. Minimum value of buckling force
is obtained for m = 1 and a/b = 0.8 using the relation
⎡
m 2 π2 ⎤ chα 2 ( b − η )
= −C5 ⎢α 23 − ( 2 −ν ) α 2
+
⎥
a 2 ⎦⎥ shα 2 b
⎣⎢
⎡
m 2 π2 ⎤ cosβ 2 ( b − η )
+C7 ⎢ β 23 + ( 2 −ν ) β 2
⎥
a 2 ⎦⎥ sinβ 2b
⎣⎢
m 2 π2 ⎤
⎥ chα1η −
a 2 ⎦⎥
k1 =
(63)
( N x )cri b 2 (σ x )cri h1b 2
=
D1 π2
D1π2
C2
C4
C5
C7
V
chαη
1
sinβη
1
shα ( b−η)
− 2
shα2b
sinβ ( b−η)
− 2
sinβ2b
VI
α1chαη
1
β1cosβη
1
⎛
⎜
⎝
⎛
m2π2 ⎞
m2π2 ⎞
−ψ⎜β12 +ν
shαη
⎟
⎟sinβη
1
1
⎜
a2 ⎟⎠
a2 ⎟⎠
⎝
⎡
⎡
⎤
m2π2 ⎤
m2π2
VIII ψ ⎢α13 −α1( 2−ν )
chαη
cosαη
−ψ ⎢β13 −β1( 2−ν )
⎥
1
1 ⎥
2
2
a ⎥⎦
a
⎢⎣
⎢⎣
⎥⎦
ψ ⎜α12 −ν
VII
α2
chα2 ( b−η)
shα2b
β2
(66)
cosβ2 ( b−η)
=0
sinβ2b
⎛ 2 m2π2 sinβ2 ( b−η) ⎞
⎛ 2 m2π2 ⎞ shα2 ( b−η)
⎜β2 −ν 2
⎟
⎜α2 −ν 2 ⎟
⎜
⎜
sinβ2b ⎟⎠
a ⎟⎠ shα2b
a
⎝
⎝
⎡ 3
m2π2 ⎤ chα2 ( b−η) ⎡ 3
m2π2 cosβ2 ( b−η) ⎤
−⎢β2 +β2 ( 2−ν )
⎢α2 −α2 ( 2−ν ) 2 ⎥
⎥
sinβ2b ⎥⎦
a ⎥⎦ shα2b
a2
⎢⎣
⎢⎣
C2
C4
C5
C7
chαη
1
sinβη
1
−shα2 ( b−η)
−sinβ2 ( b−η)
α1chαη
1
β1cosβη
1
α2 chα2 ( b−η)
β2 cosβ2 ( b−η)
=0. (65)
⎛
⎜
⎝
⎛
⎛
⎛ 2 m2π2 ⎞
m2π2 ⎞
m2π2 ⎞
m2π2 ⎞
shαη
sinβη
shα2 ( b−η)
−ψ⎜β12 +ν
−⎜α22 −ν
⎟
⎟
⎟
⎜β2 +ν 2 ⎟sinβ2 ( b−η)
1
1
⎜
⎜
⎜
a2 ⎟⎠
a2 ⎟⎠
a2 ⎟⎠
a ⎟⎠
⎝
⎝
⎝
⎡
⎡ 3
⎤ ⎡ 3
⎡ 3
m2π2 ⎤
m2π2
m2π2 ⎤
m2π2 ⎤
ψ ⎢α13 −α1( 2−ν) 2 ⎥chαη
1 −ψ ⎢β1 +β1( 2−ν) 2 cosαη
1 ⎥ ⎢α2 −α2 ( 2−ν ) 2 ⎥chα2 ( b−η) −⎢β2 +β2 ( 2−ν ) 2 ⎥cosβ2 ( b−η)
a ⎦⎥
a
a ⎦⎥
a ⎦⎥
⎣⎢
⎣⎢
⎦⎥ ⎢⎣
⎣⎢
ψ⎜α12 −ν
46 ▪ VOL. 39, No 1, 2011
FME Transactions
or the critical stress in the plate (1)
D π2 k
(σ x )cri = 1 1 ,
h1b 2
where D1 =
(67)
Eh13
for aluminium.
12 (1 −ν )
80.0
1556, National Advisory Committee for
Aeronautics, Washington, 1948.
[3] Kollbrunner, C.F. and Meister, M: Knicken,
Springer Verlag, Berlin, 1958.
[4] Волмир, А.С.: Устоичивост деформируемих
систем, Наука, Moscow, 1967.
[5] Dunjić, M.M.: Stability of Constructiv and Material
Anisotropic Plates Subjected to Thermomechanical
Load, Faculty of Mechanical Engineering,
University of Belgrade, Belgrade, 2010, (in
Serbian).
60.0
ПРОБЛЕМ СТАБИЛНОСТИ СТЕПЕНАСТИХ
ПЛОЧА У ПЛАСТИЧНОЈ ОБЛАСТИ
m=3
40.0
m=2
20.0
Момчило Дуњић
m=1
0.2
0.4 0.6
0.8 1.0 1.2
1.4 1.6
1.8
2.0 2.2
2.4 2.6
a/b
Figure 3. Final solution for critical force as a function of (a/b)
REFERENCES
[1] Илношин, А.А.: Устоичивост пластинок и оболочек
за пределами упругости, Прикладнаја математика и
механика, Vol. 8, No. 4, pp. 337-360, 1944.
[2] Stowell, E.Z.: A Unified Theory of Plastic Buckling
of Colomns and Plates, NACA Technical Note:
FME Transactions
У области стабилности правоугаоних степенастих
плоча, у последње време, обрађени су многи
проблеми у еластичној области. Та проблематика
још није обрађивана у пластичној области, било у
области малих еласто-пластичних деформација,
било у области течења. У овом раду се говори о
проблему
губитка
стабилности
степенасте
правоугаоне плоча (са две дебљине) у пластичној
зони деформисања, оптерећене притискујућом
силом у равни плоче која делује дуж ивица по
којима се мења дебљина.
VOL. 39, No 1, 2011 ▪ 47