lecture bending of plates
TRANSCRIPT
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Assumptions of Thin-plate theory
1. Before loading, the midsurface is flat.
2. Plate thickness t is small in comparison with spanwise dimensions of
the plate.
3. The plate material is homogeneous, isotropic, and linearly elastic.4. A line normal to the midsurface before loading remains normal to the
midsurface after loading.
5. Stress normal to the midsurface szis negligible compared to sxand
sy.
6. Lateral deflections are small in comparison with the plate thickness,
and the slope of the midsurface is small.
7. The midsurface is a neutral surface, i.e., the strains ex, ey, gxyare zero.
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0 Integrating, we get ( , )zw
w w x y
z
e
00 ( , )zxw u w
u z u x yx z x
g
00 ( , )yzv w w
v z v x yz y y
g
0Here ( , ) and ( , ) represent, respectively, the values
of and on the midplane.
ou x y v x y
u v
According to the last assumption, 0 0 0u v
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andw w
u z v z x y
Therefore,
2 2 2
2 22x y xy
w w wz z z
x yx ye e g
Before loading After loading
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Generalized Hookes law
x x y z Ee s s s
y y z x Ee s s s
z z x y Ee s s s
xy xy Gg
yz yz Gg
zx zx Gg
where
2 (1 )
EG
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For 0, 0, 0, we getz yz zxe g g
2 2
2 2 2 21 1x x y
E E w wz
x ys e e
2 2
2 2 2 21 1y y x
E E w wz
y xs e e
2
1xy xy
E wG z
x y g
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Stressresultants
/ 2
/ 2
x x
y yt
xy xyt
x zx
y yz
M z
M z
M z dz
Q
Q
s
s
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Substituting the expressions for , and , we getx y xys s
2 2
2 2xw wM D
x y
2 2
2 2y
w wM D
y x
2
1xy
wM D
x y
where3
212 (1 )
E tD
is the flexural rigidity of the plate
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The stresses could be written as
3
12 xx
M z
ts
3
12 yy
M z
ts
3
12 xyxy
M z
t
The maximum stresses occur on the bottom and top
surfaces (atz = t/2) of the plate.
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Load is distributed over the upper surface of the plate.
The intensity of the load is q, so that the load acting onthe element is q dx dy.
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Summation of all the forces in thez-direction
0yx
QQdx dy dy dx q dx dy
x y
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0yx
QQq
x y
Taking moments of all the forces about x-axis, we get
0xy y
y
M Mdx dy dy dx Q dx dy
x y
The moment of the load qand the moment due to change in
the force Qyare neglected, since they are small quantities of
a higher order than those retained. After simplification,
0xy y
y
M MQ
x y
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Similarly, by taking moments with respect to they-axis, we get
0
yx x
x
M M
Qy x
Substituting the expressions of Qx, Qyand noting that
Mxy=Myx, we obtain
2 22
2 22
y xyx M MM
qx yx y
Or,
4 4 4
4 2 2 42 qw w w
Dx x y y
Symbolically, 4 q
wD
2 is the Laplacian operator
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4 qwD
To determine w, it is required to integrate this equation
with the constants of integration dependent upon the
appropriate boundary conditions.
Also we can write
2 2
2
2 21 1x y
w wM M D D w
x y
Or, the moment function 2
1
x yM MM D w
Shear forcesx y
M MQ Q
x y
2 2
2 2
M Mp
x y
2 2
2 2
w w M
Dx y
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Boundary conditions
Built-in edge 0 and 0x a
x a
wwx
Simply-supported edge
2 2
2 20 and 0
x a
x a
w ww
x y
The second condition is analogous to
2
2
20 or also 0
x ax a
ww
x
which do not involve Poissons ratio .
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Boundary conditions
Free edge 0 0 0x xy xx a x ax aM M Q
Kirchhoff showed that these three boundary conditions proposed
by Poisson are too many and that two conditions are sufficient forthe complete determination of the delfection w.
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Boundary conditions
Free edge
xy
x
x a
MQ
y
0xy
x x
x a
MV Q
y
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Boundary conditions
Therefore for a free edge at x= a
3 3
3 22 1 0
x a
w w
x x y
For zero bending moments along the free edge requires
2 2
2 20
x a
w wx y
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Example
Determine the deflection and stress in a very long and narrowrectangular plate if it is simply supported at edgesy = 0 andy =b.
(a) The plate carries a nonuniform loading expressed by
0
( ) sin y
p y pb
(b) The plate is under a uniform loadp0. Letp0= 10 kPa, b = 0.4 m,
t = 10 mm,= 1/3, and E = 200 GPa.
x
y
b
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2 2
2 2x
w wM D
x y
2 2
2 2y
w wM D
y x
2
1xyw
M D
x y
where3
212 (1 )
E tD
is the flexural rigidity of the plate
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4 4 4
4 2 2 42
qw w w
Dx x y y
4
4
d w p
dx D
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The stresses could be written as
3
12x
x
M z
ts 312
yy M z
ts 312
xyxy M z
t
The maximum stresses occur on the bottom and top
surfaces (atz = t/2) of the plate.
2
1
x yM MM D w
Shear forcesx y
M MQ Q
x y
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23 2
1
2
y
yz
Q z
t t
23 2
1
2
xzx
Q z
t t
3
3 2 2 1 2
4 3 3z
p z z
t ts
The maximum shear stress, as in the case of a rectangular beam,
occurs atz= 0, and can be expressed as
,max
3
2
y
yz
Q
t ,max
3
2
xzx
Q
t
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Naviers Solution for Simply Supported
Rectangular Plates
In general, the solution of the
bending problem involves the
following Fourier Series for
load and deflection.
1 1
( , ) sin sinmnm n
m x n yp x y p
a b
1 1
( , ) sin sinmnm n
m x n yw x y a
a b
wherepmnand amnrepresent undetermined coeffiecients.
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This approach was introduced by Navier in 1820. The deflection
must satisfy the following boundary conditions.
2
20 0 ( 0, )ww x x a
x
2
20 0 ( 0, )
ww y y b
y
12
1, 2
2sin sin
m n
x yaa b
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Determination ofpmn
1 1
( , ) sin sinmnm n
m x n yp x y p
a b
Multiply each side by sin sin and integrate
between limits (0, ) and (0, ).
m x n ya b
a b
0 0
0 01 1
( , )sin sin
sin sin sin sin
a b
a b
mn
m n
m x n yp x y dx dya b
m x n y m x n yp dx dy
a b a b
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It can be shown that
0
0 ( )
sin sin / 2 ( )
a m mm x m x
dx a m ma a
0
0 ( )sin sin
/ 2 ( )
b n nn y n ydy
b n nb b
Therefore, after simplification, we can wrtite
0 0
4( , )sin sin
a b
mn
m x n yp p x y dx dy
ab a b
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Evaluation of amn
1 1
( , ) sin sinmnm n
m x n yw x y a
a b
Substitute the expression of deflection into the plate bending equation.
4 4 4
4 2 2 42
qw w w
Dx x y y
where q is the load andD is the flexural rigidity of the plate.
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After simplification,
4 2 2 4
1 1
2 sin sin 0mnmnm n
pm m n n m x n ya
a a b b D a b
This equation must apply for allxandy. Therefore,
4 2 2 4
2 0mnmnpm m n n
a
a a b b D
Or,
242 2
1
( / ) ( / )
mnmn
pa
D m a n b
And
242 2
1 1
1sin sin
( / ) ( / )
mn
m n
p m x n yw
D a bm a n b
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Special Cases
1. Uniformly distributed load 0( , )p x y p
0 0
4( , )sin sin
a b
mn
m x n yp p x y dx dy
ab a b
0
2
0
2
41 cos 1 cos
4
1 1 1 1
mn
m n
pp m n
mn
p
mn
Or,0
2
16( , 1,3, )mn
pp m n
mn
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242 2
1 1
1sin sin
( / ) ( / )
mn
m n
p m x n yw
D a bm a n b
Or,
0
262 2
16 sin( / ) sin( / )
( / ) ( / )m n
p m x a n y bw
D mn m a n b
( , 1,3, )m n
Maximum deflection occur at the centre of the plate (x=a/2, y=b/2).
( ) / 2 10
max 262 2
116
( / ) ( / )
m n
m n
pw
D mn m a n b
1 / 2 1 / 2
Note that sin( / 2) 1 and sin( / 2) 1m n
m n
( , 1,3, )m n
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2 2
2 2x
w wM D
x y
2 2
2 2y
w wM D
y x
2
1xyw
M D
x y
where3
212 (1 )
E tD
is the flexural rigidity of the plate
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2 2
0
242 2
/ ( / )16sin sin
( / ) ( / )
x
m n
m a n bp m x n yM
a bmn m a n b
2 2
0
24
2 2
/ ( / )16sin sin
( / ) ( / )y
m n
m a n bp m x n yM
a bmn m a n b
0
24 2 2
16 (1 ) 1cos cos
( / ) ( / )xy m n
p m x n yM
ab a bm a n b
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2. Patch load
( , ) / 4p x y P cd
1 1
1 1
sin siny d x c
mny d x c
P m x n yp dx dy
abcd a b
Or,
1 1
2
4sin sin sin sinmn
m x n yP m c n dp
mncd a b a b
0 0
4( , )sin sin
a b
mn
m x n yp p x y dx dy
ab a b
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3. Point load
1 1
2
4sin sin sin sinmn
m x n yP m c n dp
mncd a b a b
For patch load we wrote
The above expression could be used for point loadby putting the limiting values of c and d to zero.
1 14 sin sinmnm x n yP
p
ab a b
Deflection
1 1
242 2
sin( / ) sin( / )4sin sin
( / ) ( / )m n
m x a n y bP m x n yw
Dab a bm a n b
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When the loadP is applied at the centre of the plate (x1=a/2, y1=b/2),
the previous equation reduces to
( ) / 2 1 1 1
242 2
sin( / ) sin( / )41
( / ) ( / )
m n
m n
m x a n y bPw
Dab m a n b
( , 1,3, )m n
Furthermore, if the plate is square (a=b), the maximum deflection,
which occurs at the centre can be written as
2
max 242 2
4 1
m n
P aw
D m n
( , 1,3, )m n
2 2
maxRetaining the first nine terms 0 01142 (Exact value 0 01159 )
Pa Pa
w D D