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8/3/2019 03 Rectangular
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RECTANGULAR PLATES
Page 1
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Rectangular Plate - TopicsPage 2
Fourier Series (5-9)
Naviers Solution (10-17)
Uniform Load (18-25)
Sine Load (26-35) Uniform Load Over Subregion (36-39)
Point Load (40-42)
Levys Solution (43-53)
Uniform Load (54-64)
3 Simply Supported Edges, 1 Clamped Edge (65-69)
Opposite Edges Simply Supported, 1 Free, 1 Clamped (70-72)
Simply Supported with Load Function of x (73-78)
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Rectangular Plate - TopicsPage 3
Simply Supported Hydrostatic Load (79-80)
Simply Supported Long Rectangular Plates (81-88)
n Constant Load (89)
n Partial Line Load (90-91)
n Concentrated Load (92-93)
Symmetric Edge Moments (94-100)
One Edge Moment (101-104)
Superposition (105-112)
Strip Method (113-117) 3 Simply Supported Edges, 1 Fixed Edge (118-121)
Continuous Plates (122-128)
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Background: Expressions To KnowPage 4
( ) ( )
( ) ( )
( ) ( )
( ) ( ) ax
a
dxax
dx
duuu
dx
d
axa
dxaxdx
duuu
dx
d
axa
dxaxdx
duuu
dx
d
axa
dxaxdx
duuu
dx
d
sinh1
coshsinhcosh
cosh1
sinhcoshsinh
sin1
cossincos
cos1
sincossin
==
==
=-=
-==
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Fourier SeriesPage 5
If f(x) is continuous, bounded, periodic function of
period 2L over the range L to +L, i.e. f(x+2L) =
f(x).
Then f(x) may be represented by the Fourier series:
=
++=1
0 sincos2
)(n
nnL
xnb
L
xna
axf
pp
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Fourier SeriesPage 6
Coefficients are found through the use of
orthogonality relations:
)(
)(0sinsin
nm,ll0sincos
)(
)(0coscos
nmL
nmdxL
xn
L
xm
adxL
xn
L
xm
nmL
nmdxL
xn
L
xm
L
L
L
L
L
L
==
=
=
==
=
+
-
+
-
+
-
pp
pp
pp
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Fourier SeriesPage 7
To determine the coefficients bn of the following Fourier series:
Multiply both sides by:
To obtain:
=
=1
sin)(n
nL
xnbxf
p
L
xmpsin
L
xm
L
xn
bL
xm
xf n n
ppp
sinsinsin)( 1
==
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Fourier SeriesPage 8
Integrate both sides:
And using the orthogonality relationship, the coefficients are
found:
In the same way:
dxL
xm
L
xnbdx
L
xmxf
L
L
n
L
L
pppsinsinsin)(
+
-
+
-
=
...3,2,1sin)(1
== +
-
ndxL
xnxf
Lb
L
L
n
p
...3,2,1,0cos)(1
== +
-
ndxL
xnxf
La
L
L
n
p
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Fourier SeriesPage 9
If a function is even so that f(x) = f(-x), then
f(x)sin(nx) is odd.
And thus bn = 0 for all n.
If a function is odd so that f(x) = -f(-x), then
f(x)cos(nx) is even.
And thus an = 0 for all n.
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Navier Solution for Simply-Supported
Rectangular PlatesPage 10
Rectangular geometry:
Distributed load = p(x,y)
No closed form solutions like circular plates.
b
a
x
y
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Navier Solution for Simply-Supported
Rectangular PlatesPage 11
Solution assumes load and deflection can be
represented by Fourier series:
pmn, amn = coefficients to be determined.
( )
( )b
yn
a
xmayxw
byn
axmpyxp
n
mn
m
n
mn
m
pp
pp
sinsin,
sinsin,
11
11
=
=
=
=
=
=
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Navier Solution for Simply-Supported
Rectangular PlatesPage 12
Deflections must satisfy the governing differential equation:
Boundary conditions (due to sin terms, automatically satisfied):
b)y0,(y0y
w0
a),0(x0
x
w0
2
2
2
2
===
=
===
=
w
xw
D
p
y
w
yx
w
x
ww =
+
+
=
4
4
22
4
4
44 2
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Navier Solution for Simply-Supported
Rectangular PlatesPage 13
Physical interpretation:
Consider term m = 1 and n = 2of w:
This term represents a single sin
wave deflection in the x-direction, and a double sin wavedeflection in the y-deflection, asshown:
Note that sin terms automaticallysatisfy simply-supported
boundary conditions. Increasing the number of terms in
the series can improve accuracy.
b
y
a
xa
pp 2sinsin12
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Navier Solution for Simply-Supported
Rectangular PlatesPage 14
The general procedure:
Determine coefficients pmn:
Find pmn from p(x,y) by multiplying both sides of
equation by:
And integrate from 0 to a and 0 to b:
dxdyb
yn
a
xm pp ''sinsin
( )
dxdyb
yn
a
xm
b
yn
a
xmp
dxdyb
yn
a
xm
yxp
n
ab
mn
m
ab
pppp
pp
''
1 001
'
0
'
0
sinsinsinsin
sinsin,
=
=
=
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Navier Solution for Simply-Supported
Rectangular PlatesPage 15
Recalling orthogonality relations, in this case:
And therefore:
( ) dxdyb
yn
a
xmyxp
abp
ab
mn
ppsinsin,
4
00
=
==
==
)(if2
),(if0sinsin
)(if2
),(if0sinsin
'''
0
'''
0
nnb
nndyb
yn
b
yn
mma
mmdxa
xm
a
xm
b
a
pp
pp
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Navier Solution for Simply-Supported
Rectangular PlatesPage 16
Determine amn by substituting p(x,y) and w(x,y)
expressions into governing equation:
222
4
222
4
1
4224
1
1
0
0sinsin2
+
=
=-
+
=
-
+
+
=
=
b
n
a
m
p
Da
D
p
b
n
a
ma
b
yn
a
xm
D
p
b
n
b
n
a
m
a
ma
mn
mn
mnmn
n
mnmn
m
p
p
pppppp
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Navier Solution for Simply-Supported
Rectangular PlatesPage 17
Substituting expression for amn into w(x,y):
The series is convergent if you use enough terms, you
will approach the exact solution.
b
yn
a
xm
bn
am
p
Dw
n
mn
m
ppp
sinsin1
1
222
1
4
=
=
+
=
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 18
Uniformly distributed load: p(x,y) = p0
( )
( ) ( )
-
-=
-=
-=
=
pp
pp
pp
p
pp
p
pp
nn
bm
m
a
ab
pp
dxnn
b
a
xm
ab
pp
dxb
yn
n
b
a
xm
ab
pp
dxdyb
yn
a
xmp
abp
mn
a
mn
a
mn
ab
mn
cos1cos14
cos1sin4
cossin4
sinsin4
0
0
0
b
00
0
0
0
0
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 19
Continue evaluating:
And finally:
( )( )
( )( ) ( )( )
,...)5,3,1,(16
11114
cos1cos14
2
0
2
0
2
0
==
----=
--=
nmmn
pp
mn
pp
nmmn
pp
mn
nm
mn
mn
p
p
ppp
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 20
Substitute into w:
Since plate must deflect into a symmetrical shape for a
uniform load, knew ahead of time that m and n must be
odd.
,...)5,3,1,(sinsin116
222
6
0 =
+
=
nmb
yn
a
xm
b
n
a
m
mn
D
pw
nm
ppp
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 21
Maximum deflection occurs at center of plate:
( ) ( )
( )( )
-
+
--
-
+
=
--
+
=
=
+
=
=
12
222
6
0max
2
1
2
1
222
6
0max
2226
0
max
max
1116
11116
,...)5,3,1,(2sin2sin
116
2,
2
nm
nm
nm
nm
nm
b
n
a
mmn
D
pw
b
n
a
mmn
D
pw
nm
nm
b
n
a
mmn
D
p
w
baww
p
p
pp
p
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 22
Recall:
Thus:
,...)5,3,1,(sinsin16
222
22
4
0 =
+
+
=
nmb
yn
a
xm
b
n
a
mmn
b
n
a
m
pM
nm
x
ppn
p
y
w
x
w
2
2
2
2
+
-= nDMx
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 23
Likewise:
Mx and My are zero at x = 0, x = a, y = 0, and y = b.
However, Mxy does not vanish at edges and corners.
( ),...)5,3,1,(coscos
1116
,...)5,3,1,(sinsin16
222
4
0
222
22
4
0
=
+
--=
=
+
+
=
nmb
yn
a
xm
b
n
a
mmn
ab
pM
nmb
yn
a
xm
b
n
a
m
mn
b
n
a
m
pM
nm
xy
nm
y
ppp
n
ppn
p
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 24
Assume square plate: a = b
Use first term of Fourier series: m = n = 1
( )
( )
D
apw
D
ap
a
D
pw
b
n
a
mmn
D
p
w
nm
nm
4
0max
6
4
0
22
6
0max
12
2226
0
max
00416.0
4
12
116
1
116
=
=
=
-
+
=
-+
pp
p
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 25
Using first four terms: m,n = 1,3
Also:
Moment does not converge as rapidly as deflection.
)(00406.04
0max solutionexact
D
apw =
)(0469.0
)(0534.0
2
0max,
2
0max,
termsfourapM
termfirstapM
x
x
=
=
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 26
Loading:( )
b
y
a
xpyxp
ppsinsin, 0=
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 27
Calculate pmn:
Calculate w:
)1(4
4
sinsinsinsin4
00
0
0
0
====
=
nmpabp
abp
dxdyb
yn
a
xm
b
y
a
xp
abp
mn
ab
mn
pppp
b
y
a
x
ba
D
pw
b
yn
a
xm
b
n
a
mmn
D
pw
nm
ppp
ppp
sinsin11
1
sinsin1
2
22
4
0
222
4
0
+
=
+
=
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 28
Calculate moments:
And likewise:
y
w
x
w
2
2
2
2
+
-= nDMx
b
y
a
x
ba
ba
pM
b
y
a
x
baba
pM
y
x
ppn
p
ppn
p
sinsin1
11
sinsin1
11
222
22
2
0
222
222
0
+
+
=
+
+
=
x
w
y
w
2
2
2
2
+
-= nDMy
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 29
Calculate moments:
( )yx
w1
2
+-= nDMxy
( ) by
ax
ab
ba
pMxy ppp
n coscos111
1 2
22
2
0
+
--=
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 30
Calculate shears:
And likewise:
+
-=2
2
2
2
y
w
x
w
xDQx
by
ax
bab
pQ
b
y
a
x
baa
pQ
y
x
ppp
pp
p
cossin11
sincos
11
22
0
22
0
+
=
+
=
+
-=2
2
2
2
y
w
x
w
yDQy
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 31
Calculate total applied load:
[ ]
2
0
00
0
0
0
4
211cossin:
sinsin
p
ppp
pp
pp
abpp
aa
a
xadx
a
xNote
dxdyb
y
a
xpp
total
aa
ab
total
=
=---=
-=
=
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 32
Find reactions at edges:
Setting x = a:
( )
( )
b
y
ab
ba
p
b
y
baa
pV
b
y
a
x
ab
ba
p
b
y
a
x
baa
p
V
y
MQV
x
x
xy
xx
p
p
np
p
pp
p
npp
p
sin1
11
1sin
11
sincos
1
11
1
sincos11
22
22
0
22
0
22
22
0
22
0
+
--
+-=
+
-
+
+=
+=
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 33
Simplifying:
Likewise for Vy
at edge y = b:
( )
( )
( )a
x
ab
bab
pV
b
y
ba
baa
pV
b
y
bba
baa
pV
y
x
x
pn
p
pn
p
pn
p
sin21
11
sin21
11
sin111
11
222
22
0
222
22
0
2222
22
0
-+
+
-=
-+
+
-=
-++
+
-=
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 34
Sum of all reactions:
By simple manipulation:
( ) ( )
( )2
22
2
0
2
0
222
22
0
222
22
0
00
11
184
21
11
421
11
4
22
+
---=
-+
+
-
-+
+
-=
+
baab
pabp
ba
baa
pb
ab
bab
pa
dyVdxV
b
x
a
y
p
np
n
pp
n
pp
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 35
Note that first term equals total applied load, soreactions are larger.
From concentrated reaction at each corner:
4 reactions (one at each corner). Equilibrium is therefore satisfied.
This is physical corners tend to rise.
( ) ( )2
22
2
0
2
22
2
0
11
12
11
12,2
+
-=-=
+
--=
===
baab
pFR
baab
pF
byaxatMF
ccc
xyc
p
n
p
n
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 36
Total load P distributed uniformly over a sub-region
4cd:
( )cd
Pyxp
4, =
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 37
Determine pmn:
How? See next slide.
( )
b
dn
a
cm
b
yn
a
xm
mncd
Pp
dxdyb
yn
a
xm
abcd
Pp
dxdyb
yn
a
xmyxp
abp
mn
cx
cx
dy
dy
mn
ab
mn
ppppp
pp
pp
sinsinsinsin4
sinsin
sinsin,4
11
2
00
1
1
1
1
=
=
=
+
-
+
-
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 38
Note:
Using identities:
Obtain:
( ) ( )
( )( )
( ) ( )
a
cm
a
xm
m
a
a
cxm
a
cxm
m
a
a
xm
m
adx
a
xmcx
cx
cx
cx
ppp
bababa
bababa
bababa
pp
p
pp
p
sinsin2
sinsin2coscos
sinsincoscoscos
sinsincoscoscos
coscos
cossin
1
11
1
1
1
1
=-=--+
+=-
-=+
--+
-=
-=+
-
+
-
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 39
If expand load to cover entire plate surface by
substituting the following:
Obtain the previous solution:
2
2
2
2
11
bd
ac
by
ax ====
mn
P
pmn 216
p=
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 40
Point load at x = x1, y = y1:
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 41
Use previous derivation, with c,d 0.
b
yn
a
xm
ab
Pp
b
n
db
dn
d
a
ma
cm
a
m
c
ca
cm
c
b
dn
a
cm
b
yn
a
xm
mncd
Pp
mn
mn
11
11
2
sinsin4
sin)0lim(similarly,
1
cos)0lim(
:RulesHopital'L'use
00sin)0lim(
sinsinsinsin4
pp
pp
ppp
p
ppppp
=
=
=
=
=
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Simply-Supported Rectangular Plates
Under Various LoadingsPage 42
Leads to:
If P is applied to the center of a square plate (a=b):
Using first nine terms (m,n = 1, 3, 5):
sinsinsinsin14 11
222
4 b
yn
a
xm
b
yn
a
xm
b
n
a
mDab
Pw
nm
ppppp
+
=
( )...),,(m,n
nmD
Paw
by
ax
nm
53114
2
2
2224
2
11 =+
===
p
==
D
Paexact
D
Paw
22
max 01159.001142.0
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Levys Solution for Rectangular PlatesPage 43
Navier solution slow convergence of series forbending moments.
Levys solution:
Overcomes slow convergence. Uses single series (Navier uses double series)
Allows for more general boundary conditions (Navieronly valid for simply-supported conditions on all sides).n
Particular BC on two opposite sides (x = 0 & x = a)n Arbitrary BC on remaining edges (y = -b/2 & y = +b/2)
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Levys Solution for Rectangular PlatesPage 44
Procedure:
As before, result is sum of homogeneous and particular
solution:
Particular solution is obtained for each specific loading.
Homogeneous solution is independent of loading:
phwww +=
04 = hw
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Levys Solution for Rectangular PlatesPage 45
Homogeneous solution assumed to be:
( ) ( )
( )2
atconditionsboundaryfulfill
cossin 11
byyf
a
xmyfwor
a
xmyfw
m
mmh
mmh
=
=
=
=
=
pp
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Levys Solution for Rectangular PlatesPage 46
For example, if plate is simply-supported at x = 0
and x = a:
Use the sin term, which will automatically satisfy the
boundary conditions at these edges.
( )
=
= a
xmyfw
m
mh
psin
1
a),0(x0x
w0 2
2
===
= xw
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Levys Solution for Rectangular PlatesPage 47
Substitute into governing equation:
Term inside brackets must be zero linear
differential equation with constant coefficients.
0sin2
0
1
4
2
22
4
4
4
=
+
-
=
= a
xmf
a
m
dy
fd
a
m
dy
fd
w
m
mmm
h
ppp
024
2
22
4
4
=
+
- m
mm fa
m
dy
fd
a
m
dy
fd pp
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Levys Solution for Rectangular PlatesPage 48
Solution (from differential equations):
Can be written as:
( )AAm
yeCyeCeCeCmAm
AmAm
fa
m
dy
fd
a
m
dy
fd
AyAyAyAy
mmm
=
+++==-
=+-
=
+
-
--
,
0
02
02
4321
222
4224
4
2
22
4
4 pp
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Levys Solution for Rectangular PlatesPage 49
And thus the general solution is:
Can also be written as (usually easier to work with):
Note:
a
ym
ma
ym
ma
ym
ma
ym
mm yeDyeCeBeAfpppp
--+++= ''''
a
ymyD
a
ymyC
a
ymB
a
ymAf mmmmm
ppppcoshsinhcoshsinh +++=
( ) ( )uuuu
eeueeu--
+=-= 21
cosh2
1sinh
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Levys Solution for Rectangular PlatesPage 50
Therefore:
casesspecificfordeterminedconstants,,,
sincoshsinhcoshsinh1
+++=
=
mmmm
m
mmmmh
DCBA
a
xm
a
ymyD
a
ymyC
a
ymB
a
ymAw
ppppp
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Particular solution:
Last equation found by multiplying both sides by the sin term
below and integrating:
( )
( ) ( )
( ) ( )
=
=
=
=
=
a
m
m
m
m
mp
dxa
xmyxp
ayp
a
xmypyxp
a
xmykw
0
1
1
sin,2
sin,
sin
p
p
p
a
xm p'sin
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Levys Solution for Rectangular PlatesPage 52
Plugging into differential equation:
=
+-
=
=
= axm
Dp
axmk
am
dykd
am
dykd
D
pw
m
m
m
mmm
p
pppp sinsin211
4
2
22
4
4
4
D
pk
a
m
dy
kd
a
m
dy
kd mm
mm =
+
-4
2
22
4
4
2pp
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Levys Solution for Rectangular PlatesPage 53
The general procedure thus becomes:
From loading p(x,y), find pm.
From pm, find km and thus wp.
Using boundary conditions, find Am, Bm, Cm, Dm and thuswh.
Result is sum of wp and wh.
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Levys Solution for Rectangular PlatesPage 54
Simply-supported rectangular plate under uniform
loading.
Find pm.
( )
( )
( ) ( )...5,3,14
22cos
2sin
2
,
0
0
0
0
0
0
0
==
=
-=
=
=
mm
pyp
m
ap
aa
xm
m
ap
adx
a
xmp
ayp
pyxp
m
aa
m
p
pp
pp
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Levys Solution for Rectangular PlatesPage 55
Find km.
n Since the right hand side is not a function of y, then kmcannot be a function of y, so derivatives of km are zero:
Dm
pk
a
m
dy
kd
a
m
dy
kdm
mm
ppp 0
4
2
22
4
4 42 =
+
-
Dmapk
Dmpk
am mm 55
4
00
4
44ppp ==
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Levys Solution for Rectangular PlatesPage 56
Thus wp becomes:
This particular solution represents
the deflection of a uniformly
loaded, simply-supported strip
parallel to the y-axis (very longin the y-direction).
a
xm
mD
apw
m
p
pp
sin14
...3,1
55
4
0
=
=
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Levys Solution for Rectangular PlatesPage 57
Now must evaluate constants in homogeneous solution.
Observing that the deflection must be symmetrical with
respect to the x-axis: Am = Dm = 0.
n sinh function is not symmetric.n cosh function is symmetric.
n y*sinh is symmetric
n y*cosh is not symmetric.
sinh
cosh
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Levys Solution for Rectangular Plates
Page 58
Expression for deflection is now:
Satisfies governing equations and simply-supported
boundary conditions at x = 0 and x = a.
Remaining boundary conditions:
a
xm
Dm
ap
a
ymyC
a
ymBwww
m
mmhp
pp
ppsin
4sinhcosh
...3,155
4
0
=
++=+=
==
=
2y0
y
w0
2
2 bw
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Levys Solution for Rectangular Plates
Page 59
Substituting:
Gives:
02sinh22cosh22
04
2sinh
22cosh
55
4
0
=+
+
=++
a
bm
a
bmCa
bmCa
mB
Dm
ap
a
bmbC
a
bmB
mmm
mm
ppppp
pp
a
bmDm
apC
a
bmDm
a
bmbapmap
B
m
m
2cosh
22
cosh
2tanh4
44
3
0
55
3
0
4
0
pp
pp
pp
=
+
-=
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Page 60
Setting:
Simplifies constants:
m
m
m
mm
Dm
apC
Dm
bapmapB
ap
apap
cosh
2
cosh
tanh4
44
3
0
55
30
40
=
+-=
a
bmm
2
pa =
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Levys Solution for Rectangular Plates
Page 61
Deflection:
Maximum deflection is at x = a/2, y = 0:
axm
by
aym
b
y
mD
apw
m
m
m
m
m
mm
papa
aa
aap
sin2sinhcosh2
1
2cosh
cosh2
2tanh1
14
...3,155
4
0
+
+-=
=
2sincosh2
2tanh1
14
...3,155
4
0max
pa
aap
m
mD
apw
m
mm
m
+-=
=
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Levys Solution for Rectangular Plates
Page 62
Maximum deflection can be simplified:
Can manipulate to the following form:
Note that first term is the previous solution for a
uniformly loaded simply-supported strip.
( )( )
+-
-=
=
-
m
mm
m
m
mD
apw
aaa
p cosh22tanh
114
...3,15
2
1
5
4
0max
( )( )
+--=
=
-
m
mm
m
m
mD
ap
D
apw
aaa
p cosh22tanh14
384
5
...3,15
2
1
5
4
0
4
0max
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Levys Solution for Rectangular Plates
Page 63
For a square plate (a = b):
n Note that the series converges rapidly.
Can rewrite equation as:
( )D
ap
D
ap
D
apw
4
0
5
4
0
4
0max 00406.0...00025.068562.0
4
384
5=+--=
p
( )( )
+--==
=
-
m
mm
m
m
mD
apw
a
aa
pdd
cosh2
2tanh14
384
5
...3,15
2
1
51
4
01max
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Levys Solution for Rectangular Plates
Page 64
Can also write similar expressions for moments:
Parameters d1, d2, d3 can be determined for variouscases.
Note that as plate becomes long (b/a tends to infinity),
results equal that of simply-supported strip.
If ratio of sides is large (b/a > 4), effect of short sidesis negligible and plate can be considered as infinite
strip.
2
03max,
2
02max, apMapM yx dd ==
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Page 65
3 edges simply-supported, 1 edge clamped,
uniform pressure.
Deflection is symmetrical about x = a/2.
Thus, w uses only odd m.
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Levys Solution for Rectangular Plates
Page 66
Deflection:
Note that the particular solution is the same as before, since it is for a
uniform load with simply-supported boundary conditions at x = 0 and x
= a, and has nothing to do with the boundary conditions on the y sides.
Also, the simply-supported boundary conditions at x = 0 and x = a are
automatically satisfied due to the sin term in x.
a
xm
Dm
ap
a
ym
yD
a
ymyC
a
ymB
a
ymAw
m
m
mmm
p
p
p
ppp
sin
4
cosh
sinhcoshsinh
55
4
0
...3,1
++
++=
=
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Levys Solution for Rectangular Plates
Page 67
Remaining boundary conditions:
The boundary conditions provide 4 equations to solve
the 4 unknown constants, shown on next page.
b)(y0y
w0
)0(y0y
w0
2
2
==
=
==
=
w
w
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Page 68
After some manipulation, the constants are:
a
bm
ba
m
a
m
a
m
C
DmapB
Dm
a
Dm
apA
m
mmm
mmmm
mm
m
m
mmm
mmmmm
pb
bbb
bp
bp
bbp
b
p
pbbbbbbb
p
=
-
-
-
-=
-=
-=---
=
sinhcosh
coshsinhcoshsinh2
2
1
4
sinhcosh
sinhcosh2cosh22
2
55
40
2
55
4
0
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Levys Solution for Rectangular Plates
Page 69
For a square plate (a=b):
===
===
0,2
084.0
2,
20028.0
20max,
4
0
yaxapM
ay
ax
D
apw
y
center
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Levys Solution for Rectangular Plates
Page 70
Opposite edges simply-supported, 3rd edge free,
4th edge clamped, uniform pressure.
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Page 71
Deflection:
Note that the particular solution is the same as before, since it is for a
uniform load with simply-supported boundary conditions at x = 0 and x
= a, and has nothing to do with the boundary conditions on the y sides.
Also, the simply-supported boundary conditions at x = 0 and x = a are
automatically satisfied due to the sin term in x.
a
xm
Dm
ap
a
ym
yD
a
ymyC
a
ymB
a
ymAw
m
m
mmm
pp
p
ppp
sin
4
cosh
sinhcoshsinh
55
4
0
...3,1
++
++=
=
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Levys Solution for Rectangular Plates
Page 72
Remaining boundary conditions:
The boundary conditions provide 4 equations to solve the 4
unknown constants. Not shown here, but it is straightforwardif not tedious.
Same procedure for any similar uniformly loaded plate.
( )
shearmoment
b)(y0yx
w2
y
w0
)0(y0y
w0
2
3
3
3
2
2
2
2
==
-+
=
+
==
=
nnx
w
y
w
w
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 73
Assume rectangular plate simply-supported at x = 0
and x = a.
Assume load is a function of x only.
( )
( )
=
=
=
a
m
m
m
dxa
xmxp
ap
a
xmpxp
0
1
sin2
sin
p
p
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 74
As before, assume for the particular solution:
Plugging into governing equation:
( )
=
= a
xmykw
m
mp
psin
1
0sin21
4
2
22
4
4
4
=
-
+
-
=
= a
xmDpk
am
dykd
am
dykd
D
pw
m
mm
mm
p
ppp
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 75
Since pm is only a function of x, km cannot be a function
of y, and the derivatives with respect to y are zero.
Particular solution, which represents deflection of a strip
under load p(x), becomes:
D
pk
a
m
dy
kd
a
m
dy
kd mm
mm =
+
-4
2
22
4
4
2pp
Dm
apk
D
pk
a
m mm
mm 44
44
p
p==
a
xm
m
p
D
aw
m
mp
pp
sin1
44
4
=
=
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 76
Assume the two arbitrary edges, y = -b/2 and y =
b/2, are also simply-supported:
n Note that, as before, Am and Dm are eliminated since they
are associated with terms that are not symmetric in y.
Bm and Cm are determined from boundary conditions at
y edges:
a
xm
Dm
ap
a
ymyC
a
ymBwww
m
mmmhp
pp
ppsinsinhcosh
144
4
=
++=+=
==
=2
y0y
w0
2
2 bw
M A N
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 77
After evaluation of boundary conditions, the deflection
is:
Given p(x) find pm find w find M find .
a
bm
a
xm
a
ym
a
ym
a
ym
m
p
D
aw
m
m
m m
mmm
2
sinsinhcosh2
1
coshcosh2
2tanh1
144
4
pa
pppa
pa
aap
=
+
+-=
=
L M h d A l d N
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 78
Values of pm for
various loadings.
L M h d A li d N
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 79
Example: Hydrostatically loaded plate.
( )
( )
( ) ( )...2,11-2
00cos0
2
cossin2
sin
2
sin
2
1m0
2
2
0
0
2
2
0
0
2
0
0
0
0
==
+-
-=
-
=
=
=
=
+
mm
pp
mm
a
a
p
p
a
xm
m
xa
a
xm
m
a
a
pp
dxa
xm
xa
p
dxa
xm
a
x
pap
a
xpxp
m
m
a
m
aa
m
p
pp
pp
pp
pp
L M h d A li d N
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 80
Can now find w by substituting expression for pm.
If plate is square (a = b), for example, the maximum
deflection is:
=== 0,2
00203.04
0 ya
xD
apwcenter
L M h d A li d N
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 81
Example: Line loads on long rectangular plates.
L M h d A li d N
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 82
Only need to consider one half of plate (positive y)
due to symmetry:
Each half takes half the load:
)0(y0
y
w==
( ) )0(yy2
)( 2 =
-=-= wDxP
Qy
L M th d A li d t N
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 83
Use alternate form of governing equation:
There is no surface pressure, so the particular solution is
zero and:
Also, w and its derivatives should vanish at y = :
0'' == mm CA
a
xmyeDyeCeBeAw
m
a
ym
ma
ym
ma
ym
ma
ym
mh
pppppsin''''
1
=
--
+++=
hww =
L M th d A li d t N
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 84
Equations for w:
Evaluating symmetry boundary condition:
a
xmyeDeBw
m
a
ym
ma
ym
m
pppsin''
1
=
--
+=
==+
-
=
-+
-
==
---
a
mBDD
a
mB
ye
a
meDe
a
mB
mmmm
a
ym
a
ym
ma
ym
m
pp
pp ppp
''0''
0''
)0(y0y
w
L M th d A li d t N
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 85
Evaluating remaining boundary condition:
Need to evaluate derivatives (next slide).
( )
2)(
yxy
)0(yy2
)(
2
2
2
2
2
xPwwD
wDxP
Qy
=
+
=
-=-=
L M th d A li d t N
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 86
( )a
xmeD
a
mw
a
xmeD
a
mw
a
xmyeD
a
meD
a
meD
a
meB
a
mw
a
xmyeD
a
meDeB
a
mw
a
xmyeDeB
a
mw
a
ym
m
a
ym
m
a
ym
ma
ym
ma
ym
ma
ym
m
a
ym
m
a
ym
m
a
ym
m
a
ym
ma
ym
m
pp
pp
ppppp
ppp
pp
p
p
pppp
ppp
pp
sin'2y
sin'2
sin''''y
sin'''yy
sin''x
2
2
2
22
2
2
2
2
2
2
2
=
-=
+
-
-
=
-+
-
=
+
-=
-
-
----
---
--
L M th d A li d t N
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 87
Substituting into boundary condition:
And from previous expression:
Dm
apD
D
pD
a
my
DpeD
am
xPwwD
mm
mm
ma
ym
m
22
22
2
2
2
2
2
4'
2'20
cancel)(sin terms2
'2
2
)(
yxy
pp
pp
==
=
=
+
-
Dm
apB mm 33
3
4'
p=
L M th d A li d t N
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 88
Thus deflection becomes:
Equations for P(x):
a
xme
a
ym
m
p
D
aw
a
xme
Dm
yape
Dm
apw
aym
m
m
m
a
ym
ma
ym
m
ppp
ppp
p
pp
sin14
sin44
133
3
122
2
33
3
-
=
=
--
+=
+=
( )
=
= a
xmpxP
m
m
psin
1
Levys Method Applied to Non
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Levys Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 89
If p(x) = p0:
( )D
ap
D
ap
mD
apw
m
mD
apaww
a
xme
a
ym
mD
apw
m
m
m
a
ym
m
1536
5
1536
51
2sin
10,
2
sin11
3
0
5
4
3
0
...3,14
2
1
4
3
0max
...3,1 44
3
0
max
...3,144
3
0
pppp
p
p
ppp
p
=
=
-=
=
=
+=
=
-
=
-
=
[ ]
( )...5,3,14
cos12
cos2
sin2
0
0
0
0
0
0
==
-=
-=
=
mm
pp
mm
p
a
xm
m
pdx
a
xmp
ap
m
aa
m
p
pp
pp
p
Levys Method Applied to Non
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Levy s Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 90
If p(x) is a constant load p0 but only partially along
line:
Levys Method Applied to Non
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Levy s Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 91
And plugging into expression for w:
n For x1 = a/2 and c = a/2, get previous result.
( ) ( )
=
-+
+-=
=
+-=
+
-
a
cm
a
xm
m
pp
a
m
a
m
m
pdx
a
xmp
ap
pxp
m
m
ppp
ppp
p
sinsin4
cxcos
cxcos
2sin
2
cxtocxfrom)(
10
110
cx
cx
0
110
1
1
a
xm
a
cm
a
xm
ea
ym
mD
ap
wa
ym
m
pppp
p
p
sinsinsin1
1 1
144
3
0-
=
+=
Levys Method Applied to Non
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Levy s Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 92
For case of concentrated load:
Levys Method Applied to Non
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Levy s Method Applied to Non-
uniformly Loaded Rectangular PlatesPage 93
Substitute the following expressions into w:
a
cm
a
cm
c
PpcpP
pp
== sin2
or200
a
xm
a
xme
a
ym
mD
Paw
a
xm
a
cm
a
xme
a
ym
mD
a
c
Pw
a
ym
m
a
ym
m
ppp
p
ppppp
p
p
sinsin11
2
sinsin11
2
1
1
33
2
1
144
3
-
=
-
=
+=
+
=
Rectangular Plates Under Distributed
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Rectangular Plates Under Distributed
Edge MomentsPage 94
Simply-supported rectangular plate, symmetrically
distributed edge moments at y = +b/2 and y = -b/2:
Rectangular Plates Under Distributed
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Rectangular Plates Under Distributed
Edge MomentsPage 95
Assume a Fourier sin series for the moment distribution:
Boundary conditions:
( )
( )
=
=
=
=
a
m
m
m
dxa
xm
xfaM
by
a
xmMxf
0
1
sin
2
2sin
p
p
( )
( )
===
===
=
2by
ywD-0
,0x0x
w0
2
2
2
2
xfw
axw
Rectangular Plates Under Distributed
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Rectangular Plates Under Distributed
Edge MomentsPage 96
Since p0 = 0 and plate deflection is symmetric about x-
axis:
n Above expression satisfies governing equation and first set
of boundary conditions.
n Use other two boundary conditions to determine constants.
a
xm
a
ymyC
a
ymBw
m
mm
pppsinsinhcosh
1
=
+=
Rectangular Plates Under Distributed
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Rectangular Plates Under Distributed
Edge MomentsPage 97
First boundary condition:
a
xm
a
ymb
a
ymyCw
bCB
a
bmb
CB
w
m
mm
mmm
mmmmm
ppa
p
a
p
aaa
sincoshtanh2
sinh
tanh2
20sinh2cosh
2
by0
1
=
-=
-=
==+
==
Rectangular Plates Under Distributed
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Rectangular Plates Under Distributed
Edge MomentsPage 98
Second boundary condition:
Using basic derivation:
( )
a
xm
a
ymy
a
ymb
m
M
D
aw
Dm
aMC
a
xmM
a
xm
a
mCD
xf
m
m
m
m
m
mm
m
m
m
mm
pppa
ap
ap
ppa
p
sinsinhcoshtanh2cosh2
cosh2
sinsincosh2
2
by
y
wD-
1
11
2
2
=
=
=
-=
-=
=
-
==
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Rectangular Plates Under Distributed
Edge MomentsPage 99
If f(x) = M0 (uniformly distributed moments):
For square plate (a = b), results at center:
a
xm
a
ymy
a
ymb
mD
aMw
mm
M
a
xm
m
Mdx
a
xmM
aM
m
m
m
ax
x
a
m
pppa
ap
pp
pp
sinsinhcoshtanh2cosh
12
...3,14
cos2
sin2
...3,122
0
00
0
0
0
=
==
-=
==-==
00
20 256.0,394.0,0368.0 MMMMD
aMw yx ===
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Rectangular Plates Under Distributed
Edge MomentsPage 100
Displacement along axis of symmetry (y = 0):
For a very long strip (a >> b), i.e. center deflection ofa strip of length b subjected to two equal and opposite
bending moments at ends:
a
xm
mD
abMw
m m
m pa
ap
sincosh
tanh
...3,122
0
=
=
D
bM
D
bMw
a
xm
mD
bM
a
xm
mD
abMw
a
bm
mm
m
mmmm
842
sin1
2sin
21coshtanh
2
0
2
0
...3,1
20
...3,122
0
=
=
==
=
=
=
pp
pp
pap
paaaa
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Rectangular Plates Under Distributed
Edge MomentsPage 101
Plate with one edge moment:
n Note: This problem switches previous x and y directions.
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Rectangular Plates Under Distributed
Edge MomentsPage 102
Assume same Fourier series expression:
Deflection is in the following form:
( ) ( )
( )
=
=
=
=
b
xn
n
nx
dyb
yn
MbM
xb
ynMM
0
0
10
sin
2
0sin
p
p
b
yn
b
xn
xDb
xn
xCb
xn
Bb
xn
Aw nnnnn
pppppsincoshsinhcoshsinh 1
=
+++=
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Rectangular Plates Under Distributed
Edge MomentsPage 103
Boundary conditions:
Last two boundary conditions are automatically
satisfied, as always. First boundary condition results in: Bn = 0.
( ) ( )
( )
( )byw
aw
Mw x
===
=
==
=
==
=
,0y0y
w0
x0x
w
0
0xx
wD-0
2
2
2
2
02
2
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Rectangular Plates Under Distributed
Edge MomentsPage 104
Remaining 3 boundary conditions can be used to solve
for remaining 3 constants, resulting in:
Based on moment, can find Mn and then substitute into
w expression above.
( )b
yn
b
ynb
xna
b
xan
x
b
ann
M
D
b
wn
n p
p
pp
pp sinsinh
sinh
coshsinh
2 1
=
-
-
=
Method of Superposition Applied to
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Method of Superposition Applied to
Bending of Rectangular PlatesPage 105
Complex problem replaced by several simplersituations. Simpler solutions: Navier or Levy approach.
Superposed so that the overall governing equation and
boundary conditions are fulfilled. For example, consider a rectangular plate under any
lateral load, one edge (y=0) clamped and otherssimply-supported. Start with plate with all edges simply-supported.
Add the solution of plate with bending moment appliedalong y=0, with magnitude to eliminate rotations alongclamped edge.
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Method of Superposition Applied to
Bending of Rectangular PlatesPage 106
Example: Two edges simply-supported, two edges
clamped, uniform pressure.
Method of Superposition Applied to
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Method of Superposition Applied to
Bending of Rectangular PlatesPage 107
Plate 1 (from previous Levy approach):
For plate 2, need to determine the moment along the edges
so that the sum of the rotations of plate 1 and plate 2 alongthese edges equals zero (clamped boundary condition).
Thus, need to find the rotation along these edges in plate 1.
a
xm
b
y
a
ym
b
y
mD
apw
m
m
m
m
m
mm
papa
aa
aap
sin2
sinhcosh2
1
2cosh
cosh2
2tanh1
14
...3,155
4
01
+
+-=
=
a
bmm
2:where
pa =
Method of Superposition Applied to
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Method of Superposition Applied to
Bending of Rectangular PlatesPage 108
Rotation along edge found by taking derivative of w
with respect to y:
Substitute y = b/2:
a
xm
b
y
a
m
b
y
ba
ym
b
y
bmD
ap
y
w
m
m
mm
m
m
mm
m
mm
papa
aapa
aaa
aap
sin2
sinhcosh2
12cosh
2
cosh2
1
2sinh
2
cosh2
2tanh14
...3,155
4
01
+
+
+-=
=
a
xm
a
m
b
bmD
ap
y
w
m
m
mm
m
m
m
m
m
mm
pa
pa
aa
a
aa
a
aa
p
sinsinhcosh2
1cosh
2
cosh2
1
sinh2
cosh2
2tanh14
2
...3,1
55
4
01
+
+
+-=
=
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Method of Superposition Applied to
Bending of Rectangular PlatesPage 109
Further manipulation:
( )
( ( ))
( ( ))
abm
a
xm
mD
ap
y
w
a
xm
bmD
ap
y
w
a
xm
bbbmD
ap
y
w
m
mmm
m
m
mmm
m
mm
mmm
mmm
m
m
2:where
sin1tanhtanh12
sin1tanhtanh4
sintanh2tanhtanh14
...3,144
3
01
...3,155
4
01
2
...3,155
4
01
pa
paaaa
p
paaaa
ap
pa
aaaaa
ap
=
+-=
+-=
+++
-=
=
=
=
Method of Superposition Applied to
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Me od o Supe pos o pp ed o
Bending of Rectangular PlatesPage 110
Plate 2, from previous solution:
Edge rotation for this case:
=
=
= 2sin
1
by
a
xmMM
m
my
p
a
xm
a
ym
ya
ymb
m
M
D
a
wm
mm
m ppp
aap sinsinhcoshtanh2cosh2 12
=
-=
a
xm
a
ym
a
my
a
yma
ym
a
mb
m
M
D
a
y
w
m
m
m
m
pppp
ppa
apsincoshsinh
sinhtanh
2cosh2 1
2
--
=
=
Method of Superposition Applied to
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M p p pp
Bending of Rectangular PlatesPage 111
Further manipulation, substituting y = b/2:
(
)
( )
( )[ ] axm
m
M
D
a
y
w
a
xm
m
M
D
a
y
w
a
xm
m
M
D
a
y
w
mmmmm
m
mm
m
mmmm
mmm
m
mmm
m
m
p
aaaap
paaaaa
p
paaa
aaaap
sin1tanhtanh2
sintanhtanhtanh2
sincoshsinh
sinhtanhcosh2
1
2
1
2
1
2
--=
--=
--
=
=
=
=
Method of Superposition Applied to
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p p pp
Bending of Rectangular PlatesPage 112
Slopes must be equal and opposite:
Final result is then found by:
n Substituting Mm into w2.
n Add solutions of w1 and w2.
( )( )1tanhtanhtanh1tanh4
:gives
2
33
2
0
21
--+-
=
=
-=
mmmm
mmmmmm
apM
by
y
w
y
w
aaaaaaaa
p
S i M h d
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Strip MethodPage 113
Simple, approximate approach for computing
deflection and moment in rectangular plate with
arbitrary boundary condition.
Plate is assumed to be divided into two systems ofstrips at right angles to one another.
Each strip is regarded as functioning as a beam.
Not accurate, but this method gives conservative values
for deflection and moment. Employed in practice.
St i M th d
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Strip MethodPage 114
Based on deflections and moments of beams with
various end conditions derived from mechanics of
materials.
8
3
8
5 pLR
pLR BA ==
St i M th d
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Strip MethodPage 115
Consider the rectangular plate:
Strip of span a carries uniform load pa. Strip of span b carries uniform load pb.
St i M th d
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Strip MethodPage 116
Two conditions that must be satisfied:
For simply-supported edges (same for clamped plate):
( )00 ==+== yxpppww baba
44
4
044
4
00
44
44
384
5
384
5
baapp
babppppp
bpapww
EI
bpw
EI
apw
baba
baba
bb
aa
+=
+=+=
==
==
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Strip MethodPage 117
And thus:
Using the expressions for beams in the given tables,
once the center deflection is determined, can find the
deflection and moment anywhere along either strip.
( )
384
5
384
5
384
544
44
0
44
maxbaEI
bap
EI
bp
EI
apw ba
+===
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Strip MethodPage 118
Example: 3 edges simply-supported, 1 edge clamped,uniform p0.
Use expressions for displacement of a beam simply-
supported on each end along y, and a beam clamped onone end and simply-supported on another along x.
Determine mid-span deflection and maximum moments
St i M th d
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Strip MethodPage 119
Setting mid-span deflections equal (note: this is not
always the location of maximum deflection):
( )EIba bapEIbpw
ba
app
ba
bppppp
bpapwwEI
bp
EI
apww
bcenter
baba
baba
baba
44
44
0
4
44
4
044
4
00
44
44
521925
3845
52
2
52
5
52384
5
192
+==
+=
+=+=
==
==
Strip Method
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Strip MethodPage 120
To find maximum moments, must first find where
maximum occurs for each beam, then substitute for
appropriate load and span.
For example, for the y-span (simply-supported to
simply-supported), it is known that the maximum
moment occurs at midspan:
( )
( )( )0
524
084222
44
24
0
2
=+
=
==
-
=
yba
bapM
ybpbb
pbb
pM
y
bbby
Strip Method
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Strip MethodPage 121
In the x-direction span, must first determine where the
maximum moment occurs:
n Reaction force at simply-supported end = 3paa/8.
( )
( )
=+
=
-=
-
=
==-=
-=
axba
bapM
apaapapaM
axapxpdx
dM
axpxpxM
x
aaax
aax
aax
8
3
52128
45
128
9
8
3
8
3
8
3
2
1
8
3
8
30
8
3
8
3
2
1
44
42
0
2
2
2
Simply-Supported Continuous
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Rectangular PlatesPage 122
Continuous plate:
A uniform plate that extends over a support and has more
than one span that may be of varying length.
Intermediate supports provided by beams or columns.
Analysis is based on equilibrium of individual panels
and compatibility of displacements or force at
adjoining edges.
Assume only rigid intermediate beams: Plate has zero deflection along axis of supporting beam.
Beam does not prevent rotation, i.e. simply-supported.
Simply-Supported Continuous
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Rectangular PlatesPage 123
Consider two-paneled continuous plate, as shown:
( ) y
sin...3,1 b
mMyf
m
m
p
=
=
Simply-Supported Continuous
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Rectangular PlatesPage 124
Procedure:
Consider each panel separately.
On connecting edges, assume a moment is applied
(value unknown). Use regular plate boundary conditions to solve for
constants.
Need one additional equation since there is an
additional unknown (moment). This equation is basedon continuity at the connecting edges, i.e. the slopes
must be equal.
Simply-Supported Continuous
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Rectangular PlatesPage 125
Expressions for deflections of each plate:
(
() yxxH
xxGxFxEw
b
m
yDmbpxxD
xxCxBxAw
mmm
mmmmmmm
m
mmm
m
mmmmmm
lllll
pl
lp
l
lll
sincosh
sinhcoshsinh
sin4cosh
sinhcoshsinh
22
...3,122222
55
4
011
...3,1
11111
+++=
=
++
++=
=
=
Simply-Supported Continuous
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Rectangular PlatesPage 126
Boundary conditions:
( )
( )
( )
( )0ysinD-0
00
ysinD-0
000
2
...3,12
2
2
2
2
22
2
2
2
2
1
...3,12
1
1
2
1
12
1
1
2
1
==
=
==
=
===
==
=
=
=
xMx
ww
axx
ww
axMxww
xx
ww
m
m
m
m
m
m
l
l
Simply-Supported Continuous
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Rectangular PlatesPage 127
Additional boundary condition continuity:
Find constants Am Hm and Mm (9 unknown) using
given 9 equations. Straightforward, results on next page.
02
2
1
1
21 ==
=
xax x
w
x
w
Simply-Supported Continuous
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Rectangular PlatesPage 128
aD
M
D
MG
aD
aME
ab
pM
D
aD
Db
p
Db
pB
aa
ab
pa
b
pM
a
aa
DA
m
m
m
m
mm
m
m
mm
m
m
o
m
mmm
mm
om
mm
m
m
om
m
o
m
m
m
mm
lll
ll
lll
lll
ll
ll
llll
l
coth2H2
0F)coth1(2
)cosh1(2
2
csch
2C
4
csch2
coth4
)cosh1(2
2sinh
coth1
m
m
2
4
4
0m5
54
=-=
=-=
+-+-=
=-=
--+
+-+=
Homework Problem 6
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Homework Problem 6Page 129
A simply supported rectangular plate is under the action of
hydrostatic pressure expressed by p = p0x/a, where constant
p0 represents the load intensity along the edge x = a.
b
Homework Problem 6
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Homework Problem 6Page 130
Determine, employing Naviers approach and retaining the
first term of the series solution, the equation for the resulting
solution:
by
ax
ba
Dpw pp
psinsin
11
182
22
6
0
+
=
Homework Problem 7
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Homework Problem 7Page 131
A water level control structure consists of a vertically
positioned simply supported plate. The structure is filled with
water up to the upper edge level at x = 0.
Homework Problem 7
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Homework Problem 7Page 132
Show, using Levys method, that by taking only the first term
of the series solution, the value for the deflection at the center
for a = b/3 is:
Dapw
4000614.0=
Homework Problem 8
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Homework Problem 8Page 133
A rectangular plate is subjected to a uniform loading p0 and
supported as shown.
Determine the maximum stresses using the strip method.
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