Solution of Plate Bending Equation
Uniform Load Simply Supported Free to pull in
via sinusoidal loading
y
x
p( x, y) := po⋅sin π⋅  ⋅sin π⋅ 
a
b
loading




pxy
w=0
for
mx = my = 0
2
mx = my = 0
=>
d
2
dx
w=
x=0
2
d
dy
2
1
w=0
y=0
x=0
x=b
y=0
y=a
x=b
y=a
w( x, y) := C⋅sin π⋅
all boundary conditions satisfied if take
y
 a
⋅sin π⋅
x
 b
substitute in plate equation:
4
d
4
w(x,y) + 2⋅
dx
4
d
4
2
2
d
d
2
dx dy
w( x, y) + 2⋅
dx
2
d
2
d
2
dx dy
2
2
w(x,y) +
4
d
dy
4
d
w( x, y) +
dy
4
4
w(x,y) =
y
x
po⋅sin π⋅  ⋅sin π⋅ 
a
b




D
w( x, y) → 4⋅ C⋅sin( π⋅y ) ⋅sin( π⋅x) ⋅π
4
after collecting terms:
4
d
4
w( x, y) + 2⋅
dx
=
2
d
2
d
2
dx dy
2
4
d
w( x, y) +
dy
4
w( x, y) = C⋅sin π⋅
 a
 π2 π2 
C⋅sin π⋅ ⋅sin π⋅ ⋅ 
+
 a   b   b2 a2 
y
is a solution if
y
⋅sin π⋅
x
 π4
⋅
 b   b4
+ 2⋅
π
4
2 2
+
a ⋅b
π
a
4
4

2
x
C⋅π ⋅sin π⋅
4

y
a
⋅sin π⋅

x
b
⋅
2

 2+ 2 =
a 
b
1
1
y
x
po⋅sin π⋅  ⋅sin π⋅ 
a
b




D
is a solution if
2
po
1
1
C⋅π ⋅ 
+
=
 2
2
D
a 
b
4
w( x, y) :=
po
4
⋅
1
2
⋅sin π⋅
 1 + 1
 2
2
a 
b
 d2
mx := −D⋅
 dx2 w(x,y) +

D⋅π
D
or ....
y
 a
ν
⋅sin π⋅
x
 b
 d2

w(x,y)
 dyy 2



2
C=
po
D⋅π
4
1
⋅
 1 + 1
 2
2
a 
b
2
is the displacement for a sinusoidal loading in x and y
moments and stresses are determined from:
 d2

2
d
w(x,y) + ν
w(x,y)
2
2
dxx
 dy

my := −D⋅

and
σ x :=
mx
I
⋅zzmax
I :=
t
3
12
this result can first be generalized to:
w( x, y) :=
po
D⋅π
4
⋅
1
 n2 m2 
 +
 b2 a2


2
⋅sin mπ⋅

y
a
mx t
⋅
I 2
σ x :=
for a loading of a higher order sinusoidal loading
⋅sin nπ⋅

x
y
x
p( x, y) := po⋅sin mπ⋅  ⋅sin nnπ⋅ 
a
b
b

now consider a uniform load: po
represent po in a double fourier series:
6
σ x := mx⋅
2
t
∞
p = f(x, y) =
∞
∑ ∑
amn =
16⋅po
amn⋅ sin
odd coefficients
even = 0
2
π ⋅m⋅n
 a


 m⋅π⋅y  ⋅sin n⋅π⋅x 

 a   b 
for example on a square plate with infinity = 20 i.e. 20 terms in the series:
eachpxy
sinusoidal element (m,n) is:

amn⋅sin
m= 1 n= 1
coefficients amn can be determined and are:

N ≡ 20
M ≡ 20
⋅ sin
 b 
note here that we are using m,
and n odd
3
the displacement for each loading element
1
w( x, y) :=
D⋅π
is ...
amn
4
⋅
 n2 m2 
 +
 b2 a2


2
 m⋅π⋅y  ⋅sin n⋅π⋅x 

 a   b 
⋅sin
from above
this result can first be generalized to:
w( x, y) :=
po
D⋅π
4
⋅sin mπ⋅
1
⋅
 n2 m2 
 +
 b2 a2



2
y
a
for a loading of a higher order sinusoidal loading
⋅sin nπ⋅

x
y
x
p( x, y) := po⋅sin mπ⋅  ⋅sin nnπ⋅ 
a
b
b




so by superposition of lots of the components of the fourier expansion of po is ...
w( x, y) :=
1
4
∞
⋅
∞
∑ ∑
π ⋅D m = 1 n = 1
substituting for amn
∞
w( x, y) :=
∞
∑ ∑
m= 1 n= 1
amn
 m2 n2 

+
 a2 b2


2
 m⋅π⋅y  ⋅sin n⋅π⋅x 

 a   b 
⋅sin
16⋅po
 m2 n2 

D
+
π ⋅D⋅m⋅n⋅
 a2 b2


2
⋅sin m⋅π⋅
6

y
a
⋅sin n⋅π⋅

amn :=
16⋅po
2
n
π ⋅m⋅n
x
b
m, n, odd => substitute 2*m-1 and 2*n-1 for m, n
∞
w( x, y) :=
∞
∑ ∑
16⋅po
2
4
⋅sin(2⋅m − 1)⋅π⋅

y
x
⋅sin(2⋅n − 1)⋅π⋅ 

a
b

∑ ∑
m= 1 n= 1
 (2⋅ m − 1) 2 (2⋅ n − 1) 2 

π ⋅D⋅(2
D ⋅ m − 1)⋅(2⋅ n − 1)⋅ 
+

2
2

a
b


6
2

a

b
if we want to look at maximum deflection: x = b/2 y = a/2 expand here
from previous lecture:
 d2
w(x,y) +
2
 dx
mx := −D⋅

ν
 d2

w(x,y)
 dyy 2



 d2

2
d
w(x,y) + ν
w(x,y)
2
2
dxx
 dy

my := −D⋅

and we can solve just as above for the single half waves:
plot as a function of a/b i.e. a with b = 1
a := 1 .. 10
M := 10
b := 1
ν := 0.3
N := 10
M
mx_po_b_sq( a) :=
N
∑ ∑
m= 1 n= 1
2

(2⋅ m − 1) 
2
(m

⋅ (2⋅ n − 1) + ν⋅
⋅( −1)
2 
2 
 (2⋅ m − 1) 2
 
 a 
4
2

+ (2⋅ n − 1)  
π ⋅(2⋅ m − 1)⋅(2⋅ n − 1)⋅ 
 b 
2


a
  

  b

16
0.1
mx_po_b_sq(10) = 0.125
mx_po_b_sq( a)
0.05
0
0
5
what length (ratio) is
needed to declare a plate
long???
10
a
see how many terms we need to obtain convergence:
M
mx_po_b_sq(a , M , N) :=
N
∑ ∑
m= 1 n= 1
2

(2⋅ m − 1) 
2

⋅ (2⋅ n − 1) + ν⋅
⋅(
2 
2 
2
a
 (2⋅ m − 1)
  
4
2 

+ (2⋅ n − 1)  
π ⋅(2⋅ m − 1)⋅(2⋅ n − 1)⋅ 
 b 
2


a

 

  b

16
5
M := 1 .. 10
0.2
mx_po_b_sq(10, M , M) 0.15
0.1
0
5
10
M
the corresponding y direction moment is: ν is associated with the m term vs. the n term see above my)
M := 10
N := 10
b := 1
M
my_po_b_sq( a) :=
a := 1 .. 10
N
∑ ∑
m= 1 n= 1
ν := 0.3
max at a/2 b/2

16
⋅   ν⋅ (2⋅ n − 1)

 (2⋅ m − 1) 2
4
2



+ (2⋅ n − 1)
π ⋅(2⋅ m − 1)⋅(2⋅ n − 1)⋅

2


a

 

  b

2
2 

+
(2⋅ m − 1)
 a

 b
2
2
 ⋅(−



0.05
my_po_b_sq(10) = 0.037
0.045
my_po_b_sq( a)
0.04
0.035
mx_po_b_sq(10, 10, 10)⋅ν = 0.037
0
5
10
a
restating original form of mx/po*b^2
M
mx_po_b_sq( a) :=
N
∑ ∑
m= 1 n= 1
2

(2⋅ m − 1) 
2
(m

⋅ (2⋅ n − 1) + ν⋅
⋅( −1)
2 
2 
 (2⋅ m − 1) 2
 
 a 
4
2

+ (2⋅ n − 1)  
π ⋅(2⋅ m − 1)⋅(2⋅ n − 1)⋅ 
 b 
2


16
6



 a

 b
part of figure 9.5 in Hughes
mx_po_b_sq( a) 0.1
my_po_b_sq( a)
 


2
0.05
0
0
5
10
a
stress is related to the moment as before
σ x :=
mx
I
⋅zzmax
I :=
t
3
σ x :=
12
mx t
⋅
I 2
6
σ x := mx⋅
2
t
at maximum mid point: considering:
b
σ x = k⋅ po⋅  
t
therefore:
σx
2
  b  2
 po⋅  
 t 
=k
k=
similarly
kx( a) := 6⋅mx_po_b_sq( a)
1
6
mx⋅
2
t
  b  2
p ⋅ 
 ot 
= 6⋅ mx_po_b_sq
ky ( a) := 6⋅my_po_b_sq( a)
kx(10) = 0.748
kx( a)
ky( a)
ky (10) = 0.221
0.5
0
0
5
10
kx( 10)⋅ν = 0.224
a
the clamped situation is considerably more complicated
the results are developed in Timoshenko
results are shown in the plot:
0.1
0.5
M x_clamped 0.08
kx_clamped
7
 b

M y_clamped
ky_clamped
0.06
0.04
1
1.5
2
2.5
a_over_b
0.343
0.4
0.3
1
1.5
2
a_over_b
N.B. this NOT the same My and ky as for the simply supported case.
In the clamped case, there is an axial stress in the y (long direction) even for a long plate.
The y axis stress is the maximum at the midpoint of the short side (x = b/2 at the edge y = 0 and y = a)
the x axis stress is maximum at the midpoint of the long side (y = a/2 at the edge x = 0 and x = b)
8
2.5
the bottom line plot for simply supported and clamped/clamped plates under uniform pressure is
figure 9.6 Hughes
0.8
k in relation stress = k*p*(b/t)^2
0.7
0.6
0.5
0.4
0.3
0.2
1
1.5
2
2.5
3
ratio of a/b
3.5
4
4.5
x simply supported - > 0.75, max in center
y simply supported -> 0.225 max in center
x clamped -> 0.5 max at side mid long length
y clamped -> 0.34 max at side mid short length
Mx is max (k = 0.5) at x = a/2, i.e. at ends of short side, middle of long side
My is max (k = 0.34) at y = b/2, i.e. at ends of long side middle of short side
(think of situation in square => both are max (k = ~0.3) and equal on sides in middle)
9
5