Simply Supported Rectangular Plate
-
Upload
joshua-magat -
Category
Documents
-
view
264 -
download
2
description
Transcript of Simply Supported Rectangular Plate
Chapter 5 Simply Supported Rectangular Plate Joshua V. Magat
Simply Supported Rectangular Plate
Review:
Mx D− 2xw∂
∂
2ν 2y
w∂
∂
2+
⎛⎜⎝
⎞⎟⎠ My D− 2y
w∂
∂
2ν 2x
w∂
∂
2+
⎛⎜⎝
⎞⎟⎠
Qx D−x 2x
w∂
∂
2
2yw∂
∂
2+
⎛⎜⎝
⎞⎟⎠
∂
∂ Qy D−
y 2xw∂
∂
2
2yw∂
∂
2+
⎛⎜⎝
⎞⎟⎠
∂
∂
Mxy Myx− D 1 ν−( ) 2x 2yw∂
∂
2⎛⎜⎝
⎞⎟⎠
∂
∂
2
4xw∂
∂
4
4yw∂
∂
4+ 2 2x 2y
w∂
∂
2⎛⎜⎝
⎞⎟⎠
∂
∂
2⎡⎢⎣
⎤⎥⎦
+qD
where: D E h3
12 1 ν2−( )
So the effective transverse force per unit length for an edge parallel to y axis, Vx
Similarly, for an edge parallel to x-axis, Vy
Vy Qyx
Mxy∂
∂− Vx Qx
yMxy
∂
∂−
Vx D 3xw∂
∂
32 ν−( )
x 2yw∂
∂
2⎛⎜⎝
⎞⎟⎠
∂
∂
⎡⎢⎣
⎤⎥⎦
+⎡⎢⎣
− Vy D 3yw∂
∂
32 ν−( )
y 2xw∂
∂
2⎛⎜⎝
⎞⎟⎠
∂
∂
⎡⎢⎣
⎤⎥⎦
+⎡⎢⎣
−
PLATE UNDER SINUSOIDAL LOAD
We assume that the load distributed over the surface of the plate is given by:
q x y, ( ) qo sin π xa
⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
in which q0 represent the intensity of the load at the center of the plate.
qo 1 kN
m2:= a 5 m:= b 5 m:= q x y, ( ) qo sin π x
a⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
:=
q
The differential equation for the deflection of the surface becomes
4xw∂
∂
4
4yw∂
∂
4+ 2 2x 2y
w∂
∂
2⎛⎜⎝
⎞⎟⎠
∂
∂
2⎡⎢⎣
⎤⎥⎦
+qo
Dsin π x
a⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
Boundary Conditions:
w 0 2xw∂
∂
20 for x 0 and x a
w 0 2yw∂
∂
20 for y 0 and y b
It may be seen that all boundary condition are satisfied if we take for the deflection of surface is
w x y, a, b, ( ) C sin π xa
⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
:= Proof:
ax 2xw x y, a, b, ( )∂
∂
2lim→
0→ by 2y
w x y, a, b, ( )∂
∂
2lim→
0→
Now we all sure that w(x,y) equation could be or most likely the original deflection equation for q so lets consider the differential equation
4xw∂
∂
4
4yw∂
∂
4+ 2 2x 2y
w∂
∂
2⎛⎜⎝
⎞⎟⎠
∂
∂
2⎡⎢⎣
⎤⎥⎦
+qD
4xw x y, a, b, ( )∂
∂
4 π4 C sin π x
a⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
a4→
4yw x y, a, b, ( )∂
∂
4 π4 C sin π x
a⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
b4→
2 2x 2yw x y, a, b, ( )∂
∂
2⎛⎜⎝
⎞⎟⎠
∂
∂
2 2 π4 C sin π x
a⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
a2 b2→
4xw x y, ( )∂
∂
4
4yw x y, ( )∂
∂
4+ 2 2x 2y
w x y, ( )∂
∂
2⎛⎜⎝
⎞⎟⎠
∂
∂
2+ π
4 C sin π xa
⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
1
a21
b2+⎛
⎜⎝
⎞⎟⎠
2
It turns out that
Isolate C, yields
We conclude that the deflection of the surface is
w x y, ( )q0
π4 D 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
2sin π x
a⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
:=
π4 C 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
2 q0
D
Cq0
π4 D 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
2
we can write the q & w formula in the form of
q x y, ( ) qo sin mπ xa
⎛⎜⎝
⎞⎟⎠
sin nπ yb
⎛⎜⎝
⎞⎟⎠
w x y, ( )q0
π4 D m
a2⎛⎜⎝
⎞⎟⎠
2 nb
⎛⎜⎝
⎞⎟⎠
2+
⎡⎢⎣
⎤⎥⎦
2sin mπ x
a⎛⎜⎝
⎞⎟⎠
sin nπ yb
⎛⎜⎝
⎞⎟⎠
⎛⎜⎝
⎞⎟⎠
m n "real constant", ( )
Other Derived formula from the deflection equation above;
Mxq0
π2 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
21
a2ν
b2+⎛
⎜⎝
⎞⎟⎠
sin π xa
⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
Mxyq0 1 ν−( )
π2 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
2a b
cos π xa
⎛⎜⎝
⎞⎟⎠
cos π yb
⎛⎜⎝
⎞⎟⎠
Myq0
π2 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
2ν
a21
b2+⎛
⎜⎝
⎞⎟⎠
sin π xa
⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
Qxq0
π a 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
cos π xa
⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
\
Vxq0−
π a 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
21
a22 ν−
b2+⎛
⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
Qyq0
π b 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
sin π xa
⎛⎜⎝
⎞⎟⎠
cos π yb
⎛⎜⎝
⎞⎟⎠
4 q0 a b
π2
8q0 1 v−( )
π2 a b 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
2+ Vy
q0−
π b 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
21
b22 ν−
a2+⎛
⎜⎝
⎞⎟⎠
sin π ya
⎛⎜⎝
⎞⎟⎠
⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
R2 q0 1 ν−( )
π2 a b 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
2⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦x=a,y=b
FOURIER SERIES Joseph Fourier (21 March 1768 - 16 May 1830) was A French
mathematicians and physicist. He his best known for Fourier series: A way of writing a function as a sum of frequency component that is the sum of sine waves of different frequencies. A function must be periodic
f x t+( ) f x( ) i.e. f x( ) sin x( ) sin x 2π+( )
The basis of Fourier series is that all functions of practical significance which are defined in the interval π− x≤ π≤ can be expressed in terms of a convergent trigonometric series of the form:
f x( ) a0 a1 cos x( )+ a2 cos 2x( )+ a3 cos 3x( )+ ...+ b1 sin x( )+ b2 sin 2x( )+ b3 sin 3x( )+ ...+
when a0 a1, a2..., b1 b2..., are real numbers
f x( ) a0
1
∞
n
an cos n x( ) bn sin n x( )+( )∑=
+
where for the range π− x≤ π≤
a01
2π π−
π
xf x( )⌠⎮⌡
d an1π π−
π
xf x( ) cos n x( )⌠⎮⌡
d bn1π π−
π
xf x( ) sin n x( )⌠⎮⌡
d
n 1 2, 3,..., ( ) n 1 2, 3,..., ( )
Example of Fourier series expansion: Problem 1. Obtain a Fourier series for the periodic function f(x) defined as:
f x( ) k when − π− x< 0<
k when 0 x< π<
The function is periodic outside of this range with period 2π
Solution:
a01
2π π−
π
xf x( )⌠⎮⌡
d an1π π−
π
xf x( ) cos n x( )⌠⎮⌡
d
a0 0 an 0 n 1 2, 3,..., ( )
We see that an is always zero for all even and odd number substitute on n, but different in bn, bn
1π π−
π
xf x( ) sin n x( )⌠⎮⌡
d
n
1
2
3
4
5
⎛⎜⎜⎜⎜⎜⎝
⎞⎟⎟⎟⎟⎟⎠
bn
4 kπ
0
4 k3π
0
4 k5π
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
n 1 2, 3,..., ( ) bn
4 k sin π n2
⎛⎜⎝
⎞⎟⎠
2
π n
So, based on fourier series that expand equation of f(x) is
f x( ) 4 kπ
sin x( ) 13
sin 3 x( )+15
sin 5x( )+ ...+⎛⎜⎝
⎞⎟⎠
Show by plotting the first three partial sums of this Fourier series that as the series is added together term by term the result approximates more and more closely to the function it represents.
If k π in the Fourier series
f x( ) 4 sin x( ) 13
sin 3x( )+15
sin 5 x( )+ ...+⎛⎜⎝
⎞⎟⎠
Let
P1 x( ) 4 sin x( ):= First partial sum
P2 x( ) 4 sin x( ) 43
sin 3 x( )+⎛⎜⎝
⎞⎟⎠
:= Second partial sum
P3 x( ) 4 sin x( ) 43
sin 3 x( )+45
sin 5x( )+⎛⎜⎝
⎞⎟⎠
:= Third partial sum
3.14− 2.51− 1.88− 1.26− 0.63− 0.000.631.261.882.513.14
4.00−3.20−2.40−1.60−0.80−
0.801.602.403.204.00
P1 x( )
P2 x( )
P3 x( )
Pn x( )
x
P1 x( ) 4 sin x( )
P2 x( ) 4 sin x( ) 43
sin 3 x( )+
P3 x( ) 4 sin x( ) 43
sin 3 x( )+45
sin 5x( )+
Pn x( )1
∞
n
bnnsin n x( )( )∑
=
Example no. 2
Obtain a Fourier series for the periodic function f(x) defined as
f x( ) 2x:=
Solution:
f x( ) a0
1
∞
n
an cos n x( ) bn sin n x( )+( )∑=
+
where:
a01
2π π−
π
xf x( )⌠⎮⌡
d 0→:= bn x n, ( ) 1π π−
π
xf x( ) sin n x( )⌠⎮⌡
d:=
an x n, ( ) 1π π−
π
xf x( ) cos n x( )⌠⎮⌡
d 0→:= bni
4 sin π ni( ) π ni cos π ni( )−( )π ni( )2
:=
P200 x( )1
200
n
bnnsin n x( )( )∑
=
:=
3.14− 2.51− 1.88− 1.26− 0.63− 0.00 0.63 1.26 1.88 2.51 3.14
6.28−
5.03−
3.77−
2.51−
1.26−
1.26
2.51
3.77
5.03
6.28
f x( )
P200 x( )
x
ODD AND EVEN FUNCTION OF FOURIER SERIES
Even Functions A function y = f(x) is said to be even if f(-x) = f(x) for all values of x. Graphs of even functions are always symmetrical about y-axis(mirror image)
f x( ) x2:= and y x( ) cos x( ):=
10− 5− 0 5 10
20
40
60
80
100
f x( )
x
10− 5− 0 5 10
1−
0.5−
0.5
1
y x( )
x
Odd Functions A function y = f(x) is said to be odd if f(-x) = -f(x) for all values of x. Graphs of even functions are always symmetrical about the origin.
f x( ) x3:= and y x( ) sin x( ):=
10− 5− 0 5 10
1− 103×
500−
500
1 103×
f x( )
x
10− 5− 0 5 10
1−
0.5−
0.5
1
y x( )
x
FOURIER COSINE, SINE SERIES AND HALF RANGE
(a) Fourier Cosine Series The Fourier series of an even periodic function f(x) having period 2π contains cosine term only and may contain a constant term.
f x( ) a0
1
∞
n
an cos n x( )( )∑=
+
(b) Fourier Sine Series The Fourier series of an odd periodic function f(x) having period 2π contains sine term only.
f x( )1
∞
n
bn sin n x( )( )∑=
Half Range Fourier Series
(a) Half-Range Fourier Cosine Series The Fourier series of an even periodic function f(x) having range 0 to π, (with a period of 2π)
f x( ) a0
1
∞
n
an cos n x( )( )∑=
+ a01π 0
π
xf x( )⌠⎮⌡
d:= an2π 0
π
xf x( ) cos n x( )⌠⎮⌡
d
n 1 2, 3,..., ( )
(b) Half-Range Fourier Sine Series The Fourier series of an odd periodic function f(x) having range 0 to π, (with a period of 2π)
f x( )1
∞
n
bn sin n x( )( )∑=
bn2π 0
π
xf x( ) sin n x( )⌠⎮⌡
d
n 1 2, 3,..., ( )
FOURIER SERIES OVER ANY RANGE
u x( ) 2π xL
f x( ) a0
1
∞
n
an cos n u( ) bn sin n u( )+( )∑=
+
So, for periodic function with the period of L , f(x+L) = f(x), yields
f x( ) a0
1
∞
n
an cos 2π n xL
⎛⎜⎝
⎞⎟⎠
bn sin 2π n xL
⎛⎜⎝
⎞⎟⎠
+⎛⎜⎝
⎞⎟⎠∑
=
+
Same approach for even and odd periodic approach having a period of L
Fourier Cosine Series Fourier Sine Series
f x( ) a0
1
∞
n
an cos 2π n xL
⎛⎜⎝
⎞⎟⎠
⎛⎜⎝
⎞⎟⎠∑
=
+ f x( )1
∞
n
bn sin 2π n xL
⎛⎜⎝
⎞⎟⎠
⎛⎜⎝
⎞⎟⎠∑
=
DOUBLE SINE SERIES
The idea of a Fourier Series expansion for a function of as single variable can be extended to the case of a functions of two or more variables, For instances, we can expand p(x,y) into a double Fourier sine series:
The above represent a half-range sine series in x, multiplied by half-range sine series in y, using for a period of expansion 2a and 2b, respectively, That is
p x y, ( )1
∞
m
pm y( ) sin m π xa
⎛⎜⎝
⎞⎟⎠
⎛⎜⎝
⎞⎟⎠∑
=
p x y, ( )1
∞
m 1
∞
n
pmn sin m π xa
⎛⎜⎝
⎞⎟⎠
sin n π yb
⎛⎜⎝
⎞⎟⎠
⎛⎜⎝
⎞⎟⎠∑
=∑=
with
pm y( )1
∞
n
pmn sin n π yb
⎛⎜⎝
⎞⎟⎠
⎛⎜⎝
⎞⎟⎠∑
=
we all know that,
pmn2b
0
b
ypm y( ) sin n π yb
⎛⎜⎝
⎞⎟⎠
⌠⎮⎮⌡
d
and the coefficient of pmn
pmn4
a b0
b
y
0
a
xp x y, ( ) sin m π xa
⎛⎜⎝
⎞⎟⎠
sin n π yb
⎛⎜⎝
⎞⎟⎠
⌠⎮⎮⌡
d⌠⎮⎮⌡
d
SIMPLY SUPPORTED RECTANGULAR PLATES UNDER VARIOUS LOADINGS (Using Navier Method)
First, the load and deflection of a periodic function using double sine series;
Note: qF x y, ( )1
∞
m 1
∞
n
qmn sin m π xa
⎛⎜⎝
⎞⎟⎠
sin n π yb
⎛⎜⎝
⎞⎟⎠
⎛⎜⎝
⎞⎟⎠∑
=∑=
period along x is 2a period along y is 2b
q x y, ( ) qF x y, ( ) wF x y, ( )
1
∞
m 1
∞
n
amn sin m π xa
⎛⎜⎝
⎞⎟⎠
sin n π yb
⎛⎜⎝
⎞⎟⎠
⎛⎜⎝
⎞⎟⎠∑
=∑=
w x y, ( ) wF x y, ( )
Example (1) Rectangular plate subjected to Uniformly Distributed Load, q(x,y) = q0
q x y, ( ) q0 w x y, ( ) C sin π xa
⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
and from the previous discussion, we all know that ,
q0
Dπ
4 C sin π xa
⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
1
a21
b2+⎛
⎜⎝
⎞⎟⎠
2
then
now we can compute for the coefficient of wF(x,y) using derived equation above
amn
4a b
0
b
y
0
a
xq0
π4 D sin π x
a⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
1
a21
b2+⎛
⎜⎝
⎞⎟⎠
2⎡⎢⎣
⎤⎥⎦
sin π xa
⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
sin m π xa
⎛⎜⎝
⎞⎟⎠
sin n π yb
⎛⎜⎝
⎞⎟⎠
⌠⎮⎮⎮⎮⌡
d⌠⎮⎮⎮⎮⌡
d
amn
16 q0 sin π m
2⎛⎜⎝
⎞⎟⎠
2sin π n
2⎛⎜⎝
⎞⎟⎠
2
m n π6 D 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
2 m n 1, 2, 3, ..., ( )
then the deflection of unifomly distributed load according to double sine series is
wF x y, ( )1
∞
m 1
∞
n
16 q0 sin π m
2⎛⎜⎝
⎞⎟⎠
2sin π n
2⎛⎜⎝
⎞⎟⎠
2
m n π6 D 1
a21
b2+⎛
⎜⎝
⎞⎟⎠
2sin m π x
a⎛⎜⎝
⎞⎟⎠
sin n π yb
⎛⎜⎝
⎞⎟⎠
⎡⎢⎢⎢⎢⎣
⎤⎥⎥⎥⎥⎦
∑=
∑=
By inspection, we can see that when the values of m,n = 1,3,5...(m,n = 2,4,6.. are equal to zero), the equation above reduced to
wF x y, ( )16 q0
π6 D 1
∞
m 1
∞
n
sin m π xa
⎛⎜⎝
⎞⎟⎠
sin n π yb
⎛⎜⎝
⎞⎟⎠
m n m2
a2n2
b2+
⎛⎜⎝
⎞⎟⎠
2
⎡⎢⎢⎢⎢⎣
⎤⎥⎥⎥⎥⎦
∑=
∑=
m n 1, 3, 5..., ( )
Cq0
π4 D sin π x
a⎛⎜⎝
⎞⎟⎠
sin π yb
⎛⎜⎝
⎞⎟⎠
1
a21
b2+⎛
⎜⎝
⎞⎟⎠
2⎡⎢⎣
⎤⎥⎦
Other Derived equations
Mx16 q0
π4
1
∞
m 1
∞
n
ma
⎛⎜⎝
⎞⎟⎠
2ν
nb
⎛⎜⎝
⎞⎟⎠
2+
m n m2
a2n2
b2+
⎛⎜⎝
⎞⎟⎠
2sin m π x
a⎛⎜⎝
⎞⎟⎠
sin n π yb
⎛⎜⎝
⎞⎟⎠
⎛⎜⎝
⎞⎟⎠
⎡⎢⎢⎢⎢⎢⎣
⎤⎥⎥⎥⎥⎥⎦
∑=
∑=
My16 q0
π4
1
∞
m 1
∞
n
νma
⎛⎜⎝
⎞⎟⎠
2 nb
⎛⎜⎝
⎞⎟⎠
2+
m n m2
a2n2
b2+
⎛⎜⎝
⎞⎟⎠
sin m π xa
⎛⎜⎝
⎞⎟⎠
sin n π yb
⎛⎜⎝
⎞⎟⎠
⎛⎜⎝
⎞⎟⎠
⎡⎢⎢⎢⎢⎣
⎤⎥⎥⎥⎥⎦
∑=
∑=
Mxy16 q0 1 ν−( )
π4 a b 1
∞
m 1
∞
n
1
m n m2
a2n2
b2+
⎛⎜⎝
⎞⎟⎠
2⎡⎢⎢⎣
⎤⎥⎥⎦
cos m π xa
⎛⎜⎝
⎞⎟⎠
cos n π yb
⎛⎜⎝
⎞⎟⎠
⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
∑=
∑=
Example (2) Find the equations of the elastic surface of a simply supported rectangular plate for two particular case: (a) the plate is subjected to a load P uniformly distributed over a sub region 4cd or so called path load; (b) the plate carries a point load P at x = x1, y = y2
Solution
qmnP
a b c dy1 d−
y1 d+
y
x1 c−
x1 c+
xsin m π xa
⎛⎜⎝
⎞⎟⎠
sin n π yb
⎛⎜⎝
⎞⎟⎠
⌠⎮⎮⌡
d⌠⎮⎮⌡
d
from which
qmn4 P
π2 m n c d
sin m π x1a
⎛⎜⎝
⎞⎟⎠
sin n π y1b
⎛⎜⎝
⎞⎟⎠
sin m π ca
⎛⎜⎝
⎞⎟⎠
sin n π db
⎛⎜⎝
⎞⎟⎠
so the deflection for the given path load is;
w x y, ( ) 1
π4 D 1
∞
m 1
∞
n
qmn
m
a2⎛⎜⎝
⎞⎟⎠
2 nb
⎛⎜⎝
⎞⎟⎠
2+
⎡⎢⎣
⎤⎥⎦
2sin mπ x
a⎛⎜⎝
⎞⎟⎠
sin nπ yb
⎛⎜⎝
⎞⎟⎠
⎛⎜⎝
⎞⎟⎠
⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
∑=
∑=
(b) when c and d are made to approach zero, qmn
qmn4 Pa b
sin m π x1a
⎛⎜⎝
⎞⎟⎠
sin n π y1b
⎛⎜⎝
⎞⎟⎠
substitute the equation from the equation from the preceding discussion
w x y, ( ) 4 P
a b π4 D 1
∞
m 1
∞
n
sin m π x1a
⎛⎜⎝
⎞⎟⎠
sin n π y1b
⎛⎜⎝
⎞⎟⎠
m
a2⎛⎜⎝
⎞⎟⎠
2 nb
⎛⎜⎝
⎞⎟⎠
2+
⎡⎢⎣
⎤⎥⎦
2sin mπ x
a⎛⎜⎝
⎞⎟⎠
sin nπ yb
⎛⎜⎝
⎞⎟⎠
⎛⎜⎝
⎞⎟⎠
⎡⎢⎢⎢⎢⎣
⎤⎥⎥⎥⎥⎦
∑=
∑=
illustration of surface equation subjected to path and concentrated load;
P 1−:= x1 3:= h 1:= a 10:= d 1:=
y1 1:= D E h3
12 1 v2−( ):= b 10:= c 1:=
Deflection Curve (Path Load)
deflection curve equation of simply supported plate with concentrated load
w x y, ( ) 4 P
a b π4 D 1
200
m 1
200
n
sin m π x1a
⎛⎜⎝
⎞⎟⎠
sin n π y1b
⎛⎜⎝
⎞⎟⎠
m
a2⎛⎜⎝
⎞⎟⎠
2 nb
⎛⎜⎝
⎞⎟⎠
2+
⎡⎢⎣
⎤⎥⎦
2sin mπ x
a⎛⎜⎝
⎞⎟⎠
sin nπ yb
⎛⎜⎝
⎞⎟⎠
⎛⎜⎝
⎞⎟⎠
⎡⎢⎢⎢⎢⎣
⎤⎥⎥⎥⎥⎦
∑=
∑=
:=
Deflection Curve (Concentrated Load)
LAST EXAMPLE
A simply supported rectangular plate is subjected to a moment M at x = x1 and y = y1. Determine the equation for the deflection.
M 10:= kN-m
Deflection Curve (Concentrated Moment)