slide cơ học vật chất rắn chapter 8 1 new two dimensional problem solution

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Chapter 8: Two-dimensional problem solution
(Part 1)

TDT  University  -­‐  2015
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8.1 Two-dimensional problem solution

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8.2 Cartesian Coordinate Solutions Using Polynomials

cu

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du
o

ng

th

an

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8.3 Cartesian Coordinate Solutions Using Fourier Methods

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Institute for computational science

.c
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8.1 Two-dimensional problem solution

ng

8.2 Cartesian Coordinate Solutions Using Polynomials

cu

u

du
o

ng

th

an

co

8.3 Cartesian Coordinate Solutions Using Fourier Methods

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8.1 Two-dimensional problem solution

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Using the Airy Stress Function approach, it was shown that the plane
elasticity formulation with zero body forces reduces to a single governing

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biharmonic equation. In Cartesian coordinates it is given by

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th

an

∂ 4ϕ

∂ 4ϕ
∂ 4ϕ
+ 2 2 2 + 4 = ∇ 4ϕ = 0
4
∂x
∂x ∂y
∂y

du
o

and the stresses are related to the stress function by

cu

u

∂ 2ϕ
∂ 2ϕ
∂ 2ϕ
σ x = 2 , σ y = 2 , τ xy = −
∂y
∂x
∂x∂y

We now explore solutions to several specific problems in both
Cartesian and Polar coordinate systems
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.c
om

8.1 Two-dimensional problem solution

ng

8.2 Cartesian Coordinate Solutions Using Polynomials

cu

u

du
o

ng

th

an

co

8.3 Cartesian Coordinate Solutions Using Fourier Methods

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8.2 Cartesian Coordinate Solutions Using Polynomials

∂ 4ϕ
∂ 4ϕ
∂ 4ϕ
+ 2 2 2 + 4 = ∇ 4ϕ = 0
4
∂x
∂x ∂y
∂y

ng

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The biharmonic equation

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In Cartesian coordinates we choose Airy stress function solution of polynomial form
∞

∞

m=0 n =0

ng

th

an

ϕ ( x, y ) = ∑∑ Amn x m y n

du
o

where Amn are constant coefficients to be determined. This method produces polynomial stress
distributions, and thus would not satisfy general boundary conditions. However, we can

cu

u

modify such boundary conditions using Saint-Venant’s principle and replace a non-polynomial
condition with a statically equivalent loading. This formulation is most useful for problems
with rectangular domains, and is commonly based on the inverse solution concept where we
assume a polynomial solution form and then try to find what problem it will solve.

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∞

∞

ϕ ( x, y ) = ∑∑ Amn x y
m

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8.2 Cartesian Coordinate Solutions Using Polynomials

∂ 4ϕ
∂ 4ϕ
∂ 4ϕ
4
+
2
+
=
∇

ϕ =0
4
2
2
4
∂x
∂x ∂y
∂y

n

ng

m=0 n =0

co

Noted that the three lowest order terms with m + n ≤ 1 do not contribute to the stresses and

an

will therefore be dropped. It should be noted that second order terms will produce a constant

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stress field, third-order terms will give a linear distribution of stress, and so on for higher-order

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polynomials.

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o

Terms with m + n ≤ 3 will automatically satisfy the biharmonic equation for any choice of

cu

u

constants Amn. However, for higher order terms, constants Amn will have to be related in order
to have the polynomial satisfy the biharmonic equation. For example, the 4th-order polynomial
terms A40x4+A22x2y2+A04y4 will not satisfy the biharmonic equation unless 3A40+A22+3A04=0.
This condition specifies one constant in terms of the other two, thus leaving two constants to
be determined by the boundary conditions.
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8.2 Cartesian Coordinate Solutions Using Polynomials

∞

∑∑ m(m − 1)(m − 2)(m − 3) A

mn

m= 4 n =0
∞

x

m−4

∞

∞

y + 2 ∑∑ m(m − 1)n(n − 1) Amn x m − 2 y n − 2
n

m=2 n=2

∞

co

+ ∑∑ n(n − 1)(n − 2)(n − 3) Amn x m y n − 4 = 0

ng

∞

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Considering the general case, substituting the series into the governing biharmonic
equation yields

an

m=0 n =4

∞

th

Collecting like powers of x and y, the preceding equation may be written as
∞

ng

∑∑ ⎡⎣(m + 2)(m + 1)m(m − 1) A

m=2 n=2

m + 2, n − 2

+ 2m(m − 1)n(n − 1) Amn +

du
o

+(n + 2)(n + 1)n(n − 1) Am − 2,n + 2 ⎤⎦ x m − 2 y n − 2 = 0

cu

u

Because this relation must be satisfied for all values of x and y, the coefficient in brackets
must vanish, giving the result
(m + 2)(m + 1)m(m − 1) Am + 2,n − 2 + 2m(m − 1)n(n − 1) Amn + (n + 2)(n + 1)n(n − 1) Am − 2,n + 2 = 0

For each m, n pair, this equation is the general relation that must be satisfied to ensure that
the polynomial grouping is biharmonic.
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8.2 Cartesian Coordinate Solutions Using Polynomials
y
 

T
 

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Example 8.1 Uniaxial Tension of a Beam

Stress
 Field
 
 

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an

⎧⎪σ x (±l , y ) = T , σ y ( x, ±c) = 0
Boundary Conditions: ⎨
⎪⎩τ xy (±l , y ) = τ xy ( x, ±c) = 0

co

2l
 

u

du
o

ng

Since the boundary conditions specify

constant stresses on all boundaries, try a
second-order stress function of the form
σ x = 2 A02 , σ y = τ xy = 0
ϕ = A02 y 2

T
 
x
 

ng

2c
 

Displacement
 Field
 (Plane
 Stress)
 
 
∂u
1
T
= ex = (σ x −νσ y ) =
∂x
E
E
∂v
1

T
= ey = (σ y −νσ x ) = −ν
∂y
E
E

u=

T
T
x + f ( y ) , v = −ν y + g ( x)
E
E

τ
∂u ∂v
+ = 2exy = xy = 0 ⇒ f ′( y ) + g ′( x) = 0
∂y ∂x
µ

cu

The first boundary condition implies that A02 =
T/2, and all other boundary conditions are
f ( y ) = −ωo y + uo
identically satisfied. Therefore the stress field
g ( x) = ωo x + vo . . . Rigid-Body Motion
solution is given by
“Fixity conditions” needed to determine RBM terms
σ x = T , σ y = τ xy = 0

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u (0, 0) = v(0, 0) = u (0, c) = 0 ⇒ f ( y ) = g ( x) = 0
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8.2 Cartesian Coordinate Solutions Using Polynomials

Example 8.2 Pure Bending of a Beam

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y
 

2l
 

Boundary Conditions:

∫

c
−c

σ x (±l , y ) ydy = − M

ng

−c

σ x (±l , y )dy = 0 ,

du
o

∫

c

th

σ y ( x, ±c) = 0 , τ xy ( x, ±c) = τ xy ( ±l , y ) = 0

an

co

Stress
Field

 

σ x = 6 A03 y , σ y = τ xy = 0

cu

ϕ = A03 y 3

u

Expecting a linear bending stress distribution,
try 2nd- stress function of the form
Moment boundary condition implies that A03
= -M/4c3, and all other boundary conditions
are identically satisfied. Thus the stress field
is
3M
σ x = − 3 y , σ y = τ xy = 0
2c
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M
 

x
 

ng

2c
 

M
 

Displacement Field (Plane Stress)

∂u
3M
3M
=−
y ⇒u=−
xy + f ( y )
3
∂x
2 Ec
2 Ec3
∂v
3M
3Mν 2
=ν
y ⇒v=
y + g ( x)
3
∂y
2 Ec
4 Ec3

∂u ∂v
3M
+
=0⇒ −
x + f ′( y ) + g ′( x) = 0
∂y ∂x

2 Ec3
⎧ f ( y ) = −ω0 y + u0
⎪
⎨
3M 2
g
(
x
)
=
x + ω0 x + v0
⎪⎩
4 Ec3

“Fixity conditions” to determine RBM terms:
v(±l , 0) = 0 and u (−l , 0) = 0

u0 = ω0 = 0 , v0 = −3Ml 2 /16 Ec3
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8.2 Cartesian Coordinate Solutions Using Polynomials

Solution Comparison of Elasticity with Elementary Mechanics of Materials

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y
 

2c
 

M
 

x
 

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M
 

2l
 

an


 

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Elasticity Solution

M
y , σ y = τ xy = 0
I
Mxy
M
u=−
,v=
[4ν y 2 + 4 x 2 − l 2 ]
EI
8EI

cu

u

du
o

ng

σx = −

I = 2c 3 / 3

Mechanics of Materials Solution
Uses Euler-Bernoulli beam theory to
find bending stress and deflection of

beam centerline
M
y , σ y = τ xy = 0
I
M
v = v( x, 0) =
[4 x 2 − l 2 ]
8 EI

σx = −

Two solutions are identical, with the exception of the x-displacements
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8.2 Cartesian Coordinate Solutions Using Polynomials

Example 8.3 Bending of a Beam by Uniform Transverse Loading

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2c
 

w
 
wl
 

y
 
2l
 

ng

th

an

wl
 
Boundary Conditions:
c
⎧τ (x,±c) = 0;
∫ −c τ xy (±l, y)d y = ∓wl
⎪ xy
⎪⎪
c

σ
(x,c)
=
0;
⎨ y
∫ −cσ x (±l, y) y d y = 0
⎪
c
⎪σ (x,−c) = −w; σ (±l, y)d y = 0
∫ −c x
⎪⎩ y

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Stress Field

du
o

ϕ = A20 x 2 + A21 x 2 y + A03 y 3 + A23 x 2 y 3 −

cu

u

2 3
⎧
2
σ

=
6
A
y
+
6
A
(
x
y
−
y )
03
23
⎪ x
3
⎪⎪
3
⎨σ y = 2 A20 + 2 A21 y + 2 A23 y
⎪
2
⎪τ xy = −2 A21 x − 6 A23 xy
⎪⎩

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BC’s
 
 

A23 5
y
5

⎧
3w ⎛ l 2 2 ⎞
3w 2
2 3
⎪σ x =
⎜ 2 − ⎟ y − 3 (x y − y )
4c ⎝ c 5 ⎠
4c
3
⎪
⎪⎪
w 3w
w
y − 3 y3
⎨σ y = − +
2 4c
4c
⎪
3w
3w 2
⎪
τ
=
−
x
+

xy
3
⎪ xy
4
c
4
c
⎪⎩
/>
x
 



 

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8.2 Cartesian Coordinate Solutions Using Polynomials

Example 8.3 Bending of a Beam by Uniform Transverse Loading

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w
 

wl
 

x
 

co

y
 

ng

2c
 

wl
 

th

Elasticity Solution

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2l
 

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du
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ng

w 2
w y3 c2 y
2
σ x = (l − x ) y + ( −
)
2I
I 3
5
w ⎛ y3
2 ⎞
σ y = − ⎜ − c 2 y + c3 ⎟
2I ⎝ 3
3 ⎠

cu

w
τ xy = − x(c 2 − y 2 )
2I

Mechanics of Materials Solution

My w 2
= (l − x 2 ) y

I
2I
σy = 0

σx =

τ xy =

VQ
w
= − x (c 2 − y 2 )
It
2I

Shear stresses are identical, while normal stresses are not
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8.2 Cartesian Coordinate Solutions Using Polynomials

σx
 –
 Stress
 at
 x=0
 

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ng

σy
 -­‐
 Stress
 

an

l/c
 =
 2
 

ng

σx/w
 -­‐
 Elasticity

 
 
σx/w
 -­‐
 Strength
 of
 Materials
 

du
o

l/c
 =
 4
 

th

l/c
 =
 3
 

cu

u

Maximum differences between the two
theories exist at top and bottom of beam,

and actual difference in stress values is w/
5. For most beam problems where l >> c,
the bending stresses will be much greater
than w, and thus the differences between
elasticity and strength of materials will be
relatively small.
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 σy/w
 -­‐
 Elasticity
 
 

 σy/w
 -­‐
 Strength
 of
 Materials
 

Maximum difference between the two
theories is w and this occurs at the top of the
beam.
Again this difference will be
negligibly small for most beam problems
where l >> c. These results are generally true
for beam problems with other transverse
loadings.

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8.2 Cartesian Coordinate Solutions Using Polynomials

Example 8.3 Bending of a Beam by Uniform Transverse Loading

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w
 

wl
 

wl
 

2c
 

2l

 

co

y
 

ng

x
 

du
o

ng

th

an

⎧
w
x3
2 y 3 2c 2 y
y3
2c 3
2
2
u=

[(l x − ) y + x(
−
) +ν x( − c y +
)] + f ( y )
Displacement
 Field
  ⎪⎪
2 EI
3
3
5
3
3
⎨
4
2 2
3
2
4
2 2

 (Plane
 Stress)
 
 
⎪v = − w [( y − c y + 2c y ) + ν (l 2 − x 2 ) y + ν ( y − c y )] + g ( x)
⎪⎩
2 EI 12
2
3

2
6
5

cu

u

f ( y ) = ω0 y + u 0 , g ( x ) =

Choosing Fixity Conditions

w 4
w 2 8
x −
[l − ( + ν )c 2 ]x 2 − ω0 x + v0
24 EI
4 EI
5

u (0, y ) = v(±l , 0) = 0

5wl 4
u0 = ω0 = 0, v0 =
24 EI
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⎡ 12 ⎛ 4 ν ⎞ c 2 ⎤
⎢1 + 5 ⎜ 5 + 2 ⎟ l 2 ⎥
⎝

⎠ ⎦
⎣
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8.2 Cartesian Coordinate Solutions Using Polynomials

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Example 8.3 Bending of a Beam by Uniform Transverse Loading
⎡⎛ 2
⎛ 2 y 3 2c 2 y ⎞
⎛ y3
x3 ⎞
2c 3 ⎞ ⎤
2
−
⎢⎜ l x − ⎟ y + x ⎜
⎟ +ν x ⎜ − c y +
⎟⎥
3
3
5
3
3

⎝
⎠
⎝
⎠
⎝
⎠⎦
⎣
2
⎧ y 4 c 2 y 2 2c 3 y
⎡ 2
y4 c2 y 2 ⎤ ⎫
2 y
+
+ ν ⎢( l − x ) +
−
⎪ −
⎥⎪
12
2
3
2
6
5 ⎦⎪
w ⎪
⎣
v=−
⎨
⎬
2 EI ⎪ x 4 ⎡ l 2 ⎛ 4 ν ⎞ 2 ⎤ 2
⎪

−
+
+
+
c
x
⎜
⎟
⎢
⎥
⎪ 12
⎪
⎣2 ⎝5 2⎠ ⎦
⎩
⎭

co

ng

th

an

Displacement
 Field
 

 (Plane
 Stress)

 
 

ng

w
u=
2 EI

⎡ 12 ⎛ 4 ν ⎞ c 2 ⎤
⎢1 + 5 ⎜ 5 + 2 ⎟ l 2 ⎥
⎝
⎠ ⎦
⎣

u

du
o

5wl 4
+
24 EI

cu

v(0, 0) = vmax

5wl 4
=

24 EI

Strength of Materials: vmax
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⎡ 12 ⎛ 4 ν ⎞ c 2 ⎤
⎢1 + 5 ⎜ 5 + 2 ⎟ l 2 ⎥
⎝
⎠ ⎦
⎣

5wl 4
=
24 EI

Good match for beams where l >> c
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.c
om

8.1 Two-dimensional problem solution

ng

8.2 Cartesian Coordinate Solutions Using Polynomials

cu

u

du
o

ng

th

an

co

8.3 Cartesian Coordinate Solutions Using Fourier Methods

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Institute for computational science

8.3 Cartesian Coordinate Solutions Using Fourier Methods

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A more general solution scheme for the biharmonic equation may be found using
Fourier methods. Such techniques generally use separation of variables along with
Fourier series or Fourier integrals.

∂ 4ϕ
∂ 4ϕ
∂ 4ϕ
+2 2 2 + 4 =0
∂x 4
∂x ∂y
∂y

α = ±i β

an

αx
βy
Choosing X = e , Y = e

co

ng

ϕ ( x, y ) = X ( x)Y ( y )

th

φ = sin β x ⎡⎣( A + C β y ) sinh β y + ( B + Dβ y ) cosh β y ⎤⎦

ng

+ cos β x ⎡⎣( A′ + C ′β y ) sinh β y + ( B′ + D′β y ) cosh β y ⎤⎦

du
o

+ sin α y ⎡⎣( E + Gα x ) sinh α x + ( F + H α x ) cosh α x ⎤⎦

cu

+ φα =0 + φβ =0

e x − e− x
sinh( x) =
= −i sin(ix)
2
x
e + e− x
cosh( x) =
= cos ix
2

u

+ cos α y ⎡⎣( E ′ + G′α x ) sinh α x + ( F ′ + H ′α x ) cosh α x ⎤⎦

The general solution
includes the superposition of
the general roots plus the
zero root cases

(zero root solutions)

⎧⎪φβ =0 = C0 + C1 x + C2 x 2 + C3 x 3
where ⎨
2
3
2
2
⎪⎩φα =0 = C4 y + C5 y + C6 y + C7 xy + C8 x y + C9 xy
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8.3 Cartesian Coordinate Solutions Using Fourier Methods

qosinπx/l
 

qol/π
 

2c
 

th

(2)

σ y ( x, − c ) = 0

(3)

u

cu

∫

−c
c
−c

σ x = β 2 sin β x ⎡⎣ A sinh β y + C ( β y sinh β y + 2 cosh β y )

du
o

(1)

ng

σ x (0, y ) = σ x (l , y ) = 0
τ xy ( x, ±c) = 0
σ y ( x, c) = −qo sin(π x / l )

l
 

ϕ = sin β x ⎡⎣( A + C β y ) sinh β y + ( B + D β y ) cosh β y ⎤⎦

Boundary Conditions:

∫

x
 

an

co

Stress Field

c

qol/π
 

ng

Example 8.4 Beam with Sinusoidal Loading

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om

y
 

(4)

τ xy (0, y )dy = −qol / π

(5)

τ xy (l , y )dy = qol / π

(6)
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+ B cosh β y + D ( β y cosh β y + 2sinh β y )⎤⎦

σ y = − β 2 sin β x ⎡⎣( A + C β y ) sinh β y + ( B + Dβ y ) cosh β y ⎤⎦

τ xy = − β 2 cos β x ⎡⎣ A cosh β y + C ( β y cosh β y + 2sinh β y )
+ B sinh β y + D ( β y sinh β y + 2 cosh β y )⎤⎦
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8.3 Cartesian Coordinate Solutions Using Fourier Methods

qosinπx/l
 

qol/π
 

Example 8.4 Beam with Sinusoidal Loading

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om

y
 

qol/π
 

2c
 

co

ng

x
 

an

th

du
o

ng

Condition (2) gives

⎧⎪ A = − D ( β c tanh β c + 1)
⎨
⎪⎩ B = −C ( β c coth β c + 1)

cu

β=

π
l

πc

l
π ⎡π c
πc
πc⎤
2 2 ⎢ + sinh
cosh ⎥
l ⎣ l
l
l ⎦
2

u

Condition (3) gives C =

−qo sinh

l
 

D=

−qo sinh

l
π ⎡π c
πc
πc⎤
2 2 ⎢ − sinh

cosh ⎥
l ⎣ l
l
l ⎦
2

Condition (1) and condition (5,6) will be satisfied

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8.3 Cartesian Coordinate Solutions Using Fourier Methods
qosinπx/l
 

qol/π
 

Example 8.4 Beam with Sinusoidal Loading

qol/π
 

x
 

ng

2c
 

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om

y
 

co

Displacement Field

l
 

β

β

}

th

an

cos β x { A (1 + ν ) sinh β y + B (1 + ν ) cosh β y v = − sin β x { A (1 + ν ) cosh β y + B (1 + ν ) sinh β y
E
E
+ C ⎡⎣(1 + ν ) β y sinh β y + 2 cosh β y ⎤⎦
+ C ⎡⎣(1 + ν ) β y cosh β y − (1 +ν ) sinh β y ⎤⎦

u=−

ng

+ D ⎡⎣(1 +ν ) β y cosh β y + 2sinh β y ⎤⎦ − ωo y + uo

du
o

u (0, 0) = v(0, 0) = v(l , 0) = 0

}

+ D ⎡⎣(1 +ν ) β y sinh β y − (1 +ν ) cosh β y ⎤⎦ + ωo y + vo

ω0 = v0 = 0 , u0 =

β

⎡ B (1 + ν ) + 2C ⎤⎦
E⎣

cu

u

Dβ
sin β x ⎡⎣ 2 + (1 +ν ) β c tanh β c ⎤⎦
E
3q0l 4
π x ⎡ 1 +ν π c
πc⎤
3q0l 5
v( x, 0) = − 3 4 sin
1
+
tanh
D≈− 3 5
For the case l >> c
2c π E
l ⎢⎣
2 l
l ⎥⎦
4c π
3q0l 4
πx
v
(
x

,
0)
=
−
sin
Strength of Materials
2c3π 4 E
l
v( x, 0) =

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Institute for computational science

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Must use series representation for Airy stress
function to handle general boundary loading.
βn =

n =1

∞

th

m =1

∞

du
o

∞

ng

σ x = ∑ β n2 cos β n x ⎡⎣ Bn cosh β n y + Cn ( β n y sinh β n y + 2 cosh β n y )⎤⎦
n =1

− ∑ α m2 cos α m y [ Fm cosh α m x + Gmα m x sinh α m x ]
m =1

∞

x
 


 p(x)
 

Boundary Conditions
σ x ( ± a, y ) = 0

τ xy (± a, y ) = 0
τ xy ( x, ±b) = 0
σ y ( x, ± b ) = − p ( x )

cu

u

σ y = −∑ β n2 cos β n x [ Bn cosh β n y + Cn β n y sinh β n y ] + 2C0

a
 

b
 

an

+ ∑ cos α m y [ Fm cosh α m x + Gmα m x sinh α m x ] + C0 x 2

nπ
l

a
 

ng

ϕ = ∑ cos β n x [ Bn cosh β n y + Cn β n y sinh β n y ]

b
 


 p(x)
 

co

∞


 y
 

.c
om

8.3 Cartesian Coordinate Solutions Using Fourier Methods
Example 8.5 Rectangular Domain with Arbitrary Boundary Loading

Use Fourier series theory to handle
+ ∑ α m2 cos α m y ⎡⎣ Fm cosh α m x + Gm (α m x sinh α m x + 2 cosh α m x )⎤⎦ general boundary conditions, and this
m =1
generates a doubly infinite set of
∞
2
τ xy = ∑ β n sin β n x ⎡⎣ Bn sinh β n y + Cn ( β n y cosh β n y + sinh β n y )⎤⎦
equations to solve for unknown
n =1

constants in stress function form. See
∞
2
+ ∑ α m sin α m y ⎡⎣ Fm sinh α m x + Gm (α m x cosh α m x + sinh α m x )⎤⎦
text for details
m =1
n =1

∞

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cu

u

du
o

ng

th

an

co

ng

.c
om

Institute for computational science

CuuDuongThanCong.com

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