2-D formulation Plane theory of elasticity Att 6672

Lecture 72-D formulationPlane theory of elasticityPrint version Lecture on Theory of Elasticity and Plasticity of

Dr. D. Dinev, Department of Structural Mechanics, UACEG

7.1

Contents

1 Plane strain 1

2 Plane stress 3

3 Plane strain vs. plane stress 5

4 Airy stress function 5

5 Polynomial solution of 2-D problem 7

6 General solution strategy 14 7.2

1 Plane strain

Plane strain

Introduction

• Because of the complexity of the field equations analytical closed-form solutions to full3-D problems are very difficult to accomplish• A lot of problems into the area of engineering can be approximated by 1-D or 2-D strain

or stress state

– rods, beams, columns, shafts etc.

– Retaining walls, disks, plates, shells7.3

Plane strain

1

Problem definition• Consider an infinitely long prismatic body• If the body forces and surface tractions have no components on z-direction the deformation

field can be reduced into

u = u(x,y)

v = v(x,y)

w = 0

• This deformation is called as a state of plane strain in the (x,y)-plane• Thus all cross-sections will have same displacements

7.4

Plane strain

Field equations• The strain-displacement relations become

εxx =∂u∂x

, εyy =∂v∂y

, εxy =12

(∂u∂y

+∂v∂x

)εzz = εxz = εyz = 0

• In matrix form εxxεyy

2εxy

=

∂

∂x 00 ∂

∂y∂

∂y∂

∂x

[ uv

]

• The St.-Venant’s compatibility equation is

∂ 2εxx

∂y2 +∂ 2εyy

∂x2 = 2∂ 2εxy

∂x∂y

7.5

Plane strain

Field equations• The stress-strain relations are

σxx = (λ +2µ)εxx +λεyy

σyy = λεxx +(λ +2µ)εyy

σzz = λεxx +λεyy

σxy = µ2εxy

σxz = σyz = 0

2

• In matrix form σxxσyyσzzσxy

=

λ +2µ λ 0

λ λ +2µ 0λ λ 00 0 2µ

εxx

εyyεxy

7.6

Plane strain

Field equations• The equilibrium equations are reduced to

∂σxx

∂x+

∂σxy

∂y+ fx = 0

∂σxy

∂x+

∂σyy

∂y+ fy = 0

• In matrix form [σxx σxyσxy σyy

][∂

∂x∂

∂y

]+

[fxfy

]=

[00

]7.7

Plane strain

Field equations• The Navier’s displacement equilibrium equations are

µ∇2u+(λ +µ)

∂

∂x

(∂u∂x

+∂v∂y

)+ fx = 0

µ∇2v+(λ +µ)

∂

∂y

(∂u∂x

+∂v∂y

)+ fy = 0

where ∇2 = ∂ 2

∂x2 +∂ 2

∂y2 - Laplacian operator• The Beltrami-Michell stress equation is

∇2(σxx +σyy) =−

11−ν

(∂ fx

∂x+

∂ fy

∂y

)• The surface tractions (stress BCs)are[

txty

]=

[σxx σxyσxy σyy

][nxny

]7.8

2 Plane stress

Plane stress

Problem definition

3

• Consider an arbitrary disc which thickness is small in comparison to other dimensions• Assume that there is no body forces and surface tractions in z-directions and the surface of

the disc is stress free• Thus imply a stress field

σxx = σxx(x,y)

σyy = σyy(x,y)

σxy = σxy(x,y)

σzz = σxz = σyz = 0

7.9

Plane stress

Field equations

• The Hooke’s law εxxεyyεzzεxy

=1E

1 −ν 0−ν 1 0−ν −ν 00 0 1+ν

σxx

σyyσxy

• Relation between normal strains

εzz =−ν

1−ν(εxx + εyy)

7.10

Plane stress

Field equations

• Strain-displacement equationsεxxεyyεzz

2εxy

=

∂

∂x 0 00 ∂

∂y 00 0 ∂

∂ z∂

∂y∂

∂x 0

u

vw

εyz = εzx = 0

• St.-Venant’s compatibility equation is

∂ 2εxx

∂y2 +∂ 2εyy

∂x2 = 2∂ 2εxy

∂x∂y

7.11

Plane stress

Field equations

• Equilibrium equations - same as in plane strain[σxx σxyσxy σyy

][∂

∂x∂

∂y

]+

[fxfy

]=

[00

]7.12

4

E ν

Planestress to strain E

1−ν2ν

1−ν

Planestrain to stress E(1+2ν)

(1+ν)2ν

1+ν

Plane stress

Field equations

• The Navier’s displacement equilibrium equations are

µ∇2u+

E2(1−ν)

∂

∂x

(∂u∂x

+∂v∂y

)+ fx = 0

µ∇2v+

E2(1−ν)

∂

∂y

(∂u∂x

+∂v∂y

)+ fy = 0

• The Beltrami-Michell stress equation is

∇2(σxx +σyy) =−(1+ν)

(∂ fx

∂x+

∂ fy

∂y

)• The surface tractions (stress BCs)are[

txty

]=

[σxx σxyσxy σyy

][nxny

]7.13

3 Plane strain vs. plane stress

Plane strain vs. plane stress

Summary

• The plane problems have identical equilibrium equations, BCs and compatibility equa-tions• The similar equations show that the differences are due to different constants involving

different material constants• The field equations of plane stress can be obtained from equations of plane strain by fol-

lowing substitution• When ν = 0 plane strain ≡ plane stress

7.14

4 Airy stress function

Airy stress function

The Method

• A popular method for the solution of the plane problem is using the so called Stress func-tions• It employs the Airy stress function and reduce the general formulation to a single equation

in terms of a single unknown• The general idea is to develop a stress field that satisfies equilibrium and yields a single

governing equation from the compatibility equations.• The obtained equilibrium equation ca be solved analytically in closed-form

7.15

5


The Method

• Assume that the body forces are zero• The Beltrami-Michell stress compatibility equations are

∇2(σxx +σyy) = 0

• Equilibrium equations are

∂σxx

∂x+

∂σxy

∂y= 0

∂σxy

∂x+

∂σyy

∂y= 0

• The stress BCs are [txty

]=

[σxx σxyσxy σyy

][nxny

]7.16


The Method

• The Beltrami-Michell equation can be expanded as

∂ 2σxx

∂x2 +∂ 2σyy

∂x2 +∂ 2σxx

∂y2 +∂ 2σyy

∂y2 = 0

• The equilibrium equations are satisfied if we choose the representation

σxx =∂ 2φ

∂y2 , σyy =∂ 2φ

∂x2 , σxy =−∂ 2φ

∂x∂y

where φ = φ(x,y) is an arbitrary form called an Airy stress function• Substitution of the above expressions into the Beltrami-Michell equations lied to

∂ 4φ

∂x4 +2∂ 4φ

∂x2∂y2 +∂ 4φ

∂y4 = 0

7.17


The Method

• George Biddell Airy (1801-1892)7.18

6


The Method

• The previous expression is a biharmonic equation. In short notation

∇2∇

2φ(x,y) = 0

• Thus all equations of the plane problem has been reduced to a single equation in terms ofthe Airy stress function φ(x,y).• This function is to be determined in the 2-D region R bounded by the boundary S• Appropriate BCs are necessary to complete a solution

7.19

5 Polynomial solution of 2-D problem

Polynomial solution of 2-D problem

The Method

• The solution with polynomials is applicable in Cartesian coordinates and useful for prob-lems with rectangular domains• Based on the inverse solution concepts - we assume a form of the solution of the equation

∇2∇2φ(x,y) = 0 and then try to determine which problem may be solved by this solution7.20


The Method

• The assumed solution is taken to be a general polynomial and can be expressed in thepower series

φ(x,y) =∞

∑m=0

∞

∑n=0

Cmnxmyn

where Cmn are constants to be determined• The method produces a polynomial stress distribution and not satisfies the general BCs• We need to modify the BCs using St.-Venant principle- with statically equivalent BCs• The solution would be accurate at points sufficiently far away from the modified boundary

7.21


Example 1

• Let’s use a trial solution- first order polynomial

φ(x,y) =C1x+C2y+C3

• The solution satisfies the biharmonic equation• Go to the stress field

σxx =∂ 2φ

∂y2 = 0, σyy =∂ 2φ

∂x2 = 0, σxy =−∂ 2φ

∂x∂y= 0

Question

• What this solution mean?7.22

7


Example 2• Use a higher order polynomial

φ(x,y) =C1y2

• The solution also satisfies the biharmonic equation• The stress field

σxx =∂ 2φ

∂y2 = 2C1, σyy =∂ 2φ

∂x2 = 0, σxy =−∂ 2φ

∂x∂y= 0

7.23


Example 2• The solution fits with the uniaxial tension of a disc• The boundary conditions are

σxx(±`,y) = t

σyy(x,±h) = 0σxy(±`,y) = σxy(x,±h) = 0

• The constant C1 can be obtained from the BCs7.24


Example 3• Pure bending of a beam - a comparison with the MoM solution

7.25


Review of the beam theory• á la Speedy Gonzales• Assumptions

– Long beam- h� `

– Small displacements- u� h and v� h– Small strains- εxx� 1– Bernoulli hypothesis- εyy ≈ 0 and σxy ≈ 0

7.26

8


Review of the beam theory• Displacement field

u(x,y,z) = yθ

v(x,y,z) = v(x,y)

• Because γxy = 0, thus

θ =−∂v∂x

• The final displacements are

u(x,y,z) =−y∂v∂x

v(x,y,z) = v(x,y)

7.27


Review of the beam theory• Strain field

εxx =−y∂ 2v∂x2

• Compatibility equation

1r= κ =

dθ

ds≈ dθ

dx=

d2vdx2

• The strain field can be expressed

εxx =−yκ

• Hooke’s law

σxx = Eεxx =−yEκ

7.28


Review of the beam theory

9

• Bending moment

M =∫

AσxxydA =−EIκ

• Equilibrium equations

dVdx

=−q

dMdx

=−V

7.29


Review of the beam theory

• Differential equation

EId4vdx4 = q

• 4-th order ODE- needs of four BCs7.30


Example 3- MoM solution

• MoM solution7.31


Example 3- Elasticity solution

• Elasticity solution7.32



• Strong BCs

σyy(x,±c) = 0, σxy(x,±c) = 0, σxy(±`,y) = 0

7.33

10


Example 3- Elasticity solution• Weak BCs- imposed in a weak form (using the St.-Venant principle)∫ c

−cσxx(±`,y)dy = 0,

∫ c

−cσxx(±`,y)ydy =−M

7.34


Example 3- Elasticity solution• Based on the MoM solution (linear σxx distribution) we try a following solution

φ(x,y) = A1y3

• The function satisfies ∇4φ(x,y) = 0• The stress functions are

σxx = 6A1y

σyy = 0 → satisfies σyy(x,±c) = 0σxy = 0 → satisfies σxy(x,±c) = 0, σxy(±`,y) = 0

• This trial solution fits with the BCs7.35


Example 3- Elasticity solution• The constant A1 is obtained from the weak BC at x =±`∫ c

−cσxx(±`,y)dy≡ 0∫ c

−cσxx(±`,y)ydy = 4c3A1 =−M, → A1 =−

M4c3

• Thus the Airy stress function is

φ(x,y) =− M4c3 y3

• Corresponding stresses are

σxx =−3M2c3 y

σyy = 0σxy = 0

7.36


Example 3- Elasticity solution• Strain field-by the Hooke’s law

εxx =−3M

2Ec3 y

εyy =3Mν

2Ec3 y

εxy = 0

• Displacement field- by strain-displacement equations

u =− 3M2Ec3 xy+ f (y)

v =3Mν

4Ec3 y2 +g(x)

• The functions f (y) and g(x) have to be determined from the definition of the shear strain7.37

11


Example 3- Elasticity solution• The definition of the shear strain gives

εxy =−3M

4Ec3 x+12

∂ f (y)∂y

+12

∂g(x)∂x

• This result can be compared with the shear strain obtained from the constitutive relationsεxy = 0

− 3M4Ec3 x+

12

∂g(x)∂x

+12

∂ f (y)∂y

= 0

7.38


Example 3- Elasticity solution• The equation can be partitioned into

∂ f (y)∂y

= ω0

∂g(x)∂x

=3M

2Ec3 x+ω0

where ω0 is an arbitrary constant• Integration of the above equation gives

f (y) = uo + yω0

g(x) =3M

4c3Ex2 + xω0 + v0

• The constants u0, v0 and ω0 express the rigid-body motion of the beam7.39


Example 3- Elasticity solution• The back substitution into the displacement field gives

u(x,y) = u0 + yω0−3M

2c3Exy

v(x,y) = v0 + xω0 +3M

4c3Ex2 +

3Mν

4c3Ey2

• The constants can be found from the essential BCs7.40


Example 3- Elasticity solution• The essential BCs are concentrated at points of beam ends

u(−`,0) = 0 → u0 = 0

v(−`,0) = 0 → 3M`2

4c3E+ v0− `ω0 = 0

v(`,0) = 0 → 3M`2

4c3E+ v0 + `ω0 = 0

7.41

12



• The constants are

u0 = 0ω0 = 0

v0 =−3M`2

4c3E

• Displacement field can be completed as

u(x,y) =− 3M2c3E

xy

v(x,y) =3M

4c3E(x2 +νy2− `2)

7.42


-3 -2 -1 0 1 2 3

-2

-1

0

1

2


• Vector plot of the displacement field7.43



• Elasticity solution

u(x,y) =−MEI

xy

v(x,y) =M

2EI(x2 +νy2− `2)

• MoM solution

u(x) =−MEI

xy

v(x) =M

2EI(x2− `2)

where I = 23 c3

NoteIt is convenient to use a computer algebra system for the mathematics (Maple, Mathematica etc.)

7.44

13


Example 3

• General conclusion7.45

6 General solution strategy

General solution strategy

Selection of the polynomial order

• Step 1- Determine the maximum order of polynomial using MoM arguments

Example 1

• Normal loading-q(x)→ xn

• Shear force- V (x)→ xn+1

• Bending moment- M(x)→ xn+2

• Stress- σxx→ xn+2y• Airy function- xn+2y3

• Maximum order= n+57.46


Example 2

• Shear loading-n(x)→ xm

• Shear force- V (x)→ xm

• Bending moment- M(x)→ xm+1

• Stress- σxx→ xm+1y• Airy function- xm+1y3

• Maximum order= m+47.47

14



• Step 2- Write down a polynomial function φ(x,y) that contains all terms up to ordermax(m+4,n+5)

φ(x,y) =C1x2 +C2xy+C3y2 +C4x3 + . . .

7.48



• May use the Pascal’s triangle for the polynomial

1x y

x2 xy y2

x3 x2y xy2 y3

x4 x3y x2y2 xy3 y4

x5 x4y x3y2 x2y3 xy4 y5

• And constants

C1 C2 C3C4 C5 C6 C7

C8 C9 C10 C11 C12C13 C14 C15 C16 C17 C18

• The first three terms have no physical meaning (zero stress field)7.49



• Step 3 Compatibility condition

∇2∇

2φ(x,y) = 0

• Step 4 Boundary conditions- strong and weak. Lead to a set of equations for Ci• Step 5 Solve all equations and determine Ci

Other types of solution

• Fourier series method• . . . . . .

7.50


The End

• Any questions, opinions, discussions?7.51

15

2-D formulation Plane theory of elasticity Att 6672

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