CVEN4304 Structural Analysis & Finite ElementsCVEN9820 Computational Structural Mechanics

Summary of Elasticity Theory

Chongmin Song

School of Civil and Environmental EngineeringUniversity of New South Wales

1

Course Materials

Lecture notes

Section 3.1 of Cook

Appendix C of Logan

2

Strain-Displacement Relationship (Plane Strain)

Displacements in x-, y -directions: {u} = [ u v ]TStrains ({} = []{u})

xyxy

{}

=

uxvy

uy

+vx

=

x0

0

y

y

x

[]

{uv

} {u}

3

Stress State and Equilibrium Equations

Equilibrium in plane strain and in plane stress

x tdy xy tdx+(x +x ,x dx)tdy +(xy + xy ,y dy)tdx+Fx tdxdy = 0xx

+xyy

+ Fx = 0

xyx

+yy

+ Fy = 0

In matrix form: []T{}+ {F} = 0

x0

y

0

y

x

[]T

xyxy

{}

+

{FxFy

} {F}

= 0

Note that the differential operator [] is the same as instrain-displacement relationship.

4

Stress-Strain Relationship: Normal stresses andstrains

xx

y

y

z

z

Hookes law (E : Youngs modulus)

x =xE

y =yE

z =zE

Poissons effect (: Poissons ratio)

y = xE x = yE

x = zE

z = xE z = yE

y = zE5

Stress-Strain Relationship: Normal stresses andstrains

Superposition

x =xE y

E z

E

y = xE +yE z

E

z = xE yE+zE

Conversely

x =E

(1+ )(1 2) ((1 )x + y + z)

y =E

(1+ )(1 2) (x + (1 )y + z)

y =E

(1+ )(1 2) (x + y + (1 )z)

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Stress-Strain Relationship: Shear stresses and strains

Hookes law (G : shear modulus)

=

G

In stress components:

xy =xyG

; yz =yzG

; xz =xzG

or conversely

xy = Gxy ; yz = Gyz ; xz = Gxz

Shear modulus is not independent of E and

G =E

2(1+ )

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Stress-Strain Relationship (3D)

In matrix form ({} = [E ]{} with elasticity (constitutive) matrix [E ])

xyzxyyzxz

=

2G1 2

1 0 0 0 1 0 0 0 1 0 0 00 0 0 122 0 00 0 0 0 122 00 0 0 0 0 122

xyzxyyzxz

Conversely ({} = [E ]1{})

xyzxyyzxz

=

1E

1 0 0 0 1 0 0 0 1 0 0 00 0 0 2(1+ ) 0 00 0 0 0 2(1+ ) 00 0 0 0 0 2(1+ )

xyzxyyzxz

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Plane Strain Condition

One dimension (z) is much larger than the other two (x and y )

In-plane forces do not vary along the largest dimension z.

Dam Pipeline

z

9

Plane Strain Condition

No displacement occurs along the largest dimension z

w = 0; z = wz

= 0

Inplane displacements u and v are independent of z

u = u(x , y)v = v(x , y)

; xz =

uz

+wx

= 0

yz=vz

+wy

= 010

Plane Strain Condition: Stress-Strain Relationship

Considering z = yz = zx = 0

xyzxy00

=

2G1 2

1 0 0 0 1 0 0 0 1 0 0 00 0 0 122 0 00 0 0 0 122 00 0 0 0 0 122

xy0xy00

Elasticity matrix for plane strain condition:

[E ] =2G

1 2

1 0 1 00 0

1 22

Normal stress along the largest dimension z

z = xE yE+zE

= 0; z = (x + y )

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Plane Stress Condition

One dimension (z) is much smaller than the other two (x and y )

zz = yz = xz = 0

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Plane Stress Condition: Stress-Strain Relationship

xyzxy00

=

1E

1 0 0 0 1 0 0 0 1 0 0 0

0 0 0 2(1+ ) 0 00 0 0 0 2(1+ ) 00 0 0 0 0 2(1+ )

xy0xy00

Elasticity matrix for plane stress condition ({} = [E ]{}):

[E ]1 = 1E

1 0 1 00 0 2(1+ )

; [E ] = E1 2

1 0 1 00 0 12

Normal strain along the largest dimension z

z = E (x + y )

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Boundary Conditions

Enforced displacements

u = u; v = v ; w = w

For example, fixed boundary:

u = v = w = 0

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Boundary Conditions

Prescribed tractions

Unit outward normal (l, m) onboundary.

Equilibrium on boundary AB withlength ds

Boundary conditions

x = lx +mxyy = lxy +my

Written in matrix form{xy

}=

[l 0 m0 m l

]

[L]T

xyxy

= [L]T{}15

Boundary Conditions

Similarity with differential operator in equations of equilibrium

[L] =

l 00 mm l

; compare with [] =

x0

0

y

y

x

[L] is in the same form as [] with l , m replacing

x,

y.

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Pure Bending of a Beam

Displacement solution

u = Cxyv = 0.5C(x2 + y2)

Note that planes remain planes(u varies linearly in y and u = 0at neutral axis y = 0)

Strains

x =ux

= Cy

y =vy

= Cy

xy =vx

+uy

= Cx Cx = 017

Pure Bending of a Beam (conted)

Stresses (Plane stress condition)

x =E

1 2 (x + y ) =E

1 2 (Cy + C2y)

= CEy (Linear variation over cross section)y =

E1 2 (y + x ) =

E1 2 (Cy Cy)

= 0

xy = Gxy = 0

Equilibrium (body force Fx = Fy = 0)

xx

+xyy

= 0+ 0 = 0

xyx

+yy

= 0+ 0 = 0

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Pure Bending of a Beam (conted)

Boundary conditions

Left side: outward normal (1, 0)x = lx +mxy = CEyy = lxy +my = 0

Right side: outward normal (1, 0)

x = lx +mxy = CEyy = lxy +my = 0

x

y

hn=(1,0)n=(-1,0)

Bending moment (b: thickness of beam)

M =12(CE h

2

) h

2(

23h2

) b 2 = CE bh

3

12

Constant C for a applied bending moment M

C =MEI

(I : Moment of inertia)

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Common Civil Engineering Problems

Mathematical model in form of

[]T {}+ {F} = 0; {} = [E ]{}; {} = []{u}; {} = [L]T {}

Quantity Solid mechanics Heat conduction Seepage flow

Function Displacements {u} Temperature T Hydraulic head hField gradient Strains: {T} = []T {h} = []h

{} = []{u} where [] =[

x

y

z

]TConstitutive law Elasticity [E ] Conductivity [] Permeability []

Induced field Stresses: Fluxes: Velocities:

{} = [E ]{} {f} = []{T} {v} = []{h}Surface load Tractions: Normal flux: Normal velocity:

{} = [L]T {} fn = [L]T {f} vn = [L]T {v}Internal load Body forces: {F} Source: Q Source QEquilibrium []T {}+ {F} = 0 []T {T} Q = 0 []T {h} Q = 0

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