Chapter 5 Stress–Strain Relation

55
1 /61 Chapter 5 Stress–Strain Relation

Transcript of Chapter 5 Stress–Strain Relation

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Chapter 5

Stress–Strain Relation

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Chapter 5 Stress–Strain Relation

Contents

5.1 General Stress–Strain System

5.2 Relations Between Stress and Strain for Elastic Solids

5.3 Relations Between Stress and Rate of Strain for Newtonian Fluids

Objectives

- Understand tensor systems of stress and strain

- Study difference between displacement and deformation

- Study solid mechanics to deduce stress-rate of strain relations for fluid

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5.1 General Stress–Strain System

Parallelepiped, cube → infinitesimal C.V.

5.1.1 Surface Stress

Surface stresses: normal stress – σx

shear stress – τxy, τxz

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5.1 General Stress–Strain System

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5.1 General Stress–Strain System

0lim

( )x

xxx x A

x

x

FA

A y z

σ σ→

∆= =

∆∆ = ∆ ∆

0lim

x

yxy A

x

FA

τ∆ →

∆=

∆ 0lim

x

zxz A

x

FA

τ∆ →

∆=

0lim

( )y

xyx A

y

y

FA

A x z

τ∆ →

∆=

∆ = ∆ ∆0

limy

yyy y A

y

FA

σ σ→

∆= =

0lim

y

zyz A

y

FA

τ∆ →

∆=

0lim

( )z

xzx A

z

z

FA

A x y

τ∆ →

∆=

∆∆ = ∆ ∆

0lim

z

yzy A

z

FA

τ∆ →

∆=

∆ 0lim

z

zzz z A

z

FA

σ σ∆ →

∆= =

, ,x y zF F F∆ ∆ ∆ F∆

where = component of force vector

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σx : subscript indicates the direction of stress

τxy : 1st - direction of the normal to the face on which τ acts

2nd - direction in which τ acts

5.1 General Stress–Strain System

• subscripts

• general stress system: stress tensor

~ 9 scalar components

xx xy xz

yx yy yz

zx zy zz

σ τ ττ σ ττ τ σ

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5.1 General Stress–Strain System

[Re] Tensor

~ an ordered array of entities which is invariant under

coordinate transformation; includes scalars & vectors

~ 3n

0th order – 1 component, scalar (mass, length, pressure, energy)

1st order – 3 components, vector (velocity, force, acceleration)

2nd order – 9 components (stress, rate of strain, turbulent diffusion)

At three other surfaces,

' xx x x

xσσ σ ∂

= + ∆∂

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5.1 General Stress–Strain System

' yy y y

σ σ∂

= + ∆∂

' zz z z

zσσ σ ∂

= + ∆∂

' xyxy xy x

τ τ∂

= + ∆∂

' yxyx yx y

τ τ∂

= + ∆∂

' zxzx zx z

zττ τ ∂

= + ∆∂ (5.1)

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5.1 General Stress–Strain System

◈ Shear stress is symmetric.

→ Shear stress pairs with subscripts differing in order are equal.

→ τxy = τyx

[Proof]

In static equilibrium, sum of all moments and sum of all forces equal zero for

the element.

First, apply Newton's 2nd lawduF mdt

=∑

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where I = moment of inertia = r2mr = radius of gyration

= angular acceleration

5.1 General Stress–Strain System

Then, consider torque (angular momentum), T

2( ) ( ) ( )d d d dT rmu r m I Idt dt dt dt

ωω ω= = = =∑

ddtω

Thus,

2 dT mrdtω

=∑ (A)

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5.1 General Stress–Strain System

Now, take a moment about a centroid axis in the z-direction

( ) ( ) ( )2 2 2xy yx xy yxx y x y zy z x zLHS T τ τ τ τ∆ ∆ ∆ ∆ ∆∆ ∆ ∆ ∆= = − = −∑

2 2d dRHS dvolr x y z rdt dtω ωρ ρ= = ∆ ∆ ∆

( ) 22xy yxdx y z x y z rdtωτ τ ρ−∴ ∆ ∆ ∆ = ∆ ∆ ∆

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5.1 General Stress–Strain System

After canceling terms, this gives

22xy yxdrdtωτ τ ρ− =

2

, , 0lim 0

x y zr

∆ ∆ ∆ →→

0xy yxτ τ− =

xy yxτ τ∴ =

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5.1 General Stress–Strain System

[Homework Assignment-Special work]

Due: 1 week from today

1. Fabricate your own “Stress Cube” using paper box.

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5.1 General Stress–Strain System

5.1.2 Strain componentsMotion and deformation of fluid element

translation (displacement)

Motion

rotation

linear deformation

Deformation

angular deformation

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5.1 General Stress–Strain System

(1) Motion

1) Translation: ξ, η

, du dt udtξξ = =

, dv dt vdtηη = =

2) Rotation ← Shear flow

12x

v ux y

ω ∂ ∂

= − ∂ ∂

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5.1 General Stress–Strain System

(2) Deformation

1) Linear deformation – normal strain

x xξε ∂

=∂

y yηε ∂

=∂

i) For compressible fluid, changes in temperature or pressure cause

change in volume.

ii) For incompressible fluid, if length in 2-D increases, then length in

another 1-D decreases in order to make total volume unchanged.

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○ Strain normal strain: ε ← linear deformation

shear strain: γ ← angular deformation

5.1 General Stress–Strain System

2) Angular deformation– shear strain

xy x yη ξγ ∂ ∂

= +∂ ∂

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5.1 General Stress–Strain System

Consider a small element OABC

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i) Displacement (translation): ξ, η, ζ

ii) Deformation: due to system of external forces

(1) Deformation

Normal strain, ε

5.1 General Stress–Strain System

( ) ( )' , ,, ,O O x y zx y z ξ η ζ→ + + +

' ' ' 'OABC O A B C→

change in lengthoriginal length

ε =

( ), ,

' , ,

C x x y y z z

C x x x y x z xx x xξ η ζξ η ζ

→+ ∆ + ∆ + ∆

∂ ∂ ∂ + ∆ + + ∆ + + ∆ + + ∆ ∂ ∂ ∂

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~ ε is positive when element elongates

under deformation

5.1 General Stress–Strain System

( )0 0

O'C'lim limx x x

xxx x xOC x

OC x x

ξ ξξξε

∆ → ∆ →

∂ − − ∆++ ∆ + + ∆ − ∂∂ = = =∆ ∂

( )

0 0

O'A'lim limy y y

y y y yyyOA

OA y y

ηη ηηε

∆ → ∆ →

∂ + ∆ + + ∆ − − ∆+ ∂− ∂ = = =∆ ∂

z zζε ∂

=∂

' 'O C OC

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2) Shear strain, γ

~ change in angle between two originally

perpendicular elements

5.1 General Stress–Strain System

xyFor –plane

( ) ( ), 0 , 0lim lim tan tanxy c A c Ax y x y

γ θ θ θ θ∆ ∆ → ∆ ∆ →

= + ≅ +

, 0limx y

xx

x xx

η

ξ∆ ∆ →

∂ ∆∂=

∂∆ + ∆∂

yy

y yy

ξ

η

∂ ∆∂+

∂∆ + ∆∂

x yη ξ

∂ ∂ = + ∂ ∂

x xxξ∂ ∴ ∆ < ∆

C’D A’E

O’D O’E

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5.1 General Stress–Strain System

(5.2)

(5.4)

(5.3)

xy x yη ξγ ∂ ∂

= +∂ ∂

yz y zζ ηγ ∂ ∂

= +∂ ∂

zx z xξ ζγ ∂ ∂

= +∂ ∂

(2) displacement vector δ

i j kδ ξ η ζ= + +

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5.1 General Stress–Strain System

(3) Volume dilation

change of volume of deformed elementoriginal volume

e =

( )y y x y zx x z z

yd x zVeV x y z

ηξ ζ∂ ∂ ∂ ∆ + ∆ − ∆ ∆ ∆∆ + ∆ ∆ + ∆ ∂∂ ∂ ∆ = =∆ ∆ ∆ ∆

x y zx y zξ η ζ ε ε ε∂ ∂ ∂

≅ + + = + +∂ ∂ ∂

x y ze ε ε ε= + +

ex y zξ η ζ δ∂ ∂ ∂

= + + = ∇ ⋅∂ ∂ ∂

--- divergence

(5.5)

(5.6)

(5.7)

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in which E = Young's modulus of elasticity

= elongation in the x- dir due to normal stress, σx

5.2 Relations between Stress and Strain for Elastic Solids

5.2.1 Normal StressesHooke's law: Stress is linear with strain.

x xEσ ε=

1x xE

ε σ= (5.8)

. : yyy dir

ε− =

. : zzz dir

Eσε− =

x xσ ε∝

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where n = Poisson's ratio

5.2 Relations between Stress and Strain for Elastic Solids

Now, we have to consider other elongations because of lateral contraction of

matter under tension.

'xε .x dir− yσ''xε .x dir− zσ

= elongation in the due to

= elongation in the due to

Now, define

(5.9)' yx yn n

ε ε= − = −

'' zx zn n

Eσε ε= − = −

(5.10)

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5.2 Relations between Stress and Strain for Elastic Solids

Poisson's ratio: n = 0~0.5

water~0.5; metal~0.3

cork: Poisson's ratio is high.

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Thus, total strain εx is

5.2 Relations between Stress and Strain for Elastic Solids

(5.11)

(5.12)

( ) ( )1' '' xy z y zx x x x x

n nE E E

σ σ σ σ σε ε ε ε σ+ + = + + = − = −

( )1y z xy n

Eσ σ σε − +=

( )1x yz z n

Eσ σε σ + = −

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5.2 Relations between Stress and Strain for Elastic Solids

5.2.2 Shear Stress

~ Hooke’s law τxy = G γxy

xyxy G x y

τ η ξγ ∂ ∂= = +

∂ ∂

yzyz G y z

τ ζ ηγ ∂ ∂= = +

∂ ∂

zxzx G z x

τ ξ ζγ ∂ ∂= = +

∂ ∂ (5.13)

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where G = shear modulus of elasticity

5.2 Relations between Stress and Strain for Elastic Solids

( )2 1EG

n=

+(5.14)

■ Volume dialation

( )

( )

( )

1

1

1

y zx y z x

y z x

x yz

e nE

nE

nE

σ σε ε ε σ

σ σ σ

σ σσ

+ = + + = −

− ++

+ + −

( )( )11 2 x y znE

σ σ σ+ + = − (5.15)

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■ = arithmetic mean of 3 normal stresses

Combine Eqs. (5.12), (5.14) and (5.15)

5.2 Relations between Stress and Strain for Elastic Solids

σ

( )13 x y zσ σ σσ + += (5.16)

21 2x x

neGn

σ ε = + − (5.17)

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5.2 Relations between Stress and Strain for Elastic Solids

Therefore

23x xeGσ σ ε − = −

23y yeGσ σ ε − = −

23z zeGσ σ ε − = −

(5.18)

xy yx Gx yη ξ

τ τ∂ ∂ += = ∂ ∂

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5.2 Relations between Stress and Strain for Elastic Solids

zy yz Gy zζ η

τ τ∂ ∂ += = ∂ ∂

xz zx Gz xξ ζτ τ ∂ ∂ = = +

∂ ∂ (5.19)

[Proof] Derivation of Eqs. (5.17) & (5.18)

(A)(5.15) →

(5.12) →

(5.14) →

( ) ( )11 2 x y znE

e σ σ σ+ +−=

( )1y zx x n

Eσ σε σ + = −

2 (1 )2(1 )

EG E G nn

= → = ++

(C)

(B)

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5.2 Relations between Stress and Strain for Elastic Solids

i) Combine (A) and (B)

( ) ( )( ) ( ) ( )

( )

1 21 2 1 2

1

x y z x y z

y zx x

n n nneE En n

nE

σ σ σ σ σ σ

σ σε σ

− + + + +× = =+ −+

+ = −

( )1

1 2 x xn ne

Enε σ+

+ =−

( )1 1 2xxnE e

n nεσ +∴ = + −

(D)

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5.2 Relations between Stress and Strain for Elastic Solids

Substitute (C) into (D)

( )2

1 2xxn eG

nεσ +∴ = −

→ Eq. (5.17)

ii) Subtract (5.16) from (5.17)

( ) ( )1231 2 x y zxx

n eGn

σ σ σεσ σ + ++− = − − (E)

Substitute (A) into (E); ( )

( ) :1 2x y z

EA en

σ σ σ+ + =−

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5.2 Relations between Stress and Strain for Elastic Solids

( ) ( )1( ) 231 2 1 2

xn EeRHS of E G e

n nε +∴ = − − −

( )( )

( ) ( ) ( )

12 1 2 12 2 331 2 1 2 1 2 1 2x x

nGn G n nG e G en n n n

ε ε + + −= + = − − − − +

( )

( )

11 22 3

1 2x

nG en

ε

− −= + −

123xG eε = −

→ Eq. (5.18)

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36/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids

Experimental evidence suggests that, in fluid, stress is linear with time rate

of strain.

( )stress straint

∂∝

∂→ Newtonian fluid (Newton's law of viscosity)

[Cf] For solid,

stress strain∝

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5.3.1 Normal stressFor solid, Eq. (5.18) can be used as

Hookeian elastic solid:2

23x xeF

Lσ σ ε − = −

GBy analogy,

Newtonian fluid: 2

23x xeFt

tLσ σ ε

∂ − = − ∂

Time rate of strain

2FtL

µ ≡

(5.20)

Now set = dynamic viscosity

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Then,

223

xx

et t

εσ σ µ µ∂ ∂− = −

∂ ∂ (5.21)

By the way,

,x exξε δ∂

= = ∇⋅∂

Therefore,

x ut t x xx t

ε ξ ξ∂ ∂ ∂ ∂∂ ∂ = = = ∂ ∂ ∂ ∂∂ ∂ (5.22)

, , ( , , displacement)u v wt x tξ η ζ ξ η ζ∂ ∂ ∂

= = = =∂ ∂ ∂

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e u v wqt t x y z

δ∂ ∂ ∂ ∂ ∂= ∇ ⋅ = ∇ ⋅ = + +

∂ ∂ ∂ ∂ ∂

(5.23)

i j kδ ξ η ζ= + +

q ui v j wktδ∂

= = + +∂

u v wqx y z

∂ ∂ ∂∇ ⋅ = + +

∂ ∂ ∂

Eq. (5.21) becomes

( )223x

uqx

σ σ µ µ∂= + − ∇ ⋅

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For compressible fluid,

( )223x

uqx

σ σ µ µ∂= + − ∇ ⋅

( )223y

vqy

σ σ µ µ∂= + − ∇ ⋅

( )223z

wqz

σ σ µ µ∂= + − ∇ ⋅

(5.24)

For incompressible fluid,

0de qdt

= ∇ ⋅ =

← time rate of volume expansion=0

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0q→ ∇ ⋅ =

→ Continuity Eq.

Therefore, Eq. (5.24) becomes

2xux

σ σ µ ∂= +

2yvy

σ σ µ ∂= +

2zwz

σ σ µ ∂= +

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5.3.2. Shear stressBy following the same analogy

2xyFtG

x y x ytLη ξ η ξ

τ∂ ∂ ∂ ∂∂ + += = ∂ ∂ ∂ ∂∂

v ux yx t y t

η ξµ µ∂ ∂∂ ∂ ∂ ∂ += + = ∂ ∂∂ ∂ ∂ ∂

µ

vtη∂

=∂ u

tξ∂

=∂

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xy yx

zy yz

xz zx

v ux y

w vy z

u wz x

τ τ µ

τ τ µ

τ τ µ

∂ ∂ += = ∂ ∂ ∂ ∂ += = ∂ ∂ ∂ ∂ = = + ∂ ∂

(5.25)

[Appendix 1]

xy yxv ux y

τ τ µ∂ ∂ += = ∂ ∂

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, 'xy xyτ τⅰ)

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ⅱ) , 'yx yxτ τ

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ⅲ) composition

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1) Assume viscous effects are completely represented by the viscosity µ for

incompressible fluid

▪ Relation between thermodynamic pressure p and mean normal stress

~ minus sign accounts for pressure (compression)

2) For compressible fluid

in which µ ‘ = 2nd coefficient of viscosity associated solely with dilation

= bulk viscosity

5.3 Relations between Stress and Rate of Strain for Newtonian Fluids

σ

( )13 x y zp σ σ σσ + += − = (5.26)

( )'p qσ µ= − + ∇ ⋅

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Since, dilation effect is small for most cases

For zero-dilation viscosity effects ( µ ‘ = 0), (5.24) becomes

5.3 Relations between Stress and Rate of Strain for Newtonian Fluids

( )' 0qµ →∇ ⋅

pσ∴ = −

( )223x

up qxσ µ µ∂ = − + − ∇ ⋅ ∂

( )223y

vp qyσ µ µ∂ = − + − ∇ ⋅ ∂

( )223z

wp qzσ µ µ∂ = − + − ∇ ⋅ ∂

Normalstress pressure

Viscouseffects

(5.29)

(5.28)

(5.27)

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For zero viscous effects ( µ = 0) → inviscid fluids in motion and for all fluids

at rest

5.3 Relations between Stress and Rate of Strain for Newtonian Fluids

■ Shear stresses in a real fluid

xy yxv ux y

τ τ µ∂ ∂ += = ∂ ∂

yz zyw vy z

τ τ µ∂ ∂ += = ∂ ∂

xz zxu wz x

τ τ µ ∂ ∂ = = + ∂ ∂

(5.30)

x y z pσ σ σ σ= = = = −

0xy yz zxτ τ τ= = =

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[Appendix 2] Bulk viscosity and thermodynamic pressure

→ Boundary-Layer Theory (Schlichting, 1979) pp. 61-63

If fluid is compressed, expanded or made to oscillate at a finite rate, work

done in a thermodynamically reversible process per unit volume is

~ dissipation of energy

5.3 Relations between Stress and Rate of Strain for Newtonian Fluids

( )'p qσ µ= − + ∇ ⋅

deW p q pdt

= ∇ ⋅ =

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where µ ‘ = bulk viscosity of fluid that represents that property which is

responsible for energy dissipation in a fluid of uniform temperature during

a change in volume at a finite rate

= second property of a compressible, isotropic, Newtonian fluid

[Cf] µ = shear viscosity = first property

Direct measurement of bulk viscosity is very difficult.

5.3 Relations between Stress and Rate of Strain for Newtonian Fluids

' 0 , pµ σ= = −

' 0 , pµ σ≠ ≠ −

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[Appendix 3] Normal stress

Normal stress = pressure + deviation from it

'x xpσ σ= − +

'y ypσ σ= − +

'z zpσ σ= − +

Thus, stress matrix becomes

'0 0'0 0

'0 0

x xy xz

yx y yz

zx zy z

pp

p

σ τ ττ σ ττ τ σ

− +− −

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where λ = compressibility coefficient

5.3 Relations between Stress and Rate of Strain for Newtonian Fluids

Normal stresses are proportional to the volume change (compressibility)

and corresponding components of linear deformation, a, b, c.

Thus,

( )

( )

( )

2

2

2

x

y

z

p aa b c

p ba b c

p ca b c

σ λ µ

σ λ µ

σ λ µ

= − + ++ +

= − + ++ +

= − + ++ +

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55/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids

1. (5-3) Consider a fluid element under a general state of stress as illustrated in

Fig. 5-1. Given that the element is in a gravity field, show that the equilibrium

requirement between surface, body and inertial forces leads to the equations

yxx zxx xg a

x y zτσ τ ρ ρ

∂∂ ∂+ + + =

∂ ∂ ∂xy y xy

y yg ax y x

τ σ τρ ρ

∂ ∂ ∂+ + + =

∂ ∂ ∂yzxz z

xg ax y z

ττ σ ρ ρ∂∂ ∂

+ + + =∂ ∂ ∂

Homework Assignment # 5

Due: 1 week from today