Chapter 2 Review of Elasticity

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Chapter 2. Review of Elasticity

1. Stress

Plane-Strain - The plane-strain distribution is based on the assumption that

0 z u z ∂ = =∂ (2-1)

where z represents the lengthwise direction of an elastic elongated body of constant cross section

subjected to uniform loading (in the case of uniaxial problems,/ 0

y y u∂ ∂ = =

).

xy, , x y

u v u v x y y x

ε ε γ ∂ ∂ ∂ ∂= = = +∂ ∂ ∂ ∂ (2-2)

are functions of x and y only and the strain components

xz yz, , z

w w u w v z x z y z

ε γ γ ∂ ∂ ∂ ∂ ∂= = + = +∂ ∂ ∂ ∂ ∂ (2-3)

vanish. The corresponding strain-stress relationship becomes

( )

( )

( )

11

11

2 1

x x y

y y x

xy xy xy

E

E

G E

µ ε µ σ µσ

µ ε µ σ µσ

τ µ γ τ

+ = − − + = − −

+= =

(2-4)

Hence,

( ) ( )29,000

11,1542 1 2 1 0.3

E G

µ = = =+ + (2-5)

1



The stress-strain relationship is

( ) ( ) ( )

1 0

1 01 1 2 0 0 1 2 / 2

x x

y y

xy xy

E σ µ µ ε σ µ µ ε

µ µ τ µ γ

− = −

+ − − (2-6)

( ) z x yσ µ σ σ = + (2-7)

Plane-Stress - The plane stress distribution is based on the assumption that

0 z yz xz σ τ τ = = = . (2-8)

The stress-strain relationship is

( )2

1 0

1 01

0 0 1 / 2

x x

y y

xy xy

E σ µ ε σ µ ε

µ τ µ γ

= − −

(2-9)

and the strain-stress relationship is

( )

1 01

1 0

0 0 2 1

x x

y y

xy xy

E

ε µ σ ε µ σ γ µ τ

− = − +

(2-10)

If we assume a state of plane stress, we need not have a corresponding plane strain and conversely,

if we assume a state of plane strain, we need not have a corresponding plane stress state. These are

entirely different assumptions.

Home Work #3

2



Using the generalized Hooke’s law, prove that the maximum value of Poisson’s ratio, µ , cannot

exceed 0.5.

For an analytical determination of the distribution of static or dynamic displacements and

stresses in a structure (body) under prescribed external loading and temperature, we must obtain

a solution to the basic equations of the theory of elasticity, satisfying the imposed boundary

conditions on forces and/or displacements.

For general 3-D structures, we have

6: strain-displacement relationships; ( ), , f u v wε =

6: stress-strain relationship, generalized Hooke’s law

3: Equations of equilibrium, linear momentum balance

15

Thus, there are 15 equations available for 15 unknowns, namely

3: displacements: u, v, w

6: stresses

6: strains

15

In the case of rigid body mechanics problems, we may use 6 equations of equilibrium

including three equations of linear momentum balance and three equations of angular momentum

3



balance. Since the three equations of angular momentum balance are used up in establishing the

symmetry of shear stresses ( ij jiτ τ =

), they cannot be used again along with the symmetry of

shear stresses.

If we include thermal strains ( T e T α = ), then the total strain becomes

ij ij ije T ε δ α = + (2-11)

where ije = total strain, ijε = elastic strain, and ijδ = Kronecker delta ( = 1 when i = j and = 0

when i j). The index i and j vary along x, y, z.

Practice #1

Write the generalized Hooke’s law for the stress and strain relationship for 3-D problems in a

matrix equation including the thermal terms.

Equations of equilibrium

4



Fig. 2-1

Parallelepiped used in the derivation of internal equilibrium equations. Note that equilibrium can

only be established by forces. Hence, each stress component must be multiplied by the projected

area on which it acts and the body force must be multiplied by the infinitesimal volume,

( ) ( ) ( )dv dx dy dz =

z

xx xx

dx x

σ σ

∂+

∂

xz xz dx

x

τ τ

∂+

∂

xz τ

x yτ

xxσ

x

X

dy

dz x y

xy xy dx

x

τ τ

∂+

∂

dx

5



xy

0

0x

0

xy x xz x

y yz y

yz xz z z

X x y z

X y z

X x y z

τ σ τ

τ σ τ

τ τ σ

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

(2-12)

Equation (2-12) is the three equations of linear momentum balance written in terms of stresses

and body forces. Frequently the weight of the body is represented as an equivalent externally

applied distributed load. Note only three equations of equilibrium (not six).

Compatibility Equations

The strains and displacements in an elastic body must be continuous, and this imposes the

condition of continuity on the derivatives of displacements and strains. This can be achieved by

eliminating the displacement terms ( iu or u, v, w) in the strain-displacement relationships.

2 22

2 2

2 22

2 2

2 22

2 2

2

2

2

1

2

12

12

y xy x

y yz z

x xz z

yz xy x xz

y yz xy xz

x

yz xy xz z

y x x y

z y y z

x z x z

y z x x y z

z y x y z

x y z x y z

ε γ ε

ε γ ε

ε γ ε

γ γ ε γ

ε γ γ γ

γ γ γ ε

∂ ∂∂ + =∂ ∂ ∂ ∂∂ ∂∂+ =∂ ∂ ∂ ∂

∂ ∂∂ + =∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂= − + + ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂∂= − + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂= + −∂ ∂ ∂ ∂ ∂ ∂

(2-13)

6



For 2-D Problems

We have 8 equations for 2 displacements (u,v), 3 stresses ( , , x y xyσ σ τ ) and 3 strains ( , , x y xyε ε γ )

for 3 strain-displacement relationships, 3 stress-strain relationships, and 2 equations of

equilibrium. For 2-D plane-stress problems ( 0 z xz yz σ τ τ = = = , 0 yz xz γ γ = = , and , , , x y z xyε ε ε γ ⇐

non-zero), equation (2-13) leads to

2 22

2 2 y xy x

y x x y

ε γ ε ∂ ∂∂ + =∂ ∂ ∂ ∂(2-14)

As general 3-D problems are so very complex, particularly problems with irregular

boundaries, that it is nearly impossible, if not impossible at all, to solve analytically. Therefore,

it is advisable to refer to digital computer oriented finite element methods.

In equation (2-12), , , x y z X X X are body forces. There are two important practical cases

of body forces, viz., gravitational force and centrifugal force. If we limit our discussion to

gravitational forces only, then

x x

y y

X g

X g

ρ ρ

== (2-15)

7



where ρ =mass density of the material and , x y g =gravitational acceleration in the x and y

directions. In general, for 2-D elasticity problems, we have 3 unknowns, , , x y xyσ σ τ with only 2

equations of equilibrium. Hence, we need one more equation which is provided by the

deformation compatibility condition.

(b) Strain (at a point)

8



Fig. 2-2

Taylor series

dydx

α vuA B

β θ B ′

udu dy y

∂=∂

B ′′

du

∂=

∂

vdv dy

y

∂=

∂

C

C

′

y

A ′

C′′

9



( )2 2

2 20 0 0 2 2

0 0 0 0

1 1,

2 2 f f f f

f x dx y dy f dx dy dx dy hot x y x y

∂ ∂ ∂ ∂ + + = + + + + + ∂ ∂ ∂ ∂

( ), (at B')

u v f u dx v u dx dx hot

x x

∂ ∂+ = + + +∂ ∂ (2-16)

( ), (at C')v u

f u v dy v dy dy hot y y∂ ∂+ = + + +∂ ∂ (2-17)

An engineering shear strain is defined by the total angular change between dx and dy. Hence,

tan

1

v vdx dx v x xu u x

dx dx dx x x

α α

∂ ∂∂∂ ∂= = =∂ ∂ ∂

+ + ∂ ∂

B

tan1

u udy dy

u y yv yvdy dy dy y y

β β

∂ ∂∂∂ ∂= = =∂ ∂ ∂+ + ∂ ∂

B

as or 1.0 xy

u v u v y x x y

γ ∂ ∂ ∂ ∂= + <<< ∂ ∂ ∂ ∂

(2-18)

Neglecting higher order terms, the small strains (otherwise known as the Cauchy strains)

become

' " x

A B AB udx x

ε − ∂= =∂ (2-19)

' " y

A C AC vdy y

ε − ∂= =∂ (2-20)

Taking successive partial derivatives of the above equations, we have

2 22 3 3 3 3

2 2 2 2 2 2, , and y xy x u v u v y x y x x y x y x y x y

ε γ ε ∂ ∂∂ ∂ ∂ ∂ ∂= = = +∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

10



Hence, the compatibility equation in 2-D problems in terms of strains is

2 22

2 2 0 y xy x

y x x y

ε γ ε ∂ ∂∂ + − ≡∂ ∂ ∂ ∂(2-14)

(c) Stress-Strain Relationships

The Hooke’s law in the linearly elastic materials is

i) 1-D

x x E

σ ε =

ii) 2-D

/ / x x y E E ε σ µσ = −

/ / y y x E E ε σ µσ = −

( )/ 2 1 / xy xy xyG E γ τ µ τ = = +

(d) Summary

(1) Equilibrium

0 xy x X x y

τ σ ∂∂ + + =∂ ∂

0 y xy Y y x

σ τ ∂ ∂+ + =∂ ∂

(2) Strain-Displacement

11



x

u x

ε ∂= ∂

y

v

yε ∂=

∂

xy

u v y x

γ ∂ ∂= +∂ ∂

(3) Stress-Strain

( )1 x x y E

ε σ µσ = −

( )1 y y x E

ε σ µσ = −

( )2 11 xy y xyG E

µ γ τ τ

+= =

Recall the compatibility equation (2-14) for 2-D elasticity problems.

2 22

2 20 xy y x

y x y x

γ ε ε ∂ ∂∂ − + =∂ ∂ ∂ ∂

Rewriting (2-14) in terms of stresses leads to

( ) ( ) ( )22 2

2 22 1 0 xy x y y x y x y x

τ σ µσ µ σ µσ

∂∂ ∂− − + + − =∂ ∂ ∂ ∂(2-21)

Recall the equations of equilibrium for 2-D problems given by

0 xy x x g

x y

τ σ ρ

∂∂ + + =∂ ∂

0 y xy y g

y x

σ τ ρ

∂ ∂+ + =∂ ∂

12



These equations are automatically satisfied if one introduces a stress function, ( ), x yφ such that

2 2 2

2 2, , and x y xy x x g y g x y x x y

φ φ φ σ σ τ ρ ρ ∂ ∂ ∂= = = − − −

∂ ∂ ∂ ∂(2-22)

If the gravitational forces in equation (2-22) may be neglected (or transform as externally applied

loads). This function was first introduced by G.B. Airy in 1862 and is called the Airy’s stress

function. If one substitutes Airy’s stress function given by equation (2-22) into equation (2-21),

one obtains

4 4 4

4 2 2 42 0 x x y y

φ φ φ ∂ ∂ ∂+ + =∂ ∂ ∂ ∂ (2-23)

Since

4 4 4

4 2 2 42 0 x x y y

φ φ φ ∂ ∂ ∂+ + =∂ ∂ ∂ ∂ ( )22 2

222 2 x y

∂ ∂= + = ∇ ∂ ∂ (2-24)

Equation (2-24) may be written as 4 0φ ∇ = . The operator 2 2

22 2 x y

∂ ∂∇ = +∂ ∂ is called Laplace

operator or harmonic operator and equation (2-24) is called a biharmonic equation.

Example

13



Fig. 2-3

Assumptions

(1) h<<d

(2) P >>> g ρ ⇐neglect the gravitational force.

From the static point of view, we see that the bending moment at any point along the beam axis

will be proportional to x and the stress component, xσ , at any point of the section will be

proportional to y. Accordingly, we shall assume a trial function such that

2

12 x c xy y

φ σ ∂= =∂ (a)

where 1c =constant. Upon integration, we find

( ) ( )311 26

c xy yf x f xφ = + + (b)

where ( ) ( )1 2and f x f x are unknown functions of x only. Substituting the above expression for

φ into equation (2-23) results in

( ) ( )1 231

6

f x f xc y y

x x xφ ∂ ∂∂ = + +∂ ∂ ∂

14

lh

d

P

y



( ) ( )2 221 2

2 2 2

f x f x y

x x xφ ∂ ∂∂ = +∂ ∂ ∂

( ) ( )3 331 2

3 3 3

f x f x y

x x x

φ ∂ ∂∂ = +∂ ∂ ∂

( ) ( )4 441 2

4 4 4

f x f x y

x x xφ ∂ ∂∂ = +∂ ∂ ∂

( )2112

c xy f x

yφ ∂ = +∂

2

12 c xy y

φ ∂ =∂

3

13 c x y

φ ∂ =∂

4

4 0 y

φ ∂ =∂

( )2121

2

f xc y

x y xφ ∂∂ = +∂ ∂ ∂

( )231

2 2

f x

x y xφ ∂∂ =∂ ∂ ∂

4

2 2 0 x y

φ ∂ =∂ ∂

Hence,

( ) ( )4 4

1 24 4 0d f x d f x y

dx dx+ = (c)

15



Since ( ) ( )1 2and f x f x are functions of x only, the second term in equation (c) is independent

from y. But equation (c) must be satisfied for all values of x and y in the cantilever beam. This

is possible only if

( ) ( )4 41 2

4 40 and 0d f x d f x

dx dx= =

or

( ) 3 21 2 3 4 5 f x c x c x c x c= + + + and

( ) 3 22 6 7 8 9 f x c x c x c x c= + + +

Thus, we have

( )3 3 2 3 212 3 4 5 6 7 8 96

c xy y c x c x c x c c x c x c x cφ = + + + + + + + +

Then

( ) ( )2

2 6 3 72 6 2 y c y c x c y c x

φ σ ∂= = + + +∂

221

2 3 43 22 xy

c y c y c x c

x yφ

τ ∂= − = − − − −∂ ∂

The boundary condition requires that

0 at / 2 y y d σ = = ±

Hence,HH

16



( ) ( )2 6 3 76 / 2 2 / 2 0c d c x c d c+ + + = and

( ) ( )2 6 3 76 / 2 2 / 2 0c d c x c d c− + + − + =

These conditions must be satisfied for all x between 0 and l ; it follows that

2 6 3 7 2 6 3 7/ 2 0, / 2 0, / 2 0, / 2 0c d c c d c c d c c d c+ = + = − + = − + =

Solving these 4 equations simultaneously, we have 2 3 6 7 0c c c c= = = = .

Thus,

21

42 xy

c y cτ

= − −

To satisfy the condition that 0 on / 2 xy y d τ = = ± (no shear stress on free surface), we must have

221

4 4 108 8c d

d c c c− − = ⇒ = −

Hence,

2 21 1

2 8 xy

c c y d τ = − +

On the loaded edge of the beam, the sum of the distributed shearing stress must be equal to P .

Hence,

( )/2 /2 2 21

/2 /24

8

d d

xyd d

c hhdy y d dy P τ

− −− = − =∫ ∫ ⇒ 1 3

12 P c

hd = −==

17



Consider the reason why a negative sign is placed in front of the shear stress xyτ in the above

equation. By a consistent sign convention, the shear stress in the positive direction at the

negative face is considered as negative as shown in the sketch.

Recall the moment of inertia of a rectangular section,3

12hd

I = . The expression for the stress

components are, therefore,

22, 0,

2 4 x y xy

My Pxy P d y

I I I σ σ τ

= − = − ≡ = − −

This coincides exactly with the elementary solution. From this solution, we see that the shearing

stress distribution is parabolic and xσ varies linearly. If we introduce other boundary conditions

than those used, other solution is also feasible (recall the trial function φ ).

Recalling the strain-displacement relationships and the 2-D Hooke’s law, we have

( ) ( ) 222 1 1

4

x x

y x y

xy xy

u Pxy x E EI v Pxy

y E E EI

u v d y

y x E EI

σ ε

σ µσ µ ε

µ τ µ γ

∂ = = = −∂∂

= = = − =∂+ + ∂ ∂+ = = = − − ∂ ∂

. (d)

Integrating the first two equations, we have

negativefacef f

18



( )

( )

21

22

2

2

P u x y g y

EI P

v xy g x EI

µ

= − +

= +(e)

Substituting equation (e) into the 3rd equation of (d), yields

( ) ( ) ( )1 22 2 211

2 2 4

dg y dg x P P y x d

dy EI dx EI EI

µ µ + − + = − + −

We note that the terms on the left of the equal sign are functions of y only and the terms on the

right of the equal sign are functions of x only. A function of x can be equal to a function of y for

all x and y only if they are both equal to a constant, say, 1a . Thus,

( )1 211

2

dg y P y a

dy EI µ = + +

( ) ( )2 2 211

2 4dg x P P x d a

dx EI EI µ += − −

Integrating, we have

( ) 31 1 21

3 2 P

g y y a y a EI

µ = + + +

( ) ( )3 22 1 3

1

6 4

P P g x x d x a x a

EI EI

µ += − − +

Hence,

2 31 21

2 3 2 P P

u x y y a y a EI EI

µ = − + + + +

19



( )2 3 21 3

1

2 6 4

P P P v xy x d x a x a

EI EI EI

µ µ += + − − +

The integral constants, 1 2 3, ,a a a , can be determined by applying three boundary conditions at

the fixed support of the cantilever (at x= l ). They are

0

0

0

u

v

v x

==

∂ =∂

(3 boundary conditions)

Hence,

( ) 22

1

2

3

3

1

2 40

3

Pd P a

EI EI a

P a

EI

µ += −

=

=

l

l

and

( )2

2 3 21 12 3 2 2 P P d

u x y y y EI EI

µ µ

= − + + + − + l

2 32 3

2 6 2 3 P P P P

v xy x x EI EI EI EI

µ = + − +l l

From the elementary beam theory, the deflection curve ( ( )0v y = ) is given by

3 2 3

0 6 2 3 y

Px P x P v EI EI EI = = − +

l l

checks exactly!

The solution of a 2-D problem subjected to inplane forces may be dependent upon how to

choose a correct trial stress function, φ , that can lead to a guess work. ⇐ haphazard!!

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Chapter 2 Review of Elasticity