dz

dr r

θd

A

B

C

D

H

(1)

(2)

(3) (4)

(5)

(6) G

E

F

ES128: Homework 3 Solutions

Problem 1 Derive the equilibrium equations and the relations between strains

( rrε , θθε , zzε , θεr , rzε , zθε ) and displacements in cylindrical coordinates with

θcos1 rx = , θsin2 rx = , and zx =3 .

Solution

Fig. 1 shows an element ABCDEFGH of with 6 surfaces. In surface (1) (surface ABCD), the stresses in r, θ , and z directions are

2

)1( dr

rr

rr ∂

∂−=

σσσ ,

2

)1( dr

rr

rr ∂

∂−= θ

θθ

σσσ ,

2

)1( dr

rrz

rzrz ∂

∂−=

σσσ .

On surface (2) (surface BCFG), the stresses inθ , r, and z directions are

2

)2( θθσ

σσ θθθ

d

∂

∂−= ,

2

)2( θθσ

σσ θθθ

drrr ∂

∂−= ,

2

)2( θθσ

σσ θθθ

dzzz ∂

∂−= .

On surface (3) (surface EFGH), the stresses in r, θ , and z directions are

Fig 1

r

z θ

2

)3( dr

rr

rr ∂

∂+=

σσσ ,

2

)3( dr

rr

rr ∂

∂+= θ

θθ

σσσ ,

2

)3( dr

rrz

rzrz ∂

∂+=

σσσ .

On surface (4) (surface AEHD), the stresses in θ , r, and z directions are

2

)4( θθσ

σσ θθθ

d

∂

∂+= ,

2

)4( θθσ

σσ θθθ

drrr ∂

∂+= ,

2

)4( θθσ

σσ θθθ

dzzz ∂

∂+= .

On surface (5) (surface AEFB), the stresses in z, r, and θ directions are

2

)5( dz

zz

zz ∂

∂+=

σσσ ,

2

)5( dz

zzr

zrzr ∂

∂+=

σσσ ,

2

)5( dz

zzr

zz ∂

∂+=

σσσ θθ .

On surface (6) (surface DHGC), the stresses in z, r, and θ directions are

2

)6( dz

zz

zz ∂

∂−=

σσσ ,

2

)6( dz

zzr

zrzr ∂

∂−=

σσσ ,

2

)6( dz

zzr

zz ∂

∂−=

σσσ θθ .

The conditions that the momentums are zero yield that

rr θθ σσ = ; θθ σσ zz = ; zrra σσ = .

In the r direction, the resultant force is zero. Thus

θσ

σθσ

σ ddr

rdzdr

rd

drrdz

dr

rr

rr

r

−

∂

∂−−

+

∂

∂+

2222

∂

∂−−

∂

∂++

2cos

22cos

2

θθθσ

σθθ

θσ

σ θθ

θθ dzdr

ddzdr

d rr

rr

∂

∂−−

∂

∂+−

2sin

22sin

2

θθθσ

σθθ

θσ

σ θθ

θθ dzdr

ddzdr

d

drrddz

zdrrd

dz

zzr

zrzr

zr θσ

σθσ

σ

∂

∂−−

∂

∂++

22=0.

Since ( ) 2/2/sin θθ ≈ and ( ) 12/cos ≈θ , we obtain

01

=−

+∂

∂+

∂

∂+

∂

∂

rzrrrzrrr θθ σσσ

θσσ

.

Similarly, the condition that the resultant force is zero in the θ direction gives

021

=+∂

∂+

∂

∂+

∂

∂

rzrrrzr θθθθ σσ

θσσ

.

The condition that the resultant force is zero in the z direction gives

01

=+∂

∂+

∂

∂+

∂

∂

rzrrrzzzrz σσ

θσσ θ .

Next, let us analyze the relations between strains and displacements in cylindrical coordinates

Problem 2 Elastic bulk modulus B is defined as the stiffness under hydrostatic pressure, namely,

V

VBp∆

−=

Here ∆V / V is the volumetric strain caused by the hydrostatic pressure p. (a) Show that the volumetric strain ∆V / V relates to the axial strains as

zzyyxxV

Vεεε ++=

∆.

(b) Express the bulk modulus B in terms of Young’s modulus and Poisson’s ratio.

(c) What is the value of Poisson’s ratio when the material is incompressible? Interpret your result under uniaxial stress state.

(d) Show that Poisson’s ratio must be smaller than ½. What would happen if Poisson’s ratio were greater than ½?

Solution (a) Assume that the original dimensions are X Y Z in x, y, and z directions, respectively. After deformation, the current dimensions are X+∆X, Y+∆Y, and Z+∆Z, respectively. The volumetric strain

XYZ

XYZZZYYXX

V

V −∆+∆+∆+=

∆ ))()((,

where ∆X, ∆Y, ∆Z are assumed to be small, compared to X, Y, Z, respectively. We only keep the linear terms in the above equation

zzyyxxZ

Z

Y

Y

X

X

XYZ

XYZYXZZXY

V

Vεεε ++=

∆+

∆+

∆=

∆+∆+∆=

∆

(b) With Hooke’s law ( )( ),1zyxx v

Eσσσε +−= ( )( ),1

zxyy vE

σσσε +−= and

( )( ),1yxzz v

Eσσσε +−= we obtain that

)(21

zyxzyxE

vσσσεεε ++

−=++ ,

Under hydrostatic pressure, pzyx −=== σσσ , thus,

BppE

v

V

Vzyx /)3(

21−=−

−=++=

∆εεε

)21(3 v

EB

−=

(c) When v=0.5, the material is incompressible. For uniaxial stress state,

xxEσε

1= ; xxzy

EE

vσσεε

2

1−=−== . Thus, 0=++=

∆zyx

V

Vεεε .

(d) Since )3(21

pE

v

V

V−

−=

∆. Thus v must be smaller than 1/2. If v is greater than

1/2, with hydrostatic pressure, the volume of the body can increase.

Problem 3 Stress concentration at geometric discontinuities is the most important practical result in elasticity theory. For example, for a small circular hole in a large plate under uniaxial stress S, the elasticity solution gives the hoop stress around the hole:

( )θσ θθ 2cos21 −= S

Here the polar angle θ is measured from the loading direction. The problem is solved in many elasticity textbooks.

a) Under uniaxial tension, indicate the highest tensile stress around the hole. b) Under uniaxial compression, indicate the highest tensile stress around the

hole. c) Calculate the stress concentration at the hole when the plate is under a

pure shear stress. Use the above solution and linear superposition.

Solution (a) Under uniaxial tension, the tensile stress is the highest at 2/πθ = ( S3=θθσ ).

(b) Under uniaxial compression, the tensile stress is the highest at 0=θ

( S−=θθσ ).

(c)

In Fig. 3, case 1 is exactly same as case 2 (pure shear). The reason is as follows. Take a triangle free body from case 1 as shown if Fig.4.

θ

Fig. 3

o45

S

S

S

A

B C

Fig. 4

Based on the force balance, it is easy to show that only the shear stress S is applied on the surface AC, and the normal stress is zero. Thus case 2 can be reduced to an easier problem of case 1. Based on linear superposition,

θπθθσ θθ 2cos4))2/(2cos21()2cos21( SSS −=+−−−= ,

where θ is shown in Fig.3. The tensile stress is the highest at 2/πθ = ( S4=θθσ ).

Problem 4 An elastic layer is sandwiched between two perfectly rigid plates, to which it is

bonded. The layer is compressed between the plates, the direct stress being zσ .

Supposing that the attachment to the plates prevents lateral strain xε , yε

completely, find the apparent Young’s modulus (that is zz εσ / ) in terns of E and v.

Show that it is many times E if the material of the layer has a Poisson’s ratio only slightly less than 0.5, e.g., rubber.

Solution The shear stresses vanish, but all the three axial stresses xσ , yσ , zσ are nonzero.

By symmetry, we note that

xσ = yσ Because the elastic layer is bonded to the rigid plate, the two components of strain vanish:

0== yx εε

That is, the elastic layer is in a state of uniaxial strain: 0≠zε . Using Hooke’s law,

we obtain that

( )zyxx vvE

σσσε −−==1

0

or

zxv

vσσ

−=1

Using Hooke’s law again, we obtain that

( )( ) zyxzzEv

vvv

Eσσσσε

)1(

)21)(1(1

−−+

=+−=

The apparent Young’s modulus

)21)(1(

)1(

vv

Ev

z

z

−+−

=εσ

.

When v is only slightly less than 0.5, e.g., for rubbers, the apparent Young’s modulus is many times E.