mechanics of solids week 12 lectures

8/12/2019 mechanics of solids week 12 lectures

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Week 12 - Review 2013 (Part 1)

2

Cylindrical Pressure Vessels

Axial Stress(or Longitudinal Stress):

t

Prxx

2

Hoop Stress (or Cir cumferential Stress - higher):

tyy

Pr

Example 2For a thin-walled pressurised vessel with internal pressureP, give the equation for von Mises and

Tresca failure criteria, respectively. Determine which is more critical.

Soln:Principal stresses: Since there is no shear stress in the infinitesimal element, the direct

stressed are the principal stresses: 2Pr

1 t

yy , t

xx2

Pr2 , 03

Tresca criterion: Y 20231 or: YY 5.02/ von Mises criterion:

Y 3)20()0()2(2

1)()()(

2

1 222213

232

221

YY 5774.03/ YY 5774.05.0 , Therefore, Tresca criterion is more critical.

Thus: ,22

Pr Y

t

r

tYY

r

tP

2

2(note vm criteria: ,

2

Pr33 Y

t

r

tYP

3

2 )

Stress in any direction (3D)

Knowing direction cosines of a plane (l,m,n):

n

m

l

zzzyzx

yzyyyx

xzxyxx

nz

ny

nx

3D Stresses with coordinate rotation

Toldnew RR

T

zzzyzx

yzyyyx

xzxyxx

ttttnt

ttttnt

ntntnn

nml

nml

nml

nml

nml

nml

''''''

'''

''''''

'''

'''''''''

''''''

'''

Example 3: As in Example1, rotate the coordinate from left to right configuration

Soln:Right coordinate is rotated 180 about y-axis: The direction cosines are:

''''''

'''

00)180cos(

''''''

'''

nml

nml

nml

nml

nml

R

Toldnew RR

302515

25205

15510

''''''

'''

302515

25205

15510

''''''

''''''

T

T

nml

nml

nml

nml

nml

nml

RR

xxP

t Sectioned plane

r

x

y

xx

yy


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3

Principal stresses and maximum shear stresses in 3D

p

p

p

p

p

p

zzzyzx

yzyyyx

xzxyxx

n

m

l

n

m

l

0

p

p

p

zzzyzx

yzyyyx

xzxyxx

n

m

l

To have non-vanish solution, the coefficient determinant must be zero, i.e.

0

zzzyzx

yzyyyx

xzxyxx

or 0322

13 III

These three roots are the principal stresses: 1, 2, 3: ( 321 )

Stress Invariants

First invariant: zzyyxxI 1

Second invariant: 2222 )()()( zxyzxyxxzzzzyyyyxxI

Third invariant: 222

3 2 xyzzzxyyyzxxzxyzxyzzyyxxI Principal direction cosines

1

0

222

nml

n

m

l

p

p

p

zzzyzx

yzyyyx

xzxyxx

e.g.

1

0

0

21

21

21

1121

1111

nml

nml

nml

yzyyyx

xzyxxx

Example 4: In a plane stress problem, the stress components are MPaxx 10 ,

MPayy 20 , MPaxy 10 . Determine the stress invariants and principal stresses.

Soln: Stress tensor in a plane stress status:

000

0

0

yyxy

xyxx

3020101 yyxxzzyyxxI

100102010)()()()( 22222

2 xyyyxxzxyzxyxxzzzzyyyyxxI

02 2223 xyzzzxyyyzxxzxyzxyzzyyxxI 0)10030(010030 232332

21

3 III

82.3

18.26

2

10014)30()30( 2

0 Rank them and one can thus obtain: 0,82.3,18.26 321

Equilibrium Equation of Motion

Cartesian: 3D

zzzzzyzx

yyyzyyyx

xxxzxyxx

afzyx

afzyx

afzyx

2D

0

0

yx

yx

yyyx

xyxx


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4

Cylindrical system:

zzrzzzzrz

rzr

rrrrrzrrr

afrzrr

afrzrr

afrzrr

1

21

1

Example 5: The stress components are 22 4/,0, yhcqxy xyyyxx . Determine cand stress distribution in the boundary (if body force = 0).

Soln: Substitute the stress components into the equilibrium equation:

024

)( 2

2

cyqyy

hc

yqxy

xyx

xyxx

2

qc

To plot the boundary distribution, we separate the different stress components:

xx : qlylxx xxxx )(,0)0(

xy :

2

22

2

42)(,

42)0( y

hqlxy

hqx xyxy

0242

)2

(,0242

)2

(

2222

hhqhy

hhqhy xyxy

Chapter 2 Displacement and Strain

Definition:

Direct strain:0

0

l

ll , Shear strain:

2

1tan

2

1

Strain tensor:

zzzyzx

xzyyyx

xzxyxx

Strain Variant and Principal Strain

1stvariant: Dilatation or Volume strain: zzyyxxI 1

2nd

variant:222

2 )()()( zxyzxyxxzzzzyyyyxxI

3rd

variant: 2223 2 xyzzzxyyyzxxzxyzxyzzyyxxI Principal strains can be found from 032

21

3 III

Strain Rosette

l

lhq

2

lh

q

2

x

y

l

x

y

2

4lh

q2

4lh

q

123x

y

1

23


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5

3332

32

3

2222

22

2

1112

12

1

sincos2sincos

sincos2sincos

sincos2sincos

xyyyxx

xyyyxx

xyyyxx

Example 6: A plane stressproblem as shown, determine the strain tensor (E=200GPa and

Poissons ratio =0.3) (From the strain rosette measurement:44 102,101 BCA ) (Exam 2009, Quiz 2010)

Soln:Find xx , xy and yy

xxxyyyxx

xyyyxxnnA

0sin0cos20sin0cos

sincos2sincos)0(

22

22

4101 Axx

yyxyyyxx

xyyyxxnnC

10210

90sin90cos290sin90cos)90(

22

22

4

101

Cyy

xyyyxxxyyyxx

xyyyxxnnB

2

1

2

1

2

2

2

22

2

2

2

2

45sin45cos245sin45cos)45(

22

22

444 1011011022

1 yyxxBxy

01

211

yyxxzzzz

E

444 108571.0101101

3.01

3.0

1

yyxxzz

Thus the strain state (or namely strain tensor): 4108571.000

011

011

3D Displacement-Strain relation

Cartesian:

xw

zu

zv

yw

yu

xv

z

w

y

v

x

u

zxyzxy

zzyyxx

21;

21;

21

,,

Cylindrical:

z

w

r

uv

r

r

u

zz

rr

1

x

w

z

u

z

vw

r

r

vu

rr

v

zr

z

r

2

1

1

2

1

1

2

1

Computability Conditions

x

y

45

P A

BC

Strain Rosette


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6

xzxz

zyzy

yxyx

zxzzxx

yzyyzz

xyxxyy

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

yxzyxz

xzyxzy

zyxzyx

zzxyzxzy

yyzxyzxy

xxzyxyzx

2

2

2

2

2

2

Example 7: Check if the finite element shape function for 3D 4-node (tet) elements can

satisfy the compatibility condition.

zcycxccw

zbybxbbv

zayaxaau

3210

3210

3210

Soln:Normal strain:

33210

23210

13210

)(

)(

)(

czcycxcczz

w

bzbybxbb

yy

v

azayaxaaxx

u

zz

yy

xx

Shear strain:

1332103210

3232103210

2132103210

2

1)()(

2

1

2

1

2

1)()(

2

1

2

1

2

1)()(

2

1

2

1

cazcycxccx

zayaxaazx

w

z

u

bczbybxbbz

zcycxccyz

v

y

w

abzayaxaay

zbybxbbxy

u

x

v

zx

yz

xy

All strain are constants, the displacement should satisfy the computability equations.

Chapter 3 Stress-Strain RelationshipGeneralised Hookes law

zx

yz

xy

zz

yy

xx

zx

yz

xy

zz

yy

xx

CCCCCC

CCCCCC

CCCCCC

CCCCCC

CCCCCC

CCCCCC

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

constant matrix [C] is symmetrical and only 21 components of [C] are independent.

In principal planes:

3

2

1

333231

232221

131211

3

2

1

CCC

CCC

CCC

or

133

122

111

2

2

2

I

I

I

In general planes:

1

1

1

2

2

2

I

I

I

zzzz

yyyy

xxxx

zxzx

yzyz

xyxy

2

2

2

Isotropic materials(need twoindependent material constants)From strain to determine stress:


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7

yyxxzzzz

xxzzyyyy

zzyyxxxx

E

E

E

1211

1211

1211

zxzxzx

yzyzyz

xyxyxy

G

E

GE

GE

21

21

21

From stress to determine strain:

yyxxzzzz

xxzzyyyy

zzyyxxxx

E

E

E

1

1

1

GE

GE

GE

zxzxzx

yzyzyz

xyxyxy

2

12

12

1

Relationship between material properties

From Lames constants to find E and

)23(E

2

From E and to find Lames constants

12

E

211

E

Poissons ratio

In practice: 5.00 ; In theory 5.01 When 0 , axial tension will cause a transverse shrinkage and vice verse.When 0 , axial tension will cause zero transverse deformation.

Shear Modulus G

12

EG

When 1 G , which means that such material is impossible to shear, or in otherwords, such materials is rigid in shear.

Bulk Modulus K

11321321

)21(3)(

)21(3)(

3

1KII

EE

where constant)21(3

E

K is called bulk modulus, which is a measure of volume change

of a material when it is undergoing hydrostatic stress

10

00

0

0 )1( IV

VVV

VVzzyyxx

zzyyxx

When 5.0 K , which means that such material is incompressible, like Rubber499.0 .

Example 8: The material (rubber) is pressed in p within a

frictionless container as shown. Determine pressure qfrom the

container and volume change. If change rubber into rigid body

or incompressible body, what is the volume change?

p

q

q q

x

y

z

o

x

y

z

o

y

z

o

Rigid body

rubber


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8

Soln: From theHookes law:

xxzzyyyy

zzyyxxxx

E

E

10

10

0

0

xxzzyy

zzyyxx

0 zzyyzzyy

01)1( 2 zzyy pzzyy

11

pppzzyyxx

11 pq yyxx

1

Volume change:

)1(

)21)(1(

)1(

2

)1(

)1(1

)1(

21

100

22

10

0

E

ppp

E

pp

E

EI

V

VVyyxxzzzzzzyyxx

Rigid body,E, 0)1(

)21)(1(

p

Incompressible body, 0.5, 0)5.01(

)5.021)(5.01(

E

p

Thermal Effect Only affect the normal strain (not shear strain)

TE

T

E

TE

yyxxzzzz

xxzzyyyy

zzyyxxxx

1

1

1

zxzx

yzyz

xyxy

G

G

G

2

12

12

1

Example 9: Write the Hookes law for plane tress and plane strain problem with a temperature

change of T and thermal expansion coefficient ? (Quiz 2011)

Plan stress:

xy

yy

xx

xy

yy

xx

T

TE

2/)1(00

01

01

1 2

Plane strain:

xy

yy

xx

xy

yy

xx

T

TE

2/)21(00

01

01

)21)(1(

Example 10: As shown in a small hole in a thin solid under a uniform tension. If change the

material from mild steel (E=210GPa) to aluminium (E=70GPa) and assume their Poisson ratios

are almost the same in FEA, which of the following calculation should be right (Quiz 2012).

(a)Al

ASteel

A ,, andAl

AyySteel

Ayy ,,3

(b) AlASteel

A ,, 3 andAl

AyySteel

Ayy ,,

(c) AlASteel

A ,, 3 andAlyy

Steelyy 3

(d) AlASteelA ,, and AlyySteelyy

a

ux=0

uy=0

2m

2m

A

B


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9

Since 2cos3

12

12 4

4

2

2

r

aT

r

aTis irrelative to materials, thus replacement

technique ofEand does not affect the stress results when change from p stress to p strain.According to Hookes law, the strain is however related to Youngs modulus.

Chapter 4 Modelling and Solution

Unknowns (15):

6 stresses: zxyzxyzzyyxx ,,,,, 6 strains: zxyzxyzzyyxx ,,,,, 3 displacements: wvu ,,

Basic equations (21):

3 equilibrium equations: iijij ub ,

6 geometric equations:

x

w

z

u

z

v

y

w

y

u

x

vz

w

y

v

x

u

zxyzxy

zzyyxx

2

1;

2

1;

2

1

,,

6 strain-stress eqns:

yyxxzzzz

xxzzyyyy

zzyyxxxx

E

E

E

1211

1211

1211

zxzx

yzyz

xyxy

E

E

E

1

1

1

6 Plus compatibility equation

xzxz

zyzy

yxyx

zxzzxx

yzyyzz

xyxxyy

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

yxzyxz

xzyxzy

zyxzyx

zzxyzxzy

yyzxyzxy

xxzyxyzx

2

2

2

2

2

2

Displacement, Stress and Mixed boundary condition (B.C.)Displacement B.C.: '',' wwvvuu

Stress B.C.:

zzyzxznz

yzyyxyny

zxyxxxnx

nml

nml

nml

Mixed B.C.: Combined displacement and stress boundary

Mathematically, the requirement of B.C. in solving differential equation is obvious.

Special deformation features

Symmetric condition

Stress B.C.

Displacement

B.C.

MixedB.C.


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10

Plane stress and plane strain problems

Plane Stress Plane Strain

Model

Thin plate with in-plane loading

Thick block with perpendicular

loading to the axis

zz 0 0)( yyxxzzE

zz 0)( yyxxzzE

0

w 0w 0w

xx yyxxxxE yyxxxxE

11 2

yy xxyyyyE xxyyyyE

11 2

xy xyxyE )1( xyxyE

11

1 2

E* EE * 21*

EE

*

1

*

Unified

Hookes

Law

xyxy

xxyyyy

yyxxxx

E

E

E

*)1(*

**

**

From plant stress to plane strain soluti on:2*

*

1 PStress

PStressEE

*

*

1 PStress

PStress

,

From plant strain to plane stress solution:

2*

*

)1(

21*

Pstrain

PstrainEE

)1( *

*

Pstrain

Pstrain

Example 11: The thick cylinder is under the outer pressurepo. If we change from such a thick

cylinder to a thin disk structure, which of following result is right? (Exam 2012)

(a) cylinderrrdiskrr

E

2)1(

)21(

(b) cylinderrrdiskrr

(c) cylinderrrdiskrr

E

2*1

*

(d) cylinderrrdiskrr

1

th,b

h

b

mg

F F

po r

z

r

Ri

Ro


11/16


11

Soln: Replacing theEand in the solution of a p-strain by2)1(

)21(

Eand

1

Since

222

22

22

2 1)(

rRR

RRp

RR

Rpr

io

oio

io

oorr , )(rrr is independent on material properties.

But

222

22

22

2 1)1()1(

rRR

RRp

ERR

Rp

E io

oio

io

oorr

is inversely proportional toE.

Superposition PrinciplesKnown the stress functions in two different loading cases.

Load case 1: 0 xyyy and ayxx 6 Load case 2: 0 yyxx and axy

Thus the stress functions should be added.

ayaycasexxcasexxxx 6062,1, .

02,1, caseyycaseyyyy

acasexycasexyxy 2,1,

Solution approachesAfter define the B.C. one should solve for threegroups of unknowns:

Displacement: u,v,w Strain ,,,,,, zxyzxyzzyyxx Stress ,,,,,, zxyzxyzzyyxx

Displacement Method

Unknowns: u, v, w

Procedure: Other two sets of the variables must be eliminated. Thus we replace strain

and stress in displacement.

Stress Method

Unknowns: ,,,,,, zxyzxyzzyyxx

Procedure: Solve for stress component first and then strains and displacements.

Strain Method

Unknowns: ,,,,,, zxyzxyzzyyxx

Procedure: Solve for strain component first and then stresses and displacements.

Saint Venant Principle:(1)very small loading areacompared with the whole dimension. The affected area will

be much smaller than the unaffected area Aunaffected>>Aaffected. e.g in the gravity bar

example, L>>a, in which the affected

area will take roughly: z a.(2)Force replaced must be statically

equivalent. The replacement must not

change either the resultant force or

resultant couple.

Affected zone Affected zoneunaffected zone


12/16


12

Example 12: Different forces are applied to the same cantilever structure as shown below,resulting in different stress distributions. Explain whether or not these forces are statically

equivalent, and why? (Quiz 2013)

Soln: As per St Vanant principle, the force application affects only small area of the beam. These

forces are statically equivalent in the far field.

Example 13: A very long tube has the nominal dimensions of

inner radiusR1=25mm and outer radiusR2=50mm, as shown,

over which another long tube having nominal inner radius of

R2=50mm and outer radius ofR3=75mm, is to be thermally

shrink-fitted. The strain gauge test allow us to determine the

circumferential stress of =40MPa on the outer surface ofthe outer tube (at point A as shown). Quantify the pressure

between inner and outertubes. (Quiz 2013)

SolnStep 1: Free-body diagram: Assume that the fit pressure

between the inner and outer rings isp. We can draw free-body diagram as follows, in which

we separate the outer ring from the assembly.

Step 2: Determine p:Now we can only analyse the outer ring. The circumferential stress at

the outer surface of the outer ring under pressure of 0, oi ppp (see the right figureabove) can be written as

22

23

22

22

23

22

22

23

22

23

22

23

23

22

22

23

23

22

222

22

22

22

3

2100

1)(

RR

pR

RR

pR

RR

pR

RRR

RRp

RR

RpR

rRRRRpp

RRRpRpRr

io

oioi

io

ooii

The strain gauge determined the circumferential stress on the outer surface of the outer ring

63 1040)( Rr , i.e.

6

22

23

22

1 10402

)(

pRR

RRr i

Thus: MPaR

RRp 25

05.0

05.0075.01040

2

10402

226

2

2

22

236

(Positive here means pressure)

-1000

-1000

-2000-2000N/m

r

R1=25

R2=50

R3=75

p

p

R1i

R2i

R2oR3o

r

R1=25

R2=50

R3=75

Strain gauge

A

B


13/16


13

Chapter 5 ApplicationsTorsion of Circular Shaft

Boundary Condition

At the fixed end: Rrz 0,0 :

0 wvu Outer surface In-between: RrLz ,0

0 rzrrr

At the loading end: RrLz 0,

0 zrzz

According Saint-Venant principle: QdrdrR

z

2

0 0

2

Bending BeamAccording to St Vanant Principle, the stress BC can be expressed as follows:

Moment: MMdxdyyhh

b

bxxzz

2/

2/

2/

2/

,

2/

2/

2/

2/

0

h

h

b

b

yyzz Mdxdyx ,

Axial force:

2/

2/

2/

2/

0h

h

b

b

zzzz Fdxdy

Shear force: 0,0

00

yb

zyx

b

zx VdxVdx

Example 14 (Exam 2011): According the Saint Venant

Principle, decide the stress boundary condition on side AB.

Soln:

Stress boundary:When Lxby ,0 ,b

qyxx

Resultant force: Fdyb

xx 0

:

220

2

00

qb

b

qydy

b

qydy

bbb

xx

Moment: Mydyb

xx 0

:

bbb

xx dyqyqy

dyyb

b

qydyy

bM

0

2

00 2222

1234343422

22232

0

32

0

2qbqbqb

b

qbqb

b

qyqydy

qyqyM

bb

Shear stress: 0

0

b

yxdx

Thermal Stress AnalysisTotal strain

MM

x

y

z

L

b

h

L

Fix

End

TwistedEnd

z

Q

B A

A

r

y

0

q

L

A

b

qyxx

x

b

B


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14

mzzzz

myyyy

mxxxx

Tz

w

Ty

v

Tx

u

Equilibrium Eq

0)21(

0)21(

0)21(

21

21

21

z

TEw

z

I

y

TEv

y

I

x

TEu

x

I

Body forces:

z

TEf

y

TEf

x

TEf zyx

)21()21(,

)21(

Application example:

Dr

EC

ErdrT

rE

r

R

rr

i

22

1

)1(1

1

Dr

EC

ETErdrT

rE

r

Ri

22

1

11

1

o

i

o

i

R

Rio

iR

Rio

rdrTRR

RDrdrT

RRC

22

2

22

)1(,

)1(

r

DCrrdrT

rru

r

Ri

1

)1()(

Contact Stress

Bal-ball

*)(43

21

213

EP

RRRRa

2

21

213

*)(

169

EP

RRRR

2

2

0 1a

rqq where PE

RR

RRq

22

21

213

30 *

6

Cylinder-Cylinder

*)(

4

21

212

E

p

RR

RRb

407.0

2ln

1407.0

2ln

12 2

2

221

1

213

b

R

Eb

R

E

p

2

2

0 1b

xqq where pE

RR

RRq *

1

21

2120

Example 15 (Quiz 2013): A cylindrical bearing is loaded as shown. All

components are made of the same material.

At which part of the bearing will the highest contact stress likely occur?

Point A / Point B / Point C

Also, if loadingPis doubled, the peak contact stress P2 will be:

a.

22 P b. 5.02 P c. P2 d. Other (provide your own answer)

2RO2Ri

t

2Rs

Rf

r

R22a

P

PE1,1

E2,2

R1

R2

R1

E2,2

E1,1

pl

b b

qq0

A

B

C

R1R2

r

P


15/16


15

Solution: Peak contact stress underP: PErR

Rrq *

1

1

120

2

PErR

RrP *

1

1

1

DoubleP: PP PErR

RrPE

rR

Rr

414.1*

122*

1

1

1

1

12

Chapter 6 Stress Function MethodProcedure:

Selection of stress function (combination of several)stresses2

2

yxx

,

2

2

xyy

,

yxxy

2

B.C. to determine coefficients Fully determine stress functions

Hookes law to determine strains Strain-Displacement to derive displacement function

Displacement B.C. to fully determine displacement functionsStress function

Equilibrium Eqn: Strain-disp: Compatibility Hookes law: Stress B.C. and Displacement B.C.

To solve for the differential equations, one can assume a solution

of stress, called stress function (or Airy stress function) such that

it can satisfy the equilibrium equations

Stress2

2

yxx

,

2

2

xyy

,

yxxy

2 , which satisfy the equilibrium eqs.

Satisfy bi-harmonic equation: 022

4

22

4

2

4

yyxx

Pure Bending 3ay

2,0 h

yLx : 0 yxyy ,

22,and0 h

yh

Lxx : 0xy ,

As per S-V principle: 02/

2/

h

h

xxdy , Mah

ayayydyydyh

h

h

h

h

hxx

226

32/

2/

32/

2/

2/

2/

Cantilever Beam: 33 cybxyaxy

Ph/2

h/2

L

x

y

Ph/2

h/2

L

x

y

0)(

0

0

2yyxx

yyyx

xyxx

yx

yx

MM

h/2

h/2

L

x

y


16/16

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mechanics of solids week 12 lectures

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