mechanics of solids week 7 lectures

8/12/2019 mechanics of solids week 7 lectures

1/15

Week 7 MECH3361

1

Recap

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 tensile bar,

L>>a,in which the affected area will take roughly: z a.(2)Force replaced must be statically equivalent. The replacement must not changeeither the resultant force or resultant couple.

Combined shaft: As per Saint Venant (S-V) principle to define B.C.At z=0, z=L, Rr0 , 20 (flat end faces): ))(( drrddA

drrddrrdPR

zz

R

zz

2

0 0

2

0 0

))((

cos))((2

0 0

drrdrMMR

zzyy

0sin))((2

0 0

drrdrMR

zzxx

drdrQR

z

22

0 0

0zr

At r=R, 0 zL , 20 (cylindrical surface): 0 rzrrr

Bending BeamAt (front and back faces)

Lzh

yhb

x 0,22

,2

:

0 xzxyxx

At (top and bottom faces)

Lzh

yb

xb

0,2

,22

:

0 yzyxyy

At (side ends) Lorzh

yhb

xb

0,22

,22

: 0 zxzy ,

Moment:

2/

2/

2/

2/

h

h

b

b

zz Mdxdyy ,

2/

2/

2/

2/

0h

h

b

b

zzdxdyx ,

Resultant force

2/

2/

2/

2/

0

h

h

b

b

zzdxdy

Affected

zone

Affected

zoneunaffected zone

a

L

z z

z

y

x

dr

rdd

P

Q

MM

P

Q

y

x

MM

x

y

z

L

b

h

Symmetric line


2/15

Week 7 MECH3361

2

5.3 Thermal Stress AnalysisUse Di splacement method(u,v,w) to solve for equilibrium without mechanical body forces.

Step 1: Strain-Displacement relation

Mechanical strain mzzmyy

mxx ,,

Thermal strain: T

(Note that thermal strain does not affect the shear strains)

Total strain

mzzzz

myyyy

mxxxx

Tz

w

Ty

v

Tx

u

Tz

w

Ty

v

Tx

u

mzz

myy

mxx

Step 2: Use the Hookes law in mechanical strains:m

zzmyy

mxx ,,

mmxxzz

mmxxyy

mmxxxx

I

I

I

1

1

1

2

2

2

zxzx

yzyz

xyxy

2

2

2

where

T

z

wT

y

vT

x

uI

mzz

myy

mxx

m

1 .

Thusx-related components:

z

u

x

w

x

v

y

u

Tz

w

y

v

x

uT

x

u

xz

xy

xx

2

12

2

12

32

Step 3: Equilibrium Equation: 0

zyx

xzxyxx . Calculate different terms:

x

T

xz

w

xy

v

x

u

x

u

Tz

w

y

v

x

u

xT

x

u

xxTermst xx

322

321

22

2

2

2

2

yx

v

y

u

x

v

y

u

yyTermnd

xy

2

2

2

2

2

22

3z

u

zx

w

z

u

x

w

zzTermrd xz

So

2

222

2

2

22

2

2

2

2

322

zu

zxw

yxv

yu

x

T

xz

w

xy

v

x

u

x

u

zyxLHS xz

xyxx


3/15

Week 7 MECH3361

3

x

T

ux

I

x

T

z

u

y

u

x

u

z

w

y

v

x

u

xLHS

uI

32

32][][

21

2

2

2

2

2

2

21

)21()21)(1(

3)21(

)21)(1(3

)21)(1(

)21(

)21)(1(3

)1(2232

EEE

EEEE

Thus thermal Equilibrium Eqns:

0)21(

0)21(

0)21(

21

21

21

z

TEw

z

I

y

TEv

y

I

x

TEu

x

I

Compare with the Equilibrium equation (in terms of displacements) with body force.

0

0

0

21

21

21

z

y

x

fwz

I

fvy

I

fux

I

We can see that for the thermal deformation problems, the basic equations to be solved are

equivalent to a normal static problem with body force densities of

z

TE

f

y

TEf

x

TEf

z

y

x

)21(

)21(

)21(

Step 4: Boundary conditions

Displacement B.C. (if any): '',' wwvvuu and

Stress B.C.:

zzzyzxnz

yzyyyxny

xzxyxxnx

nmlnTE

nmlmTE

nmllTE

)21(

)21(

)21(

''

''

''

like a surface stress.


4/15

Week 7 MECH3361

4

Solution of a thermal static deformation problem without body forces and surface stresses but

with a steady state field of temperature change, T, is equivalent to a solution of a statics

problem subjected to a set of body forces and a surface stresses

When a plane stress solution is available, the corresponding plane strain solution can be

obtained by replacing E, , by21

E,

1and )1( .

When a plane strain solution is available, the corresponding plane stress solution can be

obtained by replacing E, , by2)1(

)21(

E,

1and

21

1

.

Thermal fit of a hollow disk onto a shaftSimilar to interference fit, the thermal fit

first increases the temperature of hollow

disk allowing the diameter increase to

greater than the shaft diameter, and then

inserts the shaft into the hollow disk.

Assume: Ti= temperature change at the

inner surface and To= temperaturechange at the outer surface of the disk. Consider a plane stress problem in this thin disk case.

Step 1: Boundary condition of the hollow disk (without insertion of shaft yet)

At oi RandRr : 0 rrr

At iRr : iTT and At oRr : oTT

Step 2: Analysis:

If the temperature change Tis axisymmetric (only a function of r) )(rTT , the

displacement should be axisymmetric: 0),( vruu

Step 3: Heat conductionproblem: Heat conduction equation in polar system:

0

1

dr

Tdr

dr

d

r

Integrate twice: BrAT ln Constants A and B can be determined by the thermal B.C.

iRr , iTT BRAT ii ln

oRr , oTT BRAT oo ln

ioio RRATT lnln io

io

RR

TTA

lnln

iio

ioiii R

RR

TTTRATB ln

lnlnln

Thus the temperature change is calculated as

)/ln(

)/ln(

)/ln(

)/ln(

lnln

lnln

lnln

lnln

lnln

lnln

lnln

ln)ln(lnln

lnlnln

lnlnln

0RR

rRT

RR

rRTT

RR

RrT

RR

Rr

TRR

RrT

RR

RRRr

RRR

TTTr

RR

TTT

i

ioio

oioio

iiio

o

oio

ii

io

iio

iio

ioi

io

io

2RO2Ri

t

2Rs

Rf

r


5/15

Week 7 MECH3361

5

Step 4: thermal deformationAs discussed in Week 5, the equilibrium equation as per displacement in the polar system is

dr

Tdf

r

u

r

u

rr

ux

)()1(

122

2

Rewrite it as

dr

Tdru

dr

d

rdr

d

r

u

r

u

rr

u

rr

u

rr

u

r

u

rr

u )()1(

1122

2

Integrate once the equation above

')(

)1(1

Cdrdr

Tddrru

dr

d

rdr

d

')1(

1CTru

dr

d

r

rCrTrudr

d')1(

Integrate once again the above equation

drrCrTrudrd

')1(

DCrrdrTru 2

)1( (note that Tis a function of r)

Thus r

DCrrdrT

rru

r

Ri

1

)1()(

Step 5: Strain:

22

1)1()1(

1)1(

r

DCrdrT

rT

r

DCrrdrT

rrr

u r

R

r

R

rr

ii

22

)1(01

r

DCrdrT

rr

u

r

uv

r

r

Ri

Step 6: Hookes law with thermal strain

TE

rrrr

)1(1

2, T

Err

)1(

1 2

Step 7: Stresses

Dr

EC

ErdrT

rE

TE

r

DCrdrT

r

E

r

DCrdrT

rT

E

r

R

r

R

r

R

rr

i

i

i

22

2222

222

1

)1(1

1

)1(1

)1(1

1)1()1(

1

Dr

EC

ErdrT

rE

r

R

rr

i

22

1

)1(1

1

Similarly: Dr

E

C

E

TErdrTrE

r

Ri

22

1

11

1


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Week 7 MECH3361

6

Step 8 Apply stress B.C. to determine constants C & D(At oi RandRr : 0 rrr ):

01

)1(1

1)(

22

D

R

EC

ErdrT

RERr

i

R

Ri

irr

i

i

which leads to DR

Ci2

1

1

1

. And plug it into next B.C. equation

01

)1(1

1)(

22

D

R

EC

ErdrT

RERr

o

R

Ro

orr

o

i

01

)1(

1

1

1

1

1222

D

R

ED

R

ErdrT

RE

oi

R

Ro

o

i

o

i

o

i

R

Rio

iR

Rio

rdrTRR

RDrdrT

RRC

22

2

22

)1(,

)1(

Step 8 Calculate the temperature by using fit condition: sii RRruR Thus the required temperature change for fit can be calculated.

5.4 Stress and Deformation due to Contact

There are many contact examples: gear tooth, tyres on road surface, train wheel and bearings.

When the contact area is much smaller than the characteristic dimension of a component, the

component can be considered to be semi-infinitebounded by a surface:

The governing equilibrium equations are as follows

.0

,0

,0

=f+w+z

I)+(

=f+v+y

I)+(

=f+u+x

I

)+(

z21

y21

x

21

The general solution to this equation without acceleration and body force was derived by

Papkovich in 1932 and Neuber in 1934:

dp=pdA

dpn

dpt

x

y

z

x

y

z

x

y

z

x

y

z

x

y

z

x

y

z

x

y

z

x

y

z


7/15

Week 7 MECH3361

7

)()1(4

1

)()1(4

1

)()1(4

1

0

0

0

zyxz

zyxy

zyxx

zyxz

w

zyxy

v

zyxx

u

For short )()1(4

10

Rd

grad

where: Twvu ),,(d , Tzyx ),,( are all harmonic functions that satisfy Laplace

equation 02 (2

2

2

2

2

22

zyx

)

How to select harmonic equations is skill-based. The textbook provides examples (P.190).

A Half-Space under a Normal Concentrated Load

Step 1: B.C.Displacement B.C.:

Axisymmetrical about z at 0,0 yxz : 0 vu

When R : 0 wvu (infinitely far field)

Stress B.C.:

Except at the origin, at z=0: 0 zyzxzz

0Pdxdyzz

(S-V principle)

Step 2: Select harmonic functions (refer to the textbook)

zRCR

Czyx ln,1

,0 201

Step 3: Displacement

RB

RR

zAw

zRR

yB

R

zyAv

zRR

xB

R

zxAu

1

)(

3

)(

)(

3

2

3

3

Step 4: Stresses

3

2

3

22

2

3

222

3

22

2

3

222

3

232

)()(23

2

)()(23

2

R

zBR

x

R

zA

zRR

y

zRR

zxB

R

y

R

zA

zRR

x

zRR

zyB

R

x

R

zA

zz

yy

xx


8/15

Week 7 MECH3361

8

3

2

3

3

2

3

235

232

232

)(

)2(2

6

R

xB

R

z

R

xA

R

yB

R

z

R

yA

zRR

RzxyB

R

xyzA

yz

yz

xy

Step 5: Apply B.C. to determine the constants A and B


0

4)2(4

BA

PBA

4

4

PB

PA

Step 6: Determine the stress and displacement functionsSubstitute them back to stress and displacement functions, one can determine the stress and

displacement completely.

R

P

R

zPw

zRR

yP

R

zyPv

zRR

xP

R

zxPu

1

4

)2(

4

)(44

)(44

3

2

3

3

5

2

5

2

235

5

3

22

2

3

222

3

22

2

3

222

3

2

3

2

3

)(

)2(

22

32

3

)()(23

2

)()(23

2

R

xzP

R

yzP

zRR

RzxyP

R

xyzR

zP

zRR

y

zRR

zxP

R

y

R

Pz

zRR

x

zRR

zyP

R

x

R

Pz

yz

yz

xy

zz

yy

xx

A Half-Space under a Tangential Concentrated Load

Step 1: B.C.



9/15

Week 7 MECH3361

9

0Qdxdyzx

Step 2: Select harmonic functions

zRC

zRR

xC

RC zyx

1

,)(

,0,1

3021

After derive the displacement and stress functions, apply B.C. to determine the three constants

,)1(8

)21(,

)1(8,

)1(8

)21( 2

QC

QB

QA

Step 3: Displacement

)()21(

4

)(

)21(

4

)()21(1

4

2

22

2

2

2

2

zR

x

R

xz

R

Qw

zR

xy

R

xy

R

Qv

zR

x

zR

R

R

x

R

Qu

Step 4: Stresses

5

2

5

222

2

2

5

2

2

222

2

222

2

2

2

3,

2

3,

2

)(

213

2

,2

3

23

)(

213

2

2

)(

213

2

R

Qzx

R

Qxyz

zR

RxxR

zRR

x

R

Qx

R

Qxz

zR

RxxR

zRR

y

R

Qx

zR

RyyR

zRR

x

R

Qx

yzyzxy

zz

yy

xx

Solutions to the Contact Problems

The frictionless contact can be solved by using approach to normal concentratedforce on a half-space

The frictional contact can be solved by using approach to tangential concentratedforce on a half-space

No penetration between the two contact bodyEffective Youngs Modulus:

1

2

22

1

21 11*

EEE

Two balls in contact

*)(4

3

21

213

E

P

RR

RR

a ,

2

21

213

*

)(

16

9

E

P

RR

RR

R22a

P

P

E1,1

E2,2

R1


10/15

Week 7 MECH3361

10

2

2

0 1a

rqq (distribution of contact stress)

where PERR

RRq

2

2

21

21

3

30 *

6

A sphere in contact with flat half space ( 2R )

*4

3

*)1/(4

3

*)//(4

31

1

1

2221

13

E

PR

E

P

R

R

E

P

RRRR

Ra

2

1

3

*

1

16

9

E

P

R

2

2

0 1a


where PER

q 2

2

13

30 *16

A sphere in contact with concave half space ( 22 RR )

*)(4

3

*)(4

3

12

21

21

213

E

P

RR

RR

E

P

RR

RRa

2

21

123

*

)(

16

9

E

P

RR

RR

2

2

0 1

a


where PERR

RRq

2

2

21

12

3

30 *

6

Two Parallel Cylinders

*)(

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 1bxqq (distribution of contact stress)

where pERR

RRq *

1

21

2120

Cylinders with flat half-space

*

41

2

E

pRb

2

2

0 1b

xqq where

1

20

*

R

pEq

R2

R1

E2,2

E1,1

pl

b b

qq0

P

E1,1

R2

R1

P

E1,1

R2

R1


11/15

Week 7 MECH3361

11

Chapter 6 Stress Function Method

6.1 Airy Stress Function

Basic equations: Review the basic equation in 2D plane stressproblems:

Equilibrium Eqn:

0

0

yyyyx

xxyxx

fyx

fyx

Strain-disp:

y

u

x

v

y

v

x

u

xy

yy

xx

2

1

Compatibilityyxyx

xyxxyy

2

2

2

2

2

Hookes law:

xyxyxy

xxyyyy

yyxxxx

EG

E

E

)1(21

1

1

xyxy

xxyyyy

yyxxxx

G

E

E

2

2

1

1

Stress B.C.

yyyxy

xxyxx

Fml

Fml

Displacement B.C.

*

*

vv

uu

Stress method to solve for plane stress problemuse stress as primary variableWhen 0 yx ff ,: Differentiate the 1

stequilibrium equation w.r.t.xand the 2ndw.r.t.y:

0

0

2

22

2

2

2

yyxyxy

xyxyxx

yyyxyyyx

xyxxxyxx

xyyx

xyyyxx

2

2

2

2

2

2

which is the compatibility in terms of stress. Using Hookes law leads to:

xyyyxxxxyy

EyxEyEx

1

211

2

2

2

2

2

0)1(2

2

2

2

2

2

2

2

2

2

yxyyxx

xyyyxxxxyy

0)1(2

2

2

2

2

2

2

2

2

2

2

2

yxyyxx

yyxxyyxxxxyy

0)(2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

yyxxyyxx

yyyyxxxxyyxxxxyy

0)()( 22

2

2

2

2

2

2

2

2

2

2

2

yyxxyyxxyyxxxxyy

yxyxyx


12/15

Week 7 MECH3361

12

Now the key equations need to be considered are:

0)(

0

0

2yyxx

yyyx

xyxx

yx

yx

To solve for the above differential equations, one can assume a solution to the stress, namely

stress function (or Airy stress function) such that it can satisfy the equilibrium equations. One

option can be:

2

2

yxx

,2

2

xyy

,yx

xy

2

Equilibrium equations

0

0

2

3

2

3

2

22

2

3

2

32

2

2

xyxyxyyxxyx

yxxyyxyyxyx

yyyx

xyxx

Compatibility: 02 224

4

22

4

4

4

2

2

2

22

yyxxxy

Now we need to solve for a bi-harmonicequation:

024

4

22

4

4

4

yyxx

Polar system: The stress functions can be written as:

2

2

2

11

rrrrr , 2

2

r

,

2

1

rrr

Bi-harmonic functions have been studied extensively. Tables 8.1 and 8.2 in the textbook

include some typical functions.

B.C. for stress function

To this end, all the stress functions selected for a specific problem need to satisfy B.C. as well.

We need to see what stress function could meet the requirement of B.C.

6.2 Rectangular plate under bending

Pure Bending Problem

MM

h/2

h/2

L

x

y


13/15

Week 7 MECH3361

13

Step 1: B.C.

2,0 h

yLx : 0 yxyy (no surface stresses, but 0xx )

22,and0

h

y

h

Lxx : 0xy , (pure bending)

As per S-V principle: 02/

2/

h

h

xxdy , (no axial force) Mydyh

h

xx

2/

2/

(bending moment)

Step 2: Observe the B.C. and Select stress function

Now we need to choose proper stress function to satisfy the above B.C. From Table 8.3, we

can see stress function #8 leads to a linear xxat the two endsand all other stress

components are zero. Therefore, we choose3

ay

It will be found that this stress function satisfy all the B.C. well.

Step 3: Use B.C. to determine stress functionTo determine constant a, the bending moment B.C. can be used:

ayayyy

xx 6)( 3

2

2

2

2

Mhahh

adyyaydyayydy

h

h

h

h

h

h

xx

3332/

2/

22/

2/

2/

2/ 2883

1666

Thus:3

2

h

Ma

So the stress function becomes: 33

2y

h

M

yh

My

h

Mayxx 33

12266

,

, 032

2

2

2

ay

xxyy

, 03

22

ay

yxyxxy

Cantilever Beam Problem

Step 1: B.C.

2,0 h

yLx : 0 yxyy (no surface stresses, but 0xx )

22,0

hy

hx : 0 vu (fully-clamped)

22,

h

y

h

Lx : 0xx (no axial force) Vdy

h

hxy

2/

2/ (S-V principle)

Vh/2

h/2

L

x

y

Vh/2

h/2

L

x

y


14/15

Week 7 MECH3361

14

Step 2: Observe the B.C. and Select stress functionThe clamped end can be separated (F.B.D.). We need linear normal (but vanish in total) and a

uniform shear.

Uniform shear: #4: axyall around the plates #10: uniform shear in top/bottom but non-uniform at other two ends. Thus superposition can cancel the top/bottom shear. #10 also leads to a normal stress xx at right end (x=L) but B.C. requires only normal

at left end. We therefore need other function to cancel the right end normal stress.

#8 is just what we wantThus 33 cybxyaxy

Step 3: Calculate the stress

23322

2

2

33

2

2

2

2

3

0

66

byacybxyaxyyxyx

x

cybxycybxyaxyyy

xy

yy

xx

Step 4: Applied the B.C. to determine the unknowns in the stress function:

Vhbhahbhabyaydybyady

b

h

ab

h

a

h

ba

h

yLx

cbLcybLyh

yh

Lx

h

h

h

h

h

hxy

xy

xx

33

2/

2/3

2/

2/

22/

2/

222

82823

4

3

04

3

232,0

006622

,

I

Vha

8

2

,I

Vb

6 ,

I

VLc

6

Step 5: Express stress

Thus:

2

4

42,0,)( y

h

I

VyxL

I

V

xyyyxx

6.3 Stress concentration around a circular hole

Step 1: Transfer to Polar system:

Ta

b

T


15/15

Week 7 MECH3361

15

It is convenient to use polar system. To do so, a stress needs to be transferred:

2sin

2

1cossin)sin(cos2sincos)(

2cos12

1coscossin2sincos

22

222

TT

TT

xyxxyyr

xyyyxxrr

Step 2: B.C.:

At ar : 0 rrr

At br : 2sin2

12cos1

2

1TT rrr

The superposition can be applied to the second B.C at br .

2costan

2cos2

1

2

12cos1

2

1

toonalproportatitCons

rr TTT

In this boundary ( br ):

Part 1: Trr2

1 and 0r

Part 2: 2cos2

1Trr and 2sin

2

1Tr

Step 3: Select the stress functionFrom Table 8.4, the stress function is to sum #2, 4, 15, 16 and 17. Determine cifrom BC.

2cos)(ln 2

52

4312

1 rcrccrcrc

2sin23

12

2cos3

12

12

2cos43

12

12

2

2

4

4

4

4

2

2

2

2

4

4

2

2

r

a

r

aT

r

aT

r

aT

r

a

r

aT

r

aT

r

rr

When 12cos , reaches the maximum

Ta

aT

a

aTar 3cos

31

21

2 4

4

2

2

Thus T3max

Stress concentration factor k: nominalk max (k=3 in this small hole case)

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mechanics of solids week 7 lectures

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