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Mechanics

AP200Dr. C. W. Ong

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Textbook :

Engineering Mechanics, Dynamics,by Meriam, J.L., Kraige, L.G.,John Wiley 1997 4th Ed.

Mechanics of EngineeringMaterials , by P.R. Benham andR.J. Crawford, ELBS, 1988

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Assessment method :

s Course Work 30% (Exercises and tests)

s Examination 70%

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VectorAA vector consists of two things :

1. Magnitude, represented by a

number A

2. Direction, represented by a

unit vectorAe

AeAA =

e

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In Cartesian coordination system

yAxA

yA

eAA A

)sinx(cos

yx +=

+=

=

Y

X

A

YA

XA

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jBAiBABA

jBiBB

jAiAA

YYXX

YX

YX

)()( +++=+

+=

+=

Y

X

YB

xBB

AYA

XA A

'B

CBA

=+

Addition of vectors

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kAjAiA

kAjAiA

eAA

ZYX

A

++=

++=

=

coscoscos

Direction cosine : cos , cos ,cos

1coscoscos

)coscos(cos

)coscoscos(

222

2/1222

2/1222222

222

=++

++=

++=

=++=

A

AAA

AAAAAZYX

3-D Cartesian coordination system

Y

Z

A

X

e

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Result of dot product is a scalar

Dot product :

CBA =

cosABBA

A

B

cosA

Definition is :

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Cross product

rule.handrightbydetermineddirectionwithvectors,twothetois

and,sin:Definition

vectornewaisResult

=

=

C

ABBA

CBA

AB

C

Fsin

d

F

O

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kBjBB

kCC

kAjAiAA

ZY

Z

ZYX

+=

=

++=A

B

C

CBA

)(Z

Exercise : Show that CBA

)(

is the volume of the parallelepiped

formed by the three vectors

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Presentation of vector product inrectangular coordination system

kAjAiAA zYX ++=kBjBiBB zYX

++=

ZZYYXX BABABABA ++= )(

jABkBAiBA ZXYXZY

++= iABjBAkAB ZYZXYX

ZYX

ZYX

BBB

AAA

kji

BA

=

Dot product :

Cross product :

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Fundamental of Mechanics ofMaterials

Stress and strain :

Assume F is uniformly distributed over thecross-sectional area A.

A

F

F

F

F

F

F

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Normal stress

[ ] )(/ 2 PapascalmNA

F

dA

dF

A

F

A==

0lim

F

A

FP

Normal stress at a point :

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Total force F acting on a cross-sectional

area is :

= A dAF

F F

F

F

Tensile stress (+)

Compressive stress (-)

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Shear stress :

Define shear stress :

A

A

Fixed area

F

F

A

F

dA

dF=

Force F is tangential to the area.

At a local point :

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Normal strain :A body deformed along the direction of stress. Normal strain is defined to describe the deformation :

x

x

x

F

F

l x

F

Ffixed

Shear strain :Deformation under shear stress is defined as

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Elastic modulus

Stress-strain curve

In the elastic region, (Hooke's law)

Youngs modulus (modulus of elasticity) E = /

Shear modulus : (modulus of rigidity) G = /

Stress

Straino

Yield pt.

Workhardening

breakElastic

deformation

d

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le

Find the totalextension ofthe bar. 3 The width of a cross-sectional element at x :

3 The stress in this element :

3 The strain of this element:

15mmW5mm

X

1.2m0.6mo

kN2

)(120

)105(6.0

3 mx

mm

xW ==

Paxmx

N2

7

22

31088.2

)120/(

102 =

=

dx

2

4

9

271092.1

10150

/1088.2

x

x

E

=

==

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The extension of this element :

The total extension for the whole bar becomes :

= 2.13 x 10-4 m #

dxx

dxde2

41092.1 ==

==8.1

6.0 2

41092.1dx

xdee

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Bulk modulus :)/( VV

pK

=

VV +

p

lim V 0 then

dV

dp

VK =At a local point :

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Poisson's ratio :For homogeneous isotropic materials

s normal strain :

s lateral strain :

s

Poisson's ratio :s value of : 0.2 - 0.5 d

dL

= /L

x=

F F

dd +

x

d

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Analysis of stress and strain inIsotropic Materials.

EEE

EEE

EEE

yxzz

zxy

y

zyxx

=

=

= ......(*)

z

zy

yz

yx

zx

xz

x

y

xy xy

z

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Rewrite (*) by xxx, yyy , z zz :

Eyyxxzzzz

Exxzzyyyy

Ezzyyxxxx

/)(

/)(

/)(

==

=

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)()1(

)(

)1(

)(1

zzyyxxzzzz

zzyyxxyyyy

zzyyxxxxxx

EE

EE

EE

+++

=

++

+

=

+++

=

)()21( zzyyxxzzyyxxE

++=++

Exercise : Verify)21(3

=

EK

Solution

From (*)

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For hydrostatic pressure

pzzyyxx ===

=== zzyyxx

3

33

34

3

4)(

3

4

r

rrr

V

V

=

r

r

3)(3 =

r

r

)3()21( pE

=

)21(3/

=

E

VV

pK

(symmetry)

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Similarly, take moment about x-axis, yz = zy

Take moment about y-axis, xz = zx

Shear stress and strain

Take moment about the z axis, total torque = 0,

so xy = yx

z

y

x yx

xy

x

y

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Gxyxy / =Gyzyz / =

Gzxzx / =

Shear strain :

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One better method is to select axes appropriately, such that

/2===

= yxxy

y

dx

x

dy

Shear deformation is usually defined by : .,..., etcy

dx

x

dy

However, pure rotation may cause non-zero

which cannot be used to represent any deformation

ydx

xdy

and

2/ x

y

2

x

y

xdy

x

y

xdy

Pure

rotation

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(AC)2(1+n)2 = (AD)2 + (AD)2 -2(AD)2 cos(90o+ )

2(AD)2 (1+ n)2 = 2(AD)2+2(AD)2 sin

1+2 n= 1 + sin 1 +

2 n

since = 2 yx

n= yx

Example : Show that n = yx

2/ x

y

A

C

C

D

D2

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yx (lW) sin 45o x2 =LW n (equilibrium along n-

direction)

= 2(lcos 45o) W n

Therefore yx = n

G

nn

2 =n2=== GG

nxyxy

From definition :

l

Ll

n

yx

xy

2/ x

y

A

2/

yx

xy

G

nn

2

=Example : Show that

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GE

21

=+

(from p.24))(1 zzyyxxxxxxEE

+++=

n n

- n

- n

Set xx = n = - yy ,

zz = 0,

xx = n

n = (1+ ) n /E = n /2G (previous

example)

Example : Show GE

21

=+

xamp e 3 (i) Here = 0 = 0 and

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xamp eThe cube of isotropicmaterial has side 20mm,E = 60 GPa, = 0.3.

(i) Find the force exerted bythe restraining walls upon the

block.

(ii)Find the strain in ydirection

3 (i) Here x = 0, y = 0 and

3 Since :

3 x = -9 x 106 Pa (compressive stress)

3 The force exerted by the restraining wall :

3 A x = (20 x 10-3 m)2 x (-9 x 106 Pa)

3 = -3.6 x 103 N (compressive force)

3

(ii) The strain in y direction :

z

y

12kN

x

Pam

Nz

7

23

3

103)1020(

1012=

=

)(1

zyxx

E =

)]103(3.00[1060

10 7

9

= x

#4

67

9

1095.1

)]109(3.0)103(3.00[1060

1

)(1

=

=

= xzyyE

El ti St i E

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Elastic Strain Energy3 The energy store in the material :

3 The energy store :

3 Energy density within the material :

FdxdW=dx

F Fdx

xAEFdxdU )(

==

/

/

x

AFE=

VE

AeEAEe

dxxAE

Ue

=

==

=

2

2

2

0

2

1

)()(21

21

)(

EE

V

Uu

22

2

1

2

1 ==

e=extension

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Similarly for shear strain :

F

dx

= xdFU

= Fdx

==

/

/

x

AFG

G

Gu2

2

2

1

2

1 ==

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