1

dAdF

AF

A

0lim

Normal stress = )Pa(pascalm/NA

Force 2

Mechanics of Materials

AF

F

F

F F

F Tensile stress (+)

F

F Compressive stress (-)

At a point:

2

Shear stress = = F tangential to the area / A

A F

F

dAdF

At a point,

3

Normal strain = fractional change of length= lx /

x

l

F

Ffixed

lx /Shear strain = deformation under shear stress =

F

F

l x

4

Stress-strain curve

o

Yield pt.

Work hardening

breakElasticdeformation

Hooke's law: In elastic region, , or / = E E is a constant, named as Young’s modulus or modulus of elasticity

Similarly, in elastic region, / = G, where G is a constant, named as shear modulus or modulus of rigidity.

5

Exercise set 2 (Problem 3)Find the total extension of the bar.

dxx

dxde 2

41092.1

)(120)105(6.03 mxmm

xW

Paxmx

N2

7

22

3 1088.2)120/(

102

X

15mmW5mm

1.2m0.6mo

kN2dx

2

4

9

27 1092.110150

/1088.2x

xE

Width of a cross-sectional element at x:

Stress in this element :

Strain of this element:

The extension of this element :

The total extension of the whole bar is := 2.13 x 10-4 m

8.1

6.0 2

41092.1 dxx

dee

6

Bulk modulus )/( VV

pK

VV

p

dVdpV

7

Poisson's ratio :For a homogeneous isotropic material

normal strain :

lateral strain : Poisson's ratio : value of : 0.2 - 0.5

dd

L

/L

x

F F

dd

x

d

8

Double index notation for stress and strain

1st index: surface, 2nd index: force

For normal stress components : x xx, y yy , z zz,

x xxz

x

y

x

y

z

zxzy

yzxz

xyyx

9

EEE

EEE

EEE

yyxxzzzz

zzxxyyyy

zzyyxxxx

z

x

y

xy

z

Joint effect of three normal stress components

10

Symmetry of shear stress componentsTake moment about the z axis, total torque = 0, (xy yz) x = (yx x z) y, hence, xy = yx .

Similarly, yz = zy and xz = zx

z

y

xxy

yx

x

yz

11

Define pure rotation angle rot and pure shear strain, such that the angular displacements of the two surfaces are: 1= rot+ def and 2= rot- def . Hence, rot = (1+ 2)/2 and def = (1- 2)/2

Original shear strain is “simple” strain = etc. ,..., ydx

xdy

There is no real deformation during pure rotation, but “simple” strain 0.

x

y

x

2 = -Example: 1 = 0 and 2 = - , so def = (0+)/2 = /2 and rot= (0-)/2 = -/2Pure shear strain is /2

x

y

xdy1 rot

2

def

def

12

Example: Show that)21(3

EK

)21(3E

V/VpK

3

33 )(

l

lll

VV

3/3 ll

Proof:

xx = yy = zz = , hence 3 = xx+yy+zz = (1-2v)(xx+yy+zz)/E xx =yy =zz = -p (compressive stress)

)()21(3 pE VV

For hydrostatic pressurel

l

l

13

Point C moves further along x- and y-direction by distances of AD(/2) and AD(/2) respectively.

nn = [(AD /2)2 + (AD /2)2]1/2 / [(AD)2 + (AD)2]1/2 = /2

True shear strain: yx = /2

Therefore, the normal component of strain is equal to the shear component of strain:

nn = yx and nn = /2

Example : Show that nn = /2

2/ x

y

A

C’C

D

D’2

14

yx (lW) sin 45o x2 = 2 (l cos 45o) W nn

Example : Show that nn = nn/(2G)

Consider equilibrium along n-direction:

Therefore yx = nn

From definition : = xy /G = nn /G = 2 nn

l

ln

yxxy

15

Gv

E 2 1

xx = xx/E - yy/E- v zz/E

Set xx = nn = - yy, zz = 0, xx = nn

nn = (1+) nn /E = nn /2G (previous example)

Example : Show GvE 21

16

Ex. 12 kN forces are applied to the top & bottom of a cube (20 mm edges), E = 60 GPa, = 0.3. Find (i) the force exerted by the walls, (ii) yy

z y

12kN

x(i) xx = 0, yy = 0 and

zz= -12103 N/(2010-3 m)2 = 3107 Paxx = (xx- v yy- v zz) /E0 = [xx- 0 – 0.3(- 3107)]/60109

xx = -9106 Pa (compressive)Force = Axx = (2010-3 m)2(-9106 Pa) = -3.6 103 N

(ii) yy = (yy- v zz- v xx) /E= [0 – 0.3(- 3107) – 0.3(- 9106)]/60109 = 1.9510-4

17

Elastic Strain EnergyThe energy stored in a small volume:

The energy stored :

Energy density in the material :

dxxAEFdxdU )(

VE

AeEAEe

dxxAEUe

2

22

0

21

)()(21

21

)(

EE

VUu

22

21

21

e=extension

dx

F F

x

18

Similarly for shear strain :

Fdx

xdFU Fdx

/

/x

AFGG

Gu2

2

21

21