1. Chapter 2 Mechanics of Materials
F A


 F
 Tensile stress (+)


Normal stress = [N/m ] pascal (Pa)
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s º Force 2
dF
dA
F
s d
= =
A
lim
A
® d
d 0
A
Example: Estimate the normal stress on a shin bone ( 脛骨
)

F
F
F

F
F
F
Compressive stress (-)
At a point:

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Shear stress (切應力) = t = F tangential to the area /
A

A F

F
At a point, t = dF
dA

3. 

Shear strain (?) = deformation under shear stress = x / l
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Normal strain (正應變) e = fractional change of length=x / l
x
l g
F
F
fixed
F
F
l x

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Stress-strain curve
s
Yield pt.
Work
hardening
Elastic break
deformation
o e
Hooke's law: In elastic region, s µ e, or s/ e = E
E is a constant, named as Young’s modulus or modulus of
elasticity
Similarly, in elastic region, t/g = G, where G is a constant,
named as shear modulus or modulus of rigidity.

5. W = x ´ - =
0.6 (5 10 ) 120 ( )
´ - = = 1.8
1.92 10 4 dx
0.6 2
x
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Exercise set 2 (Problem 3)
Find the total
extension of the bar.
dx
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 :
3 N
2.88 10
X
s = 2´10 = ´
7 2 1.92 10
2.88 10 /
´ - = e =
1.92 10 4
de dx 2
x
3m x m m
Pa
x m x
2
7
2 2
( /120)
15mm
5mm W
0.6m 1.2m
o
2kN
dx
2
4
9
150 10
x
x
E
´ - =
´
e = s = ´
ò ò
e de
= 2.13 x 10-4 m

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Bulk modulus º
K d
p
( d
V /V )
-
=
= -V dp
V +dV
dp
dV

7. d + Dd
d
F F
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Poisson's ratio :
For a homogeneous isotropic material

normal strain :
lateral strain :
Poisson's ratio :
x
d
e = x

e = D
L
d
n º -
e /e L value of n : 0.2 - 0.5

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Double index notation for stress and strain
1st index: surface, 2nd index: force
For normal stress components : x Þ xx, y Þ yy , z Þ zz,
sx Þ sxx
z s
x s
y s
x
y
z
szx
szy
syz sxz
sxy
syx

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Joint effect of three normal stress components
e = s - - ns
s
z ns
E E E
- ns - ns
s
E E E
zz xx yy
E E E
e =
zz
yy xx zz
yy
xx yy zz
xx
ns
e = s - ns -
x s
y s
x
y
z

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Symmetry of shear stress components
Take moment about the z axis, total torque = 0,
(sxy DyDz) Dx = (syx Dx Dz) Dy, hence, sxy = syx .
Similarly, syz = szy and sxz = szx
z
y
x
sxy
syx
Dx
Dy
Dz

11. dy
D D
Original shear strain is “simple” strain = , y ,... etc.
1 dy g qrot
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Define pure rotation angle qrot and
pure shear strain, such that the angular
displacements of the two surfaces are:
g1= qrot+ qdef and g2= qrot- qdef . Hence,
qrot = (g1+ g2)/2 and qdef = (g1- g2)/2
dx
x
There is no real deformation during pure rotation,
but “simple” strain ¹ 0.
x
y
Dx
g2 = -g
Example: g1 = 0 and g2 = - g,
so qdef = (0+g)/2 = g/2 and qrot= (0-g)/2 = -g/2
Pure shear strain is g/2
x
y
Dx
g2
qdef
qdef

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K = E
Example: Show that 3(1- 2n )
Proof:
D = @ 3Dl/l = 3e
E
3(1 2 )
l l l
V +D -
( ) 3 3
3
DV
º - D
K p
V/V
- n
=
D
l
V
exx = eyy = ezz = e, hence
3e = exx+eyy+ezz = (1-2v)(sxx+syy+szz)/E
sxx =syy =szz = -Dp (compressive stress)
= 3 (1-E2n) (-Dp)
V
For hydrostatic pressure l
l
l

13. 2 D’
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Example : Show that enn = g/2
Point C moves further along x- and y-direction by distances
of AD(g/2) and AD(g/2) respectively.
enn = [(AD . g/2)2 + (AD . g/2)2]1/2 / [(AD)2 + (AD)2]1/2 = g/2
True shear strain: eyx = g/2
Therefore, the normal component of strain is equal to the
shear component of strain:
enn = eyx and enn = g/2
g / 2 x
y
A
C’
C
D
g

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Example : Show that enn = snn/(2G)
Consider equilibrium along n-direction:
s yx (lW) sin 45o x2 = 2 (l cos 45o) W snn
l
syx
2 l cos 45o
s l n
sxy
Therefore syx = snn
From definition : g = sxy /G = snn /G = 2 enn

15. E 2 1+ =
exx = sxx/E - n syy/E- v szz/E
Set sxx = s nn = -s yy, s zz = 0, exx = enn
enn = (1+n) s nn /E = s nn /2G (previous example)
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G
E =
2
1
+
v
-snn
snn -snn
snn
Example : Show v G

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Ex. 12 kN forces are applied to the top
bottom of a cube (20 mm edges), E =
60 GPa, n = 0.3. Find (i) the force
exerted by the walls, (ii) eyy
z y
12kN
x
(i) exx = 0, syy = 0 and
szz= -12´103 N/(20´10-3 m)2 = 3´107 Pa
exx = (sxx- v syy- v szz) /E
0 = [sxx- 0 – 0.3´(- 3´107)]/60´109
sxx = -9´106 Pa (compressive)
Force = Asxx = (20´10-3 m)2´(-9´106 Pa) = -3.6 ´103 N
(ii) eyy = (syy- v szz- v sxx) /E
= [0 – 0.3´(- 3´107) – 0.3´(- 9´106)]/60´109 = 1.95´10-4

17. dU Fdx AE( x)dx
U e
AE ( x )
dx 2

AEe E e 2
A
( ) ( 
)
 
º = 1 e = s
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Elastic Strain Energy
The energy stored in a small volume:

= =
The energy stored :
= ò
0
1
= =
2
E V
2
1
= ×
2
1
2
e
F F
Energy density in the material :
E
E
u U
V
2
2
1
2
2
e=extension
dx

x

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Similarly for shear strain :
= 1 g = t
F
U = ò F × dx  = ò Fdx
G F A
= = t
dx
/
x

g
/ 
G
u G
2
2
1
2
2