Chapter 2 Mechanics of Materials

F
A

F

F

F

F

F

F

F Tensile stress (+)
Compressive stress (-)

Force
N / m2
Normal stress =  
A
F dF
  lim

At a point:
A  0  A
dA

pascal (Pa )
Example: Estimate the normal stress on a shin bone (脛骨)
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Shear stress (切應力) =  = F tangential to the area / A

F
A

F
At a point,
dF

dA
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Normal strain (正應變)  = fractional change of length= x / l


F
F
x
l
Shear strain (?) = deformation under shear stress = x / l
x
l
F

fixed
F
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Stress-strain curve
Work
hardening

Elastic
deformation
o
Yield pt.
break

In elastic region,   , or /  = E
E is a constant, named as Young’s modulus or modulus of
elasticity
Hooke's law:
Similarly, in elastic region, / = G, where G is a constant,
named as shear modulus or modulus of rigidity.
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Exercise set 2 (Problem 3)
Find the total
extension of the bar.
dx
2kN
15mm
5mm W
o
0.6m
X
1.2m
Width of a cross-sectional element at x: W  x (5 103 m)  x (m)
0.6m
120
2 103 N
2.88 107
Stress in this element :  

Pa
2 2
2
( x / 120) m
x
 2.88 107 / x 2 1.92 104
Strain of this element:   

9
E
150 10
x2
4
1
.
92

10
dx
The extension of this element : de   dx 
2
x
4
1.8 1.92  10
dx
The total extension of the whole bar is : e   de  0.6
2
x
-4
= 2.13 x 10 m
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Bulk modulus  K 
p
 (V / V )
dp
 V
dV
p
V  V
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Poisson's ratio :
For a homogeneous isotropic material
d  d
d
F
F





x
x
normal strain :  

d
L 
d
lateral strain :
Poisson's ratio :    L / 
value of  :
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  xx, y  yy , z  zz,
x  xx
z
z
zx
xz
x
zy
xy
yz
yx
y
y
x
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Joint effect of three normal stress components
 xx yy zz
 xx 


E
E
E
 yy xx zz
 yy 


E
E
E
 zz xx yy
 zz 


E
E
E
z
z
y
x
y
x
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Symmetry of shear stress components
Take moment about the z axis, total torque = 0,
(xy yz) x = (yx x z) y, hence, xy = yx .
Similarly, yz = zy and xz = zx
y
yx
xy
y
z
x
z
x
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dy dx
Original shear strain is “simple” strain = x , y ,... etc.
y
There is no real deformation during pure rotation,
but “simple” strain  0.
2
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
Example: 1 = 0 and 2 = - ,
so def = (0+)/2 = /2 and rot= (0-)/2 = -/2
Pure shear strain is /2

1
x
rot dy
x
def
def
y
2 = -
x
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x
Example: Show that K 
E
3(1  2 )
Proof: For hydrostatic pressure
( ll ) 3  l 3
V

V
l3
 3l /l  3 
l
l
l
xx = yy = zz = , hence
3 = xx+yy+zz = (1-2v)(xx+yy+zz)/E
xx =yy =zz = -p (compressive stress)
V
V

(1  2)
3
(p)
E
K
p
E

V / V 3(1  2)
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Example : Show that nn = /2
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:
y
C’
nn = yx and nn = /2
C

2
D’
A
 /2
D
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x
Example : Show that nn = nn/(2G)
Consider equilibrium along n-direction:
 yx (lW) sin 45o x2 = 2 (l cos 45o) W nn
yx
l
xy
n
l
Therefore yx = nn
From definition :  = xy /G = nn /G = 2 nn
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Example : Show
E  2G
1 v
xx = xx/E -  yy/E- v zz/E
Set xx =  nn = - yy,  zz = 0, xx = nn
nn = (1+)  nn /E =  nn /2G (previous example)
E

2G
1v
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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= -12103 N/(2010-3 m)2 = 3107 Pa
xx = (xx- v yy- v zz) /E
0 = [xx- 0 – 0.3(- 3107)]/60109
 xx = -9106 Pa (compressive)
Force = Axx = (2010-3 m)2(-9106 Pa) = -3.6 103 N
(ii) yy = (yy- v zz- v xx) /E
= [0 – 0.3(- 3107) – 0.3(- 9106)]/60109 = 1.9510-4
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Elastic Strain Energy
 The
energy stored in a small volume:
x
dU  Fdx  AE ( )dx

  The energy stored :
AE
U 
( x)dx e=extension
0 
F
1 AEe2 1 e 2

 E ( ) ( A)
2 
2 
1 2
 E V
2
e
F

dx
x

Energy density in the material :
U 1 2 1 2
u   E 
V 2
2 E
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Similarly for shear strain :
U 
 
F  dx   Fdx
F/A 
G

x/ 
dx
2
1
1

u  G  2 
2
2G
F

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