2101202 – Mechanics of Materials I By Watanachai Smittakorn
VECTORS AND TENSORS
“A beautiful story needs a beautiful language to tell. Tensor is the language of mechanics.”
scalars
tensors
vectors
matrices
VECTORS
• defined as a line with magnitude and direction • denoted by
, , , , , ...AB PQ a u Fuuur r
• two vectors are equal if (both magnitude & direction are equal) • a unit vector is (a vector with a unit magnitude) • zero vector, denoted by 0, is (a vector with zero magnitude) • | AB|, |u|, v are magnitudes of AB, u, v
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Vector addition:
• “parallelogram law” • commutative and associative
a + b = b + a
(a + b) + c = a + (b + c) Vector subtraction:
a – b = a + (-b)
Let e1, e2, e3 be the unit vectors in x1, x2, x3 directions,
x1
x2
x3
u
u1
u2
u3
1 2 3 1 2
2 2 21 2 3
( , , )u u u u u u
u u u u
= + + =
= = + +
1 2 3u e e e
u3
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Scalar product (or dot product):
1 1 2 2 3 3
cos (0 )u v u v u v
θ θ π⋅ = + +
⋅ = ≤ ≤
u vu v u v
e.g., F · s Vector product (or cross product):
2 3 3 2 3 1 1 3 1 2 2 1
1 2 3
1 2 3
( ) ( ) (
sin (0 )
u v u v u v u v u v u v
u u uv v v
θ θ π
× = − + − + −
=
× = ≤ ≤
1 2
1 2 3
u v e e ee e e
u v u v
) 3
2
e.g., r × F Properties:
1 3 3
3 3 3 1
( )( )
( )k k k
× = − ×× + = × + ×× =× = × = × =
× = × = × =
× = × = ×
1 2 2
1 2 2 1
u v v uu v w u v u wu u 0e e e e e e 0e e e e e e e e eu v u v u v
HOMEWORK: 2.5, 2.7, 2.9
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
What is mechanics? Why is tensor important?
scalars
tensors
vectors
matrices
INDICIAL NOTATION
x1, x2,…, xn is denoted as xi, i = 1,…,n
i is an index with range of 1 to n
xi vs (xi) or X member vector/matrix
Summation convention:
3
1 1 2 2 3 31
i i i ii
a x a x a x a x a x=
+ + = ≡∑ Note: The repetition of an index in a term will denote a summation with respect to that index over its range. “dummy index” – one that is summed over “free index” – one that is not summed
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Hence, aixi = ajxj = akxk
e.g., 1 2 3
1 2 32 2 2 2 2
1 2 3
1 1 2 2 3 3
i i
i i
u u uv v v
u u u u u uu v u v u v u v
= + +
= + +
= = + + =
⋅ = + + =
1 2 3
1 2 3
u e e ev e e e
uu v
(more examples: aijxj, aijxjk, aijxij, …)
Kronecker delta (δij):
1 if ,
0 if ij
ij
i j
i j
δ
δ
= =
= ≠
or
( )1 0 00 1 00 0 1
ijδ⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
then 2
i i ij i j
ij i j
u u u u u
a a
δ
δ
= =
=
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Matrices and Determinants Let A be an m×n matrix and B be an n×p matrix.
( ) 1,..., ; 1,...,
( ) 1,..., ; 1,...,ij
jk
a i m j n
b j n k p
= = =
= = =
A
B
the product of A and B is an m×p matrix defined as ( ) 1,..., ; 1,..., ; 1,...,ik ij jka b i m j n k p⋅ = = = =A B
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 21 32
11 23 32 12 21 33 13 22 31
det det( )ij
a a aa a a a
a a aa a a a a a a a aa a a a a a a a a
= =
= + +
− − −
A
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Permutation symbol (εrst):
1 if , , permute as 1,2,3,1,2,3,...1 if , , permute as 3,2,1,3,2,1,...
0 otherwise
rst
rst
rst
r s tr s t
εεε
== −=
i.e.,
123 231 312
213 321 132
111 222 333 112 121 211
1,1,
... 0
ε ε εε ε εε ε ε ε ε ε
= = == = = −= = = = = = =
then,
1 2 3det( )ij rst r s ta a a aε=
rst s t ru vε× =u v e
( ) ( ) ijk i j ka b cε⋅ × = × ⋅ =a b c a b c
HOMEWORK: 2.16, 2.19
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
TRANSLATION AND ROTATION OF COORDINATES 2-D SPACE
• Translation
x
x’
y’y A
h
k
''
x x hy y k
= += +
or ''
x x hy y k
= −= −
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
• Rotation
x
y
x’
y’ A
θ θ
'cos 'sin'sin 'cos
x x yy x y
θ θθ θ
= −= +
or ' cos sin' sin cos
x x yy x y
θ θθ θ
= += − +
Using index notation,
' (i ij jx x i 1, 2)β= = where
cos( ', )ij i jx xβ =
11 12
21 22
cos sin( )
sin cosij
β β θ θβ
β β θ θ⎛ ⎞ ⎛ ⎞
= =⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
The inverse transform is
' ( 1,2i ji jx x i )β= =
Note: (βij) or B is an orthogonal matrix:
1( ) ( ) ( )Tji ij ijβ β β −= =
1−⋅ = ⋅ =TB B B B I
Hence,
ik jk ijβ β δ=
i.e., Consider a unit vector along xi’-axis (βi1,βi2)
2 21 2( ) ( ) 1 ( 1,2i i iβ β+ = = )
)
Unit vector along xi’-axis (βi1,βi2) is perpendicular to unit vector along xj’-axis (βj1,βj2) if i≠j
1 1 2 2 0 (i j i j i jβ β β β+ = ≠
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
3-D SPACE
y
x
z
x’
y’
z’
A
In the same manner as 2-D (See proof in pp. 51-52)
'i ij jx xβ= 'i ji jx xβ=
where direction cosines
'ij i jβ ≡ ⋅e e which is a cosine of angle between xi’ and xj axes Law of transformation of any vector A:
'i ij jA Aβ= 'i ji jA Aβ= HOMEWORK: 2.34
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
ANALYTICAL DEFINITIONS OF SCALARS, VECTORS, AND CARTESIAN TENSORS
P
x1
x2
x3
x2’
x1’
x3’
Scalar (or tensor of rank 0)
• has only single component • equal in all frames of reference
e.g., temperature, moisture, mass, …
1 2 3 1 2 3( , , ) '( ', ', ')x x x x x xφ φ=
Vector (or tensor of rank 1) • has 3 components • obeys law of transformation
e.g., force, displacement, velocity, …
1 2 3 1 2 3
1 2 3 1 2 3
'( ', ', ') ( , , )( , , ) '( ', ', ')
i k
k i
u x x x u x x xu x x x u x x x
ik
ik
ββ
==
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Tensor of rank 2
• has 9 components • obeys law of transformation
e.g., stress, strain, …
1 2 3 1 2 3
1 2 3 1 2 3
'( ', ', ') ( , , )
( , , ) '( ', ', ')ij mn im jn
mn ij im jn
x x x x x x
x x x x x x
σ σ β β
σ σ β β
=
=
Tensor of rank 3
ijke
Tensor of rank 4
ijklC
: Since based on rectangular Cartesian frame
called “Cartesian tensors”
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
PARTIAL DERIVATIVES
1 21 2
,
... nn
ii
i i
f f fdf dx dx dxx x xf dxx
f dx
∂ ∂ ∂= + +
∂ ∂ ∂∂
=∂
=
Comma notation:
,
,
,
ii
ii j
j
ijij k
k
xuux
x
φφ
σσ
∂≡
∂
∂≡
∂
∂≡
∂
HOMEWORK: 2.37, 2.38
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
EIGENVALUE PROBLEM For a symmetric matrix with real elements A=(aij)
11 12 13 1 1
21 22 23 2 2
31 32 33 3 3
a a a x xa a a x xa a a x x
λ⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟=⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
or ( )ij j i ij ij ja x x a x 0λ λδ= − =
e.g., stress, strain, buckling of column, vibration,…
• There are 3 real eigenvalues λ (λ1, λ2, λ3) obtained from
det( ) 0ij ija λδ− =
• There are 3 eigenvectors X corresponding to
the eigenvalues. They are mutually orthogonal (for distinct eigenvalues) and can be found from solving
ij j ia x xλ=
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
• If the coordinate system is rotated so as to coincide with the eigenvectors, then matrix A becomes diagonal matrix
1
2
3
0 0' 0 0
0 0
λλ
λ
⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
A
• The eigenvalues are the extremum values of
aij. EXAMPLE: Find eigenvalues (and eigenvectors) for
4 2( )
2 1ijσ⎛ ⎞
= ⎜ ⎟−⎝ ⎠
Soln: 1 1
2 2
2
4 22 1
4 2det 0
2 1(4 )( 1 ) (2)(2) 0
3 8 04.70, 1.70
x xx x
λ
λλ
λ λ
λ λλ
⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥−⎣ ⎦ ⎩ ⎭ ⎩ ⎭
−⎡ ⎤=⎢ ⎥− −⎣ ⎦
− − − − =
− − == −
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
STRESS (Shames 2.2, 2.3, 2.4)
Consider a continuum (solid or fluid) of volume V with surface S under some external forcing conditions.
δA
δF
n
t
Stress vector (or traction) is defined as
20
forcelim (unit: )lengthA
dA dAδ
δδ→
= =F FT
which has components in normal and tangential directions as
0
0
lim
lim
n nn A
t tt A
F dFA dA
F dFA dA
δ
δ
δτδ
δτδ
→
→
= =
= =
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
STRESS NOTATION
P
Stress (or force intensity) at a point depends on
• position • orientation
x y
z τxx
τxy
τxz τyy
τyx
τyz
τzz
τzx
τzy
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Stress tensor (a second-order tensor)
xx xy xz
ij yx yy yz
zx zy zz
τ τ ττ τ τ τ
τ τ τ
⎛ ⎞⎜ ⎟
= ⎜ ⎟⎜ ⎟⎝ ⎠
irst index identifies the plane (normal direction).
.g., τ is a stress acting on x-plane,
τ , τ , τ are called “normal stresses”.
s”.
ote: Stress tensor is symmetric, i.e., (see proof in
FSecond index identifies the direction of the stress. e xy
and pointing in y-direction.
xx yy zzτxy, τxz, τyx,… are called “shearing stresse
Nsection 2.4)
ij jiτ τ= or
xy yx
yz zy
xz zx
τ τ
τ τ
τ τ
=
=
=
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Sign convention mponent acts
ection or
-ive: Therection or
XAMPLE:
(For normal stresses, positive means tension and
+ive: The coon +i face in +j diron –i face in –j direction component acts
on +i face in –j di on –i face in +j direction E
+τ −τ
negative means compression.) HOMEWORK: 2.36, 2.50
x y
z +τxx
+τxy +x face
−x face xx xx
−τxy
−τxx
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
STRESS TRANSFORMATION (Shames 15.2)
ΔS = area of ABC n = unit normal vector ΔS1, ΔS2, ΔS3 = areas of OBC, OCA, OAB h = distance between O and N
n
From Newton’s law (ΣFi = mai)
i
O
A
B
C
x1
x2
x3
n
N TΔS
γΔV
i iS SΔ = Δ
1 1 2 2 3 3i i i i iT S S S S V Vaτ τ τ γ ρΔ − Δ − Δ − Δ − Δ = Δ
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
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3 3i ji j i iShSn aγ ρShΔT S τ Δ
Δ − =
as h→0,
Δ −
i ji jT nτ= “Cauchy’s formula”
Then,
nn i i ji j iT n n nτ τ= =
nt i i ji j iT t n tτ τ= =
XAMPLE: Find normal stress on plane whose nit normal vector
⎟= −⎜ ⎟⎜ ⎟⎝ ⎠
13 1 3
3
31 3 1 32 3 2 33 3 3 (2000)(0.6)(0.6) (2000)(0.6)(0.8) 2 2640 kPa
nn ji j in n n n n n n n
nn n n n n n
Eu
(0,0.60,0.80)=n 3000 1000 0−⎛ ⎞1000 2000 2000 kPa0 2000 0
ijτ ⎜
τ 11 1 1 12 1 2
21 2 1 22 2 2 23 2 n n n n n
τ τ τ τ
τ τ τ
= = + +
+ + +
= + × =
τ τ τ+ + +
HOMEWORK: 15.1, 15.7
2101202 – Mechanics of Materials I By Watanachai Smittakorn
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Next, from stresses τij in xi-system, find stresses τij’ in xi’-system.
From
x1
x2
x3
xk’
x m’
'i ij jx xβ= where β is cosi eij n of angle between xi’ and xj and Cauchy’s formula
i ji jT nτ= where nj is unit normal vector to any plane If n is parallel to xk’ axis, then
jj kn β=
n
T’k
τkm’
2101202 – Mechanics of Materials I By Watanachai Smittakorn
and
'ki ji kjT τ β= The component T’k in xm’ direction which is stress (τkm’) on xk’ face in xm’ direction is
3
'
m
τ 1 1 2 2 3 3
1 1 2 2 3
' ( ' ' ' )
' ' '
'
k k kkm m
k k km m
ki mi
T T T
T T Tβ
T
β β
β
= + + ⋅
= + +
=
e e e e
Hence,
'km ji kj miτ τ β β=
x
y
z
x’
y’
z’
τ τkm’ ij
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
PRINCIPAL STRESSES & PRINCIPAL PLANES (Shames 15.3, 15.4)
rom Cauchy’s formula (for a state of stress at a
j
Fpoint)
T ni jiτ= where nj is unit normal vector to any plane, there exist three mutually perpendicular axes (called principal axes) where there is zero shear stress (i.e., Ti = τni). (See Figure 1.)
j i ij ij jnor ( ) 0ijn nτ = τ τ τδ− =
In matrix form:
or
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
11 12 13 1 1
21 22 23 2 2
31 32 33 3 3
n nn nn n
τ τ ττ τ τ ττ τ τ
⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟=⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
11 12 13 1
21 22 23 2
31 32 33 3
000
nnn
τ τ τ ττ τ τ ττ τ τ τ
−⎛ ⎞⎜ ⎟− =⎜ ⎟⎜ ⎟−⎝ ⎠
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
To solve this problem,
11 12 13
det 0τ τ τ τ⎜ ⎟21 22 23
31 32 33
τ τ τ τ
τ τ τ τ
−⎛ ⎞− =⎜ ⎟
⎜ ⎟−⎝ ⎠
Invariants (i.e., do not change with coordinate system):
gives
3 21 2 3 0I I Iτ τ τ− + − + =
yields 3 roots for τ: τ1, τ2, τ3
1 11 22 33
22 23 11 1311 122
32 33 31 3321 22
11 12 13
3 21 22 23
31 32 33
I
I
I
τ τ ττ τ τ ττ ττ τ τ ττ τ
τ τ ττ τ ττ τ τ
+
= + +
=
= +
OMEWORK: 15.14 H
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Figure 1. Stress at a point in 3-D
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
MOHR’S CIRCLE (Shames 7.5)
is a graphical method of stress transformation (for a state of stress at a point)
• Plane stress (2-D) 0zz zx zyτ τ τ= = =
y τyy
τ τyyyxτ
Sign convention
• normal stress: + for tension −
• shear+ for clockwise − for counterclockwise
for compression stress:
L
K
z
x
yx τxy
τxy
ττ xx xx
L’ K’
τxx’ τxy’
τyx’ τyy’
θ
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
EXAMPLE:
L
K
1
2
4
aτ ⎜ ⎟
2
4 2 0⎛ ⎞2 1 0 kPj
0 0 0i = −
⎜ ⎟⎝ ⎠⎜ ⎟
τnn
2 2(2.5) (2) 3.2 kPaR = + =
:
3.2 1.7 kPa
principal stresses1
2 1.51.5 3.2 4.7 kPaτ
τ = −= + =
= −
maximum shear stress:
max ns R 3.2 kPaτ = = ±
K (4, -2)
L (-1, 2)
C
τns
τ1 τ2
R
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
• General state of stress (3-D) y, z coincide with principal axes
Let x,τ
10 20 30 40-10-20normal stress
(MPa)
shearing stress(MPa)
principal stresses: , 20 MPa1 2 340 MPa, 0τ τ τ= = = −
largest shear stress:
max 30 MPansτ =
x y
z τ1
τ2
3
τ1 τ2 τ3
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
EQUILIBRIUM EQUATIONS (Shames Consider a small element of a deformable body which is in equilibrium condition.
2.6)
x
y
z
dy
dz
dx
τzx
τ
Bx
τyx+(
0;
0
x
xxxx xx
yxyx yx
zxzx zx x
F
dx dydz dydzx
dy dxdz dxdzy
dz dxdy dxdy B dxdydzz
ττ τ
ττ τ
ττ τ
=
∂⎛ ⎞+ −⎜ ⎟∂⎝ ⎠∂⎛ ⎞
+ + −⎜ ⎟∂⎝ ⎠∂⎛ ⎞+ + − + =⎜ ⎟∂⎝ ⎠
∑
yx
τxx
∂τyx/ y)dy
τxx+(
∂
∂τxx/∂x)dx
τzx+( τzx/ z)dz ∂ ∂
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Canceling terms and dividing by dxdydz, we get
0yxxx zxxB
x y zττ τ∂∂ ∂
+ + + =∂ ∂ ∂
Similarly for y and z directions, we have
0
0
xy yy zyy
yzxz zzz
Bx y z
Bx y z
τ τ τ
ττ τ
∂ ∂ ∂+ + + =
∂ ∂ ∂∂∂ ∂
+ + + =∂ ∂ ∂
as equilibrium equations (in differential form). However, using index notation, these equations can be written as
, 0ji j iBτ + =
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
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EXAMPLE:
( )( )
( )
2 3
3
2 3
2 3 10 kPa
1000 2 10 kPa
2 10 kPa
zy
zx
zz
x y
x
xy x
τ
τ
τ
= + ×
= +
y
×
= + ×
Find resultant force.
1 12
0 0
(2000 3000 )
2167 kN
(1000 2000 )
917 kN
y zy
z zz
1 12
0 0
1 1
0 0
2000 kN
(1000 2000 )x zx
F dA x y dxdy
F dA x dxdy
F dA xy x dxdy
τ
τ
= = +
=
= = +
=
∫ ∫ ∫
∫ ∫ ∫
τ= = +∫ ∫ ∫=
0
1 m
1 m
x
2101202 – Mechanics of Materials I By Watanachai Smittakorn
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STRAIN (Shames 3.1, 3.2, 3.3)
Displacement field u(x,y,z)
1. Translation 2. Rotation 3. Extension/Contraction 4. Distortion
EXAMPLE: Given the following displacement field,
u = [(x2+y) i + (3+z) j + (x2+2y) k] m what is the deformed position of a point originally at (3,1,-2)m?
C’ D’
x
y r
u
r’A B
C D A’
B’
2 2' (3,1, 2) ((3) 1,3 ( 2), (3) 2(1)) (13, 2,9)
= + = − + + + − +=
r r u
2101202 – Mechanics of Materials I By Watanachai Smittakorn 2101202 – Mechanics of Materials I By Watanachai Smittakorn
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Translation
Rotation
r
A B
C D A’
A’
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Extension/Contraction
☺ Normal strain
dy
xx
yy
zz
dxdxdy
dydz
dz
ε
ε
ε
Δ=
Δ=
Δ=
(+ for extension, − for contraction)
A’
dx dy Δdx
Δ
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Distortion
A’
B’
C’
D’
εxy
εyx 90−γxy
☺ Shear strain
2
2
2
xyxy yx
γε ε
xzxz zx
yzyz zy
γε ε
γε ε
= =
= =
= =
is called “engineering shear strain”
+ for decrease in angle − for increase in angle
γ
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
STRAIN-DISPLACEMENT RELATIONS (Shames 3.4) Assume: infinitesimal deformation or displacement
( ), ,i x y zu u u u=
• Normal strain
xxx
yyy y
zzz
ux
u
u
z
ε
ε
ε
∂=
∂∂
=∂
∂=
∂
∂ux ∂x
∂uy
∂y
x
y
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
• Shear strain
∂ux
Since
yxxy
uuy x
γ∂∂
= +∂ ∂
then,
12
yxxy yx
uuy x
ε ε∂⎛ ⎞∂
= = +⎜ ⎟∂ ∂⎝ ⎠
Also,
12
12
x zxz zx
y zyz zy
u uz xu uz y
ε ε
ε ε
∂ ∂⎛ ⎞= = +⎜ ⎟∂ ∂⎝ ⎠∂⎛ ⎞∂
= = +⎜ ⎟∂ ∂⎝ ⎠
∂uy
∂x
∂y
x
y
0−γxy 9
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Strain tensor
xx xy xz
ij yx yy yz
zx zy zz
ε ε εε ε ε ε
ε ε ε
⎛ ⎞⎜ ⎟
= ⎜ ⎟⎜ ⎟⎝ ⎠
By using index notation, εij may be written as
( ), ,12ij i j j iu uε = +
PLE: For the following displacement field, u = (sin xi + yzj + x2k) × 10-2 m,
what are the strain components at r = 2i + j − k m?
EXAM
cos cos 2100 100
1100 100
0
1 1 (0 0) 02 200
1 1 (0 2 )2 200
1 1 ( 0)2 200 200
xxx
yyy
zzz
yxxy yx
x zxz zx
y zyz zy
u xxu zy
uz
uuy x
u u xz xu u yz y
ε
ε
ε
ε ε
ε ε
ε ε
∂= = =
∂∂ −
= = =∂
∂= =
∂∂⎛ ⎞∂
= = + = + =⎜ ⎟∂ ∂⎝ ⎠∂ ∂⎛ ⎞= = + = + =⎜ ⎟∂ ∂⎝ ⎠∂⎛ ⎞∂
= = + = + =⎜ ⎟∂ ∂⎝ ⎠
4
200
1
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
COMPATIBILITY EQUATIONS Shames 3.5)
In solid mechanics, displacement field (ui) must be a single-valued & continuous function.
Also, strain field must satisfy certain requirements to be associated with the single-valued & continuous displacement field.
“compatibility equations”
(
discontinuity ≈ crack or fissure
HOMEWORK: 3.32
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
2
2
2
2 22
2 2
2 2 2
,
,
2 ,
2
yy xy yzzx
xy yz zxzz
xy yyxx
zx
z,yz xyxx zx
y z x x y
2 2
2 2
2 ,yz yy zz
2zz
z x y y z x
x y z z x y
x y y x
y z z y
z x x
ε ε
ε ε εε
ε ε εε
ε εε
ε ε ε
ε ε
∂ ∂
∂ ⎠∂ ∂ ∂⎛ ⎞∂∂
= − + +⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠∂ ∂⎛ ⎞∂∂ ∂
= − + +⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠∂ ∂∂
= +∂ ∂
∂ ∂ ∂
∂ ∂=
∂ ∂ ∂
ε ε⎛ ⎞∂ ∂∂= − + +⎜ ⎟∂ ∂ ∂ ∂ ∂⎝
∂ ∂
= +∂ ∂ ∂ ∂
2
2 .xx
zε∂
+∂
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
STRAIN TRANSFORMATION Shames 16.1, 16.2, 16.3)
The strain components at a point transform exactly as do the stress components at a point on a rotation of axes.
second-order Cartesian tensor
From strains εij in xi-system, find stresses εij’ in xi’-system.
(
x x’
y
y’
z
z’
'ij mn im jnε ε β β= where βij is cosine of angle between xi’ and xj
'i ij jx xβ=
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
EXAMPLE: The following displacement field a body under load:
x y xz z −⎡ ⎤= + + + − ×u i j k
s
describes the movement of
( ) ( )2 2 2 23 0.6 10 ft⎣ ⎦ Compute the normal strain at (0, 1, 3) in a direction s given a
0.6 0.8= +s i j
( )
2 0100
0
1.2 3.6100 100
1 1 22 2 100 100
1 1 0 0 02 2
1 1 0 02 2 100
xxx
yyy
zzz
yxxy
x zxz
y zyz
u xxuy
u zz
uu y zy x
u u
2.5
z y
ε
ε
ε
ε
ε
ε
∂= = =
∂∂
= =∂
∂ − −= = =
∂∂⎛ ⎞∂ +⎛ ⎞= + = =⎜ ⎟ ⎜ ⎟∂ ∂ ⎝ ⎠⎝ ⎠
∂ ∂⎛ ⎞= + = + =⎜ ⎟z xu u x
∂ ∂⎝ ⎠∂⎛ ⎞∂ +⎛ ⎞= + = =⎜ ⎟ ⎜ ⎟∂ ∂ ⎝ ⎠⎝ ⎠
'
2.5 3.6 (0.6)(0.8) 2 (0)(0)100 100
0.024
ij mn im jnε
ss xy sx sy yx sy sx zz sz sz
ε β β=
−⎛ ⎞ ⎛ ⎞= × +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
ε ε β β ε β β ε β β= + +
HOMEWORK: 16.1
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
Principal strains & principal axes:
Solving an eigenvalue problem
ij j in nε ε=
0xx xy xz
yx yy yz
zx zy zz
ε ε ε εε ε ε εε ε ε ε
−− =
−
yields
where A, B, and C are strain invariants:
3 2 0A B Cε ε ε− + − =
2 2 2
2 2 22
xx yy zz
xx yy xx zz yy zz xy yz xz
xx yy zz xy yz zx xy zz xz yy yz xx
A
B
C
ε ε ε
ε ε ε ε ε ε ε ε ε
ε ε ε ε ε ε ε ε ε ε ε ε
= + +
= + + − − −
= + − − −
See example 16.2) (
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
STRAIN MEASUREMENT AND STRAIN OSETTES R (Shames 8.6)
Strain gage – a device for measuring strain
Strain rosettes – used to measure the state of strain
at a point (2D)
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
INTRODUCTION TO ELASTICITY
(Hooke’s Law)
Solid Mechanics
force (F)
displacement (u)
stress (τ)
strain (ε)
equilibrium compatibility
constitutive relation
(material law)
e.g., constitutive relation for a bar in tension
τ τ
Eτ ε=
where E = modulus of elasticity or Young’s modulus
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
THREE-DIMENSIONAL HOOKE’S LAW FOR ISOTROPIC MATERIALS
) (Shames 6.2, 6.3
“isotropic materials” – have the same properties in all directions
nder τxx:
τxx τxx
τxx τxx
u
' xxxx E
τε = where E = modulus of elasticity
or Young’s modulus
and by Poisson effect
' ' xxyy xx
τ
' ' xxzz xx
E
E
ε νε ν
τε νε ν= − = −
= − = −
where ν = Poisson’s ratio
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
under τyy:
'' yyyy E
τε =
and
'' ''
'' ''
yyxx yy
yyzz yy
E
E
τε νε ν
τε νε ν
= − = −
= − = −
under τzz:
''' zzzz E
τε = and
''' '''
''' '''
zzxx zz
zzyy zz
Eτε νε ν
τE
ε νε ν= − = −
= − = −
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
τxx τxx
τyy
τyy
τzz
τzz
Hence, superposing the effects of the normal stresses
( )
( )
( )
1
1
1
xx xx yy zz
yy yy xx zz
zz zz xx yy
E
E
ε τ
E
ν τ τ
τ ν τ τ
ε τ ν τ τ
⎡ ⎤= − +⎣ ⎦
⎡ ⎤= − +⎣ ⎦
⎡ ⎤= − +⎣ ⎦
ε
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
For isotropic materials, shear stresses act independently of each other.
γxy
1
1
1
xyγ xτ yG
xγ z xz
yz yz
G
G
τ
γ τ
=
=
=
where G = shear modulus
ote: relation between the three material constants N
( )2 1EG
ν=
+
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
In matrix form: 1 0 0 0
1 0 0 0
1 0 0 0
10 0 0 0 0
10 0 0 0 0
10 0 0 0 0
xx xx
yy yy
zz zz
yz yz
xz xz
xy xy
E E E
E E E
E E E
G
G
G
ν ν
ν ν
ε τε τν νε τγ τγ τγ τ
⎤⎢ ⎥⎢ ⎥⎢ ⎥− −⎧ ⎫ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥− −⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎢ ⎥
⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
Inverse matrix:
44
66
0 0 00 0 0
0 0 0 0 00 0 0 0 0
⎡ − −
11 12 13
21 22 23
31 32 33 0 0 0
55
0 0 0 0 0
xx xx
yy yy
τ
zz zz
yz yz
xz xz
xy xy
C C CC C C
CC
ετ ετ εC C C
⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥τ
C
γτ γτ γ
⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪⎥ ⎨ ⎬
⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪
⎣ ⎦⎩ ⎭ ⎩ ⎭
Here, Cij depend on only two independent variables (E, ν, or G)
= ⎢⎨ ⎬
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
ANISOTROPIC MATERIALS: GENERALIZED HOOKE’S LAW (Shames 6.5) In general for anisotropic materials:
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
xx x
yy yy
x
zz z
yz yz
z
xz xz
xy x
C C C C C CC C C C C CC C C C C CC C C C C CC C C C C CC C C C C C y
τ ετ ετ ετ γτ γτ γ
⎧ ⎫ ⎧⎡ ⎤⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪= ⎢ ⎥⎨ ⎬ ⎨
⎫
⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎣ ⎦⎩ ⎭ ⎩
⎪⎪⎪⎪
⎪⎪⎪⎭
Special case: for materials with symmetry about planes (e.g., wood)
called “orthotropic”
0 0 00 0 00 0 0
0 0 0 0 00 0 0 0 0
⎬⎪
three orthogonal
11 12 13
21 22 23
31 32 33
44
55
660 0 0 0 0
xx xx
yy yy
τ ε
zz zz
yz yz
xz xz
xy xyC
C C CC C CC C C
CC
τ ετ ετ γτ γτ γ⎢ ⎥⎪ ⎪ ⎪ ⎪⎣ ⎦⎩ ⎭ ⎩ ⎭
⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪= ⎢ ⎥⎨ ⎬ ⎨ ⎬
⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
PLANE STRESS (Shames 7.1)
Consider a thin plate loaded in the midplane of
Hence,
the plate.
x
y y
z
0zz zx zyτ τ τ= = =
0 but 0)zx zy zz(for isotropic material: ε ε ε= = ≠
Transformation of stress and strain in 2-D can use Mohr’s circle.
small t
τyy
τxy
τxx
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
PLANE STRAIN (Shames 8.1)
A prismatic body constrained not to bend, stretch, or shorten is loaded in a direction normal to the centerline with no variation in the direction of the axis of the prism. Hence,
0zz zx zyε ε ε= = =
(for isotropic material: 0 but 0)zx zy zzτ = τ τ= ≠
Transformation of stress and strain in 2-D can use Mohr’s circle. e.g., a prismatic dam built between rigid rock supports
x
y
W H
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2101202 – Mechanics of Materials I By Watanachai Smittakorn
EXAMPLE: A gas pipeline with 600 mm inside diameter hickness is carrying a pressure of 1.5 MPa
bove atmosphere. ) Fi in
the w
and 12 mm wall ta(a nd the principal stresses and maximum shearing stress
all. (b) If the pipe is made of steel with E=200 GPa and ν=0.28, find the principal strains in three dimensions.
σ
τ
σ1 2
(a)
hoop 1
1 2max
(1.5)(300) 37.5 MPa12
(1.5)(300)
18.75 9.375 MPa2 2
prt
praxial 218.75 MPa
2 2 12t
σ σ
σ σ= = = =×
σ στ
= = = =
−= = =
(b)
( )( ) ( ) ( )( ) ( )
( )( ) ( ) ( )( ) ( )
( )( ) ( ) ( )( ) ( )
36
61 1 2 3 9
66
2 2 1 3 9
66
3 3 1 2 9
0 (plane stress)
1 10 37.5 0.28 18.75 0 161 10200 10
1 10 18.75 0.28 37.5 0 41.3 10200 10
1 10 0 0.28 37.5 18.75 78.8 10200 10
E
E
E
σ
ε σ ν σ σ
ε σ ν σ σ
ε σ ν σ σ
−
−
−
=
= − + = − + =
= − + = − + =
= − + = − + = −
σhoop
σaxial
τmax
σ
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