R&DE (Engineers), DRDO
Ramadas Chennamsetti
rd_mech@yahoo.co.in
Stress Analysis
Summary
Cauchy’s formula
Stress transformation
Principal stresses
Maximum shear stress
Octahedral stress
Stress decomposition
Equilibrium equations
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2
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R&DE (Engineers), DRDO
Introduction
Body subjected to loading – internal
resistance offered by body
Internal forces – distributed continuously
over the body
Deformation – deformable body – finite
elastic modulus
Deformation – continuous – single valued
Stress analysis – equilibrium of the body –
every point is in equilibrium
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3
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R&DE (Engineers), DRDO
Introduction
R&DE (Engineers), DRDO
Equilibrium
y
Ω
r
G
α
x
Γ
∑F
= ma
∑F
= r× m a + I G α
∑M
~
∑M
~
~
~
~
~
~
~
Dynamic equilibrium
=0
=0
∑F
∑M
=0
i
i
Static equilibrium
Ramadas Chennamsetti
=0
4
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z
Forces
R&DE (Engineers), DRDO
T
Surface forces/tractions
d Fs = T dS
~
Fs =
~
ds = infinitesimal area
dF si = Ti dS
~
∫∫ T dS
S
~
Fs = Force corresponds to
traction acting over ‘S’
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5
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Force per unit area
Acts on surfaces
Need not be perpendicular to
surface area – reactions of other parts
Traction – Surface Normal
R&DE (Engineers), DRDO
Direction of traction and normal to area different
n
θ
dA
dA – differential element
n - normal to area
~
x
T = Ti ei
~
dA = n dA
~
~
Angle between traction and area vector => θ
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T – traction on area ‘A’
y
z
T
Forces
R&DE (Engineers), DRDO
Body load
Per unit volume or mass
Distributed over the body
y
Not a surface force
Acts at center of mass
z
Gravity force, centrifugal force
B = Bx i + B y j + Bz k
~
~
~
Body force vector
~
B = B i ei
Ω
Bx
Bz
x
d F = B dV
~ B
~
~
Indices notation
Total force due to body load
Ramadas Chennamsetti
F = ∫∫∫ B dV
~ B
V
~
~
7
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By
Internal Forces
R&DE (Engineers), DRDO
Body under equilibrium – every particle
under equilibrium
y
∆F
A
I
P
II
P
∆A
x
z
A
Section A-A
Perpendicular to X - axis
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I
Internal Forces
R&DE (Engineers), DRDO
Second part of the body –
∆F
I
P
− n
n
∆A1
II
~
~
-∆F
∆F
Decomposing forces
1, x
∆A
3, z
P
∆Fz
∆Fx
∆ F = ∆ Fi e i = ∆ F1 e1 + ∆ F2 e 2 + ∆ F3 e 3
~
~
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~
~
~
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∆Fy
2, y
Internal Forces
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Normal stress –
Force perpendicular to area – σ 11 = τ 11 = lim ∆F1 = dF1
∆A→0
Shear stress –
∆A1
dA1
∆Fs acts in the plane – resolve into
components
∆F2 dF2
τ 21 = lim
=
dA1
∆A→0 ∆A1
∆F3 dF3
τ 31 = lim
=
dA1
∆A→0 ∆A1
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∆Fs dFs
=
Force parallel to area – τ s1 = lim
dA1
∆A→0 ∆A1
Internal Forces
R&DE (Engineers), DRDO
Stress components – depend on forces,
orientation of the plane
Notation – Two indices –
τ yx τ
ij
τ xx
Indices notation
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First index – plane & second index - direction
Internal Forces
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Stresses
A
P
P
A
x
z
Neighboring material exerts forces on the faces
of infinitesimal element
Decompose these forces – parallel &
perpendicular to co-ordinate axes
Divide by area,Ramadas
dA => 0Chennamsetti
12
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y
Stresses at a Point
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Stresses
Traction on each face of
the element
τ yy
τ xy
τ zy
τ zx
τ zz
dy
~
∆F
~
lim
∆A
∆A→ 0
τ xx
τ xz
y
dz
dx
z
~
~
∆ Fi
∆ F1
∆ F2
∆ F3
ei = lim
e1 +
e2 +
e3
∆A→ 0 ∆ A ~
∆A → 0 ∆ A ~
∆A ~ ∆A ~
= lim
T 1 = σ 1i ei = σ 11 e1 + τ 12 e2 + τ 13 e3
~
x
~
~
zz
~
~
~
~
~
~
T 2 = σ 2 i ei = τ 21 e1 + σ 22 e2 + τ 23 e3
~
~
~
T 3 = σ 3i ei = τ 31 e1 + τ 32 e2 + σ 33 e3
~
~
~
~
Traction on each Ramadas
plane =>
Stresses acting on that plane
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~
13
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τ yz
∆F = ∆Fi ei = ∆F1 e1 + ∆F2 e2 + ∆F3 e3
τ yx
τ
Stresses at a Point
R&DE (Engineers), DRDO
∆x, ∆y and ∆z => 0 – it becomes a point
[]
[ ] []
Nine stress components
3 – Normal stresses, 6 – Shear stresses
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τ11 τ12

T = τ 21 τ 22
~
τ 31 τ 32
 
τ13 e1 
~



τ 23  e2 => T = [σ ] e
~
~
~
τ 33 e3 
 ~ 
Stresses at a Point
R&DE (Engineers), DRDO
Shear stresses - Complementary
σyy
Static equilibrium of the element
τyx
τxy
y
σxx
τxy
τyx
x
dx
~
~
x
y
z
σyy
This gives =>
Apply same method in other directions also,
τ xy = τ yx
τ yz = τ zy , τ zx = τ xz
Shear stresses are complementary in nature
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15
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σxx
dy
∑ F = 0, ∑ M = 0
∑ F = 0, ∑ F = 0
∑M =0
Stresses at a Point
R&DE (Engineers), DRDO
Nine stress components
Three normal stresses - τxx, τyy, τzz
Six shear stresses - τyz, τzy, τxz, τzx, τxy, τyx
Complementary => τyz= τzy; τxz= τzx; τxy= τyx
Three shear stress components
Total number of stress components at a point
=> three normal + three shear = six
components
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Stresses at a point – 3D stress state
Stresses at a Point
R&DE (Engineers), DRDO
Nine (six) stress – arrange in an array –
Rows represent
First row – stresses on ‘x’ plane
Second row – stresses on ‘y’ plane
Third row – stresses on ‘z’ plane
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τ xx τ xy τ xz 


[τ ] = τ yx τ yy τ yz 
τ zx τ zy τ zz 


Sign Convention
R&DE (Engineers), DRDO
Positive or negative stress – depends on
Direction of area normal and
σxx – Force in +ve x-direction
Force direction
+n
Area normal => +ve area
=> Stress is +ve - Tensile Stress
τ xy
τ xz
+n
z
τxy – Force in +ve y – direction
σ xx
+n
x
Area => +ve area normal +ve
=> Stress is +ve
τxz – Force in +ve z – direction
Area => +ve area normal +ve
=> Stress is +ve
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18
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y
Sign Convention
R&DE (Engineers), DRDO
Positive/negative stresses
σxx – Force in -ve x-direction
y
Area normal => -ve area
=> Stress is +ve - Tensile Stress
τxz
τxy – Force in -ve y – direction
σxx
Area => -ve area normal
=> Stress is +ve
τxy
x
z
τxz – Force in -ve z – direction
Area => -ve area normal
=> Stress is +ve
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19
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-n
Sign Convention
R&DE (Engineers), DRDO
Positive/negative stresses
σxx – Force in -ve x-direction
+n
τ xy
τ xz
+n
z
Area normal => +ve area
=> Stress is -ve => Compressive
Stress
σ xx
τxy – Force in -ve y – direction
+n
x
Area => +ve area normal
=> Stress is -ve
τxz – Force in -ve z – direction
Area => +ve area normal
=> Stress is -ve
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20
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y
Sign Convention
R&DE (Engineers), DRDO
Positive/negative stresses –
σxx – Force in +ve x-direction
y
τxz
σxx
τxy – Force in -ve y – direction
-n
Area => -ve area normal
=> Stress is +ve
x
τxz – Force in +ve z – direction
τxy
z
Area => -ve area normal
=> Stress is -ve
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Area normal => -ve area
=> Stress is -ve => Compressive
Stress
Cauchy’s formula
R&DE (Engineers), DRDO
Stress components acting on an element
τ yy
τ yx
τ yz
τ zz
τ xy
τ zy
τ zx
y
τ zz
dy
On each face three stress
components are acting
x
τ xx
τ xz
A
z
dz
C
Consider a plane ABC
intersecting ‘x’, ‘y’ and ‘z’
axes
Normal has direction
cosines (l, m, n)
dx
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B
Cauchy’s Formula
R&DE (Engineers), DRDO
Six stress components are given at point –
Tractions on an inclined plane –
y
B
B
σxy
σxx
σzx
o
σxz
z
Ty
σzy
σyx
σyz
C
σzz
A
x
T
Tx
Tz
σyy
A
x
C
z
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23
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y
Cauchy’s Formula
R&DE (Engineers), DRDO
Direction cosines of ABC plane
n = l e1 + m e2 + n e3 = li ei
n
y
~
~
~
~
Equilibrium of tetrahedron
B
m
Tx A = σ xx Ax + τ yx Ay + τ zx Az
l
o
A
x
C
Ay
Ax
Az
Tx = σ xx
+ τ yx
+ τ zx
A
A
A
Tx = σ xxl + τ yx m + τ zx n
z
Ty A = τ xy Ax + τ yy Ay + τ zy Az
Tz A = τ xz Ax + τ yz Ay + σ zz Az
Ty = τ xy l + τ yy m + τ zy n
Tz = τ xz l + τ yz m + σ zz n
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n
~
Cauchy’s Formula
R&DE (Engineers), DRDO
Stress state at a point and direction cosines of a
plane => Tractions on that plane
Ty = τ xy l + σ yy m + τ zy n
Tz = τ xz l + τ yz m + σ zz n
l + m + n =1
2
2
Indices notation
Ti = σ ji l j
Tx  σ xx τ yx τ zx  l 
 
  
Ty  = τ xy σ yy τ zy m => {T} = [σ ]{α}
T  τ
n 

τ
σ
yz
zz  
 z   xz
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2
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Tx = σ xx l + τ yx m + τ zx n
Normal Stress
R&DE (Engineers), DRDO
Resultant traction –
y
TR2 = Tx2 + Ty2 + Tz2
n (l, m, n), σn
Direction cosines of TR
Ty
Tr
Tx
x
A
Tz
Ty
Tx
Tz
'
'
l = , m = , n =
TR
TR
TR
'
Projections of Ti on n
C
z
Normal stress on plane ABC
σ n = lTx + mTy + nTz
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B
Normal Stress
R&DE (Engineers), DRDO
Normal stress is given by,
σ
n
= l (l σ
m (l τ
n (l τ
xx
+ mτ
yx
+ nτ
zx
xy
+ mσ
yy
+ nτ
zy
xz
+ mτ
+ nσ
zz
yz
)+
)+
)
σ n = l 2σ xx + m 2σ yy + n 2σ zz +
2 (lm τ xy + mn τ yz + nl τ zx )
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σ n = lTx + mTy + nTz
Shear Stress
R&DE (Engineers), DRDO
Calculate shear stress from tractions and
normal stress
T = σ +τ
2
τ = T −σ
2
n
2
n, σn
T
2
Plane
P
2
σ n = l 2σ xx + m 2σ yy + n 2σ zz +
2 (lm τ xy + mn τ yz + nl τ zx )
τ = (T + T + T ) − σ
2
2
x
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2
y
2
z
2
n
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τ
2
n
Stresses in 2D
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2D stress state - σxx, σyy and τxy
n
Ty
~
Tx = lτ xx + m τ yx
T
dy
τxx
α
τxy
τyx
dl
σ n = lTx + mT y
dx
x
τyy
cos α =
dy
=l
dl
cos( 90 − α ) = sin α =
n = li + m
~
~
j
~
Ty = lτ xy + mτ yy
Tx
dx
=m
dl
σ n = τ xx l 2 + τ yy m 2 + 2lmτ xy
σ n = τ xx cos 2 α + τ yy sin 2 α + τ xy sin 2α
 τ xx + τ yy   τ xx − τ yy 
 + 
 cos 2α + τ xy sin 2α
σ n = 
2  
2 

29
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y
Stresses in 2D
R&DE (Engineers), DRDO
Normal stress in ‘t’ direction
Angle => 90+α
 τ xx + τ yy   τ xx − τ yy 
 − 
 cos 2α − τ xy sin 2α
σ t = 
2  
2 

Shear stress τ
2
=T
2
−σ
T
t
n
~
~
2
n
1
τ = − (τ xx −τ yy )sin 2α + τ xy cos 2α
2
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~
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 τ xx + τ yy   τ xx − τ yy 
 + 
 cos 2α + τ xy sin 2α
σ n = 
2  
2 

Stress Transformation
R&DE (Engineers), DRDO
Transformation of stresses from one
co-ordinate system (XYZ) to another
2D case
co-ordinate system (xyz)
Y
y
Y
P(X, Y) = P(x, y)
y
x
x
o
X
l11 = cos ∠xoX l12 = cos ∠xoY
O
z
l21 = cos ∠yoX l22 = cos ∠yoY
P(x,
Z
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31
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X
Stress Transformation
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2D Transformation –
x1 = X 1l11 + X 2l12
x2 = X 1l21 + X 2l22
 x1  l11
 =
x2  l21
l12  X1 
  => {x} = [T ]{X }

l22 X 2 
Where [T] is transformation
matrix
Writing in indices form
xi = lij X j
l11
[T ] = 
l21
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l12 

l22 
32
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x = Xl11 + Yl12 or
Writing in matrix form
Stress Transformation
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Stress is a second-order tensor – two
indices
Second order tensor => Outer product of
two first order tensors
cij = ai ⊗ b j => c = a b
~ ~
Transforming this product to xyz csys
c = a i b j = lim am l jn bn = lim l jn am bn = lim l jn cmn
'
ij
'
'
σ =l l σ
'
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Chennamsetti
ij
im jn mn
33
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≈
Dyadic product in XYZ csys
Stress Transformation
R&DE (Engineers), DRDO
Writing in terms of transformation matrix
σ
'
ij
= l im l jn σ
mn
[σ ] = [T ][σ ][T ]
T
'
τxx

[σ] = τ yx
τzx

τxy
τ yy
τzy
τxz

τ yz
τzz 
=> Stress tensor
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Where,
Stress Transformation
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Applying in 2D stress transformation –
Y
x
τxy
y
τyy
τxx
τxy
τxx
τXX
τXY
τYX
τYY
Transformation matrix
l11
[T ] = 
l21
l12 
l22 
Stress tensor in XYZ
σ 11
[σ XYZ ] = 
τ 12
τ 12 
σ 22 
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Stress tensor in xyz
[σ ] = [T ][σ ][T ]
T
xyz
XYZ
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X
Stress Transformation
R&DE (Engineers), DRDO
2D Stress transformation –
σ xx
τ
 xy
τ xy  l11
=

σ yy  l21
l12  σ XX
l22   τ XY
σ xx
τ
 xy
τ xy   l11σ XX + l12τ XY
=

σ yy  l21σ XX + l22τ XY
τ xy 
=
σ yy 

l 112 σ XX + 2l11l12τ XY + l 122 σ YY

l11l 21σ XX + l12 l 21τ XY + l11l 22τ XY + l12 l 22σ YY
σ xx = l112 σ XX + 2l11l12τ XY + l122 σ XX
τ XY  l11
σ YY  l12
l21 
l22 
l11τ XY + l12σ YY  l11
l21τ XY + l22σ YY  l12
l21 
l22 
l11l 21σ XX + l12 l 21τ XY + l11l 22τ XY + l12 l 22σ YY 

l 212 σ XX + 2 l 22 l 21τ XY + l 222 σ YY

σ yy = l212 σ XX + 2l22 l21τ XY + l222 σ XX
τ xy = l11 l 21σ XX + (l12 l 21 + l11 l 22 )τ XY + l12 l 22 σ YY
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σ xx
τ
 xy
Principal Stresses
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Maximum normal stresses which act on the
planes, where there is no shear stress
y
Tractions on ABC in x, y & z directions
n (l, m, n), σ
B
Tx = lσ xx + mτ xy + nτ xz = lσ
Ty
T
Tz = lτ xz + mτ yz + nσ zz = nσ
Tx
x
A
Tz
C
z
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Ty = lτ xy + mσ yy + nτ yz = mσ
Principal Stresses
R&DE (Engineers), DRDO
Writing in matrix form –
τ xy
σy
τ yz
τ xz  l  1
 

τ yz m = σ 0
σ z  n  0
[σ ]{α } = σ [I ]{α }
[[σ ] − σ [I ]]{α } = 0
0
1
0
0 l 
 

0m
1 n 
Homogeneous algebraic
equations – Eigenvalue
problem
σ - Eigenvector, {α} - Eigenvector
Expanding – cubic equation – three roots
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σ x

τ xy
τ xz

Principal Stresses
R&DE (Engineers), DRDO
For non-trivial solution – {α}≠ 0 –
Determinant = 0
σ x −σ τ xy
τ xz
τ xy σ y −σ τ yz = 0
τ xz
τ yz σ z −σ
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[σ ] − σ [I ] = 0
Principal Stresses
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Eigenvalue problem –
Homogeneous algebraic equations
Eigenvalues => Three principal stresses
Eigenvalues => Real values – definite matrix
Eigenvectors => Orientation of principal planes
Number of Eigenvalues => order of matrix
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Principal Stresses
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Expanding the determinant –
σ − I 1σ + I 2σ − I 3 = 0
3
2
I1 = σ x + σ y + σ z
τ xy σ x
+
σ y τ xz
σx
I3 =
τ xz σ y
+
σ z τ yz
τ xy
σy
symm
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τ yz
σz
τ xz
τ yz
σz
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σx
I2 =
τ xy
Principal Stresses
R&DE (Engineers), DRDO
Principal stresses can be obtained from the
following

I
a = − I 2 +
3

2
1
I1
a
φ 
σ 1 = + 2 − cos  
3
3
3
I1
a
φ

σ 3 = − + 2 − cos  240 + 
3
3
3

2 3 I1 I 2 

b = − I 3 +
I1 +

27
3 



−b

φ = cos 

3
 2 − a / 27 
Ramadas Chennamsetti
−1
42
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I1
a
φ

σ 2 = + 2 − cos  120 + 
3
3
3




Principal Stresses
R&DE (Engineers), DRDO
Invariant – value doesn’t change when the coordinate
system is rotated – referring stresses wrt a different
coordinate system
Same coefficients of the characteristic equation for any
orientation
Independent of orientation of the planes
The principal stresses depend only on loading
For a given stress state – three mutually perpendicular
planes - maximum stresses
Calculation of principal plane orientations – use
direction cosines relation also
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43
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Stress invariants – I1, I2 & I3
Principal Stresses
R&DE (Engineers), DRDO
n3, σ3
P
Any direction on this plane
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44
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Distinct values of σ1, σ2 and σ3 – principal
planes unique – mutually perpendicular
n1.n2 = n2.n3 = n3.n1 = 0
If σ1 = σ2 & σ3 different – any direction
perpendicular to n3 => perpendicular
direction
Principal Stresses
R&DE (Engineers), DRDO
If σ1 = σ2 = σ3 = p
every direction is principal direction –
hydrostatic state of stress – any three
orthogonal directions
p
p
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45
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p
Maximum Shear Stress
R&DE (Engineers), DRDO
Given a stress state at a point
{T}
σ2
xyz csys coincides with principal stress directions
shear forces on Ax, Ay and Az = 0
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46
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σ3
σ1
Maximum Shear Stress
R&DE (Engineers), DRDO
Tractions on plane ‘A’ => Ti = σij lj
σij = 0 for i ≠ j
Here,
T1 = lσ1 , T2 = mσ 2 , T3 = nσ 3
T = Ti ei
~
~
T = T + T + T = (lσ1 ) + (mσ 2 ) + (nσ 3 )
2
1
2
2
2
2
3
2
2
Normal Stress on plane ‘A’
σ ij = liT j
for i = j => Normal stress
σ =lT
ii
i i
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47
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2
Maximum Shear Stress
R&DE (Engineers), DRDO
Normal stress –
σ n = lT1 + mT 2 + nT3 = l (lσ 1 ) + m (m σ 2 ) + n (n σ 3 )
‘τ’ => Shear stress – Normal and shear stresses
are orthogonal
T 2 = σ 2 +τ 2
[
n
] [
τ = T − σ = (lσ 1 ) + (m σ 2 ) + (n σ 3 ) − (l σ 1 + m σ 2 + n σ 3 )
2
2
2
n
2
2
2
2
2
2
2
]
l 2 + m 2 + n 2 = 1 => n 2 = 1 − l 2 − m 2
For maximum shear stress
∂τ
∂τ
= 0,
=0
Chennamsetti
∂Ramadas
l
∂m
48
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σ n = l 2σ 1 + m 2σ 2 + n 2σ 3
Maximum Shear Stress
R&DE (Engineers), DRDO
Differentiation gives the following expressions
Solving & use l2+m2+n2 = 1 – three sets of
l = 0, m = 0, n = ±1
solutions
1
l = 0, m = ±
, n=±
2
1
l=±
, m = 0, n = ±
Ramadas
Chennamsetti
2
1
2
1
2
49
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σ1 − σ 3 
 2
2
l l (σ 1 − σ 3 ) + m (σ 2 − σ 3 ) −
=0

2 

σ 2 −σ3 
 2
2
m l (σ 1 − σ 3 ) + m (σ 2 − σ 3 ) −
=0

2 

Maximum Shear Stress
R&DE (Engineers), DRDO
Solutions – planes on which shear stresses are
minimum and maximum
1
2
3
4
5
l
±1
0
0
0
m
0
±1
0
±(1/2)0.5
n
0
0
±1 ±(1/2)0.5 ±(1/2)0.5
6
0
±(1/2)0.5
0
Min. shear planes
Maximum shear planes
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50
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±(1/2)0.5 ±(1/2)0.5
Maximum Shear Stress
R&DE (Engineers), DRDO
1
1
l = 0, m = ± , n = ±
2
2
τ max
Ramadas Chennamsetti
1
= σ 2 −σ3
2
51
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Values of shear stresses and planes –
Maximum Shear Stress
R&DE (Engineers), DRDO
1
1
l = ± , m = 0, n = ±
2
2
τ max
Ramadas Chennamsetti
1
= σ 3 − σ1
2
52
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Values of shear stresses and planes –
Maximum Shear Stress
R&DE (Engineers), DRDO
1
1
l =± , m=± , n=0
2
2
τ max
Ramadas Chennamsetti
1
= σ1 − σ 3
2
53
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Values of shear stresses and planes –
Maximum Shear Stress
R&DE (Engineers), DRDO
Maximum shear stress act at ±450 planes with
reference to principal planes
σ2
σ
σ1
τ max
1
= σ 3 − σ1
2
Ramadas Chennamsetti
1
σ = (σ 3 + σ 1 )
2
54
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σ3
τmax
Minimum Shear Stress
R&DE (Engineers), DRDO
Plane of minimum shear stress
2
Plane OBC
B
Direction cosines
σ3
C
1
o
A
σ2
Maximum principal
stress plane
Shear stress = 0
3
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55
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σ1
l = ±1, m = 0, n = 0
Minimum Shear Stress
R&DE (Engineers), DRDO
Plane of minimum shear stress
2
Plane OAC
B
σ3
C
1
o
A
σ2
l = 0, m = ±1, n = 0
Principal stress
plane - σ2
Shear stress = 0
3
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56
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σ1
Direction cosines
Minimum Shear Stress
R&DE (Engineers), DRDO
Plane of minimum shear stress
2
Plane OBC
Direction cosines
B
σ3
C
3
1
o
A
σ2
Principal stress
plane - σ3
Shear stress = 0
Maximum shear stresses are constant for a given stress
state at a point
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57
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σ1
l = 0, m = 0, n = ±1
Octahedral Stresses
R&DE (Engineers), DRDO
A plane that makes equal angles with
principal planes – Octahedral plane – eight
planes
l + m + n =1
2
2
2
1
l = ± = m= n
3
T1 = lσ1, T2 = lσ2, T3 = lσ3
σn =σoct = lT1 +lT2 +lT3 = l 2 (σ1 +σ2 +σ3 )
Ramadas Chennamsetti
Equal angle with principal planes
58
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l = m= n
Octahedral Stress
R&DE (Engineers), DRDO
Equal angles with principal axes
Normal octahedral stress => Average of
principal stresses
It depends on first invariant
For a given stress state => Octahedral
normal stress - constant
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59
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σ oct
1
I1
= (σ 1 + σ 2 + σ 3 ) =
3
3
Octahedral Stress
R&DE (Engineers), DRDO
Octahedral shear stress – shear stress from
tractions is given by,
2
2
τ 2 = T 2 − σ n2 => τ oct
= T 2 − σ oct
=> τ
2
oct
(
)
1 2
1

2
2 
=  σ 1 + σ 2 + σ 3  −  (σ 1 + σ 2 + σ 3 )
3
 3

2
=> τ oct =
1
3
(σ 1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ 1 )2
τ oct =
2
3
(I
2
1
− 3I 2
)
Octahedral shear stress
=> Chennamsetti
τoct = τ(I1, I2) => Invariant
Ramadas
60
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Simplifying this expression,
State of Pure Shear
R&DE (Engineers), DRDO
Only shear stress exist
The stress state at a point
six stress components, σij
τ
Infinite sets of planes through this point
For pure shear - Sum of main diagonal
terms in stress tensor vanish
=> Tr(σ) = 0
τxy
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61
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xy
State of Pure Shear
R&DE (Engineers), DRDO
Necessary and sufficient condition for
existence of a state of pure shear
=> First invariant of stress tensor = 0
τ xy
τ yy
τ zy
τ xz 

τ yz 
τ zz 
I1 = σ xx + σ yy + σ zz = 0
Individual normal stress components need not be zero
Ramadas Chennamsetti
62
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τ xx

[σ ] = τ yx
τ zx

Decomposition of Stress
R&DE (Engineers), DRDO
The stress tensor
In indices notation, σij
Decompose [σ]
τxx

[σ] = τyx
τzx

τxy
τ yy
τzy
τxz

τ yz
τzz 
τxx

τ yx
τzx

τxy
τ yy
τzy
τxz p

τ yz = 0
τzz  0
In indices notation
0
p
0
0 τxx − p

0 +  τ yx
p  τzx
τxy
τ yy − p
τzy
τzz − p
σ ij = pδ ij + (σ ij − pδ ij ) = pδ ij + Sij
Ramadas Chennamsetti
τxz 

τ yz 
63
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Stress state = Hydrostatic state + Deviatoric state (shear)
Decomposition of Stress
R&DE (Engineers), DRDO
For pure shear => Tr(Sij)
(σ xx − p ) + (σ yy − p ) + (σ zz − p ) = 0
1
p = (σ xx + σ yy + σ zz ) = σ mm = mean stress
3
2
3σ xx

[S] =  τ xy

τ
 xz
τ xy
2
σ yy
3
τ yz

τ xz 
σ '
 
τ yz  = τ xy
 
2  τ xz
σ zz 
xx
3

Ramadas Chennamsetti
τ xy
σ yy'
τ yz
τ xz 

τ yz 
σ zz' 
64
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Deviatoric stress tensor –
Decomposition of Stress
R&DE (Engineers), DRDO
Deviatoric stress – the first invariant is
given by, => J1 = Tr(Sij)
J1 = (σ xx − p ) + (σ yy − p ) + (σ zz − p )
But,
1
p = (σ xx + σ yy + σ zz )
3
=> J1 = 0
Use this relation
The first invariant of Deviatoric stress = 0
Ramadas Chennamsetti
65
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=> J1 = (σ xx + σ yy + σ zz ) − 3 p
Decomposition of Stress
R&DE (Engineers), DRDO
Hydrostatic stress tensor – volumetric part
of stress tensor – spherical stress tensor
Hydrostatic part – responsible for volume
change – size change
p
p
=>>>
p
p
p
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66
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p
Decomposition of Stress
R&DE (Engineers), DRDO
γ
Change in shape due to applied shear
Ramadas Chennamsetti
67
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Deviatoric part – responsible for
deformation or shape change
Change in angle between two adjacent
sides => shape change – distortion - shear
stress
Decomposition of Stress
R&DE (Engineers), DRDO
Invariants of deviatoric stress tensor
σ 'xx

[S] = τxy
τ xz

τxy
σ yy'
τ yz
τxz 

τ yz 
σzz' 
σ xx'
J2 =
τ xy
τ xy σ xx'
+
'
σ yy τ xz
σ xx'
τ xz σ yy'
+
'
σ zz τ yz
τ xz
τ yz
J3 =
'
symm
σ
Ramadas Chennamsetti zz
τ yz
σ zz'
τ xy
σ yy'
68
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'
J 1 = σ xx' + σ yy
+ σ zz' = 0
Mohr’s Circle
R&DE (Engineers), DRDO
Principal stresses => σ 1 , σ 2 , σ 3
σ1 > σ 2 > σ 3
Take principal directions as reference
Ramadas Chennamsetti
69
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Graphical technique - Normal and shear
stresses on a given plane – stress state at a
point is known
Stress state => σij – Find principal stresses
corresponding to this stress state
Mohr’s Circle
R&DE (Engineers), DRDO
Tractions on plane (l, m, n) wrt principal
directions
T
= T 1 2 + T 22 + T 32 = l 2 σ
2
2
T
=σ
2
+τ
l2 + m
But,
2
2
= l 2σ
2
1
+ m 2σ
+ m 2σ
2
1
2
2
2
2
3
+ n 2σ
+ n 2σ
2
3
2
3
+ n 2 = 1 => n 2 = 1 − l 2 − m
2
(
)σ
σ + τ = l (σ − σ ) + m (σ − σ ) + σ
(σ − σ ) + τ = l (σ − σ ) + m (σ − σ ) − − (1 )
σ
2
+τ
2
2
2
= l 2σ
2
1
2
2
1
2
2
3
2
2
+ m 2σ
2
2
2
3
2
1
+ 1− l2 − m
2
2
3
Ramadas Chennamsetti
2
2
2
2
2
3
2
3
2
2
2
3
2
3
70
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T1 = l σ 1 , T 2 = l σ 2 , T 3 = l σ
Mohr’s Circle
R&DE (Engineers), DRDO
Normal stress on this plane –
σ = lT 1 + mT 2 + nT 3 = l 2σ 1 + m 2σ 2 + n 2σ 3
σ = l 2σ 1 + m 2σ 2 + (1 − l 2 − m 2 )σ 3
2
+ (σ − σ 2 )(σ − σ 3 )
τ
2
l =
(σ 1 − σ 2 )(σ 1 − σ 3 )
σ2 +σ3 

σ 2 −σ3 
2
τ + σ −

 = l (σ 1 − σ 2 )(σ 1 − σ 3 ) + 
2
2




2
2
2
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71
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Eliminate ‘m’ from equations (1) and (2)
Mohr’s Circle
R&DE (Engineers), DRDO
The expression


τ 2 + σ −
σ2 +σ3 
2
σ 2 −σ3 
2
2
(σ 1 − σ 2 )(σ 1 − σ 3 ) + 
=
l



2
 − (1)

2
represents a circle of center & radius
2
+σ
2
3

,0

σ 2 −σ 3 
R = l (σ 1 − σ 2 )(σ 1 − σ 3 ) + 

2


on (σ, τ) plane
2
l
2
τ
Rl
C
(σ + σ )/2
2
3
Ramadas Chennamsetti
σ
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σ


2
Mohr’s Circle
R&DE (Engineers), DRDO
Similar expressions as (1) can be obtained
for direction cosines ‘m’ and ‘n’
σ2 +σ3 

σ 2 −σ3 
2
τ + σ −
 = l (σ 1 − σ 2 )(σ 1 − σ 3 ) + 
 − (1)
2
2




2
2
2
2
σ1 −σ 3 
2
2
(σ 2 − σ 1 )(σ 2 − σ 3 ) + 
=
m



2
2
 − (2)

σ1 +σ 2 

σ1 −σ 2 
2
τ + σ −
 = n (σ 3 − σ 1 )(σ 3 − σ 2 ) + 
 − ( 3)
2 
2 


2
2
2
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73
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

τ 2 + σ −
σ1 +σ 3 
Mohr’s Circle
R&DE (Engineers), DRDO
Stress state [σ]
3
 σ +σ 
σ −σ 
τ 2 +σ − 1 2  = n2(σ3 −σ1)(σ3 −σ2 ) + 1 2  −(3)
2 

 2 
2
2
n
Rn
C1
Direction cosines (l, m, n)
Say, ‘n = 0’
Circle of radius => (σ1- σ2)/2
Center at ((σ1+ σ2)/2,0)
1
One of the Mohr’s circles
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74
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2
Mohr’s Circle
R&DE (Engineers), DRDO
Similar circles can be drawn for equations
(1) and (2) – plug l = 0 and m = 0
σ −σ3
σ +σ3 
C1  2
,0 , R1 = 2
2
 2

τ
τP
P
σ −σ3
σ +σ3 
C2  1
,0 , R2 = 1
2
 2

σ −σ 2
σ +σ2 
C3  1
,0 , R3 = 1
2
 2

Draw arcs with following radii
C1
σ2 C2
σP
2
C3
σ1
σ
σ −σ 2 
R = n (σ 3 − σ 1 )(σ 3 − σ 2 ) +  1

 2 
2
n
2
2
Ramadas Chennamsetti
75
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σ3
σ −σ3 
Rl2 = l 2 (σ 1 − σ 2 )(σ 1 − σ 3 ) +  2

2


2
 σ1 −σ 3 
2
2
Rm = m (σ 2 − σ 3 )(σ 2 − σ 1 ) + 

 2 
Equilibrium Equations
R&DE (Engineers), DRDO
Body under static equilibrium – every
material point under equilibrium
y
B
σxy
x
σxx
P
z
σzx
o
σxz
C
V
Infinitesimal element,
σzy
σyx
σyz
S
σzz
A
x
σyy
z
dm = ρ d V
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76
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y
Equilibrium Equations
R&DE (Engineers), DRDO
Body and Traction forces => Fb and Ft
Newton’s second law –
d F t + d F b = dm
~
~
dV
~
dt
V = V (x , y , z , t )
~
∂ V dx
~
~
=
+
∂ x dt
dt
∂V
~
a =
Vx +
~
∂x
dV
∂ V dy
∂ V dz
∂ V dt
~
~
~
+
+
∂ y dt
∂ z dt
∂ t dt
∂V
∂V
∂V
~
~
~
Vy +
Vz +
∂y
∂z
∂t
Ramadas Chennamsetti
77
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~
Equilibrium Equations
R&DE (Engineers), DRDO
Rewrite,
~
~
∂x
Vx +
∂V
~
∂y
Vy +
∂V
~
∂z
Vz +
∂V
~
∂t
=
∂V
~
∂t
2
+ V .∇ V
~
~
Term – 1 => Local acceleration
Term – 2 => Convective acceleration
For small deformations, the particle of
mass ‘dm’ doesn’t move substantially – so
this is term – 2, which is very small
compared to local acceleration term.
Neglect term – 2
Ramadas Chennamsetti
78
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a=
∂V
1
Equilibrium Equations
This gives =>
a=
~
dt
~
∂V
d F t + d F b = ρ dxdydz
~
dV
~
,
∂t
~
≈
R&DE (Engineers), DRDO
∂V
~
∂t
d F t = T dS ,
~
~
T dS + B d V = ρ dxdydz
~
~
∂V
~
∂t
d Fb = B d V
~
~
,
∫∫ T dS +
S
~
∫∫∫
V
B dV =
~
∫∫∫
V
∂V
~
∂t
ρdV
In indices notation,
∫∫
S
T i dS +
∫∫∫
V
Bi dV =
∫∫∫
V
Ramadas Chennamsetti
∂Vi
ρdV
∂t
79
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Integrating this expression,
Equilibrium Equations
R&DE (Engineers), DRDO
But tractions on surfaces related to stresses
– Cauchy’s formula Ti = σ ij l j Use this relation
∫∫∫
S
∫∫ σ
Bi dV =
∫∫∫
V
ij
l j dS +
S
∂Vi
ρdV
∂t
V
∫∫∫
Bi dV =
∫∫∫
V
V
∂Vi
ρdV
∂t
Use Gauss divergence theorem – Volume integral to Surface integral
∫∫∫ ∇. AdV = ∫∫ A. n dS = ∫∫ A. dS
V
Indices notation
~
S
~ ~
S
∂Ai
dV = ∫∫ Ai li dS
∫∫∫
∂xi
V
S
Ramadas Chennamsetti
~
~
80
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∫∫
T i dS +
Equilibrium Equations
R&DE (Engineers), DRDO
Use Gauss Divergence theorem –
ij
l j dS +
S
∫∫∫
V
Bi dV =
∫∫∫
V
∂σ
ij
∂x
j
∫∫∫
V
This gives -
∫∫∫
dV +
V
∫∫∫
Bi dV =
V
 ∂σ

 ∂x

ij
j
∂Vi
+ Bi − ρ
∂t
∫∫∫
V
∂Vi
ρdV
∂t

d V = 0


∂σ ij
∂Vi
+ Bi − ρ
=0
∂x j
∂t
Ramadas Chennamsetti
81
rd_mech@yahoo.co.in
∫∫ σ
∂Vi
ρdV
∂t
Equilibrium Equations
R&DE (Engineers), DRDO
All three equilibrium equations –
∂τ xy
∂τ xx
∂τ xz
∂u
+
+
+ Bx = ρ 2
∂x
∂y
∂z
∂t
∂τ xy
∂τ yy
2
∂τ yz
∂v
+
+
+ By = ρ 2
∂x
∂y
∂z
∂t
∂τ xz
∂τ zz
∂w
+
+
+ Bz = ρ 2
∂x
∂y
∂z
∂t
Ramadas Chennamsetti
2
82
rd_mech@yahoo.co.in
∂τ yz
2
Equilibrium Equations
R&DE (Engineers), DRDO
Moment Equilibrium
∑ M = r× m a + I
~
~
~
G
α
~
No rotations
y
∑ M = r× m a
~
x
~
~
P
In differential form
S
r
V
d M t + d M b = r× dm
~
~
~
dV
~
dt
o
dMt and dMb moments due to traction and body loads
Ramadas Chennamsetti
83
rd_mech@yahoo.co.in
z
Equilibrium Equations
R&DE (Engineers), DRDO
Plug dMt and dMb
~
~
~
~
dt
r× T dA + r× B dV = r× dm
~
~
~
~
~
dV
~
dt
Integrating -
∫∫ r× T dA + ∫∫∫ r× B dV = ∫∫∫ r× ρ
S
~
~
V
~
~
Ramadas Chennamsetti
V
~
dV
~
dt
dV
84
rd_mech@yahoo.co.in
d M t + d M b = r× dm
dV
Equilibrium Equations
R&DE (Engineers), DRDO
Writing in indices form –
i
S
j ijk
ek dA + ∫∫∫ ri B j ε ijk ek dV = ∫∫∫ ri
~
V
Use Cauchy’s formula
~
V
∂t
ρε ijk ek dV
~
T j = σ jmlm

∂V j

ε ijk  ∫∫ riT j dA + ∫∫∫ ri B j dV − ∫∫∫ ri
ρdV  = 0
∂t
V
V
S


∂V j

ε ijk  ∫∫ riσ jm lm dA + ∫∫∫ ri B j dV − ∫∫∫ ri
ρdV  = 0
∂t
V
V
S

Use Gauss Divergence theorem
Ramadas Chennamsetti
85
rd_mech@yahoo.co.in
∫∫ r T ε
∂V j
Equilibrium Equations
R&DE (Engineers), DRDO
Surface integral to volume integral

∂V j

ε ijk  ∫∫∫ (riσ jm ), m dV + ∫∫∫ ri B j dV − ∫∫∫ ri
ρdV  = 0
∂t
V
V
V

∂V j 

dV + ε ijk ri , mσ jm = 0
riε ijk ∫∫∫  σ jm , m + B j −
∂t 
V 
σ jm ,m + B j −
∂V j
∂t
=0
So,
ε ijk ri , m σ
jm
= 0
ε ijk δ im σ
jm
= 0
ε ijk σ
Ramadas Chennamsetti
ij
= 0
86
rd_mech@yahoo.co.in
Use integration by parts - first term
Equilibrium Equations
R&DE (Engineers), DRDO
Cyclic symbol is skew-symmetric in i, j
ε ijk = −ε jik
ε ijkσ ij = 0, ε jikσ ji = 0
ε ijk (σ ij − σ ji ) = 0
=> σ ij − σ ji = 0
Shear stresses –
complementary
=> σ ij = σ ji
Ramadas Chennamsetti
Stress tensor is
symmetric -
87
rd_mech@yahoo.co.in
ε ijkσ ij + ε jikσ ji = 0
Cylindrical Co-ordinate System
R&DE (Engineers), DRDO
P
z
r
θ
To completely specify
the location of point ‘P’
(r, θ, z) required
Ramadas Chennamsetti
88
rd_mech@yahoo.co.in
Cartesian frame of reference – body
processes straight boundaries
Curved boundaries – Cylindrical, Spherical
Cylindrical csys – z(r, θ, z) – Spherical csys –
(r, θ, φ)
Cylindrical Co-ordinate System
R&DE (Engineers), DRDO
Equilibrium equations
dθ/2
τrθ
σr
Equilibrium equations
dz
dr
∂τ rz
τ rz +
dr
∂r
Ramadas Chennamsetti
~
∑ F = 0,
∑ Fθ = 0 ,
∑F =0
r
z
89
rd_mech@yahoo.co.in
∂τ rz
τ rz +
dz
∂z
∑F = 0
Cylindrical Co-ordinate System
R&DE (Engineers), DRDO
Equilibrium in radial direction –


∂σ rr 
(
)
σ
+
dr
r
+
dr
d
θ
dz
−
σ
rd
θ
dz
+

rr
 rr

∂r




  dθ 
∂τ rθ 
τ
+
dr
drdz
−
τ
drdz
cos


+
rθ
 rθ

∂r


  2 

  dθ 
∂σ θθ 
 σ θθ + ∂θ drdz + σ θθ drdz  sin  2  +



 
••
dr 
dr 


Br  r + dθdz = ρ u  r + dθdz
2
2


Ramadas Chennamsetti
90
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

∂τ rz 
dr 
dr 

τ rz + ∂z dr  r + 2 dθdr − τ rz  r + 2 dθdr  −






Cylindrical Co-ordinate System
R&DE (Engineers), DRDO
Neglect higher order terms - simplifying this
equation –
Similarly two more expression in θ and z
directions
••
∂τ rθ 1 ∂σ θθ ∂τ zθ 2τ rθ
+
+
+
+ Bθ = ρ v
∂r
r ∂θ
∂z
r
••
∂τ rz 1 ∂τ θz ∂σ zz τ rz
+
+
+
+ Bz = ρ w
∂r r ∂θ
∂z
r
Ramadas Chennamsetti
91
rd_mech@yahoo.co.in
••
∂σ rr 1 ∂τ rθ ∂τ zr σ rr − σ θθ
+
+
+
+ Br = ρ u
∂r
r ∂θ
∂z
r
Ramadas Chennamsetti
92
rd_mech@yahoo.co.in
R&DE (Engineers), DRDO