Solid Mechanics
1. Shear force and bending moment
diagrams
Internal Forces in solids
Sign conventions
• Shear forces are given a special symbol on
y
V
1
2
and
z
V
• The couple moment along the axis of the member is
given
x
M T
= =
Torque
y z
M M
= =
bending moment.
Solid Mechanics
We need to follow a systematic sign convention for
systematic development of equations and reproducibility of
the equations
The sign convention is like this.
If a face (i.e. formed by the cutting plane) is +ve if its outward
normal unit vector points towards any of the positive coordinate
directions otherwise it is –ve face
• A force component on a +ve face is +ve if it is directed
towards any of the +ve coordinate axis direction. A force
component on a –ve face is +ve if it is directed towards any of
the –ve coordinate axis direction. Otherwise it is –v.
Thus sign conventions depend on the choice of coordinate
axes.
Shear force and bending moment diagrams of beams
Beam is one of the most important structural components.
• Beams are usually long, straight, prismatic members and
always subjected forces perpendicular to the axis of the beam
Two observations:
(1) Forces are coplanar
Solid Mechanics
(2) All forces are applied at the axis of the beam.
Application of method of sections
What are the necessary internal forces to keep the segment of
the beam in equilibrium?
x
y
z
F P
F V
F M
=
=
=
0
0
0
• The shear for a diagram (SFD) and bending moment
diagram(BMD) of a beam shows the variation of shear
Solid Mechanics
force and bending moment along the length of the
beam.
These diagrams are extremely useful while designing the
beams for various applications.
Supports and various types of beams
(a) Roller Support – resists vertical forces only
(b) Hinge support or pin connection – resists horizontal and
vertical forces
Hinge and roller supports are called as simple supports
(c) Fixed support or builtin end
Solid Mechanics
The distance between two supports is known as “span”.
Types of beams
Beams are classified based on the type of supports.
(1) Simply supported beam: A beam with two simple supports
(2) Cantilever beam: Beam fixed at one end and free at other
(3) Overhanging beam
(4) Continuous beam: More than two supports
Solid Mechanics
Differential equations of equilibrium
[
]
x
F
Σ
= →+
0
y
FΣ
= ↑ +
0
V V V P x
V P x
V
P
x
Δ Δ
Δ Δ
Δ
Δ
+ − + =
= −
= −
0
x
V dV
P
x dx
lim
Δ
Δ
Δ
→
= = −
0
[ ]
A
P x
M V x M M M
Δ
Σ Δ Δ
= − + + − =
2
0 0
2
P x
V x M
M P x
V
x
Δ
Δ Δ
Δ Δ
Δ
+ − =
+ − =
2
0
2
0
2
Solid Mechanics
x
M dM
V
x dx
lim
Δ
Δ
Δ
→
= = −
0
From equation
dV
P
dx
= −
we can write
D
C
X
D C
X
V V Pdx
− = −
From equation
dM
V
dx
= −
D C
M M Vdx
− = −
Special cases
:
Solid Mechanics
Solid Mechanics
Solid Mechanics
Solid Mechanics
(
)
(
)
x
≤ ≤ −
0 2 1 1
A B
V
VV;V
− =
=
= =
5 0
5
5 5
(
)
(
)
( )
( )
( )
B C
x
V.x
V.x
V;V
.x
x.
≤ ≤ −
− + − − =
= − + −
= − =
− + − =
=
2 6 2 2
5 30 7 5 2 0
5 30 7 5 2
25 5
25 7 5 2 0
5 33
(
)
(
)
C D
x
VVV;V
≤ ≤ −
− + − − =
= +
= + = +
6 8 3 3
5 30 30 10 0
15
15 15
(
)
(
)
D E
x
VVVV;V
≤ ≤ −
− + − − + =
+ =
= −
= − = −
8 10 4 4
5 30 30 10 20 0
5 0
5
5 5
x ( ) ( )
x ( ( )
x ( ) ( )
x ( ) ( )
≤ ≤ − −
≤ ≤ − −
≤ ≤ − −
≤ ≤ − −
0 2 1 1
2 6 2 2
6 8 3 3
8 10 4 4
Solid Mechanics
Problems to show that jumps because of concentrated force
and concentrated moment
(
)
(
)
A B
x
M x
M x
M;M
≤ ≤ − −
− + =
= − +
= + =
0 2 1 1
10 5 0
5 10
10 0
(
)
(
)
( )
( )
( )
( )
E
x.
C
x
x
.x
M x x
.x
M x x
M.
M
=
=
≤ ≤ − −
−
− + − − + =
−
= − + − −
= +
=
2
2
5 33
6
2 6 2 2
7 5 2
10 5 30 2 0
2
7 5 2
10 5 30 2
2
41 66
40
(
)
(
)
[
]
( ) ( ) ( )
C
x
D
x
x C D
M x x x x
MM
=
=
≤ ≤ − − −
− + − − + − + − + =
= +
= −
6
8
6 8 3 3
10 5 30 2 30 4 10 6 20 0
20
10
[
]
(
)
(
)
( ) ( ) ( ) ( )
E
x
x D E
M x x x x x
M
=
≤ ≤ − −
− + − − + − + − + − − =
=
8
8 10 4 4
10 5 30 2 30 4 10 6 20 20 8 0
0
Solid Mechanics
We can also demonstrate internal forces at a given section
using above examples. This should be carried first before
drawing SFD and BMD.
[
]
x A B
≤ ≤ −
0 2
Solid Mechanics
A
B
V
VVV
− =
=
=
=
5 0
5
5
5
A B
M x
M x
M;M
− + =
= −
= =
10 5 0
10 5
10 0
[
]
x B C
≤ ≤ −
2 6
(
)
( )
( )
B C
V.x
V.x
V;V
.x
x.
− + − − =
= − + −
= − =
− + − =
=
5 30 7 5 2 0
7 5 2 5 30
25 5
25 7 5 2 0
5 33
( )
( )
C
E
B
x
M x x.
x
MM x..
x
M
−
− + − − + =
=
=
= =
=
=
2
2
10 5 30 2 7 5 0
2
6
40
5 33 41 66
2
0
[
]
x C D
≤ ≤ −
6 8
C D
V
VV,V
− + − − =
=
= =
5 30 10 30 0
15
15 15
Solid Mechanics
[
]
x D E
≤ ≤ −
8 10
D E
V
VV,V
− + − − + =
= −
= − = −
5 30 10 30 20 0
5
5 5
Solid Mechanics
[
]
[ ]
x Ax
y Ay
Ay
F R
F R
R kN
M M.
M k m
Δ
→+ = =
↑ + = + − =
= ↑
= + − × =
= −
0 0
0 60 90 0
30
0 60 90 4 5 0
285
(
)
( )
V x
V x
+ + − − =
= − −
= × −
= −
=
30 60 30 3 0
30 3 90
30 3 90
90 90
0
(
)
B A
B A
M M
M M
− = − −
= + = −
= −
60
60 60 285
225
Solid Mechanics
(
)
C B
C B
M M
M M
− = − −
= + = − +
= −
90
90 225 90
135
(
)
D C
D C
M M
M M
− = − −
= + = − + =
135
135 135 135 0
y
Ay Cy
Ay Cy
F
R R
R R ( )
↑ + =
+ − − =
+ =
0
200 240 0
440 1
[
]
A
Cy
Cy
Ay
M
R
R kN
R kN
=
− × − × + × =
= ↑
= ↑
0
200 3 240 4 8 0
195
245
V x
V x
V
V
+ − − =
= −
= × − = −
=
245 200 30 0
30 45
30 8 45 240 45
195
Solid Mechanics
*
M.
M.
M
− × + ×
= × − ×
=
245 3 90 1 5
245 3 90 1 5
600
[ ]
Ay By
A By
By
By
Ay
R R
M R
R
R kN
R kN
+ =
= − × + + + =
− + + =
=
=
32
0 32 2 18 8 4 0
64 16 4 0
12
20
Solid Mechanics
Problem
:
[
]
( )
x
Ax
y Ay Dy Ay Dy
F
R
F R R R R
→+ =
=
= ↑ + + − − = + =
0
0
0 60 50 0 110 1
(
)
C A
C A
M M
M M
− = − −
= + = − + =
50
50 8 25 17
V x
V x
x
x/.
+ − =
= −
− =
= =
20 8 0
8 20
8 20 0
20 8 2 5
[
]
A Dy
Dy
Ay
M.R
R kN
R kN
= − × − × + × =
= = ↑
= ↑
0 60 1 5 50 4 5 0
290
58
5
52
Solid Mechanics
( )
y
B
F V x
V x x m
= ↑ + + − =
= − ≤ ≤
0 52 20 0
20 52 0 3
[
]
( )
M
x
M x
x
M x x m
=
+ − =
= − ≤ ≤
2
2
0
20
52 0
2
20
52 0 3
2
y
B C
F
V
V kN x m
= ↑ +
+ − =
= ↑ ≤ ≤
0
52 60 0
8 3 4
[
]
(
)
( )
B C
M M x x.
M x x.x m
= − + − =
= − − ≤ ≤
0 52 60 1 5 0
52 60 1 5 3 4
Solid Mechanics
B E
B
M M.
M..
− = −
= − +
1 6
1 6 67 6
x/.m
×− =
= =
20 52 0
52 20 2 6
dM
V
dx
dV
P
dx
= −
= −
[
]
(
)
(
)
( ) ( ) ( )
M M x x.x
M x x.x x
= − + − + − =
= − − − − ≤ ≤
0 52 60 1 5 50 4 0
52 60 1 5 50 4 4 5
( )
y
F
V
V kN x
= ↑ +
+ − − =
= ≤ ≤
0
52 60 50 0
58 4 5
Solid Mechanics
D C
D C
M M
M M
− = −
= +
= − =
58
58
58 58 0
C B
C B
M M
M M
− = −
= − +
= − + =
8
8
8 66 58
B E
B E
M M.
M.M..
− = −
= − + = − +
=
1 6
1 6 1 6 67 6
66
x/.
×− =
= =
20 52 0
52 20 2 6
dM
V
dx
dV
P
dx
= −
= −
B A
M M Vdx
− = −
Solid Mechanics
2. Concept of stress
Traction vector or Stress vector
Now we define a quantity known as “stress vector” or
“traction” as
Δ
Δ
Δ
→
=
R
n
A
F
T
A
lim
0
units
a
P N/m
−
2
and we assume that the quantity
Δ
Δ
Δ
→
→
R
A
M
A
lim
0
0
(1)
n
T
is a vector quantity having direction of
R
F
Δ
(2)
n
T
represent intensity point distributed force at the point
"P"
on a plane whose normal is
ˆ
n
(3)
n
T
acts in the same direction as
R
F
Δ
Solid Mechanics
(4) There are two reasons are available for justification of the
assumption that
Δ
Δ
Δ
→
→
R
A
M
A
lim
0
0
(a) experimental
(b) as
A
Δ
→
0
,
R
F
Δ
becomes resultant of a parallel
force distribution. Therefore
R
MΔ
=
0
for
force
system.
(5)
n
T
varies from point to point on a given plane
(6)
n
T
at the same point is different for different planes.
(7)
n n
T T
′
= −
will act at the point P
(8) In general
Components of
n
T
R n t s
ˆ
ˆ ˆ
F F n v t v s
Δ Δ Δ Δ
= + +
Solid Mechanics
Δ Δ Δ Δ
Δ Δ Δ Δ
Δ Δ Δ Δ
→ → → →
= = + +
R n t s
n
A A A A
F F v v
ˆ
ˆ ˆ
T n t s
A A A A
lim lim lim lim
0 0 0 0
n nn nt ns
ˆ
ˆ ˆ
T n t s
σ τ τ
= + +
where
Δ
Δ
Δ
Δ
σ
Δ
Δ
τ
Δ
Δ
τ
Δ
→
→
→
= = =
= = =
= = =
n n
nn
A
t t
nt
A
s s
ns
A
F dF
Normal stresscomponent
A dA
v dv
Shear stresscomponent
A dA
v dv
Another shear componet
A dA
lim
lim
lim
0
0
0
σ
τ
−
−
Normal Stress
Shear stress
n nn
dF dA
σ
=
t nt
dV dA
τ
=
Notation of stress components
The magnitude and direction of
n
T
clearly depends on the
plane mm. Therefore, stress components magnitude &
direction depends on orientation of cut mm.
(a) First subscript plane on which
σ
is acting
(b) Second subscript direction
Solid Mechanics
Rectangular components of stress
Cuts
⊥
to the coordinate planes will give more valuable
information than arbitrary cuts.
Δ Δ Δ Δ
Δ
Δ Δ Δ
Δ Δ Δ Δ
→ → → →
= = + +
y
R x z
x
A A A A
v
F F v
ˆ
ˆ ˆ
T i j k
A A A A
lim lim lim lim
0 0 0 0
x xx xy xz
ˆ
ˆ ˆ
T i j k
σ τ τ
= + +
where
x
xx
A
y
z
xy xz
A A
F
Normal stress
A
v
v
Shear stress;Shear stress
A A
lim
lim lim
Δ
Δ Δ
Δ
σ
Δ
Δ
Δ
τ τ
Δ Δ
→
→ →
= =
= = = =
0
0 0
Solid Mechanics
σ
=
x xx
dF dA
y xy
dv dA
τ
=
z xz
dv dA
τ
=
Similarly,
Δ Δ Δ Δ
Δ
Δ Δ Δ
Δ Δ Δ Δ
→ → → →
= = + +
y
R x z
y
A A A A
F
F v v
ˆ
ˆ ˆ
T i j k
A A A A
lim lim lim lim
0 0 0 0
τ σ τ
= + +
y yx yy yz
ˆ
ˆ ˆ
T i j k
τ τ σ= + +
z zx zy zz
ˆ
ˆ ˆ
T i j k
xx
σ
and
xy
τ
will act only on xplane. We can see
x
σ
and
xy
τ
only when we take section
⊥
to xaxis.
The stress tensor
σ τ τ
σ τ σ τ
τ τ σ
=
xx xy xz
jj yx yy yz
zx zy zz
Rectangular stresscomponents
• This array of 9 components is called as stress tensor.
• It is a second rank of tensor because of two indices
Components
a point “P” on the xplane in x,y,z
directions
Solid Mechanics
• These 9 rectangular stress components are obtained by
taking 3 mutually
⊥
planes passing through the point
“P”
•
∴
Stress tensor is an array consisting of stress
components acting on three mutually perpendicular
planes.
τ τ τ
= + +
n nx ny nz
ˆ
ˆ ˆ
T i j k
What is the difference between distributed loading & stress?
R
A
F
q lim
A
Δ
Δ
Δ
→
=
0
yy
q
σ
=
can also be called.
No difference!
Except for their origin!
Solid Mechanics
Sign convention of stress components.
A positive components acts on a +ve face in a +ve coordinate
direction
or
A positive component acts on a negative face in a negative
coordinate direction.
Say
x xy a
;Pa P
σ τ
= − = −
20 10
and
xz
Pa
τ
=
30
at a point P
means.
Solid Mechanics
State of stress at a point
The totality of all the stress vectors acting on every possible plane
passing through the point is defined to be state of stress at a point.
• State of stress at a point is important for the designer in
determining the critical planes and the respective critical
stresses.
• If the stress vectors [and hence the component] acting
on any three mutually perpendicular planes passing
through the point are known, we can determine the
stress vector
n
T
acting on any plane “n” through that
point.
The stress tensor will specify the state stress at point.
x x x y x z
ij y x y y y z
z x z y z z
σ τ τ
σ τ σ τ
τ τ σ
′ ′ ′ ′ ′ ′
′ ′ ′ ′ ′ ′ ′
′ ′ ′ ′ ′ ′
=
can also represent state of
stress at a point.
Solid Mechanics
The stress element
Is there any convenient way to visualize or represent the
state of stress at a point or stresses acting three mutually
perpendicular planes say x plane , yplane and zplane.
xx xy xz
ij yx yy yz
P
zx zy zz
σ τ τ
σ τ σ τ
τ τ σ
+ + +
= + + +
+ + +
(
)
( )
xx xx
yy yy
x,y,z
Continuous functionsof x,y,z
x,y,z
σ σ
σ σ
=
=
Let us consider a stress tensor or state of stress at a point in a
component as
Solid Mechanics
ij
σ
− −
= −
− − −
10 5 30
5 50 60
30 60 100
Equilibrium of stress element
[
]
x
F
= →+
0
x yx zx x yx zx
dydz dxdz dydx dydz dxdz dxdy
σ τ τ σ τ τ
+ + − − − =
0
Similarly, we can show that
y
F
=
0
and
z
F
=
0
is satisfied.
y
dz
dy
z
dx
x
xy
τ
xz
τ
x
σ
Solid Mechanics
P
z
M
C.C.W ve
=
+
0
(
)
(
)
xy yx
dydz dx dxdz dyτ τ
− =
0
xy yx
τ τ
− =
0
xy yx
τ τ
=
Shearing stresses on any two mutually perpendicular planes
are equal.
P
x
M
=
0
yz zy
τ τ
=
and
P
y
M
=
0
zx xz
τ τ
=
Crossshears are equal a very important result
Since
xy yx
τ τ
=
, if
xy
ve
τ
= −
yx
τ
is also –ve
Solid Mechanics
∴
The stress tensor
xx xy xz
ij yx xy xy yz
zx xz zy yz yz
issec ondranksymmetrictensor
σ τ τ
σ τ τ σ τ
τ τ τ τ σ
= =
= =
Differential equations of equilibrium
[
]
x
F
→+ =
0
yx
x zx
x yx zx
x xy zx x
x y z y x z z y x
x y z
y z x z y x B x y z
τ
σ τ
σ τ τ
σ τ τ
∂
∂ ∂
+ Δ Δ Δ + + Δ Δ Δ + + Δ Δ Δ
∂ ∂ ∂
− Δ Δ − Δ Δ − Δ Δ + Δ Δ Δ = 0
yx
x zx
x
x y z y x z x y z B x y z
x y z
τ
σ τ
∂
∂
Δ Δ Δ + Δ Δ Δ + Δ Δ + Δ Δ Δ =
∂ ∂ ∂
2
0
Canceling
x y
Δ Δ
and
z
Δ
terms and taking limit
yx
x zx
x
xyz
lim B
x y z
τ
σ τ
Δ →
Δ →
Δ →
∂
∂ ∂
+ + + =
∂ ∂ ∂0
0
0
0
Similarly we can easily show that
Solid Mechanics
[ ]
yx
x zx
x x
B F
x y z
τ
σ τ
∂
∂ ∂
+ + + = =
∂ ∂ ∂
0 0
xy yy zy
y y
B F
x y z
τ σ τ
∂ ∂ ∂
+ + + = =
∂ ∂ ∂
0 0
[ ]
yz
xz zz
z z
B F
x y z
τ
τ σ
∂
∂ ∂
+ + + = =
∂ ∂ ∂
0 0
• If a body is under equilibrium, then the stress
components must satisfy the above equations and must
vary as above.
For equilibrium, the moments of forces about x, y and z axis
at any point must vanish.
p
z
M
=
0
xy yx
xy xy yx
yx
y
x x
x y z y z y x z
x y
y
x z
τ τ
τ τ τ
τ
∂ ∂
Δ
Δ
+ Δ Δ Δ + Δ Δ − + Δ Δ Δ
∂ ∂
Δ
− Δ Δ =
2 2 2
0
2
Solid Mechanics
xy xy yx yx
xy yx
xy yx
y x z x y z
x y z x y z
x y
yx
x y
τ τ τ τ
τ τ
τ τ
Δ Δ Δ ∂ Δ Δ Δ ∂
Δ Δ Δ Δ Δ Δ
+ − − =
∂ ∂
∂ ∂
ΔΔ
+ − − =
∂ ∂
2 2
2 2
0
2 2 2 2
0
2 2
Taking limit
xy yx
xy yx
xyz
yx
lim
x y
τ τ
τ τ
Δ →
Δ →
Δ →
∂ ∂
ΔΔ
+ − − =
∂ ∂
0
0
0
0
2 2
xy yx
τ τ
− =
0
xy yx
τ τ
=
Relations between stress components and internal force
resultants
Solid Mechanics
x xx
A
F dA
σ=
;
y xy
A
V dA
τ=
;
z xz
A
V dA
τ=
xz xy x
y dA dAz dM
τ τ
− =
(
)
x xz xy
A
M y z dA
τ τ= −
y xz
A
M dA
σ=
;
z xy
A
M dA
σ= −
Solid Mechanics
3. Plane stress and Plane strain
Plane stress 2D State of stress
If
(
)
(
)
( ) ( )
x xy
ij
xy yy
x,y x,y
plane stressis a  state of stress
x,y x,y
σ τ
σ
τ σ
= −
All stress components are in the plane
x y
−
i.e all stress
components can be viewed in
x y
−
plane.
xy
x xy
x xy
ij xy y
yx y
D Stateof stress
Stresscomponentsinplanexy
τ
σ τ
σ τ
σ τ σ
τ σ
=
−
= =
2
00
0 0 0
x xy xz
ij yx yy yz
zx zy zz
D Stateof stress
components
σ τ τ
σ τ σ τ
τ τ σ
−
= −
3
6
Solid Mechanics
This type of stressstate (i.e plane stress) exists in bodies
whose
z
 direction dimension is very small w.r.t other
dimensions.
Stress transformation laws for plane stress
The state of stress at a point P in 2Dplane stress problems
are represented by
x xy
nn nt
ij
xy y nt tt
σ τ
σ τ
σ
τ σ τ σ
= =
Solid Mechanics
* We can determine the stress components on any plane “n”
by knowing the stress components on any two mutually
⊥
planes.
Stress transformation laws for plane stress
In order to get useful information we take different cutting
planes passing through a point. In contrast to 3D problem,
all cutting planes in plane stress problems are parallel to x
Solid Mechanics
axis. i.e we take different cutting plane by rotating about z
axis.
As in case of 3D, the state of stress at a point in a plane stress
domain is the totality of all the stress. If we know the stress
components on any two mutually
⊥
planes then stress
components on any arbitrary plane mm can be determined.
Thus the stress tensor
x xy
ij
xy y
σ τ
σ
τ σ
=
is sufficient to tell about the state of stress
at a point in the plane stress problems.
dA Area of AB
dACs Areaof BC
dASin Area of AC
θ
θ
=
=
=
n
F
+ =
0
nn x xy xy
yy
dA dACos Cos dACos Sin dASin Cos
dASin Sin
σ σ θ θ τ θ θ τ θ θ
σ θ θ
− − − −
= 0
nn x xy yy
Cos Sin Cos Sinσ σ θ τ θ θ σ θ
− − − =
2 2
2 0
Solid Mechanics
nn x y xy
x y x y
nn xy
Cos Sin Sin Cos
Cos Sin
σ σ θ σ θ τ θ θ
σ σ σ σ
σ θ τ θ
= + +
+ −
= + +
2 2
2
2 2
2 2
n
F
+ =
0
nt x xy xy
y
dA dACos Sin dACos Cos dASin Sin
dASin Cos
σ σ θ θ τ θ θ τ θ θ
σ θ θ
− − + −
= 0
(
)
nt x y xy
Cos Sin Sin Cos Cos Sin
τ σ θ θ σ θ θ τ θ θ
= − + + −
2 2
(
)
(
)
( )
nt x y xy
x y
nt xy
Cos Sin Cos Sin
Sin Cos
τ θ θ σ σ τ θ θ
σ σ
τ θ τ θ
= − − + −
−
= − +
2 2
2 2
2
We shall now show that if you know the stress components
on two mutually
⊥
planes then we can compute stresses on
any inclined plane. Let us assume that we know that state of
stress at a point P is given
x xy
ij
xy y
σ τ
σ
τ σ
=
This also means that
Solid Mechanics
Solid Mechanics
If
θ θ
=
we can compute on AB
If
π
θ θ
= +
2
we can compute on BC
If
θ θ π
= +
we can compute on CD
If
π
θ θ= +
3
2
we can compute on DA
•
nn
σ
and
nt
τ
equations are known as transformation
laws for plane stress.
• They are not only useful in determination of stresses on
any plane but also useful in transforming stresses from
one coordinate system to another
• Transformation laws do not require an equilibrium state
and thus are also valid at all points of the body under
accelerations.
• These laws are true for any point P of a body.
Invariants of stress tensor
• Any quantity for which its 2D scalar components
transform from one coordinate system to another
according to
nn
σ
and
nt
τ
is called a two dimensional
Solid Mechanics
symmetric tensor of rank 2. Here in particular the tensor
is a stress tensor.
• Moment of inertia if
x xx y yy xy xy
I,I;I
σ σ τ
= = = −
• By definition a tensor is a mathematical quantity that
transforms according to certain laws, such that certain
invariant properties are maintained for all coordinate
systems.
• Tensors, as governed by their transformation laws,
possess several properties. We now develop those
properties for 2D second vent symmetric tensor.
x y x y
nn xy
Cos Sin
σ σ σ σ
σ θ τ θ
+ −
= + +
2 2
2 2
x y x y
t xy
Cos Sin
σ σ σ σ
σ θ τ θ
+ −
= + −
2 2
2 2
x y
nt xy
Sin Cos
σ σ
τ θ τ θ
−
= − +
2 2
2
Solid Mechanics
n t x y x y
I
σ σ σ σ σ σ
′ ′
+ = + = + =
1
I
=
1
First invariant of stress in 2D
n t nt x y xy x y x y
I
σ σ τ σ σ τ σ σ τ
′ ′ ′ ′
− = − = − =
2 2
2
I
=
2
Second invariant of stress in 2D
•
I,I
1 2
are invariants of 2D symmetric stress tensor at a
point.
• Invariants are extremely useful in checking the
correctness of transformation
• Of
I
1
and
I
2
,
I
1
is the most important property : the
sum of normal stresses on any two mutually
⊥
planes
(
⊥
directions) is a constant at a given point.
• In 2D we have two stress invariants; in 3D we have
three invariants of stresses.
Solid Mechanics
Solid Mechanics
Problem
:
A planestress condition exists at a point on the surface of a
loaded structure, where the stresses have the magnitudes
and directions shown on the stress element. (a) Determine
the stresses acting on a plane that is oriented at a
−
15
w.r.t.
the xaxis (b) Determine the stresses acting on an element
that is oriented at a clockwise angle of
15
w.r.t the original
element.
Solution
:
it is in C.W.
x
y
xy
Q
σ
σ
τ
= −
=
= −
= −
46
12
19
15
Solid Mechanics
Substituting
θ
= −
15
in
nt
τ
equation
x y
MPas
σ σ
+
− + −
= = = −
46 12 34
17
2 2 2
(
)
(
)
Sin Sin.;Cos Cos.
θ θ
= − = − = − =
2 2 15 0 5 2 2 15 0 866
x y x y
n xy
Cos Sin
σ σ σ σ
σ θ τ θ
+ −
= + +
2 2
2 2
n
..
σ
= − − × + ×
17 29 0 866 19 0 5
n
.MPas
σ
= −
1
32 6
x y
nt xy
Sin Cos
σ σ
τ θ τ θ
−
= − +
2 2
2
n t
MPa
τ
= −
1 1
31
x y
MPa
σ σ
−
− − −
= = = −
46 12 58
29
2 2 2
n t
..
τ
= − × − ×
1 1
29 0 5 19 0 866
Solid Mechanics
Now
As a check
t n nt
θ
σ σ τ
=
= =
2
75
n
Cos Sin
MPa
σ
= − − × − ×
= −
17 29 2 165 19 2 165
32
nt
nt
.Sin Cos
MPa
τ
τ
= −
= −
0
0 29 330 19 330
31
n t x y
..MPa s
σ σ σ σ
+ = + = − − = − = − +
32 6 1 4 34 46 12
θ
=
145
tn
Sin Cos
MPa
τ
= + × − ×
=
29 150 19 150
31
t
cos sin
σ
∴ = − − −
17 29 150 19 150
t
.MPa
σ
= −
1 4
tn n t nt
θ
τ τ τ
=
= =
2 2
75
Solid Mechanics
4. Principal Stresses
Principal Stresses
Now we are in position to compute the direction and
magnitude of the stress components on any inclined plane at
any point, provided if we know the state of stress (Plane
stress) at that point. We also know that any engineering
component fails when the internal forces or stresses reach a
particular value of all the stress components on all of the
infinite number of planes only stress components on some
particular planes are important for solving our basic
question i.e under the action of given loading whether the
component will ail or not? Therefore our objective of this
class is to determine these plane and their corresponding
stresses.
(1)
( )
n y n y
n n xy
Cos Sin
σ σ σ σ
σ σ θ θ τ θ
+ −
= = + +
2 2
2 2
(2) Of all the infinite number of normal stresses at a point,
what is the maximum normal stress value, what is the
minimum normal stress value and what are their
Solid Mechanics
corresponding planes i.e how the planes are oriented ? Thus
mathematically we are looking for maxima and minima of
(
)
n
Q
σ function..
(3)
n y n y
n xy
Cos Sin
σ σ σ σ
σ θ τ θ
+ −
= + +
2 2
2 2
For maxima or minima, we know that
(
)
n
x y xy
d
Sin Cos
d
σ
σ σ θ τ θ
θ
= = − − +
0 2 2 2
xy
x y
tan
τ
θ
σ σ
=
−
2
2
(4) The above equations has two roots, because
tan
repeats
itself after
π
. Let us call the first root as
P
θ
1
xy
P
x y
tan
τ
θ
σ σ
=
−
1
2
2
( )
xy
P P
x y
tan tan
τ
θ θ π
σ σ
= + =
−
2 1
2
2 2
Solid Mechanics
P P
s
π
θ θ= +
2 1
2
(5) Let us verify now whether we have minima or minima at
P
θ
1
and
P
θ
2
( )
( )
P
n
x y xy
n
x y P xy P
d
Cos Sin
d
d
Cos Sin
d
θ θ
σ
σ σ θ τ θ
θ
σ
σ σ θ τ θ
θ
=
= − − −
∴ = − − −
1 1
1
2
2
2
2
2 2 4 2
2 2 4 2
We can find
P
Cos s
θ
1
2
and
P
Sin s
θ
1
2
as
x y
P
x y
xy
Cos
σ σ
θ
σ σ
τ
−
=
−
+
1
2
2
2
2
2
xy xy
P
x y x y
xy xy
Sin
τ τ
θ
σ σ σ σ
τ τ
= =
− −
+ +
1
2 2
2 2
2
2
2
2 2
Substituting
P
Cos
θ
1
2
and
P
Sin
θ
1
2
Solid Mechanics
(
)
(
)
( )
P
x y x y
xy xy
n
x y x y
xy xy
x y
xy
x y x y
xy xy
x y
xy
x y
xy
d
d
θ θ
σ σ σ σ
τ τ
σ
θ
σ σ σ σ
τ τ
σ σ
τ
σ σ σ σ
τ τ
σ σ
τ
σ σ
τ
=
− − −
= −
− −
+ +
− −
= −
− −
+ +
−
−
= +
−
+
1
2
2
2 2
2 2
2
2
2 2
2 2
2
2
2
2
2
4
2
2 2
4
2 2
4
2
2
x y
n
xy
d
d
σ σ
σ
τ
θ
−
∴ = − +
2
2
2
2
4
2
(ve)
( ) ( ) ( )
( )
P P
n
x y P xy P
x y P xy P
d
Cos Sin
d
Cos Sin
π
θ θ θ
σ
σ σ θ π τ θ π
θ
σ σ θ τ θ
= = +
= − + − +
= − +
1 1
2 1
1 1
2
2
2
2 2 4 2
2 2 4 2
Substituting
P P
Cos &Sin
θ θ
1 1
2 2
m we can show that
P
x y
n
xy
d
s
d
θ θ
σ σ
σ
τ
θ
=
−
∴ = − +
2
2
2
2
2
4
2
(+ve)
Solid Mechanics
Thus the angles
P
s
θ
1
and
P
s
θ
2
define planes of either
maximum normal stress or minimum normal stress.
(6) Now, we need to compute magnitudes of these stresses
We know that,
P
x y x y
n xy
x y x y
n P xy P
Cos Sin
Cos Sin
θ θ
σ σ σ σ
σ θ τ θ
σ σ σ σ
σ σ θ τ θ
=
+ −
= + +
+ −
= = + +
1 1
1
1
2 2
2 2
2 2
2 2
Substituting
P
Cos s
θ
1
2
and
P
Sin
θ
1
2
x y x y
xy
Max.Normal stress becauseof sign
σ σ σ σ
σ τ
+ −
= + +
+
2
2
1
2 2
Similarly,
( )
( )
P P
x y x y
n P
xy P
x y x y
P xy P
Cos
Sin
Cos Sin
π
θ θ θ
σ σ σ σ
σ σ θ π
τ θ π
σ σ σ σ
θ τ θ
= = =
+ −
= = + + +
+
+ −
= − −
1
2 1
1
1 1
2
2
2
2 2
2
2 2
2 2
Substituting
P
Cos
θ
1
2
and
P
Sin
θ
1
2
Solid Mechanics
x y x y
xy
Min.normal sressbecauseof vesign
σ σ σ σ
σ τ
+ −
= − +
−
2
2
2 2
We can write
x y x y
xy
or
σ σ σ σ
σ σ τ
+ −
= ± +
2
2
1 2
2 2
(7) Let us se the properties of above stress.
(1)
P P
s
π
θ θ= +
2 1
2
 planes on which maximum normal stress
and minimum normal stress act are
⊥
to each other.
(2) Generally maximum normal stress is designated by
σ
1
and minimum stress by
σ
2
. Also
P P
;
θ σ θ σ
→ →
1 2
1 2
algebraically i.e.,
σ σ
σ
σ
>
−
− −
1 2
1
2
0
1000
Solid Mechanics
(4) maximum and minimum normal stresses are collectively
called as principal stresses.
(5) Planes on which maximum and minimum normal stress
act are known as principal planes.
(6)
P
θ
1
and
P
θ
2
that define the principal planes are known as
principal directions.
(8) Let us find the planes on which shearing stresses are zero.
(
)
nt x y xy
Sin Cos
τ σ σ θ τ θ
= = − − +
0 2 2
xy
x y
tan
directionsof principal plans
τ
θ
σ σ
=
=
=
2
2
Thus on the principal planes no shearing stresses act.
Conversely, the planes on which no shearing stress acts are
known as principal planes and the corresponding normal
stresses are principal stresses. For example the state of stress
at a point is as shown.
Then
x
σ
and
y
σ
are
principal stresses because
no shearing stresses are
acting on these planes.
Solid Mechanics
(9) Since, principal planes are
⊥
to each other at a point P,
this also means that if an element whose sides are parallel to
the principal planes is taken out at that point P, then it will
be subjected to principal stresses. Observe that no shearing
stresses are acting on the four faces, because shearing
stresses must be zero on principal planes.
(10) Since
1
σ
and
2
σ
are in two
⊥
directions, we can easily
say that
x y x y
I
σ σ σ σ σ σ
′ ′
+ = + = + =
1 2 1
Solid Mechanics
5. Maximum shear stress
Maximum and minimum shearing stresses
So far we have seen some specials planes on which the
shearing stresses are always zero and the corresponding
normal stresses are principal stresses. Now we wish to find
what are maximum shearing stress plane and minimum
shearing stress plane. We approach in the similar way of
maximum and minimum normal stresses
(1)
x y
nt xy
Sin Cos
σ σ
τ θ τ θ
−
= − +
2 2
2
(
)
nt
x y xy
d
Cos Cos
d
τ
σ σ θ τ θ
θ
= − − +
2 2
For maximum or minimum
(
)
nt
x y xy
d
Cos Sin
d
τ
σ σ θ τ θ
θ
= = − − −
0 2 2 2
(
)
x y
xy
tan
σ σ
θ
τ
− −
=2
2
This has two roots
(
)
x y
S
xy
tan
s stands for shear stress
p stands for principal stresses.
σ σ
θ
τ
−
= −
−
−
1
2
2
Solid Mechanics
( )
(
)
x y
S S
xy
tan tan
σ σ
θ θ π
τ
− −
= + =
2 1
2 2
2
S S
π
θ θ
∴ = +
2 1
2
Now we have to show that at these two angles we will have
maximum and minimum shear stresses at that point.
Similar to the principal stresses we must calculate
( )
( )
S
nt
x y xy
nt
x y S xy S
d
Sin Cos
d
d
Sin Cos
d
θ θ
τ
σ σ θ τ θ
θ
τ
σ σ θ τ θ
θ
=
= − −
= − −
1 1
1
2
2
2
2
2 2 4 2
2 2 4 2
xy
S
x y
xy
Cos
τ
θ
σ σ
τ
=
−
+
1
2
2
2
2
2
2
(
)
x y
S
x y
xy
Sin
σ σ
θ
σ σ
τ
− −
=
−
+
1
2
2
2
2
2
Substituting above values in the above equation we can
show that
Solid Mechanics
S
nt
d
d
θ θ
τ
θ
=
=
1
2
2
 ve
Similarly we can show that
S S
nt
d
d
π
θ θ θ
τ
θ
= = +
=
2 1
2
2
2
+ ve
Thus the angles
S
θ
1
and
S
θ
2
define planes of either maximum
shear stress or minimum shear stress. Planes that define
maximum shear stress & minimum shear stress are again
⊥
to each other.. Now we wish to find out these values.
(
)
( )
S
x y
nt xy
x y
nt S xy S
Sin Cos
Sin Cos
θ θ
σ σ
τ θ τ θ
σ σ
τ θ τ θ
=
−
= − +
−
= − +
1 1
1
2 2
2
2 2
2
Substituting
S
Cos
θ
1
2
and
S
Sin s
θ
1
2
, we can show that
x y
max xy
σ σ
τ τ
−
= + +
2
2
2
(
)
( ) ( )
S S
x y
nt S xy S
Sin Cos
π
θ θ θ
σ σ
τ θ π τ θ π
= = +
−
= − + + +
1 1
2 1
2
2 2
2
Substituting
S
Cos
θ
1
2
and
S
Sin
θ
1
2
x y
min xy
σ σ
τ τ
−
= − +
2
2
2
Solid Mechanics
max
τ
is algebraically
min
τ
>
, however their absolute
magnitude is same. Thus we can write
x y
max min xy
or
σ σ
τ τ τ
−
= ± +
2
2
2
Generally
max S
min S
τ θ
τ θ
−
−
1
2
Q. Why
max
τ
and
min
τ
are numerically same. Because
S
θ
1
&
S
θ
2
are
⊥
planes.
(2) Unlike the principal stresses, the planes on which
maximum and minimum shear stress act are not free from
normal stresses.
Solid Mechanics
x y x y
n xy
Cos Sin s
σ σ σ σ
σ θ τ θ
+ −
= + +
2 2
2 2
S
x y x y
n S xy S
Cos Sin
θ θ
σ σ σ σ
σ θ τ θ
=
+ −
= + +
1 1
1
2 2
2 2
Substituting
S
Cos
θ
1
2
and
S
Sin
θ
1
2
S
x y
n
θ θ
σ σ
σ σ
=
+
= =
1
2
( )
( )
S S
x y x y
n S
xy S
Cos
Sin
π
θ θ θ
σ σ σ σ
σ θ π
τ θ π
= = +
+ −
= + +
+ +
1
2 1
1
2
2
2 2
2
Simplifying this equation gives
S
x y
n
θ θ
σ σ
σ σ
=
+
= =
2
2
Therefore the normal stress on maximum and minimum
shear stress planes is same.
(3) Both the principal planes are
⊥
to each other and also the
planes of
max
τ
and
min
τ
are also
⊥
to each other. Now let us
see there exist any relation between them.
Solid Mechanics
6. Mohr’s circle
Mohr’s circle for plane stress
So far we have seen two methods to find stresses acting on
an inclined plane
(a) Wedge method
(b) Use of transformation laws.
Another method which is purely graphical approaches is
known as the Mohr’s circle for plane stress.
A major advantage of Mohr’s circle is that, the state of the
stress at a point, i.e the stress components acting on all
infinite number of planes can be viewed graphically.
Equations of Mohr’s circle
We know that,
x y x y
n xy
Cos Sin
σ σ σ σ
σ θ τ θ
+ −
= + +
2 2
2 2
This equation can also be written as
x y x y
n xy
Cos Sin
σ σ σ σ
σ θ τ θ
+ −
− = +
2 2
2 2
x y
nt xy
Sin Cos
σ σ
τ θ τ θ
−
= − +
2 2
2
( )
x y x y
n nt xy
x a y R
σ σ σ σ
σ τ τ
+ +
− + = +
↓ ↓ ↓
− + =
2
2
2 2
2
2 2
2 2
Solid Mechanics
The above equation is
clearly an equation of
circle with center at
(
)
,0
a
on
τ σ
−
plane it
represents a circle with
center at
x y
,
σ σ
+
0
2
and
having radius
x y
xy
R
σ σ
τ
−
= +
2
2
This circle on
σ τ
−
plane
Mohr’s circle.
From the above deviation it
can be seen that any point P
on the Mohr’s circle
represents stress which are
acting on a plane passing
through the point.
In this way we can
completely visualize the
stresses acting on all
infinite planes.
Solid Mechanics
(3) Construction of Mohr’s circle
Let us assume that the state of stress at a point is given
A typical problem using Mohr’s circle i.e given
x y
,
σ σ
′ ′
and
x y
τ
′ ′
on an inclined element. For the sake of clarity we
assume that,
x y
,s
σ σ
′ ′
and
x y
τ
′ ′
all are positive and
x y
σ σ
>
Solid Mechanics
• Since any point on the circle represents the stress
components on a plane passing through the point.
Therefore we can locate the point A on the circle.
• The coordinates of the plane
(
)
x xy
A,
σ τ
= + +
Therefore we can locate the point A on the circle with
coordinates
(
)
x xy
,s
σ τ+ +
• Therefore the line AC represents the xaxis. Moreover,
the normal of the Aplane makes
0
w.r.t the xaxis.
• In a similar way we can locate the point B
corresponding to the plane B.
Solid Mechanics
The coordinates of
(
)
y xy
B,s
σ τ= + −
Since we assumed that for the sake of similarity
y x
s
σ σ
<
.
Therefore the point B diametrically opposite to point A.
• The line BC represents y axis. The point A corresponds
to Q
=
0
, and pt. B corresponds to Q=
90
(+ve) of the
stress element.
At this point of time we should be able to observe two
important points.
• The end points of a diameter represents stress
components on two
⊥
planes of the stress element.
• The angle between x axis and the plane B is 90° (c.c.w)
in the stress element. The line CA in Mohr’s circle
represents x axis and line CB represents yaxis or plane
B. It can be seen that, the angle between xaxis and y
axis in the Mohr’s circle is 180° (c.c.w). Thus 2Q in
Mohr’s circle corresponds to Q in the stress element
diagram.
Stresses on an inclined element
• Point A corresponds to
0
Q
=
on the stress element.
Therefore the line CA i.e xaxis becomes reference line
from which we measure angles.
• Now we locate the point “D” on the Mohr’s circle such
that the line CD makes an angle of 2Q c.c.w from the x
axis or line CA. we choose c.c.w because in the stress
element also Q is in c.c.w direction.
Solid Mechanics
• The coordinates or stresses corresponding to point D on
the Mohr’s circle represents the stresses on the
x
′
 face or
D on the stress element.
x avg
x y
y avg
RCos
RSin
RCos
SinceD&D are planesinthe
stress element,thentheybecome
diametricallyopposite point son
thecircle,just likethe planes A&Bdid
σ σ β
τ β
σ σ β
′
′ ′
′
= +
=
= −
′
⊥
Calculation of principal stress
The most important application of the Mohr’s circle is
determination of principal stresses.
The intersection of the Mohr’s circle  with normal stress
axis gives two points
P
1
and
P
2
. Thus
P
1
and
P
2
represents
points corresponding to principal stresses. In the current
diagram the coordinates the of
P,s
P,
σ
σ
=
=
1 1
2 2
0
0
avg
R
σ σ
= +
1
avg
R
σ σ
= −
2
The principal direction corresponding to
σ
1
is now equal to
p
θ
1
2
, in c.c.w direction from the xaxis.
Solid Mechanics
p p
π
θ θ
= ±
2 1
2
We can see that the points
P
1
and
P
2
are diametrically
opposite, this indicate that principal planes are
⊥
to each
other in the stress element. This fact can also be verified from
the Mohr’s circle.
In plane maximum shear stress
What are points on the circle at which the shearing stress are
reaching maximum values numerically? Points
S
1
and
S
2
at
the top and bottom of the Mohr’s circle.
• The points
S
1
and
S
2
are at angles
θ
=
2 90
from
points
P
1
P
2
and, i.e the planes of maximum shear stress
are oriented at
±
45
to the principal planes.
• Unlike the principal stresses, the planes of maximum
shear stress are not free from the normal stresses. For
example the coordinates of
max avg
max avg
S,s
S,
τ σ
τ σ
= +
= −
1
2
max
R
τ
= ±
avg
σ σ
=
Mohr’s circle can be plotted in two different ways. Both the
methods are mathematically correct.
Solid Mechanics
Finally
• Intersection of Mohr’s circle with the
σ
axis gives
principal stresses.
• The top and bottom points of Mohr’s circle gives
maximum –ve shear stress and maximum +ve shear
stress.
• Do not forget that all these inclined planes are obtained
by rotation about zaxis.
Solid Mechanics
Mohr’ circle problem
Solution:
A
 (15000,4000)
B
 (5000,4000)
(a)
x y
MPa
σ σ
+
+
= =
15000 5000
10000
2 2
R MPa
=
6403
x y
xy
R
σ σ
τ
−
−
= + = +
= +
2
2
2 2
2 2
15000 5000
4000
2 2
5000 4000
x y
σ σ
−
=
5000
2
Solid Mechanics
Point
D
:
x
Cos.MPa
σ
′
= + =
10000 6403 41 34 14807
x y
Sin.MPa
τ
′ ′
= − = −
6403 41 34 4229
Point
D
′
:
n y
Cos.MPa
σ σ
′
= = − =
10000 6403 41 34 593
nt x y
Sin.
τ τ
′ ′
= = =
6403 41 34 4229
b)
P
.
;.
σ θ= = =
1
1
38 66
16403 19 33
2
MPa
σ
=
2
3597
c)
max S
MPa..
τ θ
= − = = −
1
6403 25 67 25 67
Solid Mechanics
(2)
θ
=
45
Principal stresses and principal shear stresses.
Solution
:
( )
x y
x y
xy
R MPa
σ σ
σ σ
τ
+
− +
= = −
−
− −
= + = + − =
2
2
2
2
50 10
20
2 2
50 10
40 50
2 2
(
)
( )
A,
B,
→ − −
→
50 40
10 40
x y
x y
p R s
p R
σ σ
σ
σ σ
σ
+
= = + = − + =
+
= = − = − − = −
1 1
2 2
20 50 30
2
20 50 70
2
Solid Mechanics
p
pp
Q.
Q.
Q.
=
=
=
1
1
2
2 233 13
116 6
206 6
s
ss
Q.
Q.
Q.
=
=
=
1
1
2
2 143 13
71 6
161 6
Solid Mechanics
Q.
x y xy
MPa,MPa and MPa
σ σ τ
= = − =
31 5 33
Stresses on inclined element
θ
=
45
Principal stresses and maximum shear stress.
Solution
:
x y
avg
MPa
σ σ
σ
+
−
= = =
31 5
13
2 2
x y
xy
R.MPa
σ σ
τ
−
= + =
2
2
37 6
2
(
)
( )
A,
B,− −
31 33
5 33
x avg
RCos s
.Cos.MPa
σ β σ
′
= +
= + =37 6 28 64 13 46
x y
RSin...
τ β
′ ′
= − = − = −
37 6 28 64 18 02
y avg
RCos
MPa
σ β σ
′
= −
= −20
Solid Mechanics
.MPa
σ
∴ =
1
50 6
.MPa
σ
= −
2
24 6
p
.
θ
=
1
30 68
max s
min
avg
.MPa.
.MPa
MPa
τ θ
τ
σ σ
= − = −
= −
= =
1
37 6 14 32
37 6
13
Solid Mechanics
7. 3DStress Transformation
3Dstress components on an arbitrary plane
Basically we have done so far for this type of coordinate
system
x x x y x z
x x x y x z
n n n D i r.c o s i n e s o f x
ˆˆ ˆ ˆ
i n i n j n k
′ ′ ′
′ ′ ′
′
−
′
= + +
y x y y y z
y x y y y z
n n n
ˆ
ˆ ˆ ˆ
j n i n j n k
′ ′ ′
′ ′ ′
′
= + +
z x z y z z
z x z y z z
n n n
ˆ ˆ
ˆ ˆ
k n i n j n k
′ ′ ′
′ ′ ′
′
= + +
Solid Mechanics
n x x x y x z
n x x x y x z
ˆ
ˆ ˆ
T T i T j T ks
ˆ
ˆ ˆ
T i j k
σ τ τ
′ ′ ′
′ ′ ′ ′ ′ ′
= + +
′ ′ ′
= + +
x x
x x
x z
ABC dA
PAB dAn
PAC dAn
PBC dAn
′
′
′
−
−
−
−
[
]
x
F
→+ =
0
x x x x x yx x y zx x z
T da dAn dAn dAn
σ τ τ
′ ′ ′ ′
= + +
x x x x x yx x y zx x z
x y xy x x y x y zy x z
x z xz x x yz x y z x z
T n n n
T n n n
T n n n
σ τ τ
τ σ τ
τ τ σ
′ ′ ′ ′
′ ′ ′ ′
′ ′ ′ ′
= + +
= + +
= + +
x x y y z
x y y y z
z x y z z
σ τ τ
τ σ τ
τ τ σ
′ ′ ′ ′ ′
′ ′ ′ ′ ′
′ ′ ′ ′ ′
x x y x z
,,
σ τ τ
′ ′ ′ ′ ′
(
)
(
)
x n x x x y x z x x x y x z
ˆ ˆ
ˆ ˆ ˆ ˆ ˆ
T i T i T j T k.n i n j n k
σ
′ ′ ′ ′ ′ ′ ′
′
= = + + + +
(1)
(
)
(
)
x y n x x x y x z y x y y y z
ˆ ˆ
ˆ ˆ ˆ ˆ ˆ
T j T i T j T k.n i n j n k
τ
′ ′ ′ ′ ′ ′ ′ ′
′
= = + + + +
(2)
(
)
(
)
x z n x x x y x z z x z y z z
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ
T k T i T j T k.n i n j n k
τ
′ ′ ′ ′ ′ ′ ′ ′
′
= = + + + +
(3)
y x x y x yx y y zx y z
y y xy y y y y y zy y z
y z xz y y yz y y z y z
T n n n
T n n n
T n n n
σ τ τ
τ σ τ
τ τ σ
′ ′ ′ ′
′ ′ ′ ′
′ ′ ′ ′
= + +
= + +
= + +
(
)
(
)
y y x y y y z y x y y y z
ˆ ˆ
ˆ ˆ ˆ ˆ
T i T j T k n i n j n k
σ
′ ′ ′ ′ ′ ′ ′
= + + + + (4)
(
)
(
)
z z x z y z z z x z y z z
ˆ ˆ
ˆ ˆ ˆ ˆ
T i T j T k n i n j n k
σ
′ ′ ′ ′ ′ ′ ′
= + + + + (5)
Solid Mechanics
(
)
(
)
y z y x y y y z z x z y z z
ˆ ˆ
ˆ ˆ ˆ ˆ
T i T j T k n i n j n k
τ
′ ′ ′ ′ ′ ′ ′ ′
= + + + + (6)
x x
x y
x z
n Cos
n Sin
n
θ
θ
′′′
=
=
= 0
y x
y y
y z
n Sin
n Cos
n
θ
θ
′′′
= −
=
= 0
z x
z y
z z
n
nn
′′′
=
=
=
0
0
1
z x z y z
z
::
σ τ τ
σ
′ ′ ′ ′ ′
= = =
=
0 0 0
( )
( )
x x y xy
y x y xy
x y x y xy
Cos Sin Sin Cos
Sin Cos Sin Cos
Sin Cos Cos Sin
σ σ θ σ θ τ θ θ
σ σ θ σ θ τ θ θ
τ σ σ θ θ τ θ θ
′
′
′ ′
= + +
= + −
= − − + −
2 2
2 2
2 2
2
2
x xy
xy y
σ τ
τ σ
0
0
0 0 0
Principal stresses
x y z
n,n,n
(
)
n x y z
n nx ny nz
ˆ
ˆ ˆ
ˆ
T n n i n j n k
ˆˆ ˆ
T T i T j T k
σ σ= = + +
= + +
Where
nx x x yx y zx z
ny xy x y y zy z
nz xz x yz y z z
T n n n
T n n n
T n n n
σ τ τ
τ σ τ
τ τ σ
= + +
= + +
= + +
x x y y z z
Tn n Tn n Tn n
σ σ σ
= = =
Solid Mechanics
(
)
( )
( )
x x yx y zx z
yx x y y zy z
xz x yz y z z
n n n
n n n Syst
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