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 built-in 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 m-m. Therefore, stress components magnitude &

direction depends on orientation of cut m-m.

(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 x-plane. We can see

x

σ

and

xy

τ

only when we take section

⊥

to x-axis.

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 x-plane 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 , y-plane and z-plane.

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

=

Cross-shears 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 stress-is 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 stress-state (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 2D-plane 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 m-m 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 plane-stress 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 x-axis (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 x-axis. Moreover,

the normal of the A-plane makes

0

w.r.t the x-axis.

• 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 y-axis or plane

B. It can be seen that, the angle between x-axis 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 x-axis 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 x-axis.

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 z-axis.

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. 3D-Stress Transformation

3D-stress 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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