Solid Mechanics

1. Shear force and bending moment diagrams

Internal Forces in sol ids

Sign convent ions

Shear forces are given a special symbol on yV12

and zV

The couple moment along the axis of the member is given

xM T= =Torque

y zM M= =bending moment.

Solid Mechanics

We need to fol low a systematic sign convention for systematic development of equations and reproducibi l ity of the equations

The sign convention is l ike 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 otherw ise 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) A ll forces are appl ied at the axis of the beam.

Appl icat ion of method of sect ions

What are the necessary internal forces to keep the segment of the beam in equi l ibrium?

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 whi le designing the beams for various appl ications.

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 rol ler supports are cal led 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 w ith 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

Di fferent ial equat ions of equi l ibrium

[ ]xFS = fi +0

yFS = + 0

V V V P x

V P x

VP

x

D DD DDD

+ - + =

= -

= -

0

x

V dVP

x dxlimDDDfi

= = -0

[ ]AP x

M V x M M MD

S D D= - + + - =2

0 02

P xV x M

M P xV

x

DD D

D DD

+ - =

+ - =

20

2

02

Solid Mechanics

x

M dMV

x dxlimDDDfi

= = -0

From equation dV

Pdx

= - we can w rite

D

C

X

D CX

V V Pdx- = -

From equation dM

Vdx

= -

D CM M Vdx- = -

Special cases:

Solid Mechanics

Solid Mechanics

Solid Mechanics

Solid Mechanics

( ) ( )x -0 2 1 1

A B

V

V

V ; 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

V

V

V ; V

-

- + - - =

= +

= + = +

6 8 3 3

5 30 30 10 0

15

15 15( ) ( )

D E

x

V

V

V

V ; 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

. xM x x

. xM x x

M .

M=

=

- -

-- + - - + =

-= - + - -

= +

=

2

2

5 33

6

2 6 2 2

7 5 210 5 30 2 0

2

7 5 210 5 30 2

241 66

40

( ) ( ) [ ]( ) ( ) ( )

C x

D x

x C D

M x x x x

M

M=

=

- - -

- + - - + - + - + =

= +

= -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 draw ing SFD and BMD.

[ ]x A B -0 2

Solid Mechanics

A

B

V

V

V

V

- =

=

=

=

5 0

5

5

5A 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

xM x x .

x

M

M x . .

x

M

-- + - - + =

=

=

= =

=

=

2210 5 30 2 7 5 0

26

40

5 33 41 66

2

0

[ ]x C D -6 8

C D

V

V

V , V

- + - - =

=

= =

5 30 10 30 0

15

15 15

Solid Mechanics

[ ]x D E -8 10

D E

V

V

V , 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 mD

fi + = =

+ = + - =

=

= + - =

= -

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 AB A

M M

M M

- = - -

= + = -

= -

60

60 60 285

225

Solid Mechanics

( )C BC B

M M

M M

- = - -

= + = - +

= -

90

90 225 90

135

( )D CD 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

[ ]ACy

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

fi + =

=

= + + - - = + =

0

0

0 60 50 0 110 1

( )C AC 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

29058

552

Solid Mechanics

( )

y

B

F V x

V x x m

= + + - =

= -

0 52 20 0

20 52 0 3

[ ]

( )

M

xM x

xM x x m

=

+ - =

= -

2

2

0

2052 0

2

2052 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

dMV

dxdV

Pdx

= -

= -

[ ] ( ) ( )( ) ( ) ( )

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

( )

yF

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

dMV

dxdV

Pdx

= -

= -

B AM M Vdx- = -

Solid Mechanics

2. Concept of stress Tract ion vector or St ress vector

Now we define a quantity known as stress vector or traction as

D

DDfi

=

Rn

A

FT

Alim0 units aP N / m-

2

and we assume that the quantity

D

DDfi

fi

R

A

MAlim0

0

(1) nT

is a vector quantity hav ing direction of RFD

(2) nT

represent intensity point distributed force at the point

" P" on a plane whose normal is n

(3) nT

acts in the same direction as RFD

Solid Mechanics

(4) There are two reasons are avai lable for justification of the

assumption that D

DDfi

fi

R

A

MAlim0

0

(a) experimental (b) as AD fi 0, RFD

becomes resultant of a paral lel

force distribution. Therefore RMD = 0

for force system.

(5) nT

varies from point to point on a given plane

(6) nT

at the same point is different for different planes.

(7) n nT T= -

w i l l act at the point P

(8) In general

Components of nT

R n t s F F n v t v sD D D D= + +

Solid Mechanics

D D D D

D D D DD D D Dfi fi fi fi

= = + +

R n t sn

A A A A

F F v v T n t sA A A Alim lim lim lim0 0 0 0

n nn nt ns T n t ss t t= + +

where

D

D

D

Ds

DD

tDD

tD

fi

fi

fi

= = =

= = =

= = =

n nnn

A

t tnt

A

s sns

A

F dFNormal stresscomponent

A dA

v dvShear stresscomponent

A dA

v dvAnother shear componet

A dA

lim

lim

lim

0

0

0

st

-

-

Normal Stress

Shear stress

n nndF dAs=

t ntdV dAt=

Notat ion of st ress components

The magnitude and direction of nT

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 s is acting (b) Second subscript- direction

Solid Mechanics

Rectangular components of st ress

Cuts ^ to the coordinate planes w i l l give more valuable information than arbitrary cuts.

D D D D

DD D DD D D Dfi fi fi fi

= = + +

yR x zx

A A A A

vF F v T i j kA A A Alim lim lim lim0 0 0 0

x xx xy xz T i j ks t t= + +

where

xxx

A

y zxy xz

A A

FNormal stress

A

v vShear stress; Shear stress

A A

lim

lim lim

D

D D

Ds

DD D

t tD D

fi

fi fi

= =

= = = =

0

0 0

Solid Mechanics

s=x xxdF dA

y xydv dAt=

z xzdv dAt=

Simi larly,

D D D D

DD D DD D D Dfi fi fi fi

= = + + yR x zy

A A A A

FF v v T i j kA A A Alim lim lim lim0 0 0 0

t s t= + +y yx yy yz

T i j k

t t s= + +z zx zy zz

T i j k

xxs and xyt w i ll act only on x-plane. We can see xs and xyt

only when we take section ^ to x-axis.

The st ress tensor

s t t

s t s t

t t s

=

xx xy xz

jj yx yy yz

zx zy zz

Rectan gular stresscomponents

This array of 9 components is cal led 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 mutual ly ^ planes passing through the point P

\ Stress tensor is an array consisting of stress components acting on three mutual ly perpendicular planes.

t t t= + +

n nx ny nz

T i j k

What is the di fference between dist ributed loading & st ress?

R

A

Fq lim

ADDDfi

=0

yyq s= can also be called.

No difference!

Except for their origin!

Solid Mechanics

Sign convent ion of st ress 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 Ps t= - = -20 10 and xz Pat = 30 at a point P

means.

Solid Mechanics

State of st ress 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 mutual ly perpendicular planes passing through the point are known, we can determine the

stress vector nT

acting on any plane n through that

point.

The stress tensor w i l l 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

s t t

s t s t

t t s

=

can also represent state of stress at a point.

Solid Mechanics

The st ress element

Is there any convenient way to visual ize or represent the state of stress at a point or stresses acting three mutual ly perpendicular planes say x- plane , y-plane and z-plane.

xx xy xz

ij yx yy yzP

zx zy zz

s t t

s t s t

t t s

+ + +

= + + + + + +

( )( )

xx xx

yy yy

x ,y ,zContinuous functionsof x ,y ,z

x ,y ,z

s s

s s

=

=

Let us consider a stress tensor or state of stress at a point in a component as

Solid Mechanics

ijs- -

= - - - -

10 5 30

5 50 60

30 60 100

Equi l ibrium of st ress element

[ ]xF = fi +0

x yx zx x yx zxdydz dxdz dydx dydz dxdz dxdys t t s t t+ + - - - = 0

Simi larly, we can show that yF = 0 and zF = 0 is satisfied.

y

dz

dy

zdx

x

xyt

xztxs

Solid Mechanics

PzM

C.C.W ve

= +

0

( ) ( )xy yxdydz dx dxdz dyt t- = 0 xy yxt t- = 0

xy yxt t=

Shearing stresses on any two mutual ly perpendicular planes are equal.

PxM = 0 yz zyt t= and PyM = 0 zx xzt t=

Cross-shears are equal- a very important result

Since xy yxt t= , if xy vet = - yxt is also ve

Solid Mechanics

\ The stress tensor

xx xy xz

ij yx xy xy yz

zx xz zy yz yz

issecondranksymmetrictensor

s t t

s t t s t

t t t t s

= = = =

Di fferent ial equat ions of equi l ibrium

[ ]xF fi + = 0

yxx zxx yx zx

x xy zx x

x y z y x z z y xx y z

y z x z y x B x y z

ts ts t t

s t t

+ D D D + + D D D + + D D D - D D - D D - D D + D D D = 0

yxx zxxx y z y x z x y z B x y zx y z

ts tD D D + D D D + D D + D D D =

2

0

Cancel ing x yD D and zD terms and taking l imit

yxx zxx

xyz

lim Bx y z

ts tD fiD fiD fi

+ + + =

000

0

Simi larly we can easi ly show that

Solid Mechanics

[ ]yxx zx x xB Fx y zts t

+ + + = =

0 0

xy yy zyy yB Fx y z

t s t + + + = =

0 0

[ ]yzxz zz z zB Fx y ztt s

+ + + = =

0 0

If a body is under equi l ibrium, then the stress components must satisfy the above equations and must vary as above.

For equi l ibrium, the moments of forces about x, y and z axis at any point must vanish.

pzM =

0

xy yxxy xy yx

yx

yx xx y z y z y x z

x y

yx z

t tt t t

t

DD+ D D D + D D - + D D D

D- D D =

2 2 2

02

Solid Mechanics

xy xy yx yx

xy yxxy yx

y x z x y zx y z x y zx y

yxx y

t t t t

t tt t

D D D D D D D D D D D D+ - - =

DD+ - - =

2 22 20

2 2 2 2

02 2

Taking l imit

xy yxxy yx

xyz

yxlim

x y

t tt t

D fiD fiD fi

DD+ - - =

000

02 2

xy yxt t - = 0 xy yxt t=

Relat ions between st ress components and internal force resul tants

Solid Mechanics

x xxA

F dAs= ; y xyA

V dAt= ; z xzA

V dAt=

xz xy xy dA dAz dMt t- =

( )x xz xyA

M y z dAt t= -

y xzA

M dAs= ; z xyA

M dAs= -

Solid Mechanics

3. Plane stress and Plane strain Plane st ress- 2D State of st ress

If

( ) ( )( ) ( )

x xyij

xy yy

x ,y x ,yplane stress-is a --- state of stress

x ,y x ,y

s ts

t s

= -

A l l stress components are in the plane x y- i.e al l stress

components can be v iewed in x y- plane.

xy

x xyx xy

ij xy yyx y

D Stateof stress

Stresscomponentsin planexy

t

s ts t

s t st s=

-

= =

2

0

0

0 0 0

x xy xz

ij yx yy yz

zx zy zz

D Stateof stress

components

s t t

s t s t

t t s

-

= -

3

6

Solid Mechanics

This type of stress-state (i.e plane stress) exists in bodies whose z- direction dimension is very smal l w .r.t other dimensions.

St ress t ransformat ion laws for plane st ress

The state of stress at a point P in 2D-plane stress problems are represented by

x xy nn ntij

xy y nt tt

s t s ts

t s t s

= =

Solid Mechanics

* We can determine the stress components on any plane n by know ing the stress components on any two mutual ly ^ planes.

St ress t ransformat ion laws for plane st ress

In order to get useful information we take different cutting planes passing through a point. In contrast to 3D problem, al l cutting planes in plane stress problems are paral lel 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 total ity of all the stress. If we know the stress components on any two mutual ly ^ planes then stress components on any arbitrary plane m-m can be determined. Thus the stress tensor

x xyij

xy y

s ts

t s

=

is sufficient to tel l about the state of stress

at a point in the plane stress problems.

dA Area of AB

dACs Areaof BC

dASin Area of AC

qq

=

=

=

nF + = 0

nn x xy xy

yy

dA dACos Cos dACos Sin dASin Cos

dASin Sin

s s q q t q q t q q

s q q

- - - -

= 0

nn x xy yyCos Sin Cos Sins s q t q q s q- - - =2 22 0

Solid Mechanics

nn x y xy

x y x ynn xy

Cos Sin Sin Cos

Cos Sin

s s q s q t q q

s s s ss q t q

= + +

+ -= + +

2 2 2

2 22 2

nF + = 0

nt x xy xy

y

dA dACos Sin dACos Cos dASin Sin

dASin Cos

s s q q t q q t q q

s q q

- - + -

= 0

( )nt x y xyCos Sin Sin Cos Cos Sint s q q s q q t q q= - + + -2 2

( ) ( )( )

nt x y xy

x ynt xy

Cos Sin Cos Sin

Sin Cos

t q q s s t q q

s st q t q

= - - + -

-= - +

2 2

2 22

We shal l now show that if you know the stress components on two mutual ly ^ planes then we can compute stresses on any incl ined plane. Let us assume that we know that state of stress at a point P is given

x xyij

xy y

s ts

t s

=

This also means that

Solid Mechanics

Solid Mechanics

If q q= we can compute on AB

If p

q q= +2

we can compute on BC

If q q p= + we can compute on CD

If p

q q= +32

we can compute on DA

nns and ntt 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 equi l ibrium state and thus are also valid at al l points of the body under accelerations.

These laws are true for any point P of a body.

Invariants of st ress tensor

Any quantity for which its 2D scalar components transform from one coordinate system to another according to nns and ntt is cal led 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 xyI , I ; Is s t= = = -

By definition a tensor is a mathematical quantity that transforms according to certain laws, such that certain invariant properties are maintained for al l 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 ynn xyCos Sin

s s s ss q t q

+ -= + +2 2

2 2

x y x yt xyCos Sin

s s s ss q t q

+ -= + -2 2

2 2

x ynt xySin Cos

s st q t q

-= - +2 2

2

Solid Mechanics

n t x y x y Is s s s s s + = + = + = 1

I =1 First invariant of stress in 2D

n t nt x y xy x y x y Is s t s s t s s t - = - = - =2 2

2

I =2 Second invariant of stress in 2D

I ,I1 2 are invariants of 2D symmetric stress tensor at a point.

Invariants are extremely useful in checking the correctness of transformation

Of I1 and I2, I1 is the most important property : the sum of normal stresses on any two mutual ly ^ 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 clockw ise angle of 15 w .r.t the original element.

Solut ion:

it is in C.W.

x

y

xy

Q

ss

t

= -

=

= -

= -

46

12

19

15

Solid Mechanics

Substituting q = - 15 in ntt equation

x y MPass s+ - + -

= = = -46 12 34

172 2 2

( ) ( )Sin Sin . ; Cos Cos .q q= - = - = - =2 2 15 0 5 2 2 15 0 866

x y x yn xyCos Sin

s s s ss q t q

+ - = + +

2 2

2 2

n . .s = - - + 17 29 0 866 19 0 5

n . MPass = -1 32 6

x ynt xySin Cos

s st q t q

- = - +

2 2

2

n t MPat = -1 1 31

x y MPas s- - - -

= = = -46 12 58

292 2 2

n t . .t = - - 1 1 29 0 5 19 0 866

Solid Mechanics

Now

As a check

t n nt qs s t = = = 2 75

n Cos Sin

MPa

s = - - - = -

17 29 2 165 19 2 165

32

nt

nt

. Sin Cos

MPa

tt

= -

= -

00 29 330 19 330

31

n t x y . . MPa ss s s s+ = + = - - = - = - +32 6 1 4 34 46 12

q = 145

tn Sin Cos

MPa

t = + - =

29 150 19 150

31

t cos sins\ = - - -17 29 150 19 150

t . MPas = - 1 4

tn n t nt qt t t = = = 2 2 75

Solid Mechanics

4. Principal Stresses Principal St resses

Now we are in position to compute the direction and magnitude of the stress components on any incl ined plane at any point, prov ided if we know the state of stress (Plane stress) at that point. We also know that any engineering component fai ls when the internal forces or stresses reach a particular value of al l the stress components on al l of the infinite number of planes only stress components on some particular planes are important for solv ing our basic question i.e under the action of given loading whether the component w i l l ai l or not? Therefore our objective of this class is to determine these plane and their corresponding stresses.

(1) ( ) n y n yn n xyCos Sins s s s

s s q q t q+ -

= = + +2 22 2

(2) Of al l 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 mathematical ly we are looking for maxima and minima of

( )n Qs function..

(3) n y n yn xyCos Sins s s s

s q t q+ -

= + +2 22 2

For maxima or minima, we know that

( )n x y xyd Sin Cosds

s s q t qq

= = - - +0 2 2 2

xy

x ytan

tq

s s=

-

22

(4) The above equations has two roots, because tan repeats itself after p . Let us cal l the first root as Pq 1

xyP

x ytan

tq

s s=

-12

2

( ) xyP Px y

tan tant

q q ps s

= + =-2 1

22 2

Solid Mechanics

P P sp

q q= +2 1 2

(5) Let us verify now whether we have minima or minima at

Pq 1 and Pq 2

( )

( )P

nx y xy

nx y P xy P

dCos Sin

d

dCos Sin

d q q

ss s q t q

q

ss s q t q

q =

= - - -

\ = - - -1 1

1

2

2

2

2

2 2 4 2

2 2 4 2

We can find PCos sq 12 and PSin sq 12 as

x yP

x yxy

Coss s

qs s

t

-=

- +

1 22

2

22

xy xyP

x y x yxy xy

Sint t

qs s s s

t t

= =- -

+ +

1 2 22 2

22

22 2

Substituting PCos q 12 and PSin q 12

Solid Mechanics

( )( )

( )

P

x y x y xy xyn

x y x yxy xy

x y xy

x y x yxy xy

x yxy

x yxy

d

d q q

s s s s t tsq s s s s

t t

s s t

s s s st t

s st

s st

=

- - -= -

- - + +

- -= -

- - + +

- - = + - +

1

2

2 2 22 2

2 2

2 22 2

22

22

2 4

22 2

4

2 2

42

2

x ynxy

d

d

s sst

q

- \ = - +

222

2 4 2 (-ve)

( ) ( ) ( )

( )

P P

nx y P xy P

x y P xy P

dCos Sin

d

Cos Sin

pq q q

ss s q p t q p

q

s s q t q

= = +

= - + - +

= - +

1 1

2 1

1 1

2

2

2

2 2 4 2

2 2 4 2

Substituting P PCos & Sinq q1 12 2 m we can show that

P

x ynxy

ds

d q q

s sst

q =

- \ = - +

2

222

2 4 2 (+ve)

Solid Mechanics

Thus the angles P sq 1 and P sq 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 yn xy

x y x yn P xy P

Cos Sin

Cos Sinq q

s s s ss q t q

s s s ss s q t q=

+ -= + +

+ -= = + +

1 111

2 22 2

2 22 2

Substituting PCos sq 12 and PSin q 12

x y x yxy

Max.Normal stress becauseof sign

s s s ss t

+ - = + +

+

22

1 2 2

Simi larly,

( )

( )P P

x y x yn P

xy P

x y x yP xy P

Cos

Sin

Cos Sin

pq q q

s s s ss s q p

t q p

s s s sq t q

= = =

+ -= = + + +

+

+ -= - -

12 1

1

1 1

22

22 2

2

2 22 2

Substituting PCos q 12 and PSin q 12

Solid Mechanics

x y x yxy

M in.normal sressbecauseof vesign

s s s ss t

+ - = - +

-

22

2 2

We can w rite

x y x yxyor

s s s ss s t

+ - = +

22

1 2 2 2

(7) Let us se the properties of above stress.

(1) P P sp

q q= +2 1 2

- planes on which maximum normal stress

and minimum normal stress act are ^ to each other.

(2) General ly maximum normal stress is designated by s 1 and minimum stress by s 2. A lso P P;q s q sfi fi1 21 2

algebraically i.e.,s sss

>

-

- -

1 2

1

2

0

1000

Solid Mechanics

(4) maximum and minimum normal stresses are col lectively cal led as principal stresses.

(5) Planes on which maximum and minimum normal stress act are known as principal planes.

(6) Pq 1 and Pq 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 xySin Cost s s q t q= = - - +0 2 2 xy

x ytan

directionsof principal plans

tq

s s=

=

=

22

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 xs and ys 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 paral lel to the principal planes is taken out at that point P, then it w i l l 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 1s and 2s are in two ^ directions, we can easi ly say that

x y x y Is s s s s s + = + = + =1 2 1

Solid Mechanics

5. M aximum shear stress Maximum and minimum shearing st resses

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 w ish to find what are maximum shearing stress plane and minimum shearing stress plane. We approach in the simi lar way of maximum and minimum normal stresses

(1) x ynt xySin Coss s

t q t q-

= - +

2 22

( )nt x y xyd Cos Cosdt

s s q t qq

= - - +2 2

For maximum or minimum

( )nt x y xyd Cos Sindt

s s q t qq

= = - - -0 2 2 2

( )x yxy

tans s

qt

- - =2

2

This has two roots

( )x yS

xytan

s stands for shear stress

p stands for principal stresses.

s sq

t

-= -

-

-

12

2

Solid Mechanics

( ) ( )x yS Sxy

tan tans s

q q pt

- -= + =

2 12 2

2

S Sp

q q\ = +2 1 2

Now we have to show that at these two angles we w i l l have maximum and minimum shear stresses at that point.

Simi lar to the principal stresses we must calculate

( )

( )S

ntx y xy

ntx y S xy S

dSin Cos

d

dSin Cos

d q q

ts s q t q

q

ts s q t q

q =

= - -

= - -1 1

1

2

2

2

2

2 2 4 2

2 2 4 2

xyS

x yxy

Cost

qs s

t

=-

+

1 22

22

22

( )x yS

x yxy

Sins s

qs s

t

- -=

- +

1 22

2

22

Substituting above values in the above equation we can show that

Solid Mechanics

S

ntd

d q q

tq =

=

1

2

2 - ve

Simi larly we can show that

S S

ntd

d pq q q

tq = = +

=

2 1

2

2

2

+ ve

Thus the angles Sq 1 and Sq 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 w ish to find out these values.

( )

( )S

x ynt xy

x ynt S xy S

Sin Cos

Sin Cosq q

s st q t q

s st q t q=

-= - +

-= - +

1 11

2 22

2 22

Substituting SCos q 12 and SSin sq 12 , we can show that

x ymax xy

s st t

- = + +

22

2

( ) ( ) ( )S S

x ynt S xy SSin Cospq q q

s st q p t q p= = +

-= - + + +

1 12 1 22 2

2

Substituting SCos q 12 and SSin q 12

x ymin xy

s st t

- = - +

22

2

Solid Mechanics

maxt is algebraical ly mint> , however their absolute magnitude is same. Thus we can w rite

x ymax min xyor

s st t t

- = +

22

2

General ly

max S

min S

t q

t q

-

-1

2

Q. Why maxt and mint are numerical ly same. Because Sq 1 &

Sq 2 are ^ planes.

(2) Unl ike the principal stresses, the planes on which maximum and minimum shear stress act are not free from normal stresses.

Solid Mechanics

x y x yn xyCos Sin s

s s s ss q t q

+ -= + +2 2

2 2

S

x y x yn S xy SCos Sinq q

s s s ss q t q=

+ -= + +

1 112 2

2 2

Substituting SCos q 12 and SSin q 12

S

x yn q q

s ss s =

+= =

1 2

( )

( )S S

x y x yn S

xy S

Cos

Sin

pq q q

s s s ss q p

t q p

= = +

+ -= + +

+ +

12 1

1

22

2 2

2

Simpl ifying this equation gives

S

x yn q q

s ss s =

+= =

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 maxt and mint are also ^ to each other. Now let us see there exist any relation between them.

Solid Mechanics

6. M ohrs ci rcle Mohrs ci rcle for plane st ress

So far we have seen two methods to find stresses acting on an incl ined plane

(a) Wedge method (b) Use of transformation laws.

Another method which is purely graphical approaches is known as the Mohrs circle for plane stress.

A major advantage of Mohrs 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 v iewed graphical ly.

Equat ions of Mohrs ci rcle

We know that, x y x yn xyCos Sins s s s

s q t q+ -

= + +2 22 2

This equation can also be w ritten as

x y x yn xyCos Sin

s s s ss q t q

+ -- = +2 2

2 2

x ynt xySin Cos

s st q t q

- = - +

2 2

2

( )

x y x yn nt xy

x a y R

s s s ss t t

+ + - + = +

fl fl fl

- + =

2 22 2

2 2 2

2 2

Solid Mechanics

The above equation is clearly an equation of circle w ith center at ( ),0a

on t s- plane it represents a circle w ith

center at x y ,s s+

02

and

hav ing radius

x yxyR

s st

- = +

2

2

This circle on s t- plane- Mohrs circle.

From the above deviation it can be seen that any point P on the Mohrs circle represents stress which are acting on a plane passing through the point.

In this way we can completely v isual ize the stresses acting on al l infinite planes.

Solid Mechanics

(3) Const ruct ion of Mohrs ci rcle

Let us assume that the state of stress at a point is given

A typical problem using Mohrs circle i.e given x y,s s and

x yt on an incl ined element. For the sake of clarity we

assume that, x y, ss s and x yt al l are positive and x ys s>

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 xyA ,s t= + +

Therefore we can locate the point A on the circle w ith

coordinates ( )x xy, ss t+ + Therefore the l ine 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 xyB , ss t= + - Since we assumed that for the sake of simi larity y xss s< .

Therefore the point B diametrical ly opposite to point A.

The l ine BC represents y- axis. The point A corresponds

to Q = 0 , and pt. B corresponds to Q = 90 (+ve) of the stress element.

A t 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 l ine CA in Mohrs circle represents x- axis and l ine CB represents y-axis or plane B. It can be seen that, the angle between x-axis and y-axis in the Mohrs circle is 180 (c.c.w ). Thus 2Q in Mohrs circle corresponds to Q in the stress element diagram.

St resses on an incl ined element

Point A corresponds to 0Q = on the stress element. Therefore the l ine CA i.e x-axis becomes reference l ine from which we measure angles.

Now we locate the point D on the Mohrs circle such that the l ine CD makes an angle of 2Q c.c.w from the x-axis or l ine 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 Mohrs 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

stresselement ,thentheybecome

diametrically oppositepoint son

thecircle, just liketheplanesA& Bdid

s s b

t b

s s b

= +

=

= -

^

Calculat ion of principal st ress

The most important appl ication of the Mohrs circle is determination of principal stresses.

The intersection of the Mohrs circle --- w ith normal stress axis gives two points P1 and P2. Thus P1 and P2 represents

points corresponding to principal stresses. In the current diagram the coordinates the of

P , s

P ,

ss

=

=1 1

2 2

0

0

avg Rs s= +1

avg Rs s= -2

The principal direction corresponding to s 1 is now equal to

pq 12 , in c.c.w direction from the x-axis.

Solid Mechanics

p pp

q q= 2 1 2

We can see that the points P1 and P2 are diametrical ly

opposite, this indicate that principal planes are ^ to each other in the stress element. This fact can also be verified from the Mohrs circle.

In- plane maximum shear st ress

What are points on the circle at which the shearing stress are reaching maximum values numerical ly? Points S1 and S2 at

the top and bottom of the Mohrs circle.

The points S1 and S2 are at angles q =2 90 from

pointsP1 P2 and, i.e the planes of maximum shear stress

are oriented at 45 to the principal planes.

Unl ike 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 ,

t s

t s

= +

= -1

2

max Rt =

avgs s=

Mohrs circle can be plotted in two different ways. Both the methods are mathematical ly correct.

Solid Mechanics

Final ly

Intersection of Mohrs circle w ith the s -axis gives principal stresses.

The top and bottom points of Mohrs circle gives maximum ve shear stress and maximum +ve shear stress.

Do not forget that al l these incl ined planes are obtained by rotation about z-axis.

Solid Mechanics

Mohr ci rcle problem

Solut ion:

A - (15000,4000)

B - (5000,-4000)

(a)

x y MPas s+ +

= =15000 5000

100002 2

R MPa= 6403

x yxyR

s st

- - = + = +

= +

2 22 2

2 2

15000 50004000

2 2

5000 4000

x ys s- = 50002

Solid Mechanics

Point D : x Cos . MPas = + =10000 6403 41 34 14807

x y Sin . MPat = - = -6403 41 34 4229

Point D: n y Cos . MPas s = = - =10000 6403 41 34 593

nt x y Sin .t t = = =6403 41 34 4229

b) P.

; .s q= = =11

38 6616403 19 33

2

MPas =2 3597

c) max SMPa . .t q= - = = -16403 25 67 25 67

Solid Mechanics

(2) q = 45

Principal stresses and principal shear stresses.

Solut ion:

( )

x y

x yxyR MPa

s s

s st

+ - += = -

- - - = + = + - =

2 222

50 1020

2 2

50 1040 50

2 2

( )( )

A ,

B ,

fi - -

fi

50 40

10 40

x y

x y

p R s

p R

s ss

s ss

+= = + = - + =

+= = - = - - = -

1 1

2 2

20 50 302

20 50 702

Solid Mechanics

p

p

p

Q .

Q .

Q .

=

=

=

1

1

2

2 233 13

116 6

206 6

s

s

s

Q .

Q .

Q .

=

=

=

1

1

2

2 143 13

71 6

161 6

Solid Mechanics

Q. x y xyMPa, MPa and MPas s t= = - =31 5 33

Stresses on inclined element q = 45

Principal stresses and maximum shear stress.

Solut ion:

x yavg MPa

s ss

+ -= = =

31 513

2 2

x yxyR . MPa

s st

- = + =

22 37 6

2

( )( )

A ,

B ,- -

31 33

5 33

x avgRCos s

. Cos . MPa

s b s= +

= + =37 6 28 64 13 46

x y RSin . . .t b = - = - = -37 6 28 64 18 02

y avgRCos

MPa

s b s= -

= - 20

Solid Mechanics

. MPas\ =1 50 6

. MPas = -2 24 6

p .q =1 30 68

max s

min

avg

. MPa .

. MPa

MPa

t q

ts s

= - = -

= -

= =

137 6 14 32

37 6

13

Solid Mechanics

7. 3D -Stress Transformation 3D-st ress components on an arbi t rary plane

Basical ly 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 . cos i n es o f x

i n i n j n k

-

= + +

y x y y y z

y x y y y z

n n n

j n i n j n k

= + +

z x z y z z

z x z y z z

n n n

k n i n j n k

= + +

Solid Mechanics

n x x x y x z

n x x x y x z

T T i T j T ks

T i j ks t t

= + +

= + +

x x

x x

x z

ABC dA

PAB dA n

PAC dA n

PBC dA n

-

-

-

-

[ ]xF fi + = 0

x x x x x yx x y zx x zT da dAn dAn dAns t t = + +

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

s t t

t s t

t t s

= + +

= + +

= + +

x x y y z

x y y y z

z x y z z

s t t

t s t

t t s

x x y x z, ,s t t

( ) ( )x n x x x y x z x x x y x z T i T i T j T k . n i n j n ks = = + + + +

(1)

( ) ( )x y n x x x y x z y x y y y z T j T i T j T k . n i n j n kt = = + + + +

(2)

( ) ( )x z n x x x y x z z x z y z z T k T i T j T k . n i n j n kt = = + + + +

(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

s t t

t s t

t t s

= + +

= + +

= + +

( )( )y y x y y y z y x y y y z T i T j T k n i n j n ks = + + + + (4)

( )( )z z x z y z z z x z y z z T i T j T k n i n j n ks = + + + + (5)

Solid Mechanics

( )( )y z y x y y y z z x z y z z T i T j T k n i n j n kt = + + + + (6)

x x

x y

x z

n Cos

n Sin

n

qq

=

=

= 0

y x

y y

y z

n Sin

n Cos

n

q

q

= -

=

= 0

z x

z y

z z

n

n

n

=

=

=

0

0

1

z x z y z

z

: :s t t

s = = =

=

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

s s q s q t q q

s s q s q t q q

t s s q q t q q

= + +

= + -

= - - + -

2 2

2 2

2 2

2

2

x xy

xy y

s t

t s

0

0

0 0 0

Principal st resses

x y zn ,n ,n

( )n x y zn nx ny nz

T n n i n j n k

T T i T j T k

s s= = + +

= + +

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

s t t

t s t

t t s

= + +

= + +

= + +

x x y y z zTn n Tn n Tn ns s s= = =

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.of linear homog.eqns.

n n n

s s t t

t s s t

t t s s

- + + =

+ - + =

+ + - =

0

0

0

x y z x y zn n n : n n n= = = + + =2 2 20 1

( )x xy zx x

xy y zy y

zx yz z z

n

n

n

s s t t

t s s t

t t s s

-

- = -

0

For non triv ial solution must be zero.

( ) ( )( )

x y z x y y z z x xy yz zx

x y z xy yz zx x yz y zx z xy

s s s s s s s s s s s t t t s

s s s t t t s t s t s t

- + + + + + - - -

- + - - - =

3 2 2 2 2

2 2 22 0

This has 3- real roots , ,s s s1 2 3

( )

( )x x yx y zx z

yx x y y zy z

x y z

n n n

n n n

and n n n

s s t t

t s s t

- + + =

+ - + =

+ + =

1

1

2 2 2

0

0

1

x y zn ,n ,n s

s s s

fi

> >1

1 2 3

St ress invariants

I I Is s s- + - =3 21 2 3 0 (1)

Solid Mechanics

x y z

x y y z x z xy yz zx

x y z xy yz zx x yz y zx z xy

I

I stress inv ar iants

I

s s s

s s s s s s t t t

s s s t t t s t s t s t

= + +

= + + - - -

= + - - -

1

2 2 22

2 2 23 2

I Is s - + =3 21 3 0

x y z x y x z y z x y y z x zI Is s s s s s s s t t t = + + = + + - - -2 2 2

1 2

I I ; I I ; I I = = =1 1 2 2 3 3

3D 2D

I

I

I s

s s ss s s s s ss s

= + +

= + +

=3

1 1 2 3

2 1 2 2 3 3 1

3 1 2

I

I

I

s ss s

= +

=

=

1 1 2

2 1 2

3 0

Principal planes are orthogonal

n n T n T .n=

x y z

x y z

n nx ny nz

n n x n y n z

n n i n j n k

n n i n j n k

T T i T j T k

T T i T j T k

= + +

= + +

= + +

= + +

Solid Mechanics

yx

n n

xy

T n T n

tt

=

=

( ) ( )n n T n T n

n n n ns s=

=1 2

( ) ( )x x y y z z x x y y z zn n n n n n n n n n n ns s + + = + +1 2 s s1 2

x x y y z zn n n n n n + + = 0

n .n must be ^ to each other.

The state of st ress in principal axis

ss

s

1

2

3

0 0

0 0

0 0

x

y

z

n x

n y

n z

T n

T n

T n

s

s

s

=

=

=

1

2

3

n x y zn n ns s s s= + +2 2 2

1 2 3

x y zn n n n

x y z

T T T T s

n n ns s s

= + +

= + +

2 2 2 2

2 2 2 2 2 21 2 3

n nTt s= -22 2

Solid Mechanics

8. 3D M ohrs ci rcle and Octahedral stress 3-D Mohrs ci rcle & principal shear st resses

x xy

ij xy y

z

s t

s t s

s

=

0

0

0 0

Once if you know ands s1 2

t

s st

s ss

-=

+=

1

2 31

1 3

2

2

t

s st

s ss

-=

+=

2

1 32

1 2

2

2

t

s st

s ss

-=

-=

3

1 23

1 2

2

2

max max , ,s s s s s s

t- - -

= 1 2 2 3 3 12 2 2

s s s> >1 2 3

Solid Mechanics

The maximum normal stress 1s and maximum shear stress maxt and their corresponding planes govern the fai lure of the engineering materials.

It is ev ident now that in many two-dimensional cases the maximum shear stress value w i l l be missed by not considering s =3 0 and constructing the principal circle.

Solid Mechanics

Problem:

The state of stress at a point is given by

x y zMPa, MPa, MPa ands s s= = - =100 40 80

xy yz zxt t t= = = 0

Determine in plane max shear stresses and maximum shear stress at that point.

Solution:

MPa, MPa MPass s s= = = -1 2 3100 80 40

MPas s

t- -

= = =1 212100 80

102 2

MPas s

t- +

= = =1 313100 40

702 2

MPas s

t- +

= = =2 32380 40

602 2

MPa

MPa

s ss

ss

+= =

=

=

1 212

13

23

902

30

20

max max , ,t t t t= 12 13 23

max MPat = 70 This occurs in the plane of 1-3

Solid Mechanics

, ,t t t fi1 2 3 Principal shear stress in 3D

( )max max , ,t t t t= 1 2 3

Solid Mechanics

Plane st ress

z

s ss s

>

= =1

3 0

x yxy

s st t

- = +

22

2 ---- in plane principal shear stresses.

maxs s s

t-

= =1 3 12 2

Solid Mechanics

Problem

A t appoint in a component, the state of stress is as shown. Determine maximum shear stress.

Solution:

ijs =

100 0

0 50 - plane stress problem

We can also w rite the matrix as ija =

100 0 0

0 50 0

0 0 0

ss

s s

=

=

- -= =

1

2

1 2

100

50

100 5025

2 2

max MPat = 25

Solid Mechanics

Now w ith , ,s s s= = =1 2 3100 50 0

max MPas s

t-

= =1 3 502

Occurs in the plane 1-3 instead of 1-2

Solid Mechanics

Some important states of st resses

(1) Uniaxial state of st ress: Only one non-zero principal stress.

ss

=

11

0 00

0 0 00 0

0 0 0

- plane stress.

(2) Biaxial state of st ress: two non-zero principal stresses.

ss

ss

=

11

11

0 00

0 00

0 0 0

- plane stress

(3) Triaxial state of st ress: A l l three principal stresses are non zero.

ss

s

-

1

2

3

0 0

0 0

0 0

3D stress

(4) Spherical state of st ress: s s s= =1 2 3 (either +ve or ve)

D

ss

s

-

0 0

0 0 3

0 0

stress-special case of triaxial stress.

Solid Mechanics

(5) Hydrostat ic state of st ress

P

P

P

+ +

+

0 0

0 0

0 0

hydrostatic tension

P

P

P

- -

-

0 0

0 0

0 0

hydrostatic compression.

(6) The state of pure shear

zy

x xy xz

ij xy y yz

zx z

s t t

s t s t

t t s

=

x y x z

ij x y y z

z x z y

t t

s t t

t t

=

0

0

0

Then we say that the point P is in state of pure shear.

I =1 0 is necessary and sufficient condi t ion for state of pure shear

Solid Mechanics

Octahedral planes and st resses

If x y zn n n= = w .r.t to the principal planes, then these planes

are known as octahedral planes. The corresponding stresses are known as octahedral stresses.

Eight number of such planes can be identified at a given point --- Octahedron

x y z

n x y z

n n n

T n n n

s s s s

s s s

= + +

= + +

2 2 21 2 3

2 2 2 2 2 2 21 2 3

x y z

x y z

n n n

n n n .

+ + =

= = = =

2 2 2

0

1

154 73

3

octs s s s

s s s

= + + + +

=

2 2 2

1 1 1

1 2 3

1 1 13 3 3

3

Solid Mechanics

1I = meanstress3

s s s+ +=1 2 3

3

oct canbeint erpreted meannormal stressat apt.s = - -

oct n octTt s= -2 2

( ) ( ) ( )octt s s s s s s= - + - + -2 22

1 2 2 3 3 113

Therefore, the state of stress at a point can be represented w ith reference to

(i) stress components of x,y,z coordinate system

(ii) stress components of x,yz coordinate system

(iii) using principal stresses

(iv) using octahedral shear and normal stresses

We can prove that:

octt is smaller than maxt (exist only on 4 planes) but can exist on 8 planes at a point.

Solid Mechanics

Decomposi t ion into hydrostat ic and pure shear st ress

x xy xz

ij yx z yz

zx zy z

s t t

s t s t

t t s

=

Mean stress x y zI

Ps s s+ +

= = 13 3

x xy xz x xy xz

yx y yz yx y yz

zx zy z zx zy z

PP

P P

P P

Hydrostatic Stateof pureshear

stat of stress Deviatoricstateof stress

Dilitational stress Stressdeviator

s t t s t t

t t t t s t

t t s t t s

- = + - -

0 0

0 0

0 0

Thus the state of the st ress at a point can alos be represented by sum of di lat ional st ress and st ress deviator

Solid Mechanics

IP

s s s+ += =1 2 3 1

3 3

P P

P P

P P

s ss s

s s

- = + -

-

1 1

2 2

3 3

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

s =1 mean stress + dev iation from the mean

The deviatoric and octahedral shear st resses are the answer for the yielding behavior of materials which is a type of fai lure of materials.

Solid Mechanics

9. Deformation and strain analysis

Two types of deformation have been observed for an infinitesimal element.

Deformation of the whole body = Sum of deformations of

Deformat ion is described by measuring two quant i t ies.

(1)Elongat ion or cont ract ion of a l ine segment

(2)Rotat ion of any two ^ l ines.

Measure of deformat ions of an infini tesimal element is known as st rain.

The strain component that measures elongation or construction normal strain -e

The strain component that measures rotation of any two ^ l ines is shearing strain- g

( )( )( )

u u x ,y ,z

v v x ,y ,z (x ,y ,z) is the point in the undeformed geometry

w w x ,y ,z

=

= =

( ) ( ) ( )= + + u u x ,y ,z i v x ,y ,z j w x ,y ,z k

Solid Mechanics

Normal st rain e - Account for changes in length between two points.

( )* * *

ns s

P Q PQ s sP lim lim

PQ sD fi D fi- D - D

= =D0 0

We can also define the same point x y z, ,

(1) By definition x is + if *s sD > D

x is - if *s sD > D

(2) It is immaterial how * *P Q is oriented final ly. However for

n we must consider PQ in the direction of n in the

undeformed geometry

(3) In general ( )n n x ,y ,z s =

(4) No units.

(5) Meaning of nn

Shearing st rain -

Accounts the change in angle

( )nY P+ Change in angle between

^ l ines in n& t direction.

( )nt ntx xy y

Y P lim limp

f a bD fi D fiD fi D fi

- = +0 00 0

2

Mm/ mm,0.5%=0.005;

,m m-= 610 1000

. mm / mm-= =61000 10 0 001

( )

( )

*n

*n

n n

s s

s s if s

s s s s

D = + D

D + D D fi

D = D

1

1 0

l im as sD fi 0

Solid Mechanics

(1)We must select two ^ l ines in the undeformed geometry.

(2)Units of ntY fi radius.

(3)By deflection nt tnY Y=

(4)Two subscripts are required for

Y - to show directions of initial

infinitesimal line segments.

(5) ntY is +ve if angle is decreased

ntY is -ve if angle is more.

By taking two ^ l ines

We can define n t nt, & Y

Rectangular st rain components

x y xy

z y yz

x z xz

, andY PQRS

, andY QABS

, andY RSCD

-

-

-

x xy xz

ij xy y yz

xz yz z

Y Y

E Y Y

Y Y

=

They represent the state of strain at a point , since we can determine strain along any direction n

- Rectangular strain components . - We then say that we have strain

computer associated w ith x ,y ,z coordinate system.

Solid Mechanics

St rain displacement relat ions: Strains are due to deformation as displacement so there must be some relation between deformational displacements and strains. So let us consider the side of the element PQRS. We shal l demonstrate that w has no impact. So it can be neglected.

P u,v

u vQ u x ; v x

x x

fi

fi + D + D

* * *

PQ x

P Q x

= D

= D

( )* xx xD + D1

( )* xxlim x x

D fiD = + D

01

* u v wx x x xx x x

u u v wx

x x x x

D = + D + D + D

= + + + + D

2 2 2

2 2 2

1

1 2

Solid Mechanics

*

xx

x

x

y

z

x xlim

x

u u v wlim

x x x x

u u v wx x x x

v u v wy y y y

w u vz z z

D fi

D fi

D - D =

D

= + + + + -

= + + + + -

= + + + + -

= + + +

0

2 2 2

0

2 2 2

2 2 2

2

1 2 1

1 2 1

1 2 1

1 2wz

+ -

2 21

So far no assumption has been made except for size of x , y& zD D D

*xy * *

yu x uCos

x yx yf

D D = + D D 1

* *yv x v

x yx y

D D + + D D 1

* *yw x w

x yx y

D D + D D

*xy xy

xyz

Y limp

fD fiD fiD fi

= -000

2

Solid Mechanics

*xy xy

xyz

SinY lim CosfD fiD fiD fi

=000

( )

( )

xy * *xyz

*x

*y

x yu u v v w wSinY lim

x y y x x y x y

x x

y y

D fiD fiD fi

D D = + + + + D D

D = + D

D = + D

000

1 1

1

1

( )( )xy x x yyz

u v u u v v w wy x x y x y x y

SinY limD fiD fiD fi

+ + + + =

+ +000

1 1

( )( )xy x y

u v u u v v w wSin

y x x y x y x yY

- + + + + =+ +

1

1 1

( )( )yz x y

u v u u v v w wy x x y x y x y

Y sin-

+ + + + =

+ +

1

1 1

( )( )xz x z

w u w w u u v vx w x z x w x zY sin-

+ + + + = + +

1

1 1

A l l bodies after the appl ication of loads under go smal l deformations

Solid Mechanics

Smal l deformat ions :

(1) The deformational displacements u ui vj wk= + +

are

infini tesimal ly smal l .

(2) The st rains are smal l

(a) Changes in length of a infinitesimal line segment are infinitesimal.

(b) Rotations of line segment are also infinitesimal.

x y zu u u v u u v

, , , ; ; ; ;x u w x x x y

<

21 1 1 1 are

negl igible compare to u v

,x x

quantities.

xux

ux

= + -

-= +

1 2 1

21

12

x

y

z

uxvy

wz

=

=

=

xy xySinY Y

Solid Mechanics

( )xy x y

u vy x v u

Yx y

+

= = +

+ +1

xz

yz

w uY

x zv w

Yz y

= +

= +

Another derivat ion : Let us take plane PQRS in xy plane.

A lso assume that ( ) ( )u u x ,y & v v x ,y= = only.

Smal l deformat ion

Displacements are small

Strains are small

* * *

xx

P Q PQ x xlim

PQ xD fi- D - D

= =D0

Strains

Solid Mechanics

*xy xy

x xy y

Y lim limp

f a bD fi D fiD fi D fi

= - = +0 00 0

2

v vx

x xtan yyx

xx

a

D

= = ++ D 11

tana a

vxuy

a

b

=

=

xyu v

Yy x

= +

u u v v, , ,

x y y x

u u v, ,

x y yx

2 22

1

We can define the state of strain at point by six components of strains

State of st rain

- Engineering strain matrix - We can find n in any

direction we can find ntY for any two arbitrary directions.

x y , z, xy xz yz

yx zx zy

, Y , Y , Y

Y Y Y

fl fl fl

x xy xz

ij xy y yz

xz yz z

Y Y

E Y Y

Y Y

=

Solid Mechanics

2D- st rain t ransformat ion

Plain st rain: In which

x xy

xy y

Y

Y

( )( )

( )

x x

y y

xy xy

x ,y

x ,y

Y Y x ,y

=

=

=

z

yz

zx

Y

Y

=

=

=

0

0

0

impl ication of these equation is that a point in a given plane does not leave that plane al l deformations are in to plane of the body.

Solid Mechanics

Given x y xy, & Y what are n t nt, & Y .

We can always draw PQRS for given n

If x y xy, & Y

As in case of stress we call these formulae as transformations laws.

x

x

x

dxSinds

dxsin

dssin cos

qa

q

q q

=

=

=

1

y ydy

cos cos sinds

a q q q= =2

xy

xy

dyY sin

dsY sin sin

a q

q q

=

=

3

Solid Mechanics

x y xy

n x y xy

x y xy

dL dxcos dysin Y dycos

dy dydL dxcos sin Y cos

dS ds ds ds

cos cos sin Y sin cos

q q q

q q q

q q q q q

= + +

= = + +

= + +2

- state of strain at a point

- stress tensor

- strain tensor

Replace

x x

y y

xyxy xy

Y

ss

t

fi

fi

fi =2

( ) ( )

x y xy

x y xy

x y xy

sin cos sin cos Y sin

cos sin cos sin Y cos

cos sin cos sin Y cos

a q q q q q

b q q q q q

q q q q q

= - + -

= - - + - -

= - -

2

2

2

( )x y xynt YY sin cosq q - = - +2 22 2 2

x xy

xy y

Y

Y

xyx

xyy

Y

Y

2

2

x xy

xy y

xyxy

Y =

2

x xy

xy y

s t

t s

x y x y xyn

Ycos sinq q

+ - = + +2 2

2 2 2

Solid Mechanics

Principal shears and maximum shear In plane- principal strains

xy xyp

x y

/tan Q

fi=

-

2 22

p pq q- - ^1 2 to each other

, >1 2 1 2

( )x ys

xy

s p

tan

/

q

q q p

- = -

= 1

22

4

s sq q- - ^1 2 to each other

x y

x y I

x y xy

y xy

xyx y

I

J

I

J

YJ

s s

s s t

+ =

+ =

- =

- =

- =

1

22

22

2

22

x ymax min xy

maxmax s

minmin s

or R

Y

Y

q

q

- = = +

= -

= -

1

2

22

2

2

2

Solid Mechanics

Mohrs Ci rcle for st rain

3D-strain transformation

xyx x y y z z xy xy

Y; ; ;s s s tfi fi fi = =

2

( )

( )( )

x xy xz

xy y yz

xz yz z

-

- =

-

0

, , 1 2 3 - > >1 2 3

* * * * *

s x y

s P Q P R

u vx y x y

x x

D = D + D

D = +

= + D + + D - D + D

2 2 2

2 2 2

2 22 21 1

x x y y,Y ,

Solid Mechanics

ny

. xx

yu vx x y

x x x

D = + D D

D = + + + D - D - D D

2

222 2 2

1

1 1

u u v vx y x y

x x y y

yx

x

u vx y x y

x y

yx

x

+ + D + + + D - D - D =D + D D

+ D + + D - D + D =D + D D

222 2 2

2

2

2 2 2 2

2

1 2 1 2

1

1 2 1 2

1

Transformation

x x x x y y y z z z xy x x x y

yz x y x z zx x z x x

n n n n n

n n n n

s s s s t

t t

= + + +

+ +

2 2 2

x x x x y x y z x z xy x x x y

yz x y x z zx x z x x

n n n n n

n n n n

= + + +

+ +

2 2 2

x yx y x y

Yt

fi fi 2

xy xy

yz yz

zx zx

t

t

t

fi

fi

fi

x x

y y

z zx

ss

s

fi

fi

fi

Solid Mechanics

Principal st rains:

( )

( )( )

x x xy y xz z

xy x y y yz z

xz x yz y z z

n n n

n n n

n n n

- + + =

+ - + =

+ + - =

0

0

0

( )

( )( )

x xy xz

xy y yz

xz yz z

-

- =

-

0

J J J - + - =3 21 1 2 3 0

x y zJ = + +1

x xyx y x z y z xy yz zx

xy y

y yz x xz

yz z xz z

J

= + + - - - +

+

2 2 22

x y z xy yz zx x yz y xz

x xy xz

z xy yx y yz

zx zy z

J = + - -

-

2 23

2

> >1 2 3

System of l inear homogeneous equations

Solid Mechanics

( )

( )x x xy y zx z

xy x y y zy z

x y z

n n n

n n n

n n n

- + + =

+ - + =

+ + =

1

1

2 2 2

0

0

1

x y zn ,n & n unique

Decomposi t ion of a st rain mat rix into state of pure shear + hydrostat ic st rain

x xy xz x xy xz

ij yx y yz yx y yz

zx zy z zx zy z

Stateof pureshear Hydrostatic

- = = - + -

0 0

0 0

0 0

where x y z + +

=3

J

J

J

= + +

= + +

=

1 1 2 3

2 1 2 2 3 3 1

3 1 2 3

Solid Mechanics

Plane strain as a special case of 3D

=3 0 is also a principal strain

z fi is a principal direction

if ; > =1 2 1 2 +ve

if 1 +ve, 2 -ve.

if +ve, -ve 1 2

P & z 1 w i l l come closer

to the maximum extent,

so that the included angle

is maxp

- 2

Solid Mechanics

Transformat ion equat ions for plane-st rain

Given state of strain at a point P.

xx xy

ijxy yy

YE

Y

=

This also means that

Now what are the strains associated w ith x ,y i.e

x x x y

i jx y y y

YE

Y

=

This also means that

deformation

Solid Mechanics

Assume that xx yy, and x yY are +ve

Applying the law of cosines to triangular P* Q* R*

( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( )

( )

xy

x x y x

y xy

P* R* P* R* Q* R* P* R* Q* R*

cos Y

x x y x

y cos Y

p

p

= + -

+

D + = D + + D + - D +

D + +

2 2 2

22 2

2

2

1 1 1 2 1

12

x x cosqD = D and y x sinqD = D

( )xy xy xycos Y sinY Yp + = - -2

( ) ( ) ( )( )( )( )

x x y

x y xy

x x cos x sin

x sin cos Y

q q

q q

D + = D + + D +

- D + + -

22 22 2 2 2 2

2

1 1 1

2 1 1

Solid Mechanics

( ) ( ) ( )( )( )( )

( ) ( )( )

x x y

x y xy

x x x x y y

xy x y x y

cos sin

sin cos Y

cos sin

sin Y

q q

q q

q q

q

+ = + + +

- + + -

+ + = + + + + +

+ + + +

22 22 2

2 2 2 2 2

1 1 1

2 1 1

1 2 1 2 1 2

2 1

( ) ( )( )

( ) ( )

x x y

xy x y

x y

xy

cos sin

Y sin

cos sin

Y sin

q q

q

q q

q

+ = + + +

+ + +

= + + +

+

2 2

2 2

1 2 1 2 1 2

2 1

1 2 1 2

2

x x y xy

xyx x y

cos sin Y sin

Ycos sin sin

q q q

q q q

+ = + + +

= + +

2 2

2 2

1 2 1 2 2 2

22

x y x y xyx

Ycos sinq q

+ - = + +2 2

2 2 2

If yQp

q = + 2

x y x y xyx

Ycos sinq q

+ - = + +2 2

2 2 2

x y x y xyy

Ycos sinq q

+ - = + -2 2

2 2 2

x y x y + = + J= =1 first invariant of strain.

Solid Mechanics

( )

x y xyx OBQ

OB x y xy

xy OB x y

Y

Y

Y

p =

+ = = +

= + +

= - +

4 2 2

2

2

( )( )

OB x y x y

x y OB x y

OB x y

Y

Y

( )

= + +

= - +

= - +

2

2

2 3

x y x y xyx OBQ Q

Ysin cosp q q = +

+ - = = - +

42 2

2 2 2- (4)

Substituting (4) in (3)

( ) ( ) ( )x y x y x y xy x yY sin Y cosq q = + - - + - +2 2

( )x y x y xyY sin Y cosq q = - - +2 2 (5) tensorial normal strain xx =engineering normal strain

xx yy z, ,=

tensorial shear strain ( ) xyxyYEngineeringshear strain

= 2 2

Solid Mechanics

( )

xzxx xy xz

ij xy yy yz

zx zy zz zz

Y =

= =

2

( )

x y x yx xy

x y x yy xy

x yx y xy

cos sin

cos sin

sin cos

q q

q q

q q

+ - = + +

+ - = - -

- = - +

2 22 2

2 22 2

2 22

Components.

- Strain tensors

Solid Mechanics

Problem:

An element of material in plane strain undergoes the fol low ing strains

x y xyY- - - = = = 6 6 6340 10 110 10 180 10

Show them on sketches of properly oriented elements.

Solut ion:

x-

= - 6340 10 ; y

- =

6110 10 ; x yY-

= 6180 10

x- = 6340 10

Solid Mechanics

Problem:

During a test of an airplane w ing, the strain gage readings

from a 45 rosette are as fol lows gage A , - 6520 10 ; gage B - 6360 10 and gage C -- 680 10

Determine the principal strains and maximum shear strains and show them on sketches of properly oriented elements.

Solut ion:

(1)

x

OB

y

-

-

-

=

=

= -

6

6

6

520 10

360 10

80 10

( )( )

xy OB x yY

rad

- - -

-

= - +

= - -

=

6 6 6

6

2

2 360 10 520 10 80 10

280 10

x y- -

- + - = = 6 6

6520 10 80 10 220 102 2

Solid Mechanics

x y- -

- - + = = 6 6

6520 10 80 10 300 102 2

xyp

x y

xyxy

etan

Y

q-

-

--

= =

-

= = =

6

6

66

2 140 102

300 10

280 10140 10

2 2

p

p p

.

. .

q

q q

\ =

= =

2 25 02

12 51 102 51

( ) ( )

x y x yxyor

.

- - -

- -

+ - = +

= +

=

22

1 2

2 26 6 6

6 6

2 2

220 10 300 10 140 10

220 10 331 06 10

.

.

-

-

\ =

= -

61

62

551 06 10

111 06 10

( ) ( )

x .

x y x yxyCos Sin

cos . Sin .

.

q

q q

=

- - -

-

+ - = + +

= + +

=

12 51

6 6 6

6

2 22 2

220 10 300 10 2 12 51 140 10 2 12 51

551 06 10

Solid Mechanics

p .q =1 12 51 and p .q =2 102 51

(b) In- plane maximum shear strains are

x yxyxymax xyminor

. -

- = +

=

22

6

2

331 06 10

( )( )

xy max

xy min

.

.

-

-

=

= -

6

6

331 06 10

331 06 10

( )x ys

xytan Q

.

-

-

- - = - =

6

6300 10

22 140 10

s

s s

Q .

Q . Q .

=

= - =

2 64 98

32 5 57 5

( )( ) ( )

x yx y xyQ .

Sin . Cos .

. . .

=

- - -

- = - +

= - - =

57 5

6 6 6

2 57 5 2 57 52

271 89 10 59 17 10 331 06 10

Solid Mechanics

and

s .q = -1 32 5

s .q = -2 32 5

x y - + = = 6220 102

min

max

Y .

Y .

-

-

= -

=

6

6

662 11 10

662 11 10

Solid Mechanics

10. Stress strain diagrams

Bar or rod the longitudinal direction is considerably greater than the other two, namely the dimensions of cross section.

For the design of the m/ c components we need to understand about mechanical behav ior of the materials.

We need to conduct experiments in laboratory to observe the mechanical behav ior.

The mathematical equations that describe the mechanical behav ior is known as constitutive equations or laws

Many tests to observe the mechanical behav ior- tensile test is the most important and fundamental test- as we gain or get lot of information regarding mechanical behav ior of metals

Tensi le test Tensi le test machine, tensile test specimen, extensometer, gage length, static test-slow ly varying loads, compression test.

St ress -st rain diagrams

A fter performing a tension or compression tests and determining the stress and strain at various magnitudes of load, we can plot a diagram of stress Vs strain.

Solid Mechanics

Such is a characteristic of the particular material being tested and conveys important information regarding mechanical behav ior of that metal.

We develop some ideas and basic definitions using s - curve of the mild steel.

Structural steel = mi ld steel = 0.2% carbon=low carbon steel

Region O-A

(1) s and l inearly proportional.

(2) A- Proportional l imit

ps - proportional ity is maintained.

(3) Slope of AO = modulus of elasticity E N / m2,Pa

(4) Strains are infinites ional.

f o

o

L L

L

- =

Solid Mechanics

Region A-B

(1) Strain increases more rapidly than s

(2) Elastic in this range

Proportional ity is lost.

Region B-C

(1) The slope at point B is horizontal.

(2) A t this point B, increases w ithout increase in further load. I.e no noticeable change in load.

(3) This phenomenon is known as yielding

(4) The point B is said to be yield points, the corresponding stress is yield stress yss of the steel.

(5) In region B-C material becomes perfectly plastic i.e which means that it deforms w ithout an increase in the appl ied load.

(6) Elongation of steel specimen or in the region BC is typical ly 10 to 20 times the elongation that occurs in region OA.

(7) s below the point A are said to be smal l, and s above A

are said to be large.

(8) s A are said to

be plastic strains = large strains = deformations are permanent.

Solid Mechanics

Region C-D

(1)The steel begins to strain harden at C . During strain hardening the material under goes changes in its crystal l ine structure, resulting in increased resistance to the deformation.

(2)Elongation of specimen in this region requires additional load,

\ s - diagram has + ve slope C to D.

(3) The load reaches maximum value ultimate stress.

(4)The yield stress and ultimate stress of any material is also known as yield strength and the ultimate strength .

(5) us is the highest stress the component can take up.

Region-DE

Further stretching of the bar is needed less force than ultimate force, and final ly the component breaks into two parts at E.

Solid Mechanics

Look of actual st ress st rain diagrams

CtoE BtoC O toA > >

(1) Strains from O to A are

so smal l in comparison to the

strains from A to E that they

cannot be seen.

(2) The presence of wel l defined

yield point and subsequent large

plastic strains are characteristics of mild steel.

(3) Metals such a structural steel that undergo large permanent strains before fai lure are classified as ductile metals.

Ex. Steel, aluminum, copper, nickel, brass, bronze, magnesium, lead etc.

Aluminum al loys Offset method

(1) They do not have clear cut yield point.

(2) They have initial straight l ine portion w ith clear proportional l imit.

(3) A ll does not have obvious

yield point, but undergoes

large permanent strains after

proportional l imit.

(4) A rbitrary yield stress is

Solid Mechanics

determined by off- set method.

(5) Off-set yield stress is not material property

Elast ici ty & Plast ici ty

(1) The property of a material by which it (doesnt) returns to its original dimensions during unloading is cal led (plasticity) elasticity and the material is said to be elastic (plastic).

(2) For most of the metals proport ional l imi t = elast ic l imi t .

(3) For pract ical purpose proport ional l imi t = elast ic l imi t = yield st ress

(4 )Al l metals have some amount of st raight l ine port ion.

Solid Mechanics

Bri t t le material in tension

(1) Materials that fai l in tension at relatively low values of strain are classified or brittle materials.

(2) Brittle materials fai l w ith only l ittle elongation (elastic) after the proportional l imit.

(3)Fracture stress = Ultimate stress for brittle materials

(4)Up to B, i.e fracture strains are elastic.

(5)No plastic deformation in case of brittle materials.

Ex. Concrete, stone, cast iron, glass, ceramics

Duct i le metals under compression

Solid Mechanics

(1) s - curves in compression differ from s - in tension.

(2)For ducti le materials, the proportional l imit and the initial portion of the s - curve is same in tension and compression.

(3)A fter yielding starts the behav ior is different for tension and compression.

(4)In tension after yielding specimen elongates necking and fractures or rupture. In compression specimen bulges out- w ith increasing load the specimen is flattened out and offers greatly increased resistance.

Bri t t le materials in compression

(1)Curves are simi lar both in tension and compression

(2)The proportional limit and ultimate stress i.e fracture stress are different.

(3)In case of compression both are greater than tension case

(4)Brittle material need not have l inear portion always they can be non-l inear also.

Solid Mechanics

11. General ized Hookes Law

(1) A material behaves elastical ly and also exhibits a l inear relationship between s and is said to be l inearly elastic.

(2) A ll most all engineering materials are l inearly elastic up to their corresponding proportional limit.

(3) This type of behav ior is extremely important in engineering al l structures designed to operate w ithin this region.

(4) Within this region, we know that either in tension or compression

E

Stressin particular direction strain inthat dir .X E

s = =

E =Modulus of elasticity Pa,N / m2

= Youngs modulus of elasticity.

(5) x xEs = or y yEs =

(6) Es = is known as Hookes law .

(7) Hookes law is val id up to the proportional l imit or w ithin the linear elastic zone.

Solid Mechanics

Poissons rat io

When a prismatic bar is loaded in tension the axial elongation is accompanied by lateral contraction.

Lateral contraction or lateral strain

f o

o

d d

d

- = this comes out to be ve

( ) lateral strainPoisson's ratio = nuaxial strain

isperpendicular to

n-

=- =

If a bar is under tension +ve, -ve and n = +

If a bar is under compression -ve, +ve and n = +

n =always +ve = material constant

For most metals . to . sn = 0 25 0 35

Concrete . to .n = 0 1 0 2

Rubber .n = 0 5

n is same for tension and compression

n is constant w ithin the l inearly elastic range.

Solid Mechanics

Hooks law in shear

(1)To plot ,Yt the test is tw isting of hol low circular tubes

(2) ,Yt diagrams are (shape of them) simi lar in shape to tension test diagrams ( )Vss for the same material, although they differ in magnitude.

(3)From Yt - diagrams also we can obtain material properties proportional l imit, modulus of elasticity, yield stress and ultimate stress.

(4)Properties are usual ly of the tension properties.

(5)For many materials, the initial part o the shear stress diagram is a st. l ine through the origin just in case of tension.

GYt = - Hookes law in shear

G =Shear modulus of elasticity or modulus of rigidity.

Pa or N / m s= 2

Proportional limit

Elastic l imit

Yield stress

Ultimate stress

Material properties

t

Proportional limit

G1

Yield point

Solid Mechanics

E,v, and G fi material properties elastic constants - elastic

properties.

Basic assumpt ions sol id mechanics

Fundamental assumptions of linear theory of elasticity are:

(a) The deformable body is a continuum

(b) The body is homogeneous

(c) The body is linearly elastic

(d) The body is isotropic

(e) The body undergoes small deformations.

Cont inuum

Completely fi ll ing up the region of space w ith matter it occupies w ith no empty space.

Because of this assumption quantities l