MECHANICS OF SOLIDS

Mechanics of SolidsMechanics of Solids

SyllabusSyllabus:- Part - A

1. Simple Stresses & Strains:-1. Simple Stresses & Strains:-

Introduction, Stress, Strain,

Tensile, Compressive & Shear Stresses,

Elastic Limit, Hooke’s Law, Poisson’s Ratio,

Modulus of Elasticity, Modulus of Rigidity,

Bulk Modulus, Bars of Varying Sections,

Extension of Tapering Rods, Hoop Stress,

Stresses on Oblique Sections.

2. Principle Stresses & Strains:-2. Principle Stresses & Strains:-

State of Simple Shear,

Relation between Elastic Constants, Compound Stresses, Principle Planes

Principle Stresses,

Mohr’s Circle of Stress, Principle Strains,

Angle of Obliquity of Resultant Stresses, Principle Stresses in beams.

3. Torsion:-3. Torsion:-

Torsion of Circular, Solid, Hollow Section Shafts

Shear Stress, Angle of Twist,

Torsional Moment of Resistance,

Power Transmitted by a Shaft,

Keys & Couplings,

Combined Bending & Torsion,

Close Coiled Helical Springs,

Principle Stresses in Shafts Subjected to Bending, Torsion & Axial Force.

Mechanics of SolidsMechanics of Solids

SyllabusSyllabus:- Part - BPart - B

1. Bending Moment & Shear Force:-1. Bending Moment & Shear Force:-Bending Moment,

Shear Force in Statically Determinate Beams Subjected to Uniformly Distributed, Concentrated & Varying Loads,

Relation Between Bending Moment,

Shear force & Rate of Loading.

2. Moment of Inertia:-2. Moment of Inertia:-

Concept Of Moment of Inertia,

Moment of Inertia of Plane Areas,

Polar Moment of Inertia,

Radius of Gyration of an Area,

Parallel Axis Theorem,

Moment of Inertia of Composite Areas,

Product of Inertia,

Principle Axes & Principle Moment of Inertia.

3. Stresses in Beams:-3. Stresses in Beams:-

Theory of Simple Bending, Bending Stresses,

Moment of Resistance,

Modulus of Section,

Built up & Composite Beam Section,

Beams of Uniform Strength.

4. Shear stresses in Beams:-4. Shear stresses in Beams:-

Distribution of Shear Stresses in Different Sections.

5. Mechanical Properties of Materials:-5. Mechanical Properties of Materials:-

Ductility, Brittleness, Toughness, Malleability, Behaviour of Ferrous & Non-Ferrous metals in Tension & Compression, Shear & Bending tests, Standard Test Pieces, Influence of Various Parameters on Test Results, True & Nominal Stress, Modes of Failure, Characteristic Stress-Strain Curves, Izod, Charpy & Tension Impact Tests,

Fatigue, Creep, Corelation between Different Mechanical Properties, Effect of Temperature, Testing Machines & Special Features, Different Types of Extensometers & Compressemeters, Measurement of Strain by Electrical Resistance Strain Gauges.

Many structural elements like bars, tubes, beams, columns, trusses, cylinders, spheres, shafts are used for the benefit of the mankind. They may be made up of timber, steel, copper, aluminium, concrete or any other materials.

The application of the laws of mechanicsthe laws of mechanics to find the support reactions due to the applied forces is normally covered under the subject of ENGINEERING MECHANICS. ENGINEERING MECHANICS.

In transferring, these forces from their point of application to supports the material of the structure develops the resistive forces and it undergoes deformation. The effect of these resisting forces, on the structural elements, is treated under the subject -- -- STRENGTH OF MATERIAL “OR” MECHANICS STRENGTH OF MATERIAL “OR” MECHANICS OF SOLIDS .OF SOLIDS .

The Strength Of Materials is an interdisciplinary subject.

Architects and civil engineers like to see that the trusses, slabs, beams, columns, etc. of the buildings and bridges are safe.

Aeronautical engineers need this subject for the design of the component of the aircraft.

Mechanical engineers and the Chemical engineers must know this subject for the design of the machine components and the pressure vessels.

Mining engineers need it to design safe mines.

Metallurgist must understand this subject well so that he can think for further improvement of the mechanical properties of the materials.

Electrical, Electronics and Computer Engineers need the basic knowledge of this subject because of several mechanical components they need in their products.

CONCEPT OF INTERNAL FORCES:-CONCEPT OF INTERNAL FORCES:-

When a member is subjected to load, it develops resisting forces; i.e. it is the force of resistance offered by the material from which the member is manufactured.

To find the resisting force developed by a member, we will use the method of section. In this method a section plane may be passed through the member and equilibrium of any part of the member can be studied.

Force/Moment can be applied in the following ways:-

• Axial ( Push / Pull )

• Flexural ( Bending)

• Torsion (Twisting )

• Shear ( Slicing )

Axis of the memberAxial Force

Axial Force:- As it’s name suggests, it is the force which is acting along the axis of the member. In other words, it’s line of action is passing through to the axis of the member.

Push / comp.

Pull / Tens.

Axis of the member

Flexural Force:- It is the force whose line of action is perpendicular to the axis of the member.

Flexural Forces

Shear Force:- Any force which tries to shear-off the member, is termed as shear force.

Torsion:- Any moment which tries to twist the member, is termed as Torsion.

Fixed end of the member

Axis of the member

Torsion.

In this subject we will derive the relationship between

FORCE, STRESS, STRAIN & DEFORMATION

To design any structure, our first aim is to find out the type, nature and magnitude of forces acting on it. Accordingly we will design the structure.

Our next aim is to ensure that the structure designed by us remain safe and serviceable.

To ensure safety, the stresses developed in the member must remain within the permissible limits specified by the standards.

To ensure Serviceability, the deformations developed in the member must remain within the permissible limits specified by the standards.

There are mainly three types of supports:

1) Simple SupportSimple Support: It restrains movement of the beam in only one direction, i.e. movement perpendicular to the base of the support. It is also known as Roller support.

Reaction

2) Hinged support: It restrains movement of the beam in two directions i.e. movement perpendicular to the base of the support and movement parallel to the base of the support.

Reactions

3) Fixed support: It restrains all the three possible

movements of the beam. i.e. movement perpendicular

to the base of the support and movement parallel to

the base of the support and the rotation at the

support.

Reactions:

There are mainly five types of beams:

Cantilever beamCantilever beam:: It is a beam which has one

end, as fixed, and the other end as free.

L

fixed endfree end

Simply- supported beam: Simply- supported beam:

It is a beam, which has it’s ends, supported freely on walls or the columns. {Out of it’s two simple supports, one support will be hinged support and the other support will be roller support, then only the beam will be determinate}

L

Over-hang beam:Over-hang beam: When the beam is continued beyond the support and behave as a cantilever then the combined beam is known as an over-hang beam.

L L1

L L1L2

4) Fixed Beam: 4) Fixed Beam:

A beam whose both the ends are fixed or built-in in the walls or in the columns, then that beam is known as the fixed beam.

L

5) Continuous Beam:5) Continuous Beam:

A beam which is supported on more than two supports that, it is called a continuous beam.

L2L3L1

POINT LOAD:-POINT LOAD:- If a comparatively large load acts on a very small area, then that load is called a point load. It is expressed in N or kN.

L

point loadW kN

UNIFORMLY DISTRIBUTED LOAD:-UNIFORMLY DISTRIBUTED LOAD:- When the load is uniformly distributed over some length, then that load is called a uniformly distributed load. It is expressed in N/m or kN/m.

=

L

ww kN/m

Total Load = w kN/m *L m = w*L kN

UNIFORMLY VARYING LOAD:-UNIFORMLY VARYING LOAD:- When the load Intensity is varying uniformly over some length, then that load is called a uniformly varying load. In this case total load will be the area covered by the triangle.

L

Total Load = ½ *w * L = (w*L)/2 kN

ww kN/m

CONCENTRATED MOMENT ( moment acting at CONCENTRATED MOMENT ( moment acting at any point):-any point):- If, at a point, a couple forms a moment, then that is called Concentrated MomentConcentrated Moment. It is expressed in Nm or kNm.

L

MM

EQUILIBRIUM OF A RIGID BODY :-EQUILIBRIUM OF A RIGID BODY :- A rigid body can be in equilibrium if the resultant force and moment of all forces at any point is zero.

A beam is noting but a rigid body.

For a rigid body in equilibrium, the condition of static equilibrium are three, Viz;

Fx = 0 ; Fx = 0 ; Fy = 0 ;Fy = 0 ;

M = 0 ;M = 0 ;

A beam is said to be statically determinatestatically determinate if the total no. of unknown reactions are equalequal to the no. of conditions of static equilibrium.

Total no. of unknown reactions will depend upon the type of beam and the type of support.

The no. of conditions of static equilibrium, for a rigid body, are 3 (three):

Fx = 0 ;Fx = 0 ; Fy = 0 ;Fy = 0 ; M = 0 M = 0

AA BB

HHAA

VVAA VVB B

conditions of static equilibrium = 3,No. of unknown reactions = 3, SoThe beam is statically determinate.The beam is statically determinate.

VVAA

HHAA

Condn. of static equilibrium = 3,No. of unknown reactions = 3, SoThe beam is statically determinate.The beam is statically determinate.

MMAA

A beam is said to be statically indeterminatestatically indeterminate if the total no. of unknown reactions are more thanmore than the no. of conditions of static equilibrium.

AA BB

HHAA

VVAA VVB B

HHBB

Conditions of static equilibrium = 3,No. of unknown reactions = 4, Sothe beam is statically indeterminate.the beam is statically indeterminate.

VVAA

HHAA

MMAA VVBB

VVAA

HHAA

MMAA

VVBB

HHBB

VVAA

HHAA

MMAA VVBB

HHBB

MMBB

We will limit our study to shear force and bending moment diagrams of STATICALLYSTATICALLY

DETERMINATEDETERMINATE BEAMS.BEAMS.

Commonly encountered statically determinate beams are,

a) Cantilever Beam,a) Cantilever Beam,

b) Simply Supported Beam,b) Simply Supported Beam,

c) Over-hanging Beamc) Over-hanging Beam.

These beams are usually subjected to the following types of loading;

a) Point Load,a) Point Load,

b) Uniformly Distributed Load,b) Uniformly Distributed Load,

c) Uniformly Varying Load,c) Uniformly Varying Load,

d) Concentrated Momentd) Concentrated Moment..

The beam transfers the applied load to the supports. The effect of applied load is to create bending moment and shear force at each cross-section. In transferring the applied load to the supports, the beam develops resistance against moments and shear force at all of it’s cross-sections.

The effect of applied load is to create bending moment and shear force at each cross-section. We will determine the shear force and bending moment caused by loads at each section, for various given loading condition. Then we will plot the Variation in the shear force and bending moment across the length of the beam.

==

10 KN/m

25 KN 5 m 25 KN

Y

Y

SFYY = 25 –20

= 5 kN

SFYY = 25 –30

= 5 kN

2 m

BMYY= 25*2 –10*2*1

= 30 kN.m

BMYY= 25*3 –10*3*1.5

= 30 kN.m

10 KN-m

25 KN 5 m 25 KN

2 m

3 m

dx

10 KN-m

25 KN

25 KN

*25-20 = 5 kN

*5

5

**30

**25.2 –10.2.1 = 30 kN.m

30

#5

##30#30-25 = 5 kN##25.3 –10.3.3/2 = 30 kN.m

5

30

5 kN 5 kN

30 kN.m 30 kN.m

The vertical force will try to shear-off the member as shown in the fig., hence it is known as the shear shear force.force.

The unbalanced-moment will try to bend the member as shown in the fig., hence it is known as the Bending Moment.Bending Moment.

##To resist the shear force, the element will develop the resisting stresses, Which is known as Shear Stresses.Shear Stresses.

##To resist the Bending Moment, the element will develop the resisting stresses, Which is known as bending Stresses.bending Stresses.

+ve Shear The shear force is positive if it tends to move the left portion in upward direction relative to the right portion.

-ve ShearThe shear force is negative if it tends to move the right portion in upward direction relative to the left portion.

SIGN CONVENTIONSSIGN CONVENTIONS

The bending moment is positive if it tends to create tension in the bottom fibers of the beam. i.e. concave face upward.

+ve B.M.

The bending moment is positive if it tends to create tension in the bottom fibers of the beam. i.e. concave face upward.

-ve B.M.

Shear force is an unbalanced force, parallel to the cross-section, mostly vertical, but not always, either the right or left of the section.

Thus, the procedure to find out the shear force, at a section is to imagine a cut in the beam at the section, consider either to the left or the right portion and find the algebric sum of all the forces normal to the axis.

Shear force diagram is the graph showing the variation of the shear force throughout the length of the beam.

Bending Moment is an unbalanced couple, either to the right or left of the section.

Thus, the procedure to find out the Bending Moment, at a section is to imagine a cut in the beam at the section, consider either the left or the right portion and find the algebric sum of the moments due to all the forces.

Bending Moment diagram is the graph showing the variation of the bending moment throughout the length of the beam.

BBAA

QP

dx

FM M

F

Section at P

F+dF

M+dM M+dM

F+dF

Section at Q

M+dMM

F

F+dF

Element PQ

w/u.l.

BBAA

QP

dx

M+dMM

F

F+dF

Element PQ

Let, F = Shear Force at P

F+ dF = Shear Force at Q

M = Bending Moment at P

M+ dM = Bending Moment at Q

w/u.l.

M+dMM

F

F+dF

Element PQ

Since the beam is in equilibrium, all the elements will be in equilibrium. So element PQ is also in equilibrium.

w/u.l.

Equating Vertical Forces at Q , upward load is considered as positive

F - (F+ dF) - wdx = 0

F - F - dF = wdx

dF/dx = -w (-ve sign indicates downward direction of force) ……………………..Eq(i)

dF/dx = W ………………………..Eq(i)

The rate of change of shear force, or in other words, the slope of the shear force diagram, will be equal to the intensity of loading at that section.

As per the sign convention, if the load is acting in downward direction it will be considered as positive

M+dMM

F

F+dF

Element PQ w/u.l.

Equating Moments @ Q

M + Fdx - wdx(dx/2) - (M+ dM) =0

Neglecting the term containing dx2

dM/dx = F ……………………..Eq(ii)

The rate of change of Bending Moment, or in other words, the slope of the Bending Moment diagram, will be equal to the intensity of Shear force at that section.

L

W

A B

W

S.F. Dia.

B.M. Dia.

+ve

-ve

WW

W.L

Sec X-X

(+ve)

11

Fx = W

Mx = -WL + Wx

x

At A, x = 0; M= -WL

At B, x = L; M= 0

5.0m

10 kN

A B

10

+ve1010

Step 1:-Step 1:- Support reaction

FY = 0,

VA= 10 kN (up)

MA=10*5=50 kN.m, (anti clock

Wise moment)

Fx = 10

S.F. DiagramS.F. Diagram

x

22

At A, x = 0; M= -50 kN.m

At B, x = 5; M= 0 kN.m

So Linear change in B.M. value

5.0m

10 kN

A B

10

B.M. Dia.

50

Step 2:-Step 2:- Bending Moment at x;

M@x

= -50+10 * x

Linear Equation

x

-ve (+ve)

-50+10 * x

5.0m

10 kN

A B

10

+ve 1010


B.M. Dia.

50 -ve (+ve)

5.0m

10 kN

A B

10

+ve 1010


FY = 0,

VA= 10 kN (up)

At B, Point load,

So sudden change in S.F. value


x

33

0

4.0m

10

B.M. Dia.

40


-ve (+ve)

5.0m

10 kN

A B

x

At A, x = 0; M = -40 kN.m

At C, x = 4; M = 0 kN.m

So Linear change in B.M. value

M@x

= - 40+10 * x

Linear Equation

C

105.0m

10 kN

A B

+ve 10100

40 -ve (+ve)

0

5.0m

10 kN

A B

30

+ve10

30


FY = 0,

VA= 30 kN (up)


20 kN

2.5m 2.5m

20

C

4 4

F(a) = 30F(c-left) = 30 F(c-right) = 30-20

= 10F(b-left) = 10 F(c-right) = 10-10

= 0

5.0m

10 kN

A B

30


20 kN

2.5m 2.5mC

Mx (AC) = -100 +30 * x Lin. Eqn

Mx (CB) = -100 + 30*(x) - 20*(x-2.5)

B.M. Dia.

100 -ve

MB = 0.0 kN.m

MC = -25 kN.mMA = -100 kN.m

25

x


+ve10

30 20

5.0m

10 kN

A B

30

20 kN

2.5m 2.5mC

B.M. Dia.B.M. Dia.

100 -ve 25

0.0

A

L

B

wL

B.M. Dia.

S.F. Dia.

+ve

-ve

wL

w.L2/2

Sec X-X

(+ve)

w/m

(-ve)

55

x

S.F.D.

Fx = wL – wx

( linear variation)

FA = wL : FB = 0

B.M.D.

Mx = -wL2/2

+ wLx - wx2/2 (parabolic variation)

MA= -wL2/2

MB= 0

Shear Diagrams by Integration of the Load:-Shear Diagrams by Integration of the Load:-

dF/dx=w

F= w dx + C1 (limit from 0 to x)

By assigning definite limits to this integral, it is seen that the shear at a section is simply an integral (i.e. sum) of the vertical forces along the beam from the left end of the beam to the section in question plus a constant of integration C1. This constant is equal to the shear on the left hand end. Between any two definite sections of a beam, the shear changes by the amount of the vertical force included between these sections. If no force occurs between any two sections, no change in shear takes place. If a concentrated force comes into the summation, a discontinuity, or a “jump” in the value of the shear occurs.

The continuous summation process remains valid nevertheless, since a concentrated force may be thought of as being a distributed force extending for an infinitesimal distance along the beam.

On the basis of the above reasoning, a shear diagram can be established by the summation process. For this purpose, the reactions must always be determined first. Then the vertical components of forces and reactions are successively summed from left end of the beam to preserve the mathematical sign convention for shear adopted. The shear at a section is simply equal to the sum of all vertical forces to the left of the section.

Moment Diagrams by Integration of the Shear:-

dM/dx = F

M= F dx + C2 (limit from 0 to x)

Where C2 is a constant of integration corresponding to boundary conditions at x=0. This is analogous to previous equation developed for construction of shear diagrams. The summation of the small areas between definite sections through a beam corresponds to an evaluation of the definite integral. If the ends of a beam are on rollers, pin jointed, or free, the starting and the terminal moments are zero. If the end is fixed, the end moment is known from the reaction calculations. If the fixed end of a beam is on the left, this moment with the proper sign is the initial constant of integration C2.

By proceeding continuously along the beam from the left-hand end and summing up the areas of the shear diagram with due regard to their sign, the moment diagram is obtained. This process of obtaining the moment diagram from the shear diagram by summation is exactly the same as that employed earlier to go from loading to shear diagrams. The change in moment in a given segment of a beam is equal to the area of the corresponding shear diagram.

A

5 m

B

50

B.M. Dia.

S.F. Dia.

+ve

-ve

50kN

125 kNm

Sec X-X

(+ve)

10 kN/m

(-ve)

66

A

5 m

B10 kN/m

77

2.5 m C

A

5 m

B

25

B.M. Dia.

S.F. Dia.

+ve

-ve

25kN

31.25 kNm

10 kN/m772.5 m

C Reactions :VA=25 kNMA=(10*2.5*1.25) = 31.25 kNmSFD:FA= 25 kNFx(AC)= 25-10xFx(CB)= 25-25=0

BMDMA = - 31.25Mx(AC) = -31.25+25x-10x2/2Mx(CB) = 0

x

31.25

A

5 m

B

10 kN/m

88

2.5 m C

A

5 mB

25

B.M. Dia.

S.F. Dia.

+ve

-ve

25kN

31.25

10 kN/m88

2.5 m

93.75Parabolabetn BC linear

C Reactions :VA=25 kNMA=(10*2.5*3.75) = 93.75 kNmSFD:FA= 25 kNFx(AC)=25Fx(CB)=25-10(x-2.5)

BMDMA = - 93.75Mx(AC) = -93.75+25xMx(CB) = -93.75+25x -10(x-2.5)2/2

x

93.75

A

7 m

B

10 kN/m

99

2 m 2 m 3 m C D

A

7 m

B

40

S.F. Dia.

+ve

40

10 kN/m99

2 m 2 m 3 m

20

xC D

140Reactions :VA =10*2 +10*2 =40kNMA=(10*2*6+10*2*1) = 140 kNm

SFD:FA= 40 kNFx(AC)= 40-10xFx(DB)=40-20-10(x-5)

A

7 m

B

40

B.M. Dia.

-ve 20

10 kN/m99

140Parabola

2 m 2 m 3 m

Parabola

80

linear

xC D

140

BMDMA = - 140Mx(AC) = -140 + 40x-10x2/2Mx(CD) -140 + 40x-20*(x-1)Mx(DB) -140 + 40x-20*(x-4)-10(x-5)2/2

A

8 m

B

10kN/m

1010

2m 2 m 2m

10 kN/m20kN

c2m

d e

A

8 m

B

60

S.F. Dia.S.F. Dia.

+ve60kN

10kN/m1010

2 m 2 m 2m

10 kN/m20kN

C

2m

D E

20

40

xReactions :VA =10*2 +20+10*2 =60kNMA=(10*2*7+20*4+ 10*2*1) = 240 kNm

SFD:FA= 60 kNFx(AC)= 60 -10x FD= 60 –20 = 40 (Left) FD= 40 –20 = 20 (Right) Fx(EB)=60-20- 20 -10(x-6)

B.M. Dia.B.M. Dia.

-ve 20

Parabola

Parabola

A

8 m

B

60

10kN/m1010

2 m 2 m 2m

10 kN/m20kN

c

2m

de

60

140 linear

linear240

x

240

xBMD

MA = - 240

Mx(AC)= -240+ 40x- 10x2/2

Mx(CD)=-240 + 40x- 20(x-1)

Mx(DE) =-240 + 40x- 20(x-1)-20(x-4)

Mx(EB) = -10x2/2

(Considering Right portion)

BAL

w/m

11 11

BA

LwL/2

wx/Lw/m

B.M. Dia.

-ve

w.L2/3 (+ve)

11 11

-ve+ve

S.F. Dia.wL/2

x

w.L2/3Reactions :VA =(1/2)Lw=wL/2MA=(1/2)L*w*(2L/3) = wL2/3SFD:FA= wL/2Fx= wL/2-{x*(wx/L)}/2 = wL/2-wx2/2LBMDMA = -wL2/3Mx = -wL2/3 + wL/2*x -

{wx2/2L*(x/3)} = -wL2/3 + wL/2*x -

wx3/6L

BA

5 m

10 kN/m

12 12

BA

5 m25kN

2x10 kN/m

B.M. Dia.

-ve

83.33 (+ve)

12 12

-ve+ve

25

x

83.33kN.m

S.F. Dia.

Reactions :VA =(1/2)*10*5 =25kNMA=(1/2)10*5*(2*5/3) = 83.33 kN.mSFD:FA= 25kNFx= 25-(x*2x/2)=25-x2

BMDMA = 83.33kN.mMx = -83.33 +25x-x2*x/3 = -83.33+25x–x3/3

BA

L

w/m

13 13

BA

LwL/2

wx/Lw/m

S.F. Dia.

B.M. Dia.

-ve w.L2/6

(+ve)

13 13

-ve

+ve

S.F. Dia.

wL/2

x

Reactions :VA =(1/2)Lw=wL/2MA=(1/2)L*w*(L/3) = wL2/6

SFD:Fx=(wx/L)*x/2 = wx2/2L (Right side)

FA= wL/2

BMDMx = (wx2/2L)*x/3 = wx3/6LMA = -wL2/6

w.L2/6

BA

14 14

10 kN/m

5 m

BA

+ve

S.F. Dia.

B.M. Dia.

-ve

(+ve)

14 14

25 kN

10 kN/m

41.67

5 m

25

-ve

+ve

S.F. Dia.

41.67Reactions :VA =(1/2)10*5 =25kNMA=(1/2)10*5*(5/3) = 41.67kN.mSFD:Fx= 2x*x/2=x2

FA= 25kN

BMDMx =x2(x/3) = x3/3MA = 41.67 kN.m

2x

x

B

15 15

A

10 m

100 kN.m

B

15 15 A

10 m

100 kN.m 100

kN.m

S.F. Dia.

B.M. Dia.

VA =0 x

100 +ve

B

16 16

A

10 m

100 kN.m

C

6 m

S.F. Dia.

B.M. Dia.

100 +ve

BA

10 m

100 kN.m

C

6 m

VA =0

AA BB

HHAA

VVAA VVB B

11

AA BB

W

L

AA BB

W

L

W/2 W/2

W/2 W/2

W/2


+ ve+ ve

- ve- ve

11

AA BB

W

L

W/2 W/2

WL/4

B.M. DiagramB.M. Diagram

+ ve+ ve

11

AA BBW

L

W/2 W/2

WL/4

B.M. DiagramB.M. Diagram + ve+ ve

11

W/2 W/2

W/2S.F. DiagramS.F. Diagram

+ ve+ ve

- ve- ve

AA BB10kN

10m5 5

25


22

5 5

5S.F. DiagramS.F. Diagram

+ ve+ ve

- ve- ve

CC5

AA BB10kN

10m

6 4

24


22

6


+ ve+ ve

- ve- ve

CC4 m

4

AA BB

10kN

10 m

8 7

32

B.M. DiagramB.M. Diagram+ ve+ ve

33

8


+ ve+ ve

- ve- ve

5kN

CC4 m 4 m

DD

77

28

AA BB

10kN 5kN

CC

3m 3 m

EE

15kN

DD

3m

10 m

4 4

AA BB

LL

W kN/m

WL/2WL/2

CC

5 5

AA BB

LL

W kN/m

WL/2

WL/2

+ ve+ ve

- ve- ve

WL2/8

+ ve+ ve

WL/2WL/2

CC

AA BB

10 m10 m

10 kN/m

CC

66

AA BB

10 m10 m

10 kN/m

50

50

+ ve+ ve

- ve- ve

125 kN.m

+ ve+ ve

5050

CC

AA BB

10 m10 m

10 kN/mCC

5 m

7 7

AA BB

10 m10 m

10 kN/m

37.5

12.5+ ve+ ve

- ve- ve

70.3125 kN.m

+ ve+ ve

12.537.5

CC

3.75 m

5 m

BBAA

10 m10 m

10 kN/mCC

88

12.5

BBAA

10 m10 m

10 kN/m

37.5

CC

37.5

12.5 - ve- ve

+ ve+ ve 3.75 m

70.3125 kN.m

+ ve+ ve

5 m

BBAA

10 m10 m

10 kN/mCC DD

3 m 3 m 4 m4 m 3 m3 m

99

20

BBAA

10 m10 m

10 kN/m

20

CC DD

3 m 3 m 4 m4 m 3 m3 m

20

20


20

BBAA

10 m10 m

10 kN/m

20

CC DD

3 m 3 m 4 m4 m 3 m3 m

B.M. DiagramB.M. Diagram

60 60 60 60

80 80

20

BBAA

10 m10 m

10 kN/m

20

CC DD

3 m 3 m 4 m4 m 3 m3 m

60 60 60 60

80 80

20

20

BBAA

10 m10 m

10 kN/mCC DD

3 m 3 m 4 m4 m 3 m3 m

1010

30

BBAA10 kN/m

30

CC DD

3 m 3 m 4 m4 m 3 m3 m

45 45 45 45

30

30

BBAA

10 m10 m

10 kN/mCC DD

3 m 3 m 4 m4 m 3 m3 m

10kN

1.51.5

1111

38.5

BBAA10 kN/m

31.5

CC DD

3 m 3 m 4 m4 m 3 m3 m

49.5 49.5

46.546.5

31.5

38.5

10kN

1.51.5

23.5

13.5

1.5

55.6155.61

55.555.5

Mmax will occur at x=2.85m from A

2.85

BBAA

LL

W kN/m

1212

BBAA

LL

W kN/m

WL/3 WL/6

WL/3

WL/6

L /L /3 3

WL2/(93)

BBAA

9 m9 m

9 kN/m

1313

BBAA

99

9 kN/m

27 13.5

27

13.5

5.25.2

46.7646.76

BBAA

9 m9 m

9 kN/m

1414

BBAA

99

9 kN/m

13.5 27

13.5

5.25.2

46.7646.76

27

BBAA

10 m10 m

1515

CC

5 m

100 kN.m

BBAA

1515CC

100 kN.m

10 10

10 10

50

-50

+ ve+ ve

- ve- ve+ ve+ ve

AA

10 m10 m

1616

50 kN.m 100 kN.m

BB

BBAA

1616CC

5 5

5 5

50

50 kN.m 100 kN.m

100+ ve+ ve

- ve- ve

AA

6 m6 m

1717

10

10

BB11 11 22 1111

10 102

30 45 3

4

AA

6 m6 m

810

BB

5 10

8.66 10 6CC DDEE FF

16.33 16.67

4.66

16.67

16.33

8

10

5

10

AA

6 m6 m

810

BB

5 10

8.66 10 6CC DDEE FF

16.33 16.67

4.66

16.3316.3316.67 16.67

22.3322.33 2525

AA

6 m6 m

BB8.66 10 6CC DDEE FF4.66

6 4.66

4

Thrust diagram

Ten:– +ve

Comp:- -ve

+ ve+ ve

- ve- ve- ve- ve

BBAA

6 kN/mCC

15m 15m 5 m5 m

DD

5 m5 m

20 kNm

Two beams AC and CB are connected by an Internal Pin to form beam ACB. Draw SF and BM diagram.

BB

AA

6 kN/m

CC

15m 15m

5 m5 m

DD

5 m5 m

20 kNm

CC

3015

15

15

170

BB

AA

6 kN/m

CC

15m 15m

5 m5 m

DD

5 m5 m

20 kNm

CC

3015

15

15

170

30 + ve+ ve

- ve- ve

S.F. will be zero at a distance of L/

3 = 8.66m from C

8.66m

BB

AA

6 kN/m

CC

15m 15m

5 m5 m

DD

5 m5 m

20 kNm

CC

3015

15

15

1708.66m

86.6

0.0 75

20 170

+ ve+ ve

- ve- ve

3.0 m 2.0m

25 kN/m100

3.0 m

25 kN/m

25 kN/m100

25 kN/mBBAA

DDCC

97.9 127.1

50

72.1

27.9

97.9

25 kN/m100

25 kN/mBBAA

DDCC

97.9 127.1181.2

50

6.0 m 2.0m

15 kN/m15 kN/m 5020

10

17.32

2 211

6.0 m 2.0m

15 kN/m15 kN/m 5020

10

17.32

6.67 48.33

CC DD

EEAA BB

FF

10.0

6.67 8.33

5

38.33

1.49 m

x

Fx = 10 - 48.33 + 30 + 7.5x2/2 = 0

X = 1.49 m

6.0 m 2.0m

15 kN/m15 kN/m 5020

10

17.32

6.67 48.33

CC DDEE

AA BBFF

1.597

1.49 m

20

26.27 35

15 6.67

6 21

50

CC DD

AA BB

EE

5050

6 21

50

CC DD

AA BB

EE

5050

83.3366.67

33.33

505016.67

50 50

16.67

6 21

50

CC DD

AA BB

EE

5050

83.3366.67

- 100- 50

6 21

50

CC DD

AA BB

EE

5015 kN/m

6 21

50

CC DD

AA BB

EE

50

86.67

53.33

5050

50 50

36.67

15 kN/m

103.332.445 m

6 21

50

CC DD

AA BB

EE

50

- 100- 50

86.67

15 kN/m

103.33- 5.17

6 21

25

CC DD

AA BB

EE

2525 kN/m

Given SFD. Draw Loading diagram and from it, BMD.

14.4

9.6

21.6

24

3 1 1 2

14.4

9.6

21.6

24

1.8 m

12 kN/m

14.4 45.6

8 kN/m

12

14.4

9.6

21.6

24

1.8 m

Given BMD. Draw Loading diagram and from it, SFD.

+ ve+ ve

CC DDAA BBEE

0.00.018.0 18.0 20.020.0

20.2520.25

0.00.0

4.54.5 0.50.5 2.02.0 3.03.0

+ ve+ ve

CC DDAA BBEE

0.00.018.0 18.0 20.020.020.2520.250.00.0

4.54.5 0.50.5 2.02.0 3.03.0

BMD Parabolic between AC – so SFD will be linear

Area of SFD between AC = 20.25 = ½ *4.5*x

X = 9 kN. ( I.e RA)

Area of SFD between CD = 20.25-20 = 0.25 = ½ *0.5*Y

Y = 1 kN.

BMD Linear = So SF constant

Area of SFD between EB = 18 = 3.0*Z

Z = 6 kN.

9 kNY = 1 kN

6 kN.

+ ve+ ve

CC DDAA BBEE

0.00.018.0 18.0 20.020.020.2520.250.00.0

4.54.5 0.50.5 2.02.0 3.03.0

9.0 6.0

2 kN/m

5.0

Make use of symmetry : If the structure is symmetric and also loaded symmetrically, then no need to calculate the support reaction. Both the reactions will be equal and it will be equal to (total load /2).

SIGN CONVENTION:

Positive Shear:

Positive Bending:

Draw the beam and locate all the important points. I.e. Supports, loading points, point at which type nature and magnitude of load changes etc.

Draw faint projection lines through all these important points.

Draw the base line for shear force diagram and Bending moment diagram. Positive values will be plotted above the base line and negative Values will be plotted below the base line.

Calculate SF and BM at all the important points and plot the ordinates on respective projection lines.

Join t he ordinates with appropriate curves. For that make use of two symmetrical equations.

dF / dx = - W and dM / dx = F

At free end of the cantilever beam, bending moment will be zero, unless a concentrated moment is given.

At Two supports of a simply supported beam bending moment will be zero, unless a concentrated moment is given.

There will be some moment, at support having overhang.

Between two points, if no other load is acting then SF diagram will be straight horizontal line and BM diagram will be an inclined line.

Between two points, if UDL is acting then SF diagram will be straight inclined line and BM diagram will be a parabolic curve.

Between two points, if UVL is acting then SF diagram will be a parabolic curve and BM diagram will be a cubical parabola.

Whenever there is a concentrated load is there will be sudden change in Shear force.

Whenever there is a concentrated Moment is there will be sudden change in Bending moment.

Bending moment will be maximum when shear force changes it’s sign.

For symmetrically loaded beams, Bending moment will be maximum at midspan.

Principal PlanesPrincipal Planes:--:--

The planes which carry only Direct Stresses and no

Tangential Stresses are called “ Principal Planes“ Principal Planes ””

Principal StressesPrincipal Stresses:--:--

The intensity of stress on the Principal Planes are

known as “ Principal Stresses ”“ Principal Stresses ”

1

1 2

2

pppn

pt

1

1 2

2

pppn

pt

All planes parallel to this planes are principal plane with zero principal stress

All planes parallel to this planes are

Principal plane with principal stress =p

Importance of Finding Principal PlanesImportance of Finding Principal Planes:--:--

1. The Principal Planes carry the Maximum Direct

Stress (Tensile or Compressive).

2. The Principal Planes helps to locate the Planes of

Maximum Shear (Inclined at 45 with Principal Planes)

45

1

1 2

2

pp

CaseCase:- 1:- 1 Element Subjected to Direct Stress & Element Subjected to Direct Stress & Shear Stresses Shear Stresses :--:--

Consider an element subjected to the direct stresses p & p’ (both tensile) alongwith a shear stress of intensity “ q ”as shown in figure. The shear stress intensity on all the sides will be “q” for equilibrium of the element. Let “ t ” be the thickness of the element.Consider a plane BE inclined at angle with BC.

A B

C

pp

p'

p'

q

q

D EForce on face BC = P1 = p * BC * tForce on face BC = P2 = q * BC * tForce on face CE = P3 = q * CE * tForce on face CE = P4 = p’ * CE * t

Force on face BC = P1 = p * BC * tForce on face BC = P2 = q * BC * tForce on face CE = P3 = q * CE * tForce on face CE = P4 = p’ * CE * t

P 2sin

P2 cos

P4

B

CE

P1

Pn

Pt

P1

cos

P1 sin

P4 cos

P4sin

P3

P2

P 3cos

P3 sin

Resolving forces parallel & perpendicular to BE, Pn = (P1+P3)cos + (P2+P4)sin

Pt = (P1+P3) sin - (P2+P4)cos

A B

C

pp

p'

p'

q

q

D E

Resolving forces perpendicular & parallel to BE,

Pn = (P1+P3)cos + (P2+P4)sin

Pt = (P1+P3) sin - (P2+P4)cos

pn = Pn / (BE.t)

=[(p*BC*t +q*CE*t) Cos+(q*BC*t + p’ *CE*t) Sin] / (BE.t)

= p cos2 + q sin cos + q sin cos + p’ sin2

= p/2 (1+cos2) + p’/2 (1-cos2) + q sin2pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2……….(I)

pt = Pt / (BE.t)

=[(p*BC*t +q*CE*t) Sin - (q*BC*t + p’ *CE*t) Cos] / (BE.t)

= p sin cos + q sin2 - q cos2 - p’ sincos

pt = {(p - p’)/2} * sin2 - q cos2 ……………….…….(II)

To locate Principal PlanesTo locate Principal Planes:--:--

pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2……….(I)

pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)

The Principal Planes are the plane which carry no Tangential

stress. Therefore pt = 0

at pt = 0, let =

{(p - p’)/2} * sin2 - q cos2 = 0

{(p - p’)/2} * sin2 = q cos2

tan2 = q / {(p - p’)/2 }

tan2 = 2q / (p - p’)

2 p - p’

2q

(p - p’)2 + 4q2

1 = [ tan-1{ 2q / (p - p’) } ]/2

2 = 90 + 1

tan2 = 2q / (p - p’) There are two

values of which will satisfy the

above eq. 1 & 2, where 2=90+ 1

sin21=2q/((p-p’)2+4q2) & cos21=(p-p’)/((p-p’)2+4q2)

sin22=-2q/((p-p’)2+4q2)& cos22= -(p-p’)/((p-p’)2+4q2)

Substituting value of sin21 & cos21, in eq.(I),

p1 = {(p + p’)/2} + {(p - p’)/2}* cos21 + q sin21…….(I)

= (p+p’)/2 + {(p-p’)/2}*{(p-p’)/((p-p’)2+4q2)} +

q* 2q/((p-p’)2+4q2)

= (p+p’)/2+ (p-p’)2*/ 2((p-p’)2+4q2) +2q2/((p-p’)2+4q2)

= (p+p’)/2 + {(p-p’)/2}2+q2

Similarly, substituting value of cos22 & sin22 in eq.(I),

p2 = (p+p’)/2 - {(p-p’)/2}2+q2

2

p - p’

2q

(p - p’)2 + 4q2

90

All planes parallel to this plane are principal planes and principal stress = p1, inclination =

All planes parallel to this plane are principal planes and principal stress = p2, inclination = + 90 deg.

p

p’

q

Let “ t ” be the thickness of the element.Consider a plane AC inclined at angle with AB.

A

BC

p1p1

p2

p2

pn

pt

p2

A

BC

p1pn

pt

p1cos

p1sin

p2cos

p2sin

To locate Planes of Maximum Shear StressTo locate Planes of Maximum Shear Stress:--:-- p1 = (p+p’)/2 + {(p-p’)/2}2+q2 …Major Principal Stress

p2 = (p+p’)/2 - {(p-p’)/2}2+q2 …Minor Principal Stress

Consider a plane AC inclined at angle with AB. Now, Force on face AB = P1 = p1*AB*t

Force on face BC = P2 = p2*BC*t

Equating all the forces along to AC,Pt = P1sin - P2cos

= p1 AB t sin - p2 BC t cos

pt = Pt / (AC. t)

p1 AB t sin - p2 BC t cos =AC. t

p1sin (AB/AC) - p2 cos (BC/AC)=

pt =

p1sincos - p2 sin cos =

(p1- p2)sincos

P2

A

BC

P1Pn

Pt

P1cos

P1sin

P2cos

P2sin

The Shear stress pt = (p1- p2)sin cos;

pt = (p1- p2)sin cos;

= (p1- p2)/2 sin2pt max. sin2 = max.

sin2 =+/- 1

2 = 90 or 270

= 45 or 135

When = 45, (EF) pt max = (p1- p2)/2

When = 135,(GH) pt max = (p1- p2)/2

Planes of maximum shear are always at Right angles toPlanes of maximum shear are always at Right angles to

each othereach other (Inclined at 45 & 135 with principal planes.)(Inclined at 45 & 135 with principal planes.)

F

E

G

45

H

135

p1 = (p+p’)/2 + {(p-p’)/2}2+q2 …Major Principal Stress


pt max = (p1- p2)/2 = {(p-p’)/2}2 + q2

(at 45 & 135 with principal planes in anticlockwise)

Location of Principal Planes; tan2 = 2q / (p - p’) (From previous slide) ( anticlockwise from vertical)

p1p2

p1p2

45

pt1

pt2

p1 = (p+p’)/2 + {(p-p’)/2}2+q2 (Major Prin. Stress )(III)

p2 = (p+p’)/2 - {(p-p’)/2}2+q2 (Minor Prin. Stress)(IV)

pt max = (p1- p2)/2 = {(p-p’)/2}2 + q2 _ _ (VI)

pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)

= (p cos2 + q sin cos + q sin cos + p’ sin2 )

pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)

Location of Principal Planes; tan2 = 2q / (p - p’) ( anticlockwise from vertical) (V)

CaseCase:- 1:- 1 Element Subjected to Direct Stress & Element Subjected to Direct Stress & Shear Stresses Shear Stresses :--:--

(at 45 & 135 with principal planes in anticlockwise)

Q.1 A rectangular bar of cross sectional area 100cm2 is subjected to an axial load of 20kN. Calculate the normal and tangential stresses on a section which is inclined at an angle of 30deg. with normal cross section of the bar. Ans. Direct stress σ = load / cross sectional area = 20000/10000 = 2MPa.Normal stress σn = (p cos2 + q sin cos + q sin cos + p’ sin2 ) as p’= q= zero we have σn = σcos230 = 2* cos230 = 1.5MPa and tangential stress σt = {(p - p’)/2} * sin2 - q cos2, as p’= q= zero

σt =(σ/2)sin2*30 = 0.866MPa

Q.2 Calculate diameter of circular bar which is subjected to an axial pull of 160kN, if the permissible shear stress on any cross section is 60MPa.

Ans. Direct stress p = 160000/(π*r2 )

Maximum shear stress q = p/2= 80000/(π*r2)

Max. q = 60 = 80000/ ( π*r2)

Gives r = 20.6mm and diameter = 41.2 mm,

say diameter = 42mm

Q.3 A rectangular bar of cross section 200mm x 400mm is subjected to an axial pull P. If permissible normal and tangential stresses are 7 and 3.7MPa, calculate safe value of force P. Ans. Direct stress p = P/ (200*400) = 7 MPa hence P = 560kN _ _ _ _(i)and Maximum shear stress q = 3.7 = p/2= P/ (2*200*400)Hence P = 592kN _ _ _ _(ii)From i and ii safe value of P = 560kN ans.

Q.4 Two wooden pieces 100mm X100mm cross section are glued to gather as shown in fig. If permissible shear stress in glue material is 1MPa, calculate maximum axial pull P.

30degP P

Glued joint

With vertical the inclination of the joint is 60deg.σt = {(p - p’)/2} * sin2 - q cos2, as p’= q= zero shear stress q = (p/2)*sin 2θSo that q = 1 = [P/(2*100*100)]sin (2*60)

Give P = 23.094kN. Ans.

Q.5 Two mutually perpendicular planes are stresses as shown in fig. calculate normal and tangential stress on an oblique plane AB.

120MPa 120MPa

100MPa

100MPa

pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2 _(I)σn = {(p - p’)/2} * sin2 - q cos2 _ _ (II) as q = zeropn = (120+100)/2 + [(120 – 100)/2 ]cos 2*30 = 115MPa and σn = [(120 -100)/2]sin 2*30 = 8.66MPa. . . .Ans

= 30deg.

Q.6 Two mutually perpendicular planes are stresses as shown in fig. calculate normal and tangential stress on an oblique plane AB.

120MPa 120MPa

60MPa

60MPa

pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2 _(I)σn = {(p - p’)/2} * sin2 - q cos2 _ _ (II) as q = zeropn = (120- 60)/2 +[ (120 + 60)/2 ]cos 2*30 = 75 MPa σn = [(120 + 60)/2]sin 2*30 = 77.94 MPa. . . .Ans

= 30deg.

Example:-7Example:-7 A point in a strained material, the intensities of Normal stress across two planes at right angles to each other are 800N/mm2 and 350N/mm2 (both tensile) and a shear stress of 400N/mm2 across the planes as shown in figure. Locate the principal Planes and evaluate the Principal Stresses (p1 & p2). Also find the planes of maximum shear stress (pt max) & its location. Find Normal(pn),Tangential(pt) stresses on a plane inclined at 50 with vertical as shown in figure.

800N/mm2

400N/mm2

400N/mm2

350N/mm2

50

SolutionSolution::

Given, p = 800 N/mm2, p’= 350 N/mm2,

q = 400 N/mm2 = 50

(i) p(i) pnn & p & ptt

pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)

= 575 - 39.07 + 393.93

= 929.85 N/mm2

pt = {(p - p’)/2} * sin2 - q cos2 ………………..……(II)

= 221.58 - ( - 69.46)

= 291.04 N/mm2


= 575 + 458.94

= 1033.94 N/mm2


= 575 - 458.94

= 116.06 N/mm2

(ii) Principal Planes:--(ii) Principal Planes:--

(iii) Location of Principal Planes:--(iii) Location of Principal Planes:-- tan2 = 2q / (p - p’) = [ tan-1{ 2q / (p - p’) } ]/2 1 = [ tan-1{ 2q / (p - p’) } ]/2 = 30.32 (ACW from p)

2 = 90 + 1 = 90 + 30.32 =120.32 (with vertical.)

(iv) p(iv) pt maxt max = (p1- p2)/2 = {(p -p’)/2}2 + q2

= 458.94 N/mm2 (at 45 with principal planes.)

Example:-8 Example:-8 A point in a strained material, the intensities of Normal stress across two planes at right angles to each other are 500N/mm2 (tensile) and 100N/mm2 (Compressive) and a shear stress of 200N/mm2 across the planes as shown in figure. Locate the principal Planes and evaluate the Principal Stresses (p1 & p2). Also find the planes of maximum shear stress (pt max) & its location. Find Normal(pn),Tangential(pt) stresses on a plane inclined at 30 with vertical as shown in figure.

500N/mm2

200N/mm2

200N/mm2

100N/mm2

30

SolutionSolution::

Given, p = 500 N/mm2, p’= -100 N/mm2,

q = 200 N/mm2 = 30


pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)

= 200 + 150 + 173.20

= 523.20 N/mm2

pt = {(p - p’)/2} * sin2 - q cos2 ………………..……(II)

= 259.80 - (100)

= 159.80 N/mm2


= 200 + 360.55

= 560.55 N/mm2


= 200 - 360.55

= -160.55 N/mm2


(iii) Location of Principal Planes:--(iii) Location of Principal Planes:-- tan2 = 2q / (p - p’) = [ tan-1{ 2q / (p - p’) } ]/21 = [ tan-1{ 2q / (p - p’) } ]/2 = 16.85 (with vertical.)

2 = 90 + 1 = 90 + 16.85 =106.85 (with vertical.)

(iv) p(iv) pt max t max = (p1- p2)/2 = {(p -p’)/2}2 + q2


Example:-9 Example:-9 A point in a strained material, the intensities of Normal stress across two planes at right angles to each other are 650N/mm2 (Compressive) and 150N/mm2 (Compressive) and a shear stress of 320N/mm2 across the planes as shown in figure. Locate the principal Planes and evaluate the Principal Stresses (p1 & p2). Also find the planes of maximum shear stress (pt max) & its location. Find Normal(pn),Tangential(pt) stresses on a plane inclined at 60 with vertical as shown in figure.

650N/mm2

320N/mm2

320N/mm2

150N/mm2

60

SolutionSolution::

Given, p = - 650 N/mm2, p’= - 150 N/mm2,

q = 320 N/mm2 = 60


pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)

= - 400 + 125 + 277.13

= 2.13 N/mm2

pt = {(p - p’)/2} * sin2 - q cos2 ………………..……(II)

= - 216.50 - ( - 160)

= - 56.50 N/mm2


= - 400 + 406.08

= 6.08 N/mm2


= - 400 - 406.08

= - 806.08 N/mm2


(iv) p(iv) pt maxt max = (p1- p2)/2 = {(p -p’)/2}2 + q2


(iii) Location of Principal Planes:--(iii) Location of Principal Planes:-- tan2 = 2q / (p - p’) = [ tan-1{ 2q / (p - p’) } ]/21 = [ tan-1{ 2q / (p - p’) } ]/2 = - 26.00 (with vertical.)

2 = 90 + 1 = 90 + 16.85 =+ 64.00 (with vertical.)

Mohr’s Stress Circle MethodMohr’s Stress Circle Method:--:--

When principal stresses & planes at a point in a two dimensional stresses are given, we can use the Mohr’s Stress Circle Method to obtain the normal and tangential stress-intensities across a given plane through the point.

In Mohr’s Circle ; The Normal Stress (pn) is plotted on X-axis. The Tangential Stress (pt) is plotted on Y-axis. Angle on Mohr’s circle is taken as twice the given angleAngle on Mohr’s circle is taken as twice the given angle

A B

C

pp

p'

p'

q

q

D E

Sign Conventions:--Sign Conventions:--1. Tensile Stresses are taken as Positive (+ve)2. Compressive Stresses are taken as Negative (-ve)3. Clockwise Shear is taken as (+ve)4. Anticlockwise Shear is taken as (-ve)

ONormal Stress (pn)

Sh

ear

Str

ess

(pt)


Case:-1Case:-1 If an element is subjected to the direct stresses (p & p’) & shear stress (q) then the principal stresses are (p1 & p2) and the normal and tangential stresses (pn & pt) across a plane inclined at angle to the plane carrying stress p are,



pt max = (p1- p2)/2 = {(p-p’)/2}2 + q2

pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)

pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)

Location of Principal Planes; tan2 = 2q / (p - p’)

C

P’(CD)

Oppnn

Q’p'

q pptt

P(BC)

Q

p q

1. Select origin O and draw axis OX.2. Draw OQ = p . On faces BC & DA the shear stress q is anticlockwise, ( -ve) , draw QP = q (downwards)3. OQ’ = p’ and on faces AB & CD the shear stress q is clockwise, hence it is +ve, so draw Q’P’ = q ( upwards as it is +ve)4. Join the points P & P’ which will intersect OX at C.

A B

Cpp

p'

q

q

DSteps to Draw Mohr’s Circle:-

5. Point C is the Centre of Mohr’s Circle. With C as centre and radius equal to CP or CP’, draw a circle, which will intersect OX at points A & B.6. Measure OA which gives maximum direct stress and no Shear stress & it is Major principal stress (p1) OA=p1

Measure OB as Minor principal stress (p2) OB = p2.

p2 A

CB

(CD)P’

O

p1

ppnnQ’

p'

q

pptt

P(BC)

Q

p q

7. From plane CP measure angle as 2 anticlockwise up to CA, therefore the principal plane will be located at anticlockwise from face BC.

p2 A

CB

(CD)P’

O

p1

ppnn2 Q’

p'

q

pptt

P(BC)

Q

p

q

A B

Cpp

p'

q

q

D

8. Measure CT which gives maximum Shear stress . Which is inclined at 90 & 270 in Mohr’s circle, so in actual case it is inclined at 45 & 135 with principal planes.

p2 A

CB

(CD)P’T

O

p1

ppnn2 Q’

p'

q

pptt

P(BC)

Q

p

q

C

(CD)P’R

O ppnn

Q’

q pptt

P(BC)

Q

q

A B

C

pp

p'

q

q

D

2

R’pn

pt

p'

p

9. From point P (BC) measure angle 2 in anticlockwise direction and draw a line CR. From R, draw RR’ on OX. Measure OR’ as pn and RR’ as pt.

Proof:- QQ’ = OQ - OQ’ = p - p’ CQ = CQ’ = (p - p’)/2 OC = OQ’ + Q’C = p’ + (p - p’)/2 = (p + p’)/2

Radius of Mohr’s Circle; CP = CQ2 + QP2 = {(p - p’)/2}2 + q2

OA = OC+CA = (p + p’)/2 + {(p - p’)/2}2 + q2 = p1

OB = OC - CB = (p + p’)/2 - {(p - p’)/2}2 + q2 = p2

since p1 & p2 are on X-axis the tangential stress is zero. Hence they represent the “Principal Planes”. Here p1 is major principal plane and p2 is minor principal plane.

p2

A

CB

P’T

O

p1

ppnn2 Q’

p'

q

pptt

p - p’

P

Q

p

q

T’

let QCP = 2 then

tan2 = QP/CQ = q / {(p - p’)/2} = 2q / (p - p’)

The max. tangential stress = max Y-ordinate in the circle

which is equal to the radius of the circle (CT = CT’)

pt max = CT = CP = {(p - p’)/2}2 + q2= (p1 - p2)/2

the position of the plane CT is at 90 from CA, hence

the plane of max. shear is inclined at 45 with principal

planes.

Example:-10 Example:-10 A point in a strained material, the intensities of Normal stress across two planes at right angles to each other (both tensile) and a shear stress of across the planes are shown in figure. Locate the principal Planes. Calculate the Principal Stresses (p1 & p2). Also find the planes of maximum shear stress (pt max) & its location. Find Normal(pn),Tangential(pt) stresses on a plane inclined at 80 with vertical as shown in figure.

800N/mm2

400N/mm2

400N/mm2

350N/mm2

80

C

P’

O ppnnQ’

350N/mm2

400

pptt

P

Q

800N/mm2

400

800N/mm2

400N/mm2

400N/mm2

350N/mm2

80

R

160

R’pn

pt

Example:-10Example:-10

p2 A

CB

P’T

O

p1

ppnnQ’

350N/mm2

400N/mm2

pptt

P

Q

800N/mm2

400

800N/mm2

400N/mm2

400N/mm2

350N/mm2

2


Solution:--

Measure:

OR’ = pn = Normal stress = 929.85MPa

RR’ = pt = Tangential Stress = 291.04MPa

OR = pr = Resultant Stress = 974.33MPa

OA = p1 = Major Principal Stress = 1033.94MPa

OB = p2 = Minor Principal Stress = 116.06MPa

Angle PCA = 2 = ___ therefore, 1 =30.32& 2=120.32

CT = pt max = Maximum Tangential Stress = 458.94MPa


Example:-11 Example:-11 A point in a strained material, the Normal stress across two planes at right angles to each other are 500N/mm2 (tensile) and 100N/mm2 (Comp.) and a shear stress of 200N/mm2 across the planes as shown in figure. Locate the principal Planes and evaluate the Principal Stresses (p1 & p2). Also find the planes of maximum shear stress(ptmax) and its location. Find Normal (pn),Tangential (pt) stresses on a plane inclined at 30 with vertical as shown in figure.

500N/mm2

200N/mm2

200N/mm2

100N/mm2

30

C

P’

O

ppnnQ’100

200

pptt

P

Q

500

200

500

200

200

100

30500

- p- pnn

R

60 R’pn

pt


p2

A

CB

P’T

O

p1

ppnn

2 Q’100

200

pptt

P

Q

500

200

500

200

200

100

500

- p- pnn


Solution:--

Measure:

OR’ = pn = Normal stress =523.20MPa

RR’ = pt = Tangential Stress = 159.80MPa

OR = pr = Resultant Stress = 547.06MPa

OA = p1 = Major Principal Stress = 560.55MPa

OB = p2 = Minor Principal Stress = -160.55MPa

Angle PCA = 2 = ___ therefore, 1 = 16.85& 2=106.85

CT = pt max = Maximum Tangential Stress=360.55MPa


Example:-12 Example:-12 A point in a strained material, the intensities of Normal stress across two planes at right angles to each other are 650N/mm2 (Compressive) and 150N/mm2 (Compressive) and a shear stress of 320N/mm2 across the planes as shown in figure. Locate the principal Planes and evaluate the Principal Stresses (p1 & p2). Also find the planes of maximum shear stress (pt max) & its location. Find Normal(pn),Tangential(pt) stresses on a plane inclined at 60 with vertical as shown in figure.

150N/mm2

320N/mm2

320N/mm2

650N/mm2

60

C

P’

Oppnn

Q’

650

320

pptt

P

Q

150

320

150

320

320

650

60

R

120

R’

pn

pt


p2 ACB

P’T

O

p1

ppnn2 Q’

650

320

pptt

P

Q

150

320

150

320

320

650


Solution:--

Measure:

OR’ = pn = Normal stress = _____

RR’ = pt = Tangential Stress = ______

OR = pr = Resultant Stress = _______

OA = p1 = Major Principal Stress = _______

OB = p2 = Minor Principal Stress = _______

Angle PCA = 2 = ___ therefore, 1 = ___& 2=___

CT = pt max = Maximum Tangential Stress = _______


CaseCase:- :- 22 Element Subjected to direct stress Element Subjected to direct stress Stresses only and no shear stressStresses only and no shear stress:--:--

Consider an element subjected to only Direct Stresses p & p’ (both tensile) as shown in figure. Here the shear stress intensity “q” is zero.

A

BC

pp

p'

p'

pn

pt



pt max = (p1- p2)/2 = {(p-p’)/2}2 + q2

CaseCase--2 Element Subjected to Direct Stresses only2 Element Subjected to Direct Stresses only:--:--

If in any case, an element is subjected to only direct

stresses & no tangential stresses as shown in figure. In that

case the pn, pt, p1, p2 & pt max can be found out by substituting

q=0 in equations of CASE- 1.

pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)

pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)


1.1. Element Subjected to Direct & Shear Stresses Element Subjected to Direct & Shear Stresses



pt max = (p1- p2)/2 = {(p -p’)/2}2 + q2

For CASE-2 Substituting q = 0 in above equations;

pn = {(p + p’)/2} + {(p - p’)/2}* cos2 ………….…….….(I)

pt = {(p - p’)/2} * sin2 …………. ……………….……(II)

p1 = p/2 + p2/4 = p …Major Principal Stress

p2 = p’/2 + p’2/4 = p’ …Minor Principal Stress

pt max = {(p-p’)/2}2 = (p - p’)/2 = (p1 - p2)/2

pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)

pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)


A

BC

p

p'

p'

pn

pt

p

CaseCase:- :- 22 Element Subjected to Direct Stress only Element Subjected to Direct Stress only and no shear stressand no shear stress:--:--

pn = (p + p’)/2 + {(p - p’)/2} cos2

pt = {(p - p’)/2} sin2

pr = pn2+ pt

2

pt max = (p - p’)/2

Example:-13Example:-13 A point in a strained material is subjected to the stresses as shown in figure. Find Normal(pn) , Tangential(pt) and Resultant (pr)stresses on a plane inclined at 50 with vertical as shown in figure.

Also find Maximum Tangential stress (pt max)

50

800 N/mm2

450 N/mm2

800 N/mm2

450 N/mm2

50

800

450 Solution:-

q = 0p = + 800 N/mm2 (Tensile)

p' = + 450 N/mm2 (Tensile)

= 50

pn = (p + p’)/2 + {(p - p’)/2} cos2 = 594.61 N/mm2

pt = {(p - p’)/2} sin2 = 172.34 N/mm2

pr= pn2+ pt

2 = 619.08 N/mm2

pt max = (p - p’)/2 = 175 N/mm2



30

500 N/mm2

200 N/mm2

500 N/mm2

200 N/mm2

30500

200 Solution:-q = 0p = - 500 N/mm2 (Comp.)

p' = + 200 N/mm2 (Tensile)

= 30

pn = (p + p’)/2+ {(p - p’)/2} cos2 = - 325 N/mm2

pt = {(p - p’)/2} sin2 = - 303.11 N/mm2

pr= pn2+ pt

2 = 444.41 N/mm2

pt max = (p - p)/2 = - 350 N/mm2



40

600 N/mm2

300 N/mm2

600 N/mm2

300 N/mm2

Solution:-q = 0p = - 600 N/mm2 (Comp.)

p' = - 300 N/mm2 (Comp.)

= 40

40600

300

pn = (p + p’)/2 + {(p - p’)/2} cos2 = - 476.04 N/mm2

pt = {(p - p’)/2} sin2 = - 443.16 N/mm2

pr= pn2+ pt

2 = 650.39 N/mm2

pt max = (p - p’)/2 = - 450 N/mm2


Case:-IICase:-II If the tangential stresses acting on element is zero, then the normal and tangential stresses across a plane inclined at angle to the plane carrying p are,

pn = (p + p’)/2 + {(p - p’)/2} cos2

pt = {(p - p’)/2} sin2

A

BC

pp

p'

p'

pn

pt

1. Set off the axis OX on this set off OA = p & OB = p’2. Bisect BA at C. 3. With C as centre and CA or CB as radius, draw a circle. This circle is known as Mohr’s circle.

Steps to Draw Mohr’s Circle:-Steps to Draw Mohr’s Circle:-

p'

A

CB

O

p

ppnn

pptt

4. Draw line CP at angle “2” with OX in anticlockwise so that the point P is on the circle.5. From P, drop perpendicular PQ on X axis. Measure OQ as pn & measure PQ as pt. 6. Join and measure OP which gives resultant stress pr.

p'

A

CB

O

p

ppnn

pptt

P

2

Qpn

pt pr

7. Join & measure CT which represents the plane of maximum shear stress. It inclined at 90 & 270 in Mohr’s circle with principal plane, hence in actual case they are inclined at 45 & 135 with principal planes.

p'

A

CB

O

p

ppnn

pptt

P

2

Qpn

pt pr

T

p'

A

CB

P T

O

p

ppnn

2

Q

pn

pt pr

pptt

p - p’

Example:-16Example:-16 A point in a strained material is subjected to following stresses. Find Normal(pn) , Tangential(pt) and Resultant (pr)stresses on a plane inclined at 50 with vertical as shown in figure. Use Mohr’s Circle Method.


50

800 N/mm2

450 N/mm2

800 N/mm2

450 N/mm2

p' = 450 N/mm2

A

CB

P T

O

p = 800 N/mm2

ppnn

pptt

100

Qpn

pt

50

800 N/mm2

450 N/mm2


pr

Solution:--

Measure:

OQ = pn = Normal stress = _____

PQ = pt = Tangential Stress = ______






30

150 N/mm2

350 N/mm2

150 N/mm2

350 N/mm2

p' = 350

ACB

PT

O

p = 150

ppnn

pptt

60

Qpn

pt

30

150 N/mm2

350 N/mm2


- ppnn

pr

Solution:--

Measure:








40

600 N/mm2

800 N/mm2

600 N/mm2

800 N/mm2

p =- 600 N/mm2

ACB

PT

O

p' = - 800 N/mm2

ppnn

pptt

80

Q

pn

pt


800 N/mm2

600 N/mm2

pr

Solution:--

Measure:






CaseCase :- :-33 Stress in One Direction OnlyStress in One Direction Only:--:--Consider an element subjected to stress p in only one direction as shown.Here, the p’ = 0 & q = 0.

1

1 2

2

pp



pt max = (p1- p2)/2 = {(p-p’)/2}2 + q2

CaseCase-3-3 Element Subjected to Direct Stresses only Element Subjected to Direct Stresses only:--:--

If in any case, an element is subjected to direct stress

in only direction & no tangential stresses as shown in figure.

Then the pn, pt, p1, p2 & pt max can be found out by substituting

p’ & q = 0 in eq. of CASE- 1.

pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)

pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)


1.1. Element Subjected to Direct & Shear Stresses Element Subjected to Direct & Shear Stresses



pt max = (p1- p2)/2 = {(p -p’)/2}2 + q2

For CASE-3 Substituting p’ & q = 0 in above equations;

pn = p /2 + p /2* cos2 = p cos2 ………….…….….(I)

pt = p /2 * sin2 ………………..……………….……(II)

p1 = p/2 + p2/4 = p …Major Principal Stress

p2 = p/2 - p2/4 = 0 …Minor Principal Stress

pt max = {(p-p’)/2}2 = p /2

pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)

pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)


CaseCase :- :-33 Stresses in One Direction onlyStresses in One Direction only :-- :--

pn = p cos2

pt = p/2 sin2

pr = pn2 + pt

2

pt max = p/2 (Inclined at 45 with Principal Planes)

1

1 2

2

pp

Example:-19Example:-19 A point in a strained material is subjected to the stress as shown in figure. Find Normal(pn) , Tangential(pt) and Resultant (pr)stresses on a plane inclined at 50 with vertical as shown in figure.


50

800 N/mm2800 N/mm2

Solution:--Solution:--

pn = p cos2 = 800 * cos2(50) = 330.54 N/mm2

pt = p/2 sin2 = (800/2) * sin(100) = 393.93 N/mm2

pr = pn2 + pt

2 = 514.23 N/mm2

pt max = p/2= 800/2 = 400 N/mm2

(Inclined at 45 with Principal Planes)

50

800 N/mm2800 N/mm2

Example:-20Example:-20 Find p1, p2 & if the stresses on two

planes are as shown in figure.

D

100 N/mm2

50 N/mm

2

30 N/mm2

60 N/mm2

C

A

B

100 N/mm2

Q

P(AB)

50

60 N/mm2

30

Q’

100 50

30

60

C

A

B

D

C’

C

B A

p1

p2

P’(CD)

2

STEPSSTEPS:-- 1. Select origin O and draw axis OX.2. On face AB, Draw pn = OQ = 100 N/mm2, the shear stress = 50 is anticlockwise, Draw QP = 50 (downwards)3. On face CD, Draw pn = OQ’ = 60 N/mm2, the shear stress = 30 is Clockwise, Draw Q’P’ = 30 (upwards)4. Join the points P & P’ . Now find the centre of line PP’ as point C’. From point C’ draw perpendicular bisector C’C , which will intersect OX at C.5. Point C is the Centre of Mohr’s Circle. Join CP & CP’. With C as centre and radius equal to CP or CP’, draw a circle, which will intersect OX at points A & B.6. Measure OA which gives maximum direct stress and no Shear stress & it is Major principal stress (p1) OA=p1

Measure OB as Minor principal stress (p2) OB = p2.7. From CP, measure the angle PCP’ in anticlockwise direction as 2. Half of this angle will give .

Example:-21Example:-21 Find p1, p2 & if the stresses on two

planes are as shown in figure.

D

900 N/mm2

1200 N/mm2

C

A

B

30

30

Principal StrainPrincipal Strain:--

If p1, p2 & p3 are principal stresses, then the principal

strains e1, e2 & e3 are given by;

e1 = p1/E - p2/E - p3/E (in direction of p1)



p1

p2

p3

Example:-22Example:-22 A rectangular block if steel is subjected to stresses of 1100 Mpa (tensile), 600 Mpa (Comp.), & 450 Mpa (tensile) across three pairs of faces. If = 0.3, and E = 2*10 5 N/mm2, Calculate strain in each direction.

Solution:-- p1 = 1100 N/mm2 (Ten)

p2 = - 600 N/mm2 (Comp.)

p3 = 450 N/mm2 (Ten)

= 0.3, and E = 2*10 5 N/mm2

e1 = p1/E - p2/E - p3/E (in direction of p1) = ______



Example:-23Example:-23 A circle of 250mm diameter is drawn on a mild steel plate. Then it is subjected to the stresses as shown in figure. Find the lengths of major & minor axis of the ellipse formed due to the deformation of the circle. Take E = 2 * 10 5 N/mm2; = 0.25

250

300 MPa

300 MPa

450 MPa

600 MPa600 MPa

450 MPa

Solution:-- E = 2 * 10 5 N/mm2; = 0.25

p = 600 Mpa , p’ = 450 MPa, q = 300 Mpa


= 525 + 309.23 =834.24 MPa


= 525 - 309.23 = 215.76 Mpa

e1 = p1/E - p2/E (in direction of p1)

= 1/E { 834.24 - 0.25(215.76)}

= 0.0039 (in direction of p1)

Increase in diameter along Major axis;

d1= e1 . d

= 0.0039 * 250 = 0.975 mm

Major axis of ellipse , d1 = 250 + 0.975 = 250.975 mm

e2 = p2/E - p1/E (in direction of p2)

= 1/E { 215.76 - 0.25(834.24) }

= 0.000036 (in direction of p2)

Increase in diameter along Major axis;

d2 = e2 . d

= 0.000036 * 250 = 0.009 mm

Minor axis of ellipse, d2 = 250 + 0.009 = 250.009 mm

Major axis of ellipse d1= 250 + 0.975 = 250.975 mm

Minor axis of ellipse d2= 250 + 0.009 = 250.009 mm

250250.975

250.009

tan2 = 2q / (p - p’) = 600 / (600 - 450) = 4 1 = 38 (Anticlockwise with vertical)

2 = 90 + 38 = 128(with vertical)

d1= 250.975 mm ; d2 = 250.009 mm

300 MPa

300 MPa

450 MPa

600 MPa600 MPa

450 MPa

38

P2

P1

d1d2

800 N/mm2

200 N/mm2

400 N/mm2

800 N/mm2

200 N/mm2

Example:-24Example:-24 Solve following 4 example using Solve following 4 example using Mohr’s circle method. Mohr’s circle method.

200 N/mm2

400 N/mm2

200 N/mm2

SOLID SHAFTS SUBJECTED TO BENDING MOMENT AND TWISTING MOMENT:(WITHOUT ANY AXIAL FORCE)Shear stress due twisting moment q = Tr/J = 16T/πd3

Bending stresses are due to bending moment = p = pbt or pbc = M/Z = 32M/ πd3( pbt and pbc = bending tensile and compressive stresses resp.) Principal stresses, p1 and p2 = p/2 ± [(p/2)2 + q2]1/2

Using values of p and q p1 and p2 = 16[ M ±( M2 + T2)1/2] / πd3

Maximum shear stress τ max = (p1 +p2)/2 = 16 [( M2 + T2)1/2] / πd3

For position of principal planes tan2 = 2q / (p - p’) gives tan2 = 2q / p as p’ is zero

For position of principal planes tan2 = 2q / (p - p’) givestan2 = 2q / p as p’ is zero = T/M1 = and 2 = + 90 deg.

HOLLOW SHAFTS SUBJECTED TO BENDING MOMENT AND TWISTING MOMENT:(WITHOUT ANY AXIAL FORCE)q = Tr/J = 32T(D/2)/π[D4 – d4] …..(1) p = pbt or pbc = M(D/2)/I = 32M(D)/ π[D4 – d4]……(2) Principal stresses are given by equations p1 and p2 = p/2 ± [(p/2)2 + q2]1/2

Inserting values of p and qp1 and p2 = 16D[ M ±( M2 + T2)1/2]/ π[D4 – d4]Maximum shear stress τ max = (p1 +p2)/2 = 16 D[( M2 + T2)1/2] /π[D4 – d4](Where D and d are outer and inner diameters of the shaft)

Q.1 A solid shaft 100mm diameter is subjected to a bending moment of 10kNm and a twisting moment of 12 kNm. Calculate principal stresses and position of plane on which they act. Ans: M = 10kNm and T = 12kNmPrincipal stresses are given by p1 and p2 = 16[ M ±( M2 + T2)1/2] / πd3

p1 and p2 = 16*106[10 ± ( 10 2 + 12 2 ] ½ / π100 3

= 130.48 MPa and 28.625MPa Maximum shear stress τ max = (p1 +p2)/2 = 16 [( M2 + T2)1/2] / πd3 =(p1 +p2)/2 = 50.928MPatan2 = T /M = 1.2, So 1 = 25.1 and 2 =115.1deg.

Q. 2 The maximum permissible shear stress in hollow shaft is 60MPa. For shaft the outer diameter shaft is twice the inner diameter. The shaft is subjected to a twisting moment of 10kNm and a bending moment of 12kNm. Calculate diameter of the shaft.Ans. Given T = 10kNm and M = 12kNmMaximum shear stress τ max = (p1 + p2)/2 = 16D[( M2 + T2)1/2] /π[D4 –d4]

60 = 16*D[ 122 + 102 ] ½ / π [ D4 - (D/2)4 ]Outer diameter D = 112.25M say 114 mmInner diameter d = 57mm

Torsion accompanied with bending and axial force: (either tensile or compressive axial force )

The following forces/moments may act of shafts:

*Axial force* Twisting moment and * Bending moment

*Direct stresses ’p’ due to axial force on shaftp = axial force/ cross sectional area of the shaft

For solid shafts, p = P/(πR2), R = radius of shaft

For hollow shafts, p = P/[π(R2 – r2) , Where R is outer radius of the hollow shaft and r is inner radius of the shaft

Shear stresses due to twisting moment on shaft:T/J = q/r = cθ/L (equation for twisting of shaft)

For solid shaft: Shear stress q = Tr/J = 16T/πd3

For hollow shaft:

Shear stress q = T(D/2)/[π(D4 – d4)/32]

= 16T/ [π(D4 – d4)]

Bending stresses due to bending moment:

Bending tensile or compressive stresses

For solid shafts:

pbt or pbc = M/Z = 32M/ πD3

= 32M/ πD3

For hollow shafts: pbt or pbc = My/I = M(D/2)/ [π(D4 – d4)/64], where D and d are outer and inner diameter of the shaft and M is bending moment.Stress at any point on the shaft (either hollow or solid) depends on the application of axial force, bending moment and twisting moment on the shaft.Principal stresses on shaft = p1 = (p+p’)/2 + [ {(p-p’)/2}2+q2]1/2 and p2 = (p+p’)/2 – [{(p-p’)/2}2+q2] 1/2

Where p = p + pbt (or pbc) i.e. algebraic sum of direct

and bending stresses and q is shear stresses due to

twisting moment

= 32M/ πD3

Principal stresses on shaft = p1 = (p+p’)/2 + [ {(p-p’)/2}2+q2]1/2 and p2 = (p+p’)/2 – [{(p-p’)/2}2+q2] 1/2

Where p = p + pbt (or pbc) i.e. algebraic sum of direct and

bending stresses and q is shear stresses due to twisting

moment. The p’ is zero in y direction.Principal stresses on shaft = p1 and p2 = (p)/2 ± [ {(p)/2}2+q2]1/2

Q. 3 A propeller shaft 200mm diameter transmits 2000kW at 300rpm. The propeller is over hanging by 500mm from support and has mass of 5000kg. The propeller thrust is 200kN. Calculate maximum direct stresses induced in the shaft. Also locate position of section where the maximum direct stresses are acting and point on cross section where the maximum direct stresses are act.

Ans.

1. Direct stress due to thrust = force / c. s. area

p = ± 200*1000/ π*1002 = 6.366MPa

2. Bending moment

M = 5000*9.81*500 Nmm (cantilever beam action )

Bending stress = M/Z = 5000*9.81*500/[ π 2003 / 32]

= 31.226MPa 3. Power, P = 2 πNT/ 60 or T = 60P/2 πN

T = 60*2000*106/2 π*300 = 63.66*106Nmm

Shear stress due to twisting moment

q = 16T/ πd3 = 16*63.66*106 / π*2003

= 40.527MPa.

Q. 3A propeller shaft 200mm diameter transmits 2000kW at 300rpm. The propeller is over hanging by 500mm from support and has mass of 5000kg. The propeller thrust is 200kN. Calculate maximum direct stresses induced in the shaft. Also locate position of section where the maximum direct stresses are acting and point on cross section where the maximum direct stresses are act.

Ans.

1. Direct stress due to thrust = force / c. s. area

σ = ± 200*1000/ π*1002 = 6.366MPa

2. Bending moment

M = 5000*9.81*500 Nmm (cantilever beam action )

Bending stress = M/Z = 5000*9.81*500/[ π 2003 / 32]

= 31.226MPa 3. Power, P = 2 πNT/ 60 or T = 60P/2 πN

T = 60*2000*106/2 π*300 = 63.66*106Nmm

Shear stress due to twisting moment

q = 16T/ πd3 = 16*63.66*106 / π*2003

= 40.527MPa.

Principal stresses on shaft = p1 and p2 = (p)/2 ± [ {(p)/2}2+q2]1/2

p = σ + pbc and p = σ - pbt p = 6.366 + 31.226 =37.592MPa at point B (comp.)p = 6.366 - 31.226 = - 24.86MPa at point A (tensile)

p = 6.366 MPa (bending stress = zero at points C & D shear stress q = 40.527MPa at circumference

shaft

A

B

C D

Principal stresses on shaft = At Point A p1 = -54.82MPa and p2 = 29,96MPa τ max = (p1 – p2)/ 2 = 42.39MPaAt Point B p1 = 65.166MPa and p2 = - 27.574MPa τ max = (p1 – p2)/ 2 = 46.37MPaAt Point C & D p1 = 43.835MPa and p2 = - 37.469MPa τ max = (p1 – p2)/ 2 = 40.652MPa

shaft

A

B

C D

Ductility is the property of the material by virtue of which it undergoes a great amount of deformation before rupture.

A material which undergoes a great amount of deformation before rupture is called ductile material. e.g. Mild Steel

Brittleness is the property of the material by virtue of which it can undergoes a little amount of plastic deformation before rupture.

A material which undergoes a very little amount of plastic deformation before rupture is called brittle material. e.g. Cast Iron

The material which is capable of recovering original size and shape on removal of load.

When the material is loaded, it will result in deformation. When the load is removed, the deformation will be disappeared. This behaviour of the material is known as elastic action.

The material which is not capable of recovering original size and shape on removal of load.

When the material is loaded, it will result in deformation. When the load is removed, the plastic deformation will remain in the material. This behaviour of the material is known as elastic action.

Strain energy:Strain energy: When load acts on the body, it undergoes deformation. The work that the body absorbs as a result of this deformation is called strain energy.

The area under the load-deformation diagram gives us the amount of energy stored.

Proof Resilience:Proof Resilience: If the body is deformed within the elastic limit, the stored strain energy is recovered on removal of deformation. This is called elastic strain energy or Proof Proof Resilience.Resilience.

The strain energy required per unit volume to rupture is called toughness of the material.

Higher Toughness - Ductile material

Lower Toughness - Brittle material

Impact test is a measure of toughness.

It is a property of the metals and alloys by virtue of which it deforms plastically under compression without rupture.

Gauge Length

Mild Steel Specimen For Tensile TestMild Steel Specimen For Tensile Test

Cup & Cone Type FailureCup & Cone Type Failure

Strain

Stress

O

A

BC

D

E

F

A - Limit of proportionality

B - Elastic Limit

C - Upper Yield Point

D - Lower Yield Point

E - Ultimate stress

F - Fracture

From point O: as the load increases, the strain(deformation) will also increases up to point C.

At point C : The load decreases slightly but the strain increases continuously up to point D.

From point D: load further starts increasing and strain also increases but at a faster rate up to point E.

IMPORTANT POINTSIMPORTANT POINTS::

From D to E, increase in length and decrease in area is substantial.

After point E, there starts forming a neck in the Centre.

The load cannot be further increased beyond point E and starts decreasing Upto point F and at point F, the specimen fails.

Strain

Stress

O

A

BC

D

E

F

better to plot P/A0 ==>L/L0

P/A0 = Load at any stage / original cross sectional area

P/A0 = Nominal stress

TRUE STRESS AND NOMINAL STRESS.TRUE STRESS AND NOMINAL STRESS.

To generalize the observations for any cross sectional area and any other gauge length,

To be more precise, it should be plotted between

P/A ==> L / L0 where,

P/ A = Load at any stage/ area measured at that load.

P/ A =True stress

PROPORTIONALITY LIMIT:PROPORTIONALITY LIMIT:

The point upto which the stress-strain curve is a straight line.

This maximum uniaxial stress Upto which the stress and strain are proportional is called proportion limit.

ELASTIC LIMIT:ELASTIC LIMIT:

For every material, a limiting value of stress is found Upto and within which the resulting strain entirely disappears when the load is removed. The value of this stress is known as the elastic limit.

YIELD POINT:YIELD POINT:

When a material is loaded beyond the elastic limit, the stress increases more quickly as the stress increased, up to point C.

The ordinate of point C, at which there is a slight increase in strain without increase in stress is known as the yield point of the material.

Permanent set :Permanent set :

Beyond point C if the load is removed there will be plastic or permanent deformation remained in the material. It is called permanent set.

Steel has two yield point, upper and lower.

STRAIN HARDENING:STRAIN HARDENING:

Loading and unloading within the elastic limit

Loading and unloading in the plastic zone.

Loading, unloading and reloading of MS in tensile test.Loading, unloading and reloading of MS in tensile test.

Strain

Stress

A

B

Strain

Stress

A

B

C

D

E

E’

F

F’

Unloading and reloading within the elastic limit causes no change and the graph returns on the same path.

Reloading again it will give us line DE, E is in continuation, with point C. i.e. curve BC, but at present still in elastic zone which means , the yield point stress is increased.

Unloading in the plastic zone will give us line CD, here AD is permanent set.

Point E’ is new yield point and we will get DE’G’ curve.

Strain

Stress

A

B

Strain

Stress

A

B

C

D

E

E’

F

F’

This raising of elastic limit (yield point) because of the process of unloading and reloading is known as the phenomenon of strain hardening.

This property of mild steel is used to increase the yield strength by cold drawing.

If L = length of the test bar at fracture.

l = length of the test bar before application of stress.

Than % elongation = (L - l) / l * 100 %

A = area before application of stress

A’ = area after application of stress

% reduction = (A- A’) / A *100 %

Measurement of Ductility:Measurement of Ductility:

(1) first method is based on the total elongation produced in the specimen.

(2) second method is based on total reduction in sectional area.

(1) Ideal Elastic Material:--The material continues to deform under the action of applied load up to failure. During any stage if the load is removed, it will regain its original shape and size.

Ideal PlasticIdeal Plastic

Linear

Non Linear

Different types of MaterialsDifferent types of Materials

(2) Ideal Plastic Material:--Up to certain stress

limit the material does not deform at all, but after that, it continuously deform to failure at constant stress.

(3) Ideal Rigid Material:--Up to the failure point the material will not deform at all.

(5) Elasto-Plastic Material:-Up to certain stress limit the material behave as perfectly elastic material then deforms as a plastic material up to failure.

RigidRigid

(4) Fluid :- The material will deform largely on the application of a very small load.

FluidFluid

Elastic Point Strain Hardened

Ideal

Strain softened

Strain HardenedStrain Hardened : After reaching the elastic limit, the material does not fail at the same value of stress, but after elastic limit as strain increases, the material is hardened and can resist still more load.

Elastic Point Strain Hardened

Ideal

Strain softened

Strain SoftenedStrain Softened : After reaching the elastic limit, the material does not fail at the same value of stress, but after elastic limit as strain increases, the material gets softened and the value of stress it can take decreases till failure.

There is no defined yield point. The breaking or rupture will be at low strain.To determine the yield strength in brittle material, it is general practice to draw a straight line parallel to the elastic portion of the curve at a predetermined strain ordinate value(0.1% to 0.2%). The point at which this line intersect the stress-strain curve, is called proof stress at 0.1% or 0.2% strain, or yield strength at 0.1% or 0.2% strain.

Linear

Failure

0.002

0.2% yield strength or0.2% proof strength

Parallel

Stress-Strain Curve for a Brittle Material:--Stress-Strain Curve for a Brittle Material:--

Stress-Strain Curve for Ferrous Metals:--Stress-Strain Curve for Ferrous Metals:--

Alloy SteelAlloy Steel

High Carbon SteelHigh Carbon Steel

Medium Carbon SteelMedium Carbon Steel

Mild Steel (Ductile)Mild Steel (Ductile)

Cast Iron (Brittle)Cast Iron (Brittle)

StressStress

StrainStrain

Stress-Strain Curve for Non-Ferrous Metals:--Stress-Strain Curve for Non-Ferrous Metals:--

Aluminium BronzeAluminium Bronze

Magnesium OxideMagnesium Oxide

BrassBrass

Annealed CopperAnnealed Copper

StressStress

StrainStrain

Hardness is the resistance of the material to permanent deformation of the surface.

It can be defined as the resistance of the metal to penetration by an indentor.

The main usefulness of hardness lies in the fact that, It seems to bear a fairly constant relationship to the tensile strength of a give material.

Empirical co-relations are developed to relate hardness with the tensile strength etc.

1.1. BRINNEL HARDNESS TEST:-BRINNEL HARDNESS TEST:-P

D

d

indentor

A standard hardened steel ball of diameter “D” is pressed into the surface of the specimen by a gradually applied load “P”, which maintained on the specimen for a definite period of time @ 10 to 30 seconds.

The impression of the steel ball so obtained is measured by a micrometer & the Brinnel Hardness Number (BHN) is found by following relation:

BHN = (Load in kg.) / (Surface area of indentation)

P (D/2) * (D - D2 - d2 )

= Kg/mm2

2.2. VICKER’SVICKER’S HARDNESS TEST:-HARDNESS TEST:-

specimen

Elevation

Plan

indentations

d1

d2

A square based diamond pyramid is used as indentor.

Vicker’s Pyramid Number,

VPN = 2P*sin(/2) / d2

P= Load in kg.

= Angle between opposite face of diamond pyramid.

d= Avg. length of two diagonals.

This method is used to determine the hardness of very thin & very hard materials.

3.3. ROCKWELL’S ROCKWELL’S HARDNESS TEST:-

The Rockwell Hardness Test is very widely used because of its speed & freedom from personal error.

This test requires much smaller penetretor than Brinnel Hardness Test.

For Soft Materials: B Scale is used,

its range is :RB 0 to RB100

For hard Materials : C Scale is used,

its range is :RC 20 to RC70

MODES OF FAILUREMODES OF FAILURE

Brittle Failure Ductile Failure

Cup & Cone Type Failure Fibrous Failure

Failure without large plastic deformation over c/s area perpendicular to the axis of the specimen.

Failure after considerable plastic stretching.

The specimen used is either cube or cylinder.

Failure takes place along an inclined plane ~ 55o.

Comp. Strength is much higher than tensile strength

e.g. Cast Iron

Failure does not take place, but the specimen deforms to the shape of a coin.

Tensile & Compressive strengths are comparable.

e.g. Mild Steel

Brittle Material Ductile Material

When one body having some momentum & energy strikes with other, the energy & momentum transfer takes place. The time of transfer is very small, hence large force is generated. This phenomenon is called Impact.

Impact test gives the measure of toughness of the material.

It is used to check the resistance offered by the material to shock or energy loads.

To design certain parts of machines, Impact test is required.

(1) IZOD IMPACT TEST

(2) CHARPY IMPACT TEST

(1) IZOD IMPACT TEST :-

This test uses a cantilever test piece of 10mm x 10mm section and 75mm length.

The “V” notch of angle 45 and depth of the notch is 2mm.

75mm

10mm

10mm

22mm

2mmV notch

The pendulum hammer strikes near its ends. The specimen is fixed in clamping vice.

IZOD IMPACT VALUEIZOD IMPACT VALUE

Energy absorbed by the specimen

C/S area below notch

Energy at Release - Final Energy

C/S area

Er - Ef

Ac

clamping vice.

=

=

=

(2) CHARPY IMPACT TEST :-

This test is most common. It uses a cantilever test piece of 10mm x 10mm section and 55mm length.

The “V” notch of angle 45 and depth of the notch is 2mm.

55mm

10mm

10mm

40mm

Hh

FATIGUE:FATIGUE:

Materials subjected to a repetitive or fluctuating stress will fail at a stress much lower than that required to cause fracture under steady loads. This behaviour is called fatigue.

Stress Cycles:Stress Cycles:

The simplest type of load is the alternate stresses.

Stress amplitude is equal to the maximum stress and the mean or average stress is zero.

-

max

min

Time

CREEP:CREEP:

In many engineering applications materials are required to sustain steady loads for long period.

If, under any conditions, deformation continues when the load Is constant, this additional deformation is known as creep.

The time dependent strain under the constant stress is also known as creep.

When the load is applied, the deformation will take place.

Most material creep to a certain extent at all temperature, although engineering materials such as steel, aluminum etc. creep very little at room temperature.

Such time dependent deformation can grow large and even result in final fracture without any increase in the load.

Strain is a fundamental physical quantity, while stress is more or less a mentalised concept.

The instrument which measure the strain is known as strain gauge.

BASIC CHARACTERSTICS OF A STRAIN GAUGE.

1 Gauge Length

2 Gauge sensitivity

3 Accuracy

4 Reproducibility

TYPES OF STRAIN GAUGES:TYPES OF STRAIN GAUGES:

The principle employed in strain gauge construction can be used as the basis for classifying the gauge into five following categories.

1 Mechanical

2 Optical

3 Electrical

4 Acoustical

5 Pneumatic

==>@ 150 different types of strain gauges are available

==> Mostly used is the electrical resistance type strain gauge.

TYPICAL ELECTRICAL RESISTANCE TYPE STRAIN GAUGE:TYPICAL ELECTRICAL RESISTANCE TYPE STRAIN GAUGE:

Wires

0.001 inch dia.

G.L.

Base of

paper

CementCover

Metal of the Wire used :- Advance , Iso elastic

Base- Acts as carrier as well as insulating agent , It transfers strain from the specimen to the gauge.

MECHANICS OF SOLIDS

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