MECHANICS OF SOLIDS

Mechanics of Solids Syllabus:Syllabus Part - A 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: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: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 Solids Syllabus:Syllabus Part - B 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: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: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:Distribution of Shear Stresses in Different Sections.

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 mechanics to find the support reactions due to the applied forces is normally covered under the subject of 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 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: 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 )

 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. Axial Force

Axis of the member Pull / Tens.

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

Flexural Forces

Axis of the member

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.

Axis of the member

Fixed end 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 Support: 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 beam: It is a beam which has one end, as fixed, and the other end as free.

free end

fixed end L

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: When the beam is continued beyond the support and behave as a cantilever then the combined beam is known as an over-hang beam. L1

L

L2

L

L1

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

L1

L2

L3

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. W kN point load

L

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

L

w kN/m

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. w kN/m L Total Load = ½ *w * L = (w*L)/2 kN

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

L

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;  Σ Fx = 0 ; Σ Fy = 0 ; Σ M=0;

A beam is said to be statically determinate if the total no. of unknown reactions are equal 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): Σ Fx = 0 ;

Σ Fy = 0 ;

A

Σ M=0 B

HA VA

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

HA MA

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

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

B

HA VA

VB

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

HB

HA MA

VA

VB

HA

HB MA

VA

VB

HA

HB MA

VA

VB

MB

We will limit our study to shear force and bending moment diagrams of STATICALLY DETERMINATE BEAMS. Commonly encountered statically determinate beams are, a) Cantilever Beam, b) Simply Supported Beam, c) Over-hanging Beam. Beam

These beams are usually subjected to the following types of loading; a) Point Load, b) Uniformly Distributed Load, c) Uniformly Varying Load, d) 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 crosssection. 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.

=

Y 25 KN

2m

10 KN/m 5m

25 KN SFYY = 25 –30

SFYY = 25 –20

= 5 kN

= 5 kN BMYY = 25*2 –10*2*1

BMYY = 25*3 –10*3*1.5

= 30 kN.m

= 30 kN.m Y

10 KN-m 25 KN

25 KN

5m 2m 3m dx

10 KN-m 25 KN **

30

25-20 = 5 kN ** 25.2 –10.2.1 = 30 kN.m

*

*

5 30

30

5

5

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

#

30

25 KN

30 kN.m

30 kN.m 5 kN

5 kN

The vertical force will try to shear-off the member as shown in the fig., hence it is known as the shear force. The unbalanced-moment will try to bend the member as shown in the fig., hence it is known as the Bending Moment. #To resist the shear force, the element will develop the resisting stresses, Which is known as Shear Stresses. #To resist the Bending Moment, the element will develop the resisting stresses, Which is known as bending Stresses.

SIGN CONVENTIONS +ve Shear

-ve Shear

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

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

+ve B.M.

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

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.

P

A Section at P

Q

w/u.l.

dx M

F

B

M F

Section at Q

F+dF M+dM

M+dM F+dF

Element PQ

F M

M+dM F+dF

P

A Element PQ

Q

w/u.l.

dx

B

F M

M+dM F+dF

Let, F = Shear Force at P F+ dF = Shear Force at Q M = Bending Moment at P M+ dM = Bending Moment at Q

Element PQ

F M

w/u.l. M+dM

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

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

Element PQ

F M

w/u.l. M+dM F+dF

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.

1

W

Sec X-X

B

A x W

L

Fx = W Mx = -WL + Wx

W

At A, x = 0; M= -WL At B, x = L; M= 0 W.L

+ve S.F. Dia. -ve θ (+ve) B.M. Dia.

W

2 Step 1:- Support reaction A ∑FY = 0, VA= 10 kN (up)

10

10 kN B x

5.0m

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

10

+ve

Fx = 10 S.F. Diagram

10

Step 2:- Bending Moment at x;

10 kN B

A x

M@x

10

= -50+10 * x Linear Equation

50

5.0m

-50+10 * x -ve θ (+ve) B.M. Dia.

At A, x = 0; M= -50 kN.m At B, x = 5; M= 0 kN.m So Linear change in B.M. value

10 kN B

A 10 10

5.0m

+ve S.F. Diagram

50

-ve θ (+ve) B.M. Dia.

10

3 Step 1:- Support reaction A ∑FY = 0, VA= 10 kN (up)

10

10 kN 4.0m x

B

5.0m

At B, Point load, So sudden change in S.F. value

10

+ve

10 0

S.F. Diagram

Step 2:- Bending Moment at x;

10 kN C

A x

M@x

10

= - 40+10 * x Linear Equation

40

5.0m

-ve θ (+ve) B.M. Dia.

At A, x = 0; M = -40 kN.m At C, x = 4; M = 0 kN.m So Linear change in B.M. value

B

10 kN B

A 10

10

40

5.0m

+ve

10 0

-ve θ (+ve)

0

4 Step 1:- Support reaction A ∑FY = 0, VA= 30 kN (up)

30

F(a) = 30 F(c-left) = 30 30 F(c-right) = 30-20 = 10 F(b-left) = 10 F(c-right) = 10-10 =0

10 kN

20 kN C 2.5m

2.5m

B

5.0m

+ve

20 10

S.F. Diagram

10 kN 20 kN C B A 2.5m 2.5m x 30 5.0m Mx (AC) = -100 +30 * x Lin. Eqn Step 2:- Bending Moment at x;

MA = -100 kN.m

MC = -25 kN.m

Mx (CB) = -100 + 30*(x) - 20*(x-2.5) MB = 0.0 kN.m 100

-ve

25

B.M. Dia.

10 kN

20 kN A

2.5m

30 30

C

2.5m

5.0m 20

+ve

10

S.F. Diagram 100

B

-ve

25

B.M. Dia.

0.0

5 S.F.D. Fx = wL – wx

Sec X-X

B.M.D.

B

A

( linear variation) wL FA = wL : FB = 0 wL

x

MA= -wL2/2 MB= 0

w.L2/2

L

θ (-ve) +ve

Mx = -wL2/2 + wLx - wx2/2 (parabolic variation)

w/m

S.F. Dia. -ve B.M. Dia. θ (+ve)

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.

6

Sec X-X

B

A 5m

50 50 kN

125 kNm

10 kN/m

θ (-ve) +ve S.F. Dia. -ve B.M. Dia. θ (+ve)

7

A

10 kN/m 2.5 m C 5m

B

7

31.25 Reactions : VA=25 kN MA=(10*2.5*1.25) = 31.25 kNm SFD: FA= 25 kN Fx(AC)= 25-10x Fx(CB)= 25-25=0 BMD MA = - 31.25 Mx(AC) = -31.25+ 25x-10x2/2 Mx(CB) = 0

A

x

25 25 kN

10 kN/m 2.5 m C 5m

+ve S.F. Dia.

31.25 kNm

-ve B.M. Dia.

B

8

10 kN/m A

2.5 m 5m

C

B

8

93.75

Reactions : VA=25 kN MA=(10*2.5*3.75) = 93.75 kNm SFD: FA= 25 kN Fx(AC)=25 Fx(CB)=25-10(x-2.5) BMD MA = - 93.75 Mx(AC) = -93.75+25x Mx(CB) = -93.75+25x -10(x-2.5)2/2

A 25 25 kN

10 kN/m 2.5 m x C 5m

B

+ve S.F. Dia. -ve

31.25

Parabola 93.75 betn BC linear B.M. Dia.

9

10 kN/m A

C 2m

D 3m 2m 7m

B

9

140

Reactions : A x VA =10*2 +10*2 2m =40kN MA=(10*2*6+10*2*1) 40 = 140 kNm

SFD: FA= 40 kN Fx(AC)= 40-10x Fx(DB)=40-20-10(x-5)

10 kN/m C D 3m

2m

7m

40 +ve

B

20 S.F. Dia.

9

140 A

10 kN/m C D x 2m

3m

B 2m

7m 40 BMD MA = - 140 Mx(AC) = -140 + 40x-10x2/2 Mx(CD) -140 + 40x-20*(x-1) Mx(DB) -140 + 40x-20*(x-4)-10(x-5)2/2 -ve 140

80

20 Parabola

linear Parabola B.M. Dia.

10

10 kN/m A

2m

20kN c

2m

d

8m

e 2m

10 kN/m B 2m

10 kN/m

10 Reactions : VA =10*2 +20+10*2 =60kN MA=(10*2*7+20*4+ 10*2*1) = 240 kNm

A 60

60 kN

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

x 2m

20kN C 2m

10 kN/m D E B 2m 2 m

8m

40 +ve 20 S.F. Dia.

10

240

BMD

A

MA = - 240

10 kN/m x 2m

20kN c

60 Mx(AC)= -240+ 40x-

2m

10 kN/m d B e x 2m 2 m

8m

10x2/2 Mx(CD)=-240 + 40x20(x-1) Mx(DE) =-240 + 40x20(x-1)-20(x-4) Mx(EB) = -10x /2 2

(Considering Right portion)

-ve

60

140

20

Parabola linear

linear 240 Parabola

B.M. Dia.

11 w/m A

L

B

11

w.L /3 A 2

wx/L Reactions : VA =(1/2)Lw=wL/2 x L MA=(1/2)L*w*(2L/3) wL/2 2 = wL /3 SFD: FA= wL/2 wL/2 Fx= wL/2-{x*(wx/L)}/2 +ve = wL/2-wx2/2L S.F. Dia. BMD MA = -wL2/3 -ve 2 Mx = -wL /3 + wL/2*x B.M. Dia. {wx2/2L*(x/3)} = -wL2/3 + wL/2*x θ (+ve) wx3/6L w.L2/3

w/m B

θ -ve

12

10 kN/m A 5m

B

12

10 kN/m

83.33kN.m

Reactions : VA =(1/2)*10*5 =25kN MA=(1/2)10*5*(2*5/3) = 83.33 kN.m SFD: FA= 25kN Fx= 25-(x*2x/2)=25-x2

BMD MA = 83.33kN.m Mx = -83.33 +25x-x2*x/3 = -83.33+25x–x3/3

A

2x

x 25kN 25

+ve

B

5m

S.F. Dia.

-ve B.M. Dia. θ (+ve) 83.33

θ -ve

13

w/m A L

B

13

w/m w.L2/6

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

A

x

wL/2 wL/2

SFD: Fx=(wx/L)*x/2 = wx2/2L (Right side) FA= wL/2 BMD Mx = (wx2/2L)*x/3 = wx3/6L MA = -wL2/6

wx/L L

θ -ve +ve

w.L2/6 -ve

S.F. Dia.

B.M. Dia. θ (+ve)

B

14

10 kN/m A

B 5m

14

10 kN/m 41.67 Reactions : VA =(1/2)10*5 A =25kN 25 kN MA=(1/2)10*5*(5/3) = 41.67kN.m SFD: Fx= 2x*x/2=x2 FA= 25kN BMD Mx =x2(x/3) = x3/3 MA = 41.67 kN.m

25

2x x 5m

θ -ve +ve S.F. Dia.

41.67

B

-ve B.M. Dia. θ (+ve)

15

100 kN.m B

A 10 m

15 100 kN.m

A VA =0

10 m

x

100 kN.m B

S.F. Dia.

+ve B.M. Dia.

100

16

100 kN.m A

C 10 m 6m

B

100 kN.m A

C

VA =0

10 m

6m S.F. Dia. +ve B.M. Dia.

100

B

A

B

HA VA

VB

1

W A

B L

1

W A

B L

W/2

W/2

W/2

+ ve

W/2 W/2

- ve

S.F. Diagram

1

W A

B L

W/2

W/2 WL/4

+ ve

B.M. Diagram

1

W

A

B

L W/2 W/2

S.F. Diagram

W/2 + ve

W/2

- ve

W/2

WL/4

B.M. Diagram

+ ve

2

5

C

A

10m

5 5

S.F. Diagram

10kN

+ ve

- ve 25

B.M. Diagram

5

5 5

+ ve

B

2

4m

A

10kN C

B

10m 6 6

S.F. Diagram

4 + ve - ve

4 24

B.M. Diagram

+ ve

4

3

4m

A

10kN 5kN C D

4m

B

10 m 8 8

7 + ve

S.F. Diagram

2 7

28

32

B.M. Diagram

- ve

+ ve

7

4

10kN A

3m

15kN

5kN

3m

C

D 10 m

3m

E

B

5

A WL/2

W kN/m C L

B WL/2

W kN/m

A

C

WL/2

L

WL/2 WL/2

B

+ ve - ve WL2/8 + ve

WL/2

6

A

10 kN/m C 10 m

B

10 kN/m

A

C

50

10 m

50 50

B

+ ve - ve 125 kN.m + ve

50

7

10 kN/m C

A

10 m 5m

B

A

10 kN/m C

37.5

10 m

37.5

B 12.5

5m

+ ve

- ve

3.75 m

70.3125 kN.m + ve

12.5

8

A

10 kN/m C 10 m

B

10 kN/m C

A 12.5 12.5

B

10 m

37.5

5m

3.75 m

+ ve

- ve 70.3125 kN.m

+ ve

37.5

9

C

A 3m

10 kN/m 4m 10 m

D 3m

B

C

A 3m 20

10 kN/m 4m 10 m

D

B

3m 20

20 20 S.F. Diagram

C

A

10 kN/m 4m 10 m

3m 20

D 3m

20

80 60

B

60

B.M. Diagram

C

A

10 kN/m 4m 10 m

3m 20

D

B

3m 20

20 20 80 60

60

10

10 kN/m C D

A 3m

4m 10 m

B 3m

10 kN/m C D

A 30

4m

3m

B 3m

30

30 30

45

45

11

10kN A

1.5

3m

10 kN/m C D 4m 10 m

B 3m

10kN A 38.5

1.5

10 kN/m C D 4m

3m

B 3m

31.5

23.5 38.5 13.5 Mmax will occur at x=2.85m from A

1.5

2.85

46.5

31.5

55.61 55.5

49.5

12

W kN/m A

B L

W kN/m A WL/3

B L

WL/6 L /√ 3

WL/3

WL/6 WL2/(9√ 3)

13

9 kN/m A

B 9m

9 kN/m A 27

B 9

13.5 5.2

27 13.5 46.76

14

9 kN/m A

B 9m

9 kN/m B

A 9

13.5

27

5.2 13.5

27 46.76

15 100 kN.m C

A

5m

10 m

B

15 A

100 kN.m C

10 10

B 10

+ ve

+ ve

50

-50 - ve

10

16

100 kN.m

50 kN.m

A

B 10 m

16

100 kN.m C

A 5 5

100

50 kN.m B 5

- ve

+ ve

5

50

17 10 10

10√ 2

10

1

1

3

45

30

A

4

2

6m

1

1

B

10 4.66

A

C

D

8.66

10

E

F

6m

16.33

16.33

8

10

5

6

B 16.67

10 5 10 8

16.67

10 4.66

A

C

D

8

10

5 8.66

10

E

F

6

6m

16.33

16.33

22.33

B 16.67

25 16.67

A

4.66

C

D

8.66

10

E

F

6

6m

Ten:– +ve Comp:- -ve

4 4.66

+ ve

- ve

- ve 6

Thrust diagram

B

Two beams AC and CB are connected by an Internal Pin to form beam ACB. Draw SF and BM diagram. 6 kN/m A

C 15m

5m

20 kNm D 5m

B

C

6 kN/m A 30

15 15m 15 C 5m

20 kNm D 5m

B 170 15

C

6 kN/m A

15 15m

30

15 C 5m

20 kNm D 5m

8.66m

30

170 15

+ ve - ve

S.F. will be zero at a distance of L/ √ 3 = 8.66m from C

B

C

6 kN/m A 30

15 15m

15 8.66m

86.6

C 5m

20 kNm D 5m

B 170 15

+ ve 0.0

75 20

- ve

170

25 kN/m

3.0 m

100

3.0 m

25 kN/m 2.0m

25 kN/m A

100 C

97.9

B

25 kN/m

127.1

97.9 27.9

50 72.1

D

25 kN/m A

100 C

97.9

B

25 kN/m

127.1 181.2

50

D

50

15 kN/m

15 kN/m 6.0 m

2

1

10

1

17.32

2.0m 2

20

50

15 kN/m

C

A x

15 kN/m

D

10

17.32

B

F

E

6.0 m

6.67

20

2.0m 48.33

1.49 m

6.67

5 8.33

Fx = 10 - 48.33 + 30 + 7.5x2/2 = 0 X = 1.49 m

10.0 38.33

50

15 kN/m

C

A

D

15 kN/m E

6.0 m 6.67

F

48.33 35

6.67

17.32

B 2.0m

1.49 m

1.597

10

26.27

15

20

20

50

50 A

C

50 B

D 1

6

E 2

50

50 A

C

B D 6

1

2

66.67 16.67 50

50

83.33

16.67 50

E

50

50 33.33

50

50 A

C

50 B

D 6

1 66.67

2 83.33

- 50 - 100

E

50 A C

15 kN/m

50 B

D 1

6

E 2

50 A C

50

15 kN/m B D 6

1 86.67

2

E

103.33 2.445 m

50

36.67 50

50

50 53.33

50 A C

15 kN/m

50 B

D 6

1 86.67

2 103.33

- 5.17 - 50 - 100

E

25 A C

25 kN/m

25 B

D 1

6

E 2

Given SFD. Draw Loading diagram and from it, BMD. 24 14.4 9.6 21.6 3

1

1

2

24 14.4 1.8 m

9.6 21.6 12

8 kN/m 14.4

12 kN/m 45.6

24 14.4 1.8 m

9.6 21.6

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

A

4.5

C

D

0.5

2.0

E

3.0

B

+ ve 0.0

20.0 20.25

18.0

0.0

A

4.5

C

D

0.5 2.0

E

3.0

B

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)

+ ve 0.0

20.25

20.0

18.0

Area of SFD between 0.0 CD = 20.25-20 = 0.25 = ½ *0.5*Y Y = 1 kN.

9 kN

Y = 1 kN 6 kN.

BMD Linear = So SF constant Area of SFD between EB = 18 = 3.0*Z Z = 6 kN.

A

4.5

C

D

0.5

2.0

E

3.0

B

+ ve 0.0

20.25

20.0

18.0

0.0

5.0 2 kN/m 9.0

6.0

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 Planes:-The planes which carry only Direct Stresses and no Tangential Stresses are called “ Principal Planes ”

Principal Stresses:-The intensity of stress on the Principal Planes are known as “ Principal Stresses ”

2

pt

θ

1

p

p pn 1

2

All planes parallel to this planes are principal plane with zero principal 2 1 stress

θ

pt p

p pn 1

2

All planes parallel to this planes are Principal plane with principal stress =p

Importance 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)

2

1 45

p

p 1

2

Case:- 1 Element Subjected to Direct Stress & 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 θ with BC. Force on face BC = P1 = p * BC * t Force on face BC = P2 = q * BC * t Force on face CE = P3 = q * CE * t Force on face CE

p'

D

q C

E

θ

p Aq

p'

p B

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

θ

P4

P3

C

θ

θ

sθ o c P1

P1

B

P2 θ

D

θ n i P 2s

qC

E

θ

p

θ

sθ co P2

Pn

P4

sin P1

Pt

θ

θ sin P3

E

θ os P 4c

θ s o P 3c

p'

Aq

p'

Force on face BC = P1 = p * BC * t Force on face BC = P2 = q * BC * t Force on face CE = P3 = q * CE * t Force on face CE

p B

Resolving forces perpendicular & parallel to BE, Pn = (P1+P3)cos θ + (P2+P4)sin θ Pt = (P1+P3) sin θ - (P2+P4)cos θ pn = Pn / (BE.t) =[(p*BC*t +q*CE*t) Cosθ +(q*BC*t + p’ *CE*t) Sinθ ] / (BE.t) = p cos2θ + q sinθ cosθ + q sinθ cosθ + p’ sin2θ = p/2 (1+cos2θ ) + p’/2 (1-cos2θ ) + q sin2θ pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ + q sin2θ ……….(I) pt = Pt / (BE.t) =[(p*BC*t +q*CE*t) Sinθ - (q*BC*t + p’ *CE*t) Cosθ ] / (BE.t) = p sinθ cosθ + q sin2θ - q cos2θ - p’ sinθ cosθ

To locate Principal Planes:-pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ + q sin2θ ……….(I) pt = {(p - p’)/2} * sin2θ - q cos2θ

……………….……(II)

The Principal Planes are the plane which carry no Tangential stress. Therefore pt = 0 at pt = 0, let θ = α {(p - p’)/2} * sin2α - q cos2α = 0 {(p - p’)/2} * sin2α = q cos2α tan2α = q / {(p - p’)/2 } tan2α = 2q-1/ (p - p’) α 1 = [ tan { 2q / (p - p’) } ]/2 α = 90 + α

2q

√ (p

- p’) 2

2α p - p’

+ 4q 2

tan2α = 2q / (p - p’) There are two values of α which will satisfy the above eq. α 1 & α 2, where α 2=90+ α

√ (p 2q 1

- p’) 2 2α

+ 4q 2

p - p’

sin2α 1=2q/(√(p-p’)2+4q2) & cos2α 1=(p-p’)/(√(p-p’)2+4q2) sin2α 2=-2q/(√(p-p’)2+4q2)& cos2α 2= -(p-p’)/(√(p-p’)2+4q2) Substituting value of sin2α 1 & cos2α 1, in eq.(I), p1 = {(p + p’)/2} + {(p - p’)/2}* cos2α 1 + q sin2α 1…….(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 cos2α 2 & sin2α 2 in eq.(I),

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

p’ 90 θ

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

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

p2 C p1

pt

pn

p2cosθ

B θ

p1 A

p2

θ

p2

C pt

p2sinθ

B

p1cosθ

θ p 1

pn θ

p1sinθ

A

Consider a plane AC inclined at angle θ 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 = P1sinθ - P2cosθ

P2cosθ

= p1 AB t sinθ - p2 BC t cosθ pt = Pt / (AC. t) = p1 AB t sinθ - p2 BC t cosθ AC. t = p1sinθ (AB/AC) - p2 cosθ (BC/AC) = p1sinθ cosθ - p2 sinθ cosθ pt = (p1- p2)sinθ cosθ

θ

P2 P2sinθ

C Pt Pn

B

P1cosθ

θ

θ

P1

P1sinθ

A

The Shear stress pt = (p1- p2)sinθ cosθ ; pt

= (p1- p2)sinθ cosθ ; = (p1- p2)/2 sin2θ

pt max.

sin2θ = max.

H

E 45 135

G

F

sin2θ =+/- 1 2θ = 90 or 270 θ = 45 or 135 When θ = 45, (EF) ptmax = (p1- p2)/2 When θ = 135,(GH) ptmax = (p1- p2)/2 Planes of maximum shear are always at Right angles to each other (Inclined at 45 & 135 with principal planes.)

p1 = (p+p’)/2 + √ {(p-p’)/2}2+q2 …Major Principal Stress p2 = (p+p’)/2 - √ {(p-p’)/2}2+q2 …Minor Principal Stress Location of Principal Planes; tan2α = 2q / (p - p’) (From previous slide) (α anticlockwise from vertical) ptmax = (p1- p2)/2 = √ {(p-p’)/2}2 + q2 (at 45 & 135 with principal planes in anticlockwise) α p1

p2

pt1

p1 α 45

pt2

p2

Case:- 1 Element Subjected to Direct Stress & Shear Stresses :-pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ + q sin2θ …….….(I) = (p cos2θ + q sinθ cosθ + q sinθ cosθ + p’

sin2θ )

p1 = (p+p’)/2 + √ {(p-p’)/2}2+q2 (Major Prin. Stress )(III) pt = {(p - p’)/2} * sin2θ - q cos2θ ……………….……(II) p2 = (p+p’)/2 - √ {(p-p’)/2}2+q2 (Minor Prin. Stress)(IV) Location of Principal Planes; tan2α = 2q / (p - p’) (α anticlockwise from vertical) (V) ptmax = (p1- p2)/2 = √ {(p-p’)/2}2 + q2 _ _ (VI) (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 cos2θ + q sinθ cosθ + q sinθ cosθ + p’ sin2θ ) as p’= q= zero we have σn = σcos230 = 2* cos230 = 1.5MPa and tangential stress σt = {(p - p’)/2} * sin2θ - q cos2θ , 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. Glued joint P

30deg

P

With vertical the inclination of the joint is 60deg. σt = {(p - p’)/2} * sin2θ - q cos2θ , as p’= q= zero shear stress q = (p/2)*sin 2θ So 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. 100MPa θ = 30deg. 120MPa

θ

120MPa

100MPa pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ + q sin2θ _(I) σn = {(p - p’)/2} * sin2θ - q cos2θ _ _ (II) as q = zero pn = (120+100)/2 + [(120 – 100)/2 ]cos 2*30 = 115MPa and σn = [(120 -100)/2]sin 2*30 = 8.66MPa. . . .Ans

Q.6 Two mutually perpendicular planes are stresses as shown in fig. calculate normal and tangential stress on an oblique plane AB. 60MPa θ = 30deg. 120MPa

θ

120MPa

60MPa pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ + q sin2θ _(I) σn = {(p - p’)/2} * sin2θ - q cos2θ _ _ (II) as q = zero pn = (120- 60)/2 +[ (120 + 60)/2 ]cos 2*30 = 75 MPa σn = [(120 + 60)/2]sin 2*30 = 77.94 MPa. . . .Ans

Example:-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. 350N/mm2 400N/mm2 50

400N/mm2

800N/mm2

Solution: Given, p = 800 N/mm2, q = 400 N/mm2

p’= 350 N/mm2, θ = 50

(i) pn & pt pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ + q sin2θ …….….(I) = 575 - 39.07 + 393.93 = 929.85 N/mm2 pt = {(p - p’)/2} * sin2θ - q cos2θ = 221.58 - ( - 69.46) = 291.04 N/mm2

………………..……(II)

(ii) Principal Planes:-p1 = (p+p’)/2 + √ {(p-p’)/2}2+q2 …Major Principal Stress = 575 + 458.94 = 1033.94 N/mm2 p2 = (p+p’)/2 - √ {(p-p’)/2}2+q2 …Minor Principal Stress = 575 - 458.94 = 116.06 N/mm2 (iii) Location of Principal Planes:-tan2α = 2q / (p - p’) α = [ tan-1 { 2q / (p - p’) } ]/2 α 1 = [ tan-1 { 2q / (p - p’) } ]/2 = 30.32° (ACW from p) α 2 = 90 + θ

1

= 90 + 30.32 =120.32 ° (with vertical.)

(iv) ptmax = (p1- p2)/2 = √ {(p -p’)/2}2 + q2 = 458.94 N/mm2 (at 45° with principal planes.)

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. 100N/mm2 200N/mm2 30

200N/mm2

500N/mm2

Solution: Given, p = 500 N/mm2, q = 200 N/mm2

p’= -100 N/mm2, θ = 30

(i) pn & pt pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ + q sin2θ …….….(I) = 200 + 150 + 173.20 = 523.20 N/mm2 pt = {(p - p’)/2} * sin2θ - q cos2θ = 259.80 - (100) = 159.80 N/mm2

………………..……(II)

(ii) Principal Planes:-p1 = (p+p’)/2 + √ {(p-p’)/2}2+q2 …Major Principal Stress = 200 + 360.55 = 560.55 N/mm2 p2 = (p+p’)/2 - √ {(p-p’)/2}2+q2 …Minor Principal Stress = 200 - 360.55 = -160.55 N/mm2 (iii) Location of Principal Planes:-tan2α = 2q / (p - p’) α = [ tan-1 { 2q / (p - p’) } ]/2 α 1 = [ tan-1 { 2q / (p - p’) } ]/2 = 16.85° (with vertical.) α 2 = 90 + θ

1

= 90 + 16.85 =106.85° (with vertical.)

(iv) ptmax = (p1- p2)/2 = √ {(p -p’)/2}2 + q2 = 360.55 N/mm2 (at 45° with principal planes.)

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. 150N/mm2 320N/mm2 60

320N/mm2

650N/mm2

Solution: Given, p = - 650 N/mm2, q = 320 N/mm2

p’= - 150 N/mm2, θ = 60

(i) pn & pt pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ + q sin2θ …….….(I) = - 400 + 125 + 277.13 = 2.13 N/mm2 pt = {(p - p’)/2} * sin2θ - q cos2θ = - 216.50 - ( - 160) = - 56.50 N/mm2

………………..……(II)

(ii) Principal Planes:-p1 = (p+p’)/2 + √ {(p-p’)/2}2+q2 …Major Principal Stress = - 400 + 406.08 = 6.08 N/mm2 p2 = (p+p’)/2 - √ {(p-p’)/2}2+q2 …Minor Principal Stress = - 400 - 406.08 = - 806.08 N/mm2 (iii) Location of Principal Planes:-tan2α = 2q / (p - p’) α = [ tan-1 { 2q / (p - p’) } ]/2 α 1 = [ tan-1 { 2q / (p - p’) } ]/2 = - 26.00° (with vertical.) α 2 = 90 + θ 1 = 90 + 16.85 =+ 64.00° (with vertical.) (iv) ptmax = (p1- p2)/2 = √ {(p -p’)/2}2 + q2 = 406.08 N/mm2 (at 45° with principal planes.)

Mohr’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 angle

D

Shear Stress (pt)

O

p'

qC

E

θ

p Aq

p'

p B

Normal Stress (pn)

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)

Mohr’s Stress Circle Method:-Case:-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 θ to the plane carrying stress p are, pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ + q sin2θ …….….(I) pt = {(p - p’)/2} * sin2θ - q cos2θ

……………….……(II)

p1 = (p+p’)/2 + √ {(p-p’)/2}2+q2 …Major Principal Stress p2 = (p+p’)/2 - √ {(p-p’)/2}2+q2 …Minor Principal Stress Location of Principal Planes; tan2α = 2q / (p - p’) ptmax = (p1- p2)/2 = √ {(p-p’)/2}2 + q2

Steps to Draw Mohr’s Circle:-

P’(CD)

pt

p

q

Aq pn

Q O

Q’

p' p

C

q

D p'

C p B

q P(BC)

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.

(CD)P’

pt q O

p2

Q

B Q’

p' p p1

C

A

q

pn

P(BC)

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 (p ) OB = p .

D

p q O

Aq

(CD)P’

pt p2

Q

B Q’

p' p p1

α

C

2α

q

q

p'

C p B

A pn

P(BC)

7. From plane CP measure angle as 2α anticlockwise

up to CA, therefore the principal plane will be located at α anticlockwise from face BC.

T pt

(CD)P’

q O

Q

B

p2

Q’ p'

p p1

C

A

2α

q

pn

P(BC)

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.

q

D

C

θ

p Aq

p'

p q

pt O

(CD)P’

B

R

2θ

Q’ pn p'

R’ p

C

pt Q q

pn

P(BC)

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

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.

T

pt

P’ q Q

B

O

Q’

T’

p' p p2

C 2α

p1

p - p’

q P

A pn

let QCP = 2α then tan2α = 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’) ptmax = 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 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. 350N/mm2

80

400N/mm2

400N/mm2 800N/mm2

350N/mm2

400N/mm2

80

800N/mm2

400N/mm2

P’

pt pt O

Example:-10

R

400

pn

160 Q Q’ R’

350N/mm2 800N/mm2

pn 400

C P

350N/mm2

Example:-10

400N/mm2

α

800N/mm2

T

400N/mm2

P’

pt 400N/mm2

O

p2

Q

B Q’

C 2α

350N/mm2 800N/mm

2

p1

P

A pn 400

Solution:--

Example:-10

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 = 2α = ___ therefore, α 1 =30.32& α 2=120.32 CT = ptmax = Maximum Tangential Stress = 458.94MPa

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. 100N/mm2 200N/mm2 30

200N/mm2

500N/mm2

100

Example:-11

200 500

30

200

pt

500

P’ 200

- pn

Q’ 100

R pt O C

60

pn

500

pn

Q R’ 200 P

100

Example:-11

200

α

500

pt

500

200

T

P’ 200

- pn

Q’

B

100

Q O C 2α

A 200

p2

500 p1

P

pn

Solution:--

Example:-11

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 = 2α = ___ therefore, α 1 = 16.85& α 2=106.85 CT = ptmax = Maximum Tangential Stress=360.55MPa

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. 650N/mm2 320N/mm2 60

320N/mm2

150N/mm2

650

Example:-12

320 60

320

150

R

pt

P’ 320 R’ Q

pn

Q’ 320

C

120

pn P

650

pt O

150

650

Example:-12

320

α

150

320

T

pt

P’ 320

pn

C

B Q’

p2 650

QO A

2α 320

P 150

p1

Solution:--

Example:-12

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 = 2α = ___ therefore, α 1 = ___& α 2=___ CT = ptmax = Maximum Tangential Stress = _______

Case:- 2 Element Subjected to direct stress Stresses 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.

p' C p

pt

pn

B θ

p A

p'

1. Element Subjected to Direct & Shear Stresses pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ + q sin2θ …….….(I) pt = {(p - p’)/2} * sin2θ - q cos2θ

……………….……(II)

p1 = (p+p’)/2 + √ {(p-p’)/2}2+q2 …Major Principal Stress p2 = (p+p’)/2 - √ {(p-p’)/2}2+q2 …Minor Principal Stress Location of Principal Planes; tan2α = 2q / (p - p’) ptmax = (p1- p2)/2 = √ {(p-p’)/2}2 + q2 Case-2 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 & ptmax can be found out by substituting q=0 in equations of CASE- 1.

pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ + q sin2θ …….….(I) pt = {(p - p’)/2} * sin2θ - q cos2θ

……………….……(II)

p1 = (p+p’)/2 + √ {(p-p’)/2}2+q2 …Major Principal Stress p2 = (p+p’)/2 - √ {(p-p’)/2}2+q2 …Minor Principal Stress Location of Principal Planes; tan2α = 2q / (p - p’) ptmax = (p1- p2)/2 = √ {(p -p’)/2}2 + q2 For CASE-2 Substituting q = 0 in above equations; pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ ………….…….….(I) pt = {(p - p’)/2} * sin2θ …………. ……………….……(II) p1 = p/2 + √ p2/4

= p …Major Principal Stress

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

Case:- 2 Element Subjected to Direct Stress only and no shear stress:-p' C p

pn = (p + p’)/2 + {(p - p’)/2} cos2θ pt = {(p - p’)/2} sin2θ

pr =√ pn2+ pt2 ptmax = (p - p’)/2

pt

pn p'

B θ

p A

Example:-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 (ptmax ) 450 N/mm2 50

800 N/mm2

800 N/mm2 450 N/mm2

Solution:-

450 50

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

800

p' = + 450 N/mm2 (Tensile) θ

= 50

pn = (p + p’)/2 + {(p - p’)/2} cos2θ pt = {(p - p’)/2} sin2θ

= 594.61 N/mm2

= 172.34 N/mm2

pr=√ pn2+ pt2 = 619.08 N/mm2 ptmax = (p - p’)/2 = 175 N/mm2

Example:-14 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 30 with vertical as shown in figure. Also find Maximum Tangential stress (ptmax ) 200 N/mm2 30

500 N/mm2

500 N/mm2 200 N/mm2

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

200 30

p' = + 200 N/mm2 (Tensile) θ

= 30 pn = (p + p’)/2+ {(p - p’)/2} cos2θ = - 325 N/mm2 pt = {(p - p’)/2} sin2θ

= - 303.11 N/mm2

pr=√ pn2+ pt2 = 444.41 N/mm2 ptmax = (p - p)/2 = - 350 N/mm2

500

Example:-15 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 40 with vertical as shown in figure. Also find Maximum Tangential stress (ptmax ) 300 N/mm2 40

600 N/mm2

600 N/mm2 300 N/mm2

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

300 40

p' = - 300 N/mm2 (Comp.) θ

= 40 pn = (p + p’)/2 + {(p - p’)/2} cos2θ = - 476.04 N/mm2

pt = {(p - p’)/2} sin2θ

= - 443.16 N/mm2

pr=√ pn2+ pt2 = 650.39 N/mm2 ptmax = (p - p’)/2 = - 450 N/mm2

600

Mohr’s Stress Circle Method:-Case:-II If the tangential stresses acting on element is zero, then the normal and tangential stresses across a plane inclined at angle θ to the plane carrying p are, pn = (p + p’)/2 + {(p - p’)/2} cos2θ pt = {(p - p’)/2} sin2θ

p

p' C pt pn p'

B θ

p A

Steps to Draw Mohr’s Circle:-

pt

O

A

B p'

C

pn

p 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.

P

pt

O

pr

pt

2θ

B p'

Q

pn

C

A pn

p 4. Draw line CP at angle “2θ ” 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

pt

O

pr

pt

2θ

B p'

T

Q

C

A pn

pn p 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.

pt P pr

pt

2θ

B

O

Q

p'

T

C

p - p’

pn p

A pn

Example:-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. Also find Maximum Tangential stress (ptmax ) 450 N/mm2 50

800 N/mm2

800 N/mm2 450 N/mm2

pt

Example:-16

50

800 N/mm2 P

450 N/mm

2

pr O

pt

100

B p' = 450 N/mm2 pn p = 800 N/mm2

T

Q

C

A pn

Solution:--

Example:-16

Measure: OQ = pn = Normal stress = _____ PQ = pt = Tangential Stress = ______ OR = pr = Resultant Stress = _______ CT = ptmax = Maximum Tangential Stress = _______

Example:-17 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 30 with vertical as shown in figure. Use Mohr’s Circle Method. Also find Maximum Tangential stress (ptmax ) 350 N/mm2 30

150 N/mm2

150 N/mm2 350 N/mm2

Example:-17

30

pt

150 N/mm2 350 N/mm

2

- pn

T

B C p' = 350

P pr

pt

60

O Q pn p = 150

A pn

Solution:--

Example:-17

Measure: OQ = pn = Normal stress = _____ PQ = pt = Tangential Stress = ______ OR = pr = Resultant Stress = _______ CT = ptmax = Maximum Tangential Stress = _______

Example:-18 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. Also find Maximum Tangential stress (ptmax ) 800 N/mm2 40

600 N/mm2

600 N/mm2 800 N/mm2

Example:-18

40

600 N/mm2

pt P

T

800 N/mm2 pr

pt B pn

C

80

Q

A p =- 600 N/mm2 pn

p' = - 800 N/mm2

O

Solution:--

Example:-18

Measure: OQ = pn = Normal stress = _____ PQ = pt = Tangential Stress = ______ OR = pr = Resultant Stress = _______ CT = ptmax = Maximum Tangential Stress = _______

Case :-3 Stress in One Direction Only:-Consider an element subjected to stress p in only one direction as shown. Here, the p’ = 0 & q = 0.

2 p

θ

1 p 1

2

1. Element Subjected to Direct & Shear Stresses pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ + q sin2θ …….….(I) pt = {(p - p’)/2} * sin2θ - q cos2θ

……………….……(II)

p1 = (p+p’)/2 + √ {(p-p’)/2}2+q2 …Major Principal Stress p2 = (p+p’)/2 - √ {(p-p’)/2}2+q2 …Minor Principal Stress Location of Principal Planes; tan2θ = 2q / (p - p’) ptmax = (p1- p2)/2 = √ {(p-p’)/2}2 + q2 Case-3 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 & ptmax can be found out by substituting p’ & q = 0 in eq. of CASE- 1.

pn = {(p + p’)/2} + {(p - p’)/2}* cos2θ + q sin2θ …….….(I) pt = {(p - p’)/2} * sin2θ - q cos2θ

……………….……(II)

p1 = (p+p’)/2 + √ {(p-p’)/2}2+q2 …Major Principal Stress p2 = (p+p’)/2 - √ {(p-p’)/2}2+q2 …Minor Principal Stress Location of Principal Planes; tan2θ = 2q / (p - p’) ptmax = (p1- p2)/2 = √ {(p -p’)/2}2 + q2 For CASE-3 Substituting p’ & q = 0 in above equations; pn = p /2 + p /2* cos2θ = p cos2θ ………….…….….(I) pt = p /2 * sin2θ ………………..……………….……(II) p1 = p/2 + √ p2/4

= p …Major Principal Stress

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

Case :-3 Stresses in One Direction only :-pn = p cos2θ pt = p/2 sin2θ pr = √ pn2 + pt2 ptmax = p/2

(Inclined at 45 with Principal Planes)

2

θ

1

p

p 1

2

Example:-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. Also find Maximum Tangential stress (ptmax ) 50

800 N/mm2

800 N/mm2

Solution:-pn = p cos2θ

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

pt = p/2 sin2θ = (800/2) * sin(100) = 393.93 N/mm2 pr = √ pn2 + pt2 = 514.23 N/mm2 ptmax = p/2 (Inclined at 45 with Principal Planes) = 800/2 = 400 N/mm2 50

800 N/mm2

800 N/mm2

Example:-20 Find p1, p2 & θ if the stresses on two planes are as shown in figure. 2

100 N/mm2

C A

θ

60 N/mm2 30 N/mm2

D

m m / N B 50

100

50 B

C A

60

θ

30 D

P’(CD)

30 C B

Q’

p2

Q A

C’

2θ

50

60 N/mm2 100 N/mm2

p1

P(AB)

STEPS:-1. Select origin O and draw axis OX. STEPS 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

Example:-21 Find p1, p2 & θ if the stresses on two planes are as shown in figure. C 900 N/mm2 30

B

θ 30

A

1200 N/mm2

D

Principal Strain:-Strain If p1, p2 & p3 are principal stresses, then the principal strains e1, e2 & e3 are given by; e1 = p1/E - µ p2/E - µ p3/E (in direction of p1) e2 = p2/E - µ p1/E - µ p3/E (in direction of p2) e3 = p3/E - µ p1/E - µ p2/E (in direction of p3) p2

p1 p3

Example:-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 µ = 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) µ = 0.3, and E = 2*10 5 N/mm2 e1 = p1/E - µ p2/E - µ p3/E (in direction of p1) = ______ e2 = p2/E - µ p1/E - µ p3/E (in direction of p2) = ______ e3 = p3/E - µ p1/E - µ p2/E (in direction of p3) = ______

Example:-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; µ = 0.25 450 MPa 300 MPa

600 MPa

300 MPa

250

450 MPa

600 MPa

Solution:-- E = 2 * 10 5 N/mm2; µ = 0.25 p = 600 Mpa , p’ = 450 MPa, q = 300 Mpa p1 = (p+p’)/2 + √ {(p-p’)/2}2+q2 …Major Principal Stress = 525 + 309.23 =834.24 MPa p2 = (p+p’)/2 - √ {(p-p’)/2}2+q2 …Minor Principal Stress = 525 - 309.23 = 215.76 Mpa e1 = p1/E - µ 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

e2 = p2/E - µ 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 250

250.975 250.009

tan2α = 2q / (p - p’) = 600 / (600 - 450) = 4 α 1 = 38° (Anticlockwise with vertical) α 2 = 90° + 38° = 128°(with vertical) d1= 250.975 mm ; d2 = 250.009 mm 450 MPa 38°

300 MPa P1

600 MPa

600 MPa d2

300 MPa

d1 P2 450 MPa

Example:-24 Solve following 4 example using Mohr’s circle method. 400 N/mm2

800 N/mm2 200 N/mm2

800 N/mm2 200 N/mm2

400 N/mm2

200 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 tan2α = 2q / (p - p’) gives tan2α = 2q / p as p’ is zero

For position of principal planes tan2α = 2q / (p - p’) gives tan2α = 2q / p as p’ is zero = T/M α 1 = α and α 2 = α + 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 q p1 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 = 12kNm Principal 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 = 50.928MPa tan2α = T /M = 1.2, So α 1 = 25.1 and α 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 = 12kNm Maximum 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 mm Inner 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 shaft p = 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

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

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.

A C

D

shaft B 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

A C

D

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

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: 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:

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

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.

Mild Steel Specimen For Tensile Test

Gauge Length

Cup & Cone Type Failure

A - Limit of proportionality B - Elastic Limit

E

C - Upper Yield Point D - Lower Yield Point

Stress

B

C A

E - Ultimate stress

F

D

F - Fracture O

Strain

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

E

Stress

B

C A

O

F

D

Strain

IMPORTANT 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.

TRUE STRESS AND NOMINAL STRESS. To generalize the observations for any cross sectional area and any other gauge length, better to plot P/A0 ==>δ L/L0 P/A0 = Load at any stage / original cross sectional area P/A0 = Nominal stress To be more precise, it should be plotted between P/A ==> δ L / L0

where,

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

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: 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: 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.

Steel has two yield point, upper and lower. 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.

STRAIN HARDENING:

E’ C

B

Stress

A

Loading and unloading within the elastic limit

E

B

Stress

Strain

F’

A

F

D Strain

Loading and unloading in the plastic zone.

Loading, unloading and reloading of MS in tensile test.

E’ C B

Stress

A

E

B

Stress

Strain

F’

A

F

D Strain

Unloading and reloading within the elastic limit causes no change and the graph returns on the same path. Unloading in the plastic zone will give us line CD, here AD is permanent set. 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. Point E’ is new yield point and we will get DE’G’ curve.

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.

Measurement of Ductility: (1) first method is based on the total elongation produced in the specimen. 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 % (2) second method is based on total reduction in sectional area. A = area before application of stress A’ = area after application of stress % reduction = (A- A’) / A *100 %

Different types of Materials (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. (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.

σ

Non Linear

Linear

ε σ Ideal Plastic

ε

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

σ

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

σ

(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.

Rigid

ε

Fluid

σ

Elastic Point

ε Strain Hardened Ideal Strain softened

ε

Strain 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. Strain 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.

σ

Elastic Point

Strain Hardened Ideal Strain softened

ε

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

σ

0.2% yield strength or 0.2% proof strength Linear Parallel 0.002

ε

Stress-Strain Curve for Ferrous Metals:-Alloy Steel High Carbon Steel Medium Carbon Steel

Stress

σ Mild Steel (Ductile) Cast Iron (Brittle)

Strain

ε

Stress-Strain Curve for Non-Ferrous Metals:-Magnesium Oxide Aluminium Bronze Brass

Stress

σ

Annealed Copper

Strain

ε

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.

BRINNEL HARDNESS TEST: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:

P

indentor D

d

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

P (π D/2) * (D - √ D2 - d2 )

Kg/mm2

2.

VICKER’S HARDNESS TEST:-

A square based diamond pyramid is used as indentor. Vicker’s Pyramid Number,

θ

VPN = 2P*sin(θ /2) / d2 P= Load in kg.

specimen

θ = Angle between opposite face of diamond pyramid.

Elevation

d= Avg. length of two diagonals. This method is used to determine the hardness of very thin & very hard materials.

d1 indentations

d2 Plan

3.

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 FAILURE Brittle Failure

Fibrous Failure

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

Ductile Failure

Cup & Cone Type Failure

Failure after considerable plastic stretching.

The specimen used is either cube or cylinder. Brittle Material

Failure takes place along an inclined plane ~ 55o. Comp. Strength is much higher than tensile strength e.g. Cast Iron

Ductile Material

Failure does not take place, but the specimen deforms to the shape of a coin. Tensile & Compressive strengths are e.g. Mild Steel

comparable.

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. 22mm V notch 2mm

75mm

10mm 10mm

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

40mm H h

10mm 10mm

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. σ + σ σ -

σ

max

min

Time

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.

CREEP: In many engineering applications materials are required to sustain steady loads for long period. When the load is applied, the deformation will take place. 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. 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: 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:

G.L.

Wires 0.001 inch dia. Base of paper Cover

Cement

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.