Mechanics of Solids [3 1 0 4] CIE 101 / 102 First Year B.E...

Mechanics of Solids [3 1 0 4]

CIE 101 / 102

First Year B.E. Degree

Mechanics of Solids

PART- IIPART- I

Mechanics of Deformable Bodies

Mechanics of Rigid Bodies

COURSE CONTENT IN BRIEF

1. Resultant of concurrent and non-concurrent coplanar forces.

2. Equilibrium of concurrent and non-concurrent coplanar forces.

3. Centroid of plane areas

4. Moment of Inertia of plane areas

5. Kinetics: Newton’s second law, D’Alembert’s principle, Work- Energy, and Impulse- Momentum principle.

PART I Mechanics of Rigid Bodies

PART II Mechanics of Deformable bodies6. Simple stresses and strains

7. Statically indeterminate problems and thermal stresses

8. Stresses on inclined planes

9. Stresses due to fluid pressure in thin cylinders

Books for Reference1.Engineering Mechanics, by Meriam & Craige, John Wiley & Sons.

2.Engineering Mechanics, by Irwing Shames, Prentice Hall of India.

3.Mechanics for Engineers, by Beer and Johnston, McGraw Hills Edition

4.Engineering Mechanics, by K.L. Kumar, Tata McGraw Hills Co.5. Machanics of Materials, by E.P.Popov

6. Machanics of Materials, by E J Hearn

7. Strength of materials, by Beer and Johnston

8. Strength of materials, by F L Singer & Andrew Pytel

9. Strength of Materials, by B.S. Basavarajaiah & P. Mahadevappa

10. Strength of Materials, by Ramamruthum

11. Strength of Materials, by S S Bhavikatti

PART - I

MECHANICS OF RIGID BODIES

Definition of Mechanics :In its broadest sense the term ‘Mechanics’ may be defined as the ‘Science which describes and predicts the conditions of rest or motion of bodies under the action of forces’.

INTRODUCTION

PART - I Mechanics of Rigid Bodies

This Course on Engineering Mechanics comprises of Mechanics of Rigid bodies and the sub-divisions that come under it.

Engineering Mechanics

Mechanics of Solids Mechanics of Fluids

Rigid Bodies DeformableBodies

Statics Dynamics

Kinematics Kinetics

Strength of Materials

Theory of Elasticity

Theory of Plasticity

Ideal Fluids

ViscousFluids

CompresFluids

Branches of Mechanics

Concept of Rigid Body :

It is defined as a definite amount of matter the parts of which are fixed in position relative to one another under the application of load.

Actually solid bodies are never rigid; they deform under the action of applied forces. In those cases where this deformationis negligible compared to the size of the body, the body may be considered to be rigid.

Particle

A body whose dimensions are negligible when compared to the distances involved in the discussion of its motion is called a ‘Particle’.

For example, while studying the motion of sun and earth, they are considered as particles since their dimensions are small when compared with the distance between them.

Force

It is that agent which causes or tends to cause, changes or

tends to change the state of rest or of motion of a mass.

A force is fully defined only when the following four

characteristics are known:

(i) Magnitude

(ii) Direction

(iii) Point of application

(iv) Sense.

Force:

characteristics of the force 100 kN are :

(i) Magnitude = 100 kN

(ii) Direction = at an inclination of 300 to the x-axis

(iii) Point of application = at point A shown

(iv) Sense = towards point A

300

100 kN

A

Scalars and Vectors

A quantity is said to be a ‘scalar’ if it is completely defined by

its magnitude alone.

Example : Length, Area, and Time.

A quantity is said to be a ‘vector’ if it is completely defined only

when its magnitude and direction are specified.

Example : Force, Velocity, and Acceleration.

Principle of Transmissibility : It is stated as follows : ‘The external effect of a force on a rigid body is the same for all points of application along its line of action’.

P PA B

For example, consider the above figure. The motion of the block will be the same if a force of magnitude P is applied as a push at A or as a pull at B.

The same is true when the force is applied at a point O.

P PO

1. RESULTANT OF COPLANAR FORCES

Resultant, R : It is defined as that single force which can replace a set of forces, in a force system, and cause the same external effect.

R

θ

=

same isA particle,on effect 321

externalFFFR ++=F3

F1

F2

A A

Resultant of two forces acting at a point

Parallelogram law of forces : ‘If two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, then the resultant of these two forcesis represented in magnitude and direction by the diagonal of the parallelogram passing through the same point.’

B C

AO

P2

P1

Rα

θ

Contd..

B C

AO

P2

P1

Rα

θ

In the above figure, P1 and P2, represented by the sides OA and OB haveR as their resultant represented by the diagonal OC of the parallelogram OACB.

It can be shown that the magnitude of the resultant is given by:R = √P1

2 + P22 + 2P1P2Cos α

Inclination of the resultant w.r.t. the force P1 is given by:

θ = tan-1 [( P2 Sin α) / ( P1 + P2 Cos α )]

Resultant of two forces acting at a point at right angle

B C

AO

P2

P1

Rα

θ

If α = 900 , (two forces acting at a point are at right angle)

222

1 PPR +=

1

2tanPP

=θθ

B C

AO

P2

P1

R

Triangle law of forces

‘If two forces acting at a point can be represented both in magnitude and direction, by the two sides of a triangle taken in tip to tail order, the third side of the triangle represents both in magnitude and direction the resultant force F, the sense of the same is defined by its tail at the tail of the first force and its tip at the tip of the second force’.

Triangle law of forcesLet F1 and F2 be the two forces acting at a point A and θ is the included angle.

θA

F1

F2

θF1

F2

R=

‘Arrange the two forces as two sides of a triangle taken in tip to tail order, the third side of the triangle represents both in magnitude and direction the resultant force R.

the sense of the resultant force is defined by its tail at the tail of the first force and its tip at the tip of the second force’.

Triangle law of forces

θF1

F2

R

θA

F1

F2

=

β

α

R

F1

F2 )180sin(sinsin21

βααβ −−==

RFF

(180 - α - β) = θ

where α and β are the angles made by the resultant force with the force F1 and F2 respectively.

Component of a force :

Component of a force, in simple terms, is the effect of aforce in a certain direction. A force can be split into infinite number of components along infinite directions.

Usually, a force is split into two mutually perpendicularcomponents, one along the x-direction and the other along y-direction (generally horizontal and vertical, respectively).

Such components that are mutually perpendicular are called‘Rectangular Components’.

The process of obtaining the components of a force is called ‘Resolution of a force’.

Rectangular component of a force

θx

F

θx

F

θx

F

Fx

Fy = Fy

Fx

Consider a force F making an angle θx with x-axis.

Then the resolved part of the force F along x-axis is given by Fx = F cos θx

The resolved part of the force F along y-axis is given by

Fy = F sin θx

Oblique component of a force

Let F1 and F2 be the oblique components of a force F. The components F1 and F2 can be found using the ‘triangle law of forces’.

β

α

F

F1

F2

The resolved part of the force F along OM and ON can obtained by using the equation of a triangle.

βα

F

F1

F2 M

O

N

F1 / Sin β = F2 / Sin α = F / Sin(180 - α - β)

Sign Convention for force components:

+ve

+vex

xyy

The adjacent diagram gives the sign convention for force components, i.e., force components that are directed along positive x-direction are taken +ve for summation along the x-direction.

Also force components that are directed along +ve y-direction are taken +ve for summation along the y-direction.

Classification of force system

Force system

Coplanar Forces Non-Coplanar Forces

Concurrent Non-concurrentConcurrent Non-concurrent

Like parallel Unlike parallelLike parallel Unlike parallel

A force that can replace a set of forces, in a force system,and cause the same ‘external effect’ is called the Resultant.

RESULTANT OF COPLANAR NON CONCURRENT FORCE SYSTEM

Coplanar Non-concurrent Force System:This is the force system in which lines of action of

individual forces lie in the same plane but act at different points of applications.

Fig. 1

F2F1

F3

Fig. 2

F1 F2

F5F4

F3

1. Parallel Force System – Lines of action of individual forces are parallel to each other.

2. Non-Parallel Force System – Lines of action of the forces are not parallel to each other.

MOMENT OF A FORCE ABOUT AN AXIS

The applied force can also tend to rotate the body about an axis in addition to motion. This rotational tendency is known as moment.

Definition: Moment is the tendency of a force to make a rigid body to rotate about an axis.

This is a vector quantity having both magnitude and direction.

MOMENT OF A FORCE ABOUT AN AXIS

Moment Axis: This is the axis about which rotational tendency is determined. It is perpendicular to the plane comprising moment arm and line of action of the force (axis 0-0 in the figure)

Moment Center: This is the position of axis on co-planar system. (A).

Moment Arm:Perpendicular distance from the line of action of the force to moment center. Distance AB = d.

Magnitude of moment:

It is computed as the product of the of the force and the perpendicular distance from the line of actionto the point about which moment is computed. (Moment center).

MA = F×d

= Rotation effect because of the force F, about the point A (about an axis 0-0)

Unit – kN-m, N-mm etc.

The sense is obtained by ‘Right Hand Thumb’ rule.

‘If the fingers of the right hand are curled in the direction of rotational tendency of the body, the extended thumb represents the sense of moment vector’.

Sense of moment:

For the purpose of additions, the moment direction may be considered by using a suitable sign convention such as +vefor counterclockwise and –vefor clockwise rotations or vice-versa.

MM

VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)

Statement: The moment of a force about a moment center or axis is equal to the algebraic sum of the moments of its component forces about the same moment center (axis).

θ

P

A

dθ

P

A

P sinθ

P cosθd1

d2

Moment of Force P about the point A,

P x d

Algebraic sum of Moments of components of the Force P about the point A,

P cosθ x d1 + P sinθ x d2

=

VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)

Proof (by Scalar Formulation):Proof (by Scalar Formulation):

Let ‘R’ be the given force.

‘P’ & ‘Q’ are component forces of ‘R’. ‘O’ is the moment center.

QR

αAO

β p

qr P

γ

Y

X

p, r and q are moment arms from ‘O’of P, R and Q respectively.

α, β and γ are the inclinations of ‘P’, ‘R’ and ‘Q’ respectively w.r.to X –axis.

We have,Ry = Py + QyR Sinβ = P Sinα + Q Sin γ ----(1)From ∆le AOB, p/AO = Sin αFrom ∆le AOC, r/AO = Sin βFrom ∆le AOD, q/AO = Sin γFrom (1),∴ R ×(r/AO) = P ×(p/AO) + Q ×(q/AO)

i.e., R × r = P × p + Q × q

Moment of resultant R about O = algebraic sum of moments of component forces P & Q about same moment center ‘O’.

Ry

Py

R

A O

Q

P

p

q r

α

βγ

Y

X

B

CD

Qy

COUPLE

Two parallel, non collinear (separated by certain

distance) forces that are equal in magnitude and opposite

in direction form ‘couple’.

d

F

F

The algebraic summation of the

two forces forming couple is zero.

=Hence, couple does not produce any

translation and produces only rotation. M = F x d

RESOLUTION OF A FORCE INTO A FORCE-COUPLE SYSTEM

Replace the force F acting at the point A to the point B F

AB

Apply two equal and opposite forces of same magnitude & direction as Force F at point B, so that external effect is unchanged

F

AB

F

Fd

Of these three forces, two forces i.e., one at A and the other oppositely directed at B form a couple. Moment of this couple, M = F × d.

Thus, the force F acting at a point such as A in a rigid body can be moved to any other given point B, by adding a couple M. The moment of the couple is equal to moment of the force in its original position about B.

Third force at B is acting in the same direction as that at P.

F

AB

F

F

AB

F

=M = F x dd

TYPES OF LOADS ON BEAMS

1. Concentrated Loads – This is the load acting for very small length of the beam.

(also known as point load, Total load W is acting at one point )

2. Uniformly distributed load – This is the load acting for a considerable length of the beam with same intensity of w kN/m throughout its spread.

Total intensity, W = w × L

(acts at L/2 from one end of the spread)

W kN

w kN/m

L

W = (w x L) kN

L

L/2

3. Uniformly varying load – This load acts for a considerable length of the beam with intensity varying linearly from ‘0’ at one end to w kN/m to the other representing a triangular distribution.

Total intensity of load = area of triangular spread of the load

W = 1/2× w × L.

(acts at 2×L/3 from ‘Zero’ load end)

w kN/m

L

L

W = ½ × L × w

2/3 ×L 1/3 ×L

EXERCISE PROBLEMS1. Resultant of force system

Q1 A body of negligible weight, subjected to two forces F1= 1200N, and F2=400N acting along the vertical, and the horizontal respectively, is shown in figure. Find the component of each force parallel, and perpendicular to the plane.

= 1200 N

X

34

YF2

F1

= 400 N

Ans : F1X = -720 N, F1Y = -960N, F2X = 320N, F2Y = -240N


Q2. Determine the X and Y components of each of the forces shown in the figure.

30º40º

12

5

300 N

390 N

400 N

X

Y

F1 =

F2 =

F3 =

(Ans : F1X = 259.81 N, F1Y= -150 N, F2X= -150N, F2Y= 360 N,

F3X = -306.42 N, F3Y= -257.12N )


Q3. Obtain the resultant of the concurrent coplanar forces shown in the figure

600N

200N

800N

20º 40º

30º

(Ans: R = 522.67 N, θ = 68.43º)


Q4. A disabled ship is pulled by means of two tug boats as shown in FIG. 4. If the resultant of the two forces T1 and T2 exerted by the ropes is a 300 N force acting parallel to the X – direction, find :

(a) Force exerted by each of the tug boats knowing α = 30º.(b) The value of α such that the force of tugboat 2 is minimum,

while that of 1 acts in the same direction.Find the corresponding force to be exerted by tugboat 2.

T2

R = 300 N

T1

α

20º

FIG. 4

X - direction

( Ans: a. T1= 195.81 N, T2 = 133.94 Nb. α = 70º, T1 = 281.91 N, T2(min) = 102.61 N )


Q5. An automobile which is disabled is pulled by two ropes as shown in the figure. Find the force P and resultant R, such that R is directed as shown in the figure.

P

Q = 5 kN

R20º

40º

(Ans: P = 9.4 kN , R = 12.66 kN)


Q6. A collar, which may slide on a vertical rod, is subjected to three forces as shown in figure. The direction of the force F may be varied Determine the direction of the force F, so that resultant of the three forces is horizontal, knowing that the magnitude of F is equal to(a) 2400 N, (b)1400N

1200 N

800 N60º

θF

RODCOLLAR

( Ans: a. θ = 41.81º ; b. The resultant cannot be horizontal.)

Q7. Determine the angle α and the magnitude of the force Q such that the resultant of the three forces on the pole is vertically downwards and of magnitude 12 kN. Refer figure


8kN5kN

Q30º

α

Fig. 7(Ans: α = 10.7 º, Q = 9.479 kN )

Q8. Determine the resultant of the parallel coplanarforce system shown in figure.

400 N

1000 No

60º

60º 30º10º

(Ans. R=800N towards left, d=627.5mm)


600 N

2000 N

Q9. Four forces of magnitudes 10N, 20N, 30N and 40Nacting respectively along the four sides of a squareABCD as shown in the figure. Determine the magnitude, direction and position of resultant w.r.t. A.

20N

10N

40N

a

a

30N

AB

C

(Ans:R=28.28N, θ=45º, x=1.77a)


D

Q10. Four parallel forces of magnitudes 100N, 150N, 25N and 200N acting at left end, 0.9m, 2.1m and 2.85m respectively from the left end of a horizontalbar of 2.85m. Determine the magnitude of resultantand also the distance of the resultant from the leftend.

(Ans: R = 125 N, x = 3.06 m)


Q11. Reduce the given forces into a single force and a couple at A.

70.7 kN200 kN

1.5m

A1m

100 N80 N

30º45º

30º

(Ans:F=320kN, θ=14.48º, M=284.8kNm)


Q12. Determine the resultant w.r.t. point A.

(Ans: R = 450 kN, X = 7.5 kNm)


A

150 N

1.5m3m1.5m

150 Nm

500 N100 N

2. EQUILIBRIUM OF FORCE SYSTEMS


EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS

Definition:-If a system of forces acting on a body, keeps the body in a state of rest or in a state of uniform motion along a straight line, then the system of forces is said to be in equilibrium.ALTERNATIVELY, if the resultant of the force system is zero, then, the force system is said to be in equilibrium.

EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS

Conditions for Equilibrium :

A coplanar concurrent force system will be in equilibrium if itsatisfies the following two conditions:

i) ∑ Fx = 0; and ii) ∑ Fy = 0

i.e. Algebraic sum of components of all the forces of the system, along two mutually perpendicular directions, is ZERO.

XY

Graphical conditions for Equilibrium

Triangle Law: If three forces are in equilibrium, then, they form a closed triangle when represented in a Tip to Tail arrangement, as shown in Fig 2.1

Fig 2.1

F 3 F2

F1

F2

F3

F1

Polygonal Law: If more than three forces are in equilibrium, then, they form a closed polygon when represented in a Tip to Tail arrangement, as shown in Fig. 2.2.

Fig 2.2 F5

F 4 F3

F2

F1

F5

F4

F3 F2

F1

LAMI’S THEOREM

If a system of three forces is in equilibrium, then, each force of the system is proportional to sine of the angle between the other two forces (and constant of proportionality is the same for all the forces). Thus, with reference to Fig.2.3, we have,

F2

β

α

γ

Fig. 2.3

F1

F3

γβα SinF

SinF

SinF 321 ==

Note: While using Lami’s theorem, all the three forces should be either directed away or all directed towards the point of concurrence.

EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCE SYSTEM

When a body is in equilibrium, it has neither translatory nor rotatory motion in any direction.

Thus the resultant force R and the resultant couple M are both zero, and we have the equilibrium equations for two dimensional force system

∑ Fx = 0; ∑ Fy = 0 ∑M = 0

These requirements are both necessary and sufficient conditions for equilibrium.

SPACE DIAGRAMS & FREE BODY DIAGRAMS

Space Diagram (SPD) : The sketch showing the physical conditions of the problem, like, the nature of supports provided; size, shape and location of various bodies; forces applied on the bodies, etc., is known as space diagram.

eg, Fig 2.4 is a space diagramWeight of sphere = 0.5 kN,Radius = 1m

Cable

P = 2kN30°

Fig. 2.4 SPDwall

3 m θ

Sphere

Free Body Diagram (FBD) :

It is an isolated diagram of the body being analyzed (called free body), in which, the body is shown freed from all its supports and contacting bodies/surfaces. Instead of the supports and contacting bodies/surfaces, the reactive forces exerted by them on the free body is shown, along with all other applied forces.

A Few Guidelines for Drawing FBD

1) Tensile Force: It is a force trying to pull or extend the body. It is represented by a vector directed away from the body.

2) Compressive Force: It is force trying to push or contract the body. It is represented by a vector directed towards the body.

3) Reactions at smooth surfaces: The reactions of smooth surfaces, like walls, floors, Inclined planes, etc. will be normal to the surface and pointing towards the body.

4)Forces in Link rods/connecting rods: These forces will be acting along the axis of the rod, either towards or away from the body.(They are either compressive or tensile in nature).

5) Forces in Cables (Strings or Chords): These can only be tensile forces. Thus, these forces will be along the cable and directed away from the body.

T = Tension in the cable

Rw = Reaction of the wall

W = self weight of the sphere

P = external load acting on the sphere

Free Body Diagrams of the sphere shown in Fig. 2.4

Fig. 2.5 F B D of Sphere

P = 2kN

30°

θ

Sphere

T

W=0.5kN

Rw

Detach the sphere from all contacts and replace that with forces like:Cable contact is replaced by the force tension = TContact with the smooth wall is replaced by the reaction Rw.

Supports: A structure is subjected to external forces and transfers these forces through the supports on to the foundation. Therefore the support reactions and the external forces together keep the structure in equilibrium.

There are different types of supports.a) Roller Support b) Hinged or pinned support c) Fixed or built in support

Supports

Types of Supports Action on body

(a) Flexible cable ,belt ,chain, ropeBODY

TBODYForce exerted by cable is always a tension away from the body in the direction of cable

(b) Smooth surfaces

Contact forces are normal to the surfaces

F F

900

900

Supports

(c) Roller support

A

A

Contact force is normal to the surface on which the roller moves. The reaction will always be perpendicular to the plane of the roller . Roller support will offer only one independent reaction component. (Whose direction is known.)

Supports

(d) pinned Support / hinged support

RvR

Rh

θ

AA

This support does not allow any translatory movement of the rigid body. There will be two independent reaction components at the support. The resultant reaction can be resolved into two mutually perpendicular components. Or it can be shown as resultant reaction inclined at an angle with respect to a reference direction.

Supports

(e) Fixed or Built-in Support

RAV

MRAH

A A

This type of support not only prevents the translatorymovement of the rigid body, but also the rotation of the rigid body. Hence there will be 3 independent reaction components of forces. Hence there will be 3 unknown components of forces, two mutually perpendicular reactive force component and a reactive moment as shown in the figure.

TYPES OF BEAMS

A member which is subjected to predominantly transverse loads and supported in such a way that rigid body motion is prevented is known as beam. It is classified based on the support conditions. A beam generally supported by a hinge or roller at the ends having one span (distance between the support) is called as simply supported beam. A beam which is fixed at one end and free at another end is called as a cantilever beam.

span

A B

span

A B

(a) Simply supported beam

TYPES OF BEAMS

B

span

A

Rv

MRH A B

(b) Cantilever beam

TYPES OF BEAMS

If one end or both ends of the beam project beyond the support it is known as overhanging beam.

A B

A B

(c) Overhanging beam (right overhang)

Statically determinate beam

Using the equations of equilibrium given below, if all the reaction components can be found out, then the beam is a statically determinate beam

the equations of equilibrium∑ Fx = 0; ∑ Fy = 0 ∑M = 0

FRICTION

Friction is defined as the contact resistance exerted by one body upon another body when one body moves or tends to move past another body. This force which opposes the movement or tendency of movement is known as frictional resistance or friction. Friction is due to the resistance offered by minute projections at the contact surfaces. Hence friction is the retarding force, always opposite to the direction of motion. Friction has both advantages & disadvantages.

Disadvantages ---- Power loss, wear and tear etc.

Advantages ---- Brakes, traction for vehicles etc.

FRICTION

NF (Friction)

P

W

Hills & Vales Magnified Surface

Frictional resistance is dependent on the amount of wedging action between the hills and vales of contact surfaces. The wedging action is dependent on the normal reaction N.

FRICTION

Frictional resistance has the remarkable property of adjusting itself in magnitude of force producing or tending to produce the motion so that the motion is prevented.

When P = 0, F = 0 block under equilibrium

When P increases, F also increases proportionately to maintain equilibrium. However there is a limit beyond which the magnitude of this friction cannot increase.

FRICTION

When the block is on the verge of motion(motion of the block is impending) F attains maximum possible value, which is termed as Limiting Friction. When the applied force is less than the limiting friction, the body remains at rest andsuch frictional resistance is called the static friction.

Further if P is increased, the value of F decreases rapidly and then remains fairly a constant thereafter. However at high speeds it tends to decrease. This frictional resistance experienced by the body while in motion is known as Dynamic friction OR Kinetic Friction.

FRICTION

Sliding friction friction experienced when a body slides over another surface.

Dynamic FrictionRolling friction friction experienced by a body when it rolls over a surface.

FRICTION

Where Fmax = Limiting Friction

N= Normal Reaction between the contact surfaces

µ =Coefficient of friction

Fmax

N

Fmax

P

W

φ

N R

F α N

Fmax = µN

µ =

Note : Static friction varies from zero to a maximum value. Dynamic friction is fairly a constant.

FRICTIONAngle of Friction

The angle between N & R depends on the value of F.

This angle θ, between the resultant R and the normal reaction N is termed as angle of friction.

Fmax

P

W

φ

N RAs F increases, θ also increases and will reach to a maximum value of φ when F is Fmax (limiting friction)

i.e. tanφ = (Fmax )/N = µ

Angle φ is known as Angle of limiting Friction.

Angle of limiting friction is defined as the angle between the resultant reaction (of limiting friction and normal reaction) and the normal to the plane on which the motion of the body is impending.

Angle of reposeWhen granular material is heaped, there exists a limit for the inclination of the surface. Beyond that angle, the grains start rolling down. This limiting angle upto which the grains repose (sleep) is called the angle of repose of the granular material.

FRICTION

FRICTION

Significance of Angle of repose:

The angle that an inclined plane makes with the horizontal, when the body supported on the plane is on the verge of motion due to its self -weight is equal to the angle of repose.

Angle of repose is numerically equal to Angle of limiting friction

FRICTIONLaws of dry friction

1. The magnitude of limiting friction bears a constant ratioto the normal reaction between the two surfaces. (Experimentally proved)

2. The force of friction is independent of the area of contactbetween the two surfaces.

3. For low velocities the total amount of friction that canbe developed is practically independent of velocity.It is less than the frictional force correspondingto impending motion.

Q1. A 10kN roller rests on a smooth horizontal floor and is held by the bar AC as shown in Fig(1). Determine the magnitude and nature of the force in the bar AC and reaction from the floor under the action of the forces applied on the roller. [Ans:FAC=0.058 kN(T),R=14.98 kN]

C

7kN

5kN

Fig(1)

A

450

300


EXERCISE PROBLEMS

Q2. A 10 kN weight is suspended from a rope as shown infigure. Determine the magnitude and direction of the least force P required to pull the rope, so that, the weight is shifted horizontally by 0.5m. Also, determine, tension in the rope in its new position. [Ans: P= 2.43 kN, θ = 14.480 ; T= 9.7kN.]

2m

10kN

Pθ

2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS

Q3. Determine the value of P and the nature of the forces in thebars for equilibrium of the system shown in figure.[Ans: P = 3.04 kN, Forces in bars are Compressive.]

60

7545 45

P2kN


Q4. A cable fixed as shown in Fig. supports three loads. Determine the value of the load W and the inclination of the segment BC. [Ans: W=25kN, θ = 54.780]

Loads are in kNW

22.520

B

C

DA

θ 60

30



Q5. Find the reactions at A,B,C and D for the beam loaded as shown in the figure. (Ans.RA=RB =34kN;RC=28.84kN;

MC=-140kNm ; θC=-33.69 ˚ )

12kN/m

4kN/m

12kN/m

4kN/m

20 kN

30kN

1m 2m 1m 1m 2m 1m 1m 2m

A BC

34

40kNm


Q6. A uniform bar AB of weight 50N shown in the figure supports a load of 200N at its end. Determine the tension developed in the string and the force supported by the pin at B. (Ans. T=529.12N;RB=807.15N, θB=64.6˚)

2.5m 2.5m200N

2.5m

A

B60˚

string


Q7. Find the position of the hinged support (x),such that the reactions developed at the supports of the beam are equal..

(Ans.x=2m.)

2.0m 1.4m1.0m 3.0m0.6

15kN18kN/m

10kN/m

x

Q8. A right angled bar ABC hinged at A as shown in fig carries two loads W and 2W applied at B &C .Neglecting self weight of the bar find the angle made by AB with vertical (Ans:θ =18.44˚)


0.5L 2W

θ

B

L m

WC

A


Q9. For the block shown in fig., determine the smallest force P required

a) to start the block up the plane b) to prevent the block moving down the plane.

Take µ = 0.20

P

25°

θ100N

[Ans.: (a) Pmin = 59.2N

(b) Pmin = 23.7N (b) θ = 11.3o]


Q10. A block of weight 2000 N is attached to a cord passing over a frictionless pulley and supporting a weight of 800N as shown in fig. If µ between the block and the plane is 0.35, determine the unknown force P for impending motion(a) to the right (b) to the left

[Ans.: (a) P = 132.8N (b) P = 1252N]

2000N P800N30°


Q11. Determine value of angle θ to cause the motion of 500N block to impend down the plane, if µ for all contact surfaces is 0.30.

500N

200N

θ = ?

[Ans.: θ = 28.4°]


Q12. A horizontal bar 10m long and of negligible weight rests on rough inclines as shown in fig. If angle of friction is 15o, how close to B may the 200N force be applied before the motion impends.

100N 200N

2 m X = ?A B

30° 60°

[Ans.: x = 3.5m]


Q13. Determine the vertical force P required to drive the wedge B downwards in the arrangements shown in fig. Angle of friction for all contact surfaces is 12o.Weight of block A= 1600 N.

AB

P

20°

[Ans.: P = 328.42N]


Q14. Determine the force P which is necessary to start the wedge to raise the block A weighing 1000N. Self weight of the wedge may be ignored. Take angle of friction, φ = 15o for all contact surfaces.

20°

A

Pwedge

[Ans.: P = 1192N]


Q15. A ladder of weight 200N, 6m long is supported as shown in fig. If µ between the floor and the ladder is 0.5 & between the wall and the ladder is 0.25 and it supports a vertical load of 1000N, determine a) the least value of α at which the ladder may be placed without slippingb) the reactions at A & B

[Ans.: (a) α = 56.3o (b) RA = 1193 N, RB = 550N]

α

1000N

5m

A

B


Q16. An uniform ladder of weight 250N is placed against a smooth vertical wall with its lower end 5m from the wall. µbetween the ladder and the floor is 0.3. Show that the ladder remains in equilibrium in this position. What is the frictional resistance on the ladder at the point of contact between the ladder and the floor?

Smooth wall

12m

5mA

B

[Ans.: FA = 52 N]


Q17. A ladder of length 5m weighing 500N is placed at 45o

against a vertical wall. µ between the ladder and the wall is 0.20 & between ladder and ground is 0.50. If a man weighing 600N ascends the ladder, how high will he be when the ladder just slips. If a boy now stands on the bottom rung of the ladder, what must be his least weight so that the man can go to the top of the ladder.[Ans.: (a) x = 2.92m (b) Wboy = 458N]

3. CENTROID OF PLANE AREA

3. CENTROID

Centre of gravity : of a body is the point at which the whole

weight of the body may be assumed to be concentrated.

A body is having only one center of gravity for all

positions of the body.

It is represented by CG. or simply G or C.

Contd.

CENTRE OF GRAVITY

Consider a block of uniform thickness and having a uniform mass m.

It is possible to support (hold) the block in stable position by a rod as shown in the figure provided rod must be positioned exactly at the point of intersection of the diagonals.

Contd.

Or the rod must be supported exactly below where the total weight of the block act.

W

R

CENTRE OF GRAVITY

The block can be supported from any position provided the support rod and the line of action of weight are in same line.

W

This indicates that the whole weight of the block act through one point. This point is called as centre of gravity.

Centre of gravity is that point about which the summation of the first moments of the weights of the elements of the body is zero.

R

Contd.

To determine mathematically the

location of the “centre of gravity”

of any body, we apply the

“principle of moments” to the

parallel system of gravitational

forces

X1

CENTRE OF GRAVITYW

W1

W2

W3W4

x

X2

Contd.

The moment of the resultant gravitational force W, about any axis

=the algebraic sum of the moments about the same axisof the gravitational forces dWacting on all infinitesimal elements of the body.

X1

CENTRE OF GRAVITYW

dW1

dW2

dW3dW4

x

X2

∫ ×=⋅ dWxWx

nn xdWxdWxdWxWx ×+×+×+×=⋅ ................dW 332211

∫ dWWhere W = Contd.

x

CENTRE OF GRAVITY

where = x- coordinate of centre of gravity

x

x

W

dWxx ∫ ⋅=

Similarly, y and z coordinates of the centre of gravity are

W

dWzz ∫ ⋅=

W

dWyy ∫ ⋅= and ----(1)

x

CENTRE OF MASS

With the substitution of W= m g and dW = g dm

(if ‘g’ is assumed constant for all particles, then )

m

dmxx ∫ ⋅=

m

dmyy ∫ ⋅=

m

dmzz ∫ ⋅=,, ----(2)

Equation 2 is independent of g and therefore define a unique point in the body which is a function solely of the distribution of mass. This point is called the centre of mass and clearly coincides with the centre of gravity as long as the gravity field is treated as uniform and parallel

Contd.

When speaking of an actual physical body, we use the term

“centre of mass”.

The term centroid is used when the calculation concerns a

geometrical shape only.

Calculation of centroid falls within three distinct categories,

depending on whether we can model the shape of the body

involved as a line, an area or a volume.

Contd.

The centroid “C” of the Volume segment,

V

dVxx ∫ ⋅=

V

dVyy ∫ ⋅=

V

dVzz ∫ ⋅=, ,

The centroid “C” of the line segment,

L

dLxx ∫ ⋅=

L

dLyy ∫ ⋅=

L

dLzz ∫ ⋅=, ,

The centroid “C” of the Area segment,

AREA: when the density ρ, is constant and the body has a small constant thickness t, the body can be modeled as a surface area. The mass of an element becomes dm = ρ t dA.

If ρ and t are constant over entire area, the coordinates of the ‘centre of mass’ also becomes the coordinates of the centroid, C of the surface area and which may be written as

A

dAyy ∫ ⋅=

A

dAzz ∫ ⋅=

A

dAxx ∫ ⋅= ,,

Contd.

Centroid of Simple figures: using method of moment ( First moment of area)

Centroid of an area may or may not lie on the area in question.

It is a unique point for a given area regardless of the choice of the origin and the orientation of the axes about which we take the moment.

The coordinates of the centroid of the surface area about any axis can be calculated by using the equn.

(A) x = (a1) x1 + (a2) x2 + (a3) x3 + ……….+(an) xn

= First moment of area

Algebraic Sum of moment of elemental ‘dA’ about the same axis

Moment of Total area ‘A’ about y-axis =

where (A = a1 + a2 + a3 + a4 + ……..+ an)

AXIS of SYMMETRY:

It is an axis w.r.t. which for an elementary area on one side of the axis , there is a corresponding elementary area on the other side of the axis (the first moment of these elementary areas about the axis balance each other)

If an area has an axis of symmetry, then the centroidmust lie on that axis.

If an area has two axes of symmetry, then the centroidmust lie at the point of intersection of these axes.

Contd.

For example:

The rectangular shown in the figure has two axis of symmetry, X-X and Y-Y. Therefore intersection of these two axes gives the centroid of the rectangle.

B

DD/2

D/2

B/2 B/2

X X

Y

Y

xx

dada

da × x = da × x

Moment of areas,daabout y-axis cancel each other

da × x + da × x = 0 Contd.

AXIS of SYMMETYRY

‘C’ must lie at the intersectionof the axes of symmetry

‘C’ must lie on the axis

of symmetry

‘C’ must lie on the axis of symmetry

To locate the centroid of simple rectangular area from first principles

D

B

To locate the centroid w.r.t. the base line x-x

X Xy

D

Let the distance of centroidfrom the base line x-x be y

Then from the Principle of Moments

∫ ⋅=⋅ dayyASum of moment of elemental area dAabout the same axis

Moment of Total area Aabout x-axis

=

Contd.

Consider a elemental area dA at a distance y from the base line (x-x)

Let the thickness of the element be ‘dy’

Area of small element = dA = B .dy

x xy

y

dyB

dAMoment of this elemental area about x-x axis

= (area) x (distance)

= (B.dy) . (y)

Contd.

Sum of Moment of all such elemental areas comprising the total area =

∫ ⋅= day

∫ ⋅⋅= ydyBD

By

0

2

2 ⎥⎦

⎤⎢⎣

⎡= ⎥

⎦

⎤⎢⎣

⎡=

2

2BD

2

2BDyA =Then from the Principle of Moments

ABDy2

2

=BD

BDy2

2

=2Dy =,,

Contd.

Similarly we can show

x

Y

Y

B

xdx B

Dx

0

2

2 ⎥⎦

⎤⎢⎣

⎡=∫ ⋅=⋅ daxxA ( )∫ ⋅⋅= dxDx

x = B/2

To locate the centroid of simple right angle triangular area from first principles

B

H

To locate the centroid w.r.t. the base line x-x.

Let the distance of centroid from the base line x-x be y

H

x xy


∫ ⋅=⋅ dayyAMoment of Total area A aboutx-axis

Algebraic Sum of moment of elementary area ‘dA’ about the same axis.

=

Contd.

Consider a small elemental area ‘dA’ at a distance ‘y’from the base line (x-x)

Let the thickness of the element be ‘dy’

Area of small element = dA = b .dy

x xy

y

dy

b

dAMoment of this small elemental area about x-x axis

= (area) x (distance)

= (b.dy) . (y)Contd.


( ) dyH

yHBdybda ×−

=×=∫ ⋅=⋅ dayyA

( )H

yHBb −=∫ ⋅⋅=⋅ ydybyA

( )∫ ⋅⋅

−=⋅ ydy

HyHByA

y

y = H/3

X X

HH/3

BContd.

Similarly we can prove that x = B/3( ) dx

BxBHdxhda ⋅

−=⋅=∫ ⋅=⋅ daxxA

( )∫ ⋅⋅

−=⋅ xdx

BxBHxA

∫ ⋅⋅=⋅ xdxbxA( )

BxBHb −

=

x = B/3

Y

xH

YB/3 B

dx

xb

Contd.

The centroid of simple right angled triangle area from the base

B

H

B/3

Centroid

H/3

To locate the centroid of Semi Circular Area w.r.t. the diameter AB from first principles

Consider a semicircle of radius R,

A = area = 8

2Dπ D = 2R

Let ‘G’ be the centroid of the Semicircle, and y is its distance from the diameter AB.

Contd.

drdrdA ××= θ

Consider a small elemental area da, located at distance yfrom the diameter AB,

Let ‘r’ =radial distance of area ‘da’ from centre of the semi circle.

drdrda ××= θ

θsin×= ry

Moment of this elemental area about the diameter AB =

θθ ddrr ⋅⋅= sin2

Contd.

Moment of all such elemental area ‘da’ about the diameter AB

∫ ∫ ⋅⋅=π

θθO

R

O

ddrr .sin2

∫ ⋅⋅⎥⎦

⎤⎢⎣

⎡=

π

θθO

R

O

dr .sin3

3

[ ] )cos(3

3πθ O

R−=

3.2 3R

= Contd.


∫ ⋅=⋅ dayyA

∫ ∫ ⋅⋅π

θθO

R

O

ddrr .sin2

=⋅ yA

=⋅ yA

3.2 3R

π34Ry =

xx

y

y

Centroid

R

π34Ry =

Because of symmetry x = 0

R

To locate the centroid of Quarter Circular Area w.r.t. the boundary radial line AB from first principle

r

dr

dθθ

R

∫ ⋅=⋅ dayyA

y

∫ ∫ ⋅⋅2

2 .sin

π

θθO

R

O

ddrr=⋅ yA

π34Ry =

A B

Contd.

π34Ry =

π34Ry =

π34Rx =

Centroid of Quarter Circular Area w.r.t. the boundary radial lines

Centroid

To locate the centroid of Circular Sector w.r.t. the y-axis shown from first principle

Consider a triangle of differential area = da =

heightbase××=21

RdR ×××= θ21

Distance of the differential area ‘da’, from y-axis = xc =

θcos32

××= R

y

daXc=(2/3)Rcosθ

Contd.

To locate the centroid of Circular Sector w.r.t. the y-axis shown from first principle

Consider a triangle of differential area = da =

heightbase××=21

RdR ×××= θ21

Distance of the differential area ‘da’, from y-axis = xc =

θcos32

××= RContd.


∫ ×=× cxdaxAαθ

α

α×=××= ∫−

221 RRdRA

( ) θθαα

α

dRRxR 22

21cos

32

×⋅=× ∫−

( ) αα sin32 32 ××=× RxR

ααsin

32 ×

=Rx

ααsin

32 ×

=Rx

For a semicircular area 2α = π, if we use this value in the above formula we get

π34 Ry =

To locate the centroid of area under the curve x = k y3

from x = 0 to x = a from first principle

Consider a vertical element of area da = y dx at a distance x from the y-axis.

xx

da

To find x- coordinate,

∫ ×=× xdaxAX

At x = a, y = b,

i.e. a = k b3, k = a/b3∫ ⋅=a

dxyA0

Contd.

Substituting y = (x/k)1/3 and k = a/b3

∫∫ =aa

xydxydxx00

dxkxxdx

kxx

aa

∫∫ ⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

0

31

0

31

73

43 2baxab

=⋅

ax74

= Contd.

To find y, Coordinate of centroid of the rectangular element is yc = y/2

dAyyA c ×=× ∫

( )∫∫ ⎟⎠⎞

⎜⎝⎛=

aa

ydxyydxy00 2

Substituting, y = b( x/a)1/3

103

43 2abyab

=⋅

by52

=

EXERCISE PROBLEMS3. Centroid of plane area

Problem No.1:

Locate the centroid of the shaded area shown

50

40

10

10

Ans: x=12.5, y=17.5


Problem No.2:

Locate the centroid of the shaded area shown

300

300

500

1000 m

m

1000 mm

500

r=600D=600

Ans: x=474mm, y=474mm


Problem No.3:Locate the centroid of the shaded area w.r.t. to the axes shown

x-axis

y-axis

90

20

20

60

120

r=40

Ans: x=34.4, y=40.3


Problem No.4:

Locate the centroid of the shaded area w.r.t. to the axes shown

250 mm2010

380

10

200 mm x-axis

y-axis

10

Ans: x= -5mm, y=282mm


Problem No.5


3050

40

20 20

40

x

y

r=20

30

Ans:x =38.94, y=31.46


Problem No.6


2.4

m

1.0

1.5

1.5

1.0 1.0

r=0.6x

y

Ans: x=0.817, y=0.24


Problem No.7


Ans: x= -30.43, y= +9.58


Problem No.8

Locate the centroid of the shaded area.

20

Ans: x= 0, y= 67.22(about base)


Problem No.9

Locate the centroid of the shaded area w.r.t. to the base line.

Ans: x=5.9, y= 8.17

2


Problem No.10


Ans: x=21.11, y= 21.11


Problem No.11


Ans: x= y= 22.22


Problem No.12


R=25

X

Y

R=25

75

50

50

80

Ans: x= 24.33 y= 4.723

4. MOMENT OF INERTIA OF PLANE AREA

4. MOMENT OF INERTIA

Moment of Inertia( Second moment area)

The product of the elemental area and square of the perpendicular distance between the centroid of area and the axis of reference is the “Moment of Inertia” about the reference axis

y

x

ydA

o

Iox= da1 y12 + da2 y2

2+ da3 y32+ --

= ∑ da y2

Ioy = da1 x12 + da2 x2

2 + da3 x32+ ----

= ∑ da x2

Radius of Gyration

kk

kk

kk

Elemental areaElemental area

rr11

rr22

rr33

AA AA

BB BBRadius of gyration is defined as a constant distance of all Radius of gyration is defined as a constant distance of all elemental areas which have been rearranged with out elemental areas which have been rearranged with out altering the total moment of inertia.altering the total moment of inertia.

IIABAB= da k2 + da k2 + ----------- IIABAB= A k2

IIAB AB = ∑ da k2 k=k=√√IIABAB/A/A

Polar moment of Inertia (Perpendicular Axes theorem)

The moment of inertia of an area about an axis perpendicular to the plane of the area is called “Polar Moment of Inertia” and it is denoted by symbol Izz or J or Ip. The moment of inertia of an area in xy plane w.r.to z. axis is Izz = Ip = J = ∫r2dA = ∫(x2 + y2) dA = ∫x2dA + ∫y2dA = Ixx +Iyy

O

y

x

r

z

Y

x

Polar moment of Inertia (Perpendicular Axes theorem)

Hence polar M.I. for an area w.r.t. an axis perpendicular to its plane of area is equal to the sum of the M.I. about any two mutually perpendicular axes in its plane, passing through the point of intersection of the polar axis and the area.

Parallel Axis Theorem

x x

dA

y*G

d

A BMoment of inertia of any area about an axis AB is equal to the M.I. about parallel centroidal axis plus the product of the total area and square of the distance between the two axes.

IAB =∑dA (d +y)2

= ∑dA (d2 + y2 + 2 × d × y)

=∑dA. d2 +∑dA y2 + ∑ 2×d×dA y

= ∑dA. d2 +∑dA y2 + 2×d. ∑ y dA

In the above term (2×d) is constant & ∑ y dA = 0

IAB = Ixx + A.d2

MOMENT OF INERTIA BY DIRECT INTEGRATION

Moment of inertia of rectangular area about centroidalhorizontal axis by direct integration

M.I. about its horizontal centroidal axis :

12

)(

3

2/

2/

2

22/

2/

BD

ydyB

ydAID

D

D

Dxx

=

∫ ××=

×∫=

+

−

+

−

G

B

xx

dy

yD

D/2

.

Moment of Inertia of rectangular area about its base-

(about the line AB) using Parallel Axis Theorem

IAB = IXX + A(d)2

Where d = D/2, the distance between axes xx and AB

BA

G

B

xx

dy

yD

D/2

=BD3/12+(BD)(D/2)2

=BD3/12+BD3/4

=BD3/3

Moment of inertia of Triangular area about the base by direct integration

hx

Bh/3

x

b

xy

(h-y)dy

AA

From similar triangles b/h = x/(h-y)

∴ x = b . (h-y)/h

hx

Bh/3

x

b

xy

(h-y)dy

AA

IAB = ∫ dA.y2 = ∫ (x.dy)y2

hIxx = ∫ (b . (h-y) y2.dy) /h

0

= b[ h (y3/3) – y4/4 ]/h= bh3/12

Moment of inertia of Triangular area about the centroidal horizontal axis

Using Parallel axis theorem .MI about any line(AB) = MI about cenroidal parallel axis + Ad2

hx

Bh/3

x

b

xy

(h-y)dy

AA

CentroidalCentroidal horizontal axishorizontal axis

IAB = Ixx + Ad2

Ixx = MI about centroidal axis x xIAB= MI about the Base line AB

Ixx = IAB – Ad2

= bh3/12 – bh/2 . (h/3)2

= bh3/36

Moment of inertia of Circular area about the centroidalhorizontal axis

Ixx = ∫ dA . y2

R 2π= ∫ ∫ (x.dθ.dr) r2Sin2θ

0 0

R 2π=∫ ∫ r3.dr Sin2θ dθ

0 0

R 2π=∫ r3 dr ∫ {(1- Cos2θ)/2} dθ

0 R

0 2π

=[r4/4] [θ/2 – Sin2θ/4] 0 0

= R4/4[π - 0] = πR4/4

IXX = π R4/4 = πD4/64

BA

xxRθ

dθ

y=rSinθr

Moment of inertia of Semi-circular area about the Base & centroidal horizontal axis

IAB = ∫ dA . y2

R π= ∫ ∫ (r.dθ.dr) r2Sin2θ

0R

0π

=∫ r3.dr ∫ Sin2θ dθ0 0

R π=∫ ∫ r3 dr (1- Cos2θ)/2) dθ

0 0 π

=[R4/4] [θ/2 – Sin2θ/4]0

= R4/4[π/2 - 0] = πR4/8 = (πD4/32)

4R/3π

y0

y0

BAxx R

Moment of inertia of Semi-circular area about the centriodal horizontal axis

using parallel axis theorem:

IAB = Ixx + A(d)2

Ixx = IAB – A(d)2

= π R4/8 πR2/2 . (4R/3π)2

Ixx = 0.11R4

IAB = ICD

R π/2IAB = ∫ ∫ (r.dθ.dr). r2Sin2θ

0 0

R π/2=∫ r3.dr ∫ Sin2θ dθ

0 0

R π/2 =∫ r3 dr ∫ (1- Cos2θ)/2) dθ

0 0 π/2

=[R4/4] [θ/2 – (Sin2 θ)/4] 0

= R4 (π/16 – 0) = πR4/16

A B

x x

C

D y

y

4R/3π

Moment of inertia of Quarter-circular area about the base & centroidal horizontal axis

4R/3π

Moment of inertia about Centroidal axis,Ixx = IAB - Ad2

= πR4/16 - πR2. (0. 424R)2

= 0.055R4

Sl.No Figure I x0-x0I y0-y0

I xx I yy

1bd3/12 - bd3/3 -

2bh3/36 - bh3/12 -

3πR4/4 πR4/4 - -

40.11R4 πR4/8 πR4/8 -

50.055R4 0.055R4 πR4/16 πR4/16

b

dx0

x

x0

xd/2

b

h

xxx0

x0h/3

x0x0

y0

y0

OR

y0

y0xxx0

x0

4R/3π

x0

y y0

4R/3π

4R/3π

Y

Y Xo

EXERCISE PROBLEMS

Q1. Determine the moment of inertia about the centroidal axes.

20

30mm

30mm

30mm

100mm

[Ans: Y = 27.69mm Ixx = 1.801 x 106mm4

Iyy = 1.855 x 106mm4]

EXERCISE PROBLEMS

Q2. Determine second moment of area about the centroidalhorizontal and vertical axes. 300mm

[Ans: X = 99.7mm from A, Y = 265 mm

Ixx = 10.29 x 109mm4, Iyy = 16.97 x 109mm4]

200mm

200

300mm

900mm

EXERCISE PROBLEMS

Q3. Determine M.I. Of the built up section about the horizontal and vertical centroidal axes and the radii of gyration.

200mm20

60

100mm

140mm

20

[Ans: Ixx = 45.54 x 106mm4, Iyy = 24.15 x 106mm4

rxx = 62.66mm, ryy = 45.63mm]

EXERCISE PROBLEMS

Q4. Determine the horizontal and vertical centroidal M.I. Of the shaded portion of the figure.

60X X2020

60 60

[Ans: X = 83.1mm

Ixx = 2228.94 x 104mm4, Iyy = 4789.61 x 104mm4]

EXERCISE PROBLEMS

Q5. Determine the spacing of the symmetrically placed vertical blocks such that Ixx = Iyy for the shaded area.

[Ans: d/2 = 223.9mm d=447.8mm]

200mm

600mm

d400mm

200mm

200mm

200mm

EXERCISE PROBLEMS

Q6. Find the horizontal and vertical centroidal moment of inertia of the section shown in Fig. built up with R.S.J. (I-Section) 250 x 250 and two plates 400 x 16 mm each attached one to each.

Properties of I section are

Ixx = 7983.9 x 104mm4

Iyy = 2011.7 x 104mm4

[Ans: Ixx = 30.653 x 107mm4, Iyy = 19.078 x 107mm4]

4000mm

2500mm

160mm

160mmCross sectional area=6971mm2

EXERCISE PROBLEMS

Q7. Find the horizontal and vertical centroidal moment of inertia of built up section shown in Figure. The section consists of 4 symmetrically placed ISA 60 x 60 with two plates 300 x 20 mm2.

[Ans: Ixx = 111.078 x 107mm4, Iyy = 39.574 x 107mm4]300mm

Properties of ISA

Cross sectional area = 4400mm2

Ixx = Iyy ;Cxx = Cyy =18.5mm

18.5mm18.5mm

20mm

200mm

Q8. The R.S. Channel section ISAIC 300 are placed back to back with required to keep them in place. Determine the clear distance d between them so that Ixx = Iyy for the composite section.

[Ans: d = 183.1mm]

Properties of ISMC300

C/S Area = 4564mm2

Ixx = 6362.6 x 104mm4

Iyy = 310.8 x 104mm4

Cyy = 23.6mm X X

Y

Y

Lacing

d

380mm

23.6mm

EXERCISE PROBLEMS

EXERCISE PROBLEMS

Q9. Determine horizontal and vertical centroidal M.I. for the section shown in figure.

[Ans: Ixx = 2870.43 x 104mm4, Iyy = 521.64 x 104mm4]90mm

160mm

40mm

40mm

40mm

5. Kinetics of rectilinear motion

5. Kinetics of rectilinear motion

In this chapter we will be studying the relationship between forces on a body/particle and the accompanying motion

Newton’s Second law of motion:

Newton’s first and third law of motion were used extensively in the study of statics (the bodies at rest) whereas Newton’s second law of motion is used extensively in the study of the kinetics.

Work done by force:

F

F

s

α αA B

F cosθ F cosθ

F sinθF sinθ

Work done by a force is the product of the force and the distance moved by the point of application in the direction of the force. It is a scalar quantity.

Work done = (F cosα ) × s

X- component of force F moves through distance a S,

S = displacement of force from A to B Unit: Nm ( Joule )

POWER:-

It is defined as the time rate of doing work.

Power = work done /Time= (force × distance) /Time

= force × velocity

Unit: (Nm)/s = [watt] (kN m)/s = [kilo watt]

1 metric H.P=735.75 watts.

Energy:-

It is defined as the capacity to do work. It is a scalar quantity.

Unit :- N m (Joule)

Momentum:-Quantity of motion possessed by a body is called momentum. It is the product of mass and velocity. It is a vector quantity.

Unit:- N s.

Impulse of a Force:-It is defined as the product of force and the time over which itacts. It is a vector quantity.

Unit:- N s.

Newton’s second law of motion.

“If the resultant force acting on a particle is not zero , the particle will have an acceleration proportional to the magnitude of the resultant force and its direction is along that of the resultant force.”

F α aF =Resultant of forcesa = Acceleration of the particle.

F = mam= mass of the particle.

The constant value obtained for the ratio of the magnitude of the force and acceleration is characteristic of the particle and is denoted by ‘m’. Where ‘m’ is mass of the particle

Since ‘m’ is a +ve scalar, the vectors of force ‘F’and acceleration ‘a’ have the same direction.

UnitsForce in Newtons (N) N = 1 Kgm/s2

Acceleration in m/s2

F1

F2

R = m a

Body will m

ove in th

e

directio

n of resultant

F3

R = Resultant of forces F1,F2 and F3

Using the rectangular coordinate system we have components along axes as,

ΣFx = max

ΣFy = may

ΣFz = maz

where Fx ,Fy Fz and ax , ay ,az are rectangular components of resultant forces and accelerations respectively.

Newton’s second law may also be expressed by considering a force vector of magnitude ‘ma’ but of sense opposite to that of the acceleration. This vector is denoted by (ma)rev. The subscript indicates that the sense of acceleration has been reversed and is called the inertia force vector.

F1

F2

R = m a

F3

R = Resultant of forces F1,F2 and F3

Body w

ill mov

e in th

e dire

ction

of

resulta

nt

F1

F2

R = m a

F3

Inert

ia For

ce

F1

F2

F = R = m

a

F3

Inertia Forc

e If the inertia force vector is added to the forces acting on the particle we obtain a system of forces whose resultant is zero.

Resultant of forces F1,F2, F3 and Inertia force = 0

0321 =+++ maFFF

The particle may thus be considered to be in equilibrium. (THIS IS DYNAMIC EQUILIBRIUM)

It was pointed out by D’Alembert (Alembert, Jean le Rond d’ (1717-1783), French mathematician and philosopher) that problems of kinetics can be solved by using the principles of statics only (the equations of equilibrium) by considering an inertia force in a direction directly opposite to the acceleration in addition to the real forces acting on the system

D’Alembert’s principle states thatWhen different forces act on a system such that it is in

motion with an acceleration in a particular direction, the vectorial sum of all the forces acting on the system including the inertia force (‘ma’ taken in the opposite direction to the direction of the acceleration) is zero.

The problem under consideration may be solved by using the method developed earlier in statics. The particle is said to be in dynamic equilibrium.

If

ΣFx = 0

ΣFy= 0 including inertia force vector

ΣFz = 0

This principle is known as D’Alembert’s principle

From Newton’s second law of motion∑ F = m × a = -----------(1)

Also a = dv/dt =( dv/dt) ×( ds/ds) = v × dv/dssubstituting in (1)

∑ F = m × v × dv/ds

ΣF × ds = m × v × dv ------------------(2)

Let the initial velocity be u and the final velocity after it moves through a distance ‘ s’ be v

Work-Energy relation for translation

Integrating both sides, we get

Therefore work done by a system of forces acting on a body while causing a displacement is equal to the change in kinetic energy of the body during the displacement.

( )∑ −=×

∑ =×

∫=∑ ∫

22

2

v

u0

21

)2

(

dv

uvmsF

vmsF

vmdsFs

v

u

Impulse-momentum relationship

F = m × aF = m × (v - u)/t = (mv – mu)/t

Force = Rate of change of momentum

F × t = mv – mu

Impulse = final momentum – Initial momentum

The component of the resultant linear impulse along any direction is equal to change in the component of momentum in that direction.

EXERCISE PROBLEMS5. Kinetics

Q1. Blocks A and B of mass 10 kg and 30 kg respectively are connected by an inextensible cord passing over a smooth pulley as shown in Fig. Determine the velocity of the system 4 sec. after starting from rest. Assume coefficient of friction =0.3 for all surfaces in contact.

B A

60o 30o

Ans: v=13.6m/s


Q2. A tram car weighs 150kN. The tractive resistance being 1% of the weight of car. What power will be required to move the car at uniform speed of 20 kmph

(i) Up an incline 1 in 300

(ii) (ii) Down an incline 1 in 250. Take efficiency 75%.

Ans: Pull = 2 kN

a) Output power=11.12kW


Q3. Two masses of 5 kg and 3 kg rest on two smooth inclined plane, each of inclination 30º and are connected by a string passing over a common apex. Find the velocity of 3 kg mass after 2 sec when released from rest. Find the distance it will cover before changing direction of motion, if 5kg mass is cut off after two sec of its release from rest.

30º 30º

5kg 3kg

V = 4.45 m/s

s = 0.61 m


Q4. A locomotive weighing 900 kN pulls a train of 10 coaches each weighing 300 kN at 72 Kmph on a level track against a resistance of 7 N/kN. If the rear 4 coaches get snapped from the train, find the speed of the engine and the remaining coaches after 120 secs. Assume no change in resistance and draw bar pull. Find also distance traveled by detached coaches before coming to rest.

4x300=1200kN 6x300=1800kN 900kN

P

V = 23.66 m/s

s = 2.9 km


Q5. Find the tension in the cord supporting body C in Fig. below. The pulleys are frictionless and of negligible weight.

150 kN

450 kN

A

B

C

Assume all blocks moving either downward or upward and accordingly draw FBD

0=aA +2aB +cC

300 kN

Ans : T=211.72 kN


Q6. Two blocks A and B are released from rest on a 30o

inclined plane with horizontal, when they are 20m apart. The coefficient of friction under the upper block is 0.2 and that under lower block is 0.4. compute the time elapsed until the block touch. After they touch and move as a unit what will be the constant forces between them.

(Ans : t = 4.85 s, contact force=8.65 N)


Q7. An elevator cage of a mine shaft weighing 8kN when empty is lifted or lowered by means of rope. Once a man weighing 600N entered it and lowered at uniform acceleratin such that when a distance of 187.5 m was covered, the velocity of the cage was 25m/s. Determine the tension in the cable and force exerted by man on the floor of the cage.

(Ans: T=7139 N and R=498 N)


Q8. A small block starts from rest at point a and slides down the inclined plane. At what distance along the horizontal will it travel

before coming to rest . Take µk=0.3 [Ans :s=6m ]

A3

4

5m

B Cs


Q9. The system starts from rest in the position shown . How much further will block ‘A’ move up the incline after block B hits the ground . assume the pulley to be frictionless and massless and µ is 0.2 .WA=1000N, WB=2000N. [ Answer s =1.27m]

3m

A

B34


Q10. A 1500Kg automobile travels at a uniform rate of 50kmph to 75kmph . During the entire motion, the automobile is traveling on a level horizontal road and rolling resistance is 2 % of weight of automobile . Find (i) maximum power developed (ii) power required to maintain a constant speed of 75kmph.

[ ANSWER: power developed = 6.131KN]


Q11. Two bodies A and B weighing 2000N and 5200N are connected as shown in the figure . find the further distance moved by block a after the block B hits Wall. µ=0.2 .[ Answer s=1.34m]

512

A

B 3m


Q12. A spring is used to stop 60kg package which is sliding on a horizontal surface . the spring has a constant k = 20kN/m and is held by cable such that it is initially compressed at 120mm. knowing that the package has a velocity of 2.5m/s in position shown and maximum additional displacement of spring is 40mm . Determine the coefficient of kinetic friction between package and surface. (Answer µk=0.2)

600m

60kg

2.5 m/s


Q13. The system shown in figure has a rightward velocity of 4m/s, just before force P is applied. Determine the value of P that will give a leftward velocity of 6m/s in a time interval of 20sec. Take µ = 0.2 & assume ideal pulley. [Answer P=645.41N]

1000N

400N

P


Q14. A locomotive of weight 500kN pulls a train of weight of 2500kN. The tractive resistance, due to friction is 10N/kN. The train can go with a maximum speed of 27kmph on a grade of 1in100. Determine (a) Power of the locomotive. (b) Maximum speed it can attain on a straight level track with the tractive resistance remaining same. [Answer (a) Power= 450kN (b) v=15m/s]

Q15. A wagon weighing 400kN starts from rest, runs 30m down a 1% grade & strikes a post. If the rolling resistance of the track is 5N/kN, find the velocity of the wagon when it strikes the post.

If the impact is to be cushioned by means of one bumper string, which compresses 1mm per 20 kN weight, determine how much the bumper spring will be compressed. [Answer v=1.716m/s, x=77.5mm]

Q16. A train whose weight is 20kN moves at the rate of 60kmph. After brakes are applied, it is brought to rest in 500m. Find the force exerted, assuming it to be uniform

a) Use work-energy relation

b) Use D’Alemberts equation.

Ans: F = 5.663 kN



Q17. The blocks A and B having weights 100 N and 300 N start from rest. The horizontal plane and the pulleys are frictionless. Determine the acceleration and the tension in the string.

A

B

string

string

Frictionless pulley

Frictionless pulley

Ans: aA=8.403m/s2

aB=4.201 m/s2

T= 85.71 N


Q18. The blocks A and B having weights 100 N and 300 N start from rest. The horizontal plane and the pulleys are frictionless. Determine the velocity of block B after 0.5 seconds and the tension in the string. Use impulse-momentum relation.

A

B

string

string

Frictionless pulley

Frictionless pulley


Q19. The blocks A and B having weights 100 N and 300 N start from rest when a load of 100 N is applied on the block A as shown in the figure. The horizontal plane and the pulleys are frictionless. Determine the acceleration and the tension in the string.

A

B

string

string

Frictionless pulley

Frictionless pulley

100 N


Q20. Two blocks A and B are connected as shown in the figure. At the instant of their release if the block A, which is on smooth horizontal plane has a left ward velocity of 2 m/s, what would be its velocity 5 seconds after their release. The blocks A and B weigh 100 N and 300 N respectively.

A

B

string

string

Frictionless pulley

Frictionless pulley


Q21. An engine weighing 500 kN drags carriages weighing 1500kN up an incline of 1 in 100 against a resistance of 5N/kN starting from rest. It attains a velocity of 36 kmph (10m/s) in 1 km distance with a constant draw bar pull supplied by the engine. What is the power required for the same ? What is the tension developed in the link connecting the engine and carriages?

1500N500N

1001

P

Ans:

Pull=40.19kN, power=401kW, T=30.15kN

Q22. what velocity the block A will attain after 2 seconds starting from rest? Take µ = 0.2. WA = 1500N, WB = 2000N. Use impulse-momentum relation.

4

3

34

AB


Mechanics of Deformable Bodies

PART - II

COURSE CONTENT IN BRIEF

6. Simple stresses and strains

7. Statically indeterminate problems and thermal stresses

8. Stresses on inclined planes


6. Simple stresses and strains

The subject strength of materials deals with the relations between externally applied loads and their internal effects on bodies. The bodies are no longer assumed to be rigid and the deformations, however small, are of major interest

Alternatively the subject may be called the mechanics of solids.

The subject, strength of materials or mechanics of materials involves analytical methods for determining the strength , stiffness (deformation characteristics), and stability of various load carrying members.

GENERAL CONCEPTS

STRESS

No engineering material is perfectly rigid and hence, when a material is subjected to external load, it undergoes deformation.

While undergoing deformation, the particles of the material offer a resisting force (internal force). When this resisting force equals applied load the equilibrium condition exists and hence the deformation stops.

These internal forces maintain the externally applied forces in equilibrium.

STRESS

The internal force resisting the deformation per unit area is

called as stress or intensity of stress.

Stress = internal resisting force / resisting cross sectional area

AR

=

STRESS

SI unit for stress

N/m2 also designated as a pascal (Pa)

Pa = N/m2

kilopascal, 1kPa = 1000 N/m2

megapascal, 1 MPa = 1×106 N/m2

= 1×106 N/(106mm2) = 1N/mm2

1 MPa = 1 N/mm2

gigapascal, 1GPa = 1×109 N/m2

= 1×103 MPa

= 1×103 N/mm2

STRESSAXIAL LOADING – NORMAL STRESS

Consider a uniform bar of cross sectional area A, subjected to a tensile force P.

Consider a section AB normal to the direction of force P

P

P

P

R

BA

R

P

Let R is the total resisting force acting on the cross section AB.

Then for equilibrium condition,

R = P

Then from the definition of stress, normal stress = σ = R/A = P/A

σ = Normal StressSymbol:

STRESSAXIAL LOADING – NORMAL STRESS

Direct or Normal Stress:

Intensity of resisting force perpendicular to or normal to the section is called the normal stress.

Normal stress may be tensile or compressive

Tensile stress: stresses that cause pulling on the surface of the section, (particles of the materials tend to pull apart causing extension in the direction of force)

Compressive stress: stresses that cause pushing on the surface of the section, (particles of the materials tend to push together causing shortening in the direction of force)

STRESS

• The resultant of the internal forces for an axially loaded member is normalto a section cut perpendicular to the member axis.

• The force intensity on that section is defined as the normal stress.

AP

AF

aveA

=∆∆

=→∆

σσ0

lim

STRAIN

STRAIN :

when a load acts on the material it will undergo deformation. Strain is a measure of deformation produced by the application of external forces.

If a bar is subjected to a direct load, and hence a stress, the bar will changes in length. If the bar has an original length L and change in length by an amount δL, the linear strain produced is defined as,

LLδε =

Original lengthChange in length

=Linear strain,

Strain is a dimensionless quantity.

Linear Strain

LL

AP

δδε

σ

==

=

22

L

AP

AP

δε

σ

=

==22

strain normal

stress

==

==

L

AP

δε

σ

STRESS-STRAIN DIAGRAM

In order to compare the strength of various materials it is necessary to carry out some standard form of test to establish their relative properties.

One such test is the standard tensile test in which a circular bar of uniform cross section is subjected to a gradually increasing tensile load until failure occurs.

Measurement of change in length over a selected gauge length of the bar are recorded throughout the loading operation by means of extensometers.

A graph of load verses extension or stress against strain is drawn as shown in figure.


Proportionality limit

Typical tensile test curve for mild steel


Typical tensile test curve for mild steel showing upper yield point and lower yield point and also the elastic range and plastic range

Stress-strain Diagram

Limit of Proportionality :

From the origin O to a point called proportionality limit the stress strain diagram is a straight line. That is stress is proportional to strain. Hence proportional limit is the maximum stress up to which the stress – strain relationship is a straight line and material behaves elastically.

From this we deduce the well known relation, first postulated by Robert Hooke, that stress is proportional to strain.

Beyond this point, the stress is no longer proportional to strain

APP

P =σ Load at proportionality limitOriginal cross sectional area

=


Elastic limit:

It is the stress beyond which the material will not return to its original shape when unloaded but will retain a permanent deformation called permanent set. For most practical purposes it can often be assumed that points corresponding proportional limit and elastic limit coincide. Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will be thus some permanent deformation when load is removed.

APE

E =σ Load at proportional limitOriginal cross sectional area=


Yield point:

It is the point at which there is an appreciable elongation or yielding of the material without any corresponding increase of load.

APY

Y =σ Load at yield pointOriginal cross sectional area=

Ultimate strength:

It is the stress corresponding to maximum load recorded during the test. It is stress corresponding to maximum ordinate in the stress-strain graph.

APU

U =σ Maximum load taken by the materialOriginal cross sectional area=


Rupture strength (Nominal Breaking stress):

It is the stress at failure.

For most ductile material including structural steel breaking stress is somewhat lower than ultimate strength because the rupture strength is computed by dividing the rupture load (Breaking load) by the original cross sectional area.

APB

B =σ load at breaking (failure)Original cross sectional area=

True breaking stress = load at breaking (failure)Actual cross sectional area


After yield point the graph becomes much more shallow and covers a much greater portion of the strain axis than the elastic range.

The capacity of a material to allow these large plastic deformations is a measure of ductility of the material

Ductile Materials:

The capacity of a material to allow large extension i.e. the ability to be drawn out plastically is termed as its ductility. Material with high ductility are termed ductile material.

Example: Low carbon steel, mild steel, gold, silver, aluminum


Percentage elongation

A measure of ductility is obtained by measurements of the percentage elongation or percentage reduction in area, defined as, increase in gauge length (up to fracture)

original gauge length×100=

Percentage reduction in area original area

×100Reduction in cross sectional area of necked portion (at fracture)

=

Cup and cone fracture for a Ductile Material


Brittle Materials :A brittle material is one which exhibits relatively small extensions before fracture so that plastic region of the tensiletest graph is much reduced.

Example: steel with higher carbon content, cast iron, concrete, brick

Stress-strain diagram for a typical brittle material

HOOKE’S LAW

Hooke’s Law

For all practical purposes, up to certain limit the relationship between normal stress and linear strain may be said to be linear for all materials

stress (σ) α strain (ε)stress (σ) strain (ε) = constant

Thomas Young introduced a constant of proportionality that came to be known as Young’s modulus.

stress (σ) strain (ε) =

Young’s ModulusE =Modulus of Elasticity

or

HOOKE’S LAW

Young’s Modulus is defined as the ratio of normal stress to linear strain within the proportionality limit.

stress (σ) strain (ε) LA

PLLL

AP

δδ

=÷==E

The value of the Young’s modulus is a definite property of a material

From the experiments, it is known that strain is always a very small quantity, hence E must be large.

For Mild steel, E = 200GPa = 2×105MPa = 2×105N/mm2

Deformations Under Axial Loading

AEP

EE ===

σεεσ

• From Hooke’s Law:

• From the definition of strain:

Lδε =

• Equating and solving for the deformation,

AEPL

=δ

• With variations in loading, cross-section or material properties,

∑=i ii

iiEALPδ

Consider an element of length, δx at a distance x from A

B

WW

Axd1 d2dx

( ) xL

ddd ×−

+= 121 c/s area at x, ( )21

21

44 kxdd +==ππ

xkd ×+= 1

Diameter at x,

Change in length over a length dx is ( ) ⎟⎟

⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

×+=⎟

⎠⎞

⎜⎝⎛=

Ekxd

WdxAEPL

dx 214

π

Change in length over a length L is ( )

∫⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

×+=

L

Ekxd

Wdx0 2

14π

Consider an element of length, δx at a distance x from A

( )∫

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

×+=

L

Ekxd

Wdx0 2

14π

( )∫

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

×=

L

Et

kdtW

0 2

4π

Change in length over a length L is

Put d1+kx = t,

Then k dx = dt

LLL

kxdEkW

tEkWt

EkW

0100

12

)(1414

14

⎥⎦

⎤⎢⎣

⎡+

−=⎥⎦

⎤⎢⎣⎡−=⎥

⎦

⎤⎢⎣

⎡−

=+−

πππ

EddWL

dEdWL

×==

4

42121

ππ

Derive an expression for the total extension of the tapered bar AB of rectangular cross section and uniform thickness, as shown in the figure, when subjected to an axial tensile load ,W.

WW

A B

L

d1d2

bb

W W

A B

x

d1d2

bb

dxConsider an element of length, δx at a distance x from A

( ) xL

ddd ×−

+= 121 c/s area at x, ( )bkxd += 1

xkd ×+= 1

depth at x,

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛×+

=⎟⎠⎞

⎜⎝⎛=

EbkxdWdx

AEPL

dx 1

Change in length over a length dx is

Change in length over a length L is ( )∫ ⎟⎟

⎠

⎞⎜⎜⎝

⎛×+

=L

EbkxdWdx

01

( )12 loglog ddkEb

Pee −

××=

( ) ( )1212

loglog302.2 ddddEbLP

−−××××

=

Derive an expression for the total extension produced by self weight of a uniform bar, when the bar is suspended vertically.

L

Diameter d

P1x

dx dx

elementP1 = weight of the bar below

the section,= volume × specific weight= (π d2/4)× x × γ= A× x ×γ

Diameter d Extension of

the element due to weight of the bar below that,

AEdxxA

AEdxP

AEPL

dx

)(1 ρ××==⎥⎦

⎤⎢⎣⎡=

Hence the total extension entire bar

EL

Ex

AEdxxA

LL

22)( 2

0

2

0

γγγ=⎥

⎦

⎤⎢⎣

⎡=

××= ∫

AEPL

AELAL

AA

EL

×=×

=×=21

2)(

2

2 γγThe above expression can also be written as

Where, P = (AL)×γ= total weight of the bar

SHEAR STRESS

Consider a block or portion of a material shown in Fig.(a) subjected to a set of equal and opposite forces P. then there isa tendency for one layer of the material to slide over another to produce the form failure as shown in Fig.(b)

PP

R RP P

Fig. cFig. a Fig. b

The resisting force developed by any plane ( or section) of the block will be parallel to the surface as shown in Fig.(c).

The resisting forces acting parallel to the surface per unit area is called as shear stress.

Shear stress (τ)

=Shear resistanceArea resisting shear

τ

If block ABCD subjected to shearing stress as shown in Fig.(d), then it undergoes deformation. The shape will not remain rectangular, it changes into the form shown in Fig.(e), as AB'C'D.

B

Shear strain

AP

=

This shear stress will always be tangential to the area on whichit acts

τD

C

A

τB'

D

C'

Aτ

B C

Fig. d Fig. e

τ Shear strain is defined as the change in angle between two line element which are originally right angles to one another.

The angle of deformation is measured in radians and hence is non-dimensional.

D

B' C'B

A

C

φ

τFig. e

φφ ≈=′

= tanstrain shear ABBB

The angle of deformation is then termed as shear strain φ

SHEAR MODULUS

For materials within the proportionality limit the shear strain is proportional to the shear stress. Hence the ratio of shear stress to shear strain is a constant within the proportionality limit.

Shear Modulus or

Modulus of Rigidity

Shear stress (τ) Shear strain (φ) G= == constant

The value of the modulus of rigidity is a definite property of a material

For Mild steel, G= 80GPa = 80,000MPa = 80,000N/mm2

example: Shearing Stress

• Forces P and P‘ are applied transversely to the member AB.

AP

=aveτ

• The corresponding average shear stress is,

• The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P.

• Corresponding internal forces act in the plane of section C and are called shearing forces.

• The shear stress distribution cannot be assumed to be uniform.

State of simple shear

Consider an element ABCD in a strained material subjected to shear stress, τ as shown in the figure

τBA

D Cτ

Force on the face AB = P = τ × AB × t

Where, t is the thickness of the element.

Force on the face DC is also equal to P


Now consider the equilibrium of the element.(i.e., ΣFx = 0, ΣFy = 0, ΣM = 0.)

P

The element is subjected to a clockwise moment

P × AD = (τ × AB × t) × AD

BAFor the force diagram shown,ΣFx = 0, & ΣFy = 0, But ΣM = 0 D C

Pforce

But, as the element is actually in equilibrium, there must be another pair of forces say P' acting on faces AD and BC, such that they produce a anticlockwise moment equal to ( P × AD )


P BAP ' × AB = P × AD= (τ × AB × t)× AD ----- (1) P ' P '

CDP

If τ1 is the intensity of the shear stress on the faces AD and BC, then P ' can be written as, P ' = τ ' × AD × t

τ BA

τ ' τ ' CD

τEqun.(1) can be written as

(τ ' × AD× t ) × AB = (τ × AB × t) × AD ----- (1)

τ ' = τ


Thus in a strained material a shear stress is always accompanied by a balancing shear of same intensity at right angles to itself. This balancing shear is called “complementary shear”.

τ

The shear and the complementary shear together constitute a state of simple shear

BA

τ'= τ τ'= τ

CDτ

Direct stress due to pure shear

Consider a square element of side ‘a’ subjected to shear stress as shown in the Fig.(a). Let the thickness of the square be unity.

a

A B

CD

τ

τ

τ

τa

τAB

Fig.(b) shows the deformed shape of the element. The length of diagonal DB increases, indicating that it is subjected to tensile stress. Similarly the length of diagonal AC decreases indicatingthat compressive stress.

C

τaa

τ

Dτ

Fig.(a). Fig.(b).


Now consider the section, ADC of the element, Fig.(c).

Resolving the forces in σn direction, i.e., in the X-direction shown

a

Fig.(c).

aa

A

CD

( )a2

For equilibrium

A σn

C

a

X

τ

Dτ

( ) ( )τσ

τσ=

××−×××=

=∑

n

n aa

Fx

45cos212

0


Therefore the intensity of normal tensile stress developed on plane BD is numerically equal to the intensity of shear stress.

Similarly it can be proved that the intensity of compressive stress developed on plane AC is numerically equal to the intensity of shear stress.

POISSON’S RATIO

Poisson’s Ratio:

Consider the rectangular bar shown in Fig.(a) subjected to a tensile load. Under the action of this load the bar will increase in length by an amount δL giving a longitudinal strain in the bar of

ll

lδε =

Fig.(a)

The associated lateral strains will be equal and are of opposite sense to the longitudinal strain.

POISSON’S RATIO

The bar will also exhibit, reduction in dimension laterally, i.e. its breadth and depth will both reduce. These change in lateral dimension is measured as strains in the lateral direction as given below.

dd

bb

latδδε −=−=

Provided the load on the material is retained within the elasticrange the ratio of the lateral and longitudinal strains will always be constant. This ratio is termed Poisson’s ratio (µ)

POISSON’S RATIOLateral strainLongitudinal strain

=l

ld

d

δ

δ )(−=

ll

bb

δ

δ )(−OR

Poisson’s Ratio = µ

For most engineering metals the value of µ lies between 0.25 and 0.33

In general y

Lx

LyLz

PP x

z

x

x

y

y

ll

ll

δ

δ−

=

x

x

z

z

ll

ll

δ

δ−

=Lateral strain

Poisson’s Ratio

OR= Strain in the direction of load applied

Poisson’s Ratio = µ

In general

Strain in X-direction = εx

z

y

xPxPx

Lx

LyLz

x

x

llδ

=

Strain in Y-direction = εy

x

x

y

y

ll

ll δµδ

==

x

x

z

z

ll

ll δµδ==Strain in Z-direction = εz

Load applied in Y-direction

z

y

x

Py

Lx

LyLz

Py

y

y

x

x

ll

ll

δ

δ−

=

y

y

z

z

ll

ll

δ

δ−

=Lateral strain

Poisson’s Ratio


y

y

x

x

ll

ll δ

µδ==Strain in X-direction = εx

Load applied in Z-direction

y

z

x

Pz

Lx

LyLz

Pz

z

z

y

y

ll

ll

δ

δ−

=

z

z

x

x

ll

ll

δ

δ−

=Lateral strain

Poisson’s Ratio


z

z

x

x

ll

ll δµδ==Strain in X-direction = εx

Load applied in X & Y direction


z

y

xPxPx

Lx

LyLz

Py

EEyx σ

µσ−=

Py

EExy σµ

σ−=Strain in Y-direction = εy

EExy σµ

σµ −−=Strain in Z-direction = εz

General case:


Py

Strain in Y-direction = εy

Strain in Z-direction = εz

z

y

xPxPx

Pz

PyPzEEE

zyxx

σµσ

µσε −−=

σx

σzσy

σx

σz σy

EEEzxy

yσµσµ

σε −−=

EEExyz

zσµ

σµσε −−=

Bulk Modulus

Bulk Modulus

A body subjected to three mutually perpendicular equal direct stresses undergoes volumetric change without distortion of shape. If V is the original volume and dV is the change in volume, then dV/V is called volumetric strain.

Bulk modulus, K

A body subjected to three mutually perpendicular equal direct stresses then the ratio of stress to volumetric strain is calledBulk Modulus.

⎟⎠⎞

⎜⎝⎛

=

VdVσ

Relationship between volumetric strain and linear strain

Consider a cube of side 1unit, subjected to three mutually perpendicular direct stresses as shown in the figure.

Relative to the unstressed state, the change in volume per unit volume is

( )( )( )[ ] [ ]

eunit volumper in volume change

1111111

=

++=

+++−=+++−=

zyx

zyxzyxdV

εεε

εεεεεε

Relationship between volumetric strain and linear strain

Volumetric strain

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

EEEzyx σµ

σµσ

⎟⎟⎠

⎞⎜⎜⎝

⎛−−+

EEEzxy σµσµ

σ⎟⎟⎠

⎞⎜⎜⎝

⎛−−+

EEExyz σµ

σµσ

zyxVdV εεε ++=

( )zyxEσσσµ

++−

=21

For element subjected to uniform hydrostatic pressure,

σσσσ === zyx

( )

( ) 321

21

σµ

σσσµ

EVdV

EVdV

zyx

−=

++−

=

⎟⎠⎞

⎜⎝⎛

=

VdV

K σ

( )

( )µ

µ

2-13KEor

modulusbulk 213

=

=−

=EK

Relationship between young’s modulus of elasticity (E) and modulus of rigidity (G) :-

A

D τ

B

τ

aa

45˚A1

φφ

B1

C

H

Consider a square element ABCD of side ‘a’ subjected to pure shear ‘τ’. DA'B'C is the deformed shape due to shear τ. Drop a perpendicular AH to diagonal A'C.Strain in the diagonal AC = τ /E – µ (- τ /E) [ σn= τ ]

= τ /E [ 1 + µ ] -----------(1)Strain along the diagonal AC=(A'C–AC)/AC=(A'C–CH)/AC=A'H/AC

In ∆le AA'HCos 45˚ = A'H/AA'A'H= AA' × 1/√2AC = √2 × AD ( AC = √ AD2 +AD2)Strain along the diagonal AC = AA'/ (√2 × √2 × AD)=φ/2 ----(2)Modulus of rigidity = G = τ /φ

φ = τ /GSubstituting in (2) Strain along the diagonal AC = τ /2G -----------(3)Equating (1) & (3)τ /2G = τ /E[1+µ]

E=2G(1+ µ)

Relationship between E, G, and K:-

We have

E = 2G( 1+ µ) -----------(1)

E = 3K( 1-2µ) -----------(2)

Equating (1) & (2)

2G( 1+ µ) =3K( 1- 2µ)

2G + 2Gµ=3K- 6Kµ

µ= (3K- 2G) /(2G +6K)

Substituting in (1)

E = 2G[ 1+(3K – 2G)/ (2G+6K)]

E = 18GK/( 2G+6K)

E = 9GK/(G+3K)

Working stress: It is obvious that one cannot take risk of loading a member to its ultimate strength, in practice. The maximum stress to which the material of a member is subjected to in practice is called working stress.

This value should be well within the elastic limit in elastic design method.

Factor of safety: Because of uncertainty of loading conditions, design procedure, production methods, etc., designers generally introduce a factor of safety into their design, defined as follows

Factor of safety =Allowable working stress

Maximum stress Allowable working stress

Yield stress (or proof stress)or

Malleability: A property closely related to ductility, which defines a material’s ability to be hammered out in to thin sheets

Homogeneous: A material which has a uniform structure throughout without any flaws or discontinuities.

Isotropic: If a material exhibits uniform properties throughout in all directions ,it is said to be isotropic.

Anisotropic: If a material does not exhibit uniform properties throughout in all directions ,it is said to be anisotropic or nonisotropic.

Exercise Problems

Q1. An aluminum tube is rigidly fastened between a brass rod and steel rod. Axial loads are applied as indicated in the figure. Determine the stresses in each material and total deformation. Take Ea=70GPa, Eb=100GPa, Es=200GPa

500mm 700mm600mm

steelaluminum

brass20kN 15kN 15kN 10kN

Ab=700mm2Aa=1000mm2

As=800mm2

Ans: σb=28.57MPa, σa=5MPa, σs=12.5MPa, δl = - 0.142mm

Q2. A 2.4m long steel bar has uniform diameter of 40mm for a length of 1.2m and in the next 0.6m of its length its diameter gradually reduces to ‘D’ mm and for remaining 0.6m of its length diameter remains the same as shown in the figure. When a load of 200kN is applied to this bar extension observed is equal to 2.59mm. Determine the diameter ‘D’ of the bar. Take E =200GPa

Ф = 40mm

Ф = D mm

200kN200kN

500mm500mm1000mm

Q3. The diameter of a specimen is found to reduce by 0.004mm when it is subjected to a tensile force of 19kN. The initial diameter of the specimen was 20mm. Taking modulus of rigidity as 40GPa determine the value of E and µ

Ans: E=110GPa, µ=0.36

Q.4 A circular bar of brass is to be loaded by a shear load of 30kN. Determine the necessary diameter of the bars (a) in single shear (b) in double shear, if the shear stress in material must not exceed 50MPa.

Ans: 27.6, 19.5mm

Q.5 Determine the largest weight W that can be supported by the two wires shown. Stresses in wires AB and AC are not to exceed 100MPa and 150MPa respectively. The cross sectional areas of the two wires are 400mm2 for AB and 200mm2 for AC.

Ans: 33.4kN

WA

CB300 450

Q.6 A homogeneous rigid bar of weight 1500N carries a 2000N load as shown. The bar is supported by a pin at B and a 10mm diameter cable CD. Determine the stress in the cable

Ans: 87.53MPa

3m

A CB

2000 N

3m

D

Q.7. A stepped bar with three different cross-sectional areas, is fixed at one end and loaded as shown in the figure. Determine the stress and deformation in each portions. Also find the net change in the length of the bar. Take E = 200GPa

250mm 270mm320mm

300mm2 450mm2

250mm2

10kN40kN20kN

Ans: -33.33, -120, 22.2MPa, -0.042, -0.192, 0.03mm, -0.204mm

Q.8 The coupling shown in figure is constructed from steelof rectangular cross-section and is designed to transmit a tensile force of 50kN. If the bolt is of 15mm diameter calculate:

a) The shear stress in the bolt;b) The direct stress in the plate;c) The direct stress in the forked end of the coupling.

Ans: a)141.5MPa, b)166.7MPa, c)83.3MPa

Q.9 The maximum safe compressive stress in a hardened steel punch is limited to 1000MPa, and the punch is used to pierce circular holes in mild steel plate 20mm thick. If the ultimate shearing stress is 312.5MPa, calculate the smallest diameter of hole that can be pierced.

Ans: 25mm

Q.10 A rectangular bar of 250mm long is 75mm wide and 25mm thick. It is loaded with an axial tensile load of 200kN, together with a normal compressive force of 2000kN on face 75mm×250mm and a tensile force 400kN on face 25mm×250mm. Calculate the change in length, breadth, thickness and volume. Take E = 200GPa & µ=0.3

Ans: 0.15,0.024,0.0197mm, 60mm3

Q.11 A piece of 180mm long by 30mm square is in compression under a load of 90kN as shown in the figure. If the modulus of elasticity of the material is 120GPa and Poisson’s ratio is 0.25, find the change in the length if all lateral strain is prevented by the application of uniform lateral external pressure of suitable intensity.

180

3030

90kN

Ans: 0.125mm

Q.12 Define the terms: stress, strain, elastic limit, proportionality limit, yield stress, ultimate stress, proof stress, true stress, factor of safety, Young’s modulus, modulus of rigidity, bulk modulus, Poisson's ratio,

Q.13 Draw a typical stress-strain diagram for mild steel rod under tension and mark the salient points.

Q.14 Diameter of a bar of length ‘L’ varies from D1 at one end to D2 at the other end. Find the extension of the bar under the axial load P

Q.15 Derive the relationship between Young’s modulus and modulus of rigidity.

Q.16 Derive the relationship between Young’s modulus and Bulk modulus.

Q.17 A flat plate of thickness ‘t’ tapers uniformly from a width b1at one end to b2 at the other end, in a length of L units. Determine the extension of the plate due to a pull P.

Q.18 Find the extension of a conical rod due to its own weight when suspended vertically with its base at the top.

Q.19 Prove that a material subjected to pure shear in two perpendicular planes has a diagonal tension and compression of same magnitude at 45o to the planes of shear.

Q.20 For a given material E=1.1×105N/mm2& G=0.43×105N/mm2 .Find bulk modulus & lateral contraction of round bar of 40mm diameter & 2.5m length when stretched by 2.5mm.

ANS: K=83.33Gpa, Lateral contraction=0.011mm

Q.21 The modulus of rigidity of a material is 0.8×105N/mm2 , when 6mm×6mm bar of this material subjected to an axial pull of 3600N.It was found that the lateral dimension of the bar is changed to 5.9991mm×5.9991mm. Find µ & E. ANS: µ=0.31, E= 210Gpa.

7. STATICALLY INDETERMINATE MEMBERS

&

THERMAL STRESSES

STATICALLY INDETERMINATE MEMBERS

Structure for which equilibrium equations are sufficient to obtain the solution are classified as statically determinate. But for some combination of members subjected to axial loads, the solution cannot be obtained by merely using equilibrium equations. The structural problems with number of unknowns greater than the number independent equilibrium equations are called statically indeterminate.

The following equations are required to solve the problems on statically indeterminate structure.

1) Equilibrium equations based on free body diagram of the structure or part of the structure.

2) Equations based on geometric relations regarding elastic deformations, produced by the loads.

COMPOUND BAR

L2

Material(2)

L1Material(1)

A compound bar is one which is made of two or more than two materials rigidly fixed, so that they sustain together an externally applied load. In such cases

(i) Change in length in all the materials are same.

(ii) Applied load is equal to sum of the loads carried by hb

W

(dL)1 = (dL)2

(σ1/ E1)L1 = (σ2 /E2)L2

σ1 = σ2 ×( E1/E2)(L1/L2) (1)

E1/E2 is called modular ratio

Total load = load carried by material (1) + load carried by material(2)

W = σ1 A1 + σ2 A2 (2)

From Equation (1) & (2) σ1 and σ2 can be calculated

Temperature Stress

L

A B

L

AB

L

A B

B´P

αTL

Any material is capable of expanding or contracting freely due to rise or fall in temperature. If it is subjected to rise in temperature of T˚C, it expands freely by an amount ‘αTL’ as shown in figure. Where α is the coefficient of linear expansion, T˚C = rise in temperature and L = original length.

From the above figure it is seen that ‘B’ shifts to B' by an amount ‘αTL’. If this expansion is to be prevented a compressive force is required at B'.

Temperature strain = αTL/(L + αTL) ≈ αTL/L= αTTemperature stress = αTE

Hence the temperature strain is the ratio of expansion or contraction prevented to its original length.

If a gap δ is provided for expansion then Temperature strain = (αTL – δ) / LTemperature stress = [(αTL – δ)/L] E

Temperature stress in compound bars:-

Material(2)

Material(1)

α2TL

∆

α1TL

(dL)1

P1

(dL)2P2

x

xWhen a compound bar is subjected to change in temperature, both the materials will experience stresses of opposite nature.

Compressive force on material (1) = tensile force on material (2)

σ1A1 = σ2A2 (there is no external load) σ1=( σ2A2)/A1 (1)

As the two bars are connected together, the actual position of the bars will be at XX.

Actual expansion in material (1) = actual expansion in material (2)

α1TL – (dL)1 = α2TL + (dL)2

α1TL – (σ1 / E1) L =α2TL + (σ2 / E2) L

αT – (σ1 / E1) = α2T + σ2 / E2 --------------------------(2)

From (1) and (2) magnitude of σ1 and σ2 can be found out.

Exercise problems

Q.1 A circular concrete pillar consists of six steel rods of 22mm diameter each reinforced into it. Determine the diameter of pillar required when it has to carry a load of 1000kN. Take allowable stresses for steel & concrete as 140Mpa & 8Mpa respectively. The modular ratio is 15 ANS: D=344.3mm

Q.2 Determine the stresses & deformation induced in Bronze & steel as shown in figure. Given As=1000mm2, Ab=600mm2, Es= 200Gpa, Eb= 83Gpa ANS: ( σb=55Mpa, σs=93.5Mpa, dLs=dLb=0.093mm)

160kN

Bronze BronzeSteel

Q..3 A cart wheel of 1.2m diameter is to be provided with steel tyre. Assume the wheel to be rigid. If the stress in steel does not exceed 140MPa, calculate minimum diameter of steel tyre & minimum temperature to which it should be heated before on to the wheel.

ANS: d=1199.16mm T=58.330C Q.4 A brass rod 20mm diameter enclosed in a steel tube of 25mm internal diameter & 10mm thick. The bar & the tube are initially 2m long & rigidly fastened at both the ends. The temperature is raised from 200C to 800C. Find the stresses in both the materials.

If the composite bar is then subjected to an axial pull of 50kN, findthe total stress. Es=200GPa, Eb=80GPa, αs=12×10-6/0C, αb=19×10-6/0C.

ANS: σb=8.81N/mm2 ( C ) , σs=47.99N/mm2( T )

8. STRESSES ON INCLINED PLANES

INTRODUCTION

The state of stress on any plane in a strained body is said to be ‘Compound Stress’, if, both Normal and Shear stresses are acting on that plane. For, example, the state of stress on any vertical plane of a beam subjected to transverse loads will, in general, be a Compound Stress. In actual practice the state of Compound Stress is of more common occurrence than Simple state of stress.

In a compound state of stress, the normal and shear stress may have a greater magnitude on some planes which are inclined (or, Oblique) to the given stress plane.

Hence in compound state of stresses it is necessary to find the following

(i) The normal and shear stress on a plane which is inclined (Oblique) to the given stress plane;

(ii) The inclination of max. and min. normal stress planes and values of the normal stress (max. / min.) on them;

(iii) The inclination of max. shear stress planes and the values of the shear stress (max.) on them.

EQUATIONS, NOTATIONS & SIGN CONVENTIONS :

(i) Normal & Shear stress on plane inclined (Oblique) to given stress plane:

Consider an element ABCD subjected to a state of compound stress as shown in the Fig. Let:σx Normal Stress in x- directionσy Normal Stress in y- directionτ Shear Stresses in x & y – directionsθ Angle made by inclined plane AE wrt verticalσθ Normal Stress on inclined planeτθ Shear Stress on inclined plane

A

E

D

C B

σy

τ

τ

σx σxθ

σy

σθ

τθ

Normal Stress, σθ, and Shear stress, τθ, on inclined plane are given by:

)1(2sin2cos22

−−−+⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟⎟

⎠

⎞⎜⎜⎝

⎛ += θτθ

σσσσσθ

yxyx

)2(2cos2sin2

−−−−⎟⎟⎠

⎞⎜⎜⎝

⎛ −= θτθ

σστ θ

yx

(ii) The inclination of max. and min. normal stress planes and the values of the normal stress (max. / min.) on them

Let, θP be the inclination of the plane of max. or min. normal stress and σP be the value of the max. or min. normal stress on that plane, then, from Eqn. (1):

(1)--- 2sin2cos22 PP

yxyxP θτθ

σσσσσ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=

( ) ( )

( ) ( )

0

02cos2sin2

02cos22sin22

0dθ

dσ min.,or max. be toσFor PP

=⇒

=−⎟⎟⎠

⎞⎜⎜⎝

⎛ −⇒

=+−⎟⎟⎠

⎞⎜⎜⎝

⎛ −⇒

=

P

PPyx

PPyx

θτ

θτθσσ

θτθσσ

Thus, the condition for max. or min. normal stress to occur on a plane is, shear stress on that plane should be zero.These planes on which shear stress is zero and the normal stress on them being either the max. or the min. are called ‘PRINCIPAL PLANES’.

( ) ( ) (3) ---

2

2tan02cos2sin2

have, we,0,

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=⇒=−⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

=

yxPPP

yx

PFrom

σστθθτθ

σσ

τθ

The above Eqn. (3), gives two values for θP, which differ by 900. Thus, there are two mutually perpendicular Principal planes, on which there are only normal stresses, shear stress being zero onthem.

On one of them, the value of the normal stress is the max.; it is called the ‘Major Principal plane’, the max. normal stress on it is called the ‘Major Principal Stress’.

On the other principal plane, the value of the normal stress is the min.; it is called the ‘Minor Principal plane’, the min. normal stress on it is called the ‘Minor Principal Stress’.

( )

2

2

2

2

2)(

2/2cos

2)(

2sin

get, we,

2

2tan From

τσσ

σσθ

τσσ

τθ

σστθ

+⎥⎦⎤

⎢⎣⎡ −

−±=

+⎥⎦⎤

⎢⎣⎡ −

±=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

YX

YXP

YX

P

yxP

(σx-σy)/ 2

2θP

τ

±√[(σ

x - σ

y/2]2

+ τ2

Substituting for sin 2θP and cos 2θP in Eqn. (1), and simplifying, we get the equation for principal stresses as:

( ))4(

222

2

−−−+⎟⎟⎠

⎞⎜⎜⎝

⎛ −±

+= τ

σσσσσ yxyx

P

The above equation (4) gives two values for principal stresses. The numerically max. of the two values (+ ve or − ve) is the Major Principal Stress, (σMajor or σMax); The numerically min. (+ ve or − ve) is the Minor Principal Stress (σMinoror σMin).

(iii) Inclination of max. shear stress planes, Max. shear stress Equation. Let, θS be the inclination of the plane of max. or min. shear stress and τS be the value of the max. or min. shear stress on that plane, then, from Eqn. (2):

)2( 2cos2sin2

−−−−⎟⎟⎠

⎞⎜⎜⎝

⎛ −= SS

yxS θτθ

σστ

( ) ( )

12tan 2 thave We:

(5) --- 2

2tan

02sin22cos22

0dθd min.,or max. be toFor

S

SS

−=×

⎟⎟⎠

⎞⎜⎜⎝

⎛ −

−=⇒

=−−⎟⎟⎠

⎞⎜⎜⎝

⎛ −⇒

=

SP

yx

S

SSyx

anNOTE θθτ

σσ

θ

θτθσσ

ττEqn. (5) gives two values for θS, which differ by 900. Thus, there are two mutually perpendicular planes, on which shear stress are max.; numerically equal but opposite in sense.The planes of Max. Shear stresses are inclined at 450 to the Principal planes.

( )2

2

2

2

2)(

2/2sin

2)(

2cos

get, we,2

2tan From

τσσ

σσθ

τσσ

τθ

τ

σσ

θ

+⎥⎦

⎤⎢⎣

⎡ −

−±=

+⎥⎦

⎤⎢⎣

⎡ −±=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −

−=

yx

yxS

yx

s

yx

S

(σx-σy)/ 2

2θS

τ

±√[(σ

x − σ

y)/2]

2 + τ2

Substituting for sin 2θS and cos 2θS in Eqn. (2), we get the equation for Max. shear stresses as:

)6(

2

2

.max

22

.max

−−−

⎥⎥⎥⎥⎥

⎦

⎤

⎟⎟⎠

⎞⎜⎜⎝

⎛ −±=⇒

+⎟⎟⎠

⎞⎜⎜⎝

⎛ −±=

MinorMajor

yx

σστ

τσσ

τ

The above equation (6) gives two values for Max. shear stresses,which are numerically equal but opposite in sense.

NOTATIONS & SIGN CONVENTIONS

σx Normal Stress in x- directionσy Normal Stress in y- directionτ Shear Stresses in x & y – directionsθ Angle made by inclined plane wrt verticalσθ Normal Stress on inclined plane AEτθ Shear Stress on inclined plane AE

A

E

D

C B

σy

τ

τ

σx σxθ

σy

σθ

τθ

θP Inclination of Principal planesσP Principal stresses θS Inclination of Max. shear stress planes [θS = θP + 450].

All the parameters are shown in their +ve sense in the Fig.

Normal stresses, σ Tensile stresses +ve.Shear Stresses, τ, in x – direction & Inclined Plane Clockwise +ve.Shear Stresses, τ, in y – direction Anti-Clockwise +ve.Angle, θ measured w r t vertical, Anti-Clockwise +ve.

Exercise Problems:

Q.1 The principal stresses at a point in a strained material are 80 MPa(C) and 40 MPa(T). Find the normal, tangential and resultant stress on a plane inclined at 50o to the major principal plane. [Ans: - 9.58MPa, -59MPa].

Q.2 The stresses in a strained material is as shown in Fig. Find the normal and shear stresses on plane inclined at 30o to the horizontal. Also determine the intensity and position of the plane upon which there is only shear stress. Sketch the plane.

Ans:σ = − 22.5MPa, τ = − 64.95MPa. Pure shear plane σ = 0, Ө = 39.23o w.r.t. horizontal.τpure = − 73.48MPa.

60MPa

30o

90 MPa

Q.3 A plane element is subjected to the system of stresses as shown in Fig. Determine (i) the principal stresses and inclination of their planes (ii) maximum shearing stresses and inclination of their planes. Represent your answers in neat sketches.

160 MPa

200 MPa

80 MPa

Ans:σ1 = 262.46MPa, σ2 = 97.54MPa. Өp = 37.98 or 127.98 with horizontalτtmax = 82.46, Өs = 82.98 or 172.98 with horizontal

Q.4 At a point in a structural member subjected to stresses asshown in fig. determine the principal stresses and the maximum shear stress. Also determine and sketch planes on which these stressesact.

100

80

40

[Ans:131.23, 48.77, − 37.98o, − 127.98o

41.23, 7o, 970 , Angles w.r.t. horizontal]

Q.5 At a point in a material under stress, the intensity of resultant stress on a certain plane is 60 MPa, directed outwards and inclined at 30o to the normal of that plane. The stress on the plane at right angles to this has a normal stress component of 40 Mpa (T). Find (i) the principal stresses and inclination of their planes, (ii) the maximum shear stresses and inclination of their planes .

[Ans: 76.57MPa, 15.39MPa, 39.36o , 129.36o and 30.59 MPa, 84.36o , 174.36o, Angles w.r.t. horizontal]

9 -THIN CYLINDERS

They are,

1. Hoop or Circumferential Stress (σC) – This is directed along the tangent to the circumference and tensile in nature. Thus, there will be increase in diameter.

In many engineering applications, cylinders are frequently used for transporting or storing of liquids, gases or fluids.

Eg: Pipes, Boilers, storage tanks etc. These cylinders are subjected to fluid pressures. When

a cylinder is subjected to internal pressure, at any point on the cylinder wall, three types of stresses are induced on three mutually perpendicular planes.

INTRODUCTION:

2. Longitudinal Stress (σL) – This stress is directed along the length of the cylinder. This is also tensile in nature and tends to increase the length.

3. Radial pressure (σ r) – It is compressive in nature. Its magnitude is equal to fluid pressure on the inside wall and zero on the outer wall if it is open to atmosphere.

σC σL

1. Hoop/Circumferential

Stress (σC)

2. Longitudinal Stress (σL) 3. Radial Stress (σr)

Element on the cylinder wall subjected to these three stresses

σL

σL

σL

p pσr

σCσC

σC

p

σLσL

σCσr

σrσC

A cylinder or spherical shell is considered to be thin when the metal thickness is small compared to internal diameter.

i. e., when the wall thickness, ‘t’ is equal to or less than ‘d/20’, where ‘d’ is the internal diameter of the cylinder or shell, we consider the cylinder or shell to be thin, otherwise thick.

Magnitude of radial pressure is very small compared to other two stresses in case of thin cylinders and hence neglected.

)1....(....................t2dpσ stress, ntialCircumfere c ×

×=

)2 .........(..........t4dpσ stress, alLongitudin L ×

×=

Wherep = internal fluid pressured= internal diameter, t = thickness of the wall

Maximum Shear stress, τmax = (σc- σL) / 2

τmax = ( p x d) / (8 x t)

EVALUATION OF STRAINS

σ L=(pd)/(4t)

σ C=(pd)/(2t) σ C=(pd)/(2t)

σ L=(pd)/(4t)

A point on the surface of thin cylinder is subjected to biaxial stress system, (Hoop stress and Longitudinal stress) mutually perpendicular to each other, as shown in the figure. The strains due to these stresses i.e., circumferential and longitudinal are obtained by applying Hooke’s law and Poisson’s theory for elastic materials.

µ)2(Eσ

Eσµ

Eσ2

Eσµ

Eσε

:ε strain, ntialCircumfere

L

LLLCC

C

−×=

×−×=×−=

µ)21(Eσ

E)σ2(µ

Eσ

Eσµ

Eσε

:ε strain, alLongitudin

LLLCLL

L

×−×=×

×−=×−=

)3..(..............................µ)2(Et4

dp dδdε i.e., C −×

×××

==

C=(pd)/(2t)

L=(pd)/(4t)

L=(pd)/(4t)

C=(pd)/(2t)

)4..(..............................µ)21(Et4

dpLδl ε i.e., L ×−×

×××

==

µ)2(Et4

dp2µ)21(Et4

dp

dd2

LdL

Ld4π

dd24πLdLd

4π

Vdv

dd24πLdLd

4πdV

L.d4πV volume,have We

2

2

2

2

−×××

×+×−×××

=

×+=××

××××+××=

××××+××=

××=

dd

d

= εL+2× εCVdV

)5.......(..........µ)45(Et4

dpVdv i.e., ×−

×××

=

σC=(pd)/(2t)σC=(pd)/(2t)

σL=(pd)/(4t)

σL=(pd)/(4t)

Vdv STRAIN, VOLUMETRIC

JOINT EFFICIENCY

The cylindrical shells like boilers are having two types of joints namely Longitudinal and Circumferential joints. Due to the holes for rivets, the net area of cross section decreases and hence the stresses increase. If the efficiencies of these joints are known, the stresses can be calculated as follows.

Let ηL=Efficiency of Longitudinal jointand ηC=Efficiency of Circumferential joint.

...(1).......... ηt2

dpσ L

C ×××

=

Circumferential stress is given by,

Longitudinal stress is given by,

...(2).......... ηt4dpσ

CL ××

×=

Note: In longitudinal joint, the circumferential stress is developed and in circumferential joint, longitudinal stress is developed.

Exercise Problems

Q.1Calculate the circumferential and longitudinal strains for a boiler of 1000mm diameter when it is subjected to an internal pressure of 1MPa. The wall thickness is such that the safe maximum tensile stress in the boiler material is 35 MPa. Take E=200GPa and µ= 0.25.

(Ans: ε C=0.0001531, ε L=0.00004375)

Q.2A water main 1m in diameter contains water at a pressure head of120m. Find the thickness of the metal if the working stress in the pipe metal is 30 MPa. Take unit weight of water = 10 kN/m3.

(Ans: t=20mm)

Q.3A gravity main 2m in diameter and 15mm in thickness. It is subjected to an internal fluid pressure of 1.5 MPa. Calculate the hoop and longitudinal stresses induced in the pipe material. If a factor of safety 4 was used in the design, what is the ultimate tensile stress in the pipe material?

(Ans: σC=100 MPa, σL=50 MPa, σU=400 MPa)Q.4At a point in a thin cylinder subjected to internal fluid pressure, the value of hoop strain is 600×10-4 (tensile). Compute hoop and longitudinal stresses. How much is the percentage change in the volume of the cylinder? Take E=200GPa and µ= 0.2857. (Ans: σC=140 MPa, σL=70 MPa, %age change=0.135%.)

Q.5A cylindrical tank of 750mm internal diameter and 1.5m long is to be filled with an oil of specific weight 7.85 kN/m3 under a pressure head of 365 m. If the longitudinal joint efficiency is 75% and circumferential joint efficiency is 40%, find the thickness of the tank required. Also calculate the error of calculation in the quantity of oil in the tank if the volumetric strain of the tank is neglected. Take permissible tensile stress as 120 MPa, E=200GPa and µ= 0.3 for the tank material.

(Ans: t=12 mm, error=0.085%.)

Mechanics of Solids [3 1 0 4] CIE 101 / 102 First Year B.E...

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