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Chapter: 1 Fundamental Principles of Mechanics

Mechanics of Solids

1

Mechanics Study of Force & Motion

Mechanics of Solids

2

Mechanics Study of Force & Motion Gross/Overall Motion (Dynamics)

KINETICS Mechanics of Solids

KINEMATICS 3

Mechanics Motion Localized Motion (Deformation) (STATICS)

Rigid Bodies Mechanics of Solids

Deformable bodies 4

Topics to be addressed for Ch-1 • • • • • • • •

Force Moment Types of loading Types of Support & related Reactions Free Body Diagrams Equilibrium Conditions Trusses & Frames Friction

Mechanics of Solids

5

Definitions Matter: Matter is actually made up of atoms and molecules. Since this is too complex, matter is taken to be continuously distributeda continuum; it may be rigid or deformable. Particle: A material with negligibly small dimensions is called a 'particle', the mass being concentrated at a point. The concept is analogous to that of a point in geometry, which has position but no dimensions. Mechanics of Solids

6

However, there is a difference in that a particle, being a material, has mass while a point, being only a geometrical concept, has none. Body: A collection of particles is called a 'body'. It may be a rigid body or an elastic or deformable body. Rigid Body: The particles in a rigid body are so firmly connected together that their relative positions do not change irrespective of the forces acting on it. Thus the size and shape of a rigid body are always maintained constant Mechanics of Solids

7

Elastic Body: A body whose size and shape can change under forces is a deformable body. When the size and shape can be regained on removal of forces, the body is called an elastic body. Scalar Quantity: A quantity which is fully described by its magnitude only is a scalar. Arithmetical operations apply to scalars. Examples are: Time, mass, area and speed.

Mechanics of Solids

8

Vector Quantity: A quantity which is described by its magnitude and also its direction is a vector. Operations of vector algebra are applicable to vectors. Examples are: Force, velocity, moment of a force and displacement . Mass: The quantum of matter in a body is characterized by its 'mass', which is measure of its inertia, or its resistance to change of velocity. It is the property of a body by virtue of which it can experience attraction to other bodies. The symbol used is 'm'. Mechanics of Solids

9

Analysis of Engineering systems Study of forces • (Tensile, Compressive, Shear, Torque, Moments) Study of motion • (Straight, curvilinear, displacement, velocity, acceleration.) Study of deformation • (Elongation, compression, twisting) Application of laws relating the forces to the deformation In some special cases one or more above mentioned steps may become trivial e.g. For rigid bodies deformation will be negligible. If system is at rest, position of system will be independent Mechanics of Solids 10 of time.

Step-By-Step procedure to solve problems in mechanics of solids Select actual/real system of interest. Make assumptions regarding desired characteristics of the system Develop idealized model of the system (Structural and Machine elements) Apply principles of mechanics to the idealized model to compare these calculated results with the behaviour of the actual system Mechanics of Solids

Contd2

11

If the results (calculated and actual) differ widely repeat above steps till satisfactory idealized model is obtained.

Mechanics of Solids

12

Force Force: In physics, a net force acting on a body causes that body to accelerate; that is, to change its velocity. The concept appeared first in the second law of motion of classical mechanics. There are three basic kinds of forces as mentioned below

•Tensile force or pull •Compressive force or push •Shear force Mechanics of Solids

13

Tensile Force or Pull •When equal and opposite forces are applied at the ends of a rod or a bar away from the ends, along its axis, they tend to pull the rod or bar. This kind of a force is called a tensile force or tension.

Mechanics of Solids

14

Compressive Force or Push •When equal and opposite forces act at the ends of a rod or a bar towards the ends along its axis, they tend to push the rod or the bar. This kind of force is called a compressive force or compression.

Mechanics of Solids

15

Shear Force When equal and opposite forces act on the parallel faces of a body, shear occurs on these planes. This tends to cause an angular deformation as shown.

Mechanics of Solids

16

• One Newton is force which gives an acceleration of 1 m/s2 to a mass of 1 kg. • A mass of 1kg on earth’s surface will experience a gravitational force of 9.81 N. • Hence a mass of 1kg has because of gravitational force of the earth, a weight of 9.81 N.

Mechanics of Solids

17

System of Forces

Coplanar

Non-Coplanar

Collinear Concurrent

Concurrent

Parallel, Nonconcurrent

Parallel

Non-concurrent & Non-Parallel Mechanics of Solids

Non-concurrent & Non-Parallel 18

System of Force • Terms to be familiar with – Concurrent Forces (a) – Parallel Forces (b) – Line of Action (c) – Coplanar Forces

(c)

Mechanics of Solids

19

Forces Coplanar & Non-Concurrent F

R

Mechanics of Solids

20

Resultant Force If a force system acting on a body can be replaced by a single force, with exactly the same effect on the body, this single force is said to be the 'resultant' of the force system.

=

F1

F2

Mechanics of Solids

F3

R 21

Law of parallelogram of forces: If two forces acting at a point are represented in magnitude and direction by the adjacent sides of a parallelogram, then their resultant is represented by the diagonal passing through the point of intersection of the two sides representing the forces. F1 F

P Mechanics of Solids

F = (F12 + F22 + 2F1 F2 cosθ)1/2 F2 22

Lami’s Theorem γ P

Q

β

O R

α

P Q R = = sin α sin β sin γ

If three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two. Mechanics of Solids

23

Superposition Principal If a system is acted up on a set of forces, then the net effect of these forces is equal to the summation net effect of individual force. This principle is applicable for linear systems. F2

F2

= F1

F3 Mechanics of Solids

F1

+

+ F3 24

The moment of force

The moment of force F about a point O is r X F. Mechanics of Solids

25

Mechanics of Solids

26

Moment = r×F The moment itself is a vector quantity. Its direction is perpendicular to the plane determined by OP and F. The sense is fixed by the right hand rule From calculus the magnitude of the cross product r×F = F r sinφ, Where r sinφ is the perpendicular distance between point O and vector F. Mechanics of Solids

27

Moment M = F X x F

x Mechanics of Solids

28

Ex. 1 on Moment 15 kN

30o O

4m

Mechanics of Solids

29

Solution for Ex. 1 on Moment 15 sin 30 kN 15 kN

30o O

4m

Mo = 15sin30 kN X 4 m = 30 kN-m (CCW) Mechanics of Solids

30

Ex. 2 on Moment 10 kN

O

4m

Mechanics of Solids

31

Solution for Ex. 2 on Moment 10 kN

O

4m

Mo = 10 kN X 2 m = 20 kN-m (CW) Mechanics of Solids

32

Ex. 3 on Moment

10 kN 2m

O

4m

Mechanics of Solids

33

Solution for Ex. 3 on Moment

10 kN 2m

O

4m

Mo = 10 kN X 2 m = 20 kN-m (CW) Mechanics of Solids

34

Ex. 4 on Moment y F

x

z

d

Mz = F kN X d m = Fd kN-m Mechanics of Solids

35

Ex. 5 on Moment y F

x

z

d

My = F kN X d m = Fd kN-m Mechanics of Solids

36

Ex. 6 on Moment y F1

F2

x

z

d

MR = (Mz2 + My2)1/2 kN-m Mechanics of Solids

37

CALCULATE THE MOMENT 400 N FORCE ABOUT THE POINT O 40 mm 400 cos 60

400 sin 60

Let, CCW moments are +ve and CW -ve

MO = 400 cos 60 (40)x10-3 – 400 sin 60 (120)x10-3 MO = - 33.6 Nm Mechanics of Solids

MO = 33.6 Nm (CW) 38

Couple A special case of moments is a couple. A couple consists of two parallel forces that are equal in magnitude, opposite in direction. It does not produce any translation, only rotation. The resultant force of a couple is zero. BUT, the resultant of a couple is not zero; it is a pure moment. a

d A F

B

F

Sum of the moments of both the forces about B, we have total moment MB = F * (a+d) - (F) * a = Fd (CCW) d - distance between the forces. Mechanics of Solids

39

About couple…. • This result is independent of the location of B. • Moment of a couple is the same about all points in space. • A couple may be characterized by a moment vector without specification of the moment center B, with magnitude Fd. • Encircling arrow indicates moment of a couple. Mechanics of Solids

40

Mechanics of Solids

41

Types of Loading

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42

Types of Loading 1. Concentrated Load / Point Load F

Mechanics of Solids

F

43

Types of Loading 2. Uniformly Distributed Load (UDL)

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44

Types of Loading 3. Uniformly Varying Load (UVL)

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45

Types of Loading 4. Uniformly Varying Load (UVL)

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46

Types of Loading 5. Moment (Pure Bending Moment) y

x MB

Mechanics of Solids

47

Types of Loading 6. Moment y y

x

z

MT

z

MT Twisting Moment Mechanics of Solids

48

Support & Reaction at the Support

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49

Types of Supports & Reactions 1. Simple Support F

A

B

F

Mechanics of Solids

RA

RB

50

Types of Supports & Reactions 2. Simple Support with hinge at A & B F

A

B

F HB

HA Mechanics of Solids

RA

RB

51

Types of Supports & Reactions 3. Roller Support

F

A

B

F

Mechanics of Solids

RA

RB

52

Types of Supports & Reactions 4. Roller support & Simple support F B

A

F

Mechanics of Solids

RA

RB

53

Types of Supports & Reactions 5. Fixed support & roller support F

B

A

F (RA)H MA

(RA)V

Mechanics of Solids

(RB)V 54

Types of Supports & Reactions 6. Smooth/Frictionless Support

N

Mechanics of Solids

55

Types of Supports & Reactions 7. Frictional Support

FS = µN N

Mechanics of Solids

56

Force Transmitting properties of some idealized mechanical elements

Mechanics of Solids

57

Mechanics of Solids

58

Mechanics of Solids

59

Mechanics of Solids

60

Equilibrium Conditions

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61

Equilibrium Static Eqlbm

Dynamic Eqlbm

If the resultant force acting on a particle is zero then that can be called as equilibrium Dynamic Equilibrium The body is said to be in equilibrium condition when the acceleration is zero

Static Equilibrium The static body is in equilibrium condition if the resultant force acting on it is zero Mechanics of Solids

62

Necessary and sufficient condition for body to be in Equilibrium Summation of all the FORCES should be zero

ΣF=0 Summation of all the MOMENTS of all the forces about any arbitrary point should be zero

ΣM=0 Mechanics of Solids

63

Two Force Member • Two forces can’t have random orientation, must be along AB & FA=FB

FA

FA

FB

A B FB

Non-equilibrium condition

Mechanics of Solids

Equilibrium Condition

64

Three Force Member • Three forces can’t have random orientation

FA

FB

FA

FB

FC FC • All must lie in the plane ABC if total moment about each of the points A,B & C is to vanish. • They must all intersect in a common point O. Mechanics of Solids

65

GENERAL COPLANER FORCE SYSTEM • External forces acting on a system in equilibrium all lie in the plane of sketch. • Three of six general scalar equations of equilibrium are satisfied: no force components perpendicular to the plane , & If moments are taken about a point O lying in the plane, the only moment components will be perpendicular to the plane. • Hence for 2-dimensional problem, three independent scalar conditions of equilibrium remains. i.e

∑FX = 0

∑FY =0 and ∑M =0 Mechanics of Solids

66

Statically Determinate System • If it is possible to determine all the forces involved by using only the equilibrium requirements without regard to deformations, such systems are called

statically determinate determinate.. • Use of equilibrium conditions to solve for unknown forces from known forces.

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67

Statically Indeterminate Systems

Mechanics of Solids

68

Ex. 1.8 Crandal Statically Indeterminate Situation 1 kN 3.0 m A

B

2.2 m

Mechanics of Solids

69

Engineering Application • STEPS IN GENERAL METHOD OF ANALYSIS 1. SELECTION OF SYSTEM 2. IDEALIZATION OF SYSTEM CHARACTERISTICS • FOLLOWED BY 1. STUDY OF FORCES & EQUILIBRIUM REQUIREMENTS 2.STUDY OF DEFORMATION & CONDITIONS OF GEOMETRIC FIT 3.APPLICATION OF FORCE-DEFORMATION RELATIONS Mechanics of Solids

70

Free –body diagram The sketch of the isolated component from a system / and all the external forces acting on it is often called a free- body diagram

Mechanics of Solids

71

B B

X

W

Y

A

RX

Z

W

RY

A

RBA RZ

RX

W W

RZ

RY

RAB

FBD for A Mechanics of Solids

FBD for B

72

B W

A Z

W X Y

30o

RBA RZ

W W

RY N

RBA RX

FBD for A Mechanics of Solids

FBD for B

73

Example

Mechanics of Solids

74

Determine the forces at B and C Idealization 1. Neglect the friction at pin joints 2. Ignore the self weight of rods

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75

Free body diagram of the frame

Mechanics of Solids

76

Free body diagrams of bar BD and CD i.e. Isolation from the system

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77

Σ Fy= 0 FB sin 45- 20 = 0 FB = 28.28 kN Σ Fx= 0 FC - FB cos 45 = 0 FC = 20 kN

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78

Trusses

Mechanics of Solids

79

Trusses • Trusses are designed to support the loads and are usually stationary, fully constrained structures. All the members of a truss are TWO FORCE members

Mechanics of Solids

80

Use of truss

1. Construction of roof

2. Bridges

Mechanics of Solids

81

3. Transmission line towers.

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82

Relation between number of joints and member in trusses

m = 2j - 3

If a truss obeys the equation, is called Perfect truss (just rigid truss)

If no. of members are more than this equation then it is called Redundant truss (over rigid truss)

If no. of members are less than this equation then it is called Deficient truss (under rigid truss)

Mechanics of Solids

83

Method of analysis of truss 1.Joint equilibrium method 2.Section equilibrium method

Mechanics of Solids

84

Joint Equilibrium Method In this method using FBD for each joint, equilibrium of the joints of the truss is consider one at a time. At each joint it is necessary to satisfy the static equilibrium equations. This method is suitable, when it is asked to calculate forces in all the members of the truss Only two static equilibrium equations will be satisfied at a time because Σ M = 0 will always be satisfied for concurrent force system. Only two unknown forces will be determined at a time. Mechanics of Solids

85

Step by step procedure of joint equilibrium method 1. Draw a FBD of an entire truss showing the external forces acting on it. (i.e. applied forces/ loads and reactions) 2. Obtain the magnitude and direction of reactions by using the static equilibrium equations. 3. Select a joint where only two unknown forces exit and draw its FBD assuming all unknown forces tensile. 4. Apply the equilibrium equations to the joint, which will give the forces in the members. 5. If the obtained value is –ve then our assumed direction is wrong. Change the nature of force. 6. Apply the above steps to each joint and determine the forces. 7. At the end, redraw the truss and show the correct direction as well as its magnitude of force in each member. Mechanics of Solids

86

W

F

A

W

B

F

C

AX AY

Mechanics of Solids

CY

87

B FAB

FBC

Compression FAB

A

Mechanics of Solids

FBC Tension FAC

FAC

C

88

Calculate the force in each member of the loaded truss.

Mechanics of Solids

89

2 KN

CX AY

CY

Σ FX = 0 = CX – 2

CX = 2 KN

Σ MC = 0 = 2 x 3 – AY x 6

AY = 1 KN

Σ FY = 0 = AY - CY

CY = 1 KN

Mechanics of Solids

90

FAE

FAB

A AY

At Joint A Σ FY = 0 = AY - FAE sin 45 --- FAE = 1.414 kN (C) Σ FX = 0 = FAB – FAE cos 45 ---- FAB = 1 kN (T) Mechanics of Solids

91

E

FAE

FDE

FBE

FAB

A AY

At Joint E Σ FY = 0 = FAE cos 45 - FBE --- FBE = 1 kN (T) Σ FX = 0 = FAE sin 45 - FDE ---- FDE = 1 kN (C) Mechanics of Solids

92

E

FAE

FDE

FBE

FAB

A

FBD

FBC

B AY

At Joint B Σ FY = 0 = FBE - FBD sin 45 --- FBD = 1.414 kN (C) Σ FX = 0 = FBC – FAB – FBD cos 45 -- FBC = 2 kN (T) Mechanics of Solids

93

E

2 kN

FDE

D FAE

FBE

FAB

A

FBD

CX

FBC

B AY

FCD

C

CY

At Joint C Σ FY = 0 = FCD - CY --- FCD = 1 kN (T)

Mechanics of Solids

94

Sr. No. 1 2 3 4 5 6 7

Mechanics of Solids

Member AE AB BE BC BD CD DE

Force 1.414 kN 1 kN 1 kN 2 kN 1.414 kN 1 kN 1 kN

Nature of force Compressive Tensile Tensile Tensile Compressive Tensile Compressive

95

Section equilibrium method This method is also based upon the conditions of static equilibrium. This method is suitable when forces only in few of the members of the truss, particularly away from the supports are required. Conditions of equilibrium are Σ FX = 0

Σ FY = 0

ΣM=0

Where Σ M is the moments of all the forces about any point in the plane of the forces.

Mechanics of Solids

96

Since we have only three equilibrium equation, only maximum three unknown forces can be found at a time. Therefore a care should be taken while taking a section that section line does not cut more than three members in which the forces are unknown.

Mechanics of Solids

97

Step by step procedure of section equilibrium method 1. Draw a FBD of an entire truss showing the external forces acting on it. (i.e. applied forces/ loads and reactions) 2. Obtain the magnitude and direction of reactions by using the static equilibrium equations. 3. Draw an imaginary line which cuts the truss and passes through the members in which we need to find the forces, care must be taken to see that the section line must cut maximum of three members in which forces are unknown. 4. Prepare FBD of either part showing loads, reactions and internal forces in the members, which are cut by section line. 5. Apply the three conditions of equilibrium. If more than three unknowns are to be calculated, then two section lines can be selected and then the truss is to be solved by the above procedure.

Mechanics of Solids

98

Determine the forces in members BC, BF, FJ J

A

F

B

E

C

100 kN

50 kN

D

100 kN

9m

Mechanics of Solids

99

J

F

E 50 kN

AX AY

D B

A

100 kN

C DY

100 kN

Σ MA = 0 = DY X 9 – 100 X 3 – 100 X 6 – 50 X 3 sin 60 DY = 114.43 kN Σ FX = 0 = 50 – AX ------------

AX = 50 kN

Σ FY = 0 = AY – 100 – 100 + DY ------Mechanics of Solids

AY = 85.57 kN 100

J FFJ

F

E 50 kN

FBF FBC

AX AY

D A

B 100 kN

C 100 kN

DY

Taking moment about point B Σ MB = 0 = - AY X 3 + FFJ X 3 sin 60 ----- FFJ = 98.80 kN (C) Σ FY = 0 = AY + FBF sin 60 – 100 ------ FBF = 16.66 kN (T) Σ FX = 0 = FBC + FBF cos 60 – FFJ – AX ------- FBC = 140. 47 kN (T) Mechanics of Solids

101

Sr. No. 1 2 3

Mechanics of Solids

Member BC BF FJ

Force 140.47 kN 16.66 kN 98.80 kN

Nature of force Tensile Tensile Compressive

102

Friction Friction forces are set up whenever a tangential force is applied to a body pressed normally against the surface of another

(a) Body A pressed against B Mechanics of Solids

103

FBD of A

FBD of B

Mechanics of Solids

104

Friction force is the result of interaction of the surface layers of bodies in contact Two separate friction forces are given by FS = fs N Where FS is static friction force FK = fk N Where FK is kinetic friction force fs & fk are static and kinetic coefficients of friction and are intrinsic properties of the interface between the materials in contact. Mechanics of Solids

105

Main Properties of Friction Force (F) 1. If there is no relative motion between a & b, then the frictional force (F) is exactly equal & opposite to the applied tangential force (T). 2. This condition can be maintained for any magnitude of T between zero & certain limiting value Fs, called the static friction force force.. 3.

IF T > FS, sliding will occur.

Mechanics of Solids

106

4. If body A slides on body B, then F acting on body A will have a direction opposite to velocity of A relative to B, & its magnitude will be Fk, called the kinetic friction force.

• FS and FK are proportional to normal force N.

»FS= fS N, FK =fk N

Mechanics of Solids

107

Both coefficients, Depends on materials of bodies in contact Depends on state of lubrication/contamination at the interface Independent of the area of the interface Independent of the roughness of the two interfaces Independent of time of contact and relative velocity respectively

Mechanics of Solids

108

It is of interest to find the range of values of W which will hold the block of weight WO = 500 N in equilibrium of the inclined plane if the coefficient of static friction is fS = 0.5.

Mechanics of Solids

109

Mechanics of Solids

110

Estimate the force in the link AB when a man of weight equal to 800 N stands on top of the ladder.

Mechanics of Solids

111

Mechanics of Solids

112

Mechanics of Solids

113

Draw FBD of blocks A & B

µ1

A (1000 N) B µ2

Mechanics of Solids

(2000 N)

P

114

Draw FBD of blocks A & B

µ1

µ2

Mechanics of Solids

θ 115

Draw FBD of blocks A & B µ1 µ2

A W kN

30o

Mechanics of Solids

116

Problem: A ladder of weight 390 N and 6 m long is placed against a vertical wall at an angle of 300 as shown in figure. The coefficient of friction between the ladder and the wall is 0.25 and that between ladder and floor is 0.38. Find how high a man of weight 1170 N can ascend, before the ladder begins to slip.

Mechanics of Solids

117

Problem A cantilever truss is loaded as shown in figure. Find the value of W which will produce a force of 150 kN magnitude in the member AB.

Mechanics of Solids

118

Problem By using the method of sections find the forces in the members BC, BF and BE of the cantilever truss shown in figure.

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119

What we studied in Ch-1 • • • • • • • •

Force Moment Types of loading Types of Support & related Reactions Free Body Diagrams Equilibrium Conditions Trusses Friction

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120

Numericals are based on • • • •

Free Body Diagrams Force/Reaction Determination Trusses Friction

Problems List: 6, 9, 11, 12, 13, 15, 16, 17, 21, 31, 37, 43, 44, 45.

Mechanics of Solids

121

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Mechanics of Solids

1

Mechanics Study of Force & Motion

Mechanics of Solids

2

Mechanics Study of Force & Motion Gross/Overall Motion (Dynamics)

KINETICS Mechanics of Solids

KINEMATICS 3

Mechanics Motion Localized Motion (Deformation) (STATICS)

Rigid Bodies Mechanics of Solids

Deformable bodies 4

Topics to be addressed for Ch-1 • • • • • • • •

Force Moment Types of loading Types of Support & related Reactions Free Body Diagrams Equilibrium Conditions Trusses & Frames Friction

Mechanics of Solids

5

Definitions Matter: Matter is actually made up of atoms and molecules. Since this is too complex, matter is taken to be continuously distributeda continuum; it may be rigid or deformable. Particle: A material with negligibly small dimensions is called a 'particle', the mass being concentrated at a point. The concept is analogous to that of a point in geometry, which has position but no dimensions. Mechanics of Solids

6

However, there is a difference in that a particle, being a material, has mass while a point, being only a geometrical concept, has none. Body: A collection of particles is called a 'body'. It may be a rigid body or an elastic or deformable body. Rigid Body: The particles in a rigid body are so firmly connected together that their relative positions do not change irrespective of the forces acting on it. Thus the size and shape of a rigid body are always maintained constant Mechanics of Solids

7

Elastic Body: A body whose size and shape can change under forces is a deformable body. When the size and shape can be regained on removal of forces, the body is called an elastic body. Scalar Quantity: A quantity which is fully described by its magnitude only is a scalar. Arithmetical operations apply to scalars. Examples are: Time, mass, area and speed.

Mechanics of Solids

8

Vector Quantity: A quantity which is described by its magnitude and also its direction is a vector. Operations of vector algebra are applicable to vectors. Examples are: Force, velocity, moment of a force and displacement . Mass: The quantum of matter in a body is characterized by its 'mass', which is measure of its inertia, or its resistance to change of velocity. It is the property of a body by virtue of which it can experience attraction to other bodies. The symbol used is 'm'. Mechanics of Solids

9

Analysis of Engineering systems Study of forces • (Tensile, Compressive, Shear, Torque, Moments) Study of motion • (Straight, curvilinear, displacement, velocity, acceleration.) Study of deformation • (Elongation, compression, twisting) Application of laws relating the forces to the deformation In some special cases one or more above mentioned steps may become trivial e.g. For rigid bodies deformation will be negligible. If system is at rest, position of system will be independent Mechanics of Solids 10 of time.

Step-By-Step procedure to solve problems in mechanics of solids Select actual/real system of interest. Make assumptions regarding desired characteristics of the system Develop idealized model of the system (Structural and Machine elements) Apply principles of mechanics to the idealized model to compare these calculated results with the behaviour of the actual system Mechanics of Solids

Contd2

11

If the results (calculated and actual) differ widely repeat above steps till satisfactory idealized model is obtained.

Mechanics of Solids

12

Force Force: In physics, a net force acting on a body causes that body to accelerate; that is, to change its velocity. The concept appeared first in the second law of motion of classical mechanics. There are three basic kinds of forces as mentioned below

•Tensile force or pull •Compressive force or push •Shear force Mechanics of Solids

13

Tensile Force or Pull •When equal and opposite forces are applied at the ends of a rod or a bar away from the ends, along its axis, they tend to pull the rod or bar. This kind of a force is called a tensile force or tension.

Mechanics of Solids

14

Compressive Force or Push •When equal and opposite forces act at the ends of a rod or a bar towards the ends along its axis, they tend to push the rod or the bar. This kind of force is called a compressive force or compression.

Mechanics of Solids

15

Shear Force When equal and opposite forces act on the parallel faces of a body, shear occurs on these planes. This tends to cause an angular deformation as shown.

Mechanics of Solids

16

• One Newton is force which gives an acceleration of 1 m/s2 to a mass of 1 kg. • A mass of 1kg on earth’s surface will experience a gravitational force of 9.81 N. • Hence a mass of 1kg has because of gravitational force of the earth, a weight of 9.81 N.

Mechanics of Solids

17

System of Forces

Coplanar

Non-Coplanar

Collinear Concurrent

Concurrent

Parallel, Nonconcurrent

Parallel

Non-concurrent & Non-Parallel Mechanics of Solids

Non-concurrent & Non-Parallel 18

System of Force • Terms to be familiar with – Concurrent Forces (a) – Parallel Forces (b) – Line of Action (c) – Coplanar Forces

(c)

Mechanics of Solids

19

Forces Coplanar & Non-Concurrent F

R

Mechanics of Solids

20

Resultant Force If a force system acting on a body can be replaced by a single force, with exactly the same effect on the body, this single force is said to be the 'resultant' of the force system.

=

F1

F2

Mechanics of Solids

F3

R 21

Law of parallelogram of forces: If two forces acting at a point are represented in magnitude and direction by the adjacent sides of a parallelogram, then their resultant is represented by the diagonal passing through the point of intersection of the two sides representing the forces. F1 F

P Mechanics of Solids

F = (F12 + F22 + 2F1 F2 cosθ)1/2 F2 22

Lami’s Theorem γ P

Q

β

O R

α

P Q R = = sin α sin β sin γ

If three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two. Mechanics of Solids

23

Superposition Principal If a system is acted up on a set of forces, then the net effect of these forces is equal to the summation net effect of individual force. This principle is applicable for linear systems. F2

F2

= F1

F3 Mechanics of Solids

F1

+

+ F3 24

The moment of force

The moment of force F about a point O is r X F. Mechanics of Solids

25

Mechanics of Solids

26

Moment = r×F The moment itself is a vector quantity. Its direction is perpendicular to the plane determined by OP and F. The sense is fixed by the right hand rule From calculus the magnitude of the cross product r×F = F r sinφ, Where r sinφ is the perpendicular distance between point O and vector F. Mechanics of Solids

27

Moment M = F X x F

x Mechanics of Solids

28

Ex. 1 on Moment 15 kN

30o O

4m

Mechanics of Solids

29

Solution for Ex. 1 on Moment 15 sin 30 kN 15 kN

30o O

4m

Mo = 15sin30 kN X 4 m = 30 kN-m (CCW) Mechanics of Solids

30

Ex. 2 on Moment 10 kN

O

4m

Mechanics of Solids

31

Solution for Ex. 2 on Moment 10 kN

O

4m

Mo = 10 kN X 2 m = 20 kN-m (CW) Mechanics of Solids

32

Ex. 3 on Moment

10 kN 2m

O

4m

Mechanics of Solids

33

Solution for Ex. 3 on Moment

10 kN 2m

O

4m

Mo = 10 kN X 2 m = 20 kN-m (CW) Mechanics of Solids

34

Ex. 4 on Moment y F

x

z

d

Mz = F kN X d m = Fd kN-m Mechanics of Solids

35

Ex. 5 on Moment y F

x

z

d

My = F kN X d m = Fd kN-m Mechanics of Solids

36

Ex. 6 on Moment y F1

F2

x

z

d

MR = (Mz2 + My2)1/2 kN-m Mechanics of Solids

37

CALCULATE THE MOMENT 400 N FORCE ABOUT THE POINT O 40 mm 400 cos 60

400 sin 60

Let, CCW moments are +ve and CW -ve

MO = 400 cos 60 (40)x10-3 – 400 sin 60 (120)x10-3 MO = - 33.6 Nm Mechanics of Solids

MO = 33.6 Nm (CW) 38

Couple A special case of moments is a couple. A couple consists of two parallel forces that are equal in magnitude, opposite in direction. It does not produce any translation, only rotation. The resultant force of a couple is zero. BUT, the resultant of a couple is not zero; it is a pure moment. a

d A F

B

F

Sum of the moments of both the forces about B, we have total moment MB = F * (a+d) - (F) * a = Fd (CCW) d - distance between the forces. Mechanics of Solids

39

About couple…. • This result is independent of the location of B. • Moment of a couple is the same about all points in space. • A couple may be characterized by a moment vector without specification of the moment center B, with magnitude Fd. • Encircling arrow indicates moment of a couple. Mechanics of Solids

40

Mechanics of Solids

41

Types of Loading

Mechanics of Solids

42

Types of Loading 1. Concentrated Load / Point Load F

Mechanics of Solids

F

43

Types of Loading 2. Uniformly Distributed Load (UDL)

Mechanics of Solids

44

Types of Loading 3. Uniformly Varying Load (UVL)

Mechanics of Solids

45

Types of Loading 4. Uniformly Varying Load (UVL)

Mechanics of Solids

46

Types of Loading 5. Moment (Pure Bending Moment) y

x MB

Mechanics of Solids

47

Types of Loading 6. Moment y y

x

z

MT

z

MT Twisting Moment Mechanics of Solids

48

Support & Reaction at the Support

Mechanics of Solids

49

Types of Supports & Reactions 1. Simple Support F

A

B

F

Mechanics of Solids

RA

RB

50

Types of Supports & Reactions 2. Simple Support with hinge at A & B F

A

B

F HB

HA Mechanics of Solids

RA

RB

51

Types of Supports & Reactions 3. Roller Support

F

A

B

F

Mechanics of Solids

RA

RB

52

Types of Supports & Reactions 4. Roller support & Simple support F B

A

F

Mechanics of Solids

RA

RB

53

Types of Supports & Reactions 5. Fixed support & roller support F

B

A

F (RA)H MA

(RA)V

Mechanics of Solids

(RB)V 54

Types of Supports & Reactions 6. Smooth/Frictionless Support

N

Mechanics of Solids

55

Types of Supports & Reactions 7. Frictional Support

FS = µN N

Mechanics of Solids

56

Force Transmitting properties of some idealized mechanical elements

Mechanics of Solids

57

Mechanics of Solids

58

Mechanics of Solids

59

Mechanics of Solids

60

Equilibrium Conditions

Mechanics of Solids

61

Equilibrium Static Eqlbm

Dynamic Eqlbm

If the resultant force acting on a particle is zero then that can be called as equilibrium Dynamic Equilibrium The body is said to be in equilibrium condition when the acceleration is zero

Static Equilibrium The static body is in equilibrium condition if the resultant force acting on it is zero Mechanics of Solids

62

Necessary and sufficient condition for body to be in Equilibrium Summation of all the FORCES should be zero

ΣF=0 Summation of all the MOMENTS of all the forces about any arbitrary point should be zero

ΣM=0 Mechanics of Solids

63

Two Force Member • Two forces can’t have random orientation, must be along AB & FA=FB

FA

FA

FB

A B FB

Non-equilibrium condition

Mechanics of Solids

Equilibrium Condition

64

Three Force Member • Three forces can’t have random orientation

FA

FB

FA

FB

FC FC • All must lie in the plane ABC if total moment about each of the points A,B & C is to vanish. • They must all intersect in a common point O. Mechanics of Solids

65

GENERAL COPLANER FORCE SYSTEM • External forces acting on a system in equilibrium all lie in the plane of sketch. • Three of six general scalar equations of equilibrium are satisfied: no force components perpendicular to the plane , & If moments are taken about a point O lying in the plane, the only moment components will be perpendicular to the plane. • Hence for 2-dimensional problem, three independent scalar conditions of equilibrium remains. i.e

∑FX = 0

∑FY =0 and ∑M =0 Mechanics of Solids

66

Statically Determinate System • If it is possible to determine all the forces involved by using only the equilibrium requirements without regard to deformations, such systems are called

statically determinate determinate.. • Use of equilibrium conditions to solve for unknown forces from known forces.

Mechanics of Solids

67

Statically Indeterminate Systems

Mechanics of Solids

68

Ex. 1.8 Crandal Statically Indeterminate Situation 1 kN 3.0 m A

B

2.2 m

Mechanics of Solids

69

Engineering Application • STEPS IN GENERAL METHOD OF ANALYSIS 1. SELECTION OF SYSTEM 2. IDEALIZATION OF SYSTEM CHARACTERISTICS • FOLLOWED BY 1. STUDY OF FORCES & EQUILIBRIUM REQUIREMENTS 2.STUDY OF DEFORMATION & CONDITIONS OF GEOMETRIC FIT 3.APPLICATION OF FORCE-DEFORMATION RELATIONS Mechanics of Solids

70

Free –body diagram The sketch of the isolated component from a system / and all the external forces acting on it is often called a free- body diagram

Mechanics of Solids

71

B B

X

W

Y

A

RX

Z

W

RY

A

RBA RZ

RX

W W

RZ

RY

RAB

FBD for A Mechanics of Solids

FBD for B

72

B W

A Z

W X Y

30o

RBA RZ

W W

RY N

RBA RX

FBD for A Mechanics of Solids

FBD for B

73

Example

Mechanics of Solids

74

Determine the forces at B and C Idealization 1. Neglect the friction at pin joints 2. Ignore the self weight of rods

Mechanics of Solids

75

Free body diagram of the frame

Mechanics of Solids

76

Free body diagrams of bar BD and CD i.e. Isolation from the system

Mechanics of Solids

77

Σ Fy= 0 FB sin 45- 20 = 0 FB = 28.28 kN Σ Fx= 0 FC - FB cos 45 = 0 FC = 20 kN

Mechanics of Solids

78

Trusses

Mechanics of Solids

79

Trusses • Trusses are designed to support the loads and are usually stationary, fully constrained structures. All the members of a truss are TWO FORCE members

Mechanics of Solids

80

Use of truss

1. Construction of roof

2. Bridges

Mechanics of Solids

81

3. Transmission line towers.

Mechanics of Solids

82

Relation between number of joints and member in trusses

m = 2j - 3

If a truss obeys the equation, is called Perfect truss (just rigid truss)

If no. of members are more than this equation then it is called Redundant truss (over rigid truss)

If no. of members are less than this equation then it is called Deficient truss (under rigid truss)

Mechanics of Solids

83

Method of analysis of truss 1.Joint equilibrium method 2.Section equilibrium method

Mechanics of Solids

84

Joint Equilibrium Method In this method using FBD for each joint, equilibrium of the joints of the truss is consider one at a time. At each joint it is necessary to satisfy the static equilibrium equations. This method is suitable, when it is asked to calculate forces in all the members of the truss Only two static equilibrium equations will be satisfied at a time because Σ M = 0 will always be satisfied for concurrent force system. Only two unknown forces will be determined at a time. Mechanics of Solids

85

Step by step procedure of joint equilibrium method 1. Draw a FBD of an entire truss showing the external forces acting on it. (i.e. applied forces/ loads and reactions) 2. Obtain the magnitude and direction of reactions by using the static equilibrium equations. 3. Select a joint where only two unknown forces exit and draw its FBD assuming all unknown forces tensile. 4. Apply the equilibrium equations to the joint, which will give the forces in the members. 5. If the obtained value is –ve then our assumed direction is wrong. Change the nature of force. 6. Apply the above steps to each joint and determine the forces. 7. At the end, redraw the truss and show the correct direction as well as its magnitude of force in each member. Mechanics of Solids

86

W

F

A

W

B

F

C

AX AY

Mechanics of Solids

CY

87

B FAB

FBC

Compression FAB

A

Mechanics of Solids

FBC Tension FAC

FAC

C

88

Calculate the force in each member of the loaded truss.

Mechanics of Solids

89

2 KN

CX AY

CY

Σ FX = 0 = CX – 2

CX = 2 KN

Σ MC = 0 = 2 x 3 – AY x 6

AY = 1 KN

Σ FY = 0 = AY - CY

CY = 1 KN

Mechanics of Solids

90

FAE

FAB

A AY

At Joint A Σ FY = 0 = AY - FAE sin 45 --- FAE = 1.414 kN (C) Σ FX = 0 = FAB – FAE cos 45 ---- FAB = 1 kN (T) Mechanics of Solids

91

E

FAE

FDE

FBE

FAB

A AY

At Joint E Σ FY = 0 = FAE cos 45 - FBE --- FBE = 1 kN (T) Σ FX = 0 = FAE sin 45 - FDE ---- FDE = 1 kN (C) Mechanics of Solids

92

E

FAE

FDE

FBE

FAB

A

FBD

FBC

B AY

At Joint B Σ FY = 0 = FBE - FBD sin 45 --- FBD = 1.414 kN (C) Σ FX = 0 = FBC – FAB – FBD cos 45 -- FBC = 2 kN (T) Mechanics of Solids

93

E

2 kN

FDE

D FAE

FBE

FAB

A

FBD

CX

FBC

B AY

FCD

C

CY

At Joint C Σ FY = 0 = FCD - CY --- FCD = 1 kN (T)

Mechanics of Solids

94

Sr. No. 1 2 3 4 5 6 7

Mechanics of Solids

Member AE AB BE BC BD CD DE

Force 1.414 kN 1 kN 1 kN 2 kN 1.414 kN 1 kN 1 kN

Nature of force Compressive Tensile Tensile Tensile Compressive Tensile Compressive

95

Section equilibrium method This method is also based upon the conditions of static equilibrium. This method is suitable when forces only in few of the members of the truss, particularly away from the supports are required. Conditions of equilibrium are Σ FX = 0

Σ FY = 0

ΣM=0

Where Σ M is the moments of all the forces about any point in the plane of the forces.

Mechanics of Solids

96

Since we have only three equilibrium equation, only maximum three unknown forces can be found at a time. Therefore a care should be taken while taking a section that section line does not cut more than three members in which the forces are unknown.

Mechanics of Solids

97

Step by step procedure of section equilibrium method 1. Draw a FBD of an entire truss showing the external forces acting on it. (i.e. applied forces/ loads and reactions) 2. Obtain the magnitude and direction of reactions by using the static equilibrium equations. 3. Draw an imaginary line which cuts the truss and passes through the members in which we need to find the forces, care must be taken to see that the section line must cut maximum of three members in which forces are unknown. 4. Prepare FBD of either part showing loads, reactions and internal forces in the members, which are cut by section line. 5. Apply the three conditions of equilibrium. If more than three unknowns are to be calculated, then two section lines can be selected and then the truss is to be solved by the above procedure.

Mechanics of Solids

98

Determine the forces in members BC, BF, FJ J

A

F

B

E

C

100 kN

50 kN

D

100 kN

9m

Mechanics of Solids

99

J

F

E 50 kN

AX AY

D B

A

100 kN

C DY

100 kN

Σ MA = 0 = DY X 9 – 100 X 3 – 100 X 6 – 50 X 3 sin 60 DY = 114.43 kN Σ FX = 0 = 50 – AX ------------

AX = 50 kN

Σ FY = 0 = AY – 100 – 100 + DY ------Mechanics of Solids

AY = 85.57 kN 100

J FFJ

F

E 50 kN

FBF FBC

AX AY

D A

B 100 kN

C 100 kN

DY

Taking moment about point B Σ MB = 0 = - AY X 3 + FFJ X 3 sin 60 ----- FFJ = 98.80 kN (C) Σ FY = 0 = AY + FBF sin 60 – 100 ------ FBF = 16.66 kN (T) Σ FX = 0 = FBC + FBF cos 60 – FFJ – AX ------- FBC = 140. 47 kN (T) Mechanics of Solids

101

Sr. No. 1 2 3

Mechanics of Solids

Member BC BF FJ

Force 140.47 kN 16.66 kN 98.80 kN

Nature of force Tensile Tensile Compressive

102

Friction Friction forces are set up whenever a tangential force is applied to a body pressed normally against the surface of another

(a) Body A pressed against B Mechanics of Solids

103

FBD of A

FBD of B

Mechanics of Solids

104

Friction force is the result of interaction of the surface layers of bodies in contact Two separate friction forces are given by FS = fs N Where FS is static friction force FK = fk N Where FK is kinetic friction force fs & fk are static and kinetic coefficients of friction and are intrinsic properties of the interface between the materials in contact. Mechanics of Solids

105

Main Properties of Friction Force (F) 1. If there is no relative motion between a & b, then the frictional force (F) is exactly equal & opposite to the applied tangential force (T). 2. This condition can be maintained for any magnitude of T between zero & certain limiting value Fs, called the static friction force force.. 3.

IF T > FS, sliding will occur.

Mechanics of Solids

106

4. If body A slides on body B, then F acting on body A will have a direction opposite to velocity of A relative to B, & its magnitude will be Fk, called the kinetic friction force.

• FS and FK are proportional to normal force N.

»FS= fS N, FK =fk N

Mechanics of Solids

107

Both coefficients, Depends on materials of bodies in contact Depends on state of lubrication/contamination at the interface Independent of the area of the interface Independent of the roughness of the two interfaces Independent of time of contact and relative velocity respectively

Mechanics of Solids

108

It is of interest to find the range of values of W which will hold the block of weight WO = 500 N in equilibrium of the inclined plane if the coefficient of static friction is fS = 0.5.

Mechanics of Solids

109

Mechanics of Solids

110

Estimate the force in the link AB when a man of weight equal to 800 N stands on top of the ladder.

Mechanics of Solids

111

Mechanics of Solids

112

Mechanics of Solids

113

Draw FBD of blocks A & B

µ1

A (1000 N) B µ2

Mechanics of Solids

(2000 N)

P

114

Draw FBD of blocks A & B

µ1

µ2

Mechanics of Solids

θ 115

Draw FBD of blocks A & B µ1 µ2

A W kN

30o

Mechanics of Solids

116

Problem: A ladder of weight 390 N and 6 m long is placed against a vertical wall at an angle of 300 as shown in figure. The coefficient of friction between the ladder and the wall is 0.25 and that between ladder and floor is 0.38. Find how high a man of weight 1170 N can ascend, before the ladder begins to slip.

Mechanics of Solids

117

Problem A cantilever truss is loaded as shown in figure. Find the value of W which will produce a force of 150 kN magnitude in the member AB.

Mechanics of Solids

118

Problem By using the method of sections find the forces in the members BC, BF and BE of the cantilever truss shown in figure.

Mechanics of Solids

119

What we studied in Ch-1 • • • • • • • •

Force Moment Types of loading Types of Support & related Reactions Free Body Diagrams Equilibrium Conditions Trusses Friction

Mechanics of Solids

120

Numericals are based on • • • •

Free Body Diagrams Force/Reaction Determination Trusses Friction

Problems List: 6, 9, 11, 12, 13, 15, 16, 17, 21, 31, 37, 43, 44, 45.

Mechanics of Solids

121

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