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

Mechanics of Solids [3 1 0 4]

CIE 101 / 102CIE 101 / 102

First Year B.E. Degree

Mechanics of Solids

PART- IIPART- I

Mechanics of Deformable

BodiesMechanics 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

and Impulse- Momentum principle.

PART II Mechanics of Deformable bodies

6. 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 Reference

1.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

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

or motion of bodies under the action of forces’.

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 Deformable Ideal Viscous Compressible

Branches of Mechanics

Bodies

Statics Dynamics

Kinematics Kinetics

Strength of

Materials

Theory of

ElasticityTheory of Plasticity

Ideal

Fluids

Viscous

Fluids

Compressible

Fluids

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.

Concept of Rigid Body :

application of load.

Actually solid bodies are never rigid; they deform under the

action of applied forces. In those cases where this deformation

is 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 fourA 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(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

θθθθ

=

same isA particle,on effect

321

external

FFFR ++=F3

F1

F2

AA

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 forces

is represented in magnitude and direction by the diagonal of

Resultant of two forces acting at a point

is represented in magnitude and direction by the diagonal of

the parallelogram passing through the same point.’

BC

AO

P2

P1

Rα

θ

Contd..

In the above figure, P1 and P2, represented by the sides OA and OB have

BC

AO

P2

P1

Rα

θ

In the above figure, P1 and P2, represented by the sides OA and OB have

R 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 = √P12 + 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 )]

BC

AO

P2

P1

Rα

θ

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

Resultant of two forces acting at a point at right angle

222

1PPR +=

1

2tanP

P=θ

θ

B C

AO

P2

P1

R

If α = 90 , (two forces acting at a point are at right angle)

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 intip to tail order, the third side of the triangle represents both in

magnitude and direction the resultant force F, the sense of the same ismagnitude 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’.

Let F1 and F2 be the two forces acting at a point A and θ is the

included angle.


θ

F1

F

θ

F1R

=

A F2F2

‘Arrange the two forces as two sides of a triangle taken in tip totail 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’.


θ

A

F1

F2

θ

F1

F2

R=

ββββ

αααα

R

F1

F2

(180 - α - β) = θ

)180sin(sinsin

21

βααβ −−==

RFF

where α and β are the angles made by the resultant force

with the force F1 and F2 respectively.

Component of a force, in simple terms, is the effect of a

force 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 perpendicular

components, one along the x-direction and the other along y-

Component of a force :

components, 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

Fyθx

F

θθθθx

F

Fx

= Fyθθθθx

F

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

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’.

ββββF F2

Oblique component of a force

FF2 M

N

ααααF1

F2

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

The resolved part of the force F along OM and ON can

obtained by using the equation of a triangle.

ββββαααα

F1

O

Sign Convention for force components:

+ve

+vex

x

yy

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

ConcurrentNon-concurrent

Concurrent Non-concurrentConcurrent Non-concurrent

A force that can replace a set of forces, in a force system,

and cause the same ‘external effect’ is called the Resultant.

Like parallel Unlike parallel

Like parallel Unlike parallel

Numerical Problems & Solutions(Q1.1)

Resolve the forces shown in figure along x and y

directions.

20 kN

250

25

35 kN

60 kN

3

2


solution:

20 kN

250

20 cos θx

20 sin θx = 20 sin65= 20 cos65

35 kN

60 kN

3

2

60 cos θx

60 sin θx

= 60 cos33.7

= 60 sin33.7

(Q1.1)

solution:

20 kN

250

35 kN

3

2

20 cos θx

60 sin θx

20 sin θx = 20 sin65= 20 cos65

= 60 sin33.7Answer:

60 kN

3

60 cos θx

60 sin θx

= 60 cos33.7

= 60 sin33.7

Force X-comp Y-comp

35kN - 35 0

20kN - 20 cos 65 -20 sin 65

60kN - 60 cos 33.7 + 60 sin 33.7

Answer:


Resolve the forces shown in figure along x and y

directions.

15 kN

105 kN

150

75 kN

45 kN

400

60 kN

350

solution:

15 cos150

15 sin150

15 kN

150

105 kN

(Q1.2)

45 cos550

45 sin550

60 sin400

60 cos400

150

75 kN

45 kN

400

60 kN

350

550

15 cos150

45 cos550

45 sin550

60 sin400

60 cos400

15 sin150

15 kN

150

105 kN

75 kN

45 kN

400

60 kN

350

550

(Q1.2)

45 sin550

Force X-comp. Y-comp

105 0 +105

75 -75 0

15 + 15 cos15 + 15 sin15

45 - 45 cos55 - 45 sin55

60 + 60 cos40 - 60 sin40


Obtain the resultant of the concurrent coplanar forces acting

as shown in figure

15 kN

105 kN

15 kN

150

75 kN

45 kN

400

60 kN35

0

Numerical Problems & Solutions

solution:

15 cos150

15 sin150

15 kN

150

105 kN

(Q1.3)

45 cos550

45 sin550

60 sin400

60 cos400

150

75 kN

45 kN

400

60 kN

350

550


15 cos150

45 cos550

45 sin550

60 sin400

60 cos400

15 sin150

15 kN

150

105 kN

75 kN

45 kN

400

60 kN

350

550


105 0 +105

75 -75 0

15 + 15 cos15 + 15 sin15

45 - 45 cos55 - 45 sin55

(Q1.3)

45 sin550 45 - 45 cos55 - 45 sin55

60 + 60 cos40 - 60 sin40

------- --------------- ----------------

R ΣFx = ΣFy =

- 40.359 + 33.453


∑ Fx = – 75 + 15 cos 15 – 45 cos 55 + 60 Cos 40

= - 40.359 kN = 40.359 kN

∑ Fy = + 105 + 15 Sin 15 – 45 sin 55 – 60 Sin 40

= + 33.453 kN

(Q1.3)

= + 33.453 kN

ΣFx = 40.359kN

ΣFy = 33.453 kN

R

θx

0

1-

69.39

tan ;tan

42.52

=

Σ

Σ=

Σ

Σ=

=Σ+Σ=

θ

θθy

x

x

x

y

x

yx

F

F

F

F

kNFFR

Answer:

Obtain the resultant of the concurrent coplanar forces

acting as shown in figure.

120

100kN50kNº

(Q1.4)

75kN

1202

3

3012

25kN

º

Solution:

(Q1.4)

120

100kN50kN

º 100 sin θx

100 cos θx

= 100 cos 33.750 sin θx = 50 sin 26.3

75kN

1202

3

301

2

25kN

º

º50 cos θx

= 50 cos 26.3

= 100 sin 33.7

75 cos θx

= 75 cos 30

25 cos θx

= 25 cos 63.43

75 sin θx

= 75 sin 30

25 sin θx

= 25 sin 63.43

Solution:

(Q1.4)

75kN

1202

3

3012

25kN

100kN50kN

º50 cos 26.3 100 sin 33.7

100 cos 33.750 sin 26.3

75 cos 3025 cos 63.43

75 sin 3025 sin 63.43


100 -100 cos33.7 -100 sin33.7

75 -50 cos26.3 +50 sin26.3

15 -25cos 63.43 -25 sin63.43 75 sin 3025 sin 63.4315 -25cos 63.43 -25 sin63.43

45 +75 cos30 -75 sin30

------- --------------- ----------------

R ΣFx = ΣFy =

- 74.26 kN -93.17 kN

∑Fx = -50 Cos 26.31- 100 Cos33.69 – 25 Cos 63.43 + 75 Cos 30

74.26kN= -74.26kN =

∑FY = 50sin26.31- 100sin 33.69 – 75sin30 – 25sin63.43

(Q1.4)

Contd..

= 93.17kN= -93.17kN

∑Fx

∑FyR

θ

Answers:

(Q1.4)

R = √(∑Fx) 2 + (∑Fy) 2 = 119.14 kN

Θ = tan-1(∑Fy / ∑Fx ) = 51.44o

A system of concurrent coplanar forces has five forces of which

only four are shown in figure. If the resultant is a force of

magnitude R = 250 N acting rightwards along the horizontal,

find the unknown fifth force.

150N200N

(Q1.5)

120N

150N

50N

200N

45º

50°

110º

�- Assume the fifth force F5 in the first quadrant, at an angle α, as

shown.

�The 150 N force makes an angle of 20o w.r.t. horizontal

�R is the resultant of Five forces including F5

Solution:

(Q1.5)

150N

50N

200N

120N

45°

50°110 º

F5

α

R =250 N

20º

�- Resolve the forces along X & Y axis

150N 200N

Solution:

(Q1.5)

F5y=F5 sin α

50N120N

45°

50°110 º

F5

α

R =250 N

20º

F5y=F5 sin α

F5x=F5 cos α

Solution:(Q1.5)

150N

50N

200N

120N

45°

50°

F5

α

R =250 N

20º

F5y=F5 sin α

F5x=F5 cos α


F5 +F5 cosα +F5 sinα

50 -50 cos45 +50 sin45

200 +200cos 50 +200 sin 50

120 0 -120

150 -150 cos20 +150 sin20

------- --------------- ----------------

R ΣFx = ΣFy =

+250 kN 0

∑FX = R = + 250 & ΣFy = 0

ΣFx = + 250 = 200 cos 50 – 150 cos 20 – 50 cos 45 + F5 cos α

∴ F5 cos α = +297.75 N

because the resultant is acting along x-direction

∑FY = 0 = F5 sin α + 200sin 50 + 150 sin 20 – 120 + 50 sin 45

(Q1.5)

Y 5

F5 sin α= -119.87 N

α = 21.90º

119.87N

297.75N

F5 = 320.97N

tan α = F5sin α /F5cos α

=0.402

α = 21.90º

F5= 320.97N

F5cosα =

F5sinα =

Answers

A system of concurrent coplanar forces has four forces of which

only three are shown in figure. If the resultant is a force R =

100N acting as indicated, obtain the unknown fourth force.

75N

(Q1.6)

60°

45°

R=100N

50N

25N

70°

40°

(Q1.6)

75N

75 sin70

25 sin30 F4

F4 sin α

� Assume the fourth force (F4) in the first quadrant, at an angle α, as shown.

�The 25 N force makes an angle of 30o w.r.t. horizontal

�R is the resultant of Four forces including F4

60°

45°

R=100N

50N

25N

70°

40°

30°

75 cos70

25 cos30

50 cos45

R cos40

R sin40

α

F4 cos α

α

(Q1.6)

60°

45°

50N

25N

70°

40°

75N

30°

75 cos70

25 cos30

50 cos45

R cos40

R sin40

75 sin70

25 sin30


F4 +F4 cosα +F4 sinα

50 - 50 cos45 +50 sin45

25 - 25 cos30 +25 sin30

75 +75 cos70 +75 sin70

R=100N

50N50 cos45R sin40------- --------------- ----------------

R ΣFx = - R cos40 ΣFy = -R sin40

= -100 cos40kN = - 100 sin40kN

ΣFx = -Rcos40 = F4cosα + 75cos70 – 50cos45 – 25sin60

F4cosα = - 45.25N

ΣFx = -Rcos40+ve

ΣFy = -Rsin40 = F4sinα + 75sin70+25cos60+50sin45

∴ F4sinα = -182.61N ;

+ve ΣFy = -Rsin40

(Q1.6)

α= 76.08º

45.25N

F4=188.13N

182.61N

F4cosα =

F4sinα =

Answers:

tan α = (F4sinα /F4cosα)

α = 76.08º

& F4 =188.13N

The resultant of a system of concurrent coplanar forces is a force

acting vertically upwards. Find the magnitude of the resultant, and the

force F4 acting as shown in figure.

10 kN

70°

F4

(Q1.7)

60°

30°

15 kN

5 kN

70°

45°

Contd..

Solution:

60°

30°

10 kN

70°

45°

F4

R

(Q1.7)

F4 sin70 – 10cos 60 – 15cos 45 – 5cos 30 = 0; or, F4sin70 = 19.94

∑Fx = 0+ve

30°

15 kN

5 kN

45°

∴F4 = 21.22kN Contd..

Solution:

60°

30°

5 kN

10 kN

70°

45°

F4R

(Q1.7)

F4cos70 + 10sin60 – 15sin45 + 5sin30 = +R

∴ +R - 0.342F4 = 0.554

Substituting for F4 , R= +7.81kN

∑Fy = +R

+ve

15 kN

5 kN

Fig. 4

Answers:

F4 = 21.22 kN

R= +7.81kN

Obtain the magnitudes of the forces P and Q if the resultant of the

system shown in figure is zero .

40°

Q

70°

100N

(Q1.8)

40°

60°

P

50N

70°

45°

Contd..

40°

60°

P

50N

Q

70°

45°

100N

(Q1.8)

Contd..

For R to be = zero,

∑Fx = 0 and ∑ Fy = 0

∑Fx = 0 :

-Psin45 – Qcos40 + 100cos70 + 50cos60 = 0

0.707P + 0.766Q = 59.2

+ve

∑Fy = 0

-Pcos45 + Qsin40 + 100sin70 – 50sin60 = 0

or, -0.707P + 0.642Q = -50.67 ------(b)

+ve

(Q1.8)

Answers:

Solving (a) & (b)

P = 77.17 N & Q = 6.058N

100N

Forces of magnitude 50N and 100N are the oblique components of a

force F. Obtain the magnitude and direction of the force F.

Refer figure.

(Q1.9)

30°

100N

50N

Contd..

30°

100N

X - AXIS

Y-AXIS

30°

100N

50N50N

(Q1.9)

Rotating the axes to have X parallel to 50N,

∑Fx = +50 + 100cos30 = +136.6N

∑ Fy = +100sin30 = +50N

+ve

+ve

Contd..

Fig. 1.9

Fig. 6

30°

100N

X - AXIS

Y-AXIS

30°

100N

50N

50N

(Q1.9)

F = 145.46N

θ = 20.1º w r t X direction (50N force)

50N

F= √(∑Fx)2+(∑Fy)2

= tan-1[(∑Fx)2+(∑Fy)2]

Fig. 6

X - AXISF

θ

Y-AXISθ

Resolve the 3kN force along the directions P and Q. Refer figure.

Q3kN

(Q1.10)

P

45°60°

20°

Contd..

3kN

P

45ºQ

45º

Q

60°

30° X – Axis

3kN

45º

(Q1.10)

Move the force P parallel to itself to complete a triangle. Using

sine rule,

P/sin45 = Q/sin90 = 3/sin45

Answer :

P = 3kN, and Q = 4.243kN

30° X – Axis

P

Resolve the 5kN force along the directions P and Q. Refer Fig. 1.11.

Q3kN

(Q1.11)

P

45°60°

20°

Fig. 1.10 Contd..

Q

60°

20°X – AxisP

5kN

45º

(Q1.11)

Q

60°

20° X – Axis

5kN

45ºQ

60°

P

5kN45º

55°

P 20° X – Axis

P55°

60°

X

(Q1.11)

X – Axis

Using sine rule,

P/sin45 = Q/sin80 = 5/sin55

Answer :

Q

60°

P

45º

55°

800

5N

X – Axis

P = 4.32 kN, and Q = 6.01 kN

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.

RESULTANT OF COPLANAR NON CONCURRENT

FORCE SYSTEM

Fig. 1

F2F1

F3

Fig. 2

F1 F2

F5

F4

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

Definition: Moment is the

tendency of a force to make a

The applied force can also tend to rotate the body about

an axis in addition to motion. This rotational tendency is

known as moment.

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 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.

It is computed as the product of the of the force and

the perpendicular distance from the line of action

to the point about which moment is computed.

(Moment center).

Magnitude of moment:

MA = F×dMA = 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’.

For the purpose of additions,

the moment direction may be

Sense of moment:

the moment direction may be

considered by using a suitable

sign convention such as +ve

for counterclockwise and –ve

for clockwise rotations or vice-

versa.M

M

A 100N vertical force is applied to the end of a lever at ‘A’,

which is attached to the shaft at ‘O’ as shown in the figure.

A

F=100 N

(Q1.12)

Determine,

1. The moment of 100N force about ‘O’.

2. Magnitude of the horizontal force applied

60º

O

F=100 N2. Magnitude of the horizontal force applied

at ‘A’, which develops same moment

about ‘O’.

3. The smallest force at ‘A’, which develops

same effect about ‘O’.

4. How far from the shaft a 240N vertical

force must act to develop the same effect?

1) Perpendicular distance from the line of

action of force F to the moment center ‘O’

= d

SOLUTION:

(Q1.1)

60º

A

F

1. The moment of 100N force about ‘O’.

= d

= 240 cos 60º = 120 mm.

Moment about ‘O’ = F × d = 100 × 120

= 12,000 N-mm

(Clockwise)

60ºO

d

If force F is acting horizontally then

the perpendicular distance between the line of

action of horizontal force F at A , to moment

center ‘O’

(Q1.12)

d

F

2. Magnitude of the horizontal force applied at ‘A’, which develops same

moment about ‘O’.

center ‘O’

= d = 240 sin 60º = 207.85 mm.

Moment about ‘O’ = F × d

= F × 207.85

= 12,000 N-mm (Clockwise)

Therefore,

F = 12,000 / 207.85 = 57.73 N

OA

60º

3) Solution:

F = M/d

Force is smallest when the perpendicular

distance is maximum so as to produce

same M.

(Q1.12)

A

F

3. The smallest force at ‘A’, which develops same effect about ‘O’.

same M.

Maximum distance between the point, 0

and the point A is 240 mm.

If the line of action of the force is such that

d = 240 mm

i.e., d = 240 mm.

Therefore, Fmin = 12,000/240 = 50N.

O

60º

F

4) Solution:

Distance along x-axis, X = M/F

= 12,000/240

A

(Q1.12)

4. How far from the shaft a 240N vertical force must act to

develop the same effect?

= 12,000/240

= 50 mm.

Distance along the shaft axis

d = X/cos 60

= 50/cos 60

= 100 mm

O

60º

X F

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

dθ

PP sinθ

P cosθ

d

Moment of Force P about the

point A,

P x d

θ

A

d

Algebraic sum of Moments of

components of the Force P

about the point A,

P cosθ x d1 + P sinθ x d2

=

A

d1

d2

VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)

Y

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.

Q

R

AO

ααααβ p

qrP

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.

γ

Y

X

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)

Ry

Py

RQ

P

p

q r

αααα

βγ

Y

B

CD

Qy

∴ 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’.

A O

αααα

X

COUPLE

Two parallel, non collinear (separated by certain

distance) forces that are equal in magnitude and opposite

in direction form ‘couple’.

d

F

The algebraic summation of the d

FHence, couple does not produce any

translation and produces only rotation. M = F x d

The algebraic summation of the

two forces forming couple is zero.

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 & Apply two equal and opposite forces of same magnitude &

direction as Force F at point B, so that external effect is

unchangedF

AB

F

Fd

Of these three forces, two forces i.e., one at A and the other

F

AB

F

F

AB

F

=

M = F x dd

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.

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

W kN

w kN/m

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)

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

w kN/m

L

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)

L

W = ½ × L × w

2/3 ×L1/3 ×L

A 100N force acts on the corner of a 4m x 3m box as

shown in the Fig. Compute the moment of this force about

A by a) Definition of Moment

b) Resolving the force into components along CA

and CB.

(Q1.13)

and CB.

A B

CD60º

3m

4m

F=50 kN

A B

CD60º

36.87º

60º

d

a) By Definition of Moment:

To determine ‘d’:

AC =

CAD = tan-1(3/4) = 36.87º

ECD = 60º

3m

4m

F=50 kN

m534 22 =+ 23.13

(Q1.13)

A BdE

ECD = 60º

ACE = 60º – 36.87º = 23.13º

From ∆le ACE, d = AC × sin ( ACE)

= 5 × sin 23.13º = 1.96 m.

Moment about A = 50 × 1.96 = 98.20 kNm.

CD 60º

b) By Components:

Fx = 50 × cos 60 = 25kN.

Fy = 50 × sin 60 = 43.30kN.

+ ΣMA = - Fx × 3 + Fy × 4

= - 25 × 3 + 43.3 × 4

3m

4m

F=50kN

Fx

Fy

(Q1.13)

A B

= - 25 × 3 + 43.3 × 4

= + 98.20kNm.

An equilateral triangle of sides 200mm is acted upon by 4

forces as shown in the figure. Determine magnitude and

direction of the resultant and its position from point ‘D’.

50kN

(Q1.14)

30º

80kN

30kN

60kN

60º

200mmD

Resultant & its inclination:

Resolving forces

+ ΣFx = Rx = +30 + 60 cos30º – 50 cos60º

= +56.96kN.

+ ΣFy = Ry = -80 + 60 sin30º + 50 sin60º

50kN

60kN

60 cos30

60 sin30

50 sin 60

50 cos 60

(Q1.14)

+ ΣFy = Ry = -80 + 60 sin30º + 50 sin60º

= -6.69kN.

R=

Inclination w.r.to horizontal = θR = tan-1(Ry/Rx)

= tan-1(6.69/56.96) = 6.7º

30º

80kN

30kN60º

200mm

b) Position w.r.to D:

Moment of the component forces about D:

+ MD = - 60 × 100 + 80 × 100 = 2000kNmm.

= R × d

where ‘d’ = perpendicular distance from point D to the line

of action of R.

∴

(Q1.14)

of action of R.

= 2000 × d.

∴ d =2000/57.35 = 34.87mm

Find the resultant and its position w.r.to ‘O’ of the non-

concurrent system of forces shown in the figure.

F2=500NF1=2500N

F5=2000N

Ө2Ө41

(Q1.15)

F3=1000N

F4=1500N1m

1m

Ө2Ө4

1

1

Ө5

O

A) To find the resultant –

Ө2 = tan-1(1/2) = 26.56°

Ө4 = tan-1(3/2) = 56.31°

Ө5 = tan-1(1/1) = 45°

+ ΣFx = Rx = F2 cosӨ2 +F3

F3=1000N

F2=500NF1=2500N

F5=2000N

1m

1m

Ө2Ө4

1

1

Ө5

O

(Q1.15)

+ ΣFx = Rx = F2 cosӨ2 +F3-F4 cosӨ4-F5 cos Ө5

= 500 × cos26.56 + 1000 –1500 × cos56.31-2000 × cos45

= -799.03N = 799.03N←+↑ΣFy = Ry= F1+F2 sin Ө2-F4 sin Ө4+F5 sin Ө5

= 2500+500 sin26.56-1500 sin56.31+2000 sin45

=2889.70N ↑

F4=1500N1m

∴ Resultant R =

= 2998.14N

ӨR = tan-1 = tan-1(2889.7/799.03) = 74.54°

B) Position of Resultant w.r.to ‘O’:

By Varignon’s theorem, Moment of the resultant about ‘O’

x

y

R

R

Rx

RyR

ӨR

(Q1.15)

By Varignon’s theorem, Moment of the resultant about ‘O’

= Algebraic sum of the moments of its components

about ‘O’.

+ Mo =R×d = +2500×2 + 500×sin26.56×5 – 500×cos26.56×3 - 1000×1- 1500× cos56.31×0–1500×sin56.3×1+2500× cos45×1-2500×sin45×0= 2998.14 × d

d = 1.43 m from O.

Determine the resultant of three forces acting on a dam

section shown in the figure and locate its intersection with

the base. Check whether the resultant passes through the

middle one-third of the base.

(Q1.16)

60º

30 kN

120 kN

50 kN

6 m

1 m2 m

3 m

A B

60º

30 kN

120 kN

50 kN

6 m

1 m

2 m3 m 30º

A B

(Q1.16)

+ ∑Fx = Rx = 50 – 30 × cos30 = 24.02 kN

+ ∑Fy = Ry = -120 – 30 × sin 30= -135 kN

Resultant, R=

6 m

kNRR yx 12.13713502.242222 =+=+

θR= tan-1(Ry/Rx) = tan

-1(135/24.12) = 79. 91º

Location of the resultant w.r.t. B:

MB= 30×1 + 120 × (6-2) - 50 × 3 = Ry × X

360 = 135× X

Therefore, X = 360/135 = 2.67m from B.

From A, X = 6 –2.67= 3. 33 m.

(Q1.16)

From A, X = 6 –2.67= 3. 33 m.

Middle 1/3rd distance is between 2m and 4m.

2m

A 50 N force is applied to the corner of a plate as

Shown in the fig. Determine an equivalent

force-couple system at A. Also determine an equivalent

system Consisting of a 150 N force at B and another

force at A.

(Q1.17)

50 N

100 mm

50 mm

30 mm

30º

B

A

Force – Couple System at A:

Fx = 50 ×sin 30= 25 N.

Fy = 50 × cos 30= 43.3 N50 N

100 mm

50 mm

30 mm

30º

50N

50 sin 30

B

A

60º

(Q1.17)

50 cos 30

50N

a) Force – Couple System at A:

Fx = 50 ×sin 30 = 25 N.

Fy = 50 × cos 30 = 43.3 N

These forces can be moved to

A by adding the couple.

Moment of the couple about A

+ ∑M = F ×50-F ×100

100 mm

50 mm

30 mm

30º

50N

Fx=50 sin 30

B

A

(Q1.17)

+ ∑MA= Fx×50-Fy×100

= 25×50 - 43.3×100

= -3080 N-mm.

= 3080 N-mm

Fy=50 cos 30

50N

100 mm

50 mm

A

Fy=50 cos 30

Fx=50 sin 30

MA=3080N-mm

b) Forces at A and B :

The couple MA is because of two

equal and opposite forces at A

and B.

i.e., MA = 150 × cosθ × 30

= 3080

Therefore, θ = 46.8º.

50 mm

30 mmB

A

θ

150 N

(Q1.17)

Therefore, θ = 46.8º.

Resultant force at A:

FX=50×sin30-150×cos46.8

= -77.68N = 77.68N

FY=-50×cos30-150×sin46.8

= -152.65N = 152.65N

100 mm

100 mm

50 mm

A

Fy=50 cos 30

Fx=50 sin 30θ=46.8º

150N

EXERCISE PROBLEMS

Q1.18 A body of negligible weight, subjected to two forces F1=1200N, and F2=400N acting along the vertical, and thehorizontal respectively, is shown in figure. Find the componentof each force parallel, and perpendicular to the plane.

= 1200 NF

1. Resultant of force system

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

= 1200 N

34

Y

F2

F1

= 400 N

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

12

5

390 N

X

YF2 =

FIG.2.

EXERCISE PROBLEMS1. Resultant of force system

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

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

30º40º

5

300 N400 N

F1 =F3 =

FIG. 2

600N

800N

20º40º

3. Obtain the resultant of the concurrent coplanar forces shown in FIG.3


200N

20º40º

30º

FIG. 3

(Ans: R = 522.67 N, θ = 68.43º)

4. 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.


while that of 1 acts in the same direction.

Find the corresponding force to be exerted by tugboat 2.

( Ans: a. T1= 195.81 N, T2 = 133.94 N

b. α = 70º, T1 = 281.91 N, T2(min) = 102.61 N )

T2

R = 300 N

T1

α

20º

FIG. 4

X - direction

5. An automobile which is disabled is pulled by two ropes as

shown in Fig. 5. Find the force P and resultant R, such that R is

directed as shown in the figure.

P


P

Q = 5 kN

R20º

40º

Fig. 5

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

6. A collar, which may slide on a vertical rod, is subjected to three forces

as shown in Fig.6. 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


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

1200 N

800 N60º

θ

F

ROD

COLLAR

Fig.6

7. 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 Fig. 7.

8kN5kN


Q30º

α

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

8. Determine the resultant of the parallel coplanar

force system shown in figure.

1000 N

2000 N

600 N o

60º

60º 30º

10º


400 N2000 N 60º10º

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

9. Four forces of magnitudes 10N, 20N, 30N and 40N

acting respectively along the four sides of a square

ABCD as shown in the figure. Determine the

magnitude, direction and position of resultant w.r.t. A.


10N

40N

a

a

30N

20N

AB

CD

(Ans:R=28.28N, θ=45º, x=1.77a)

10. 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 horizontal

bar of 2.85m. Determine the magnitude of resultant

and also the distance of the resultant from the left

end.


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

11. Reduce the given forces into a single force and a

couple at A.

70.7 kN


100 N80 N

1m

1.5m

70.7 kN200 kN

A

30º45º

30º

(Ans:F=320kN, θ=14.48º, M=284.8kNm)

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


150 N

1.5m3m1.5m

150 Nm

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

A

1.5m

500 N100 N

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

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