ENGINEERING MECHANICS (17CA03302) - crectirupati.com · Problem :Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured ... Problem : Three forces

LECTURE NOTES

ON

ENGINEERING MECHANICS

(17CA03302)

2018 – 2019

II B. Tech I Semester

Dr. G. Maruthi Prasad Yadav

Associate Professor

Mechanical Engg Dept.

CHADALAWADA RAMANAMMA ENGINEERING COLLEGE (AUTONOMOUS)

Chadalawada Nagar, Renigunta Road, Tirupati – 517 506

Department of Mechanical Engineering

ENGINEERING MECHANICS

Engineering Mechanics

It is defined as that branch of science, which describes and predicts the conditions of rest or

motion of bodies under the action of forces. Engineering mechanics applies the principle of

mechanics to design, taking into account the effects of forces.

Statics

Statics deal with the condition of equilibrium of bodies acted upon by forces.

Rigid body

A rigid body is defined as a definite quantity of matter, the parts of which are fixed in position

relative to each other. Physical bodies are never absolutely but deform slightly under the action

of loads. If the deformation is negligible as compared to its size, the body is termed as rigid.

Force

Force may be defined as any action that tends to change the state of rest or motion of a body to

which it is applied.

The three quantities required to completely define force are called its specification

characteristics. So the characteristics of a force are:

1. Magnitude

2. Point of application

3. Direction of application

Composition of two forces

The reduction of a given system of forces to the simplest system that will be its equivalent is

called the problem of composition of forces.

Parallelogram law

If two forces represented by vectors AB and AC acting under an angle α are applied to a body at

point A. Their action is equivalent to the action of one force, represented by vector AD, obtained

as the diagonal of the parallelogram constructed on the vectors AB and AC directed as shown in

the figure.

Force AD is called the resultant of AB and AC and the forces are called its components.

Resolution of a force

The replacement of a single force by a several components which will be equivalent in action to

the given force is called resolution of a force.

Action and reaction

Often bodies in equilibrium are constrained to investigate the conditions.

Free body diagram

Free body diagram is necessary to investigate the condition of equilibrium of a body or system.

While drawing the free body diagram all the supports of the body are removed and replaced with

the reaction forces acting on it.

Draw the free body diagram of the body, the string CD and the ring.

Equilibrium of colinear forces:

Equllibrium law: Two forces can be in equilibrium only if they are equal in magnitude,

opposite in direction and collinear in action.

Resolution of a force

Replacement of a single force by several components which will be equivalent in action to the

given force is called the problem of resolution of a force.

By using parallelogram law, a single force R can be resolved into two components P and Q

intersecting at a point on its line of action.

Law of superposition

The action of a given system of forces on a rigid body will no way be changed if we add to or

subtract from them another system of forces in equilibrium.

Problem 3: Two spheres of weight P and Q rest inside a hollow cylinder which is resting on a

horizontal force. Draw the free body diagram of both the spheres, together and separately.

Equilibrium of concurrent forces in a plane

� If a body known to be in equilibrium is acted upon by several concurrent, coplanar forces,

then these forces or rather their free vectors, when geometrically added must form a closed

polygon.

� This system represents the condition of equilibrium for any system of concurrent forces in a

plane.

Lami’s theorem

If three concurrent forces are acting on a body kept in an equilibrium, then each force is

proportional to the sine of angle between the other two forces and the constant of proportionality

is same.

P/Sinα =Q/Sinβ= R/Sinγ

Theory of transmissibility of a force:

The point of application of a force may be transmitted along its line of action without changing

the effect of force on any rigid body to which it may be applied.

Method of moments

Moment of a force with respect to a point:

� Considering wrench subjected to two forces P and Q of equal magnitude. It is evident that

force P will be more effective compared to Q, though they are of equal magnitude.

� The effectiveness of the force as regards it is the tendency to produce rotation of a body about

a fixed point is called the moment of the force with respect to that point.

� Moment = Magnitude of the force × Perpendicular distance of the line of action of force.

� Point O is called moment centre and the perpendicular distance (i.e. OD) is called moment

arm.

� Unit is N.m

Parallel forces on a plane

Like parallel forces: Coplanar parallel forces when act in the same direction.

Unlike parallel forces: Coplanar parallel forces when act in different direction.

Resultant of like parallel forces:

Let P and Q are two like parallel forces act at points A and B.

R = P + Q

Resultant of unlike parallel forces:

R = P - Q

R is in the direction of the force having greater magnitude.

Couple:

Two unlike equal parallel forces form a couple.

The rotational effect of a couple is measured by its moment.

Moment = P × l

Sign convention: Anticlockwise couple (Positive)

Clockwise couple (Negative)

Theorem of Varignon:

The moment of the resultant of two concurrent forces with respect to a centre in their plane is

equal to the algebraic sum of the moments of the components with respect to some centre.

Problem :Determine the magnitude of the resultant force FR = F1 + F2 and its direction,

measured

Counter clockwise from the positive x axis.

ψ = 90 deg − β + α

FR = 867 N

θ = 63.05 deg

φ = θ + α

φ = 108 deg

Problem: Two identical rollers each of weight Q = 445 N are supported by an inclined plane and

a vertical wall as shown in the figure. Assuming smooth surfaces, find the reactions induced at

the points of support A, B and C.

(Ra /sin120) = (S/ Sin 150) = (445 / Sin90)

Ra = 385.38N.

S= 222.5N

Resolving vertically

Fy =0

Rb cos 60 = 445+ S sin 30

Rb = 642.302 N

Resolving horizontally

Fx = 0

Rc = Rb Sin30 + S Cos30

Rc = 513.84N

Problem: A weight Q is suspended from a small ring C supported by two cords AC and BC. The

cord AC is fastened at A while cord BC passes over a frictionless pulley at B and carries a

weight P. If P = Q and α = 50°, find the value of β.

Fx=0

S Sin50=Q Sinβ

Fy=0

S Cos50 +Q Sinβ= Q

SCos50 =Q (1-Cosβ)

Putting S from eqe 1,

Q Sinβ. Cos 50/Sin50= Q (1-Cosβ)

Cot50 =(1-Cosβ)/Sinβ

0.839 Sinβ = 1- Cosβ

Squaring both sides,

0.703 Sin2β =1+ Cos

2β-2 Cosβ

1.703Cos2β-2Cosβ+0.297=0

β= 63.130

Problem : The four wheels of a locomotive produce vertical forces on the horizontal girder AB.

Determine the reactions Ra and Rb at the supports if the loads P = 90 KN each and Q = 72 KN

(All dimensions are in m).

Fv = 0

Ra + Rb = 3P+ Q

Ra + Rb = 342N

Ma=0

RbX9.6 = PX1.8 +PX3.6+PX5.4+QX8.4

Rb = 177.75KN.

Problem : The resultant of two forces when they act at an angle of 600 is 14N. If the same forces

are acting at right angle, their resultant is √136 N. Determine the magnitude of two forces.

Given

R1 =14N

α = 600

R2 =√136N

α= 900

R1 =√(P2+Q

2+2PQCosα)

14 =√(P2+Q

2+2PQCosα)

196 = (P2+Q

2+2PQCosα)-------(1)

R2 = √(P2+Q

2)

136= (P2+Q

2--------(2)

Sub (2) in (1)

PQ = 60

2PQ=120N-------(3)

(P+Q)2 = P

2+Q

2+2PQ

P+Q =16

P=16-Q-------(4)

Sub (4) in (3)

Q =10N

P=6N

Problem : Two Forces acting at a point O as shown in fig.Determine the resultant in magnitude

and direction.

P=50N

Q=100N

α=300

R =√(P2+Q

2+2PQCosα)

=145.4N

Tanθ=Qsinα/(P+Qcosα)

θ =20.10

Problem : The resultant of two forces is 1500N and angle between forces is 900. The resultant

makes an angle of 360 with one of the force. Find the magnitude of each force.

Tanθ=Qsinα/(P+Qcosα)

Q/P = Tan36

Q =0.726P-----(1)

R =√(P2+Q

2+2PQCosα)

P = 1213.8N

Q =0.726P =881.2N

Problem : If the strings and bulb are attached as shown in fig. Find the tensions in the strings.

Weight at C =15N

∟OAC =600

∟CBD=450

T1 =Tension in BC

T2=Tension inAC

Lamis Theorem

15/Sin ∟BCA = T1/Sin∟ACE = T2/Sin∟BCE

T1 = 7.76N

T2=10.98N

Problem : Three forces 40N, 15N and 20N are acting at a point O shown in fig. The angles

made by forces with X-axes are 600, 120

0 and240

0.determine resultant force and its direction.

H= R1 Cosθ1 + R2 Cosθ2+ R3 Cosθ3

H= 2.5N

V=R1Sinθ1+R2Sinθ2+R3Sinθ3

V=30.3N

R =√(H2+V

2)= 30.4N

Tanθ=V/H

θ = 85.280

Problem : Four forces 10N, 20N, 30N and 40N are acting along four sides of square ABCD

shown in fig. Determine resultant moment about point A. Each side of square is 2m.

Moment about A = 40X0 + 10X0+30X2+20X2

=100N-m.

Problem : Three forces F1, F2, F3 acting on a body shown in fig and body is in equilibrium.If

magnitudeof F3 is 400N, Find F1 and F2.

FH=0

F1cos30 = F2 Cos30

F1 = F2

Fv =0

F1 Sin30 + F2 Sin30 =F3

F1 =400N

F1=F2 =400N

Problem : Three parallel forces F1, F2 and F3 acting on a body shown in fig and body is in

equilibrium. If F1 =250N, F3 = 1000N and distance between F1 and F2 is 1m, determine

magnitude of F2 and distance of F2 from F3.

FH=0

F1+F3=F2

F2 =1250N

MA=0

F2 X1 =F3(1+x)

X = 0.25m

Problem : Five forces F1,F2, F3, F4 and F5 acting at a point on body shown in fig and body in

equilibrium. If F1=18N, F2 = 22.5N, F3 = 15N, F4 =30N.Find F5 in magnitude and direction.

FH=0

F1+F2cos45 =F4cos30+F5Cosθ

F5cosθ=7.92-----(1)

Fv=0

F2sin45+F3=F4sin30+F5sinθ

F5sinθ=15.6N-----(2)

(2)/(1) = 2.007

θ = 63.50

Sub in (1)

F5cosθ=7.92

F5 = 17.76N

Problem : Fig shows coplanar system of forces acting on a flat plate. Determine (i) Resultant (ii)

x and Y intercepts of resultant.

FH = F1cos60+F2cos33.6-F3cos63.4

=1250.3N (Left)

Fv = F1sin60-F2sin33.6-F3sin63.43

=1705.1N

R =√(FH2+FV

2) =2114.4N

Angle

Tanθ =Ry/Rx

=1.360

θ= 53.70

Mo =F3sin63.43X4-F2cos33.6X3+F2sin33.6X3-F1cos60X3-F1sin60X2

=1659.55N-m.

From Fig

Mo=1659.55=Ry X x

x = 0.97m

Mo=1659.55=Rx.Y

Y=1.32m.

Problem : A circular roller of 100N and 10cmradius hangs by tie rod AB=20cm and rests

against a smooth vertical wall at C. Determine (i) Force F (ii) Rc(Reaction)

Fx=0

Rc=Fsinθ

Triangle ACB

Sinθ=BC/AB=1/2

θ=300

Rc = F/2

Fy=0

F cosθ=100

F=115.4N

Rc =F/2=57.7N.

Problem : A ball 120Nrests in right angled groove shown in fig. The sides of groove are inclined

to angle 300 and60

0 to horizontal. If all surfaces are smooth, determine reaction Ra and Rc.

Fx=0

Rc sin30 =Ra sin60

Rc=1.73Ra-----(1)

Fy=0

Rc cos30+Racos60 =120

Ra=60N

Sub in (1)

Rc =103.9N

Problem : A circular roller 5cm radius and 100N weight rests on smooth horizontal surface and

held in position by inclined bar AB 10cm length shown in fig. A horizontal force 200N action at

B. Find tension in bar AB and vertical reaction at C.

Triangle ABC

Sinθ=BC/AB=5/10=0.5

θ= 300

Fx=0

Fcos30 =200

F = 230.9N

Fy =0

Fsin30+ 100= Rc

Rc =215.47N

Problem : Two identical rollers P and Q each weight W, are supported by an inclined plane and

a vertical wall shown in fig. Assume smooth surfaces. Draw free body diagram of (i) roller Q (ii)

roller P.

Also if W=1000N,find reactions at point of supports A, B and C.

Roller P

Fx=0

Rd cos30=Racos60

Rd =0.577Ra-----(1)

Fy=0

Rd sin30+Rasin60=W

Ra=866.17N

Rd=499.78N

Roller Q

Fx=0

Rc =Rd cos30+Rbcos60

Rc=0.5Rb+432.8-----(2)

Fy=0

W+Rd sin30=Rbsin60

Rb=1443.3N

Sub in (2)

Rc =1154N

Problem : Two spheres each weight 1000N and 25cm radius rest in a horizontal channel of

width 90cm shown in fig. Find reactions at A, B and C.

Roller 2

Fy =0

Rd cosθ=W

Rd =1666.6N

Fx=0

Rd Sinθ=Rc

Rc = 1333.3N

Roller 1

Fx=0

Ra=Rd Sinθ

Ra=1333.3N

Fy=0

Rb =W+RdCosθ

=2000N

Problem : Two smooth cylinders W=1000N and radius 15cm are connected at centres by string

AB length =40cm and rest upon a horizontal plane, supporting above them a third cylinder

weight=2000N and radius 15cm shown in fig. Find force S in string AB and pressure Produced

on floor at points of contact D and E.

Roller 3

Fx=0

Rf Sinθ =Rg Sinθ

Rf = Rg

Fy =0

Rf Cosθ+ Rg Cosθ =2000N

Rf =1342.2N = Rg

Roller 1

Fy=0

Rd =1000+ Rf Cosθ

=2000N

Fx=0

S =Rf Sinθ

=895.2N

Rollers 1,2 and 3

Fy=0

Rd +Re =1000+2000+1000

Re=2000N

Problem : A simply Supported beam AB of length 9m carries UDL 10KN/m for a distance of

6m from left end. Calculate reactions at A and B.

Fy=0

Ra+Rb =10X6

Ra+Rb = 60N----(1)

Ma=0

10X6X3 = RbX 9

Rb =20KN

Ra =40KN.

Problem : A simply supported beam of10m length carries UDL and two point loads shown in

fig. Calculate Ra and Rb.

Fy=0

Ra+Rb = 50+ 10X4 + 40

Ra + Rb = 130N-----(1)

Ma=0

Rb X 10 =50X2 +40X6 +10X4X4

Rb = 50KN

Sub in (1)

Ra = 80KN

Problem : A simply supported beam span 9m carries UVL from zero at end A to 900N/m at end

B. Calculate reactions at two ends.

Total load = 9X900X1/2

=4050N

Fy=0

Ra +Rb =4050---- (1)

Ma=0

Rb X 9 =400X (2/3) X 9

Rb = 2050N

Sub in (1)

Ra = 2000N

Friction

� The force which opposes the movement or the tendency of movement is called Frictional

force or simply friction. It is due to the resistance to motion offered by minutely projecting

particles at the contact surfaces. However, there is a limit beyond which the magnitude of this

force cannot increase.

� If the applied force is more than this limit, there will be movement of one body over the other.

This limiting value of frictional force when the motion is impending, it is known as Limiting

Friction.

� When the applied force is less than the limiting friction, the body remains at rest and such

frictional force is called Static Friction, which will be having any value between zero and the

limiting friction.

� If the value of applied force exceeds the limiting friction, the body starts moving over the

other body and the frictional resistance experienced by the body while moving is known as

Dynamic Friction. Dynamic friction is less than limiting friction.

� Dynamic friction is classified into following two types:

a) Sliding friction

b) Rolling friction

� Sliding friction is the friction experienced by a body when it slides over the other body.

� Rolling friction is the friction experienced by a body when it rolls over a surface.

� It is experimentally found that the magnitude of limiting friction bears a constant ratio to the

normal reaction between two surfaces and this ratio is called Coefficient of Friction.

Coefficient of friction = F/N.

where F is limiting friction and R is normal reaction between the contact surfaces.

Coefficient of friction is denoted by µ.

Thus, µ = F/R

Laws of friction

1. The force of friction always acts in a direction opposite to that in which body tends to move.

2. Till the limiting value is reached, the magnitude of friction is exactly equal to the force which

tends to move the body.

3. The magnitude of the limiting friction bears a constant ratio to the normal reaction between

the two surfaces of contact and this ratio is called coefficient of friction.

4. The force of friction depends upon the roughness/smoothness of the surfaces.

5. The force of friction is independent of the area of contact between the two surfaces.

6. After the body starts moving, the dynamic friction comes into play, the magnitude of which is

less than that of limiting friction and it bears a constant ratio with normal force. This ratio is

called coefficient of dynamic friction.

Angle of friction

Consider the block shown in figure resting on a horizontal surface and subjected to horizontal

pull P. Let F be the frictional force developed and N the normal reaction. Thus, at contact surface

the reactions are F and N. They can be graphically combined to get the reaction R which acts at

angle θ to normal reaction. This angle θ called the angle of friction is given by

tan θ = F/R

As P increases, F increases and hence θ also increases. θ can reach the maximum value

α when F reaches limiting value. At this stage,

tan α = F/R= µ

This value of α is called Angle of Limiting Friction. Hence, the angle of limiting friction may be

defined as the angle between the resultant reaction and the normal to the plane on which the

motion of the body is impending.

Angle of repose

Consider the block of weight W resting on an inclined plane which makes an angle θ with the

horizontal. When θ is small, the block will rest on the plane. If θ is gradually increased, a stage is

reached at which the block start sliding down the plane. The angle θ for which the motion is

impending, is called the angle of repose. Thus, the maximum inclination of the plane on which a

body, free from external forces, can repose is called

Angle of Repose.

Resolving vertically,

N = W. cos θ

Resolving horizontally,

F = W. sin θ

Thus, tan θ = F/R

If ɸ is the value of θ when the motion is impending, the frictional force will be limiting

friction and hence,

tan ɸ = F/R

= µ = tan α

ɸ = α

Thus, the value of angle of repose is same as the value of limiting angle of repose.

Cone of friction

• When a body is having impending motion in the direction of force P, the frictional

force will be limiting friction and the resultant reaction R will make limiting angle α with the

normal.

• If the body is having impending motion in some other direction, the resultant reaction makes

limiting frictional angle α with the normal to that direction. Thus, when the direction of force

P is gradually changed through 360°, the resultant R generates a right circular cone with semi-

central angle equal to α.

Problem : A body 300N resting on rough horizontal table. A pull 100N applied at an angle 150

with horizontal just causes body to slide over the table. Make calculations for normal reaction

and coefficient of friction.

Fx=0

F=100Cos15=96.59N

Fy=0

R=W-100Sin15

R=274.2N

µ = F/R= 0.352.

Problem : A body 100N weight rests on a horizontal surface (µ = 0.3) and is acted by a force

applied at an angle of 300 to horizontal. What force is required to just cause body to slide over

the surface?

Also determine inclination and magnitude of minimum force required to set block into

impending motion.

Fx =0

F=PCosθ----(1)

Fy=0

R+ PSinθ =100----(2)

µ= F/R

Therefore, µR =PCosθ

R = P Cosθ/0.3

Sub in (2)

P= 29.5N

Now let Angle is α

P= µW/ (Cosα+µSinα)

Here P is minimum when denominator is maximum

=0

Tanα= 0.3=µ

α= 16.7o

Pmin =28.7N

Problem : A wooden block 50N weight rests on a horizontal Plane. Determine force required to

just (a) pull it (b) Push it. Take µ=0.4 between mating surfaces.

Fx=0

µR=F=P Cosθ-----(1)

Fy=0

R+PSinθ=W

R = (W-P Sinθ)-----(2)

Sub (2) in (1)

P =18.7N

Fx =0

F = P Cosθ---- (1)

Fy =0

R=W+P Sinθ---- (2)

Sub (2) in (1)

P = 23.17N

Pulling is easy than pushing.

Problem: A body of 100N placed on a horizontal plane. Determine µ if horizontal force 60N just

causes body to slide over plane.

W=100N

P=60N

Fx=0

P=F

F=60N =µR----(1)

Fy=0

R=W=100N----(2)

Sub (2) in (1)

60 = µX100

Μ=0.6

Problem: A pull 20N inclined at 250 to horizontal plane, is required just to move a body placed

on a rough horizontal plane. But the push required to move body is 25N. If push is inclined at

250 to horizontal find weight of body and co efficient of friction.

Body is pulled

Fx=0

F =µR

=P Cos25

F =18.126---(1)

Fy=0

R+P Sin25 = W

R = W-8.45-----(2)

Sub (2) in (1)

µ(W-8.45) =18.12----(2’)

Body is pushed

Fx=0

F=P’Cos25

F = 22.65 =µR’-----(3)

Fy=0

R’ =W+P’ Sin25

R’ =W+ 10.56----(4)

Sub (4) in (3)

Μ(W+10.56) = 22.65-----(5)

(2’)/(5)

Solving

W=84.54N

Sub W in(2’)

µ(W-8.45) = 18.12

µ=0.24

Problem: An effort 200N is required just to move body up an inclined plane angle 150, the force

acting parallel to plane. If angle of inclination of plane is made 200, the effort required again

applied parallel to plane, is found to be 230N. Find weight of body and coefficient of friction.

P1=200N

P2=230N

θ1=150

θ2 =200

Case 1

Fx=0

F1+WSIn15 =P

µR1+W Sin15 =P----(1)

Fy=0

R1=W Cos15

Sub R1 in (1)

W (Sin15+µCos15) =200----(2)

Case 2

Fx=0

F2+W Sin20=230

µR2+ W Sin20=230----(3)

Fy=0

R2=W Cos20

Sub R2 in (3)

W(Sin20+µCos20) =230----(4)

(4)/(2) =1.15

µ=0.26

Sub µ in (4)

W(Sin20+0.26Cos20) =230

W= 230/0.586

=392.3N

Problem : A block weighing 500N just starts moving down a rough inclined plane when

supported by a force of 200N acting parallel to the plane in upward direction. The same block is

on the verge of moving up the plane when pulled by a force of 300N acting parallel to the plane.

Find the inclination of the plane and coefficient of friction between the inclined plane and the

block.

Fy=0

R = 500.cosθ

F =µR=µ.500cosθ

Fx=0

200+F1 ==500 Sinθ

200+µX500.Cosθ=500 Sinθ

Fy=0

R =500 Cosθ

F2=µR=µ.500.Cosθ

Fx=0

500.Sinθ+µ.500.Cosθ=300

Add (1) and(2)

500=1000.Sinθ

Sinθ=0.5

θ=300

Sub in (2)

500Sin30 +µ.500.Cos30 =300

µ=0.1154

Problem : A block in shape of rectangular prism rest on a inclined plane, shown in fig. The

block is tied up by a horizontal string which has tension 10N. If block weight 35N, determine (i)

Frictional Force on block (ii) Normal Reaction (iii) µ

F+T Cos30 =W Sin 30

F =µR = 8.84N----(1)

Fy=0

R = WCos30 + T Sin30

R = 35.3N----(2)

Sub (2) in(1)

µ= 0.25

Problem : Two Blocks W1 and W2 rest on a inclined plane and are connected by a string. The

coefficient of friction between these two blocks and plane are µ1 and µ2. Show that blocks will be

on point of motion when inclination of plane with horizontal is

Tanθ =(µ1W1+µ2W2)/(W1+W2)

What will be inclination if blocks are of equal weight and µ1=1/2 and µ2=1/3.

Block 1

Fx=0

T+F1=W1 Sinθ

T+µ1R1=W1Sinθ----(1)

Fy=0

R1=W1 Cosθ----(2)

Sub (2) in (1)

T+ µ1W1 Cosθ = W1 Sinθ----(3)

Block 2

Fx=0

F2 = T+ W2 Sinθ

µ2R2= T+W2 Sinθ-----(4)

Fy =0

R2 =W2 Cosθ----(5)

Sub (5) in (4)

µ2W2 Cosθ = T+ W2 Sinθ-----(6)

From (3)

T= W1 Sinθ-µ1W1 Cosθ-----(7)

From (6)

T = µ2W2 Cosθ-W2 sinθ----(8)

(7) = (8)

W1 Sinθ + W2Sinθ = µ1W1 Cosθ+ µ2W2 Cosθ

Sinθ/Cosθ =(µ1W1+µ2W2)/(W1+W2)

Tanθ=(µ1W1+µ2W2)/(W1+W2)

Problem : Two blocks 200N and 300N connected by a string passing over a frictionless pulley

rest on rough surfaces; Block of 200Non horizontal surface and other on inclined surface shown

in fig. for both surfaces µ=0.25. Find minimum force for motion to impend.

Block 1

Fx=0

T+F1 =P Cosθ

T+µR1 = PCosθ------(1)

Fy =0

R1+P Sinθ=W1----(2)

Sub (2) in (1)

P =347.3/ (Cosθ+0.25 Sinθ)

For min P, Denominator should be maximum.

=0

θ=140

P=347.3/( Cos14 +0.25 Sin14)

P=337.2N

Centre of gravity

Centre of gravity: It is that point through which the resultant of the distributed gravity force

passes regardless of the orientation of the body in space.

• As the point through which resultant of force of gravity (weight) of the body acts.

Centroid: Centrroid of an area lies on the axis of symmetry if it exits.

Centre of gravity is applied to bodies with mass and weight and centroid is applied to plane

areas.

Xc =( A1 x1 +A2 x2)/ ( A1+ A2 )

yc =( A1 y1+ A2 y2)/ ( A1+ A2 )

xc = yc =Moment of area/ Total area

Problem : Consider the triangle ABC of base ‘b’ and height ‘h’. Determine the distance of

centroid from the base.

Let us consider an elemental strip of width ‘b1’ and thickness ‘dy’.

Therefore, yc is at a distance of h/3 from base.

Problem : Find the centroid of the T-section as shown in figure from the bottom.

A1 =100X20=2000m2

A2=20X100=2000 m2

X1=0

X2=0

Y1=110m

Y2=50m

A1X1=10000m3

A2X2=10000m3

A1Y1=220000 m3

A2Y2=100000 m3

yc =( A1 y1+ A2 y2)/ ( A1+ A2 )

=320000/4000

=80

Due to symmetry, the centroid lies on Y-axis and it is at distance of 80 mm from the bottom.

Problem : Locate the centroid of the I-section.

As the figure is symmetric, centroid lies on y-axis. Therefore, x = 0

A1=100X20=2000

A2=100X20=2000

A3=150X30=4500

X1=0

X2=0

X3=0

Y1=140

Y2=80

Y3=15

A1X1=0

A2X2=0

A3X3=0

A1Y1=280000

A2Y2=160000

A3Y3=67500

yc =( A1 y1+ A2 y2+ A3 y3)/ ( A1+ A2 + A3 )

=59.71mm

Thus, the centroid is on the symmetric axis at a distance 59.71 mm from the bottom.

Problem : Determine the centroid of the composite figure about x-y coordinate. Take x = 40

mm.

A1 = Area of rectangle = 12x.14x=168x2 =268800mm

2

A2 = Area of rectangle to be subtracted = 4x.4x = 16 x2 = 25600 mm

2

A3 = Area of semicircle to be subtracted =πR2/4 = 25.13 x

2 = 40208 mm

2

A4 = Area of quatercircle to be subtracted = πR2/4 = 12.56 x

2 =20096 mm

2

A5 = Area of triangle = 6 x X 4 x/2 = 12 x2=19200mm

2

X1=7x = 280mm

X2 =2x= 80mm

X3= 6x= 240mm

X4= 10x + (4x –(4X4x)/3π)= 492.1mm

X5 = 14x + 6x/3 = 16x = 640mm

Y1 = 6x= 240mm

Y2= 10x = 400mm

Y3= (4X4x)/(3π) = 67.906mm

Y4 = 8x + (4x- (4X4x)/3π)= 412.093

Y5= 4x/3 = 53.33mm

Xc =( A1 x1- A2 x2- A3 x3- A4 x4+ A5 x5)/ ( A1- A2 - A3 - A4 + A5 ) =326.404mm

yc =( A1 y1- A2 y2- A3 y3- A4 y4+ A5 y5)/ ( A1- A2 - A3 - A4 + A5 )=219.12mm

Problem : Determine the centroid of the following figure.

A1 = Area of triangle = 1X 80 X80 /2 = 3200m2

A2 = Area of semicircle = (πd2 / 8) –(πR

2/2) = 2513.274m

2.

A3 = Area of semicircle = πD2 / 2 = 1256.64m

2.

X1= 2X(80/3)=53.33mm

X2 = 40

X3 = 40

Y1= 80/3=26.67mm

Y2= -4X40/3π=-16.97mm

Y3= 0

Xc =( A1 x1+ A2 x2- A3 x3)/ ( A1+ A2 - A3 ) = 49.57mm

yc =( A1 y1+ A2 y2- A3 y3)/ ( A1+ A2 - A3)= 9.58mm.

Problem : Determine the centroid of the following figure.

A1 = Area of the rectangle=30000

A2 = Area of triangle= 3750

A3 = Area of circle = 7853.98

X1=100

X2=100+200/3 = 166.67

X3 = 100

Y1= 75

Y2= 75+150/3= 125

Y3= 75

Xc =( A1 x1- A2 x2- A3 x3)/ ( A1- A2 - A3 ) = 86.4 mm

yc =( A1 y1- A2 y2- A3 y3)/ ( A1- A2 - A3)= 64.8 mm.

Problem : Find the centroid of the given area

A1= 3X4X1/2= 6m2

A2 = 5X4 = 20m2

A3= π22/2=6.28 m

2

X1= 5+ 3X (1/3) = 6m

Y1= 4X (1/3)=1.33m

X2=2.5m

Y2=2m

X3= -4R/(3π)= - 0.85m

Y3 = 2m

Xc =( A1 x1 +A2 x2 +A3 x2)/ ( A1+ A2 +A3)

=2.5

yc =( A1 y1+ A2 y2+ A3 x2)/ ( A1+ A2 +A3)

=1.875m

Problem : A triangular plate in form of isosceles triangle ABC has base BC=10cm and

altitude=12cm.From this a portion of isosceles triangle OBC is removed. If O is mid point of

altitude of triangle ABC, Determine distance of CG of remainder section from base.

A1 = ½ BC. AD

= 60cm2

A2 = ½ BC.OD

=30cm2

Y2= 1/3 X6 =2cm

yc =( A1 y1- A2 y2)/ ( A1- A2)

=6cm.

Problem : A rectangular lamina ABCD 20cmX25cm has rectangular hole 5cm X 6cm shown in

fig. Locate centroid of section.

A1 = 20X25=500cm2

A2 = 5X6=30cm2

X1= 10cm

Y1=12.5cm

X2= 12+2.5=14.5cm

Y2=3+ (6/2) =6cm

Xc =( A1 x1 -A2 x2 )/ ( A1- A2 )

=9.71cm

Yc =( A1 y1- A2 y2)/ ( A1- A2)

=12.9cm

Problem : From a circular lamina of diameter d, a square hole has been punched out. If one

diagonal of square coincides with radius of circle, determine distance of centre of remainder

from centre of circle.

A1= πd2/4

X1=d/2

d2=l

2+l

2 =2l

2

A2=l2= d2/8

X2= d/2 +d/4 = 3d/4

X =( A1 x1 -A2 x2 )/ ( A1- A2 )

=d(π-(3/4))/(2π-1)

Distance from circle is

= d/2-x

=d/(4(2π-1))

Problem : Find centroid of semicircular section having outer and inner radii 200mm and

160mm.

A1= π2002/4

X1=0

Y1 =4X200/(3π)

=84.92mm

A2= π1602/4

X2=0

Y2 = 4X160/(3π)

=67.94mm

X =( A1 x1 -A2 x2 )/ ( A1- A2 )

=0

Y =( A1 y1- A2 y2)/ ( A1- A2)

=115.11mm

Problem : Determine coordinates of centroid of lamina shown in fig. The shaded area is opening

in lamina.

Circle-R

R=BG = √(BC2+CG

2)=4.243m

A1 =πR2

=56.56m2

Y1=0

Triangle BDG

A2 = 6X3X(1/2)= 9m2

Y2= 2/3 h= 2/3 X 3 = 2m

Triangle-BDE

A3 = ½ X6X4 = 12m2

Y3 = 3+ 1/3 X 4 = 4.33m

Semi circle (Minus)

A4 = ½ π22

=6.28m2

Y4 = -(2+ 4X2/3π) =-2.85m

Sector BGD (Minus)

A5 = πR2/4 = 14.14m

2

α=45Xπ/180 = 0.785

Y5 = 2/3 X R Sinα/α = 2.55m

yc =( A1 y1+ A2 y2+ A3 x3- A4 x4- A5 x5)/ ( A1+ A2 +A3- A4- A5)

= 0.907m

Problem : An area of semi circle 10cm radius has been punched from a square plate 50cm X

50cm shown in fig. If centroid of remainder piece is to kept same as for original piece, make

calculations for width w of area, shown hatched that needs to be removed.

A1= 50X50 = 2500cm2

A2 = π102 /2 = 157cm

2

A3 = 50w

Centroid is at X = 50/2 = 25cm

X1 = 25cm

X2 = 4X10/3π

X3 =50w/2

X =( A1 x1 -A2 x2- A2 x2 )/ ( A1- A2 - A2) =25cm

w = 2.76cm

Problem : The frustum of right circular cone has bottom radius 5cm top radius 3cm and height

8cm. A co-axial cylinder hole 4cm diameter is made throughout frustum. Locate position of

centere of gravity of remaining solid.

Cone EAB

Similar triangles EAB & EDC

5/3 = (8+h)/h

H = 12cm

V1 = 1/3 X π52X20 = 523.3cm

3

Y1= 20/4 = 5cm

Cone EDC,

H=12cm

V2= 1/3X π32X12 =11cm

3

Y2 = 8+ (12/4) = 11cm

Cylindrical hole,

V3 = π/4 X 42X8

Y3 = 8/2=4

Yc =( V1 y1- V2 y2- V3 x3)/ ( V1- V2 -V3)

= 3.139cm

Problem : Consider a semi-circle of radius R. Determine its distance from diametral

axis.

Due to symmetry, centroid ‘yc’ must lie on Y-axis.

Consider an element at a distance ‘r’ from centre ‘o’ of the semicircle with radial width dr.

Area of element = (r.dθ)×dr

Moment of area about x =

yc = Moment of area/ Total area

Problem : A body consists of solid hemisphere radius 4cm and a right circular solid cone height

12cm. The hemisphere and cone have a common base and are made of same material. Locate CG

of composite body.

Cone ABC,

V1= 1/3X πr2h= 200.9cm

2

Y1= 12 – 12/4 =9cm

Hemisphere BCD,

V2 = ½ (4/3) πr3 = 133.97cm

3

Y2 = 12 + (3/8)r = 13.5cm

Yc =( V1 y1+ V2 y2)/ ( V1+ V2 )

=10.8cm from A

Problem : A wooden plank of 10cm size width and 1cm thick is fixed on top of semi circular

steel plate of 5cm radius and 1cm thickness shown in fig. Determine maximum height h of

wooden plank so that CG of composite body lies in junction of two. Take density of wood equal

to 1/10 that of steel.

Mw =ρwhtb

= 1000hρw

Ms =ρsπr2t/2 = 39250ρs

Yw = r + h/2

Ys =4r/3π

Yc =( Mw yw+ Ms ys)/ ( Mw+ Ms )

Y=0

And

ρs= ρw

Therefore , h = 129.09mm

Moment of Inertia:

Moment of force =F.x

Second moment of Force = F.x.x = F.x2

If force is replaced by area / mass then it is called MOI.

Area MOI = A.x2

Mass MOI =m.x2

MOI and Radius of Gyration(k):

Ixx =A1Y12

+ A2Y22+……

= A y2

Here y = k= Radius of gyration

Similarly Iyy = Ax2 = Ak

2

Parallel axis theorem

MOI of a plane about any axis is equal to sum of MOI about a parallel centroidal axis and

product of area (mass) and square of distance between two axes.

MOI of dA about AA

= dA (h+y)2

MOI of total plane about AA

IAB = € dA(h+y)2

= €dAh2+ €dA y

2 + €dA2hy

IAB =h2€dA + €dAy

2 + 2h €dAy

= Ah2 + IG+ 0

IAB = IG +Ah2

Perpendicular axis theorem

MOI of dA about perpendicular axis

= dAr2

= dA (x2+y

2)

MOI of total plane about Z axis

=€dA x2 + €dAy

2

IZZ = IXX + IYY\

Therefore, the centroid of the semicircle is at a distance of 4R/3π from the diametric axis.

Problem: Moment of inertia of Plane Figure (Rectangle):

Moment of inertia of element about centroidal axis XX

Ixx =Y2dA

=Y2bdy

Moment of Inertia of total Area

Ixx =

Ixx =bd3/12

Similarly,

Iyy =db3/12

Problem: Moment of Inertia of Triangle about its base

Consider a small element strip at a distance y from base of thickness dy. Let dA is area of the

strip.

dA = b1dy

b1 = (h-y)Xb/h

Moment of Inertia of strip about base AB,

= y2 dA

= y2 b1dy

=y2 (h-y).b.dy/h

Moment of Inertia of triangle about AB,

IAB =

= d(h3/3- h

4/4h)

=bh3/12

IAB = bh3/12

Problem: Moment of Inertia of Circle about its centroidal axis

Consider an elementary strip of thickness dr, the side ofstrip is r.dθ.

Moment of Inertia of strip about XX,

= y2dA

=(rSinθ)2.rdθ.dr

=r3sin

2θdθdr

Therefore Moment of Inertia of Circle about XX axis,

Ixx =

=

= R4X2π/8

=π.R4/4

Ixx = π.R4/4 = π.D

4/64

Problem: Moment of Inertia of Rectangle about its Base,

MOI of rectangle about centroidal axis,

Ixx = bd3/12

Now MOI of rectangle about axis-AB can be obtained by applying Parallel axis theorem,

IAB = Ixx + A h2

= bd

3/12 + (bd)(d/2)

2

= bd3/12 + bd

3/4

= bd3/3

IAB = bd3/3

Problem: Moment of Inertia of hollow rectangular section:

Moment of Inertia of hollow rectangle section,

Ixx = BD3/12 – bd

3/12

= (BD3-bd

3)/12

Problem: Moment of Inertia of Triangular Section about centroidal Axis:

Applying Parallel axis theorem,

IAB = Ixx +Ah2

Bh3/12 = Ixx + (bh/2)(h/3)

2

Ixx = bh3/12 –bh

3/18

Ixx = bh3/36

Problem: Moment of Inertia of semi circle (about diametral axis):

Moment of Inertia of Semi circle about AB = (1/2)(πd4/64)

=πd4/128

MOI of semi circle about centroidal axis XX,

h =4R/3π =2d/3π

Area, A = (1/2)(πd2/4)

= πd2/8

Using Parallel Axis theorem,

IAB = Ixx + A h2

πd4/128 = Ixx + (πd

2/8)( 2d/3π)

2

= Ixx + d4/18π

Ixx = πd4/128- d

4/18π

Problem: Determine Moment of Inertia Composite section about axis passing through centroidal

axis. Also determine MOI about axis of symmetry.

Divide section into two areas,

A1 and A2

A1=150X10 =1500mm2

A2=140X10 =1400mm2

Distance of centroid from base of the composite figure

Y = (A1y1+A2y2)/(A1+A2)

= 108.8mm

Moment of Inertia of area about xx axis,

Ixx = {(150X103/12) + 1500X(145-108.8)

2}+ {(10X140

3/12) +1400X(108.8-70)

2}

Ixx= 6372442.5mm4

Similarly,

Iyy = (10X1503/12) +(140X10

3/12)

Iyy= 2824166.667mm4

Radius of Gyration, Kxx = √(Ixx/A)

=√(6372442.5/2900)

=46.8mm

Similarly, Kyy = √(Iyy/A)

= √(2824166.6/2900)

=31.2mm

Problem: Determine MOI of L section about its centroidal axis parallel to the legs. Also find the

polar moment of Inertia.

A1 = 125X10 =1250mm2

A2=75X10 =750mm2

A = A1+A2 = 2000mm2

Distance of centroid from axis 1-1,

Y = (A1y1+A2y2)/(A1+A2)

=(1250X62.5 +750X5)/2000

=40.9mm

Distance ofcentroidal axis from 22,

X = (A1X1+A2X2)/(A1+A2)

= (1250X5 +750X(75/2 +10))/2000

=20.9mm

Moment of inertia about XX axis,

Ixx = { (10X1253/12) + 1250X(62.5-40.93)

2}+ { (75X10

3/12) + 750X(40.9-5)

2}

Ixx = 3183658.8mm4

Similarly about YY centroidal axis,

Iyy = {(125X103)/12 +1250X(20.9-5)

2}+ { (10X75

3/12 + 750X(47.5-20.9)

2}

= 1208658.9mm4

Polar moment of Inertia,

Izz =Ixx + Iyy

=4392317.8mm4

Problem: Determine MOI of I section about its centroidal axis XX and YY. Also determine

polar MOI of the section.

A1= 200X9 =1800mm2

A2 =232X6.7 =1554.4mm2

A3 = 200X9 =1800mm2

MOI abou XX Axis,

Ixx = {(200X93/12 )+ 1800X(125-4.5)

2}+ {6.7X232

3/12) + 1554.4X(0)} + { (200X9

3/12) +

1800X(125-4.5)2}

Ixx = 59269202.13mm4

MOI about Y axis,

Iyy = (9X2003/12) + (232X6.7

3/12) + (9X200

3/12)

= 12005814.7mm4

Polar MOI

Izz =Ixx + Iyy

= 71275016.8mm4

Problem: Calculate Moment of Inertia of Shaded Rea about XX axis

MOI of Shaded section about XX = MOI of triangle about XX

+ MOI of Semicircle ACS about XX

– MOI of Circle

= (100X1003/12) +(πX100

4/128) –(πX50

4/64)

= 8333333.3 + 2454369.2 – 306796.1

=1.048X107 mm

4

Problem: The flat surface of hemisphere of radius R is connected to one flat surface of cylinder

of radius R and length L and made of same material. If total mass be M, show that the moment of

inertia of combination about axis of cylinder is

MR2(L/2+4R/15)/(L+2R/3)

Let ρ be the density of solids. Then mass of system,

M = (2/3)(πR2ρ) + πR

2Lρ

Therefore, ρ = M/(πR2(L+2R/3))

MOI of the combination,

Ixx = Ixx of hemisphere + Ixx of cylinder

For cylinder of mass m and radius r: I = mr2/2

For sphere of mass m and radius r: I = 2mr2/5

Therefore,

Ixx = (2/5)(2πR3ρ/3)R

2+ (πR

2ρL)R

2/2

= MR2(L/2+4R/15)/(L+2R/3)