Chapter 5 Centre of Gravity 5.1 Centre of Gravity Everybody is attracted towards the centre of the earth due gravity. The force of attraction is proportional to mass of the body. Everybody consists of innumerable particles, however the entire weight of a body is assumed to act through a single point and such a single point is called centre of gravity. Every body has one and only centre of gravity. 5.2 Centroid In case of plane areas (bodies with negligible thickness) such as a triangle quadrilateral, circle etc., the total area is assumed to be concentrated at a single point and such a single point is called centroid of the plane area. The term centre of gravity and centroid has the same meaning but the following d ifferences. 1. Centre of gravity refer to bodies with mass and weight whereas, centroid refers to plane areas. 2. centre of gravity is a point is a point in a body through which the weight acts vertically downwards irrespective of the position, whereas the centroid is a point in a plane area such that the moment of areas about an axis through the centroid is zero Plane area ‘A’ G g 2 g 1 1
Transcript
Chapter 5 Centre of Gravity
5.1 Centre of Gravity
Everybody is attracted towards the centre of the earth due gravity. The force of
attraction is proportional to mass of the body. Everybody consists of innumerable particles,
however the entire weight of a body is assumed to act through a single point and such a single
point is called centre of gravity.
Every body has one and only centre of gravity.
5.2 Centroid
In case of plane areas (bodies with negligible thickness) such as a triangle
quadrilateral, circle etc., the total area is assumed to be concentrated at a single point and such
a single point is called centroid of the plane area.
The term centre of gravity and centroid has the same meaning but the following d ifferences.
1. Centre of gravity refer to bodies with mass and weight whereas, centroid refers to
plane areas.
2. centre of gravity is a point is a point in a body through which the weight acts vertically
downwards irrespective of the position, whereas the centroid is a point in a plane area
such that the moment of areas about an axis through the centroid is zero
Plane area ‘A’
Gg2 g1
d2 d1
a2 a1
Note: In the discussion on centroid, the area of any plane figure is assumed as a force equivalent to the centroid referring to the above figure G is said to be the centroid of the plane area A as long as
a1d1 – a2 d2 = 0.
1
5.3 Location of centroid of plane areas
Y
Plane area
X G
Y
The position of centroid of a plane area should be specified or calculated with respect to
some reference axis i.e. X and Y axis. The distance of centroid G from vertical reference axis
or Y axis is denoted as X and the distance of centroid G from a horizontal reference axis or X
axis is denoted as Y.
While locating the centroid of plane areas, a bottommost horizontal line or a horizontal line
through the bottommost point can be made as the X – axis and a leftmost vertical line or a
vertical line passing through the leftmost point can be made as Y- axis.
2
In some cases the given figure is symmetrical about a horizontal or vertical line such that
the centroid of the plane areYa lies on the line of symmetry.
X G
Y
Xb
b/2
The above figure is symmetrical about a vertical line such that G lies on the line of
symmetry. Thus
X= b/2.
Y=?
The centroid of plane geometric area can be located by one of the following methods
a) Graphical methods
b) Geometric consideration
c) Method of moments
The centroid of simple elementary areas can be located by geometric consideration. The
centroid of a triangle is a point, where the three medians intersect. The centroid of a square is a
point where the two diagonals bisect each other. The centroid of a circle is centre of the circle
itself.
4.4 METHOD OF MOMENTS TO LOCATE THE CENTROID OF PLANE AREAS
3
Let us consider a plane area A lying in the XY plane. Let G be the centroid of the plane area. It is required to locate the position of centroid G with respect to the reference axis like Y- axis and Xi- axis i.e, to calculate X and Y. Let us divide the given area A into smaller elemental
areas a1, a2, a3 …….. as shown in figure. Let g1 ,g2, g3…… be the centroids of elemental
areas a1, a2, a3 …….. etc.
Let x1, x2, x3etc be the distance of the centroids g1 g2 g3 etc. from Y- axis is =A X --(1) The sum of the moments of the elemental areas about Y axis is
a1 . x1 + a2 . x2 + a3 . x3 + ……………….(2)
Equating (1) and (2)
A .X = a1 . x1 + a2 . x2 + a3 . x3 + ………..
a 1. x 1 a a ..X =A
(ax) x.dAX = or X =
A A
Where a or dA represents an elemental area in the area A , x is the distance of elemental area from Y axis.
Similarly
a1. y1 a2 . y2 a3 . y3 ..Y =
A
(a. y) y.dAY = or Y =
A A
4
TO LOCATE THE CENTROID OF A RECTANGLE FROM THE FIRST PRINCIPLE
(METHOD OF MOMENTS)
Y
b
dy
X Gd
y
Y
X
Let us consider a rectangle of breadth b and depth d. let g be the centroid o f
the rectangle. Let us consider the X and Y axis as shown in the figure.
Let us consider an elemental area dA of breadth b and depth dy lying at a distance of y
from the X axis.
5
.W.K.T d
y.dA
=
0
YA
A = b . ddA = b .dy
.d
y.(b.dy)Y = 0
b.d
Similarly
1 d
y.dyY = d 0
1 y 2d
=Y d 2 0
= 1 d 2
Y d 2
= dY 2
6
Triangle :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’.
AEF ABC
b1 h y bh
b bh y
1 h
b b1y
1 h Area of element EF (dA) = b1×dy
b 1y dy
h
y y.dA
c Ah y
y.b1h dy
0 1 b.h
2
y2b
y3h
7
h2 3
ú2 3h 01
b.h2
2 h2
h3
2
h2
h 6
h3
Therefore, yc is at a distance of h/3 from base.
8
3
Semi circle: 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.
Let us consider a quarter circle with radius r. Let ‗O‘ be the centre and ‗G‘ be the centroid of
the quarter circle. Let us consider the x and y axes as shown in figure. Let us consider an
elemental area ‗dA‘ with centroid ‗g‘ as shown in fig.
Let ‗y‘ be the distance of centroid ‗g‘ from x axis. Neglecting the curvature, the elemental area becomes an isosceles triangle with base r.dθ and height ‗r‘.
Here y =2r
. sin 2r3 . sin .dA = 2r0cos / 23
W K T Y = 3A
= 4r [0+1]y.dA
dA =1
.r.d .r3
2Y = 4rAr 2 Y =
.r2dA = .d
3A = 2
2 Similarly=Y
4ry.dA X2r r 2 = 3
Y =A 3 .sin . 2 .d
.r2
4
11
12
5.5 Centroid of Composite Sections
In engineering practice, use of sections which are built up of many simple sections is very common. Such sections may be called as built-up sections or composite sections. To locate the centroid of composite sections, one need not go for the first principle (method of integration). The given composite section can be split into suitable simple figures and then the centroid of each simple figure can be found by inspection or using the standard formulae listed in the table above. Assuming the area of the simple figure as concentrated at its centroid, its moment about an axis can be found by multiplying the area with distance of its centroid from the reference axis. After determining moment of each area about reference axis,
13
the distance of centroid from the axis is obtained by dividing total moment of area by total area of the composite section.5.6 PROBLEMS:
Q1. Locate the centroid of the T-section shown in fig.
Q10: Determine the centroid of the following figure.
A1 = Area of the rectangle A2 = Area of triangleA3 = Area of circle
Area (Ai) xi yi Ai xi Aiyi
30,000 100 75 3000000 22500003750 100+200/3
= 166.6775+150/3=125
625012.5 468750
7853.98 100 75 785398 589048.5
x Aixi A1x1 A2 x2 A3x3 86.4mm
c A AAAi
y Aiyi
1 2 3
A1 y1 A2 y2 A3 y3 64.8mm
c A AAA
21
22
42
II
Numerical Problems (Assignment)
1. An isosceles triangle ADE is to cut from a square ABCD of dimension ‘a’. Find the altitude ‘y’ of the triangle so that vertex E will be centroid of remaining shadedarea.
2. Find the centroid of the followingfigure.
3. Locate the centroid C of the shaded area obtained by cutting a semi-circle of diameter ‘a’ from the quadrant of a circle of radius‘a’.
4. Locate the centroid of the compositefigure.
23
Chapter -6
Simple Machine or Lifting Machine:
A machine a device by which heavy load can be lifted by applying less effort as compared to the load.
e.g. Heavy load of car can be lifted with the help of simple screw jack by applying small force.
Compound Machine:
Compound machine is a device which may consists of number of simple machines. A compound machine may also be defined as a machine which has multiple mechanisms for the same purpose.
Compound machines do heavy work with less efforts and greater speed.
e.g. In a crane, one mechanism (gears) are used to drive the rope drum and other mechanism (pulleys) are used to lift the load. Thus, a crane consists of two simple machines or mechanisms i.e. gears and pulleys. Hence, it is a compound machine.Effort:
It may be defined as, the force which is applied so as to overcome the resistance or to lift the load.
It is denoted by „P‟.
Magnitude of effort (P) is small as compared to the load (W).Load:
The weight to be lifted or the resistive force to be overcome with the help of a machine is called as load (W).Velocity Ratio (V.R.):
It is defined as the ration of distance traveled by the effort (P) to the distance traveled by the load (W)
V.R. Distance travelled by effortDistance travelled by load
Velocity ratio will be always more than one and for a given machine, it remains constant.Mechanical Advantage:
It is defined as the ratio of load to be lifted to the effort applied.
24
Load (W) WM .A.
Effort (P) P
Input:
The amount of work done by the effort is called as input and is equal to the product of effort and distance travelled by it.
Input = P x X, travelled by the effort
where, P – Effort and X – distance
Output:
The amount of work done by the load is called as output and is equal to the product of load and distance travelled by it.
Output = W x Y travelled by the load
where, W – Load and Y – distance
Efficiency:
The ratio of output to input is called as efficiency of machine and it is denoted by Greek letter eta (η)
Generally, efficiency is expressed in percentage%η = (Output/Input)*100
It is always less than 100 because of friction, therefore output <input.
But Output = W.Y and Input = P.X
%η Output
x100 W x Y
x100Input P x X
WM .A.
x100(Since M .A.
Wand V.R.
X)
% η
Px100
PX V.R. YY
25
Therefore, efficiency of a machine is also defined as the ratio of mechanical advantage (M.A.) to the velocity ratio (V.R.). It is also expressed in percentage.
%η M.A./
V.R. x100
It is always less than 100 because of friction, therefore M.A. < V.R.Actual Machine:
The machine whose efficiency is always less than 100 % due to frictional resistance offered by the different moving component parts of the machine is called as actual machine.
For such machines, η < 100 % and hence M.A. < V.R.Ideal Machine:
The machine whose efficiency is 100 % and in which friction is totally absent or zero, is called as ideal machine.
For ideal machines, η = 100 % and hence M.A. = V.R.Ideal Effort (Pi):
The effort which is required to lift the load when there is no friction is called as an ideal effort (Pi)
Ideal Effort Pi =
WV.R.
Where, Pi = Ideal Effort, W = Load to be lifted, V.R. = Velocity Ratio Ideal Load (Wi):
The load which can be lifted by an effort (P), when there is no friction, is called as an ideal load (Wi)
Ideal Load Wi = P x V.R.Where, P = Effort applied, Wi = Ideal Load, V.R. = Velocity Ratio
LAW OF MACHINE:
The equation which gives the relation between load lifted and load applied in the form of a slope and intercept of a straight line is called as Law of a machine i.e. P = mW + C
Where, P = effort applied, W = load lifted, m = slope of the line and C = y – intercept of the straight line.
To draw the graph of Load (W) V/s Effort (P), effort is applied on
26
a machine and the corresponding values of the loads are noted down. The graph of Load (W) V/s Effort (P) is drawn by taking load (W) on the x-axis and the effort (P) on the y-axis as shown in the figure.
mtanθ Y2Y
1
X2 X1
Effort(P) (X2, Y2)
O(X1, Y1)
C
OLoad (W)
It has been observed that, the graph of load v/s effort is a straight line cuts the Y-axis givingthe intercept „C‟ which indicatesthe effort lost on friction, when no load is applied.
Is must be noted that, if the machine is an ideal machine, the straight line of the graph will pass through the origin.
Comparing to the equation of straight line i.e. y = mX + C, we getP = mW + C
Where, P =Effort applied, W = Load applied, m = slope of the line and C = Y-intercept of the line.
27
Note:
If we know the law of machine i.e. values of „m‟ and „C‟ areknown, then for a given load effort can be found out or for a given effort corresponding value of the given load can be found out. The law of machine also indicates the friction in the machine and maximum M.A.Maximum Mechanical Advantage (Max. M.A.):
We know that,
M.A. W/
PBut, P mW C
M.A. W
mW CDividing the numerator & Denominator by ' W' we getM.A.
1
m CW
In the above equation if ' W' is more, the ratio WC
will be very small
Neglectingthe ratio WC
, M.A. will be maximum
Max. M.A. m1
Maximum Efficiency:
The ratio of maximum M.A. to the V.R. is called as maximum efficiency.
It is also expressed in percentage as
% Maximum η Max M .A. x 100 1 x 1 x 100 ( SinceMaxM .A. 1 )V.R.V.R. m m
28
Reversible Machine:
When a machine is capable of doing some work in the reverse direction even on removal of effort, it is called as reversible machine.Condition for Reversible Machine:
The efficiency of the machine should be more than 50%.Irreversible Machine / Non-reversible Machine / Self Locking Machine:
When a machine is not capable of doing some work in the reverse direction even on removal of effort, it is called as irreversible machine or non-reversible machine or self locking machine.Condition for Irreversible Machine:
The efficiency of the machine should be less than 50%.Friction in Machines in terms of Effort and Load:
In any machine, there are number of parts which are in contact with each other in their relative motion. Hence, there is always a frictional resistance and due to which the machine is unable to produce 100 % efficiency.
Let, P = Actual Effort, Pf = Effort Lost in friction, Pi = Ideal Effort Effort Lost in friction (Pf) = Actual Effort (P) – Ideal Effort
(Pi)
P PPP W (Since P W )f i V.R. i V.R.
Let, W = Actual load lifted,Wf = Load Lost in friction, Wi =
IdealLoad
Load Lost in friction (Wf) = Ideal Load (Wi) – Actual load lifted
(W)
Wf Wi W (P x V.R.) W (Since WiP x V.R.)
29
STUDY OF SOME SIMPLE MACHNES:
We know that, velocity ratio of a machine is the ratio of distance travelled by the effort to the distance travelled by the load. It is observed that, the distance travelled by the effort is greater than the distance moved by the load. Experimentally, it has been found that, the velocity ratio remains constant for all loads. The velocity ratio changes from machine to machine but remains constant for a given machine.1. Simple Wheel and Axle:
In simple wheel and axle, effort wheel and axle are rigidly connected to each other and mounted on a shaft. A string is wound round the axle so as to lift the load (W) another string is wound round the effort wheel in opposite direction so as to apply the effort (P) as shown inthe figure.
Let,
W = Load lifted, P = Effort Applied D =
Diameter of the
effort wheel d = diameter of the load axle.
When, the effort wheel
completes one
revolution,the effort moves througha distance equal to the circumference of the effortwheel (πD)
and simultaneously the load moves up through a distance equal to thecircumference of the load axle (πd)
V.R.
Distance travelled by the Effort
πD D
Distance travelled by the Load
π d d
30
2.Differential Axle and Wheel:
Differential axle and wheel is a further modification and improvement over the simple axle and wheel.It consists of a loadaxle made up of bigger axle of diameter d1and smaller axle of diameter
31
d2rigidly connected to each other and an effort wheel of diameter „D‟.Since, the load axle is made up of two axles of different diameters; it is called as a differential axle. Differential axle and effort wheel aremounted on the same shaft which is supported on ball bearings as shown in the figure.
A string is wound round the effort wheel so as to apply theeffort „P‟. Another string is wound round the bigger axle further passing over the pulley carrying the load „W‟ attached to the snatch block. The same string is further wound round the smaller axle in the opposite direction to that of bigger axle. The winding of string on effort wheel and smaller axle is done in the same direction; the string unwinds from the effort wheel & smaller axle and winds over the bigger axlesimultaneously, when the effort „P‟ is applied.
When effort wheel complete one revolution, the differential axle also completes one revolution.
Distance travelled by the effort = πD
Length of the string woundover the bigger axle = πd1
Length of the string unwoundover the smaller axle = πd2
Total winding over the bigger axle = πd1- πd2 = π(d1 – d2) 2 D
d1 d2
Where, D = Diameter of effort wheel, d1 = diameter of bigger axle, d2 = diameter of smaller axle.
If radius of effort wheel, bigger axle and smaller axle are giventhen,
32
V.R.
2 RR1 R2
Where, R = Diameter of effort wheel, R1 = diameter of bigger axle, R2 = diameter of smaller axle.
33
3. A Simple Screw Jack:
A screw jack is commonly used for lifting and supporting the heavy load. A very small effort can be applied at the end of the lever or handle or tommy bar for lifting the heavy loads. This effort is very small as compared to the load to be lifted. As jack has a simple mechanism, it is most commonly used in repair work of vehicles.
When the effort is applied to the handle or lever arm to complete one revolution then load is lifted through one pitch of the screw (p), therefore the distance moved by the load is equal to the pitch of the screw and the distance moved by the effort is equal to 2πL
Let, L = length of the handle or lever arm and p = pitch of the thread or screw, then
V.R.
Distance travelled by the effort
2π L
Distance travelled by the load p
If the effort wheel is used at the place of handle or lever arm for applying the effort, then
V.R.
Distance travelled by the effort
πD
Distance travelled by the load p
Where, D = diameter of the effort wheel and p = pitch of thread or screw.
4. Weston’s Differential Pulley Block:
It consists of upper and lowerblock. Upper block is having biggerpulley and smaller pulley of different diameter mounted on same common axle and that of lower block is having asingle pulley. The weight „W‟ is attachedto the lower block. Upper block is fixed and lower block is movable.
34
An endless chain passes over the bigger and pulley and single pulley in lower block as shown in the figure.
When there is one complete revolution of the bigger pulley, distance moved by the effort is equal to the circumference of the bigger pulley (πD). When the bigger pulley completes one revolution, the smaller pulley also completes one revolution because both pulleys are fixed on the same axle. Therefore, for one complete revolution, the length of the chain wound round the bigger pulley is equal to the circumference of the bigger pulley (πD) and at the same time, the length of the chain unwound from the smaller pulley is equal to the circumference of the smaller pulley (πd).
Net winding = πD – πdAs the load (W) attached to the lower block, is equally
distributed between the two parts of the chain, distance moved by load is equal to the half of the net winding.
Distance moved by the load = ½ ( πD – πd )
V.R. 2D D d
Where, D = diameter of bigger or upper pulley, d = diameter of smaller or lower pulley
If radii of bigger and smaller pulley are given, then
V.R. 2R R r
Where, R = radius of bigger or upper pulley, r = radius of smaller or lower pulley
If number of teeth or cogs of the bigger and smaller pulley are given, then
35
V.R.
Distance travelled by the effort πDDistance travelled by the load 1 ( πD π
d)2
V.R.
2 T1T T1 2
Where, T1 = number of teeth or cog on bigger or upper pulley, T2 = number of teeth or cog on smaller or lower pulley5. Worm and Worm Wheel:
This machine is made of toothed wheel known as worm wheel and a square threaded screw known as worm. Worm and worm wheel is geared with each other. The effort wheel is attached to the worm. A load drum is centrallyfixed to the worm wheel as shown in the figure. The effort can be either by wheel or handle. The worm may be single threaded or multi-threaded.
Let, w = load lifted, T = number of teeth on the worm wheel, P = effort applied, r = radius of load drum, R = radius of effort
wheel.Consider the worm is single threaded.For one complete revolution of effort wheel, distance moved by
the effort P = 2 π R and load drum performs (1/T) revolution. Distance moved by the load = 2πr
T
We know that,
36
V.R.Distance travelled by the effort 2πRR TDistance travelled by the load 2 π r r
T
V.R. RrT
If handle is used in place of effort wheel,
V.R. LrTWhere, L = length of the handle
In general, if worm is „n‟ threaded,
V.R.
R T
L T
Where, n = number of thread on worm
nr
nr
If the worm is double threaded, V.R.
R T
L T since, n = 2
2 r
2 r
6. Single Gear Crab or Single Purchase Winch Crab:
This machine consists of mainly the larger gear wheel known as Spur or main gear and smaller gear known as pinion. Spur or main gear is mounted rigidly on the load drum or load axle and the spur is geared to the pinion which is further mounted rigidly on the shaft to which the effort wheel or handle is attached.
37
A rope or string is wound round the load drum so as to lift the load
„W‟ and an another string is wound round the wheel so as to apply the effort „P‟ as shown in the figure.
Let, W = Load lifted, P = Effort applied, D = diameter of effort wheel, T1 = number of teeth on spur or main gear, T2 = number of teeth on pinion d = diameter of load drum or load axle.
For one complete revolution of effort wheel, distance moved bythe effort = πD and the pinion also completes one revolution, that time
spur performs T2 revolutions and as load drum and spur being rigidlyT
1
connected, the load drum also performs T2 revolutions.T1
Distance moved by the load = T2 x circumference of load drumT1
= T2x πdT1
We know that,
V.R.
Distance travelled by the effort πDDistance travelled by the load
T2
xπ d
T1
V.R. D x T2 d T1
If handle of length „L‟ is used in place of effort wheel,
V.R. 2d
L x T1where, L =length of the handle.
6.Double Gear Crab or Double Purchase Winch Crab:T2
This machine consists of two larger gear wheels A and C called as spurs or main gears and smaller gear wheels B and D called as pinions as
38
shown in the figure. A load drum or load axle is rigidly connected to the spur A, which is further geared with pinion B. Pinion B and spur C are rigidly mounted on same shaft called as intermediate shaft or axle and spur C is further geared with pinion D mounted on the shaft. So called as effort axle to which the effort wheel is attached.Let, W = Load lifted, P = Effort applied, D = diameter of effort wheel,
d = diameter of load drum or load axle.T1 = number of teeth on spur A, T2
=pinion B mounted on intermediate shaft,on spur C mounted on intermediate shaft,
number of T3 = number
teeth on of teeth
T4 = number of teeth on pinion D mounted on effort axle
For one complete revolution of effort wheel,Distance moved by effort = πDThe pinion D on the effort axle also makes one revolution and
therefore spur C and pinion B on the intermediate shaft performsT4 revolution.T3
39
That time, spur A and load drum performs same revolutions as T
4 xT2 because both are rigidly connected to each other.
T3 T
1
Distance moved by the load=T4 x T2 revolutions xT T
3 1
circumference of load drum =
T4 x T2 x πdT T
3 1
We know that,V.R.
Distance travelled by the effort πDDistance travelled by the load T T
4
x2 x π
dT T3 1
V.R. D x
T1 x
T3
T Td2 4
If handle of length „L‟ is used in place of effort wheel,
![Diameter - cisco.com · Diameter EAP application is based on the Diameter Network Access Server Application [NASREQ] and is intended for environments similar to NASREQ. In the Diameter](https://static.documents.pub/doc/80x56/6027b4060df5ad7d7c575ab9/diameter-ciscocom-diameter-eap-application-is-based-on-the-diameter-network-access.jpg)
