Centre of Gravity and Moment of Inertia

UNIT 4 CENTRE OF GRAVITY AND MOMENT OF INERTIA Structure 4.1

Introduction Objectives

4.2

Concept of Centre of Gravity 4.2.1 4.2.2 4.2.3 4.2.4 4.2.5 4.2.6 4.2.7 4.2.8 4.2.9

4.3 4.3

Centre of Gravity of Thin-Wires Theorems of Pappus and Guldinus 4.4.1 4.4.2 4.4.3 4.4.4 4.4.5 4.4.6

4.5

System of Two Equal Masses Centre of Gravity of a l'hin Uniform Rod Centre of Gravity of a Uaifonn Recta~gularPlate Mass-centre of a System of Two Unequal Masses Centre of Gravity of a System of Collinear Masses Centre of Gravity of a Unifam Triangular Plete Mass-centre of a System of Coplanar Particles Centroid of Ircegular Shaped Area First Moment of A m

Theorem I Theorem II Cylinder Cone Sphere Hemisphere

Moment of Inertia of Area 4.5.1 Definition 4.5.2 Perpendicular Axis Theorem 4.5.3 Parallel Axis Theorem

4.6

Principal Axes 4.6.1 Rodua of Inertia 4.6.2 Rotated Set of Axes 4.6.3 Principal Moment of Inertia

Mass Moment of Inertia 4.8 Summary 4.9 Key Words 4.10 Answers to SAQs 4.7

INTRODUCTION This unit will help you in understanding the concepts of centre of gravity of wires, thin uniform plates of various shapes with or without openings and simple solids such as cone, sphere and cylinders. The concept of centroid of plane areas is similar to centre of gravity of thin uniform plates of same shape as the plane areas. Moment of Inertia of plane areas about a given axes are mathematically referred to as second-moment of areas. The knowledge of moment of inertia of plane areas are useful in the analysis of bending stresses of beams under flexure. The moment of inertia of rods, discs, or spheres are useful in studying the problems in dynamics of rigid-bodies.

Objectives After studing this unit, you will be able to determine position of centre of gravity of uniform-wires of various shapes, position of centre of gravity of masses of plates of various shapes with or without openings, the volume as well as surface areas of various simple solids and their location of centre of gravity by using two theorems given by Pappus and Guldinus, and compute moment of inertia of various areas & masses, about the given axes.

4.2

CONCEPT OF CENTRE OF GRAVITY

A body can be considered as an agglomeration of a large number of particles each of which is adhering to its adjacent particles. These particles may have different sizes, density or may have uniform size and density. Each of these particle is subjected to gravitational force directed towards the centre of the earth. The magnitude of the gravitational force will depend on its mass. For all earthly bodies whose sizes are very srnall as compared to its distance upto the centre of the earth, the forces of gravitational attraction on various particles of the body can be assumed to be parallel. Hence the total weight of the body is the summation of all these parallel forces acting vertically down towards the centre of the earth. The line of action of the resultant R of these forces can therefore be found. Consider a case of a triangular metal plate ABC placed in a vertical plane with its side AB vertical. If you divide the area of triangle in large number of imaginary vertical strip parallel to BA for each of the strip the gravitational force can be computed and resultimt R of all these forces will be along a vertical line (1-G) as shown in Figure 4.1 (a), where G is a point in which entke weight of the plate is concentrated.When the plate is oriented at right angle to its previous position, line BC is placed vertically as shown in Figure 4.1 @). New vertical strips can be formed and resultant R, which is obviously of same magnitude R as in the case of first orientation (a), will be along vertical line (2-G). The location of G will be the intersection of lines of action of resultants obtained in cases (a) and (b). The mass centre is a unique point G in the plate ABC through which resultant weight of all the strip-masses will always pass through irrespective of the orientation of the plate. Figure 4.1 (c) shows the third orientation of triangle ABC where one of the medians BGB ' in the triangle is kept vertical. It will be proved later on that the point of intersection of all the medians represent the mass-centre of the triangular plate.

Figure 4.1 Position of ResultantR in Different Orientation of Plate ABC

Centre of gravity or Mass-centre is a point in the body where entire mass or weight is assumed to be concentrated and for convenience single resultant gravity load R can be used as a replacement for distributed gravity loads at various locations of its particles.

4.2.1 System of Two Equal Masses Consider a system consisting of two masses mAand m, with their centres at locations A and B which are rigidly connected by a thin rod of negligible weight as shown in Figure 4.2. The resultant force R due to combined effect of the weight of the two masses is (mAg + m, g) and the location G on the thin rod of this resultant of a single equivalent force (mA+ m,) x g is the centre of gravity of the two masses. When the two masses are of equal magnitude, mA= mB = m, the resultant R of magnitude 2mg will act at mid-point of AB as shown in Figure 4.2. Hence, the two equal masses m at A and B can be looked upon as a single equivalent mass 2m at centre of gravity G where GA = GB. The case of two unequal masses will be dealt with later in this unit.

Centre of Gravity and Moment of Lnertia

Figure 4.2

4.2.2 Centre of Gravity of a Thin Uniform Rod Let All = L be the length of a unifcxm rod or a uniform strip with its mid point G. The mass per unit length of this rod or strip is uniform at all points. For any elemental mass (Sm) at point P located at distance x from mid-point G as shown in Figure 4.3, there is an equal mass (Sm)at p' at the same distance x' = x on the other side of centre - line C of the bar. The centre of gravity of these two equal masses at P and P is at their midpoint G. This is true for all other masses in the portion GB of the bar and the corresponding masses in the counterpart GA. Thus G, the midpoint of AR, happens to be the centre of gravity of the uniform rod. However, if the bar is of non-uniform weight, the position of C will shiit from the centre point.

Figure 4.3

4.2.3 Centre of Gravity of a Uniform Rectangular Plate Let the dimensions of uniform rectangular plate ABCD be L x B, where AB = L and BC = B as shown in Figure 4.4. It is already seen from the previous section that the centre of gravity or masscentre of any uniform strip of length L parallel to AB is at its midpoint G.Since the plate ABCD has auniform mass all over its area, the centre of gravity of the first strip from AB or DC is indicated by GI, while those of second strip, third strip and so on, can be indcated by G, ,G, etc. depending upon the number of even strips chosen for the rectangular plate. Further, if the width of the strips chosen are equal, the masses associated with G,, G,, C, etc. will he of same magnitude, say m.In fact, you can realise that all these

L- L4--Figure 4.4

masses (m) are spread uniformly on line M, M, where M, and M2 are midpoints of AB and DC. Obviously the C.G. of all these masses will be at G which is the midpoint of M, M2.

4.2.4 Mass-centre of a System of Two Unequal Masses Consider two spherical bodies of masses m, and mz with centres at points A and B respectively where distance AB = L. Let us assume that these masses are connected by a massless thin rod so that distance AB is not changed. Let AB be a horizontal line along X direction. Let the weight W , of mass m, act vertically along A A, and weight W2of mass n, along BB, as shown in Figure 4.5.

Let G be the centre of gravity of the two masses at a distance

from A.

Resultant weight R = (m, + m2).x g acts vertically downward at G. Using the theorem of moments, we have,

[

Moment of Resultant (R) about a point (A)

where R = (ml

+ mz) x

I

=

[

Algebraic sum of moments of all the masses in the system about (A)

g = [Vm)]x g

Similarly, by considering equilibrium of moments about mass at B,

Case I

As a special case, when m1 =

+= m

Thus, midpoint of line AB is the location of C.G. for two equal masses.

Case 11 When AB = 6 m and masses are as follows m, = 2kg at point A

This implies that distances of the C.G. from the two particle masses are inversely proportional to their masses.

Moment of mass m, about G just balanws the moment of mass m, about G. 'Ihus G, the centre of gravity of two masses, happens to be fulcrum of an imaginary level AB of negligible weight having two masses (m,) and (mJ placed at El and % from the fulcrum F as shown in Figure 4.6.

The resultant mass (ml + m,) is concentrated at location G so that total resultant weight R = (m, + mJ g acts downward at G and if a fulcrum or support F is provided to the lever at G, the vertical reaction V offered by this support will balance the resultant weight R to keep the system of masses in equilibrium.

4.2.5 Centre of Gravity of a System of Collinear Masses 1t is well understood by now, that centre of gravity of several masses is a point through which resultant weight R of the masses, passes. Consider the case of two unequal masses with c e d e s A and B as given in previous section. If the line AB is kept vertical, the two weights (m, g) and (m, g) will be collinear since they act along same vertical direction AB. Hence, the resultant R = (m, + m,) x g will also be along the line AB passing through the centre of gravity G as shown in Figure 4.7 (a). 'Ihe distance % has been worked out in the previous section. This is also true for a system of masses m, ,mB,mc ,mD etc. all having their centres A,B,C,D lying on a rigid massless rod along line AD. Hence, all these collinear masses can be placed along a single vertical line AD as shown in Figure 4.7 (b).

Cemtre d Gravity and

M~tdIne*

Since the resultant force R of all these weights must also act along the line AD, the centre of gravity of the collinear pusses is always on the line joining their centres. This logical thinking is even otherwise appealing to our commonsense. The same can be applied to determine the location of centre of gravity of a triangular plate of uniform mass per unit area.

4.2.6 Centre of Gravity of a Uniform Triangular Plate The centre of gravity of a uniform triangular plate (i.e. with uniform thickness) lies at its centroid which is the point of intersection of its medians. This can be proved on the basis of following log1c. Consider the triangular plate ABC as shown in Figure 4.8.

Figure 4.8

Let M , and M, be the midpoints of sides BC and AB of the triangle. Consider a typical small strip B,C, parallel to side BC. The midpoint M,' of this strip lies on the median AM,. Since the strip has uniform mass per unit of length, mass centre of this strip is at MI'. This result is based on again the,old logic that for every elemental mass 6m at B', there is an equal mass 6m at c,' which are equidistant from M,' . Similarly mass-centre of any other strip parallel to BC will lie on this median AM, . Further, since the mass-centre for all such strips (parallel to BC) in the triangle lie on the median AM,, it can be concluded that the mass-centre of all these masses must lie on the median AM,. Similarly, considering the strips, like A, B, parallel to AB, it can be proved that mass-centre of Ihe trirngdirr plMe shmld lie on the median CM,. Hence, mass-centre of the triangular pl* &be attibe iatmwYonof(he medkm.

4.2.7 Mass-centre of a System of Coplanar Particles Consider a system of coplanar particles A, ,A, ,A, etc. having masses m,, m, ,m,, ... (or weight w, ,w, , w, etc) which have their co-ordinates (x, ,yl), (q,YZ),(-% ,~ 3 and ) so on9 as shown in Figure 4.9.

I

For simplicity of understanding, assume #at all the particles lie in a horizontal plane XOY and weighty w, ,w, ,w, of these particles A, ,A, ,A, are vertically downward along Z direction. Let the resultant W of all these weights act at G whose co-ordinates are (,i,y ). W = (w, +

'Ihen,

W,

+ w, +

...) =

C wi

Hence, applying the Varignon's theorem which states that "moment of the resultant face at G is equal to the algebraic sum of the moment of all the individual forces at A, ,A, ,A, etc. about the horizontal axis OY ".

1

Similarly, considering the theorem about axis OX.

wy = w, y, + w,

Y

y,

+ w,

y,

+ ...

=

WiY, .'. -y = C w; 1

P I

1

where C implies the sum of d l terms where i varies from 1 onwards for all masses. I

The concept of centre of gravity or mass-centre of thin plates can be extended to that of centroid of plane areas. Thus replacing the masses by corresponding areas, we have centroid of a rectangular area ABCD at its geometric centre or the point of intersection of diagonals. The centroid of a triangular area is at it? point of intersection of the medians. Note that centroid of a plane area of a given shape and centre of gravity of thin plate of the same shape are at the same location provided plate has uniform-mass per unit area.

4.2.8 Centroid of Irregular Shaped Area Consider an irregular areaA as shown in Figure 4.10 which can be divided into large number of elemental areas 6 A,

Centre of Gravity and Moment of Inertia

Figure 4.10

Considering centroid of the area at C ( 35, ) the principle of moments can be apphed on the same lines as that of centre of gravity of weights of various elemental masses 6m, .

:.

A 35 = X (A, x,) I

or

-x = - x A, XI I

X Ai I

Similarly

-y =

XA, Yi

z. Ai 1

co&dering thin plate of uniform mass m per unit area i.e., S mi = 6 A, x m M = Xm, = m X 6A,

Total mass

I

I

= mA.

:.

( mgA) 5 5 = Zm, gx,

Hence, centre of gravity of thin uniform plate is given by -

Cm,x, x=---

x

-

ZA,x, EAl

'"1

Hence, numerically the position of centroid of a given area is the same as centre of gravity of thin uniform plate of the same area.

4.2.9 First Moment of Area The terms A, x, and Ai y, are the first moment of area Ai about y and x axes respectively, considering XOY as a horizontal plane. It is to be noted that xi and y, co-ordinates of centres of small area A, . Similarly, the terms (m, g) x, is the fust moment of the weight of mass m, about y axis where x, is the x co-ordinate of centre of the small mass m,. The numerator (X A, xi) or (Z m, xJ in the expression for Z referred in previous section is obtained from the concept of summation of First Moment of area (A ,) or mass (mi). Example 4.1 Determine the centroid of the area OBCD as shown in Figure 4.1 1.

-

Solution

The trapezium area is divided into two simple areas, (i) Al = Rectangle OLDC with centroid at G1 (1.5,3) (ii)

- 9 m2 with its centroid at G2 (4,2) A~ = ~riangulararea D L .= --

2

-

The computation of co-ordinates ( ?, j ) for G and other aspects are given id following Table :

nen,

-

x =

63 = 2.33 m 27

H e m , the Cordinates of centroid G are

j =

72 n -- 2.67m

x = 2.33 m and

j = 2.67 m.

Example 4.2 Determine the centroid of an area shown in Figure 4.12 (a).

'Ihisproblem can be attempted in two ways as shown in Figure 4.12 (b)atad Figure 4.12 (c). ?he total area A is divided into tbe three parts A, , A2 and A3 in Figure 4.12 (b) while the same areaA can be divided in two parts A'1 and A'2 (shown shaded) in Figure 4.12 (c).

-

Let G be the centroid of area A as shown in Figure 4112 (a) with @ , ;J) co-ordinates.

statics

It is to be noted from above referred Table that whichever way, we consider the division of regular area, either all positive or some of them negative, we should fmaly get a unique value of (i) X A ,

(ii) C A, xi and (iii) C Ai yi

For a given area A , the co-ordinates, Y and 7 will be a unique value. Note : If the problem of C,G. (Centre of gravity) of a uniform plate is considered where out of a square plate OACB a triangular part M I C M2 is cut out, Cle position of centre of gravity of such a plate is the same as position of centroid of area in Figure 4.12 (a). Example 4 3 A square plate of uniform thickness and density is bent along M IM 2till comer C coincides with centre C' as shown in Figures 4.13 (a) & (b). Determine its centre of gravity.

Solution

Let w be the uniform weight of the plate per unit area. The entire plate after it is bent can be considered to be made up of three parts. (1)

W l = weight corresponding to a square plate OACB = (36 w) at location (3,3)

(2)

W2 = weight corresponding to overlapped portion M l C M 2 = (4.5 w) at location (4,4)

(3)

(-

W3)= portion (M1CM2)which is removed = (- 4.5 w) at location (53)

I

i

I

C q t r c ef Gravity and Moment of Inertia

:.

x- = -- 2.88 m and 36

= 2.88 m.

This example can also be solved with other alternative way by considering total plate as follows : (1) WIf= weight of rectangular plate (BM2 M30) 18 w with its mass-centre at (1.5.3)

Wi = weight of square plate (CMI AM3)

(2)

= 9 w with its mass-centre at (4.5, 1.5)

W3' = weight of two triangular plates (MI M2 C) 9 w with its mass-centre at (4,4) Note that resultant weight W = I; w, = (18 + 9 + 9) w = 36 w as before. (3)

Example 4.4

Determine the centroid of the shaded area show11in Figure 4.14.

Solution

Net area of shaded portion of Figure 4.j4

= The area of full circle of radius r (A,) - The area of cut out circle of radius I: (Az) 2

Area A2 is to be regarded as negative area. Consider moment of areas a b u t G; ZAi xi = 0 since the lever - arm of the Resultant areaA about G is zero, A1x,

+

(-Az)+=A

x (0) = 0

n

Positive sign of indicates that with respect to the origin 0 of reference axes x and y, = OG is along positive direction of x axis. Since the centres 0 a d A of the two area (Al) and (- AZ)are taken along x axis; G lies on AO.

x

Example 4.5

Determine the centroid of an area of a semicircle shown in Figure 4.15. Solution You will note that it is not possible to divide a semicircle into a few suitable standard - areas for each of which magnitude as well as location of its centroid is known. Hence, in such a situation, we should consider a very large number of very thin strips of width dx and assume that its centroid is at the centre of the strip.

Consider the semicircle with centre C and its straight boundary ECD along y axis. I h e axis CX is therefore the axis of symmetry ofthe area. This means that portion of quadrant CFE is the mirror-image of quadrant CFD. Consider element of an area as a strip AB at distance x from C with width dx tending to zem. If a is the radius of semicircle. AB = 2 a s i n 0

d4 = 2 a s i n 0 x dx

Elemental Area

x = acos0

where

Since

X:

Since value of 0 = 2'

:.

M2 is already a known value of a semicircle since m'a of full circle is (?r 2). A =2

Considering the first moment d elemntal areas about y axis, we have

C a t r e of Gravity rmd

Moment d Inertia

Example 4.6 Determine the centroid of the shaded area shown in Figure 4.16 (a). Solution The shaded area A can be considered as algebraic sum of three areas A, ,A2 and as shown in Figure 4.16 (b).

where A, = 200 x 200 = 40,000 mm2 with centroid at (100,100)

' 2'Oo2 = 15,700mm2 with centroid at (242.4.100). and A3 = [-' 602) = 5655 mm2 with centroid at (100,25.44)

A2 =

;

Sr. No.

&ea Ai

Ai xi (mm3)

Yi (mm)

(mm3)

1.

A, = 40,000

100

400 x lo4

100

400 x 104

2.

A2 = 15,700

242.4

380 x 104

3.

A, =

100

- 56 x 104

100 25.44

157 x 104 - 14 x lo4

- 5655

Xi

Ai

4

(mm)

(d)

4.3 CENTRE OF GRAVITY QE THIN WIRES A Ulin uniform wire bent in a given shape ABCDE can be divided into a suitable number of straight lengths for each of which is known to be at its mid-point. Referring Figure 4.17.

If wire of total length L is formed of L, = AB, 4 = BC and so on with their masscentres (or C.G.) located at their mid-points G, .G, e k , then L = L,

+ L,+ ., . =

x

L,.You can

I

apply the Varignon's theorem again as Ibllows Consider that plane XOY is horizontal s o that axes OX and OY are horizontal. Assuming the weight of wire as w newtons/metre length. We have, weight of portion

AB = wL, = W,

weight of portion BC = Total Weight,

w k = W2and

W = Wl + W, + ... =

C

W j=

WCL~

1

i

Considering first moment of all these weights about axis OY,if P, y are co-ordinates of mass-centre,

Similarly, considering first moment of weights about y axis

Note : Similar to C.G. of a thin uniform wire, the concept of centroid of a length can be defined. Thus, the above referred point ( Z, 9 can be said to be a centroid of length

ABCDE. Example 4.7

Determine the C.G. of the thin uniform wire bent into shape ABC as shown in Figure 4.18. Solution

Let G be the location of C.G. of the bar as shown in Figure 4.18,

Centre of Gravity d

Moment dlnertia

Taking moments about x axis,

L 7 = 5 x 2.5

t I

+5x

6.5 = 5 x (2.5+ 6.5)

Example 4.8

Determine the C.G. of a wire of uniform cross-section bent into shape of a semicircle as shown in Figure 4.19.

X

Solution Refer Figure 4.19 showing length of wire nu where a = radius of semicircle with centre at 0.

For element (GL)= a d 0

"i'

= a2

cos 0 dB =

2

[sin el

2a2 2 a :. Z = = - = 0.637~ Ica

~c

,-,,

(+ f l )

SAQ 1 Determine the centmid of a plate with uniform mass pet unit area having a shape given in Figure 4.20.

SAQ 2 Determine the centtoid of the area as shown in Figure 4.21 where D is the contact point of the circle to the edge GF.

SAQ 3 A thin uniform triangular plate OAB where L AOB = 90' is bent along M IM2 where M Iand M2are midpoints of AB and A 0 respectively till apex A is made to coincide on point 0 . Here OA = 12 cm and OB = 6 cm.

Determine the location of C.G. of such a plate with respect to axes OA (x axis) and OB ( y axis).

SAQ 4 Determine the position of centroid of a quadrant OAB of a circle, where arc AB subtends an angle of 90' at the centre 0 and radius OA = a.

F

SAQ 5 Determine the centroid of the area shown in Figure 4.22 where portion BCE is a quadrant of circle of radius = 10 c m

SAQ 6 Determine the C.G.of the plate area shown in Figure 4.22, if the area of a circle of radius 4 cm with its centre at ($5) is cut out from this plate.

SAQ 7 Determine the C.G.of the chair-frame made of rod bent along ABCDE as shown in Figure 4.17. Here AB = 40 cm, BC = 40 cm and CD = 30 cm (DE = 26 cm with its horizontal projection = 10 cm and vertical projection 24 cm).

SAQ 8 Determine the centroid of length in form of a channel ABCD as shown in Figure 4.23.

SAQ 9 Determine the centroid of the semicircular c w e of radius a = 10 cm as shown in Figure 4.24.

cmtn of ~ m v i t sod y Moment of Inertia

statics

4.4 THEOICEMS OF PAPPUS AND GULDINUS There are two theorems established by Pappus and Guldinus. First theorem is useful in computing the surface of revolution of a given curve rotating about a given axis which does not intersect with the curve. 'me second theorem relates with determination of the solid-volume of revolution of a given area rotating about a given non-intersecting axis. The details are given below.

4.4.1 Theorem I The area of surface generated by revolving a plane curve about a non-intersecting axis in the plane of the curve is equal to the product of (i) the length of the curve and (ii) the distance travelled by the centroid G of the curve during the revolution.

4.4.2 Theorem I1 The volume of the solid generated byrevolving a plane area APBA in Figure 4.26 about a non-intersecting axis in its plane is equal to the product of (i) the area APBA and (ii) the length of the path travelled by the centroid G of the area during the rotation about the axis. Proof (Theorem I) Let AB be a curve in a vertical plane ppsing through the vertical axis z as shown in Figure 4.25. The x and y axes are in the horizontal plane. Rotate the curve about z axis till it is in W X plane. Let L be the length of the curve AB. Consider a small elemental length PQ = (6L) on this curve as shown in the Figure 4.25. If the curve AB is rotated about z axis through an angle 8, the surface S generated by it is shown as AA'B'B. Element PQ will generate a surface PPQ'Q given by

where 8 is common rotation for any element on AB.

:.

~ o t asurface l

s = / ~ s =1 6 L X x x e

where X is the x coordiiate of the centroid G of the length AB .

= Product of (i) Land (ii) the distance (Z 8) travelled by the centroid G during the rotation through L8 about the axis of rotation. Proof (Theorem 11) Let A be the area of the plane curve APB in W X plane as shown in Figure 4.26. Consider a small elemental area dA at distance x from the z axis, which is considered as the axis of rotation for the area A. During the rotation 8 about z axis, elemental area dA moves through distance x 8. Hence, the solid volume 6V generated by this small area is given by

Total volume generated by the plane area A is therefore given by

Centre of Gravity aod Momont.pfInertia

where 0 is the common rotation for all such elernelltal area d4. :. V = 0 [ A i ]

where iis the x co-ordinate of the centroid of the are.a A

where Z 0 is the distance travelled by the centroid G of the areaA during the rotation L 0 about the given axis z . The sirnple solids considered in Engineering Mechanics can be cylinder, cone, sphere, hemisphere etc. The surface areas as well as the volunies of some of these solids are computed below.

4.4.3 Cylinder Let a be the radius of a cylinder and H be its height. Assume the axis of cylinder along z axis as shown in Figure 4.27.

ZA

In this case, we can consider generator as straight line AB of length H, which when rotates about z axis forms a cylinder as shown in the above figure. In this case, the centroid G of length AB is at mid-point of AB.

EatG=a From the First Theorem of Pappus and Guldinus, Surface of cylindcr = L x ( Z 0)

:.

S=Lxnt)

For full revolution t) = 2 .lr: Surface S = 2 .lr: a 1,

(11) To determine the volume of cylinder consider the plane area OBAC shown shaded where 0 and C are the centres of circles traced by points B and A. The volume of cylinder is generated when this area rotates about z axis (i.e. OC ).

Area OBAC = A = H x a

GI

x

The centroid of this area is GAwhere co-ordinate is .-

.Using theorem I1

Volume of Cylinder = A x ( Z 0 )

4.4.4 Cone ( I ) A cone is generated when an inclined length AB = L revolves about z axis which is taken along OA, the axis of the cone.

a The centroid GLof the Length L is at the mid height where ZL = - as observed in 2 Figure 4.28.

1

Cone surface generated by the lengthAB h u t = L x ( X L O ) the axis OA where 0 = 2z

;.. Cone surface = L x -a2 x 2% Solid volume 0f-e

is obtained when plane area AOB is revolved about axis

OA. The centroid GA of the plane area ABO is at

a

FA = 3'

Volume of cone = Area (AOB) x

8

4.4.5 Sphere A spherical surface is generated when a semicircular arc ACB of diameter AB = 2 a revolves about the axis AB.

'Ibe radius of semicircle is a as shown in Figure 4.29.

(I)

As determined in Example 4.8, the centroid of the semicircular arc is at GL

where 2 =

2u from central diameter AB. Length of semicircle ACB = nu. 7t

Figure 4.29

in one revolution 2a =nu x x 27t 7t

(11)

Volume of sphere :The area A of the plane figure ACBA =

Pt1.m \

/

distance of the centroid GA of this area is (z)fromAB, as computed in \ / Example 4.5. Volume of the solid Sphere = A x 2 x 2n

4.4.6 Hemisphere The curved surface of hemisphere =

The volume of Hemisphere =

-

1 (volume of 2

sphere)

Example 4.9 Determine the centre of gravity of a right circular cone of height H and the base radius a. Solution

Refer Figure 4.30, where radius of the base circle is shown as a and semivertical angle of the cone is a

:.

a = Htana

At any level z from apex 0, radius of thin slice of volume dV is given by x =

Since dz is very small thickness of the slice,

z tan a.

CPLI~ of~GIPV~Q C pnd

Moment of Inertia

about 0 Applying the theorem of moments about 0, if G is the centre of gravity of cone where OG = H

V x iT=j(dV)xz 0

- 3 z = - H from apex 0 4

:.

Location of G fiom base of Cone =

H 4 '

Example 4.10 '~eterminethe position of centre of gravity of a hemisphere of radius a.

Solution Figure 4.31 shows the cross-section of the hemisphere where DE = 242,the diameter of the hemisphere. Consider a thin slice AB of volume dV taken at a distance x fiom centre 0 of the hemisphere. Radius of slice = a sin 0 If dx is the small thickness of the slice,

d~ = x(asin0IL dx x = acos0

dx = - a sin0de

Considering equation of moments about 0,

Example 4.11 Locate the position of C.G. of the fiustnun of a cone witb height H = 12 cm and having diameter 8 cm and 12 6x11 at top and bottom of tbe frusbnun of the cone respectively as shown in Figure 4.32 (a). Solution Generators BITl and B2T2are extended to meet at A.

'Ihe volume V of tbe frustrum TI T2 B2 BLof a cone can be considered as (Vl - V2), where V,= volume of cone B1 B2 A and V2 = voiume of cone T , T A as shown in figure 4.32 @). Considering similar trianglesATT, and ABB2

:.

H, = Height of full cone = 2A + 12 = 36 cm

1 V2 = x 42 x 24 = 128 x cm3

Similarly,

3

:.

V = (432 - 128) x = 304 n cm3 Considering moment about axis perpendicular to B, I

304n x

F = 432x

(T)-

128n ( b + 12)

:. 5= 5.21 cm. Example 4.12 Determine the C.G. of a body formed of solid sphere over the solid cylinder of same radius a as shown in Figure 4.33, where a = 20 cm. Solution Consider y ,axis to be vertically along the axis of cylinder with centre O of its base as origin.

Vl = Volume of cylinder = x a2 x 242

V1= Volume of sphere

--

3

M' .

-

If G is the centre of gravity of the combined body where OG = y

SAQ 10 Determine the C.G. of a body formed of solid volumes as shown in the Figure 4.34.

Hemisphere I

SAQ 11 Determine the volume of glass of water whose inner dimensions of cross-section through central axis of revolution is shown in Figure 4.35.

Axis of t o v o l u t ~on

SAQ 12 Determine volume of water in a conical shaped water tank as shown in Figure 4.36 where central shaft of 4 metre diameter is provided for locating circular stairs to have access to the roof of the tank the water portion is shown shaded. axis

axis

4.5 MOMENT OF INERTIA OF AREA 4.5.1 Definition Consider an area of a surface consists of large number of small elements of area& each. ?he area integral of all such elements can be written mathematically as under :

Referring Figure 4.37,the Area Moment of Inertia of elemental area dA about x axis, in its plane is defined as

Since, the axis x lies in the plane of element, these are also called as axial moment of inertia of the element dA.

Figure 437

Polar moment of inertia of dA about z axis perpendicular to plane of A,

1

Product of Inertia of element dA with respect to axes x and y,

Moment of Inertia of Area A

Moment of Inertia (M.I.) of full areaA about centtoidal axis CX,where C is the centroid of the areaA is given by surface integral as follows:

Similarly, about axis CY

I,,,

=

Jd

~ 2. x

4.5.2 Perpendicular Axis Theorem Polar Moment of Ineda ofA about z axis passing through C.Refening Figure 4.37,where axis ZC is perpendicular to the plane of area A, we have

Cemtre of Gravity and Moment d h e r t i a

Moment of ~nerti'aof the area A about any axis XlXlshown in Figure 4.37 is given by Zqx,,

x (square of distance from XlX,axis)

=

where y, is the perpendicular distance between XlXIand CX.Thus for a given axis X,X,, y, is amstant. Similarly,

4.53 Parallel Axis Theorem With reference to Figure 4.37 it is noted that fust moment of elemental area dA about centroidal axis is given by (dA x y). By the defmition of centroid C of the area, it is further noted that

.

m

This means that horizontal plate of areaA gets balanced about a;xis CX.When Moment of Inertia of areas are computed about any random axis XIXI, then

Similarly,

Zxs2 =

L +A( ~ 3 ' .

'Ibis eqwtion is tenned as parallel axis theorem whereby it is observed that outaf all axes parallel to centroidal axis, CX, the Moment of Inertia about the centroidal iuis is minimum, for a given direction of the axis. Similarly. Referring Figure 4.37,

where x, is the perpendicular distance between the axes y, y, ahd CY.

where the perpendicular distance between axes z,z, passing through point K ( q , yl)and CZ

is

Example 4.13 Determiae the axial moment of inertia of a rectangular area of base b and height d about centroidd axis GX &.the tase Bl 4. Referring Figure 4.38, where centroidal axis GX divides the area at mid-depth i.e. d/2.

For a thin strip shown shaded, of width b and thickness (very small) dy, all points on it are at a constant distance y from axis GX,

Considering y as positively upward from centtoidal axis GX, for elements below GX, y will be treated as negative.

.

JMY = 0

Similarly, referring toy axis through centroid G,

Moment of inertia a h t base B1& can be computed either directly or by using parallel axis theorem. Direct approach is as follow.Referring Figure 4.39,

Alternatively, using theorem of parallel axis, we have

where yl = = perpendicular distance between GX and B,&. 2

Example 4.14 Determine the Moment of Inertia of a triangular area ABC having base b and height d about its base BC. Hence or otherwise determine the Moment of Inertia about centroidal axis parallel to the base. Solution

Figure 4.40 shows the

area ABC with its centroid G where perpendicular

distance from G to BC

Consider a thin strip LM of thickness dy and at distance y from the base BC. Considering similar triangles, ALM and ABC,

:.

Cmtre of Gravity Pnd

Elemental Area a!! = b 1-

dy

Moment of InertiP

Using theorem of parallel axes,

Example 4.15 Determine Moment of Inertia of an I shaped area about its centroidal axis as shown below in Figure 4.41 (a).

Solution

(A) The area of I section can be divided into three parts namely A,, A, and A3.

A = Al+A2+A3 = ( 4 0 0 ~ 1 5 0 ~ 2 + 3 0 0 ~ ~ ) = 1 8 ~ l d l 1 1 ~ 1 1 ~ The centroid G is at mid-depth as shown in Figure 4.41 (a).

where,

IA3(x) -

1 = -x 400 X 150' + 400 x 150 x (2251,

12

(B) Alternatively, the M.I. of I-section about x axis can be obtained by

subtracting the M.I. of area [2 x (A'2)] from the M.I. of area (A'1) as shown in Figure 4.41 (b). Here, A;= 400x600 mm2 and A ; = 100x300 mm2

:.

A=A;-~A;

=24~10~-(~~1~=18~10~mm~ Then

l ~ ( x )= Z(A;)- [QX

21

Example 4.16 Determine Moment of Inertia of circular area of radius a = 10 cm about its centroidal axis OX as shown in Figure 4.42.

Solution Consider a thin strip of width By at distance y from axis OX where, y = asin0 and B y = 2 ( a c o s 0 )

:.

dy = a cos 0 d0 dA.= Bydy=(2acos0)acos0d0

A

All elements have to be considered for values of 0 ranging from - 90' to

+ 90'.

cos 20 = cos20 - sin20 = 2 cos20 - 1

Note that

:.

2 cos20 = 1 + cos 20

.'.

Ix =

d 2

(7

( 2 s i n 0 c 0 ~ 0 )d0 ~

(-

M)

We know cos 40 = 1- 2 (sin 20)~

:.

2 (sin 2012 = 1- cos 40

- -lro4

-

64

where D = diameter of circle = 2u

It is to be noted that circular cross-section is axis-symmetrici.e. it is symmetric about both x aqd y axes or any other radial direction and also has same nature of shape about all centroidal axes.

where A = Area of the circle of radius a. Considering perpendicular axis theorem,

Example 4.17 Determine the M.I.of an annular area between two circles where difference of radius ' is quite small, about any centroidal axis. Solution The centroid-6 of such a thin circular strip of small thickness t is at its geometrical centre as shown in Figure 4.43.

We have, li4 = d r x t -where ds = arc length = a x d0 Considering plane of area in XOY plane,

I

statics

Figure 4.43

= 2natxa2 = (Area of annular strip) x a2

= A'a2

Since this annular strip of

area^' is also axis symmetric, then I,

= I,, ,

Hence, we have

Example 4.18 Determine the M.1 of a semicircleABM about its periphery AB and about centrojdal axes x and y as shown in Figure 4.44.

Figare 4.44

Solution The position of centroid C of the semicircle is already computed earlier. It is

,,

for semicircle -

I

+A

I

GY

1

-3

m4

[IABfor full circle ] = 8

-2 1 ( x ) = I(AB)for semicircle = 2I

for full circle

Ceatre of Gravity md

Moment of Inertia

SAQ 13 Determine the area mornent of inertia of a T-section about its centroidal axis as shown in Figure 4.45. Determine also the radius of gyration about its centroidal axis

ex.

SAQ 14 Determine the moment of inertia of the semicircle about axis AT4vhich is tangential to the circle as shown in Figure 4.46.

Figure 4.46

SAQ 15 Compute the M.I. of a hollow section shown in Figure 4.47 about its centroidal axis

ox.

4.6 PRINCIPAL AXES Moment of inerti of any area is related with the axis. Even if, we consider a case of a centroidal axis, d el axis can have any orientation L0 passing through centroid G of the area. Thus, value of moment of inertia of the area about the centroidal axis will vary and will be dependent on LO, the orientation of the axis. It is found that M.I. about axis passing through a given point has a maximum value for one value of LO = 61 while M.I. reaches a minimum value for LO = 62 = ( + 90' ). The maximum value of moment of inertia for axis passing through the given point on the area is called Major Principal Moment of Inertia and generally denoted as I, un~espondingto 0 = 8,. The minimum value generally denoted as I, of M.I. about axis through the same point at orientation 0 = 8, is referred to as Minor Principal Moment of Inertia. Values of moment of inertia for any other orientation 0 will be in between the values I, and I,. The axes along direction 0 = and 82 are referred to as principal axes of the area. In case area has two axes of symmetry, these k e also the principal axes. If area has only one axis of symmetry, this axis is one of the principal axes and the other one is at right angle to this.

4,6.1 Product of Inertia Product of Inertia of a typical channel section about centroidal axes CX and CY are the principal axes even though channel section is symmetrical about one axis CX only. B,ydefinition, product of inertia with respect to X and Y axes,

We frnd that for symmetrical section shown in Figure 4.48 for any element P of area dA and m d i n a t e s (x ,y), there is an element P' of saqoe area dA with co-ordinates (x , -y). I

Catre of Gravity and Moment d InerLia

Hence, products of inertia for these two elements are equal and opposite .

The product of inertia of areas which do not have any axis of symmetry are non-zero. Also product of inertia for the symmetrical channel section exists with respect to any other rotated axes X' Y' as shown in Figure 4.48. This aspect is explained in next section.

I !

I i

Extension of Parallel A x i s Theorem

Consider an u.s);mmetricalareaA shown in Figure 4.49 where CX and CY are the centroidal axes and KX,and KY,are axes parallel to centroidal axes through a point K with co-ordinate (x,, yl). It can be proved that product of inertia Zxl ylwithrespect to axes KX,and KY,is always higher than I,, with respect to axes CX and CY and is computed as given under :

t L

r

1

'Ihis equation is similar to parallel axes theorem for moment of inertia of an area and you can note that the product of inertia is therefore minimum about centroidal axes for given directions X and Y.

i

4.6.2 Rotated Set of Axes

L

Moment of inertia and product of inertia of any area shown in Figure 4.50 with respect to rotated set of axes x' and y 'through the centroid C may be expressed in terms of moment and product of inertia with respect to standard x and y-axes and are as follows.

4.6.3 Principal Moment of Inertia Moment of Inertia of an area depends on L O of the axes. It can be proved that the maximum and minimum moment of inertia of any area A at point C occur with respect to principal axes X and for which tan (2G) =

- 2zxy zx - zy

and values of principal moment of inertia are

The last result indicates that product of inertia of an area with respect to principal axes is zero. Example 4.19 Determine product of inertia of a rectangular area b x d with respect to its sides as shown in Figure 4.51.

Solution

Let sides OA and OB be 'the reference axes of area for product of inertia. Z,(,

= IdAXY

Cmtrc d Gravity pod Moment of I n 4

If G is the centroid of the area, then Z ,(G) = 0 by virtue of symmetry. Alternatively,

b

where obviously

xc =

2

However, product of inertia w.r.t. axes BC and Y is negative as shown below. '2.J

(B)

= Z*G) + A 1;; Y;;

where with respect to B,

4.7

MASS MOMENT OF INERTLA

Concept of mass moment of inertia is very useful in solving problems in kinetics of bodies, Inertia force which is indeed due to rotational acceleration of the body is proportional to "mass moment of inertia" of the body about the axis of rotation. The concept of mass moment of inertia about a given axis KK, for a thin plate of uniform mass m per unit area is defmed similar to area moment of inertia of similar area. Aasmall element of mass dm replaces the small elements area a!!. men, if M is the total mass of a plate area shown in Figure 4.52.

where r and r,, are defrned as the radii of gyration about x and y axes respectively.

If KK is an axis parallel to YY at a distance x,,

= M( iY)l+ M (x1)z

= M(,:

+x3

For a rectangular plate ( b x d ) and total mass M for area A = b x d

Replacing M by A ,we get, Mass moment of inertia

4.7.1 Radius of Gyration of an Area A or Mass M The radius of gyration of an area A with respect to a given axis KK is defmed as rm and is given by

Similarly, radius of gyration of a thin plate of area A and mass M is given by

zucn

= M( rm)L

For rectangular area b x d, if rGxis the radius of gyration about centroidal axis GX then

where

Considering a case of a. circular area of radius a, we have,

Centre of Gravity Pad

Moment d Ine&

Also

SAQ 16 Determine the product of inertia of the equal angle section 150 x 150 x 10 mm about centroidal axes parallel to the two sides.

SAQ 17 Detennine the principal moment of inertia of the angle section 150 x 150 x 10 mm and locate theii orientation.

SAQ 18 Detennine radius of gyration of a circular area of radius a about a tangent line at peripheral point of the circle in its own plane and also about axis passing through this point and perpendicular to its plane.

SAQ 19 Determine mass moment of inertia of the circular thin plate (Disc) of radius a and mass M about a centroidid axis. Mass is uniformly distributed all over the area .

Statics

SAQ 20 Determine the mass moment of inertia of ring of radius a about an axis

4.8

(a)

passing through center in the plane of the ring, and

(b)

passing through centre in a direction perpendicular to the plane of the ring.

SUMMARY Centroid C of an area A is a point in the plane of the area where entire area can be assumed to be concentrated. Centre of gravity G of a thin plate of same surface area A referred above is a point where the entire mass or weight of the plate isassumed to be concentrated. In fact, it is possible to support this plate in a horizontal plane by a tip of the vertically placed single pin at point G. The geometric centre of a rectangular area or a circular area also represents its centroid. Centroid lies on an axis of symmetry of the area. When there are two axes of symmetry, the centroid is the point of intersection of the two axes. For the triangular areaA (or triangular plate of same area A with mass M and of small uniform thickness), the centroid of A (or centre of gravity of the plate of mass M ) is at the point of intersection of the medians of the triangle. Generally, any given area can be divided into parts A, to An for each of area A, can,be computed along with the location of its centroid (xi , y,) then the position @, 3 of centroid can be computed as A i=

x

Ai xi and

For a uniform plate of mass M ,the entire mass can be divided into individual masses Mi for which centre of gravity is known as (xi, yi ), the values of n , 5 for centre of gravity is given by

x n

M Z=

Mi xi and

i=l

42

Distance of centroid of a semicircular area of radius a from the centre is -.

7n

Distance of centre of gravity of a uniform wire bent into a shape of semicircular arc 2a

of radius a from its centre is -.

R

Ceotre of Gravity md Moment of Inertia

meorem I of Pappus and Guldinus : The area of the surface generated by revolving a plane curve about a nonintersecting axis in the plane of the curve is equal to the product of (i) length of the curve and (ii) the distance travelled by the centroid G of the c w e during the revolution Theorem 11 of Pappus and Guldinus : The volume of solid generated by revolving a plane area A about a non-intersecting axis in its plane is equal to the product of (i) Area A ,and (ii) the length of the path travelled by the centroid G of the area during the revolution about the axis. Specific parameters of some solids :

I dA ( y ) = Area moment of inertia about y axis = I dA ( x ') =

Area moment of inertia a b u t x axis =

Im(A)

A

I,(,

A

Parallel Axis Theorem :Area moment of inertia about X,X,axis at distance y, from XX = I,+ A ( Y , ) ~ . Perpendicular Axis Theorem :If z axis is perpendicular to the plane of area A, then Iz,=Ixx+~yy

Mass moment of inertia of plate of uniform thjclcness abouCX axis

M

For Circular area of radius a,

For a circular plate of mass M and radius a, .

'

Product of Inertia of an area is zero if it has at least one axis of symmetry. Principal Moment of inertia with respect to rotated axes Z and

-

and

y at < 8

L-=O XY

4.9

KEY WORDS

Centre of Gravity

Centre of Mass

The Centre of Gravity is a point in the body where entire mass or weight is assumed to be concentrated and for convenience single resultant gravity load R can be used as a replacement for distributed gravity loads at various locations of its particles. The Centre of Mass, for a system of particles, is defined as the point where the entire mass of the system is assumed to be concentrated for consideration of its translational motion. In a uniform gravitational field, the centre of mass and the centre of gravity of a system are coincident.

Rigid Bodies

Bodies of a ftnite size can be regarded as ma& up of a collection of a large number of point masses, constituting a body, does not change under external forces, the body is said to be a rigid body. No object is a perfectly rigid body. Generally, solids can be regarded as rigid bodies.

Pappus Theorem I

The area of surface generated by revolving a plane curve about a non-intersecting axis in the plane of the curve is equal to the product of (i) the length of the curve and (ii) the distahce travelled by the centroid G of the curve during the revolution. The volume of the solid generated by revolving a plane area about a non-intersecting axis in its plane is equal to the product of (i) the area and (ii) the length ~f the path travelled by the centroid G of the area during the rotation about the axis. The moment of inertia of an element of the area with respect to any axis is defined as the product of the area of the element and the square of the distance from the axis to the element. It is also called second moment of area. The radius of gyration of an area with respect to an axis is defrned as the length whicb, when squared and multiplied by the area, will give the moment of inertia of the area with respect of the given axis.

Pappus Theorem 11

Moment d Inertia

Radius of Gyration

PrincipalAxes

Two perpendicular axes passing through the wntroid about which the moment of inertia is maximum and

minimum.

4.10 -

ANSWER3 TO SAQs /

SAQ 1

Total area is divided into three parts Al, A2 and Ag as shown in Figure below.

mare for Answer bSAQ 1

and SAQ 2

Referring to Figure for Answer to SAQ 2, total ma is divided into tbree parts Al, A2 and A,.

Fiye for Ansacr bSAQ 2

Ccobe dCm*

.ad

Molnetdloeliia

Area mark

Area cm

A, A2

4

20 20 50.26

Total

90.26

-

-

xi

Y; cm

cm

cm3

1 7

5 1 6

8

100 20 301.59

20 140 402.12

421.59

562.12

y = - 562.12 - 6.23 cm. 90.26

.

SAQ3

Refer Figure for Answer to SAQ 3. Since portion A M2 M1 takes up a new position A' M, M I , weight of plate A' M2 M1 add to weight of plate OM2 M 1 , then portion OM2 M I B is considered as areas A , and 2.4,

Figure for Answer to SAQ 3

I

I

Total

36

4

I

I

I

108

72

:. x= 3 and y = 2 SAQ 4

We have already determined position of centroid of semicircular area A in Example 4.5 which is as under,

Let TI be distance of centroid of upper quarter of circle of area A! shown shaded in Figure for Answer to SAQ 4 (b).

1

(a)

Figwe for Answer to SAQ 4

Area of lower quarter of the circle

By observation,

By considering moments about OB

.

Similarly, we can prove that

SAQ 5

Dividing the area into two parts (1) Square of area A, = 10 x 10 cm2 and (2) Quadrant of circle of A2 =

Area Mark

Area cm2

A, A2

100 78.54

Total

178.54

1Fa2

= 2% = 78.54 cm2

-

-

Xi

Yi

AiZi

Aiyi

5 4.24

500 1118.4

500 333.3

1618.4

833.3

5 14.24

.:

Z = 9.06 crn

and y=4.67cm SAQ 6

Adding negative area of circle to Table of SAQ 5 :

SAQ 7

Considering B as the origin, x3= 30 and y3 = 40

-y = - 3352 -

136

24.44 crn

SAQ 8

If C is the centroid of the channel-shaped length, it is noted that this channel-shaped length is symmetrical about axis CX which is parallel to flanges AB and CD. It is also to be noted that this length is not symmetrical about other centroidal axis CY which is parallel to the web BC as seen in the Figure for Answer to SAQ 8,

Figwe tor Anrwer to SAQ 8

Let n be the perpendicular distance.of C from BC. Total lenght L = L1 + &

+ L,

= 40 cm

Considering mement about BC

LX~=CZ~X~

'Ibe position of centriod G of a semicircleAMD with centre C is already determined in ~ x a r n ~4.8. l e Now referring the figure shown below. -

x, = distance C G = 0.637 a. x, = distance MG = a - 0.637 a

= 0.363 a

SAQ 10

The entire body of solid volume can be divided into three parts.

-

Volume V, = Volume of cone 1 = -xBase areax 3a = 3

-

Xi

vixi

3a

0.75 n d

4

xd

4a

V2 = Volume of cylinder = d x 2a = 2 x a3 4 V3 = Volume of hemisphere = - m3 6

a( 5 +

3 i) = 5.375~

Total V = ZVi = x a3 (3.67)

8 m4 3.58

d

12.33 x

d'

SAQ 11

Referring the Figure shown below, volume of the lower conical portion L'M'ML of cone can be computed as (V, - V2 ), where V, = (Volume of full cone ILM) - (shaded volume IL'M' ) Let z be the vertical axis of symmetry of the glass. Let H be the height of shaded portion of cone IL'M. From similar triangles IL'M' and ILM, we have

= (269 x ld - 98 x

ld)mm3

m f o r A n r r r r b SAQ 11

L

Volume V, = 7 ~ 3 5 ~ x 6=0231 x 10, mm3

Considering origin at I , 3 630 for volume Vl ; z1 = - (60 + 150) = -mm = 157.5 mm 4 4 3 (150) = 450 = 112.5 m m for volume V2 ; z2 = 4 4

Portion

Voluny (mm

zi

Vi Zi

vt

269 x lo3

157.5

42.36 x lo6

v2

- 98 x lo3

112.5

- 11.02 x lo6

v3

231 x lo3

240

55.41 x lo6

402 x lo3

-

86.75 x lo6

vi

SAQ 12

Refer to Figure for Answer to SAQ 12.

Figure for Answer to SAQ 12

V = Volume of shaded portion of water tank = Vl - V2 - V3

where Vl = Volume of cone OUV

V2 = Volume of cone OLM

,

V = Volume of cylinder above LM.

Height H of lower portion O W is given hy

;. Considering origid at 0

6+H Volume V, = -n x lo2 = 250 n m3 3

Centre of Gravity m d Moment of Inertia

:.

With respect to 0 as origin,

Note :The volume can also be computed using theorem of Pappus and Guldinus

V = (Area of triangle ULL' ) x 2m

SAQ 13

Let G be the centroid at distance y~ from top face as shown in figure.

--l200cmC Figure for Amwrr to SAQ 13

Zabout CX = 5o

203 12

+

1000 x (13.3)~+ 2o

;2403 + 800 x (6.71~

SAQ 14 Referring the Figure for Answer to SAQ 14

.

ZAT = Zox

+ A x a2... where

Figure for Answer to SAQ 14

SAQ 15 Referring the Figure for Answer to SAQ 15.

Figure for Answer to SAQ 15

xuL A =2

Ccatre of Gravity ;ad Moment of InertiP

SAQ 16

Centroid C and centroidal axes are shown in the Figure for Answer to SAQ 16. Co-ordinates of centre of areas A, andA2 with respect to C are given below to further compute I, .

Figure for Answer to SAQ 16

with respect to C Area

Ai xi Yi Xic

Yic

A,= 15cm2

7.5 - 4.12 = 3.38

- 3.62

- 183.53

A, = 14 cm2

- (4.12 - 0.5) = - 3.62

8 - 4.12 = 3.88

- 196.64

,

'

2 = - 380.17 SAQ 17

~ e f e m n gthe Section 4.5.4. the angle 8 of principal axis C%with respect to CX is given by tan (28) =

-21 1, - zy

As per the Answer to SAQ 16 1 , =- 380.17

and

1, = Zy = 637.24 tan (28)

+

rn

SAQ 18 Radius of gyration of a circular area of radius a about axis OX as shown in figure is given by

Considering tangent line OX',

nu4

2

2

A x (rofd)2 = lo, = --- + m ( a ) 4

Ffgwe for Answer to SAQ 18

d ( r o r d ) 2=

, 5

(4

r 0 ,= ( 4 1 . 2 5 ) ~ Considering O% as axis perpendicular to the plane of area,

SAQ 19

Mass moment of inertia of circular disc of radius a about centroidal axis is similar to corresponding area moment of inertia. If m is mass b r unit area, total mass M is given by

.:

Centre of Gravity and Moment of Inedia

Mass moment of inertia d disc is a2

whereby

2

IGx(Aw = (W 7 = M a ~ G X=

=~ ( r c x ) ~

5

SAQ 20

Considering a ring of radius a and mass M where dM is an elemental mass of its peripheral element of length a dB as shown in Figure for Answer to SAQ 20. Consider initialy the mass moment of inertia about Z axis which is perpendicular to the plane of the ring. If m is mass per unit length,

&we for Answer to SAQ 20

Then,

ZGz =

J(

=

d ) a2 ~

m(a dB) a2 [now

1

a dB = 2za]

= ma2 [ 2m]

=~

Since, by SynImeW, IGX = ICY,

( 2since )

2xaxm=M

F U ~ T H E RREADING 1. 2.

3. 4.

5.

Timoshenko, S. P. and Young, D. H., Engineering Mechanics, McGraw Hill, New York. Beer, F. P. and Johnston, E. P. ,Vector Mechanicsfor Engineers, McGraw Hill. New York. Shames, I., EngineeringMechanics, Prentice Hall, New Jersey. Thadani, B. N., EngineeringMechanics (Revised by J.P. Desai), Weinall Book Corporation, Bombay, India. Mclean and Nelson (Schaum's outline Series), Engineering Mechanics (3rd Edition).

1t