Strength of Materials by R.K.Bansal_text

m

186 . STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA 187

The C.G. of this triangular element is at G =!•"

where OG - —2 x OC = o xR

—

o3 (v dA^y.dx)

The distance of C.G. of area dA from y-axis is given by, = 2- ydx

x* = OG x cos 0 = — R x cos 0 ^y-.dx

U 2

Now using equation (5.2 A), AMoment of total area about x-axis is obtained by integrating the above equation.
A.•. Moment of total area about x-axis
ra/2 ( 9 \( r2 de
.. ..2
jx*dA 2 ^/ecose
13 . dx
1

Jo

jdA a/2 #2

f2 de
Jo

—r3 r'2 r - a/2 [R 2 dx (v x varies from - R to R)
cos 6 cZ0
O JQ sin 0 y

Zr r*o/2 2R 0 J-R

C?0 -ja/2 But equation of semi-circle is

2 Jo 9 Rx2 + y2 = 2 or y2 = R2 - x2
Jo
Substituting this value of y2 in the above equation, we get
a•
Moment of total area A about x-axis
Sln
22? l 2 = 4R s.m fcO Ans-
^r *
f-' -1-:j? 2
2

,2j 0-1 R

The area OA8 is symmetrical about the x-axis, hence —R02 . x - X

y = 0. Ans. J-fl

For a semi-circle, a = it = 160°, hence R R2 3

— —R- —4 R. #};R 2 (- R) - 1

x
= . fn'] 3
sin

3a {2J

R4 - . f 180 J = 4R R3 - Rc

sin Ans. .#3

3xjc 2 J 3it

^„. _ rohiem 510‘ Determine the centre of gravity of a semi-circle of radius R as shown in 2ft 3 ft‘ 2R3 2R 3 >

rig. 5.10 (a). 3 |
'
Sol. This problem can also be solved by the - - 3 +
method given in problem 5.9. The following other ft 3
J.
methods can also be used. Due to symmetry, x = 0. 3

The area AOB is symmetrical about the Y-axis, hence 2R3 2 ft 3 1 4ft 3 2R3
~2* 3
x = 0. The value of y is obtained by taking the mo- 3 -..(*)
ments of small areas and total area about x-axis. ...(«)
Let y = Distance of C.G. of the total area of semi-circle from x-axis.
1. Considering the strip parallel to Y-axis
—2
Area of strip* dA - y, dx
• reft
The distance of the C.G. of the area dA from
The total area of semi-circle is also equal to

jL

Moment of this total area about s-axis

—r-axis is equal toy
.. 2

= v_ x ni?

Moment of area dA about x-axis 2

= dA. I Equating the two values given by equations (i) and (ii), we get

2 —_ xR 2 2 R3

Jy x
3

STRENGTH OF MATERIALS

-2R3 2 4R

3 2 3ji

jzR

Hence the location of C.G. of semi-circle is ( 0,

2. Considering the strip parallel to x-axis
Area of strip, dA = 2x , dy

The distance of the C.G. of this area from x-axis is y
Moment of this area about x-axis

-y. dA

= y. 2xdy ...(j)
= 2xy dy

But, we know x2 +y2 = R2
x2 = R 2 -y 2

or x=^R 2 - 2 *

y ^F.
*g
Substituting the above value of x in equation (i), we get 5 10

“Moment of area dA about x-axis,=2 Jr 2- 2 .y .dy

y

from 0 A°f t0taI area ab°Ufc *'axis wil1 be obtained by integrating the above equation

.'. Moment of area A about x-axis

j\jR2 -y2 .ydy (v y varies from O to R)

=

— (R 2 2 3/2

- yR 2 ~ 2 (- 2y) dy = ~ - y)

y
Jo 3/2

|= - CO - =
3 3

Also the moment of total area A about x-axis = A x y

Ae — Total area of semi-circle —

2

y = Distance of C.G. of area A from x-axis

_r>2

A.•. Moment of total area about x-axis = xy

2

Equating the two values given by equations (ii) and (Hi),

2 * _ _ 2R3
ixR ~~ir
y
2

— — ——R 4Rr
- 23 2 .
y= x 25- =
. Ans.

3 jtR 3ji

, —

190 STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA ~A, nab
_ ba 2
rv= £[ - x 2 )™ .dbcT = -[ X ••• see equation (vi)
3A
[

a |_Jo J a[ 4

\;^••• The co-ordinates of the C.G. of given area are

ALet y = the distance of C.G. of the total area from x-axis. - 4a , _ _ 4b
Then moment of total area A about x-axis X= y
,
-Axy 3;t

ixab 5.5.4. Centroid of Volume. Centroid of z dv
volume is the point at which the total volume Iv
of a body is assumed to be concentrated. The
volume is having three dimensions i.e., length, Sj 1
width and thickness. Hence volume is meas-
ured in [length]3 . The centroid li.e., or centre of (j c.oj
gravity] of a volume is obtained by dividing the
The equations ( iii ) and (vii) give the moment of total area about x-axis. Hence equating given volume into a large number of small vol- z* Z
these equations, we get umes as shown in Fig. 5.10 id). Similar method
was used for finding the centroid of an area in n /
itab _ ab 2 which case the given area was divided into large /
number of small areas. The centroid of the vol- y*
ume is hence obtained by replacing dA by dv in / X* / Y
equations (5.2A) and (5.2B).
—_ ab 2

v
= 4 4b A.ns.

. =.

3 nab 3n

To find x take the moment of small area dA about y-axis.

The C.G. of area dA is at a distance of x from y-axis.

Moment of area dA about y-axis = x.dA Fig. 5.10 (d)

= x.y.dx ('. dA = ydx)

Moment of total area A about y-axis is obtained by integration Then these equations becomes as
Now moment of total area A about y-axis

x.y.dx O( v x varies from to a) ...(5.3 A)

J"

: fr.V-r2 )1 y = —(a2 - x2 y2 from equation (v) I
JO a
)

= -b frx.(ra„ 2 -x„8!\)i/2 rbra (-2) . x(a 2 -x 2 2 .dx and ...(5.3 B)

a Jo a Jo (-2) \dv

V-* T_ & As volume is having three dimensions, hence third equation is written as

“ - 2a |

[
2 3/2 ^6 “] 6a 2 ...(.viii) Jz* dv ...(5.3 C)
3 z = —(
) C°
3a
3/2 Jo

Also the moment of total area A about y-axis dv
\
=A x x where x* - Distance of C.G. of small volume dv from y-z plane (i.e., from axis oy)
Awhere x - Distance of C.G. of total area from y-axis.
y* = Distance of C.G. of small volume dv from x-z plane (i.e., from axis ox)
Equating the two values given by equations (viii) and (jx),
z* = Distance of C.G. of small volume dv from x-y plane

Ax, _ ba 2 and x y, z - Location of centroid of total volume.
Xm ,

3 Note. If a body has a plane of symmetry, the centre of gravity lies in that plane. If it has two
planes of symmetry, the line of intersection of the two planes gives the position of centre of gravity. If it
*Please refer some standard TextBook of Mathematics. has three planes of symmetry, the point of intersection of the three planes gives the position of centre of

2 - X2 dx~ ^ xja2 - x2
-Ja
L2
~a — -f

Jo
+ 2 sin' 1 - 0 + a 2 sin"1 (1) gravity.

2 a J0 L 2

2 1 2j 4

: CENTRE OF GRAVITY AND MOMENT OF INERTIA

STRENGTH OF MATERIALS 2 Ch x*
rRit x
Problem 5.12. A right circular cone of radius R at the base and of height h is placed as
h2 Jo
shown in Fig. 5.10 (e). Find the location of the centroid of the volume of the cone.
Sol. Given [y Limits of integration are w.r.t. x. And x varies from 0 to h]

Radius or cone = R

Height of cone = h

AProblem 5.13. hemisphere of radius R is placed as shown in Fig. 5.10 (f). The axis of

symmetry is along z-axis. Find the centroid of the hemisphere.

Sol. The hemisphere is placed as shown in -

Fig. 5.10 (f). The axis of symmetry is taken as Z-axis.
The centroid will be at the Z-axis. Hence x = 0 and

Fig. 5.10 (e) y =0.
Radius of hemisphere = R.

In the Fig. 5.10 (e), the axis of the cone is along x-axis. The centroid will be at the x-axis. To find z , consider a small volume dv of the I
hemisphere. For this, take a thin circular plate at a
Hence, y = 0 and z - 0. height z and thickness dz. Let is the radius of this

To find x, consider a small volume dv. For this, take a thin circular plate at a distance plate.
x from O. Let the thickness of the plate is dx as shown in figure and radius of the plate is r. The ''
Then dv = Area of section x thickness
Vcentroid of the plate is at a distance from O. Hence x* = x.

Now volume of the thin plate, = xy2 x dz ...(i )

do = ar2 x dx •••© ( v Area of any section for sphere *X

or hemisphere = nr2 Here r= y) _. _ , „ ,
, Fig. 5.10 if)

Let us find the value of r in terms of x. The centre of gravity of the small volume is at

From similar triangles, we get a distance z from O.

Rh Let us now, find the value of y in terms of z.

From Fig. 6.10 (/>, we have —Hi)

R2 ~z2 + y2
Ror y2 = 2 - z2

Substituting the value of y2 in equation (i), we get
dv = ir[R 2 - z2 ] x dz

As in this case, the axis of symmetry is Z-axis. Hence x and y are zero. The distance of
the centroid from x-y plane is given by equation (5.3C) as

I z* dv

where z* = Distance of centroid of the small volume dv from x-y plane.
= z [In the present ease]

—

194 STRENGTH OF MATERIALS

CENTRE OF GRAVITY AND MOMENT OF INERTIA 195

Z dv The product of the area (or mass) and the square of the distance of the centre of gravity
of the area 'or mass) from an axis perpendicular to the plane of the area is known as polar
^Z =
moment of inertia and is represented by J.
Jdo

z x k(R 2 - z 2 )dz Consider a plane area which is split up into small areas av o2, a3, ... etc. Let the C.G. of
j the small areas from a given axis be at a distance of r v r2 , r3> ... etc. as shown in Fig. 5.12.
= p [v F'rom equation (£i), dv = n(R z - z 2 x dz]
Then the moment of inertia of the plane area about the given axis is given by
)

jn(R 2 -z2 )dz

f k(R 2 z - z3 ) dz 5.7. a^I = 2 + a r 2 + a r 2 + ... -(5.4)
_ Jo or 23 -(5.5)

I = Zar2.

-z2 2 RADIUS OF GYRATION
)cte
f jt(JJ

Jo

[The limits of integration are according to dz. Here z varies from 0 to R\ Radius of gyration of a body (or a given lamina) about an axis
is a distance such that its square multiplied by the area gives mo-
ment of inertia of the area about the given axis.

5.6. For the Fig. 5.12, the moment of inertia about the given axis is

given by equation (5.4) as

I= ar 2 + ar2 + ar 2 + ... -(i)
L 22
3

Let the whole mass (or area) of the body is concentrated at a

AREA MOMENT OF INERTIA distance k from the axis of reference, then the moment of inertia of

Consider a think lamina of area A as shown in Fig. 5.11. fY Lamina of the whole area about the given axis will be equal to Ah2 .
ALet * = Distance of the C.G. of area from the axis OY.
aA 5.8. If Ak2 = I, then h is known as radius of gyration about the
y*

y = Distance of the C.G. of area A from the axis OX. given axis. Fig. 5.12
Then moment of area about the axis OY
f ^)
= Area x perpendicular distance of C.G. of area from
—' CA
. ..

axis OY \
[

I"'- ' y

= Ax ...(5.3D) *x THEOREM OF THE PERPENDICULAR AXIS

l

Equation (5.3D) is known as first moment of area about o X Theorem of the perpendicular axis states that if I and lyy be the moment of inertia of

}:x
the axis OY. This first moment of area is used to determine the X Xa plane section about two mutually perpendicular axis
Pi 5 11 and Y-Y in the plane of the section,

centre of gravity of the area. XXthen the moment of inertia of the section Izz about the axis Z-Z, perpendicular to the plane

If the moment of area given by equation (5. 3D) is again multiplied by the perpendicular and passing through the intersection of and Y-Y is given by

OYdistance between the C.G. of the area and axis {i.e. , distance*), then the quantity (Ax), x -Ax 1 ^zz = Ixx + Arr

is known as moment of the moment of area or second moment ofarea or area moment of inertia The moment of inertia Izz is also known as polar moment of inertia.
about the axis OY. This second moment of area is used in the study of mechanics of fluids and Proof. A plane section of area A and lying in plane x-y is
shown in Fig. 5.13. Let OX and OY be the two mutually per-
mechanics of solids.
pendicular axes, and OZ be the perpendicular axis. Consider a
OXSimilarly, the moment of area (or first moment of area) about the axis = Ay.

OXAnd second moment of area (or area moment of inertia) about the axis = (Ay) . y = Ay2 . small area dA.

If, instead of area, the mass (m) of the body is taken into consideration then the second Let * = Distance of dA from the axis OY
y = Distance of dA from axis OX
moment is known as second moment of mass. This second moment of mass is also known as r = Distance of dA from axis OZ

mass moment of inertia.

OYHence moment of inertia when mass is taken into consideration about the axis = mx 2 Then r2 =x2 4 y2
and about the axis OX = my2 . .

Hence the product of the area (or mass) and the square of the distance of the centre of - Now moment of inertia of dA about x-axis

gravity of the area (or mass) from an axis is known as moment of inertia of the area (or mass) = dA x (Distance of dA from x-axis) 2

about that axis. Moment of inertia is represented by I. Hence moment of inertia about the axis = dA x 2

OX OYis represented by I whereas about the axis y.
**
. by /„

yy

:

STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA

Moment of inertia of total area A about x-axis, , = YdAy2 . But dA . y represents the moment of area of strip about X-X axis. And YdAy represents
the moments of the total area aboutX-X axis. But the moments of the total area aboutX-X axis
ASimilarly, moment of inertia of total area about y-axis, IyY = YdAx2 is equal to the product of total area (A) and the distance of the C.G. of the total area from X-X

Aand moment of inertia of total area about 2 -axis, Izz - YdAr2 axis. As the distance of the C.G. of the total area from X-X axis is zero, hence YdAy will be

= ZdA[x 2 + y2] r2 = x2 + y2) equal to zero.

= YxlA x2 + YdA y2 Substituting this value in equation (ii), we get

= Iyy + IXX 7ab =F.A + /g + 0

Or ^1ZZ == IIjXqXt + Ilyy.- ...(5.7) or 7as = /g + AA 2 -(5.8)

The above equation shows that the moment of inertia of an area about an axis at origin Thus if the moment of inertia of an area with respect to an axis in the plane of area
(and passing through the C.G. of the area) is known, the moment of inertia with respect to any
normal to x, y plane is the sum of moments of inertia about the corresponding x and y-axis.

In equation (5.7), I77 is known as Polar Moment of Inertia. parallel axis in the plane may be determined by using the above equation.

5.9. THEOREM OF PARALLEL AXIS 5.10. DETERMINATION OF AREA MOMENT OF INERTIA

It states that if the moment of inertia of a plane area about an ^Plane -v The area moment of inertia of the following sections will be determined by the method
\
axis in the plane of area through the C.G. of the plane area be repre- area /tK '/ of integration

sented by /0, then the moment of the inertia of the given plane area j G j* 1. Moment of inertia of a rectangular section,

about a parallel axis AS in the plane of area at a distanced from the x x. ~/ f x 2. Moment of inertia of a circular section,

C.G. of the area is given by 3. Moment of inertia of a triangular section,

I^Ia + Ah2. f 4. Moment of inertia of a uniform thin rod.
g
where *AB = Moment of inertia of the given area about AS ^ 5.10.1. Moment of Inertia of a Rectangular Section
1st Case. Moment of inertia of the rectangular section about the X-X axis pass-
Ic = Moment of inertia of the given area about C.G.
Fig. 5.14, ing through the C.G. of the section.
A = Area of the section
Fig. 5.15 shows a rectangular section ABCD having width = b and depth = d. Let X-X is
h = Distance between the C.G. of the section and the axis AB. the horizontal axis passing through the C.G. of the rectangular section. We want to deter-

Proof. A lamina of plane area A is shown in Fig. 5.14. mined the moment of inertia of the rectangular section about X-X axis. The moment of inertia

Let X-X = The axis in the plane of area A and passing through the C.G. of the area.

AAB - The axis in the plane of area and parallel to axis X-X. of the given section about X-X axis is represent by I

h = Distance between AB and X-X. Consider a rectangular elementary strip of thickness dy at H b*

Consider a strip parallel to X-X axis at a distance y from the X-X axis. a distance y from the X-X axis as shown in Fig. 5.15. ^ ® "y
Let the area of the strip = dA
Area of the strip =b . dy. mdy A nnnninrnTr -A
?
Moment of inertia of the area of the strip about X-X axis = “j

Moment of inertia of area dA about X-X axis = dAy2. Area of strip x 2 -± -¥-

Moment of inertia of the total area about X-X axis, y x xT

= (5 . dy) x y2 = by2dy.

lyj. or Iq = YdAy 2 ...(i) Moment of inertia of the whole section will be obtained by -jT

Moment of inertia of the area dA about AB • ., - —d d J sI~

= dA(h 2 integrating the above equation between tihe limits , jfc. L

+ y) to ^ .

= dA[h2 +y2 + 2hy]. d> Fig. 5.15

(
^=L Lfd/2
Moment of inertia of the total area A about AB, by„dy=b

Iju = YdA[h2 + y2 + 2hy\ 2
= YcLAh2 + YdAy2 + YdA 2hy.
b is constant and can be taken outside the integral sign)

As h or h2 is constant and hence they can be taken outside the summation sign. Hence

the above equation becomes

IAB = h2YdA + YdAy2 + 2hYdAy.

But YdA = A. Also from equation (0, YdAy2 = IQ . Substituting these values in the above
equation, we get

1A# = h2 A + IG + 2h YdAy. ...Hi)
.

— H|

198 STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA 199

b 2d 3 bd 3 The moment of inertia of the main section ABCD aboutX-X axis is given by equation (5.7),

3' 8 ' ...(5.9) _~ bd
12
12
where b = Width of main section
Similarly, the moment of inertia of the rectangular section about Y-Y axis passing through
d = Depth.
the C.G. of the section is given by
The moment of inertia of the cut-out section EFGH about
db 3 ...(5.10)
12 ' X-X axis

_ Mi3

12

where b - Width of the cut-out section, and Fig. 5.17
x

d = Depth of the cut-out section.
1

Then moment of inertia of hollow rectangular section about X-X axis,

lyj, Moment of inertia of rectangle ABCD aboutX-X axis —moment of inertia of rectangle
EFGH about X-X axis
1

bd3 Mt3

12 ~ 12

Fig. 5.15 (a) 5.10.2. Moment of Inertia of a Circular Section. Fig. 5.18

shows a circular section of radius R with O as centre. Consider an

Velementary circular ring of radius and thickness ‘dr’. Area of cir-

2nd Case. Moment of inertia of the rectangular section about a line passing cular ring

through the base. = 2:rr. dr.

Fig. 5.16 shows a rectangular section ABCD having width = b —N b - In this case first find the moment of inertia of the circular
and depth -d. We want to find the moment of inertia of the rectangu-
TA 1 —" B Osection about an axis passing through and perpendicular to the
lar section about the line CD, which is the base of the rectangular |
plane of the paper. This moment of inertia is also known as polar Fig. 5.18
section.
moment of inertia. Let this axis be Z-Z. (Axis Z-Z is not shown in

Consider a rectangular elementary strip of thickness dy at a Fig. 5.18). Then from the theorem of perpendicular axis, the moment of inertia about X-X axis

distance y from the line CD as shown in Fig. 5.16. or Y-Y axis is obtained.

Area of strip = b . dy. WWWUUh ^ Moment of inertia of the circular ring about an axis passing through O and perpendicu-
_ dy
Moment of inertia of the area of strip about the line CD lar to the plane of the paper

= Area of strip . y2 y = (Area of ring) x (radius of ring from O)2
= b . dy . y2 = by2 dy.
DC = (2itr . dr) . r2
Moment of inertia of the whole section about the line CD is ob-
Fig. 5.16 = 2xrsdr ...(()
tained by integrating the above equation between the limits o to d.
Moment of inertia of the whole circular section is obtained by integrating equation (i)
Moment of inertia of the whole section about the line CD. between the limit 0 to R.

= by2dy = b j"y 2dy Moment of inertia of the whole section about an axis passing through O and perpen-

dicular to the plane of paper is given as

...(5.11) ^I =J 2 k3 dr - 2it r a dr
/-
J

3rd Case. Moment of inertia of a hollow rectangular section.

EpB^GrHr Fig 5 ' 17 Sh°WS a hoIlow rectangular section in which ABCD is the main section and

', cut-out section.

is the

D

CENTRE OF GRAVITY AND MOMENT OF INERTIA 201

But b-y (v BC = b)

^A ft

Dwhere - Diameter of the circular section DESubstituting this value of in equation (t), we get

, it ( = by ,

2(2i2772 — X — Area of strip 4<iy '

~h

Polar moment of inertia = Distance of the strip from the base = (ft - y)

32 ... Moment of inertia of the strip about the base
= Area of strip x (Distance of strip from base)2
But from the theorem of perpendicular axis given by equation (5.7), we havelzz =1^ + lyy.
^ ^= 2
But due to symmetry, = lyy
y)
. dy . (ft -y)2 = (ft - . dy.

IlXX~-T1 YY~- Izz ft n

The moment of inertia of the whole triangular section about the base (IBC) is obtained by
integrating the above equation between the limits O to ft.
_ JiP 4 1 4
= jtD
32 *2 ...(5.13)
64
'Wo T“-
Moment of Inertia of a hollow circular section

Fig. 5.19 shows a hollow circular section. =~ h

DLet = Diameter of outer circle, and y(h~y)2 dy
h Jfo
d = Diameter of cut-out circle.
( / b and h are constants and can be taken outside the integral sign)

Then from equation (5.13), the moment of inertia of the outer

— Dcircle about X-X axis = 4 |= y (h? + y2 - 2hy ) dy = £ (yhz + y3 - 2fty2) dy

. J*

64

And moment of inertia of the cut-out circle about X-X axis b f y2h2 4

ft 2 y
4

Moment of inertia of the hollow circular section, about X-X axis, Fig. 5.19 ]Vl_3
]xx - Moment of inertia of outer circle— moment of inertia of cut-out circle
2ft.ft
22 4 4 4
b <i ft
+6 [ft . ft ft
+

ft 2 4 3 ft 2 4

—= JL £>4_ JL d4 = [D4 -d4] = b A4[6±3-8l bh* 1
64 64 64
12
— ID4 ft L 12 J

Similarly, - d4
).

_ bjf_ ...(5.14)

5.10.3. Moment of Inertia of a Triangular Section 12

1st Case. Moment of inertia of a triangular section about its base. 2nd Case. Moment of inertia of the triangular section about an axis passing

Fig. 5.20 shows a triangular section ABC of base A_„ through the C.G. and parallel to the base. t a
Consider a triangular section of base = b and height = ft
width = b and height = h. Consider a small strip of thickness 7\"f” ~Ty ] /\
as shown in Fig. 5.21. Let X-X is the axis passing through the
dy at a distance y from the vertex A. Vf//////, /x
h C.G. of the triangular section and parallel to the base.
DEArea of the strip = . dy ...(j)
,

DEThe distance in terms ofy, b and it is obtained from / The distance between the C.G. of the triangular section h / c.G

two similar triangles ADE and ABC as _±_ [_ and base AB_ - —n

BC h u Now from the theorem of parallel axis, given by equa-
DE = RC y
Fig. 5.20 tion (5.8), we have

Fig. 5.21

:

CENTRE OF GRAVITY AND MOMENT OF INERTIA 203

STRENGTH OF MATERIALS

Moment of inertia about

BC = Moment of inertia about C.G. + Area x (Distance between X-X and BC)2

aJ^ibBCc =~ ig+ 2

I]

hc~ “(!/

bh 3 “he = and Area =

12 " (¥)•(!)'

33
3-2)r-O
6ft 6ft (
'
_

12 18 36

bh 3 ...(5.15)
36

Problem 5.13 (A). Determine the moment of iner-

tia of the section about an axis passing through the base

BC of a triangular section shown in Fig. 5.21 (a).

(U.P. Tech. University, 2002-2003)

Sol. Given

mmBase, b = 100 ; height, ft = 90 mm.

Moment of inertia of a triangular section about an — 100 mm

axis passing through the base is given by equation (5. 14) as Fig. 5.21 (a)

1bc ~_ bh 3
~12

— mm= 100 x903 = 6.075 x 10® 4 Ans.
.

5.10.4. Moment of Inertia of a Uniform Thin Rod. Y
Consider a uniform thin rod AB of length L as shown in
t
Fig. 5.22. A—

mLet = Mass per unit length of rod, and «-

M = Total mass of the rod „.(i)

= mxL

Suppose it is required to find the moment of inertia of Fig. 5.22

the rod about the axis Y-Y. Consider a strip of length dx at a
distance * from the axis Y-Y.

Mass of the strip — Length of strip x Mass per uni t length

m m= dx . or . dx.

Moment of inertia of the strip about Y-Y axis

= Mass of strip x x2

= (m . dx) . x2

= mx 2dx.

' 2^

STRENGTH OF MATERIALS -CENTRE OF GRAVITY AND MOMENT OF INERTIA

^' oment °f inertia of the total area about y -axis is obtained by integrating the above a 1 yl + a2 y2 20 x 9 + 16 x 4 _ 180 + 64
‘ + U2 36
equation between the limits O to a. Of v * varies from to a) — _ 20+16 = 6.777 cm.
"36Uusing tohne relation, jy

Hence the C.G. of the given section iies at a distance of 6.777 cm from GF. Now find the

moment of inertia of the T-section.

Now, Let Iq - Moment of inertia of rectangle (1) about the horizontal axis and passing

2 =-baa„7/2 2 2 through its C.G.
='
.
Ja 7/2 J0 7
IG = Moment of inertia of rectangle (2) about the horizontal axis and passing
J mdeitl-ige.m5ec.n2t2r?a(/bao)uctaxtn-habfxeimcsooinmsseeinqdteuraeoldftitonotebhreetiama oromefcettnhaetngoglfeivioenfnetrthaiiracekaonfeatsbhsoeudrtx.ecxTt-haaxneigslm,eotmahbeeonuettloeiftmseibnneatrstesi.ha oowfnthiisn
through the C.G. of the rectangle (2)
Moment of inertia of the element about x-axis
k = The distance between the C.G. of the given section and the C.G. of the
x
rectangle (1)

—_ dx.y3 rv ... bd 3 . . = . and d = 1 = y1 - y = 9.0 - 6.777 = 2.223 cm
2 it is
— where b dx y h = The distance between the C.G. of the given section and the C.G. of the rec-
* 2

tangle. (2)

The moment of inertia of the = y-y2 = 6-777 - 4 -° = 2.777 cm-

equation between the limits O given area about x-axis is obtained by integrating the —10 x 2 3
to a.
above Now Ig, - • = 66'.66f6>7 cm4

rr _ dx.y3 3

y,

IG = = 85.333 cm4.

A .4 From the theorem of parallel axes, the moment of inertia of the rectangle (1) about the

rLv« j horizontal axis passing through the C.G. of the given section

’ *> —j= Jx from equation (its) = I + aji 3 = 6.667 + 20 x (2.223)2

Vet Gl

= 6.667 + 98.834 - 105.501 cm4.

Similarly, the moment of inertia of the rectangle (2) about the horizontal axis passing

b3 through the C.G. of the given section
3 . a 3/2
_2 s - a = = IGi + aji* = 85.333 + 16 x (2.777)2

"Is* = 85.333 + 123.387 = 208.72 cm4 .
.*. The moment of inertia of the given section about the horizontal axis passing through
Problem 5.14. Fig. 5.23 shows a T-section of dimensions * 10 cm H
20 x 20 x 2 cm. Determine the moment of inertia of the section
about the horizontal and vertical axes, passing through the cen- HsT" rT the C.G. of the given section is,

tre ofgravity of the section. Also find the polar moment of inertia I Ti I = 105.601 + 208.72 = 314.221 cm4. Ans.

of the given T-section. 0H i The moment of inertia of the given section about the vertical axis passing through the

c

Sol. First of all, find the location of centre of gravity of the C.G. of the given section is, k-10cm+l

given T-section. The given section is symmetrical about the axis + ~ 2 x 103, + 8 x 23, A Y, B .
Y-Y and hence the C.G. of the section will lie on Y-Y axis. The
© *yy 12 12 L® 1 2-Em
given section is split up into two rectangles ABCD and EFGH for i
^ DH e c j
calculating the C.G. of the section. -W2 cmw- = 166.67 + 5.33 = 172 cm4. Ans.
; I
Let y - Distance of the C.G. of the section from the bot- Fig. 5.23
Now the *polar moment of inertia (/44 ) is obtained from ! tn ptri
tom line GF
equation (5.7) as *

A

hz ^ Ixx* I >4 2om

o-i ~ Area of rectangle ABCD - 10 x 2 = 20 cm2 yy : K^
= 314.221 + 172 = 486.221 cm4 . Ans.
oj ; |p I 2-Sm
y2 = Distance of C.G. of the area o from the bottom line GF = 8 + 1 = 9 cm
1 © HI
Problem 5.15. Find the moment of inertia of the section i
a = Area of rectangle EFGH = 8 x 2 = 16 cm2 j
shown in Fig. 5.24 about the centroidal axis X-X perpendicular ;
2 M
20 c^,
(AMIE, Summer 1977)
y2 - Distance of C.G. of rectangle EFGH from the bottom line GF = ~ = 4 cm to the web.
Fig. g5.224
Sol. First of all find the location of centre of gravity of the

given figure. The given section is symmetrical about the axis

206 STRENGTH OF MATERIALS .iCENTRE OF GRAVITY AND MOMENT OF INERTIA 207

Y-Y and hence the C.G. of the section will lie on Y- Y axis. The given section is split up into Now WlG, = ~ = 6 ' 667 Cm4
three rectangles ABCD, EFGH and JKLM. The centre of gravity of the section is obtained by yf

using

- _ <*i >i + + ^3^3 ^ = 166.667 cm4

Cl2 + Q>2 + G3

where MLy = Distance of the C.G. of the section from the bottom line Iai = = 13 333 cm4-

a, = Area of rectangle ABCD - 10 x 2 = 20 cm2 Prom the theorem of parallel axes, the moment of inertia of the rectangle (1) about the

y x = Distance of the C.G. of the rectangl? ABCD from the bottom line ML horizontal axis passing through the C.G. of the given section

„„ '2 + . = 13 cm = 7Gi + aji-f = 6.667 + 20 x (7.51 2

= 2 + 10 + 1
12

a = Area of rectangle EFGH = 10 x 2 = 20 cm2 = 6.667 + 1125 = 1131.667 cm4 .

2

MLy2 - Distance of the C.G. of rectangle EFGH from the bottom line Similarly, the moment of inertia of the rectangle (2) about the horizontal axis passing

—10 through the C.G. of the given section

= 2 + = 2 + 5 = 7 cm 1.5 2

At = Ig + a h 2 = 166.667 + 20 x
2
a = Area of rectangle JKLM = 20 x = 40 cm2 2 2
a
2 = 166.667 + 45 = 211.667 cm4 .

y3 = Distance of the C.G. of rectangle JKLM from the bottom line ML And moment of inertia of the rectangle (3) about the horizontal axis, passing through

= -2 = 1.0 cm. \ the C.G. of the given section

I = IGs + a3h32 = 13.333 + 40 x 4.5 2

Substituting the above values in equation (i), we get

_ _ 20 x 13 + 20 x 7 + 40 x 1 • = 13.333 + 810 = 823.333 cm4
7 ~ 20 + 20 + 40
Now moment of inertia of the given section about the horizontal axis, passing through

260 + 140 + 40 440 the C.G. of the given section
80
^= = = 5.50 cm. = Sum of the moment of inertia of the rectangles (1), (2) and (3) about

The C.G. of the given section lies at a distance of 5.50 cm from the bottom line ML. We the horizontal axis, passing through the C.G. of the given section

want to find the moment of inertia of the given section about a horizontal axis passing through = 1131.667 + 211.667 + 823.333 = 2166.667 cm4. Ans.

the C.G. of the given section. Problem 5.15(A). Determine the polar moment of inertia of 1-section shown in

Let Iqj = Moment of inertia of rectangle (1) about the horizontal axis passing through Fig. 5.24 (a). (All dimensions are in mm). (U.P. Tech. University, 2001-2Q02)

its C.G. Sol. Let us first find the location of C.G. of the given

Ig = Moment of inertia of rectangle (2) about the horizontal axis passing through section. It is symmetrical about the vertical axis, hence C.G.

2 lies on this section.

the C.G. of rectangle (2) Now, A = Area of first rectangle
t
Ig = Moment of inertia of rectangle (3) about the horizontal axis passing through mm= 80 x 12 = 960 2

3

the C.G. of rectangle (3) A2 = Area of second rectangle

h = The distance between the C.G. of the rectangle (1) and the C.G. of the given [(150 - 12 - 10) x 12]
x
mm= 128 x 12 = 1536 2
section

= y x - y = 13.0 - 5.50 = 7.50 cm A = Area of third rectangle
a
h = The distance between the C.G. of rectangle (2) and the C.G. of the given mm= 120 x 10 = 1200 2
2

section yl = Distance of C.G. of area A.
from bottom line
Fig. 5.24 (a)

=- y = 7.0 - 5.50 = 1.50 cm
y-i

h - The distance between the C.G. of the rectangle (3) and the C.G. of the given —= 150 - = 144 mm

3

section

= y -LV3 = 5.50- 1.0 = 4.5 cm Ay2 = Distance of C.G. of area from bottom line

2

= 10 + —12r8— = 74 mm

208 STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA 209

Ay3 = Distance of C.G. of area from Polar moment of inertia (/zz ) is given by,

3

bottom line = 10 = 5„ cm. ^zz ~ Ixx + At
mm= 12.46 x 10® + 1.979 x 10s
4

y = Distance of C.G. of the given section mm= 14.439 x 10® 4 Ans.

.

from bottom line. Problem 5.16. Find the moment of inertia of the area shown
shaded in Fig 5.25, about edge AS.
The C.G. of the section is obtained by using,

A A A- Sol. Given :

^ Radius of semi-circle, R = 10 cm
_ il'i + 2 y2 + 3 y3

+ A2 + Ag

_ 960 X 144 + 1536 x 74 + 1200 x 5 Width of rectangle, b = 20 cm
960 + 1536 + 1200
Depth of rectangle, d = 25 cm

138240 + 113664 + 6000 257904 Moment of inertia of the shaded portion about AB

3696 ~ 3696 = M.O.I. of rectangle ABCD about AS

= 69.779 = 69.78 cm DC- M.O.I. of semi-circle on about AS Fig. 5.25

Location of centroidal axis is shown in Fig. 5.24 (6). M.O.I. of rectangle ABCD about AS

(i) Moment of inertia of the given section about X-X [see equation (5.11)]
M.O.I. of the rectangle © about centroid axis X-X is given by,

= A(fG + x hf where h =(y - y)
, 1 1
Iyyi )x l

mm80 x 12s 4 20 x 25“ = 104.167 cm4
+ 960(144 - 69.78) 2 = 5.3x 10 6
DCM.O.I. of semi-circle about
M.O.I. of rectangle ® about centroid axis X-X is given by,

VAxnI where h (y2 - = — x [M.O.I. of a circle of radius 10 cm about a diameter]
= <Ici '>x + x 2 = y)
2
i

^12 X 283 ~-x[—=
di = i x x 20 4 = 3.925 cm4
= - + 1536 x (74 - 69.78)2

mm= 2.12 x 106 4 2 164 J 2 64

^3-^Ixxs DCDistance of C.G. of semi-circle from
and = ( Ia3 \x + ^a xh 2 where ^3 =
3

mm120 x 10 3 • 1200 x (5 - 69.78)2 = 5.04 x 10 s 4 4r 4x10
= 4.24 cm
Fig. 5.24 (6)
3it 3 ji

?XX - IXXI + ^XX2 + ^XXi AxAnroeua uojf. osceum-ii-ic0i1r1coliev,) •» —= g = 157.1— c— m2
mm= 5.3 X 106 + 2.12 X 106 + 5.04 x 10s •
4
2

mm= 12.46 x 10® 4 CDM.O.I. of semi-circle about a line through its C.G. parallel to
= M.O.I. of semi-circle about CD - Area x [Distance of C.G. of semi-circle from DC}2
(ii) M.O.I. of the given section about Y-Y

mm mm12 x 803 4 = 3925 - 157.1 x 4.242
: 521 X 103
4 = 0.521 x 106 = 3925 - 2824.28 = 1100.72 cm4

12

128 X 12 mm mm18.432 x 103 4 Distance of C.G. of semi-circle from AB
- 4 = 0.018432 x 10®
V~

^YY2
^0 = 25 - 4.24 = 20.76 cm
2

mm10 X 120s 4 M.O.I. of semi-circle about AS = 1100.72 + 157.1 x 20.762

IYY3~^G3 >y = 12 = 1.44 x 106 = 1100.72 + 67706.58 = 68807.30 cm4

lyy = fyTl + ^YY2 + ^YY3 .'. M.O.I. of shaded portion about AS
mm mm= 0.521 x 106 + 0.018432 x 10® + 1.44 x 106 4 = 104.167 - 68807.30 = 35359.7 cm 4. Ans.
4 = 1.979 x 106

V

210

Problem 5.16 (A). Find the moments of inertia about the centroidal XX and YY axes of

the section shown in Fig. 5.25 (a). (U.P. Tech. University, 2002-2003)

Sol. First find the location of the C.G. of the given figure:

Let a, = Area of complete rectangle y*

= BxD H-B/2-H

a = Area of removed rectangle portion
2

_B DD BD ©
- 2* 2" 4
fjf !
BD
V*1 = •J'l = 2 a m!§!

* “ 2 + - f-1 = Fig. 5.25 (a)
2
2 L2J

U rD •1if— BD 3 BD f 4dY* _ BD 3 16BD 3
2
X
where (Xp 33 ) and (.r2, yf) are the co-ordinates of the C.G. of the complete rectangle and cut out + +

rectangle respectively. Area a 2 is negative. 192 4 V 12 J 192 4 x 144

_ ai*i ~ ^2^2 BD x B —BxD ZB = BD 3 + BD3 ~ 3BD3 + 16BD3 _ 19BD 3
x “
2 44 T92" ~36~ 576 576

—j.

~ a2 -BD XX= M.O.I. of given section about centroidal axis

4 ~ ~~

—B2 *D _ 3 B z D ^XXl ^XX2 ^

BD 2 IggOj .lgjOl, 52Bg'- 19BO;_33Sgi j0 0573BD,
16 - 5 xD 144 576 576 576
2 16

— BD bd 12 Similarly, the M.O.I. of the given section about centroidal axis Y-Y is given by
!
4 mlyy = / - /yyj

OlYl ~ “2^2 BD D BD 3D > = M.O.I. of rectangle ® about centroidal axis Y-Y
“2
4 4

Similarly, “l -«2 +A= IGly -xxlx 2

-BD 11 ]

4 _DB3 DB 2 BD x
+
—BD~ 3 BD 2 BD2 = ~12~ + x [B ~ 5Bf = 13 nn3
["2"
2 16 16 5 xD
"12J

— BD -BD 12 At2 ~ ^G2y + -^2^*2 -c J

4 4 D (B

XXNow draw the centroidal axes and YY as shown in Fig. 5.25 (5). DB3 DB 3 _ 19DB 3

Let IXX1 = M.O.I. of complete rectangle CD about centroidal axis X-X 192 36 576
= M.O.I. of complete rectangle ® about horizontal axis passing through its C.G.

+ Area of complete rectangle ® -^-.DB 3 19DB 3 DB3 = 0.0573 DB3. Ans.

x Distance between X-X axis and horizontal axis passing through the C.G. of 144

rectangle ® 1^ AJ(By theorem of parallel axis) [v 1
= IGlXX + 1

\

+ (B x D) bh-yl2 5.11. MASS MOMENT OF INERTIA

* MConsider a body of mass as shown in Fig. 5.26.
MLet x = Distance of the centre of gravity of mass from axis OY
+ BDp-“ 1
[2 12 J My = Distances of the C.G. of mass from axis OX
MThen moment of the mass about the axis OY = x.
—+ BD "Dl2

12

.. d
CENTRE OF GRAVITY AND MOMENT OF INERTIA
212 STRENGTH OF MATERIALS 213

The above equation is known as first moment of
mass about the axis OY.

If the moment of mass given by the above equa-

tion is again multiplied by the perpendicular distance

between the C.G. of the mass and axis OY, then the

Mquantity (M r) . x = . x2 is known as second moment

of mass about the axis OY. This second moment of the
Mmass (i.e. quantity
, . x2 ) is known as mass moment of

OYinertia about the axis •

Similarly, the second moment of mass or mass

moment of inertia about the axis OX

- (M.y ) . y - M.y 2

Hence the product of the mass and the square of

the distance of the centre ofgravity of the mass from an Fig. 5.28 Fig. 5.29
axis is known as the mass moment of inertia about that
.-. Mass of the element = Density x Volume of element
axis. Mass moment of inertia is represented by/,.. Hence

mass moment of inertia about the axis OX is represented ='p x [Area x thickness of element]

OYby (/m )^ whereas about the axis by (/m ) . = px[6xdyx(l [ v p = Density and t = thickness]

Consider a body which is split up into small = pbt dy

mm mmasses v 2, 3 etc. Let the C.G. of the small Mass moment of inertia of the element about X-X axis
= Mass of element x y2
areas from a given axis be at a distance of rv r r = (pbt dy) xy2 = pbt y2 dy
2<
3 Mass moment of inertia of the plate will be obtained by integrating the above equation

etc. as shown in Fig. 5.27. Then mass moment of Ld

inertia of the body about the given axis is given by — —between the limits - to .

m m m1m - 2 + r 2+
r+ r
2 2 S3
\\

= Xmr2

If small masses are large in number then the sum- rd/2 rd/2 y2„ dy
mation in the above equation can be replaced by inte-
(/„)« = f pbt y2 dy = pbt\
gration. Let the small masses are replaced by dm in- Fig. 5.27 J-d/2 J-d/2
...(5.16)
stead of ‘m’, then the above equation can be written as [ v p, b, t are constant, and can be taken outside the integral sign]

7m = J r2 dm ULr M df2

5.12. DETERMINATION OF MASS MOMENT OF INERTIA =p ,, \lX' .fiWffdV f
3v
The mass moment of inertia of the following bodies will be determined by the method of
pbt d ds + d3 1 pbt_ 2d
integration : j
pbt. f 3* 8
1. Mass moment of inertia of a rectangular plate,
2. Mass moment of inertia of a circular plate, 88
3. Mass moment of inertia of a hollow circular cylinder,
—-pbt ,3 =pK x bdA ...(5.17)
5.12.1. Mass Moment of Inertia of a Rectangular Plate
12 12
A(a) Mass moment of inertia of a rectangular plate aboutA- axis passing through
—But is the moment of inertia of the area of the rectangular section about X-X axis.
the C.G. of the plate.
12
Fig. 5.28 shows a rectangular plate of width b, depth ‘d’ and uniform thickness V. Con-
sider a small element of width ‘6’ at a distance ‘y’ from X-X axis as shown in Fig. 5.29. This moment of inertia of the area is represented by /„.

Here X-X axis is the horizontal line passing through the C.G. of the plate. (/„)„ = px/xJ** ...(5.18)
Area of the element = b xdy
where = Mass moment of inertia of the plate about X-X axis . through C.G. of the

passing

plate.

/ = Moment of inertia of the area of the plate about X-X axis.

.

214 CENTRE OF GRAVITY AND MOMENT OF INERTIA

Again from equation (5.5), we have (c) Mass moment of inertia of a hollow rectangular plate.

Fig. 5.31 shows a hollow rectangular plate in ^ b ^
A
which AB CD is the main plate and EFGH is the cut-out 6

= p6 x d x t x section.

The mass moment of inertia of the main plate X-~

ABCD about X-X is given by equation

~= Md2

M( v = Mass of the plate = p x Volume of the plate = p x [b x d x t]) The mass moment of inertia of the cut-out sec-

~= Md2 ...(5.19) tion EFGH about X-X axis

Similarly, the mass moment of inertia of the rectangular plate about Y-Y axis passing —= md, 2 ^
12 1
through the C.G. of the plate is given by
Mwhere = Mass of main plate ABCD

= p .b.d.t 3 .31

<Uy=T2 Mb2' ...(5.20) m - Mass of the cut-out section EFGH
b.
(b) Mass moment of inertia of the rectangular ^ — P by. . d-y . f
B
plate about a line passing through the base. Then mass moment of inertia of hollow rectangular plate about X-X axis is given by

Fig. 5.30 shows a rectangular plate ABCD, having —Md=-i 2

width = b, depth = d and uniform thickness = t. We want to "(/m')xx 12 1.
md2 - - ...(5.22)
find the mass moment of inertia of the rectangular plate
12

about the line CD, which is the base of the plate. Consider 5.12.2. Mass Moment of Inertia of a Circular Plate

a rectangular elementary strip of width b, thickness t and d RFig. 5.32 shows a circular plate of radius and thick- Y

depth ‘dy’ at a distance y from the line CD as shown in ness t with O as centre. Consider an elementary circular

Fig. 5.30. uumimuummiimm j dy Vring of radius and width dr as shown in Fig. 5.32 (a).

Area of strip, dA = b.dy Area of ring, dA = 2nr . dr y' j
Volume of ring = Area of ring x t = dA . t
/* r

Volume of strip -dAxt-b.dy.t = b.t.dy , y
, |
Mass of the strip, dm = Density x Volume of strip = 2«r . dr . t / l/ —\ —
D C
Mass of ring, dm = Density x Volume of ring \-
= p{b . t . dy) = p . b .t . dy Fig. 5 _ 30
—fl
Mass moment of inertia of the strip about the line CD
= p(2-xr.dr.t) \ 'X JJ
= Mass of strip . y2
In this case first find the mass moment of inertia \ /trd'
= dm . y2 = y2 . dm
about an axis passing through O and perpendicular to the

Mass moment of inertia of the whole rectangular plate about the line CD is obtained by plane of the paper i.e., about axis Z-Z.

integrating the above equation between the limits 0 to d. .-. Mass moment of inertia of the circular ring about Y
axis Z-Z
Mass moment of inertia the rectangular plate about the fine CD Fig. 5.,32
= (Mass of ring) x (radius of ring)2

dm= f
2 y= f 2 . (p . b . t . dy) [ v dm = p .b.t. dy] dm= x r2 = (p . 2itr dr . t) x r2 = p . t 2jws dr

y. The mass moment of inertia of the whole circular plate will be obtained by . integrating
JO jo
the above equation between the limits O to R.
= p b t. . I y2 dy [v p, b and t are constant]
JO

Mass moment of inertia of circular plate about Z-Z axis is given by

—V s =p.b.t.d—3 d2 p , t . 2JW3 dr r3
== f drp2ji .
p.b.t. =p.b.t.d.~^ .t f

Jo Jo

...(5.21) = 2itp . t

[ p . b . t . d = Mass of rectangular plate = M] lo

.. <• 1H

216 STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA

= 2k . p . t =n p.t -r- S?' —^ —= p x 2it x L x 2_ 21
.
Fig. 5.32 (a) " [i?0 2 + Rf]

-^

Now mass of circular plate, TV Rt[ V - = (K0* - R, 2 )(R (t + 2
7?i )]
M = p x Volume of plate
- 2 + 2
R= p x jt 2 x t (77?po
L /= P X k[rRrq„ - i?;] X 7R?j; )
[Volume of plate - Area xt = itR2 x t]
X

2

hM( R
-
R<?+ pxiz Rp - R?2))
Substituting this value in above equation, we get = x x 2 = M]
[•.•.•. it (TT? 0fl

p —2 jVfT>2

=
22
— —xtxm =(v 7m')Z„Z rp x jtT?2 ...(5.23) Now (/_)„ M(Ro 2 2

+Ri )

(7ra)„ = (7m)„ = -4

But from the theorem of perpendicular axis given by equa- 5.12.4. Mass Moment of Inertia of a Right Circular Cone of base Radius R, Height

tion (5.7), we have MH and Mass about its Axis

Ia ~ Ixx + ^yy Let R = Radius of the base of the cone,

0r = <?m>a + ^Jyy H - Height of the cone,
And due to symmetry, we have (7^)^ = (7m )jy
M = Mass of the cone
am )xx = am)yy = (im\j2
= Density x Volume of cone = p x — itR2 x 77 — ?x

Consider an elemental plate of thickness dy and of radius \ T-
j j\
x at a distance y from the vertex (as shown in Fig. 5.32 (6 )). JAj j“\ ^

5.12.3. Mass Moment of Inertia of a Hollow Circular Cylinder xVViH T/z-n*jnnxpm - i

R I

Let J?„ = Outer radius of the cylinder -We have, tana = -X = R x=jj*y ^\ Td;

R = Inner radius of the cylinder Mass of the elemental plate, /J V
t
dm - p x Volume A/ 1
L = Length of the cylinder
2 x dy) A \
M - Mass of cylinder (itx
— — B|
= Density x Volume of cylinder = p x
KH ~-Rr—

L= - L RP*-= „ v Jt y2 x 2 Rr. . _„ _ 7(*xZy~1]
p x jt[R 2 2 x ...(i) .. FigV. 5.32 ((6b))
0 TJj ]

Vdm = Mass of a circular ring of radius width ‘dr’ and length L [Refer to Fig. 5.32]

= Density x Volume of ring = p x Area of ring x L The mass moment of inertia of the circular elemental plate about the axis of the cone
= p x 2itr dr x L (here axis of the cone is Z-Z axis of the circular elemental plate) is given by equation (5.23) as
Now mass moment of inertia of the circular ring about Z-Z axis
^ njzz ~_ Mass of plate x 2

radius

= Mass of ring x (radius) 2 2

= (p x 2%rdr x L) x r2 = {dm) x 2 _ dm x x2 v. r = xx))
”
r 2 (.

The mass moment of inertia of the hollow circular cylinder will be obtained by integrat- 2

R Ring the above equation between the limits xx—“ —dm,
to 0. j j?y - idly ,
~ffT
t xdy p x ay

.'. Mass moment of inertia of the hollow circular cylinder about Z-Z axis is given by, = P * II

(p *2xrdr.L)r> PiL^lxdyx Ry

r hJ

= p x 2n x L r3 dr = p x 2ji x Z, pxaRr>4 x 4

y
4y
2T7

= px 2*xix Now the total mass moment of inertia of the circular cone will be obtained by integrat-

ing the above equation between the limits O to 77.



STRENGTH OF MATERIALS

[Here x' is + ve, bul y' is -ve]

(a) (6)

Fig. 5.36

Let now the axes are rotated anticlockwise by 90° as shown in Fig. 5.36 (6) keeping the

Atotal area in the same position. Let x and y x are the new axes. The co-ordinates of the same
1

small area dA with respect to new axes are x’ and y'.

AHence the product of inertia of the total area with respect to new axes Xj and y 1

becomes as

y^jx'y'dA - (h)

Now let us find the relation between old and new co-ordinates. From Figs. 5.36 (a) and Fig. 5.37

5.36 (6), we get

x-=-y’ and y = x'
or y' = - x and x' = y

Substituting the values of x' and y' in equation (ii), we get

l (y)(- x) dA = xydA =-lxy xydA = Ixy x' = y sin 0 + x cos 0 —d)

Xiyi = J - J ('•’ J j

and y' -y cos 0 - x sin 0 ...(ii)

The above result shows that by rotating the axes through 90°, the product of inertia has AThe moment of inertia and product of inertia of area with respect to old axes are

become negative. This means that the product of inertia which was positive previously has = J y2 dA, I = x2 dA and I = J xy dA. ...(5.27)
now become negative by rotating the axes through 90°. Hence product of inertia has changed yy
J xy
its sign. It is also possible that by rotating the axes through certain angle, the product of
AAlso the moment of inertia and product of inertia of area with respect to new axes
inertia will become zero. The new axes about which product of inertia is zero, are known as

principal axes. will be

Note. (0 The product of inertia is zero about principal axes. =| 2 dA, /Wl =J (xl2 dA and I = x’y' dA
(it) As the product of inertia is zero about symmetrical axis, hence symmetrical axis is the princi- Cv') J
Xiy ,
pal axis of inertia for the area.
Letlls substitute the values of x', y' from equation (i) and (ii) in the above equations,

(t ii) The product of inertia depends upon the orientation of the axes. we get

5.15. PRINCIPAL MOMENTS OF INERTIA fVl =J (y'fdA
= J (y cos 0 - * sin 0) 2 dA
AFig. 5.37 (a) shows a body of area with respect to old axes (x, y) and new axes (x v y^). [v y' =y cos 0-x sin 0]

The new axes x and y, have been rotated through an angle 6 in anticlockwise direction. Con-
l
dA
sider a small area dA, The co-ordinates of the small area with respect to old axes is (x, y) =J 2 cos 2 0 + x2 sin 2 8 - 2xy cos 0 sin 0)

(y

whereas with respect to new axes, the co-ordinates are x' and y' The new co-ordinates (x', y')

are expressed in terms of old co-ordinates (x, y) and angle 9 as [Refer to Figs. 5.37 (5) and = J y2 cos2 0 dA + J x2 sin2 0 dA - J 2xy cos 0 sin 0 dA

5.37 (c)]

1 11

222 STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA 223

= cos2 0 J y2 dA + sin2 0 J x2 dA - 2 cos 0 sin 0 j xy dA 44<4 * 4j*-4 cos 20 - sin 20 ...(5.27E)
( v After rotation, the angle 0 is constant and hence
,
cos 2 0, sin2 0 and 2 cos 0 sin 0 are constant)
22

To find the values of Iy # , substract equation (5.272?) from (5.27C). Now substracting

= (cos2 0)4 + (sin2 0)4 - (2 cos 0 sin 0)4 ...(5.27A) equation (5.27 2?) from equation (5.27C), we get

4 4 4 424, =
2 x 2 dA = Iyy and J + ) - 1(4 - } cos 20 “ 2 sin 26]
4fv
y dA = 1^, J xy dA =

J j ““• 29 * '« 26

...(5.27F)

4nSimilarly, = J (x')2 dA Product of Inertia about New Axes

= (y sin 0 + x cos 0) 2 dA [ v x' = y sin 0 + x cos 0] 4Let and angle 0.
J us now find the value of I in terms of
Xl y t

=J 2 sin 2 0 + x2 cos 2 0 + 2xy sin 0 cos 0) dA We know that 4^ = J (x'Xy') dA

(y

= J y 2 sin2 0 dA + j" x 2 cos 2 0 dA + j" 2xy sin 0 cos 0 dA Substituting the values ofx' and y', we get

- sin2 0 J y2 dA + cos2 0 J x 2 dA + 2 sin 0 cos 0 J xy dA 4r, = J (y sin 0 + x cos 0)(y cos 0 - x sin 0) dA
(v x' = y sin 0 + x cos 0 andy' =y cos 0 - x sin 9)

(v 0 is constant and hence sin 0 and'cos 0 are constants)

6.4= sin2 + cos 2 0 I + 2 sin 0 cos 0 Ixy ...(5.27B) 4 =?1 J" 2 sin 0 cos 0 - xy sin 2 0 + xy cos 2 0 - x2 cos 9 sin 0) dA
(y
yy

4(v 2 x2 dA xy sin2 QdA + j xy cos2 0 dA - J x2 cos 0 sin 6 dA
J dA = Ixx> j dA = I and J xy = - y2 sin 0 cos 0 dA - J
y yy J
)

Adding equations (5.27A) and (5.27B), we get - sin 0 cos 0 J 2 dA - sin 2 xy dA + cos2 0 xy dA - cos 0 sin 0 J x2 dA
J"
y J

44,*i [sin2 0 + cos 2 0J + Iyy [sin2 0 + cos 2 0]
+ Iy = (7 0 is constant and hence sin 0, cos 0 are constants)

iyi

+ 2 sin 0 cos 0 I - 2 sin 0 cos 0 I^ 2sin0cos0 d„A- 2 cos 0 sin 0 r x-, t2A
2 ^
xy 04r f y,2 2 + cos2 0 1^
j
4>,
=4+4 ['’ sin 2 0 + cos 2 0 = 1] ...(5.27 C) = . j

sin

The equation (5.27C) shows that sum of moments of inertia about old axes (x,y) and new >

Aaxes xydA = Ixy
(x v y) are same. Hence the sum of moments of inertia of area is independent of orien- J
5 j

Nowtation of axes. let us find the value of I*1*1 - I„ . ~ ~ - ~-4sin 20
42
> 1^1 T T f „„ - s.m 0) ~ sin 20 T

Substracting equation (5.27B) from equation (5.27A), we get + (cos 0 2

4*i " 4y, = cos2 9 4. + sil> 2 0 l ~ 2 cos 0 sin 0 I j/^=4-j X2 dA = ly

yy xy

04- [sin2 . + cos 2 0 Iyy + 2 cos 0 sin 0 Ixy\

= Ixrxr (cos 2 0 - sin2 0) + / (sin 2 0 - cos2 0) - 4 cos 0 sin 0 . 4sin 20 + c°s2 0 - sin2 0 )

yy -v [ ^ sin 0

4= (cos 2 0 - sin2 0) - I (cos2 0 - sin2 0) - 4 cos 0 sin 0 I
yy
xy

= (/** - 1 ) (cos 2 0- sin2 0) - 2 x 2 cos 0 sin 0 xI ——^ sin 20 +

yy *y (cos2 0 - sin2 0)

= (I^ - Iyy) cos 20 - 2I sin2 0 ......((55..22772?) 2y
xy

0= 1 + cos 20 j. = 1 - cos 20 (Ixx Iyy)
, sin
v cos 0 = sin 20 + I cos 20 ...(5.27G)
2 xy
22 ( / cos 2 0 - sin2 0 = cos 20)

.. cos2 0 - sin2 0 = cos 20 and 2 sin 0 cos 0 = sin 20

Direction of Principal Axes

Now let us find the values of and ly^ in terms of 1^, I and 0. We have already defined the principal axes. Principal axes are the axes about which the
yy
Now
Adding equations (5.27C) and (5.272?), we get product of inertia is zero. the new axes (x v y) will become principal axes if the product of
t

424*1 “ 14 + 43 + [[4 - 24) cos 20 - sin 20 ] inertia given by equation (5.27G) is zero (i.e., I = 0).

Xiyi

——

:

STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA 225

For principal axes, lXlyl = 0 (a) Consider rectangle (1)

~ sin 20 + Ixy cos 26 = 0 The C.G. of rectangle (1) is at a distance of 20 cm from x-axis and at a distance of 25 cm

: from y-axis.

V~ lyy* sin 26 = - / cos 29 VJl-Voh, +*!»*?

where (Ixx)1 = M.O.I. of rectangle (1) about x-axis passing through the centroid of the given
figure of the given section.

0(I ) L* = M.O.I. of rectangle (1) about an axis passing through C.G. of rectangle (1) and

parallel to

bd 3

tan 26 = ...(5.27H)

The above equation will give the two values of 29 or 6. These two values of 0 will differ ^x^
by 90°. By substituting the values of0 in equations (5.27#) and (5.27#), the values of principal
- (Here b = 10 and d = 30)

12

= 2.25 x 104 cm4

moments of inertia (Ix,Xt and IWl ) can be obtained. If from equation (5.27#), the values of Aj = Area of rectangle (1) = 10 x 30 = 300

sin 26 and cos 20 in terms of I and I are substituted in equation (5.27#), we get (Ajx) = Distance of C.G. of rectangle (1) from x-axis
yy
xy,

= 20

These are the values of principal moment of inertia. Substituting the above values in equation (1), we get
Problem 5.18. For the section shown in Fig. 5.38 (a) determine : (IJy = 2.25 x 104 + 300 x 20 2
= 2.25 x 104 + 12 x 10 4

= 14.25 x 104 cm4 ...(A)

(i) Moment of inertia about its centroid along (x, yj axis. Similarly, the M.O.I. of rectangle (1) about y-axis passing through the centroid of the
(ii) Moment of inertia about new axes which is turned through an angle of 30° anticlock- given figure is given by,

wise to the old axis. (yi = (/G)i,+Ai(V)2

(iii) Principal moments of inertia about its centroid. bd 3 30 x 10 3 = n0.n25. x 1in044 cm4

All dimensions are in cm.. wh, ere ,, , = 12

{#G,)l,y„ 12

Sol. Given (kjy) = Distance of C.G. of rectangle (1) from y-axis = 25

The Fig. 5.38 (a) shows the given section. It is symmetrical about x-axis. The C.G. of the (Iyy\ = 0.25 x 104 + 300 x 25 (v A = 300)
1
Osection lies at (origin of the axes). To find moment of inertia of the given section, it is divided
into three rectangles as shown in Fig. 5.38 (5). First the moment of inertia of each rectangle = 0.25 x 104 + 18.75 x 10 4

about its centroid is calculated. Then by using parallel axis theorem, the moment of inertia of = 19 x 104 cm4 ...(B)

the given section about its centroid is obtained. (6) Consider rectangle (2)
The C.G. of this rectangle coincides with the C.G. of the given section. Hence
x1 +y —: H 10 U

HI fflT “, = bd 3 = 60 x 10 3 = 0..5, x 1ln04 cm4 n, ..
12
«i Td ? i (/„),2 12 ...(C)
i

C.G. 3 10 V2 = 10 x 60 4 = 18 x 104 cm4

T30-1—'* (c) Consider rectangle (3)

iUi 30 The C.G. of rectangle (3) is at a distance of 20 cm from x-axis and at a distance of 25 cm
from y-axis. Hence k.pc = 20 cm and k.y = 25 cm.
*j 10 |«l t
Now d A(/„)8 = (k^x)2
— JL G\x + a
! io

10 x 30 4 + (10 x 30X20)2 = 2.25 x 104 + 12 x 104 = 14.25 x 10 4 cm4

Fig. 5.38

1

CENTRE OF GRAVITY AND MOMENT OF INERTIA 227

29 x 10 4 + 56 x 4 29 x 10 4 - 56 x 10' 60° - 30 104 ) 60°
-
and 0^3 = ^G^3y + ^3^3 y) 10 cos (- x sin

———an x in3 = 42.5 x 104 - 13.5 x 10 x - + 30 x 104 x 0.866

= + 300 x 25 2 = 0.25 x 104 + 18.75 x 104 = 19 x 104 cm4 . 99—= 35.75 x 104 + 26 x 104 = 61.75 x 104 cm4. Ans.

12

(i) Moment of inertia of complete section about its centroid

= 14.25 x 104 + 0.5 X 104 + 14.25 x 104 cm4 = Ixx + Iyy - Iyy
= 29 x 10* cm4. Ans.
-ytfi 2 ' cos 0 + sin 26
2 *y

and dl = yy ) 1 + (I )2 + (Iyy)3 1_ 29 x 104 56 x x ° 4 _ 29 x 104 ~ 56 x 104 cos 60° + (- 30 X 10 4 ) sin 60°
yy yy
22
= 19 x 10 4 + 18 x 10 4 + 19 x 104
cm= 42.5 x 10 4 + 6.75 x 10 4 - 26 x 104 = 23.25 x 104 4 Ans.

.

= 56 x 104 cm4. Ans. (Hi) Principal moments of inertia about the centroid
The principal moments of inertia are the moments
(ii) Moment of inertia of complete section y
about new axes which is turned through an angle of inertia about the principal axes.
The direction of principal axes is given by equa- f
of 30° anticlockwise.
tion (5.27H) as t
Here 0 = 30°. v
s Z s;
Let us first calculate the product of inertia of
whole area about old axes x, y. X

(a) Consider rectangle (1) Il -I‘n

A = 10 x 30 = 300. yy
1
2 X (- 30 X 4
The C.G. of rectangle (1) is at a distance of 20 cm
above x-axis and at a distance of 25 cm from v-axis. 10 )

Hence co-ordinates of this C.G. are 56 x 10 4 - 29 x 10 4

Fig. 5.38 (e)

Xj = - 25 cm and yx = 20 cm. = —-- 60 x 10 4 2.222St
(6) For rectangle (2) .
ft S t S.t

27 X io 4 Fig. 5 - 38

A2 = 10 x 60 = 600 cm2 . The C.G. of rectangle (2) lies on the origin (O). Hence x = 0 and As 20 is negative, hence it lies in 2nd and 4th quadrant.
20 = tan-1 (- 2.222)
2

y2 = 0.

(c) For rectangle (3) = - 65.77° and 114.23°

A = 10 x 30 = 300 cm 2 or 0 =- 32.88° and 57.12°
a

The C.G. of rectangle (3) is at a distance of 20 cm below x-axis and at a distance of 25 cm The +ve angle is taken anti-clock and - ve angle is taken clockwise to the existing axes

from y-axis. x andy. The principal axes are shown asx and y , in Fig. 7.38 (e). The moment of inertia along
x
these axes is the principal moment of inertia. Hence by substituting 0 = — 32.88° and 57.12°, in
Hence co-ordinate of this C.G. are : x = 25 cm and y3 = (- 20 cm).
a
equations (5.271?) and (5.27F), we get principal moment of inertia.
The product of inertia (T ) of the whole figure is given by equation (5.26A) as

A AI x + AgX^y
xy 3
— ,x ,y + 3 2y 2 ( Y"“ I +1hy_ *I** -Il
3
= jy_
= 300 x (- 25) x 20 + 600 x 0 x 0 + 300 x 25 x (- 20) cos 20 _ l sin 20
y
22
= - 15 x 104 + 0 + (- 15 x 10 4 )
v Amin.

= - 30 x 10 4 cm4 29 x 10 4 + 56 x 10 4 29 x 10 4 - 56 x 4

= 10

Now the moment of inertia of the complete section +
about the new axes (Xj.y.) can be obtained from equations
22
(5.27E) and (5.27F) as
X cos (- 2 X 32.88) - (- 30 x 10 4) sin (- 2 x 32.88)

[v 0 =-32.88°]

Uxx+Iyy) = 42.5 x 104 - 13.5 x 10 4 x 0.41 + 30 x 104 x (- 0.912)
= 42.5 x 104 - 5.535 x 10 4 - 27.36 x 10 4
g + 2 C°S 29 " -V sm 20 = 9.605 x 10 4 cm4

where 1^ = 29 x 10 4 cm4 1 - 56 x 104 cm 4
, ,

I = - 30 x 10 4 cm4 and 0 = 30°

xy

Fig. 5.38 (d)

1

CENTRE OF GRAVITY AND MOMENT OF INERTIA 229

228 STRENGTH OF MATERIALS

4 10.

—= 1%r + l>* _ hs. I™. cos 20 + 7 sin 20 Radius of gyration of a body (or a given lamina) is the distance from an axis of reference where
the whole mass (or area) of the given body is assumed to be concentrated so as not to aftei the

/** 22 yy moment of inertia about the given axis. It is represented by k. Mathematically, k -

V 'min. -

= 42.5 x 10 + 5.535 x 10 4 + 27.36 x 10 4 = 75.395 x 104 cm4 11. According to theorem of perpendicular axis lzz = Ixx + iyy whei'e !xx and !yy = Moment of inertia

Hence principal moment of inertia are axes X-X and Y-Y in the plane of the section,
and passing through the inter-
cmImax = 75.395 x 104 4 Ans. of a plane section about two mutually perpendicular to the plane
. of the section perpendicular
Izz = Moment of inertia
cmImin = 9.605 x 104 4 Ans.
section of X-X and Y-Y axes.
.

Alternate Method the theorem of parallel axis IAB = Ia + Ah2 where
,
12. According to
The principal moments of inertia can also be obtained by
Moment of inertia of a given area about an axis passing through C.G. of the area
-2 IG = Moment of inertia of the given area about an axis AB, which is parallel to the axis passing
'vv ) =
Ixx ^+ *yy 1^

through G

h - Distance between the axis passing through G and axis AB

-_ 29 X 10 4 -+ 56 X 10 4 1(29 x 10 4 - 56 x 10 4 ) 2 + (. -O3A0x1l0n4 , 2 A = Area of the section.
)

= 42.5 x 104 ± y](- 13.5 x 10 4 ) 2 + (- 30 x 10 4 ) 2 13. Moment of inertia of a rectangular section :

bd s
(i) about an horizontal axis passing through C.G. =

= 42.5 x 104 ± 104 x 32.89

= (42.5 + 32.89) x 104 and (42.5 - 32.89) x 104 (ii) about an horizontal axis passing through base =

= 75-39 x 10 4 and 9.61 x 104 cm4

Imal - 75.39 x 104 cm4 and I . = 9.61 x 10 4 cm4 14. Moment of inertia of a circular section = -g
min. 15. Moment of inertia of a triangular section :

Now Imax and Imin are the required principal moment of inertia. Ans.

HIGHLIGHTS (i) about the base

1. The point, through which the whole weight of the body acts, is known as centre of gravity. (ii) about an axis passing through C.G. and parallel to the base -
2. The point, at which the total area of a plane figure is assumed to be concentrated, is known as
where b = Base width, and h - Height of the triangle.
centroid of that area. The centroid and centre of gravity are at the same point. 16. The C.G. of an area by integration method is given by
3. The centre of gravity of a uniform rod lies at its middle point.
4. The C.G. of a triangle lies at a point where the three medians of a triangle meet. and yJjpM
5. The C.G. of a parallelogram or a rectangle is at a point where its diagonal meet each other.
6. The C.G. of a circle lies at its centre. \dA ^ dA
where x* = Distance of C.G. of area dA from y-axis
7. The C.G. of a body consisting of different areas is given by
y* = Distance of C.G. of area dA from r-axis.
- _ aix i + a x> '/ + ^3X3 + • and y = + g2T2 + °3y3 + -
*1 + «2 + a3 + al + 0-2 + a3 + 17. The C.G. of a straight or curved line is given by

where x find y = Co-ordinates of the C.G. of the body from axis of reference _ lx* dL . _ y* dL
IdL
\

av a2, a = Different areas of the sections of the body y IdL
s,

*x x = Distances of the C.G. of the areas av a a from Y-axis
1, 2, 3 2, 3

yv y2, y3, .= Distances of the C.G. of the areas cq, a a from X-axis. EXERCISE 5
z, 3,

8. If a given section is symmetrical about X-X axis or Y- Y axis, the C.G. of the section will lie on the

axis symmetry. (A) Theoretical Questions

9. The moment of inertia of an area (or mass) about an axis is the product of area (or mass) and 1. Define centre of gravity and centroid. of a plane area using method of moments.
2. Derive an expression for the centre of gravity
square of the distance of the C.G. of the area (or mass) from that axis. It is represented by I.

) —H

230 STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA

3. What do you understand by axes of reference ? Find the centre of gravity of the /-section shown in Fig. 5.40. k- 8 cm -+l I
4. Define the terms : moment of inertia and radius of gyration,
[Ans. 6.44 cm] El ! III 2
i
T
5. State the theorem of perpendicular axis. How will you prove this theorem ? [Hint, a, = 8 x 2 = 16 cm2 a = 12 x 2 = 24 cm2 fl
,2 , 1
6. State and prove the theorem of parallel axis.
a = 16 x 2 = 32 ; = 2 + 12 + 1 = 15, f
3 i

7. Find an expression for the moment of inertia of a rectangular section : y2 = 2 + 6 = 8, y3 = 1 t12 cm
(i) about an horizontal axis passing through the C.G. of the rectangular section, and
_ - atfi + a 2y2 + -i
(ii) about an horizontal axis passing through the base of the rectangular section. y
a-i + a2 + c&3 crjn

(AMIE Summer, 1985) 16 X 15 + 24 X 8 + 32 X 1 %i 1
8. Prove that the moment of inertia of a circular section about an horizontal axis (in the plane of 16 + 24 + 32 i
i T
1

H - 16 cm

* “ _ rj4 = 240 + 192 + 32 = 464 . FigB. 5.40

the circular section) and passing through the C.G. of the section is given by = 6.44 cm.]

. 72 72

64

9. Prove that moment of inertia of a triangular section about the base of the section 3. (a) Find the centre of gravity of the L-section shown in Fig. 5.41. [Ans. x = 1.857, y = 3.857]

bh3 —M2 cmH

12

where b = Base of triangular section, and

h = Height of triangular section.

10. Derive an expression for the moment of inertia of a triangular section about an axis passing
through the C.G. of the section and parallel to the base.

11. Show that 10 = Ia + Ah 2 where I0 is the moment of inertia of a lamina about an axis through its
,

centroid and lying in its plane and h is the distance from the centroid to a parallel axis in the

Asame plane about which its moment of inertia is IQ, being the area of the lamina.

12. State and prove the parallel axes theorem on moment of inertia for a plane area. —M 6 cm

13. Prove that the moment of area of any plane figure about a line passing through its centroid is Fig. 5.41

zero.

14. Show that the product of inertia of an area about two mutually perpendicular axis is zero, if the (5) Find the moment of inertia of ISA 100 x 75 x 6 about the centroidal XX and YY axis, shown

area is symmetrical about one of these axis. (U.P. Tech. University, 2002-2003) in Fig. 5.41 (a). (U.P. Tech. University, 2001-2002)

15. Determine an expression for mass moment of inertia of hollow steel cylinder of mass M, outer -H 6 M-

R Rradius 0, inner radius i and length L about its axis. The hole in the cylinder is concentric.

(U.P, Tech. University, 2002-2003

H M16. Derive an expression for mass moment of inertia of a right circular cone of base radius R, height
and mass about its axis. ( U.P. Tech. University, 2001-2002)

(B) Numerical Problems

1. Find the centre of gravity of the 7-section shown in Fig. 5.39. [Ans. 8.272 cm]

H 75 mm H T

Fig. 5.41 (a)

[Hint. Locate first x and first y

mma = 100 x 6~ 600 2 x = 3 mm, y = 50
y 1 l
,

mm —a2 = 69 x 6 = 414 2 69 = 40.5

, x2 = 6 +

y2 = 3 mm

600 x 3 + 414 x 40.5 = 1- 8D.31
mm_

x
Fig. 5.39 = —a\LX-ii + aoXo

mm- a iyi + a2y2 600 x 50 + 414 x 3
: 30.81

) )

232 STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA

mma = 2 = 45 mm,
4
Now find the moment of inertia about centroidal X-X axis '• = 10 x 10 100 , jc4

iV7xxi = /G( i )I + a mmy, = 10 + 5 = 15 T'
‘

6 X 10 °3 _ apt! + 02^2 + <*3*3 + 04X4 20 I
12
6 X 10

-

12
-= + x - 2 = + 600(50 - 30.81)2 (n 3 + a2 + U3 + n4 ) (T) @^10
600 (y 1 yJ 1500 + 12000 + 1000 + 4500
) ^ T10
1000 Lt »
= 720.95 x 103 mm* ^so ^

O

^ VTXX2 = = "69 6 414^2 - 2 = 1.5 + 12 + 1 + 4.5 = 19 rom. Ans. 4 40 !

+“ 2' + y > ®8.

>x 2 1 - _ Qil1 ! + gzj2 4 03^3 + 0^4

= + 414(3 - 3.0.81) = 321.428 x 103 (a-i + a2 + + a4 ) [___

mm4. _ 4500 + 2000 - 2000 + 1500
!xx = !xxi + !XX2 - 720 -95 x 103 + 321.428 x 103 = 1042.378 x 103 4 Ans. 1000 Fig. 5.43

.

To find M.O.I. about centroidal axis Y-Y = 4.5 + 2 - 2 + 1.5 = 6 mm. Ans.]

10 63 A thin homogeneous wire is bent into a triangular shape ABC such that AB = 240 mm, BC =

°^
m mmI 4
(/Gl )y «!(*! - 2 = + 600(3 - 18.31)2 = 142.437 x 103
= + x mm260 and AC = 100 mm. Locate the C.G. of the wire with respect to co-ordinate axes. Angle at
)

m (hl = 2 )y + a2(*2 - x = mm+ 414(40.5 - 18.31)2 = 368.1 x 103 4 A is right angle.

[Hint. First determine angles a and (3. Use sine rule Yf

m m mmIyy = 7 + / = (142.437 + 368.1) x 103 4 = 510.537 x 10s nun4. Ans.3 BC AC AB
_
From a rectangular lamina ABCD 10 cm x 14 cm a rectangular hole of 3 cm x 5 cm is cut as
sin 90° sin a sin P .//901

shown in Fig. 5.42. Find the centre of gravity of the remainder lamina. AC x sin 90° 100

[Ans. 3c =4.7 cm, y = 6.444 cm]

—a = 22.62°. Also sin p = x sin 90° =

P = 67.38° FFig. 5.44

Using equation 5.2 (c) and 5.2 (d

_ _ Ljy 1 + ^2*2 + L3 X3 =AB = 240,

(Li + L2 + I3) ’ 1

x = distance of C.G. of AB fromy-axis
t
— mm940
= x cos a = 120 x cos 22.62° = 110.77
2

L = BC = 260 mm, x = Distance of C.G. of BC from y-axis = 130
2 2

Fig. 5.42 L, = AC = 100 mm, x3 = Distance of C.G. of AC from y-axis

5. For the T-section shown in Fig. 5.39, determine the moment of inertia of the section about the —= BD + — cos p = 240 cos a + 50 cos P

horizontal and vertical axes, passing through the centre of gravity of the section.

[Ans. 567.38 cm 4 294.67 cm 4 ) = 240 x cos 22.62° + 50 cos 67.38° = 240.77
,

6. For the /-section shown in Fig. 5.40, find the moment of inertia about the centroidal axis X-X _ 240 X 110.77 + 260 x 130 + IPO x 240.77 _ 140.77 mm. Ans.
[Ans, 2481.76 cm4] 1 “ 240 + 260 + 100
perpendicular to the web,

7. Locate the C.G. of the area shown in Fig. 5.43 with respect to co-ordinate axes. All dimensions - Ltfi + ^2^2 + ^3^3 = -r- sin a = 120 x sin 22.62° = 46.154
J __ la+Lg + Ls
are in mm. where y 2

mm[Hint. a = 10 x 30 = 300 2 x = 5 mm, yt = 15.
t t
,

mma 2 = 40 x 10 = 400 2 x2 = 10 + 20 = 30 mm, sin ^ ~ 50 sin 67.38° = 48.154

, y2 = 0 ,y3 =

y2 = 5 mm

mma3 = 10 x 20 = 200 2 x = 5 mm, 240 x 46.154 + 260 x 0 + 100 x 46.154
y = 600
, 3
= 26.154 mm. Ans.]
y3 = - 10 mm

—

9. Determine the C.G. of the uniform plane lamina shown in Fig. 5.45. All dimensions are in cm.

[Hint. The Fig is symmetrical about y-y axis. 6

y _ atfl * a2^2 + + <*4^4 4Y Shear Force and Bending Moment

^1 + + aZ + a4 |

“where
a = 40 x 30 = 1200 cm2 y l = ~ 15 cm
1 ,

—a2 30
= cm 2a - 45 cm
= 30 x 20 600 , y2 = 30 +

— —a~ = - 6.1. INTRODUCTION
n X 102 = — „„ — 4r * 4 x 10 «= —40
The algebraic sum of the vertical forces at any section of a beam to the right or left of the
50jc. Vo section is known as shear force. It is briefly written as S.F. The algebraic sum of the moments
2 3 3k 3rc 3k of all the forces acting to the right or left of the section is known as bending moment. It is
written as B.M. In this chapter, the shear force and bending moment diagrams for different
—a types of beams (i.e., cantilevers, simply supported, fixed, overhanging etc.) for different types
=-^20-x=10-100,^ = 60- 10 170
- of loads (i.e., point load, uniformly distributed loads, varying loads etc.) acing on the beams,
T4
will be considered.
~ —1200 x 15 + 600 x 45 - 50ji x - 100 x

y 3s 3_

1200 + 600 - 50n - 100

18000 + 27000 - 666.7 - 5666.7 Fig. 5.45 6.2. SHEAR FORCE AND BENDING MOMENT DIAGRAMS
1700 - 50rc
A shear force diagram is one which shows the variation of the shear force along the

length of the beam. And a bending moment diagram is one which shows the variation of the

25.06 cm from Origin 0. Ans. ] bending moment along the length of the beam.

1542.92

mm mm10. From a circular plate of diameter 100 is cut as shown in —Before drawing the shear force and bending moment) diagrams, we must know the
a circular part of diameter 50

Fig. 5.46. Find the centroid of the remainder. {U.P. Tech. University, 2002-2003 ) different types of beams and different types of load acting -on the beams.' .

6.3. TYPES OF BEAMS ''

The following are the important types of beams :

1. Cantilever beam, 2. Simply supported bearn^
3. Overhanging beam, 4. Fixed beams, and

5. Continuous beam.

A6.3.1. Cantilever Beam. beam which is fixed at one end and free at the other end, is

known as cantilever beam. Such beam is shown in Fig. 6.1.

. Fig. 5.46

[Hint. Fig. 5.46 is symmetrical about x-axis. Hence centroid lies on x-axis.

y = 0.6. The value of x is given by = ClXl + 02X2

.'. *

a ~ “2
l

mm—But a = x 1002 = 7853.98 2x = = 50 mm
l ,t

Fig. 6.1 Fig. 6.2

5°2 1
mm« = - 2
mm2
=~ x 1963.5 , x = 100 - 25 = 75 6.3.2. Simply Supported Beam. A beam supported or resting freely on the supports at

[f 2 its both ends, is known as simply supported beam. Such beam is shown in Fig. 6.2.

-

mmX
7853.98 x 50 - 1963.5 x 75
~ 4167
7853.98 - 1963.5

Hence centroid is at (41.67 mm, 0). Ans.l

235

236 STRENGTH OF MATERIALS SHEAR force and bending moment

6.3.3. For solving numerical problems the total load is equal to the area of the triangle and

Overhanging Beam. If the end portion of a beam is extended beyond the support, this total load is assumed to be acting at the C.G. of the triangle i.e., at a distance of |-rd of
such beam is known as overhanging beam. Overhanging beam is shown in Fig. 6.3. total length of beam from left end.

Simply supported Overhanging v&z 6.5. SIGN CONVENTIONS FOR SHEAR FORCE AND BENDING MOMENT

Vportion portion V N(i) Shear force. Fig. 6.9 shows a simply supported beam AS, carryring a load of 1000
R Rat its middle point. The reactions at the supports will be equal to 500 N. Hence A = B
Fig. 6.3 Fig. 6.4
= 500 N.
A6.3.4. Fixed Beams. beam whose both ends are fixed or built-in walls, is known as fixed
Abeam. Such beam is shown in Fig. 6.4. fixed beam is also known as a built-in or encastred beam. Now imagine the beam to be divided into two portions by the section X-X. The resultant

6.3.5. Continuous Beam. A beam which is pro- 1 | Nof the load and reaction to the left of X-X is 500 vertically upwards. (Note in this case, there

vided more than two supports as shown in Fig. 6.5, is || || is no load to the left of X-X). And the resultant of the load and reaction to the right of X-X is
known as continuous beam. HI HI
N(1000 l - 500 | = 500 1 N) 500 downwards. The resultant force acting on any one of the parts
6.4. TYPES OF LOAD Fig. 6.5
normal to the axis of the beam is called the shear force at the section X-X. Here the shear force
A beam is normally horizontal and the loads acting on the beams are generally vertical. at the section X-X is 500 N.

The following are the important types of load acting on a beam : The shear force at a section will be considered positive when the resultant of the forces
to the left to the section is upwards, or to the right of the section is downwards. Similarly the
1. Concentrated or point load,
shear force at a section will be considered negative if the resultant of the forces to the left' of
2. Uniformly distributed load, and the section is downwards, or to the right of the section is upwards. Here the resultant force to
the left of the section is upwards and hence the shear force will be positive.
3. Uniformly varying load.
Convexity
6.4.1. Concentrated or Point Load. A concentrated load is one which is considered to

act at a point, although in practice it must realty be distributed over a small area. In Fig. 6.6,

W shows the point load.

1000 N Concavity

SOON i Convexity Concavity
(a) Positive B.M. (b) Negative B.M.
< 500 N

Fig. 6.6 Fig. 6.7 fig- 6- 10

Fig. 6.9

6.4.2. Uniformly Distributed Load. A uniformly distributed load is one which is spre'ad (ii) Bending moment. The bending moment at a section is considered positive if the
over a beam in such a manner that rate of loading w is uniform along the length {i.e., each unit bending moment at that section is such that it tends to bend the beam to a curvature having
concavity at the top as shown in Fig. 6.10 (a). Similarly the bending moment (B.M.) at a section
length is loaded to the same rate) as shown in Fig. 6.7. The rate of loading is expressed as is considered negative if the bending moment at that section is such that it tends to bend the
beam to a curvature having convexity at the top as shown in Fig. 6.10 b( ). The positive B.M. is
w N/m run. Uniformly distributed load is, represented by u.d.l.

For solving the numerical problems, the total ^"fTT often called sagging moment and negative B.M. as x i iooo N
uniformly distributed load is converted into a point load,
hogging moment. £ I
acting at the centre of uniformly distributed load. Consider the simply supported beam AB, |

6.4.3. Uniformly Varying Load. A uniformly vary- i

ing load is one which is spread over a beam in such a man- 1 ,, Ncarrying a load of 1000 at its middle point. ,; ; ~ i
Fig 6.8
ner that rate of loading varies from point to point along the I RReactions IiA and B are equal and are having m—«— *-| g. y
magnitude 500 N as shown in Fig. 6.11. Imagine 1
beam as shown in Fig. 6.8 in which load is zero at one end || i a _-
and increases uniformly to the other end. Such load is known ^ill x
the beam to be divided into two portions by the r* Fig. 6.11
as triangular load. m
RA = 500 N
section X-X. Let the section X-X is at a distance of

m1 from A.

238 STRENGTH OF MATERIALS SHEAR FORCE AND BENDING MOMENT 239

6.

The moments of all the forces (i.e., load and reaction) to the left of X-X at the section The bending moment at the two supports of a simply supported beam and at the free

R NmX-X is a x 1 = 500 x 1 = 500 (clockwise). Also the moments of all the forces {i.e., load and end of a cantilever will be zero.

reaction) to the right of X-X at the section X-X y.-.Rd x 3 (anti-clockwise) - 1000 x 1 (clockwise) 6.7. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR A CANTILEVER
Nm Nm Nm= 500 x 3 WITH A POINT LOAD AT THE FREE END
- 1000 x 1 = 1500 - 1000 = 500 (anti-clockwise).
Fig. 6.14 shows a cantilever AB of length L fixed at A and free at B and carrying a point
Hence the tendency of the bending moment at X-X is to bend the beam so as to produce
Wload at the free end B.
concavity at the top as shown in Fig. 6.12.

X X

Clockwise Anticlockwise I

|

Fig. 6.12 Fig. 6.13

The bending moment at a section is the algebraic sum of the moments of forces and
reactions acting on one side of the section. Hence bending moment at the sectionX-X is 500 Nm.

The bending moment will be considered positive when the moment of the forces and
reaction on the left portion is clockwise, and on the right portion anti-clockwise. In Fig. 6.12,
the bending moment at the section X-X is positive.

Similarly the bending moment will be considered negative when the moment of the
forces and reactions on the left portion is anti-clockwise, arid on the right portion clockwise as
shown in Fig. 6.13. In Fig. 6.13, the bending moment at the sectionX-X is negative.

6.6. IMPORTANT POINTS FOR DRAWING SHEAR FORCE AND BENDING MOMENT C B.M. diagram

DIAGRAMS Fig. 6.14

In Art. 6.2, it is mentioned that the shear force diagram is one which shows the varia- Let F = Shear force at X, and
x
tion of the shear force along the length of the beam. And a bending moment diagram is one Mx = Bending moment at X.
which show the variation of the bending moment along the length of beam. In these diagrams,
the shear force or bending moment are represented by ordinates whereas the length of the XTake a section at a distance x from the free end. Consider the right portion of the
beam represents abscissa.
section.
The following are the important points for drawing shear force and bending moment
The shear force at this section is equal to the resultant force acting on the right portion
diagrams :
X Wat the given section. But the resultant force acting on the right portion at the section is
1. Consider the left or the right portion of the section.
and acting in the downward direction. But a force on the right portion acting downwards is
2. Add the forces (including reaction) normal to the beam on one of the portion. If right
portion of the section is chosen, a force on the right portion acting downwards is positive while Xconsidered positive. Hence shear force at is positive.
a force acting upwards is negative.
WF = +
If the left portion of the section is chosen, a force on the left portion acting upwards is x
positive while a force acting downwards is negative.
AThe shear force will be constant at all sections of the cantilever between and B as
3. The positive values of shear force and bending moments are plotted above the base there is no other load between A and B. The shear force diagram is shown in Fig. 6.14 (6).
line, and negative values below the base line.
Bending Moment Diagram
4. The shear force diagram will increase or decrease suddenly i.e., by a vertical straight
line at a section where there is a vertical point load. XThe bending moment at the section is given by

5. The shear force between any two vertical loads will be constant and hence the shear M = -Wxx
force diagram between two vertical loads will be horizontal. x

(Bending moment will be negative as for the right portion of the section, the moment of

W Xat is clockwise. Also the bending of cantilever will take place in such a manner that

convexity will be at the top of the beam).

: )

240 STRENGTH OF MATERIALS 'SHEAR FORdE AND BENDING MOMENT 241

From equation (£), it is clear that B.M. at any section is proportional to the distance of S.F. at D, Fd = + 800 N
NS.F. at C, Fc = + 800 + 500 = + 1300
the section from the free end. N1 S.F. at B, Fb = + 800 + 500 + 300 = 1600

At x = 0 i.e., at B, B.M. = 0 S.F. at A, EA = + 1600 N.

WxLAt* = L i.e., at A, B.M. = The shear force, diagram is shown in Fig. 6.15 (6) which is drawn as :

Hence B.M. follows the straight line law. The B.M. diagram is shown in Fig. 6.14 (c). At

WxLpoint A, take AC = in the downward direction. Join point B to C. Draw a horizontal line AD as base line. On the base line mark the points B and C below

The shear force and bending moment diagrams for several concentrated loads acting on Nthe point loads. Take the ordinate DE = 800 in the upward direction. Draw a EFline parallel

a cantilever, will be drawn in the similar manner. to AD. The point F is vertically above C. Take vertical line FG = 500 N. Through G, draw a
GH Hhorizontal line
mProblem 6.1. A cantilever beam of length 2 carries the point loads as shown in in which point is vertically above B. Draw' vertical line HI = 300 N.

Fig. 6.15. Draw the shear force and B.M. diagrams for the cantilever beam. From I, draw a horizontal line IJ. The point J is vertically above A. This completes the shear

Sol. Given force diagram.

Refer to Fig. 6.15. Betiding Moment Diagram

The bending moment at D is zero :

300 N 500 N 800 N (i ) The bending moment at any section between C and D at a distance x and D is given by,

Mx = - 800 x x which follow's a straight line law.

At C, the value of x = 0.8 m.

.-. B.M. at C, Mc = - 800 x 0.8 = - 640 Nm.

D(it) The B.M. at any section between B and C at a distance * from is given by

(At C, x = 0.8 and at B, x = 0.8 + 0.7 = 1.5 m. Hence here x varies from 0.8 to 1.5). ...(i)

Mx -~ 800 x - 500 {x - 0.8)

Bending moment between B and C also varies by a straight line law.

mB.M. at B is obtained by substituting x = 1.5 in equation (i),

Mb = - 800 x 1.5 - 500 (1.5 - 0.8)

= - 1200 - 350 = - 1550 Nm.

A D( Hi ) The B.M. at any section between and B at a distance x from is given by
m(At B. x = 1.5 and at A, x = 2.0 m. Hence here x varies from 1.5 to 2.0 m)

Mx = - 800 x - 500 (x - 0.8) - 300 (x - 1.5) ...( ii

Bending moment between A and B varies by a straight line law.

A mB.M. at is obtained by substituting x = 2.0 in equation (ii),

Ma = - 800 x 2 - 500 (2 - 0.8) - 300 (2 - 1.5)

= - 800 x 2 - 500 x 1.2 - 300 x 0.5

_ _ 1600 - 600 - 150 = - 2350 Nm.

Hence the bending moments at different points will be as given below :

MD = 0
Fig. 6.15 Mc 640 Nm
MB = - 1550 Nm
Shear Force Diagram = -

D DThe shear force at is + 800 N. This shear force remains constant between and C. Mand A = - 2350 Nm.

At C, due to point load, the shear force becomes (800 + 500) = 1300 N. Between C and B, the The bending moment diagram is shown in Fig. 6.15 (c) which is drawn as.
shear force remains 1300 N. At B again, the shear force becomes (1300 + 300) = 1600 N. The
Draw a horizontal line AD as a base line and mark the points B and C on this line. Take
Ashear force between B and remains constant and equal to 1600 N. Hence the shear force at Nm Nmvertical lines CC' = 640 Nm, BB' = 1550
and AA' = 2350 in the downward direction.
different points will be as given below :
-Join points D, C', B' and A' by straight lines. This completes the bending moment diagram.

OTDCMfttU nr MATPR1AI 5? : 243

6.8. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR A CANTILEVER M SHEAR FORCE AND BENDING MOMENT
WITH A UNIFORMLY DISTRIBUTED LOAD
(The bending moment will be negative as for the right portion of the section, the moment
AFig. 6.16 shows a cantilever oflength L fixed at and carrying a uniformly distributed of the load at x is clockwise. Also the bending of cantilever will take place in such a manner
load of w per unit length over the entire length of the cantilever.
that convexity will be at the top of the cantilever).

From equation (i), it is clear that B.M. at any section is proportional to the square of the

distance of the section from the free end. This follows a parabolic law.

MAt B, x = 0 hence x-0

M —L2
At A, x = L hence x=-w . .

z

The bending moment diagram is shown in Fig. 6.16 (c).

mProblem 6.2. A cantilever oflength 2.0 carries a uniformly distributed load ofl kN/m
mrun over a length of 1.5 from the free end. Draw the shear force and bending moment diagrams

for the cantilever.

Sol. Given

U.D.L., w = 1 kN/m run

Refer to Fig. 6.17.

Fig. 6.16

XTake a section at a distance of x from the free end B.

MLet Fx = Shear force at X, and
x = Bending moment at X.
Here we have considered the right portion of the section. The shear force at the section

X will be equal to the resultant force acting on the right portion of the section. But the result-

ant force on the right portion = w x Length of right portion = w.x.

This resultant force is acting downwards. But the resultant force on the right portion

Xacting downwards is considered positive. Hence shear force at is positive.

F = + w.x
x
A' Straight line B.M. diagram
The above equation shows that the shear force follows a straight line law.

At B, x = 0 and hence F = 0 Fig. 6.17
x

At A, x = L and hence F = w.L
x

The shear force diagram is shown in Fig. 6.16 b( ). Shear Force Diagram

Bending Moment Diagram Consider any section between C and B a distance of x from the free end B. The shear

It is mentioned in Art. 6.4.3 that the uniformly distributed load over a section is con- force at the section is given by

verted into point load acting at the C.G. of the section. F = w.x (+ve sign is due to downward force on right portion of the section)
x
XThe bending moment at the section is given by
= 1.0 x x (v w = 1.0 kN/m run)
Mx = - (Total load on right portion)
At B, x = 0 hence F - 0
Xx Distance of C.G. of right portion from x

FAt C, x = 1.5 hence x - 1.0 x 1.5 = 1.5 kN.

= ~{w.x). — = -w.x.— = -w.— The shear force follows a straight line law between C and B. As between A and C there

2 22 ,..(£) . Ais no load, the shear force will remain constant. Hence shear force between and C will be

represented by a horizontal line.



.

246 STRENGTH OF MATERIALS SHEAR FORCE AND BENDING MOMENT 247

The equation (i) shows that the B.M. varies according to the parabolic law. From In equation (i), x varies from 0 to 0.5. The equation (i) shows that shear force varies by

equation (i), we have a straight line law between B and C.

At B, x = 0 hence MB = — (3x0 + 02) = 0 At B, x = 0 hence FB = 1.5 x 0 = 0

mAt A, x = 2 hence M, = -(3x2 + 22) = - 10 kN/m At C, x = 0.5 hence Fc = 1.5 x 0.5 = 0.75 kN

Now the bending moment diagram is drawn as shown in Fig. 6.18 (c). In this diagram, A(ii) Now consider any section between and C at a distance x from free end B. The
AA' = 10 kNm and points A' and J3 are joined by a parabolic curve.
shear' force at the section is given by

A mProblem 6.4. cantilever of length 2 carries a uniformly distributed, load of 1.5 kN/m F - + w.x + 2 kN (+ve sign is due to downward force
mrun over the whole length and a point load of2 kN at a distance of 0.5 from the free end. Draw x on right portion of the section) -

the S.F. and B.M. diagrams for the cantilever. = 1.5x + 2 ...(ii)

Sol. Given : L=2m In equation (ii), x varies from 0.5 to 2.0. The equation (ii) also shows that shear force
Length,
Avaries by a straight line law between and C.

U.D.L., w = 1.5 kN/m ran At C, x = 0.5 hence Fc = 1.5 x 0.5 + 2 = 2.75 kN
Point load,
W = 2 kN At A, x = 2.0 hence FA = 1.5 x 2.0 + 2 = 5.0 kN

mDistance of point load from free end = 0.5 Now draw the shear force diagram as shown in Fig. 6.19 (b) in which CD = 0.75 kN,

Refer to Fig. 6.19. DK = 2.0 kN or CE = 2.75 kN and AF = 5.0 kN. The point B is joined to point D by a straight

E Fline whereas the point is also joined to point by a straight line.

Bending Moment Diagram

(i) The bending moment at any section between C and B at a distance x from the free
end B is given by

M —x = - (w.x) .
2

—= - (1.5 x x). (v w = 1.5 kN/m)

2 ...(Hi)

= - 0.75x2

(The bending moment will be negative as for the right portion of the section the moment

at the section is clockwise).

In equation (Hi), x varies from 0 to 0.5. Equation (Hi) shows that B.M. varies between

C and B by a parabolic law.

MAt B, x = 0 hence B - - 0,75 x 0 = 0

MAt C, x = 0.5 hence c = - 0.75 x 0.52 = - 0.1875 kNm.

(ii) The bending moment at any section between A and C at a distance x from the free

end B is given by

Mxr = ~ (w.x.) —— - 2(x - 0.5) = - (1.5 x x) . - 2(x - 0.5)

2' 2

(v w = 1.5 kN/m)

= - 0.75 x 2 -2(x- 0.5) ...(iv)

Fig. 6.19 In equation (iv), x varies from 0.5 to 2.0. Equation (iv) shows that B.M. varies by a

Shear Force Diagram parabolic law between A and C.

(i) Consider any section between C and B at a distance x from the free end. The shear MAt C, x = 0.5 hence c = - 0.75 x 0.5 2 - 2(0.5 - 0.5) = - 0.1875 kN/m
MAt A, x = 2.0 hence A = - 0.75 x 2 2 - 2(2.0 - 0.5) kNm = - 3.0 - 3.0 = - 6.0 kNm
force at the section is given by,

F = + w.x (+ve sign is due to downward Now the bending moment diagram is drawn as shown in Fig. 6.19 (c). In this diagram
x force on right portion)
CO Cline
- 0.1875 and AA' = 6.0. The points A', and B are on parabolic curves.

= 1.5 x x ...»

:

248 STRENGTH OF MATERIALS 249

A mProblem 6.5. cantilever 1.5 long is loaded with a uniformly distributed load of Bending Moment Diagram
m2 kN/m run over a length of 1.25 from the free end. It also carries a point load of 3 kN at a B.M. at B = 0

distance of 0.25 mfrom the free end. Draw the shear force and bending moment diagrams of the DB.M. at = - 2 x 0.25 x 0 25 = - 0.0625 kNm

cantilever.

Sol. Given mL = 1.5 : i ok -(3xl) = - 4.563 kNm
Length,
U.D.L., w = 2 kN/m DB.M. at = - 2 x 1.25 x
Point load,
W = 3 kN AB.M. at = - 2 x 1.25 x + 0.25 - 3 x (1 + 0.25) = - 5.94 kNm.

Refer to Fig. 6.20. j

The bending moment between B and C and between C and D vanes by a parabolic law.

But B.M. between A and D varies by a straight line law.

Now the bending moment diagram is drawn as shown in Fig. 6.20 (c). In this diagram

1: line CC' - 0.0625, DD' = 4.563 and AA' = 5.9. The points B, C' and D' are on parabolic curve

whereas points A’ and D' are joined by a straight line.

mAProblem 6.6. cantilever of length 5.0 is loaded as shown in Fig. 6.21. Draw the

S.F. and B.M. diagrams for the cantilever.

Sol. The shear force at B is 2.5 kN and remains constant between B and C.

2x1The shear force increases by a straight line law to 2.5 + = 4.5 kN at D. The shear

Dforce remains constant between and E. At point E, the shear force suddenly increases to
4.5 + 3 = 7.5 kN due to point load at£. Again the shear force remains constant between A and£.

Now the shear force diagram is drawn as shown in Fig. 6.21 (6).

Fig. 6.20

Shear Force Diagram

The shear force at B is zero.

The shear force increases to 2 x 0.25 = 0.5 kN by a straight line at C. Due to point load

of 3 kN, the shear force suddenly increases to 0.5 + 3 = 3.5 kN at C.

2x1=The shear force further increases to 3.5 + 5.5 kN by a straight line at D. The

Ashear’ force remains constant between and D as there is no load between A and D.

Now the shear force diagram is drawn as shown in Fig. 6.20 ( b ). In this diagram line
= 5.5 kN. The point B is joined to £ by a
AHCE = 0.5 kN, CF = 3.5 kN, DG = 5.5 kN and
G GHFstraight line. The point is also joined to by a straight line. Line
is horizontal. Fig. 6.21

— .

:.

/ SHEAR FORCE AND BENDING MOMENT

250 STRENGTH OF MATERIALS XThe shear force and the section at a distance x from free end is given by,

Bending Moment Diagram F Bx = Total load on the cantilever for a length x from the free end
B.M. at B = 0 = Area of triangle BCX

B.M. at C = - 2.5 x 0.5 = - 1.25 kNm XB.XC v XB = x, XC = :
DB.M. at = - 2.5 x 2.5 - 2 x 1 x 1 = - 8.25 kNm
B.M. at£ = -2.5x4-2xlx (1.5 + 1.0) = - 10 - 5 = - 15 kNm The equation (i ) shows that the S.F. varies according to the parabolic law.
A -3x1B.M. at = - 2.5 x 5 - 2 x 1 x (1 + 1.5 + 1.0)
w x 02
= - 12.5 - 7.0 - 3 = - 22.5 kNm. 2L
At B, * = 0 hence —FB = =0
Now the bending moment diagram is drawn as shown in Fig. 6.21 (c). In this diagram,
At A, x ~ L hence —FA. ~ _ W '^J
Dthe B.M. varies according to parabolic law between points C and only. Between other points 2L
2
B.M. varies according to straight line law.
XThe bending moment at the section at a distance x from the free end B is given by,
6.9. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR A CANTU.EVER CAR-
RYING A GRADUALLY VARYING LOAD M Xx = ~ (Total load for a length x) x Distance' of the load from

AFig. 6.22 shows a cantilever of length L fixed at and carrying a gradually varying load

from zero at the free end to w per unit length at the fixed end.

A= - (Area of triangle BCX) x Distance of C.G. of the triangle from

The equation (ii) shows that the B.M. varies according to the cubic law.

At B, x = 0 hence MB" = - W*^ = 0

6L

MA.t..A,x = Lr ,hence,, ui.l? w.l'}
A
61 .

6

mProblem 6.7. A cantilever of length 4 carries a gradually varying load, zero at the

free end to 2 kN/m at the fixed end. Draw the S.F. and B.M. diagrams for the cantilever.

Sol. Given L=4m
Length,

Load at fixed end, w = 2 kN/m

Shear Force Diagram

The shear force is zero at B. The shear force at C will be equal to the area of load

diagram ABC.

^ ^ = 4 kN

Shear force at C =

2

Fig. 6.22 The shear force between A and B varies according to parabolic law.

XTake a section at a distance x from the free end B. Bending Moment Diagram

Let F = Shear force at the section X, and j
x
——,2
Mx - Bending moment at the section X.
The B.M. at B is zero. The bending moment at A is equal to -

Let us first find the rate of loading at the section X. The rate ofloading is zero at B and M.A - - = = _ 5.33 kNm.
is w per metre run at A. This means that rate of loading for a length L is w per unit length.
6 6

—Hence rate of loading for a length of x will be x x per unit length. This is shown in Fig. 6.22 (a) The B.M. between A and B varies according to cubic law.

by CX, which is also known as load diagram. Hence CX = ~' W
-.

252 STRENGTH OF MATERIALS . .
C
SHEAR force and bending moment 253

A S.F. diagram

B.M. diagram Base line

Fig. 6.23 Fig. 6.24

Now consider any section between C and B at distance x from end A. The resultant force

on the left portion will be

6.10. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR A SIMPLY This force will also remain constant between C and B. Hence shear force between C and B

SUPPORTED BEAM WITH A POINT LOAD AT MID-POINT —is equal to -
W —WAt the section C the shear force changes from + -— to -
Fig. 6.24 shows a beam AB of length L simply supported at the ends A and B and carry-

Wing a point load at its middle point C.

—The reactions at the support will be equal to as the load is acting at the middle point

—Wof the beam. Hence RA - RB - A

The shear force diagram is shown in Fig. 6.24 b( ). _

X ATake a section at a distance x from the end between A and C. Bending Moment Diagram

Let F = Shear force at X, (i) The bending moment at any section betweenA and C at a distance of.r from the end A,
x
Mx = Bending moment at X.
and is given by

XHere we have considered the left portion of the section. The shear force at will be M MRx = a.x or = +—W.x -rw\

equal to the resultant force acting on the left portion of the section. But the resultant force on x

—the left portion is acting upwards. But according to the sign convention, the resultant force M X(B will be positive as for the left portion of the section, the moment of all forces at

Xon the left portion acting upwards is considered positive. Hence shear force at is positive and is clockwise. Moreover, the bending of beam takes place in such a manner that concavity is at

W the top of the beam).

its magnitude is At A, x = 0 hence M —A = IT x 0 = 0

2'

W —At C, x - hence W WxX L _ x L

AHence the shear force between and C is constant and equal to + 22 4

2'



257
256

0 0

258 STRENGTH OF MATERIALS SHEAR FORCE AND BENDING MOMENT 259

The values of B.M. at different points are : Shear Force Diagram

— ——M, ,, w. A AConsider any section at a distance x from between and C. The shear force at the
w.L
At A, x = 0 hence = . = section is given by,

0

.

F = + ha ~ 10 * = + 40 - 1 * •••(*)
x
MAt B, x = L hence —L - .L2 - 0
B= Equation (z) shows that shear force varies by a straight line law between A and C.

w.L A w_ (L_ ~w.L2 w.L2 At A, x = 0 hence FA = + 40 - 0 = 40 kN
48
— MAt C, x = hence c - 2 2 2 1.2 At C, x = 6 m hence Fc = + 40 - 10 x 6 = - 20 kN

A — —Thus the B.M. increases according to parabolic law from zero at to + - at the The shear force at A is + 40 kN and at C is - 20 kN. Also shear force between A and C

middle point of the beam and from this value the B.M. decreases to zero at B according to the Avaries by a straight line. This means that somewhere between and C, the shear force is zero.

parabolic law. FLet the S.F. is zero at* metre from A. Then substituting the value of S.F. (i.e., x) equal to zero

Now the B.M. diagram is drawn as shown in Fig. 6.27 (c). in equation (i), we get
Problem 6.9. Draw the shear force and bending moment diagram for a simply supported
0 = 40 - 10*
m mbeam of length 9 and carrying a uniformly distributed load of 10 kN/m for a distance of 6
—x = 40 = 4A m
from the left end. Also calculate the maximum B.M. on the section.
10
R RSol. First calculate reactions A and B .
mHence shear force is zero at a distance 4 from A.

The shear force is constant between C and B. This equal to - 20 kN.

Now the shear force diagram is drawn as shown in Fig. 6.28 (6). In the shear force

mDdiagram, distance AD = 4 m. The point is at a distance 4 from A.

B.M. Diagram

The B.M. at any section between A and C at a distance x from A is given by,

nA y m nB M= RA, x* - 10 . * . — = 40* - 5* 2 ...(til
40 kN
20 kN 2

Base line Equation (it) shows that B.M. varies according to parabolic law between A and C.

i M
A
40 ; C /' At A, * = 0 hence = 40x0-5x0 = 0
(6) 1 :.
L' B mAt C, * = 6 hence Mc = 40 x6-5x 62 = 240 - 180 = + 60 kNm

A . Z3120 mAt D, x - 4 hence Mp = 40 x 4 - 5 x 42 = 160 - 80 = + 80 kNm

S.F. diagram

The bending moment between C and B varies according to linear law.

B.M. at B is zero whereas at C is 60 kNm.

The bending moment diagram is drawn as shown in Fig. 6.28 (c).

D Maximum Bending Moment

B.M. diagram The B.M. is maximum at a point where shear force changes sign. This means that the

Fig. 6.28 point where shear force becomes zero from positive value to the negative or vice-versa, the
B.M. at that point will be maximum. From the shear force diagram, we know that at point D,
Taking moments of the forces about A, we get
the shear force is zero after changing its sign. Hence B.M. is maximum at point D. But the
I?Box9 = 10x6x — = 180 B.M. at D is + 80 kNm.

2 Max. B.M. = + 80 kN. Ans.

Problem 6.10. Draw the shear force and B.M. diagrams for a simply supported beam of

mlength 8 and carrying a uniformly distributed load of 10 kN/m for a distance of 4 mas shown

in Fig. 6.29.

RSol. First calculate the reactions RA and B.

Taking moments of the forces about A, we get

Ra = Total load on beam - RB = 10 x 6 - 20 = 40 kN. R b x 8 = 10 x 4 x f 1 + 1 j = 120

.

260 STRENGTH OF MATERIALS SHEAR FORCE AND BENDING MOMENT 261

wV B.M. Diagram
Ra U-
B.M. at A is zero

B.M. at B is also zero

B.M. at C = Ra x 1 = 25 x 1 = 25 kNm
DThe B.M. at any section between C and at a distance x from A is given by,

rvv / / * J M= A 10(x- 1). =25xx-5(x-l)2 ...(H)
x
w T25 : 2

4- At C, x = 1 hence M c = 25 x 1 - 5(1 - l)2 = 25 kNm

: At D, x = 5 hence MD = 25 x 5 - 5(5 - l) 2 = 125 - 80 = 45 kNm
At E, x = 3.5 hence
AC M kNmE = 25 x 3.5 - 5(3.5 - l)2 = 87.5 - 31.25 = 56.25

S.F. diagram 15 < B.M. will increase from 0 at A to 25 kNm at C by a straight line law. Between C and D

the B.M. varies according to parabolic law as is clear from equation (a). Between C and D, the
B.M. will be maximum at E. From D to B the B.M. will decrease from 45 kNm atD to zero at B

Parabolic ^rrrrr. TOT? Straight according to straight line law.

Straight \ p ^line Problem 6.11. Draw the S.F. and B.M. diagrams of a simply supported beam of length
>r
line

+ se 25 45 m7 carrying uniformly distributed loads as shown in Fig. 6.30.

7\ ....

C E

B,M. diagram

Fig. 6.29

RBK = 120 = 15 kN
8

fiA = Total load on beam - RB

= 10x4- 15 = 25 kN

Shear Force Diagram

The shear force at A is + 25 kN. The shear force remains constant between A and C and

equal to + 25 kN. The shear force at B is - 15 kN. The shear force remains constant between
B and D and equal to - 15 kN. The shear force at any section between C and Data distance

x from A is given by,

F = + 25- 100r- 1) —G)

At C,x = 1 hence Fc = + 25- 10(1 - 1) = + 25 kN

At D, x = 5 hence FD = + 25 - 10(5 - 1) = - 15 kN

DThe shear force at C is + 25 kN and at is — 15 kN. Also shear force between C and D

Dvaries by a straight line law. This means that somewhere between C and the shear force is

,

Fzero. Let the S.F. be zero at x metre from A. Then substituting the value of S.F. (i.e., x ) equal Fig. 6.30

to zero in equation (i ), we get Sol. First calculate the reactions RA and RB .

0 = 25 - 10(x - 1) Taking moments of all forces about A, we get

or 0 = 25 - 10* + 10 or lOx = 35

— mx = 35 —Rb x 7 = 10 x 3 x | + 5 x 2 x ^3 + 2 + |j = 45 + 60 = 105
= „„
3.5

10

mHence the shear force is zero at a distance 3.5 from A. RB = = 15 kN

A mHence the distance E = 3.5 in the shear force diagram shown in Fig. 6.29 (6). 7

j

STRENGTH OF MATERIALS SHEAR FORCE AND BENDING MOMENT 263

Ra = Total load on beam - RB
= (10 x 3 + 5 x 2) - 15 = 40 - 15 = 25 kN

S.F. Diagram.

The shear force at A is + 25 kN

The shear force at C = RA — 3 x 10 = + 25 — 30 = - 5 kN
AThe shear force varies between and C by a straight line law.

The shear force between C and D is constant and equal to - 5 kN

The shear force at B is - 15 kN

DThe shear force between and B varies by a straight line law.

The shear force diagram is drawn as shown in Fig. 6.30 b( ).

AThe shear force is zero at points between and C. Let us find the location of E from A.

Let the point £ be at a distance x from A.

The shear force at E = JfA - 10 x x = 25 - lOx
But shear force at E = 0

25 - 10* = 0 or lOx = 25

— mx = 25 = 2n.5r

B.M. Diagram B.M. diagram

AB.M. at is zero Fig. 6.31

B.M. at B is zero

B.M. at C, Mro - R.a x 3 - 10 X 3 X - = 25 x 3 - 45 = 75 - 45 = 30 kNm

2 S.F. Diagram

At f,i = 2.5 and hence The S.F. at A, FA =£A = + 80kN

B.M. at E, M —Re = a x 2.5 • 10 x 2.5 x 2 =; 25 x 2.5 — 5 x 6.25 The S.F. will remain constant between A and C and equal to + 80 kN ;!
!;
25 = 31.25 kNm The S.F. just on R.H.S. of C = RA — 50 = 80 — 50 = 30'kN
The S.F. just on L.H.S. of D = RA - 50 - 10 x 4 = 80 - 50 - 40 = - 10 kN §
B.M. at D, Md = 25(3 + 2) - 10 x 3 x + 2 = 125 - 105 = 20 kNm DThe S.F. between C and varies according to straight line law.
|
The B.M. between AC and between BD varies according to parabolic law. But B.M. The S.F. just on R.H.S. of D =RA - 50- 10 x 4- 40 = 80 - 50 - 40- 40 = - 50 kN
Dbetween C and varies according to straight line law. Now the bending moment diagram is ;ii
The S.F. at B = - 50 kN ja
drawn as shown in Fig. 6.30 (c).
The S.F. remains constant between D and B and equal to - 50 kN ,a
AProblem 6.12. simply supported beam of length 10 m, carries the uniformly distrib- sj
The shear force diagram is drawn as shown in Fig. 6.31 (b). ?l
uted load and two point loads as shown in Fig. 6.31. Draw the S.F. and B.M. diagram for the
beam. Also calculate the maximum bending moment. The shear force is zero at point E between C and D.

R RSol. First calculate the reactions A and B. AELet the distance of from point is x.

Taking moments of all forces about A, we get Now shear force at E = RA - 50 - 10 x (x - 2)

= 80 - 50 - lte + 20 = 50 - lOx

Rb x 10 : 50 X 2 + 10 x 4 x |2 + ~ + 40(2 + 4) But shear force at E = 0

: 100 + 160 + 240 = 500 — 5m50 - lOx = 0 or i = =

10

500 kN B.M. Diagram
B 10
AB.M. at is zero
Ra = Total load on beam - R
;i B.M. at B is zero

= (50 + 10 x 4 + 40) - 50 = 130 - 50 = 80 kN

•

264 STRENGTH OF MATERIALS SHEAR FORCE AND BENDING MOMENT 265

B.M. at C, Mc = ra x 2 = 80 x 2 = 160 kNm

B.M. at D, -- R,A x 6 - 50x4-10x4x—

2
= 80 x 6 - 200 - 80 = 480 - 200 - 80 = 200 kNm

mAt E, x = 5 and hence B.M. at E,

M Fe = a x 5 ~ 50(5 - 2) - 10 x (5 - 2) x j

= 80 x 5 - 50 x 3 - 10 x 3 x | = 400 - 150 - 45 = 205 kNm

DThe B.M. between C and varies according to parabolic law reaching a maximum value
at E. The B.M. between A and C and also between B and D varies according to linear law. The

B.M. diagram is shown in Fig. 6.31 (c).

Maximum B.M.

The maximum B.M. is at E, where S.F. becomes zero after changing its sign.

Max. B.M. - Mg = 205 kNm. Ans.

6.13. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR A SIMPLY SUP-
PORTED BEAM CARRYING A UNIFORMLY VARYING LOAD FROM ZERO AT
EACH END TO w PER UNIT LENGTH AT THE CENTRE

Fig. 6.32 shows a beam of length L simply supported at the ends A and B and carrying a

uniformly varying load from zero at each end to ui per unit length at the centre. The reactions
at the supports will be equal and their magnitude will be half the total load on the entire

length, as the load is symmetrical on the beam.

But total load on the beam = Area of load diagram ABO

AB x CO Lx w w.L

_ 2 -2 or
2

Ra = Rb = Half the total load

_ 1 (w.L'

y2 V. 2

XConsider any section between A and C at a distance x from end A.

XThe rate of loading at

= Vertical distance XD in load diagram

CO XD x x CO 2x . w\

Now load on the length AX of the beam = Area of load diagram AXD

2w

XX
~L

—= Ul .X 2

L

); .

266 STRENGTH OF MATERIALS SHEAR FORCE AND BENDING MOMENT 267

B.M. Diagram

The bending moment is zero at A and B.

XThe B.M. at is given by,

Mx = Ra . x - Load of length AX . —

W.L W X 2 . —% W.L W 3 —W)
L3 -.x
x, x wL
4 4

Equation (ii) shows that B.M. between A and C varies according to cubic law.

MAt A, x = 0 hence x= 0

^ MAt C, x = hence c= 4 it _ -i£-
2
2 3L {2

w.L2 wL2 _ Zw.L2 - wL2 _ wL2

~
8 24"
24 12

The maximum B.M. occurs at the centre of the beam, where S.F. becomes zero after

changing its sign.

.-. Max. B.M. is at C, M,, = wL2
cr
12

But total load on the beam, W=

Max. B.M. wL L w.L

=

~T'6 ~1T'

6.14. SHEAR FORCE AND B.M. DIAGRAMS FOR A SIMPLY SUPPORTED BEAM
CARRYING A UNIFORMLY VARYING LOAD FROM ZERO AT ONE END TO w
PER UNIT LENGTH AT THE OTHER END

Fig. 6.33 shows a beam AB of length L simply supported at the ends A and B and carry-
ing a uniformly varying load from zero at end A to w per unit length at B. First calculate the
R Rreactions A and r .

Taking moments about A, we get

Rs k L = sAUl ATotal load = w.L\ is acting — L from

R,A - Total load on beam - RB = -U-— - = u> ^
:
2
3 6

X XConsider any section at a distance x from end A. The shear force at is given by,

F=R,, ~ loadi on length AX = w.L - w.x x

a -z~- -

Load on AX AX . CX wx x. .

_ wL wx2
~~6~~~2L

Equation (i) shows that S.F. varies according to parabolic law.

|

268 STRENGTH OF MATERIALS SHEAR FORCE AND BENDING MOMENT 269

AEquation (ii) shows the B.M. varies between and B according to cubic law. Now calculate the reactions RA and RB .

Max. B.M. occurs at a point where S.F. becomes zero after changing its sign. Taking the moments about A, we get

- - of

2 f
URBr = x
^That point is at a distance of from A. Hence substituting* = in equation (ii), we x 6 4000 + 2000 x 6
|
J

get maximum B.M. NRb = 2000 + 1333.33 = 3333.33

Max. B.M. = ~w.L~‘ L W Ra = Total load on beam - Ru

6 73 6L N= (4000 + 2000) - 3333.33 = 2666.67

Consider any section X-X at a distance x from A.

_~ wL2 wl? _ 3w.I? - wL 2 __ wl} ' Rate of loading at the section X-X
6V? ” 1873 ~ 1873 973
= Length CE = CD + DE

mProblem 6.13. A simply supported beam of length 5 carries a uniformly increasing = 800 + - x 800 = 800 + 160*

load of 800 N/m run at one end to 1600 N/m run at the other end. Draw the S.F. and B.M. 5
diagrams for the beam. Also calculate the position and magnitude ofmaximum bending moment.
Total load on length AX
Sol. The loading on the beam is shown in Fig. 6.34. The load may be assumed to be
consisting of a uniformly distributed load of 800 N/m over the entire span and a gradually = Area of load diagram ACDEF

varying load of zero at A to 800 N/m at B. = Area of rectangle + Area of ADEF
Then load on beam due to uniformly distributed load of 800 N/m = 800 x 5 = 4000 N
= 800 x * + 16°? * * = 800* + 80x 2
NLoad on beam due to triangular loading = x 800 x 5 = 2000
2

Now the S.F. at the section X-X is given by,

Fx = Ra - load on length AX' ...(i)

= 2666.67 - (800* + 8Q* 2) = 2666.67 - 800* - 80* 2

A BEquation (i) shows that shear force varies between and according to parabolic law.

At A, * = 0 hence F = 2666.67 - 800 x 0 - 80 x 0 = + 2666.67 N
At B, x = 5 hence x

F = 2666.67 - 800 x 5 - 80 x 52
x

= 2666.67 - 4000 - 2000 = - 3333.33 N

Let us find the position of zero shear. Equating the S.F. equal to zero in equation (i), we get
0 = 2666.67 - 800* - 80*2

2666.67 *2 + 10* - 33.33 = 0
*2 + 10*

The above equation is a quadratic equation. Its solution is given by,
- 10 ± 7l0 2 + 4 x 33.33 - 10 ± 7233.33

_ - 10 + 15.274 (Neglecting - ve root)

2

= 2.637 m

B.M. Diagram
The B.M. at the section X-X is given by

B.M. diagram Mx =Ra xx- 800 x * x — -.*.160*.
Fig. 6.34 ZJ
23

= 2666.67* -400*2 - 80 3

~~x

o

270 STRENGTH OF MATERIALS ) 271

SHEAR FORCE AND BENDING MOMENT

Equation (ii) shows that B.M. between A and B varies according to cubic law. ' RBr ~ =9 kN

At A,x = 0, Mx = 0 and 4

At B, x = 5, =0 Ra = Total load - RB =2x6-9=3 kN

mMaximum B.M. occurs where S.F. is zero. But S.F. is zero at a distance of 2.637 Shear Force Diagram
mfrom A. Hence maximum B.M. is obtained by substituting x = 2.637 in equation (ii).
Shear force at A = + RA = + 3 kN
—Oft
Max. B.M. = 2666.67 x 2.637 - 400 x 2.637 z - x 2.637 3 = 3761.5 Nm. Ans. A A(i) The shear force at any section between and B at a distance x from is given by,

u FA ^RA -2x (y Ra = 3)

6.15. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR OVER-HANGING = 3-2x ...(£)

BEAMS At A, x = 0 hence FA = 3 kN
At B, x = 4 hence Fs = 3- 2x4 = -5kN
If the end portion of a beam is extended beyond the support, such beam is known as The shear force varies according to straight line law between A and B. At A, the shear
overhanging beam. In case of overhanging beams, the B.M. is positive between the two sup- Aforce is positive whereas at B, the shear force is negative. Between and B somewhere S.F. is
ports, whereas the B.M. is negative for the over-hanging portion. Hence at some point, the
B.M. is zero after changing its sign from positive to negative or vice-versa. That point is known F;ero. The point, where S.F. is zero, is obtained by substituting x = 0 in equation (i).

as the point of contraflexure or point of inflexion. m0 = 3 - 2x or x = —g = 1.5

6.15.1. Point of Contraflexure. It is the point where the B.M. is zero after changing 2

its sign from positive to negative or vice-versa. m AHence S.F. is zero at a distance of 1.5 from (or S.F. is zero at point D).

Problem 6.14. Draw the shear force and bending moment diagrams for the over-hanging (ii) The S.F. at any section between B and C at a distance x from A is given by,
beam carrying uniformly distributed load of2 kN/m over the entire length as shown in Fig. 6.35.
R RFx = + a - 4 x 2 + B - (x - 4) x 2 = 3 - 8 + 9 - 2frc - 4)
Also locate the point of contraflexure.
= 4-2(*-4) ...(ii)
R RSol. First calculate the reactions A and B
AtB,r = 4m hence FB = 4 - 2(4 - 4) = + 4 kN
Taking moments of all forces about A, we get mAt C, * = 6 hence Fc = 4 — 2(6 - 4) = 0

— (v Total load on beam = 2 x 6 = 12 kN. This Between B and C also S.F. varies by a straight line law. At B, S.F. is + 4 kN and at C,
mR„Bx4= 2 x 6 x = 36
2
load is acting at a distance 3 from A) S.F. is zero.

The S.F. diagram is shown in Fig. 6.35 (b).

B.M. Diagram

AThe B.M. at is zero.
A( i The B.M. at any section between and B at a distance x is given by,

Mx = Ra x x - 2 x x x —X

= 3x -x2 ...(Hi)

At A, x = 0 hence MA = 0
At B, x = 4 hence MB = 3x4-42 = -4 kNm

Max. B.M. occurs at D, where S.F. is zero after changing its sign.

M kNmAt D,x = 1.5 hence D - 3 x 1.5 - 1.5 = 4.5 - 2.25 = 2.25

The B.M, between A and B varies according to parabolic law.

(ii) The B.M. at any section between B and C at a distance x is given by,

Mx = Ra x x - 2 x x x ^ + Rb x (x - 4)

= 3jc-x2 + 9(x-4) ...( iv )

At B, x = 4 hence MB - 3 x 4 - 42 + 9(4 - 4) = 4 kNm
At C, x = 6 hence
M^
c = 3 x 6 - 6 2 + 9(6 - 4) = 18 - 36 + 18 = 0

The B.M. diagram is shown in Fig. 6.35 (c).

F

STRENGTH OF MATERIALS SHEAR FORCE AND BENDING MOMENT
272

Point of Contraflexure S.F. Diagram

This point will be between A and B where B.M. is zero after changing its sign. But B.M. AS.F. at = + = + 2 kN
AAat any section at a distance x from between and B is given by equation (iii) as A A(i) The S.F. at any section between and B at a distance x from is given by,

M = 3x - x2 F = + Ra - 2 x x
x x

MEquation x to zero for point of contraflexure, we get = 2 - 2x -U)

0 = 3x- x2 = x(3 - x) At A, x = 0 hence FA = 2-2x0 = 2 kN
At B, x = 4 hence FA = 2~2x4 = - 6 kN
or 3 _ x = o x cannot be zero as B.M. is not

changing sign at this point) AThe S.F. between and B varies according to straight line law. At A, S.F. is positive
Aand at B, S.F. is negative. Hence between and B, S.F. is zero. The point of zero S.F. is
x=3

mHence point of contraflexure will be at a distance of 3 from A. Fobtained by substituting x = 0 in equation (t).

Problem 6.15. Draw the S.F. and B.M. diagrams for the overhanging beam carrying m0 = 2 - 2x or x = —2 = 1

uniformly distributed load of 2 kN/m over the entire length and a point load of 2 kN as shown 2

in Fig. 6.36. Locate the point of contraflexure. The S.F. is zero at point D. Hence distance of D from A is 1 m.
U( ) The S.F. at any section between B and C at a distance x from A is given by,
R RSol. First calculate the reactions A and B .
Fx = + Ra - 2 x 4 + Rb - 2(x - 4)
Taking moments of all forces about A, we get
= 2 - 8 + 12 - 2(x - 4) = 6 - 2(x - 4)
Rb x 4 = 2x6x3 + 2x6 = 36 + 12 = 48 ...(«)

—= = 12 kN At B, x = 4 hence FB = 6 - 2{4 - 4) = + 6 kN
At C, x = 6 hence Fc = 6 - 2(6 - 4) = 6 - 4 = 2 kN
B
4

and RA = Total load- B = (2 x 6 + 2)- 12 = 2 kN The S.F. diagram is drawn as shown in Fig. 6.36 6( ).

B.M. Diagram

B.M. at A is zero
A(i) B.M. at any section between A and B at a distance x from is given by,

MX =F,a xx~ 2 xxx — = 2x-x 2 ...(Hi)

2

The above equation shows that the B.M. between A and B varies according to parabolic

law.

At A, x = 0 hence MA = 0
At B, x = 4 hence MB = 2x4-42 = -8 kNm

DMax. B.M. is at where S.F. is zero after changing sign

MAt D, x - 1 hence D = 2xl-l2 = l kNm

The B.M. at C is zero. The B.M. also varies between B and C according to parabolic law.

Now the B.M. diagram is drawn as shown in Fig. 6.36 (c).

Fig. 6.36 Point of Contraflexure

AThis point is at E between and B, where B.M. is zero after changing its sign. The

E MAdistance of from is obtained by putting x = 0 in equation (iii).

0 = 2x - x2 = x (2 - x)

2-x = 0

and x = 2m. Ans.

A m mProblem 6.16. beam of length 12 is simply supported at two supports which are 8
mapart, with an overhang of 2 on each side as shown in Fig. 6.37. The beam carries a concen-
Ntrated load of 1000 at each end. Craw S.F. and B.M. diagrams.

274 STRENGTH OF MATERIALS SHEAR FORCE AND BENDING MOMENT 275

Sol. As the loading on the beam is symmetrical. Hence reactions RA andf?B will be equal Problem 6.17. Draw the S.F. and B.M. diagrams for the beam which is loaded as shown
in Fig. 6.38. Determine the points of contraflexure within the span AB.
and their magnitude will be half of the total load.
R RSol. First calculate the reactions A and B .

Taking moments about A, we have

RB x 8 + 800 x 3 = 2000 x 5 + 1000(8 + 2)

or 8RB + 2400 = 10000 + 10000

B 20000 - 2400 ^ 17600 = 22QQ N

gg

and Ra = Total load ~RB = 3800 - 2200 = 1600

B.M. diagram

Fig. 6.37

S.P. at C = - 1000 N

AS.F. remains constant (i.e., = - 1000 N) between C and

AS.P. at =- 1000 + Ra = - 1000 + 1000 = 0

S.F. remains constant (i.e., = 0) between A and B

S.F. at B = 0 + 1000 = + 1000 N

S.F. remains constant (i.e., = 1000 N) between B and D

S.F. diagrams is drawn as shown in Fig. 6.37 b( ). Fig. 6.38

B.M. Diagram S.F. Diagram

B.M. at C = 0 S.F at C = - 800 N

B.M. at A = - 1000 x 2 = - 2000 Nm (- ve sign is due to hogging moment) S.F. between C and A remains - 800 N

B.M. between C and A varies according to straight line law. AS.F. at = - 800 + RA = - 800 + 1600 = + 800 N

The B.M. at any section in AB at a distance of x from C is given by, S.F. between A and D remains + 800 N

Mx = - 1000 x x + Af? (a; - 2) DS.F. at = + 800 - 2000 = - 1200 N

Nm= - 1000 x jc + 100G(x - 2) = - 2000 S.F. between D and B remains - 1200 N

Hence B.M. between A and B is constant and equal to - 2000 Nm. S.F. at B = - 1200 + RB = - 1200 + 2200 = + 1000 N

B.M. atD = 0. S.F. between B and E remains + 1000 N

B.M. diagram is shown in Fig. 6.37 (c). S.F. diagram is shown in Fig. 6.38.

Note. In this particular case, the S.F. is zero between AB and B.M. is constant. Hence length AB
is subjected to only constant B.M. The length between A and B is absolutely free from shear force.

R-

276 STRENGTH OF MATERIALS

B.M. Diagram

B.M. at C =0
B.M. at A
= - 800 x 3 = - 2400 Nrn
B.M. at D
= - 800 x (3 + 5) + RA x 6
B.M. at B
B.M. at £ = - 800 x 8 + 1600 x 5

= - 6400 + 8000 = + 1600 Nm
= - 1000 x 2 = - 2000 Nm

• =0

The B.M. diagram is drawn as shown in Fig, 6,38 (c).

Points of Contraflexure

0There will be two points of contraflexure and 0 where B.M. becomes zero after
1 2,
Dchanging A
its sign. Point 0 lies between and D, whereas the point 09 lies between and B.
1

(i) Let the point Oj is x metre from A.

Then B.M. at O = - 800(3 + x) + RA x * = - 800(3 + x) + 1600a:
l
= - 2400 - 800* + 1600x = - 2400 + 800*

But B.M. at O is zero
x

O - - 2400 + 800* * = 2400 = 3„ m. .
(ii) Let the point O hex metre from B.
Ans.

800

Then B.M. at 0 = 1000(x + 22))--RB XX = 1000* + 2000 - 2200 x * = 2000 - 1200*
2 ==00

But B.M. at 0„ O = 2000 - 1200*

m* = - — = 1.67 from B. Ans.
1200 3

mProblem 6.18. A horizontal beam 10 long is carrying a uniformly distributed load of
m1 kN/m. The beam is supported on two supports 6 apart. Find the position of the supports, so

that B.M. on the beam is as small as possible. Also draw the S.F. and B.M. diagrams.

m mSol. The beam CD is 10 long. Let the two supports 6 apart are at A and B.

Let * = Distance of support A from C in metre

Then distance of support B from end D

m= 10 - (6 + x) = (4 - *)

RRFirst calculate the reactions A and H .

Taking moments about A, we get

——lxxx
— + ffB„x6 = (10 - jc) x 1 x ——

2 2

(10 - xY or x2 + 12£ n = (10 x*)2 = 100 +*2 - 20*

+ 6i? n =

12Rb = 100 + x2 - 20* - x2 = 100 - 20x

D 100-20* 4(25 -5x) 1 r, 5 ,
/cr
B 12 3
12 3

278 STRENGTH OF MATERIALS , 279

SHEAR FORCE AND BENDING MOMENT

For the condition that the B.M. shall be as small as possible, the hogging moment at the

support A and the maximum sagging moment in the span AB should be numerically equal.

Equating equations (ii) and (Hi) and ignoring the - ve sign of B.M. at A, we get

—5 (-x2„ + 4x + 5)=±r-2

18 2

or - 5x 2 + 20x + 25 = 2 or 14r2 - 20* - 25 = 0
9jc

The above equation is a quadratic equation. Hence its solution is given by

Xv ~_ 20 ± z + 4 x 14 x 25 _ 20 ± f/t400 + 1400 20 ± 42.42

y]20

2 x 14 28 “ 28

_ 20 ± 42.42 (Neglecting - ve value)
28

m= 2.23

Substituting this value of x in equation (i), we get

y = |o (1 + 2.23)-^!^ = 5.38 m

3

Now the values of reactions RA and Ru are obtained as :

Ra = ^ (1 + x) = | (1 + 2.23) = 5.38 kN

and Rb = ^(5-*) = ~ (5 - 2.23) = 4.62 kN

o 3

Now the S.F. and B.M- diagrams can be drawn as shown in Fig. 6.40.

S.F. Diagram

S.F. at C =0

AS.F. just on L.H.S. of = - 1 x 2.23 = - 2.23 kN Fig. 6.40

AShear force varies between C and by a straight line law. EB.M. at (i.e. mat a distance y = 5.38 from point C)

RAS.F. just on L.H.S. of = - 2.23 + a —= - 1 x 5.38 x 2 + R,A x (5.38 - 2.23)
= -2.23 + 5.38 = + 3.15 kN

S.F. just on L.H.S. of B = + 3.15 - 1 x 6 = - 2.85 kN = - + 5.38 x 3.15 = 2.49 kNm

Shear force between A and B varies by a straight line law. 2

S.F. just on R.H.S. of B = - 2.85 + RB B.M. atB x^=-=- 1 x 1.77 1.06 kNm
A
= - 2.85 + 4.62 = + 1.17 kN
DS.F. at =1.17-1x1.77 = 0 The B.M. between C and A ; between A and B ; and between B and D varies according to

parabolic law. B.M. diagram is shown in Fig. 6.40 (c).

DS.F. between B and varies by a straight line law.

S.F. diagram is drawn as shown in Fig. 6.40 (c). 6.16. S.F. AND B.M. DIAGRAMS FOR BEAMS CARRYING INCLINED LOAD

B.M. Diagram. =0 The shear force is defined as the algebraic sum of the vertical forces at any section of a
beam to the right or left of the section. But when a beam carries inclined loads, then these
B.M. at C
inclined loads are resolved into their vertical and horizontal components. The vertical compo-
B.M. at A = - 2.49 kNm nents only will cause shear force and bending moments.

280 STRENGTH OF MATERIALS SHEAR FORCE AND BENDING MOMENT 281
.

The horizontal components of the inclined loads will introduce axial force or thrust in DSimilarly the inclined load at is having horizontal component
the beam. The variation of axial force for all sections of the beam can be shown by a diagram
= 200 x cos 45° = 141.4 N,
known as thrust diagram or axial force diagram.
Nwhereas the vertical component = 200 x sin 45° = 141.4
In most of the cases, one end of the beam is hinged and the other end is supported on The inclined load at B is having horizontal component
rollers. The roller support cannot provide any horizontal reaction. Hence only the hinged end
= 300 x cos 30° = 300 x 0.866 = 259.8 N,
will provide the horizontal reaction.
Nwhereas the vertical component = 300 x sin 30° = 150
mProblem 6.19. A horizontal beam AB of length 4 is hinged at A and supported on
N Nrollers at B. The beam carries inclined loads of 100 N, 200 and 300 inclined at 60°, 45° and The horizontal and vertical components of all inclined loads are shown in Fig. 6.41 (5).

30° to the horizontal as shown in Fig. 6.41. Draw the S.F., B.M. and thrust diagrams for the As beam is supported on rollers at B, hence roller support at B will not provide any

beam. horizontal reaction. The horizontal reaction will be only provided by hinged end A.

Sol. First of all, resolve the inclined loads into their vertical and horizontal components. HLet a = Horizontal reaction at A

The inclined load at C is having horizontal component - Sum of all horizontal components of inclined loads

whereas the vertical component = 100 x cos 60° = 100 x 0.5 = 50 N, = 50 + 141.4 + 259.8

N= 100 x sin 60° = 100 x 0.866 = 86.6 (All horizontal components are acting in the same direction)

= 451.20 N
R ARTo find the reactions A and B, take the moments of all forces about ,

Rb x 4 = 86.6 x 1 + 141.4 x 2 + 150 x 3 = 819.4

or R„ = = 204.85 N

R Ra - Total vertical load - B
N’ = (86.6 + 141.4 + 150) - 204.85 = 173.15

86.6 N 141.4 N 150 N S.F. Diagram

The S.F. is due to vertical loads (including vertical reactions) only

AS.F. at = + Ra = + 173.15 N

NS.F. remains constant between A and C and equal to 173.15

S.F. suddenly changes at C due to point load and S.F. at C

= 173.15 - 86.6 = 86.55 N

D NS.F. remains constant between C and and is equal to 86.55

DS.F. at - 86.55 - 141.40 = - 54.85 N

The S.F. remains constant between E and D and is equal to - 54.85 N

The S.F. at B = - 54.85 - 150.00 = - 204.85 N

The S.F. diagram is shown in Fig. 6.41 (c).

B.M. Diagram

The B.M. is only due to vertical loads (including vertical reactions) only

The B.M. at A =0
B.M. at C
B.M. at D NmR= a x 1 = 173.15 x 1 = 173.15

= Ra x 2 - 86.6 x 1

Nm= 173.15 x 2 - 86.6 = 259.7

B.M. at E =i?A x 3- 86.6x2- 141.4x1

Nm= 173.15 x 3 - 86.6 x 2 - 141.4 = + 204.85

B.M. at B =0

The B.M. diagram is shown in Fig. 6.41 d( ).

Pig. 6.41

SHEAR FORCE AND BENDING MOMENT 283

282 STRENGTH OF MATERIALS

Thrust Diagram or Axial Force Diagram

The thrust diagram is due to horizontal components including horizontal reaction.

AAxial force at H= + A = 451.20 N

NThe axial force remains constant between A and C and is equal to 451.20

Axial force at C H= NA - 50 = 451.20 - 50 = 401.2

NAxial force remains constant between C and D and is equal to 401.2

DAxial force at N= 401.2 - 141.40 = 259.8

NDAxial force remains constant between and E and is equal to 259.8

EAxial force at = 259.8 - 259.8 = 0

Axial force between E and B is zero.

Thrust diagram or axial force diagram is shown in Fig. 6.41 (e). A CDE B

Problem 6.20. A horizontal beam AB of length 8 mis hinged at A and placed on rollers Axial force diagram

at B. The beam carries three inclined point loads as shown in Fig. 6.42. Draw the S.F., B.M.
and axial force diagrams of the beam.

Sol. First resolve the inclined loads into their vertical and horizontal components.

Vertical component of force at C
= 4 sin 30° = 4 x 0.5 = 2 kN

Horizontal component of force at C

—= 4 x cos 30° = 4 x 0.866 = 3.464 kN

DVertical component of force at

= 8 x sin 60° = 8 x 0.866 = 6.928 kN

DHorizontal component of force at
= 8 x cos 60° = 8 x 0.5 = 4 kN <—

Vertical component of force at E
= 6 x sin 45° = 6 x 0.707 = 4.242 kN

Horizontal component of force at E
= 6 x cos 45° = 6 x 0.707 = 4.242 kN

The horizontal and vertical components of all inclined loads are shown in Fig. 6.42 (5).

The horizontal reaction will be provided by the hinged end A.

.'. Horizontal reaction at A,

Ha = - 3.464 + 4 + 4.242 = 4.778 kN Fig. 6.42

RTo find vertical reactions RA and B , take the moments of all forces about A. S.F. remains 6.0245 kN between A and C
Rb x 8 = 2 x 2 + 6.928 x 4 + 4.242 x 6 = 57.164
S.F. at C = + 6.0245 - 2 = + 4.0245 kN

_ 57.164 = . S.F. remains 4.0245 kN between C and D
8
RBr = 7. 1455 kN DS.F. at = + 4.0245 - 6.928 = - 2.9035 kN

Now R Ra = Total vertical loads - B S.F. remains - 2.9035 kN between D and E

= (2 + 6.928 + 4.242) - 7.1455 = 6.0245 kN S.F. at E = - 2.9035 - 4.242 = - 7.1455 kN

S.F. Diagram S.F. remains constant between E and B and equal to - 7.1455

S.F. is due to vertical loads S.F. diagram is shown in Fig. 6.42 (c).

AS.F. at = + Ra = + 6.0245 kN

284 STRENGTH OF MATERIALS SHEAR FORCE AND BENDING MOMENT 285

B.M. Diagram S.F. Diagram

B.M. is only due to vertical loads S.F. atA =NA =-4kN

B.M. at A =0 AThe S.F. remains constant ( i.e ., equal to - 4 kN) between and B.

B.M. at C =fiA x2 = 6.0245 x 2 = 12.049 kNm The S.F. diagram is shown in Fig. 6.43 (c).
B.M. at D = 6.0245 x 4 - 2 x 2 = 20.098 kNm

B.M. at E = 6.0245 x 6 - 2 x 4 - 6.928 x 2 = 14.291 kNm

B.M at B =0

B.M. diagram is shown in Fig. 6.42 (d).

Axial Force Diagram

Axial force is due to horizontal components including horizontal reaction.

Axial force at A H= + A = + 4.778 kN

Axial force remains 4.778 kN between A and C

Axial force at C = + 4.778 + 3.464 = + 8.242

Axial force remains 8.242 kN between C and D

DAxial force at = 8.242 - 4.0 = + 4.242

Axial force remains 4.242 kN between D and E

Axial force at E = + 4.242-4.242 = 0

Axial force remains zero between E and B

Axial force diagram is shown in Fig. 6.42 (e).

6.17. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR BEAMS SUB- j

JECTED TO COUPLES

When a beam is subjected to a couple at a section, only the bending moment at the Fig. 6.43

section of the couple changes suddenly in magnitude equal to that of the couple. But the S.F.

does not change at the section of the couple as there is no change in load due to couple at the i
section. But while calculating the reactions, the magnitude of the couple is taken into account.
B.M. Diagram

The sudden change in B.M. at the section of the couple can also be obtained by calculat- I B.M. atA =0

ing B.M. separately with the help of both the reactions. B.M. just on the L.H.S. ofC =i?A x2 = -4x2=-8 kNm

mProblem 6.21. A simply supported beam AB of length 6 Ais hinged at and B. It is B.M. just on the R.H.S. of C =i?s x4 = 4x4 = + 16 kNm
{B.M. just on the R.H.S. of C can also be calculated as the sum of moments due to RA
mkNmsubjected to a clockwise couple of24
at a distance of2 from the left end A. Draw the S.F.

and B.M. diagrams. and moment due to couple. But moment due to RA is anti-clockwise whereas due to couple is

Sol. Fig. 6.43 (a) shows the simply supported beamAB, hinged atA andB. The clockwise clockwise. Hence net B.M. on R.H.S. of C — — 8 + 24 = + 16 kNm).

couple at C will try to lift the beam up at the support A, and to depress the beam down at the There is a sudden change in B.M. at C due to couple.

supports. Hence the reaction atA will be downwards and atS the reaction will be upwards as B.M. at B =0

shown in Fig. 6.43 (6). B.M. diagram is shown in Fig. 6.43 (d).

To find reactions of RA and RB , take the moments about A. mProblem 6.22. A beam 10 long and simply supported at each end, has a uniformly
Rb x 6 - 24 = 0 ('• Moment due to R B is anti-clockwise and moment
Cat is clockwise) distributed load of 1000 N/m extending from the left end upto the centre of the beam. There is
mkNmalso an anti-clockwise couple of 15
at a distance of 2.5 from the right end. Draw the S.F.

R^a = SZ = 4 kN t1 and B.M. diagrams.

6 Sol. The reaction at A will be upwards. To find whether the reaction at B is upwards or
ASince there is no external vertical load on the beam, therefore the reaction at will be
downwards, take the moments about A.
the same, as that of B, but in opposite direction.
The following are the moments at A :
Ra = Load on beam - R„ (v('•' Load on beam = 0)
= - Rb = - 4 kN.
Nm(i) Moment due to U.D.L. = 1000 x 5 x -g = 12500 (clockwise)