Strength of Materials by R.K.Bansal_text

86 STRENGTH OF MATERIALS PRINCIPAL STRESSES AND STRAINS 87

FG.-. Normal stress and tangential stress across the section are obtained as.

J Normal stress, Force normal to section FG
Area of section FG
/
/ F
/
/ (90-0) ^ p P COS0
ar 'i
“0V/

Fig. 3.1 (a) Fig. 3.2 (b ) — —p p

Then area of section, EF = EF x 1 = A. A A= cos 0 . COS 0 = COS2 0

The stress on the section EF is given by : a COS2 0 H("

Force P Tangential stress (i.e., shear stress), P( v P, = sin 0)
—o= =
Area of EF A ...(l) _ Tangential force across section FG
‘ Area of section FG
EFThe stress on the section is entirely normal stress. There is no shear stress (or tan-
P sin 0
gential stress) on the section EF.
ar ^
Now consider a section FG at an angle 6 with the normal cross-section EF as shown in

Fig. 3.1 (a).

Area of section FG = FG x 1 (member is having unit thickness)

v In A EFG,

A sin 0 . cos 0 = a sin 0. cos 0

=- (v £Fxl=A)

cos 0

FG.'. Stress on the section, —= x 2 sin 0 cos 0 [Multiplying and dividing by 2]

PP
—FG A= Force =r* -
Area of section
= y~ CcoOSs 0 = ~ sin 20 (y 2 sin 0 cos 0 = sin 20)
AI
2
j

[cos 9 j From equation (3.2), it is seen that the normal stress (on ) on the section FB will be
maximum, when cos2 0 or cos 0 is maximum. And cos 0 will be maximum when 0 = 0° as
= o cos 0 £fv = oj ...(3.1)
FGcos 0° = 1. But when 0 = 0°, the section
will coincide with section EF. But the section

This stress, on the section FG, is parallel to the axis of the member (i.e., this stress is EF is normal to the line of action of the loading. This means the plane normal to the axis
along .r-axis). This stress may be resolved in two components. One component will be normal
of loading will cany the maximum normal stress.
to the section FG whereas the second component will be along the section FG (i.e., tangential
Maximum normal stress, - o cos2 0 = o cos2 0° = a ...(3.4)
to the section FG). The normal stress and tangential stress (i.e shear stress) on the section
From equation (3.3), it is observed that the tangential stress (i.e., shear stress) across
FG are obtained as given below [Refer to Fig. 3.1 (6)].
the section FG will be maximum when sin 20 is maximum. And sin 20 will be maximum when
Let Pn - The component of the force P, normal to section FG
sin 20 = 1 or 20 = 90° or 270°
= P COS 0
or 0 - -td° or 135°.

P = The component of force P, along the surface of the section FG (or tangential This means the shear stress will be maximum on two planes inclined at 45° and 135° to
t the normal section EF as shown in Figs. 3.1 (c) and 3.1 d( ).

to the surface FG)

= P sin 0

a = Normal stress across the section FG ^.•. Max. value of shear stress = w sin 20 = ^m sin 90° = Z ...(3.5)
n

a = Tangential stress (i.e., shear stress) across the section FG.
t

88 STRENGTH OF MATERIALS

First plane of maximum a = Shear stress (or tangential stress) across the section FC.
shear stress 0 = 45b t

aJ,n - Force normal to section FC

Then normal stress, Area of section FC

_ Pn ( bar is of unit thickness)

~ FCxl

oPC_ x x cos 6 (v p = a x BC x cos 0)
( x

FC

second plane of maximum = X cos 0 x cos 0 --In triangle FBC, = cos 0j
shear stress 0 = 1 35°

Fig. 3.1 (c) Fig. 3.1 (d) = o x cos 2 0 ...(3.5A)
x
From equations (3.4) and (3.5) it is seen that maximum normal stress is equal to a
whereas the maximum shear stress is equal to a/2 or equal to half the value of greatest normal Similarly, tangential (or shear) stress.

Force along section FC _ P
;

stress. ° Area of section FC FC x 1
f

Second Method c, x BC x lx sin 0 (v P^o^BCx 1)

A member subjected to a direct stress in one plane. Fig. 3.2 shows a rectangular =

Amember of uniform cross-sectional area and of unit thickness. The bar is subjected to a FC

ADprincipal tensile stress o on the faces and BC. = cos 0
ECD a: x cos 0 x sin 0 In triangle FBC,
x l

= Oj. x cos B x sin 0 (Multiplying and dividing by two)

~= x 2 x cos 0 x sin 0

= x sin 20 ...(3.5B) <(v 2 sin 0 cos 0 = sin 20)

A FB EmcoaFsx0ii°mFnu=ormr1o,mmawBlehuqtetuonawtthcihoeoesnn2l(i003n.eo=5rAo)f0c,°o,msiottth9iueoninssseeomecfatntixhotienhmalFutomCat.dhiweniAglnnl.odrcTomhciaionslscismd0teerwaeiwnslislsth(btoeh„s)eemcoaptnilxoaitnnhmeeEuFnsm.oercwBtmhiuaoetlnnttPhoCethwsiee0lcltaixboiens

Fig. 3.2

Area of cross-section = BC x Thickness of bar

= BC x 1 of loading will carry the maximum normal stress.

Let the stresses on the oblique plane FC me to be calculated. The plane FC is inclined at the secFtiroonmMFaeCxquiwaimtluliombnen(mo3r.a5mxBa)il,misuttmriseswoshb=esen0rjvsiecndos2t20h0aits=tmohaexxctioamsnugme0n.t=iAanoldxstsrienss20(iw.iellsbheeamrasxtiremsus)•m••awcross

an angle 0 with the normal cross-section EF (or BC). This can be done by converting the stress

BCo acting on face into equivalent force. Then this force will be resolved along the inclined

planes FC and perpendicular to FC. (Please note that it is force and not the stress which is to sin 20 = 1 or 26 = 90“ or 270° or 0 = 45° or 135
This means the shear stress will be maximum on two planes inclined at 45
be resolved). and 135 to

Tensile stress on face BC = 0 BCthe normal section EF or as shown in Figs. 3.2 (a) and 3.2 (6).
,
Second plane of
Now, the tensile force on BC,

P, = Stress (a, ) x Area of cross-section (v Area = BC x 1) ' maximum shear
stress, 0 = 135°
= o x BC x 1 First plane ot maximum
1 \ NP13 3
shear stress, 6 = 45° 1
The above tensile force P is also acting on the inclined section FC, in the axial direction
1
as shown in Fig. 3.2. This force is resolved into two component, i.e., one normal to the plane
P
1

FC and other along the plane FC. FC \\K —

Let P = Component of the force Pv normal to the section A 135°V.
n

- P cos 0 ( Pi - Oj x BC x 1)
1

= Oj x BC x 1 x cos 0

P = Component of the force P, , along the section FC
t

= P sin 0 Fig. 3.2
l

- a x BC x 1 x sin 0
x

o = Normal stress on the section FC
n.

.

90 STRENGTH OF MATERIALS PRINCIPAL STRESSES AND STRAINS 91

Max. value of shear stress = -^ — —sin 20 = - sin 90° = ...{3.5 D) —Dn, = =640000 = 1567
Z x jt x 65
From equations (3.5C) and (3.5D) it is seen that maximum normal stress is equal to a, D = 39.58 nun. Ans.

mmProblem 3.3. A rectangular bar of cross-sectional area of 11000 2 is subjected to a
tensile load P as shown in Fig. 3.3. The permissible normal and shear stresses on the oblique
whereas the maximum shear stress is equal to or equal to half the value of greatest normal plane BC are given as 7 N/mm2 and 3.5 N/mm2 respectively. Determine the safe value of P.

stress. Sol. Given :

Note. It is the force which is resolved in two components. The stress is not resolved. mmAArea of cross-section, = 11000 2 c

mmProblem 3.1. Arectangular bar of cross-sectional area 10000 2 is subjected to an Normal stress, o = 7 N/mm2 ”7
n P.
axial load of 20 kN. Determine the normal and shear stresses on a section which is inclined at p
o = 3.5 N/mm2 +/
Shear stress, t
/A 60°
an angle of 30° with normal cross-section of the bar. Angle of oblique plane with the axis of bar = 60°.

Sol. Given : Angle of oblique plane BC with the normal cross- B

Cross-sectional area of the rectangular bar, section of the bar,

A = 10000 mm2 p-jg g g

Axial load, P = 20 kN = 20,000 N 6 = 90° - 60° = 30°

Angle of oblique plane with the normal cross-section of the bar, Lot p = Safe value of axial pull

0 = 30° a = Safe stress in the member.

Now. . , . ... =—o P = 20000 = „ .N./. mm2 Using equation (3.2),
direct stress, 10000 2
A
cr = 0 cos2 0 or 7 = o cos2 30°
Let on = Normal stress on the oblique plane n

a = Shear stress on the oblique plane. = a (0.866)2 (v cos 30° = 0.866)
t .

Using equation (3.2) for normal stress, we get a ~ 0O.R8B6S6 x^ n0.r8k6r6 = 9-334 N/mm2

on = o cos2 0 (•/ a = 2 N/mm 2 Using equation (3.3),
= 2 x cos2 30°
)

= 2 x 0.8662 (v cos 30° = 0.866) a = — sin 20
12
= 1.5 N/mm2. Ans.

Using equation (3.3) for shear stress, we get

*2—a. ~ sin 20 = — x sin (2 x 30°) ~ —or 3.5 = “ sin 2 x 30° = ^2 sin 60° = 2 x 0.866
2
= 1 X sin 60° = 0.866 N/mm2 . Ans. 083NWa= 3.5 x 2
= 8-
Problem 3.2. Find the diameter of a circular bar which is subjected to an axial pull of O86(r -
.

N/mm160 kN, if the maximum allowable shear stress on any section is 65 2 N/mmThe safe stress is the least of the two, i.e., 8.083 2

. .

Sol. Given : Safe value of axial pull,
Axial pull,
P = 160 kN = 160000 N P = Safe stress x Area of cross-section

Maximum shear stress = 65 N/mm2 = 8.083 x 11000 = 88913 N = 88.913 kN.

Let D = Diameter of the bar Problem 3.4. Two wooden pieces 10 cm x 10 cm

= — D2 ABin cross-section are glued together along line as

.'. Area of the bar shown in Fig. 3.3(a) below. What maximum axial *p P

640000 N/mm2 force P can be applied ifthe allowable shearing stress ^/Vf3s0i°
^
Direct stress. alongABis 1.2 N/mm2 ? (AMIE, Summer 1987)

Maximum shear stress is given by equation (3.5). Sol. Given ;
:
= 10 x 10 = 100 cm2
Area of cross-section

.•. Maxi. mum shear stress = —a = 6400—00 mm mm= 100 x 100
D2 2 x re 2 2 = 10000 2

But maximum shear stress is given as = 65 N/mm2 . Allowable shear stress, a = 1.2 N/mm2

t

Hence equating the two values of maximum shear, we get ABAngle of. line with the axis of axial force = 30°

6„5„ — 640000 Angle of line AB with the normal cross-section,

•- 2 x kD92 0 = 90° - 30° = 60°

92 93

Pacting P
P = Maximum axial force The tensile forces and 2 are also acting on the oblique section FC. The force P is
ct = Maximum allowable stress in the direction of P. - acting l
in the axial 1 whereas the force P2 is
Using equation (3.3), downwards as shown in Fig. 3.4 (a).
direction,

Two forces P and P each can be resolved into two components i.e., one normal to the plane FC
t 2

and other along the plane FC. The components of P are P cos 0 normal to the plane FC and
1l

AJ of P
—<x = . Pj sin 0 along the plane in the upward direction. The components 2 are P sin 0 normal to
2
sin 20

'2 the plane FC and P cos 0 along the plane in the downward direction.
2

—1.2 = x sin (2 x 60°) = x sin 120° Let P = Total force normal to section FC
^2 2 n

1-2 = Component of force P normal to section FC
i
a=
sin 120 = 2.771 N/mm2 + Component of force P normal to section FC
2
0.866
= P cos 0 + P sin 0
Maximum axial force, l2
P = Stress in the direction of P x Area of cross-section
= a, x BC x cos 0+ o x BF x sin 0 (v P = a x BC, P = o x BF)
N= 0 x 10000 = 2.771. x 10000 = 27710 = 27.71 kN. 2 1 t 2 2

Ans. P = Total force along the section FC
t

= Component of force Pj along the section FC

A3.4.2. Member Subjected to like Direct Stresses in two Mutually Perpendicu- + Component of force P along the section FC
2
lar Directions. Fig. 3.4 (o) shows a rectangular bar ABCD of uniform cross-sectional area A
P= Pj sin 0 + (- 2 cos 0) (-ve sign is taken due to opposite direction)
and of unit thickness. The bar is subjected to two direct tensile stresses (or two-principal ten- P= Pjj sin 0 - cos 0

sile stresses) as shown in Fig. 3.4 (a). 2

= o x BC x sin 6 - a x BF x cos 0
t2
(Substituting the values P, and P2 )

on = Normal stress across the section FC
_ Total force normal to the section FC
Area of section FC

P o t x BC x cos 0 + o 2 x BF x sin 0

=n
FC x 1 "
FC

BC— BF—
— —= a, ._
x x cos O0 + Oo x
1 FC 2 FC x sin 0

= x cos 0 x cos 0 + 0 x sin 0 x sin 0
2

Fig. 3.4 (a) v In triangle FBC, BC BF ^
FC = cos 0, FC sin 0
Let FC be the oblique section on which stresses are to be calculated. This can be done by (
)
= 0 cos 2 0 + 0 sin2 0
BCconverting the stresses a (acting on face and o (acting on face AB) into equivalent forces. j2
l 2
)

Then these forces will be resolved along the inclined plane FC and perpendicular to FC. Con- _ a ri + c°s20V +a fl-cos26V*

sider the forces acting on wedge FBC. [v cos2 0 = (1 + cos 20)/2 and sin 2 8 = (1 - cos 20)/2]

Let 6 - Angle made by oblique section FC with normal cross-section BC ^ ^_^= „0J1 + 04 + 01 —

a, = Major tensile stress on face AD and BC _

AB CD cos29

a = Minor tensile stress on face and
2
0 = Tangential stress (or shear stress) along section FC
P = Tensile force on face BC (

1

——— — — FC—
P = Tensile force on face FB. Total force alon•g—the section (0
2
- . I bl

The tensile force on face BC, Area of section FC l

P = x Area of face BC - 0 x BC x 1 ( v Area = BC x 1) * cos 20 = cos 2 0 - sin2 0 ** cos 20 = cos2 0 - sin 2 0
1j = cos 2 0 - (1 - cos2 0) = 2 cos 2 0 — 1 = (1 - sin2 0) - sin2 8 ;
(1 + cos 20)
The tensile force on face FB,

P, = Stress on FB x Area of FB = o x FB x 1. e=. , (1 - cos 20)
2
sin

2

: 95

STRENGTH OF MATERIALS principal stresses and strains
94

P q, x BC x sin 6 - a 9 x BF x cos 6 Principal planes. Principal planes are the planes on which shear stress is zero. To
FC locate the position of principal planes, the shear stress given by equation (3.7) should be equated
= t= r
to zero.
FC x 1
.-. For principal planes,
— —BC= Cj x
BFx . „ — sin 20 = 0
. - cr2 x
0 x cos 6
sin

= a x cos 0 x sin 6- ct x sin 6 x cos 0
x 2
sin 20 = 0 - a
—BC BF— = sin 0 [v (^ 2 ) cannot be equal to zero]
, = cos 0,
v In triangle FBC, ^ 20 = 0 or 180°

= (a, - o cos 0 sin 0 0 = 0 or 90°
2
)

—q„ = - q2
= (gl ~ 0 } x 2 cos 6 sin 0 (Multiplying and dividing by 2) when 0 = 0, q,-+ q~2 + q, cos 26
? -
...(3.7) 2
2
^= — ~
sin 20 = cos0°
2 2
2
g + °2 P ~ q2
The resultant stress on the section FC will be given as l 1

= , , ( v cos 0° = 1)

= V°»2 + 2 22

°i

Obliquity [Refer to Pig. 3.4 (6)3. The angle made D

by the resultant stress with the normal of the oblique plane, when 0 = 90° 21M= qI± o1 + qi -a2 cqs2x! 2
is known as obliquity. It is denoted by <j>. Mathematically,

~tan =T<[> = + cos 180°
r-t
...[3.8 (A)] 22

= x(- d (v cos 180° = - 1)
2
Maximum shear stress. The shear stress is given ^ fb 2

by equation (3.7). The shear stress (a ) will be maximum Fig. 3.4 (6)
(

when Note. The relations, given by equations (3.6) to (3.9), also hold good when one or both the stresses

sin 20 = 1 or 20 = 90° or 270° (v sin 90° = 1 and also sin 270° = 1) are compressive.

or 0 = 45° or 135° Problem 3.5. The tensile stresses at a point across two mutually perpendicular planes

are 120 N/mm2 and 60 N/mm2 Determine the normal, tangential and resultant stresses on a
.
—And
maximum shear stress, (o = 1 — ...(3.9) plane inclined at 30° to the axis of the minor stress.
t
)max „

The planes of maximum shear stress are obtained by making an angle of 45° and 135° Sol. Given
with the plane BC (at any point on the plane BC) in such a way that the planes of maximum
Major principal stress, = 120 N/mm2
shear stress lie within the material as shown in Fig. 3.4 (c).
Minor principal, q = 60 N/mm2
2

Plane of maximum shear stress Angle of oblique plane with the axis of minor principal stress,

/C XC 0 = 30°.

Y ^135 a Normal stress

The normal stress (q ) is given by equation (3.6),

q _ q + q2 + q ~ q2 cqs28
l l

n2 2

Fig. 3.4 (c) 120 + 60 + 120-60 cos 2„ x 3„„0„

Hence the planes, which are at an angle of 45° or 135° with the normal cross-section BC 22
[see Fig. 3.4 (c)], carry the maximum shear stresses.
= 90 + 30 cos 60° = 90 + 30 x T

= 105 N/mm2. Ans.

)

STRENGTH OF MATERIALS principal stresses and strains 97

Tangential stress o2 = 60 N/mm Using equation (3.8) for resultant stress,

The tangential (or shear stress) o is given Or = jo n Z +O 2 -
( t

by equation (3.7).

———a = - = 715625 + 16874 = 180.27 N/mm2. Ans.

f2 sin 20 Axis Of / The inclination of the resultant stress with the normal of the inclined plane is given by

minor stress equation [3.8 (A)] as

120 - 60 fAxis of
sin (2 x 30°)
major stress

= 30 x sin 60° = 30 x 0.866 (j> = tan-1 1.04 = 46° 6'. Ans.

= 25.98 N/mm2. Ans. ct2 = 60 N/mm Maximum shear stress

Resultant stress Fig. 3.5 Maximum shear stress is given by equation (3.9)

The resultant stress o( R ) is given by equa- 200 - 100) = a 150 N/mm2. Ans.

tion (3.8)

(v o'£)'mmaax , o2 2O 92

°r = \!°n + <V = V 105 2 + 25.98 s Problem 3.7. At a point in a strained material the principal tensile stresses across two

= ^11025 + 674.96 = 108.16 N/mm2. Ans. perpendicular planes, are 80 N/mm2 and 40 N/mm2. Determine normal stress, shear stress and

Problem 3.6. The stresses at a point in a bar are 200 N/mm2 (tensile) and 1 00 N/mm2 the resultant stress on a plane inclined at 20° with the major principal plane. Determine also
the obliquity. What will be the intensity of stress, which acting alone will produce the same
(compressive). Determine the resultant stress in magnitude and direction on a plane inclined at
maximum strain if Poisson’s ratio = -j.
60° to the axis of the major stress. Also determine the maximum intensity of shear stress in the

material at the point. (AMIE, Winter 1984) Sol. Given 2
:
Sol. Given : o = 80 N/mm2 60 N/mm
Major principal stress, y a**
Minor principal stress,
Major principal stress, Oj = 200 N/mm2 a = 40 N/mm2 D c
2
| A
Minor principal stress, a = - 100 N/mm2 The plane CE is inclined at angle 20° with “e
2 Major principal
plane ,
(Minus sign is due to compressive stress) major principal plane (i.e., plane BC). g*
s _§
Angle of the plane, which it makes with the major principal stress = 60° '0=20° -
rg

Angle 6 = 90° - 60° = 30°. —Poisson’s ratio, n -

Resultant stress in magnitude and direction 4

First calculate the normal and tangential stresses. Let a = Normal stress on inclined plane
CE
“22Using equation (3.6) for normal stress, 60 N/mm

-O4j + a, 0,-00 “. '. ‘1“0°“N/4n»“n“““ a = Shear stress and
2. + 1 t
=—a cos 29 = Resultant stress. Fig. 3.7

200 + T(—- 100) + 200 - (- 100) wA ! LL, W Using equation (3.6), we get
|*
22 J-JJ- . >T E n 01+02 0i - 02 cos 20 = 80 + 40 + 80 - 40 cos (2 x 20°)
cos (2 x 30°) g *- E|
fAxis of / MA r* zJj —.
* :H
major stress As°1
= 30( '•'0 § 22 22
”< 7 j~'
) § = 60 + 20 x cos 40° = 75.32 N/mm2. Ans.
/}
-~^-200-100 200 + 100 *.«'

= + cos60° The shear stress is given by equation (3.7)

II I w8-0-.40

f

: 50 + 150 X j cos 60° = ± 00i,--0.o, . „_„_ . (2 x 20°) = 20 sin 40°
.
sin
1 00 N/mm
= 50 + 75 = 125 N/mm2 = 12.865 N/mm2. Ans.
. Fig. 3.6

Using equation (3.7) for tangential stress, The resultant stress is given by equation (3.8)

200 (~100) = 2 +a 2
2. gin 20 = sin (2 x 30°) l
°fl \°n
2

200 + 100 N/mm= 775.32 s + 12.856 s = 76.4 2 Ans.

sin 60° = 150 x 0.866 = 129.9 N/mm2 . .

2

: a

98 STRENGTH OF MATERIALS 99
Obliquity (<j>) is given by equation [3.8 (A)]
principal stresses and strains

100 + 60 sin 80° = 80 x 0.9848 = 78.785 N/mm2. Ans.

4> = tan" 9° 41'. Ans. • Resultant stress ( R)
Using equation on (3.8),

Let c = stress which acting alone will produce the same maximum strain. The maxi- °r = -jol + a2 = 2 + 78.785 2

mum strain will be in the direction of major principal stress. V33.89

Maximum strain E EOi (ia 2 (oi - = Jl 148.53 + 6207.07 = 85.765 N/mm2 . Ans.

Maximum shear stress

Using equation (3.9),
Oj - o 2 100 -(-60)

The OstUrUaUi1n dUUuUe VtoU ksJtUrVeUsUs

Cj

Equating the two strains, we get EE 100 + 60 = 80 N/mm2. Ans.
o = 70 N/mm2. Ans.
Problem 3.9. At a point in a strained material, the principal stresses are 100 N/mm2

Problem 3.8. At a point in a strained material the principal stresses are 100 N/mm2 tensile and 40 N/mm2 compressive. Determine the resultant stress in magnitude and direction

N/mm(tensile) and 60 2 (compressive). Determine the normal stress, shear stress and resultant on a plane inclined at 60° to the axis of the major principal stress. What is the maximum

stress on a plane inclined at 60° to the axis of major principal stress. Also determine the maxi- intensity of shear stress in the material at the point ? (AMIE, Winter 1982)

mum shear stress at the point. (AMIE, Summer 1982) Sol. Given :

Sol. Given The major principal stress, Oj = 100 N/mm2

Major principal stress, a = 100 N/mm2 The minor principal stress, o = - 40 N/mm2 (Minus sign due to compressive stress)
x 2

Minor principal stress, cr2 = — 60 N/mm2 (Negative sign due to compressive stress) Inclination of the plane with the axis of major principal stress = 60°

Angle of the inclined plane with the axis of major principal stress = 50°

.-. Angle of the inclined plane with the axis of minor principal stress, Inclination of the plane with the axis of minor principal stress,

0 = 90-50 = 40°. 0 = 90 - 60 = 30°.

Normal stress (on ) Resultant stress in magnitude
Using equation (3.6), The resultant stress (crR) is given by equation (3.8) as

+ 09 0t — 09
1
-o„= Of + cos 20<-»r\ °fl= o,2

22 where on = Normal stress and is given by equation (3.6) as

100 +(-60) 100 -(-60) -V^ 2^X= a, +a2 + -Oi *-09 COs26
cos (2 x 40°)

= 100 - 60 + 100 + 60 cos 80 M-^= 100^ + cos(2x30O)
22
100-40 + 100 + 40 cos 60°
= 20 + 80 x cos 80° = 20 + 80 x .1736
22
= 20 + 13.89 = 33.89 N/mm2. Ans.

Shear stress (<J )
t

= 30 + 70 x 0.5 (v cos 60° = 0.5)

Using equation (3.7), o = 65 N/mm2
t

100 -(-60) and a = Shear stress and is given by equation (3.7) as
t
sin (2 x 40°)
10 °- -
sin29= 40) sin (-2 x 30°)
(

AA

—2 -

100 STRENGTH OF MATERIALS

100 + 40 sin 60° = 70 x .866 = 60.62 N/mm2 .-. Stress along x-axis Force along x-axis
=
2
~Area normal to x-axis
N/mmaR = ij65 2 + 6 0.6 2 = 88.9 2 Ans.
= = 800 N/cm2
.
1.5
Direction of resultant stress
Oj = 800 N/cm2
Let the resultant stress is inclined at an angle $ to the normal of the oblique plane. Then
Force along y-axis
using equation [3.8 (A)].

Stress along y-axis, a = Area normal to y-axis
2
60.62
65

= 250 N/cm2

60.62 43°. Ans.
<j> = tan-1
tan 6 = — = 1.33
65
0
Maximum shear stress Also

—Using 0 = tan- 1 1.33 = 53.06°

equation (3.9), (a = - Let o = Normal stress on diagonal AB
( ;i
) mai, a = Shear stress on diagonal AB
t
= 100 -(-40) = 100 + 40 = 70 N/mm2. A.ns.
— —01 "f Go CJ« “ On
22 Using equation (3.6), a = —+ 1 -cos 20
n
Problem 3.10. A small block is 4 cm long, 3 cm high and 0.5 cm thick. It is subjected to
800-250
N Nuniformly distributed tensile forces of resultants 1200 and 500 as shown in Fig. 3.7 (a)
-
below. Compute the normal and shear stresses developed along the diagonal AB. ——=
800 + 250 + cos (2 x 53.06)
(AMIE, Summer 1987)
= 525 + 275 x cos 106.12° = 525 + 275 x (- 0.2776)

= 525 - 76.35 = 448.65 N/cm2. Ans.

Now using equation (3.7), a = —01 — 0O sin 20

(

= 800 ~ 25- sin (2 x 53.06°)

= 275 sin 106.12° = 275 x 0.96 = 264.18 N/cm2. Ans.

3.4.3. AMember Subjected to a Simple Shear Stress. D *C
Fig. 3.8 shows a rectangular bar ABCD of uniform cross-sectional L
Aarea and of unit thickness. The bar is subjected to a simple 1
BCshear stress (q) across the faces and AD. LetFC be the oblique P‘/
| i
section on which normal and tangential stresses are to be
A /C
calculated.
/ P„
Let 0 = Angle made by oblique section F_C- . .
F/
with normal tB

Fig. 3.7(a) cross-section BC, *

Sol. Given : x = Shear stress across faces BC and AD.

Length = 4 cm, height = 3 cm and width = 0.5 cm It has already been proved (Refer Art. 2.9) that a shear stress is always accompanied by

NForce along x-axis = 1200 CDan equal shear stress at right angles to it. Hence the faces AB and will also be subjected to
NForce along y-axis = 500
a shear stress q as shown in Fig. 3.8. Now these stresses will be converted into equivalent
Area of cross-section normal to x-axis = 3 x 0.5 = 1.5 cm2
Area of cross-section normal to y-axis = 4x 0.5 = 2 cm2 forces. Then these forces will be resolved along the inclined surface and normal to inclined

surface. Consider the forces acting on the wedge FBC of Fig. 3.9.

—

102 STRENGTH OF MATERIALS

Let Q j = Shear force on face BC L -£ o
g
= Shear stress x Area of face BC
Q, cos e/o 8

=xxBCxl r"
<& / t x BC x 1
BC( .* Area of face 3.4.4. A Member Subjected to Direct Stresses in two Mutually Perpendicular
= -BC x 1)
Directions Accompanied by a Simple Shear Stress. Fig. 3.10 (a) shows a rectangular bar
=txbc °y ^ ;t 0g
ABCD of uniform cross-sectional area A and of unit thickness. This bar is subjected to .
QA0 = Shear force on face FB \A r 6 x FBo x .
= x x Area of FB 1

Q“

•? °os&

= x x FB x 1 =' t . FB pig. 3.9

P = Total normal force on section FC
n

P = Total tangential force on section FC.
t

The force is acting along face CB as shown in Fig. 3.9. This force is resolved into two

components i.e., Q cos 6 and Q, sin 6 along the plane CF and normal to the plane CF respectively.
x

i.e., The Qforce is acting along the face FB. This force is also resolved into two component
2
Q sin the plane FC respectively.
? 6 and Q2 cos 0 along the plane FC and normal to

.-. Total normal force on section FC,

QP + 2 cos 6
n
= Qj sin 6

= x x BC x sin 0 + x x FB x cos 0. (v = x x BC and Q2 — x x FB) Fig. 3.10

And total tangential force on section FC. BC AD(i)
and
P Q Q= (-ve sign is taken due to opposite direction) tensile stress a on the face
t 1
sin 0 - cos 0.
a on the face AB and CD
2 1 2

= x x FB x sin 0 - x x BC x cos 0 ( v Q2 = x . FB and = x . BC) (ii) tensile stress

Let on = Normal stress on section FC (Hi) a simple shear stress x on face BC and AD.
6n
a = Tangential stress on section FC But with reference to Art. 2.9, a simple shear stress is always accompanied by an equal
t
shear stress at right angles to it. Hence the faces AB and CD will also be subjected to a shear
Total normal force on section FC
°n _ Area of section FC stress x as shown in Fig. 3.10 (a).

P„ We want to calculate normal and tangential stresses on oblique section FC, which is

inclined at an angle 0 with the normal cross-section BC. The given stresses are converted into

FC x 1 equivalent forces.

BC FBx . . sin 8 + x . . cos 0 Area - FC x 1) The forces acting on the wedge FBC are :

: FC x 1 P BC= Tensile force on face due to tensile stress

—BC— ——x= .
FB. si. n a0 + x . „ = OjX Area of BC (
FC FC . cos 0 = a x BC x 1

t Area = BC x 1)

- x . cos 0 . sin 0 + x . sin 0 . cos 0 = a x BC
t
— ———In triangle FBC, = sin 0
= cos 0, P = Tensile force on face FB due to tensile stress o
FC t C 2 2

= 2t cos 0 . sin 0 = o x Area of FB = o x FB x 1
= x sin 20 2 2

( v 2 sin 0 cos 0 = sin 20) ...(3.10) = a x FB
2
_ Total tangential force on section FC
‘ Area of section FC Q = Shear force on face BC due to shear stress x

= x x Area of BC

Pt = x x BC x 1 = x x BC

~ FCx 1 Q2 = Shear force on face FB due to shear stress x
= x x Area of FB
_ x x FB x sin 0 - x x BC x cos 0
“ FC x 1 = x x FB x 1 = x x FB.

FB .' BC Pv P normal to the oblique section FC,
=xx x sm 0 - x x x cos 0 the above four forces 2> an<^
Resolving {i.e., Qi Q?)

FC FC we get

= xx sin 0 x sin 0 - x x cos 0 x cos 0

1 .. j 5

104 105

Total normal force, . sin 20 — x cos 20 (v cos 2 6 — sin2 0 = cos 20) ...(3.13)

P p Qp~
n
cos 9 + 2 sin 6 + Q, sin 0 + cos 0

i 2

Substituting the values of Pv P Q. and Qv we get Position of principal planes. The planes on which shear stress (i.e., tangential stress)
2, is zero, are known as principal planes. And the stresses acting on principal planes are known

BC BCP ~
n
°i FB- cos 9 + o2 . . sin 0 + x . . sin 0 + x . FB. cos 6

Similarly, the total tangential force (P ) is obtained by resolving P,, P,„ Q, and Q, along principal stresses.
t
The position of principal planes are obtained by equating the tangential stress [given by
the oblique section FC.

Total tangential force, equation (3.13)] to zero.

P = P sin 9 - P cos 0 - Q cos 0 + sin .-. For principal planes, a =0
t y 2 t t
Q, 0
—or sin 20 - x cos 20 = 0
BC FB BC= Oy .
. sin 0 - o . cos 0 - t . FB. cos 0 + x . . sin 0

2.

the of Pv P
2,
Q(substitute
values Qy and 2) Pl °2

Now, Let on = Normal stress across the section FC, and or _ sin 20 = x cos 29

a 2
t
= Tangential stress across the section FC. sin 29 x

Then normal stress across the section FC, or cos 20 ~ (Pi -p
2)

a _ Total normal force across section FC Pn 2

Area of section FC FC x 1 tan 20 - 2x

BC FB BC FB_ cr -oor .
t
. . cos 9 + <j % . sin 0 + x . sin 0 + x . . cos 0 (CJi 2)

FCxl But the tangent of any angle in a right angled triangle

^ — —= °i BC
BC sm0FBc°s„ FB Height of right angled triangle
9 + p2 +. „ +x . . cos© Base of right angled triangle
.
• sin 6 x .
•

= Oj . cos 0 . cos 0 + o sin 0 . sin 0 + x . cos 0 . sin 0 + x sin 0 cos 0
2
.

Height of right angled triangle _ 2x

jv In triangle FBC, = cos 0 and = sin Base of right angled triangle (a-, - a 2 )

0

= a, cos2 0 + o, sin2 0 + 2t cos 0 sin 0 Height of right angled triangle = 2x

' (- Base of right angled triangle = (a, - a2 ).
1+
= a cos 20 j + °2 [“ cos 20 Now diagonal of the right angled triangle I
i (“ 2
J + x sin 20 ML (a, - a2)

cos 2 „ 1+ smcos 20 . J, „ 1 - cos 20 =± yj(ay - a2 f 2 = ± yjiay - a2 2 + 4x 2 Fig. 3.11
9 0,
= = and 2 cos 0 sin 0 + (2x) )

= sin 20

+ Og — CTo = ^(ctj - p2 + 4x 2 and - ^(cq -o 2 2 + 4x 2
~~~~2 2 2)
01 )

+ cos 20 + x sin 20 ...(3.12)

and tangential stress (i.e. , shear stress) across the section FC, 1st Case. -pDiagonal = J(ctx 2 + 4x z
2)

FCTotal tangential force across section —P, —Height 2x
1
a, Then sin 29 = 7- i a2 r
Diagonal 2) 4x 2
Area of section FC FC x 1 - +

—BC_ Pi
FBFBo, sin 9 - 2 .
. cos 9 ~x.BC . cos 0 + x . (oy-a2 )
sin 0 — —Base

PC x l and cos 20 = r- j o2 + 4x 2y •
Diagonal 2)
^(o -
^
JBcCBC FCFB~ 01 FB
„
FC’ 8
S.m 9 “ ° C°S _X C°S 8 + T Sin 6 The value of major principal stress is obtained by substituting the values of sin 20 and
''
2' FC' ' '
’

= Pj - cos 0 sin 0 - o sin 0 cos 0 - x cos 0 cos 6 + x sin 0 sin 0 cos 26 in equation (3.12).
2
. . . . ,

. .

—BC FB .-. Major principal stress

v In triangle FBC, = cos 0 and = sin 0 _ CTl + —+ Hi Hi. cos 29 + x sin 20
( FC FC
22
= {Oj - o cos 0 sin 0 - x cos2 0 + x sin2 0
2 ) . _(°i~€2) 2t

°i + ga °i ~ °g , T„ +4t z

= • 2 cos 0 sin 0 - x (cos 2 0 - sin2 0) , -J(0i - p 2 )^ + 4x 2 J(oy P2

^ 22 - 2)

)

106 STRENGTH OF MATERIALS

0 + 02 1 q( l ~ g 2 )
t

||

-a +42 2
+4x02
/( ! .
2 t 2 yj(a 1 - o2 ) 2
2)

01 + O2 (0i — 0g) + 4x
2
2^0! -0 2 +4t 2
2
)

01+02 + 1 _a \2 +41x 2"
[/ a 2)
l
~\/(

g +g g ~g
l 2 . If l 2 I . -.2

2nd Case. Diagonal =- ^/(a, - a2 + 4x 2
2)

-V(°i -°2) +4t Fig. 3.11 (a)

—cos 29 = (0. - 0,2 Maximum shear stress. The shear stress is given by equation (3.13). The shear stress
will be maximum or minimum when
-a- \(o-[ z +4x 2
2) A(O() = 0

Substituting these values in equation (3.12), we get minor principal stress. ——— sin 20 - x cos 20 = 0

Minor principal stress

——O] + Go + 0 1 — (Jn .

cos 20 + x sin 20

22 or 0l (cos 20) x 2 — x (— sin 20) x 2 = 0

g +g 2 gl - a 2 .. °1 ~ °2 2x
_l
t 0( j - a cos 20 + 2t sin 20 = 0
2
) .

2 2 - “ 02 2 +4t z -^(0! -o 2 +4t 2 =

V(0i ) 2 ) or 2x sin 20 - (o - o2 ) cos 20
L

(01 -02 ) 2 = (a - ot) cos 20

_ 01 +02 2t^ 2

2 sin 26 a-2 - 0
2 4r 2 2 4x z —or = X

2i/(0i - 02 ) + 7( 0 i - 0a) + cos 20 2t

0j + q 2 (01 - o2 2 2 —or tan 29 = ...(3.17)

2 ) +4-C

2-^(0! - 02 + 4r 2
2)

= 01 2°^ - v '' fl - fi 2 l? 'k ~
' |T

2

Equation (3.15) gives the maximum principal stress whereas equation (3.16) gives mini- Fig. 3.12

mum principal stress. These two principal planes are at right angles.

The position of principal planes is obtained by finding two values of 6 from equation (3. 14).
Fig. 3.11(a)shows the principalplanesin which Sj^andQgare the values from equation (3.14).

a

1 08 STRENGTH OF MATERIALS PRINCIPAL STRESSES AND STRAINS

109

Substituting the values of sin 26 and cos 20 in equation (3.13), the maximum and mini- Problem 3.11. At a point within a body subjected to two mutually perpendicular direr

mum shear stresses are obtained. tions, the stresses are SO N/mm• tensile and 40 N/mm * tensile. Each of the above stress! is
the normal stress, shear stress and
Maximum shear stress is given by accompanied by a shear stress of 60 N/mm2 Determine of 45° with the axis of minor tensile
.

resultant stress on an oblique plane inclined at an angle

g ~ ^2 sii*GSs .
l
sin 20 - % cos 20
Sol. Given :

a ~ g2 (cr 2 - Pi) _ Major tensile stress, o. = 80 N/mm2
i J(o 2 - Oj} 2 + 4x 2 Minor tensile stress,
Shear stress,
— 2 + 4t a = 40 N/mm2
%
-J(o 2 ^1 )

2 r = 60 N/mm2
2)
-a(aj Angle of oblique plane, with the axis of minor tensile stress,
0 = 45°.
2^(cr - 2 + 4t z ^(a 2 -a z + 4x
1)
2 Oj)

— 7T=: ± -|(=042 -0ri1)2 +44xt2 (i) Normal stress (on)
= ± —1 f - "TTii ++ 4^ Using equation (3.12),
J(dn
0, )

2^/(o 2 - oj) 92 + 4t 2, ^

4- On — Go
2~
—°n = Oi Cft

+ C°S + T S*n 29

<°Pmax = | \/(02 2 +4r2 g

-0i) — — — —-

80 + 40 + 80 - 40 cos (2 x 45 °) + 60 sin (2 x 45°)
2
= |V(ct i-°2) 2 + 4t 2 ...(3.18) jj

The planes on which maximum shear stress is acting, are obtained after finding the two = 60 + 20 cos 90° + 60 sin 90°

values of 0 from equation (3.17). These two values of 0 will differ by 90°. -60 + 20x0 + 60x1 ( v cos 90° = 0)
= 60 + 0 + 60 = 120 N/mm2. Ans.
The second method of finding the planes of maximum shear stress is to find first princi-

pal planes and principal stresses. Let Oj is the angle of principal plane with planeBC of Fig. 3.11
plane BC
(a). Then the planes of maximum shear wall be at 0 + 45° and 0, + 135° with the as

shown in Fig. 3.12 (a). t

Fig. 3.13

(ii) Shear (or tangential) stress (oj

Using equation (3.13),

— —o = a, - a2 sin 20 - t cos 20
t

= —80-40 . (2 x 4,5°), - 60 x cos (2 x 45°)

sin

Fig. 3.12 (a) = 20 x sin 90° - 60 cos 90°
= 20 x 1 - 60 x 0
Note. The above relations hold good when one or both the stresses are compressive.
= 20 N/mm2. Ans.

(Hi) Resultant stress ( R)

Using equation,

110 STRENGTH OF MATERIALS b

- PRINCIPAL STRESSES AND STRAINS

= v 14800 = 121.655 N/mxn2. Ans. 110 + 47 110-47
2 + 63 2 =78.5-70.436
AProblem 3.12. rectangular block of material is subjected to a tensile stress of 110
= 8.064 N/mm2. Ans.
N/mm 2 on one plane and a tensile stress of 47 N/mm2 on the plane at right angles to the former.
Each of the above stresses is accompanied by a shear stress of 63 N/mm2 and that associated The directions of principal stresses are given by equation (3.14).
Using equation (3.14),
with the former tensile stress tends to rotate the block anticlockwise. Find :

(i) the direction and magnitude of each of the principal stress and

(ii) magnitude of the greatest shear stress. (AMIE, Summer 1983) ZT _Z x 63

Sol. Given : * T t, an 2„0„

Major tensile stress, a, = 110 N/mm2 Oj - o2 110 - 47
^= = 2.0
Minor tensile stress, a2 = 47 N/mm2

Shear stress, x = 63 N/mm2 63

(i) Major principal stress is given by equation (3.15). 20 = tan' 1 2.0 = 63° 26' or 243° 26'

9 = 31° 43’ or 121° 43 . Ans.
(ii) Magnitude of the greatest shear stress
Greatest shear stress is given by equation (3,18).
Using equation (3.18),

= |\/(a1 -a 2 2 + 4t 2

)

= /(100-47) 2 + 4x63 2

|A

=— + 4 x 63 2 = 1 x 63 x -M

* 2

= 70.436 N/mm2. Ans.

two Problem 3.13. Direct stresses of 120 N/mm2 tensile and 90 N/mm2 compression exist on

perpendicular planes at a certain point in a body. They are also accompanied by shear
N/mmstress on the planes. The greatest principal stress at the point due to these is 150
2

.

(a) What must be the magnitude of the shearing stresses on the two planes ?

( ) What will be the maximum shearing stress at the point ?

Sol. Given :

Major tensile stress, = 120 N/mm2
Minor compressive stress,
Greatest principal stress a = - 90 N/mm2 (Minus sign due to compression)
2

= 150 N/mm2

(a) Let t = Shear stress on the two planes.

Using equation (3.15) for greatest principal stress, we get

—g + °2 ~ °2 + 2
-
Greatest principal stress = - -+ j -r

= 78.5 + 70.436 = 148.936 N/mm 2 Ans.

.

Minor principal stress is given by equation (3.16).

120 + (-90) 120 -(-90)

2 2

Minor principal stress,

120 - 90

5

StRENGTH OF MATERIALS 113
112

Major tensile stress, cij = 20 N/mm2
Minor tensile stress,
= 15 + 2 + x2 Shear stress, cr = 10N/mm 2
2
-Jl05

qj’ 150 — 15 ~ -Jl05^^T x = 10 N/mm 2

Location of principal planes

+^or 135 = ^1052 The location of principal planes is given by equation (3.14).
Using equation (3.14),
Squaring both sides, we get

— —t. a. n 26 = 2t
135 2 = 105 2 + t2 « 2 x 10 2 x 10

or • x2 = 1352 - 105 2 = 18225 - 11025 = 7200 = = 20
Oj - a 2 20-10 10

x= J7200 = 84.853 N/mm 2 Ans. 26 = tan' 1 2.0 = 63° 26' or 243° 26'
.

(b) Maximum shear stress at the point or 6 = 31° 43' or 121° 43'. Ans.
Using equation (3.18) for maximum shear stress,
Magnitude of principal stresses

<O = %/(Ol-0 2 ) +472 The major principal stress is given by equation (3,15)
t Major principal stress
)m ax \

= — J[120 - (- 2 + 4 x 7200 (v = 7200) _ gi .+ q 2 f Qi -q r 20 + 10 20 - 10 2
2 1
90)] 4 + t2
jf +

|

= 2 J21Q 2 + 28800 = - 744100 + 28800 = -^ x 270

i 2 * = 15 + y 2 + 100 = 15 + x/~25 + 100 = 15 + /l25 = 15 + 11.18

2 = 26.18 N/mm2. Ans.

= 135 N/mm2. Ans. The minor principal stress is given by equation (3.16).
Minor principal stress
Problem 3.14. At a certain point in a strained material, the stresses on two planes at
They are accompanied by
anles Teach other are 20 N/mm* and 10 Nlmm2 60//, tensile.
right
ashear stress ofa magnitude of 10 N/mm2. Find graphically or otherwise,

planes and evaluate the principal stresses. a + O2 q p~
i _ If i 2

Sol. Given : 2 VI 2

t = 10 N/mm : 20+10 f20-10r +1°2
\{~2
2 "~
j

• 15 - 11.18 = 3.82 N/mm2 . Ans.

Problem 3.15. A point in a strained material is subjected to the stresses as shown in

Fig. 3.15.

Locate, the. principal planes, and evaluate, the principal stresses.

Principal^/ . x = 10 N/mm
planes

v //
o*,'

Fig. 3.14 (a)

Fig. 3.15

STRENGTH OF MATERIALS principal stresses and strains
114

ADstress on the face BC or is not normal. It is inclined at an angle °^0 wth face = 45.98 + 30.6
into two components i.e. , norm oe
The This stress can be resolved = 76.58 N/mm2. Ans.
BC or AD.
The minor principal stress is given by equation (3.16).
and along the face BC (or AD). Minor principal stress

Stress normal to the face BC or AD N/mm2 Pi + q 2 °1 ~ q
2
= 60 x sin 60° = 60 x 0.866 = 51.96 2

BC or AD iV 2 .

Stress along the face = 60 x cos N/mm2

60° =60 x0.5 = 30 51.96 + 40 51.96 - 40

n2 2 j

mof 30 N/mm2 . Now the stresses acting on the material are shown = 45.98-30.6

g. • = 15.38 N/mm2. Ans.

Problem 3.16. The normal stress in two mutually perpendicular directions are 600 N/mm2
and 300 N/mm2 both tensile. The complimentary shear stresses in these directions are of intensity
450 N/mm2. Find the normal and tangential stresses on the two planes which are equally inclined

to the planes carrying the normal stresses mentioned above.

Sol. Given : o =. 600 N/mm2
Major tensile stress, 5
Minor tensile stress,
Shear stress, o = 300 N/mm2
2

x = 450 N/mm2

The normal and tangential stresses are to be calculated on the two planes which are
equally inclined to the planes of major tensile stress and of minor tensile stress. This means
0 = 45° and 135°.

Fig. 3.16

Angle 6 = 45° and 135°.

Major tensile stress, = 51.96 N/mm (0 Normal stress (on) is given by equation (3.12).
Minor tensile stress,
Shear- stress, o.2 = 40 N/mm2 — — —an = a, - a 2- 0, - a,
t = 30 N/mm2 -
h — cos 20 + t sin 20

Location of principal planes make with the stress of 40 N/m . (a) When 0 = 45°, the normal stresses (o ) becomes as
Let 6 = Angle, which one
planes .

of the principal

The location of the principal planes is given by the equation (3.14). 600 + 300 600 - 300

n2 cos (2 x 45°) + 450 sin (2 x 45°)
'
2
Using equation (3.14), we get
= 450 + 150 cos 90° + 450 sin 90°
—— 2x30
tan 042609 =- = = 4.999 = 450 + 150 x 0 + 450 x 1 (v cos 90° = 0 and sin 90° = 1)
51 g6 _ 40
CTi _ a2 = 900 N/mm2. Ans.

26 = tan’ 1 4.999 = 78° 42' or 258° 42' (5) When 0 = 135°, the normal stress o( n ) becomes as

or 6 = 39° 21' or 129° 21'. Ans. ~—0„ = 600 + 300 + 600-300 cos (2 x 135) + 450 sin (2 x 135°)

Principal stress

The major principal stress is given by equation (3.15).

.-. Major principal stress = 450 + 150 cos (270°) + 450 sin 270°
= 450 + 150 x 0 + 450 x (- 1) (v cos 270° = 0 and sin 270° = - 1)
CTj + Oj o1 ~ q2
= 450 — 450 = 0. Ans.
2
(ii) Tangential stress (a ) is given by equation (3.13)
(

51.96 + 40 51.96 - 40 sin 20 - x cos 20
2

r

STRENGTH OF MATERIALS PRINCIPAL STRESSES AND STRAINS

ABThe shear stress on plane is, i.e., = 400 N/cm2 then to maintain the equilibrium
,
When = becomes
() 0 45°, the tangential stress (o ) as on the wedge ABC, another shear stress of the same magnitude, i.e., xBC = 400 N/cm2 must act
4

600 - 300 sin 90° - 450 cos 90° on the plane BC. The free body diagram of the element ABCD is shown in Fig. 3.16 (a), showing

0( = normal and shear stresses acting on different faces.

= 150 x 1 — 450 x 0 = 150 N/mm2. Ans. BC(i) Resultant stress on plane

() When 0 = 135°, the tangential stress (a ) becomes as On plane BC, from Fig. 3.16 (a),
t
N/cm2
a = 600
2
a, = 600 ~ 300 sin 270° - 450 cos 270°
Shear stress, r = 400 N/cm2
-
Ans. nResultant stress on plane BC
= 150 x (— 1) - 450 x 0 = - 150 N/mm2.
2
a~ 2
Problem 3.17. The intensity of resultant stress on a 600 N/cm - +T
V.... 2

ABplane [Fig. 3.16 (a)} at a point in a material under stress is = ^600 2 + 400 2 = 721 N/cm2. Ans.

800 N/cm.2 and it is inclined at 30° to the normal to that plane, 'f -rrrrT* 2

The normal component of stress on another plane BC at right ; 800 N/cm The resultant will be inclined at an angle 0 with the horizontal given by,
I
angles to plane AB is 600 N/cm2. ‘ s' —tan 0 = o2 600 = 1.5
i 400
Determine the following : 0 = tarn1 1.5 = 56.3°. Ans.

(i) the resultant stress on the plane BC,

(ii) the principal stresses and their directions, 'A (ii) Principal stresses and their directions
The major principal stress is given by equation (3.15).
(iii) the maximum shear stresses and their planes. pj 3 16 (a)
(AMIE, Summer 1989)
Major principal stress

Sol. Given : = 800 N/cm2 q +02 q ~ °2
L l
Resultant stress on plane AB

Angle of inclination of the above stress = 30° 2 ,
.

Normal stress on plane BC = 600 N/cm2 692.82 + 600 692.82-600"

The resultant stress 800 N/cm2 on plane AB is resolved into normal stress and tangential 2

stress. = 646.41 - 402.68

= 600 N/cm = 1049.09 N/cm2 (Tensile). Ans.

The minor principal stress is given by equation (3.16)

.-. Minor principal stress

t = 400 N/cm Oj + q2 q ~ q2
cr, = 692.82 N/cm l
2
<j, - 692.82 N/cm' 2

i = 400 N/cm _ 692,82 + 600 _ 692.82 - 600 2

x = 400 N/cm' + 40q2

Fig. 3.16 (6) j

The normal stress on plane AB = 646.41 - 402.68

= 800 x cos 30° = 692.82 N/cm2 . = 243.73 N/cm2 (Tensile). Ans.

The tangential stress on plane AB The directions of principal stresses are given by equation (3.14), as

= 800 x sin 30° = 400 N/cm2 . fan 20 = 2t = 2x400 800 =8.618

(cq-oa) (692.82-600) 92.82

26 = tan" 1 8.618 = 83.38° or 263.38°

9 = 41.69° or 131.99°. Ans.

(iii) The maximum shear stress and their planes.

The maximum shear stress is given by equation (3.18).

118 STRENGTH OF MATERIALS

= 402.68 N/cm2. Ans.

Problem 3.18. At a certain point in a material under stress the intensity of the resultant

stress on a vertical plane is 1000 N/cm2 inclined at 30 ° to the normal to that plane and the

stress on a horizontal plane has a normal tensile component of intensity 600 N/cm2 as shown in

Fig. 3.16 (c). Find the magnitude and direction of the resultant stress on the horizontal plane

and the principal stresses. (AMIE, Winter 1990)

Z

600 N/cm

Fig. 3.16 (e) .
,

Sol. Given :

Resultant stress on vertical plane AS = 1000 N/cm2

Inclination of the above stress = 30“

Normal stress on horizontal plane BC = 600 N/cm2
The resultant stress on plane AB is resolved into normal and tangential component.

The normal component

= 1000 x cos 30° = 866 N/cm2

Tangential component

= 1000 x sin 30° = 500 N/cm2 = 733 ± 517.38
.

Hence a shear stress of magnitude 500 N/cm2 is acting on plane AB. To maintain the = (733 + 517.38) and (733 - 517.38)
wedge in equilibrium, another shear stress of the same magnitude but opposite in direction
= 1250.38 and 215.62 N/cm2.
must act on the plane BC. The free-body diagram of the element ABCD is shown in Fig. 3.16 id),
Major principal stress = 1250.38 N/cm2. Ans.
showing normal and shear stresses acting on different faces in which :
Minor principal stress = 215.62 N/cm2 . Ans.
= 866 N/cm2
, Problem 3.19. At a point in a strained material, on plane BC there are normal and
shear stresses of 560 N/mm2 and 140 N/mm2 respectively. On plane AC, perpendicular to plane
a = 600 N/cm2
2 NBC, there are normal and shear stresses of 280 N/mm2 and 140 fmm~ respectively as shown in

and x = 500 N/cm2

(i) Magnitude and direction of resultant stress on horizontal plane BC. Fig. 3.16 (e). Determine the following :

Normal stress on plane BC, o = 600 N/cm2 (t principal stresses and location of the planes on which they act,
2 )

(ii) maximum shear stress and the plane on which it acts. (AMIE, Summer 1990)

120 STRENGTH OF MATERIALS 121

PRINCIPAL STRESSES AND STRAINS

The plane on which maximum shear stress acts is given by equation (3.17) as

Fig. 3.16 (e) Oo 01
tan 28 = 2„t

Sol. Given : 560 -(-280) 840 3p
=
On plane AC, o = ~ 280 N/mm2 2x140 280
On plane BC, 1 (- Ve sign due to compressive stress)
t = 140 N/mm2
20 = tan-1 3.0 = 71.56° or 251.56°

a = 560 N/mm 2 8 = 35.78° or 125.78°. Ans.
2
mmProblem 3 20. On a mild steel plate, a circle of diameter 50
x = 140 N/mm2 is drawn before the

(i) Principal stresses and location of the planes on which they act. plate is stressed as shown in Fig. 3.17. Find the lengths of the major and minor axes of an
Principal stress are given by equations (3.15) and (3.16)
ellipse formed as a result of the deformation of the circle marked.

Principal stresses

= 140 ± 442.7

= 582.7 and (140 - 442.7) N/mm 2

= 582.7 and - 302.7 N/mm2

Ma.i°r principal stress = 582.7 N/mm2 (Tensile). Ans.

Minor principal stress = - 302.7 N/mm2. Ans.

The planes on which principal stresses act, are given by equation (3.14) as

tt „a_n 2„0„ 2t —_2x140 — = 280 - _ o 3'

Oi - o 2 - 280 - 560 - 840

28 = tan" 1 - 0.33 = -18.262 Fig. 3.17
- ve sign shows that 28 is lying in 2nd and 4th quadrant

20 = (180 - 18.26°) or (360 - 18.26°)

= 161.34° or 341.34° ^Take E = 2 x 105 N/mm2 and = j-

8 = 80.67° and 170.67°. Ans. Sol. Given :
(“) Maximum shear stress and the plane on which it acts.
Maximum shear stress is given by equation (3.18). Major tensile stress, a = 80 N/mm2
t

principal stresses and strains 123

122 STRENGTH OF MATERIALS

Minor tensile stress, a = 20 N/mm2 —_ - (_ ve sign shows that there is a decrease in length)
2
" 8000
Shear stress, r = 40 N/mm2

Value of £ = 2 x 10s N/mm2 AC.-. Decrease in length of diameter along

Major principal stress is given by equation (3.15). = Strain along AC x Dia. of hole

/. Major principal stress — mm= x 50 = 0.00625
8000
q q q~*
mm... The circle will become an ellipse whose major axis will be 50 + 0.025 = 50.025
°ii
2. II 2I a and minor axis will be

2 V2 50 - 0.00625 = 49.99375 mm.

80 + 20 (80- 20 3.5. GRAPHICAL METHOD FOR DETERMINING STRESSES ON OBLIQUE SECTION

- 50 + 2 + 40 2 = 50 + 50 = 100 N/mra2 (tensile) Two cases are considered :

tJ30 A(i) body is subjected to direct stresses in two mutually perpendicular directions when
the stresses me unequal and alike.
Minor principal stress
A(jj) body is subjected to direct stresses in two mutually perpendicular directions when
°l tq2 II q ~ q2 2
l the stresses are equal and alike.
2
Vl 2 3 5 1 A body is Subjected to Direct Stresses in two Mutually Perpendicular

80 + 20 F80-20 Directions when the Stresses are Unequal and alike. Fig. 3.18 shows a rectangular bar

of uniform cross-sectional area A. The bar is subjected to two tensile stresses. It is required to
find the normal and tangential stresses graphically on the oblique plane FC.

= 50-50 = 0.

BD ACFrom Fig. 3.17, it is clear that diagonal
will be elongated and diagonal will be

BDshortened. Hence the circle will become an ellipse whose major axis will be along and
minor axis along AC as shown in Fig. 3.17.The major principal stress acts along BD and minor

principal stress along AC.

BD.'. Strain along

_ Major principal stress Minor principal stress

E mE

2 x 10 5 2 x 10 5 x 4

2000 Fig. 3.18

BD.•. Increase in diameter along
= Strain along BD x Dia. of hole

= x5° = a025mm Let a, = Major principal tensile stress,

2000 a = Minor principal tensile stress, and
2
Strain along AC
0 = Angle made by the oblique section with the axis of minor principal stress.

Minor principal stress Major principal stress Procedure.

_ mE 1. Draw two mutually perpendicular lines meeting at O.

E

0 100 2. Take OA = Stress a and OB = a to some scale.
2 x 105 4 x 2 x 10 5 1 2

!,

. ©,

3. Draw two concentric circles with centre O and radii equal to OA and OB as shown in principal stresses and strains 125

Fig. 3.19. In right angled triangle CGF, Z.GCF -90-0

CG = CF cos (90 - 0) = CF sin 0 CF['•" = (o - a2) sin 0]
1
= (cjj - <j,j) sin 0. sin 0

= (a - a ) sin2 0
2
t

= a- (CTj - 2 ) sin2 0
^OG CG= OC -
[v OC = ap CG = (a, - oz) sin2 0]

= a - Oj sin 2 0 + o2 sin2 0 = ajl - sin 2 0) + o,, sin2 0
x

= a cos 2 0 + a sin 2 9 (v 1 - sin2 0 = cos 2 0) —
x 2

But from equation [3.5 (A)], normal stress across the oblique section is given by

°n = °1 COs2 6 + °2 S2n2 ®
Equating equation (i) and Hi), we get

OG = on = Normal stress -

Tangential stress, a, = GF = CF sin (90 - 0) = CF cos 0

= (o - a2 ) sin 0 cos 0
t

—_ fh g sjn q cog g
2

= sin 26.
2

GE.-. represents the tangential stress.

Fig. 3.19 3.5.2. Important Points. The normal stress tangential stress and resultant stress on

plane by the above method (if any one or both of Oj and a2, are compressive) are
the same manner. Only the
Fposition of point will change. The position of point
Through O, draw a line AfN making an angle 0 with the axis of minor principal stress Fthe oblique
,
obtained in
(i.e., axis of stress a2 ).
will be as follows :
D5. Through O, draw a line ODC at right angles to MN, meeting the two circles at and C.
F(i) The point will be in first quadrant if Oj and o are tensile stresses (i.e., a and a
2 l 2

6. From C, draw a line CE perpendicular to OA. are +ve).

7 . Prom D, draw a line DF perpendicular to CE. F(ii) The point will be in second quadrant if Oj is compressive and o is tensile (i.e., o is
2 1

8. Join OF Then OF represents the resultant stress on the oblique plane. - ve and o is +ve).
2
a
F(iii) The point will be in third quadrant if Oj and o are compressive (i.e., 1 and o are
2 2
9. From F, draw a line FG perpendicular to OC. Then OG represents the normal stress
on the oblique plane. And GF represents the tangential stress. -ve).

(iv) The point F wilt be in fourth quadrant if Oj is tensile and o2 is compressive (i.e. o is
t

Normal stress - OG and +ve and a is -ve).
2

Tangential stress = GF. Problem 3.21. Solve the problem. 3.5 by graphical method.

Proof. (See Fig. 3.19). Sol. The data given in problem 3.5, is

a = 120 N/mm2 o = 60 N/mm2 0 = 30°.
t , 2 ,
CD = OC - OD

= - OC OA OD Scale

a (v = = a and = OB = a2 ) Take 1 cm = 20 N/mm2
2 1

In right angled triangle OEC, LEOC = 9.

LOCE = 90° - 0. Then ^a = = 6 cm and a = = 3 cm.
1 2

In right angled triangle DCF, LDCF = 90° - 9. (j) Draw two mutually perpendicular lines meeting at O as shown in Fig. 3.20.
(ii) Take OA = 6 cm and OB = 3 cm.
LCDF = 0 O(iii) Draw two concentric circles with centre and radii equal to OA = 6 cm and

DF = CD cos 9 = (Oj - o2 ) cos 0 (-.- CD = Oj - a2) OB - 3 cm.
CF = CD sin 0 = (a, - a„) sin 0.

principal stresses and strains

As is + ve and a is - ve, the point F will be in fourth quadrant (see Art. 3.5.2) on
2

page 125).

Scale. Take 1 cm = 20 N/mm2
Then
~ ~0l = 200 100
cm , = 5 cm.
= 10 a =
and 2

(i) Draw two mutually perpendicular lines meeting at O as shown in Fig. 3.21.

Fig. 3.20

MN(iy) Draw a line through O, making an angle 30° with OB.

OC DCv ) Through O, draw a line
at right angles to MN, cutting the two circles at and C.

From C, draw a line CE perpendicular to OA.

(vi) From D, draw a line DF parallel to OA, meeting the line GE at F.

(vii) Join OF. Then OF represents the resultant stress on the oblique plane.

(viii) From F, draw a line FG perpendicular to line OC. Then FG represents the tangential

OGstress and represents the normal stress. (it) Take OA = 10 cm and OB - 5 cm.
(iii) Draw two concentric circles with centre O and radii equal to OA = 10 cm and
(ix) Measure the lengths OF, FG and OG.

By measurements, we get OB = 5 cm.

Length OF = 5.411 cm MN(to) Draw a line Othrough at an angle of 30° with the line OB.

Length FG = 1.30 cm (o) Through O, draw a line OC at right angles to MN, cutting the two circles at C and D.

Length OG ~ 5.25 cm. (ui) From C, draw a line CE perpendicular to OA. Produce the line CE upto C'. Join the

OF.-. Resultant stress, aR = Length OBline OC', cutting the circle of radius at D‘.

x Scale F(vii) As the point F will lie in the fourth quadrant, the point will be obtained by drawing

= 5.41 x 20 (v 1 cm = 20 N/mm 2 a line D'F parallel to OA.

)

N= 108.2 /mm2. Ans. OF(viii) Join OF. Then represents the resultant stress.

Normal stress, an = Length OG x 20 N/mm2 = 5.25 x 20 — 105 N/mm 2 Ans. OG(lx) From F, draw FG perpendicular to line OC. Then represents the normal stress,

.

Tangential stress, a = Length FG x 20 N/mm 2 = 1.30 x 20 = 26 N/mm2. Ans. and GF represents the tangential stress on the oblique plane.
t

Problem 3.22. Solve the problem 3.6 by graphical method. (x) Measure the lengths OF, OG and GF.

Sol. Given : From Fig. 3.21, by measurements,

The data given in problem 3.6, is Length OF = 9.0 cm

Oj = 200 N/mm2 Length OG = 6.25 cm

a = - 100 N/mm2 Length GF = 6.50 cm.
2
6 = 30°

128 STRENGTH OF MATERIALS PRINCIPAL STRESSES AND STRAINS

Resultant stress = Length OF x Scale AO = AC + CO

Normal stress = 9.0 x 20 ( v 1 cm = 20 N/mm2) Oj — Og = ^02 + 0^ — 02 Oy 4- 02
+ 2
Tangential stress = 180 N/mm2. Ans. =q22

= Length OG x Scale = 6.25 x 20 = 125 N/mmz. Ans. OD = OE cos 29

= Length GF x Scale = 6.50 x 20 = 130 N/mm2. Ans. (v OE^^^j

3.6. MOHR’S CIRCLE

Mohr’s circle is a graphical method of finding normal, tangential and resultant stresses AD = AO + OD
on an oblique plane. Mohr’s circle will be drawn for the following cases :
01 + 0o 0T — On
A(t) body subjected to two mutually perpendicular principal tensile stresses of unequal = + C0S2°

intensities. = on or Normal stress

A(it) body subjected to two mutually perpendicular principal stresses which are un- and ED = OE sin 20

equal and unlike (t.e., one is tensile and other is compressive). = £fJpL sin28

A( iii ) body subjected to two mutually perpendicular principal tensile stresses accompa- o.- or Tangential stress.
t
nied by a simple shear stress.

3.6.1. Mohr’s Circle when a Body is Subjected to two Mutually Perpendicular Important points. (See Fig. 3.22)

Principal Tensile Stresses of Unequal Intensities. Consider a rectangular body subjected (i) Normal stress is along the line ACB. Hence maximum normal stress will be when

to two mutually perpendicular principal tensile stresses of unequal intensities. It is required point E is at B. And minimum normal stress will be when point E is at C. Hence maximum

to find the resultant stress on an oblique plane. AB = and minimum normal = AB = o2 .

normal stress = a stress
L
Let Oj = Major tensile stress .
(ii) Tangential stress (or shear stress) is along a line which is perpendicular to line CB.
o = Minor tensile stress, and
2 Hence maximum shear stress will be when perpendicular to line CB is drawn from point O.
0 = Angle made by the oblique plane with the axis of minor tensile stress.
Then maximum shear stress will be equal to the radius of the Mohr’s circle.
Mohr’s circle is drawn as : (See Fig. 3.22).
q ~ q2
Take any point A and draw' a horizontal l
line through A. TakeAB = Oj and AC = o towards
E(Hi) When the point Bis at or at C, the shear stress will be zero.
2
(iv) The angle <)> (which is known as angle of obliquity) will be maximum, when the line
right from A to some suitable scale. With BC as
Odiameter describe a circle. Let is the centre of AE is tangent to the Mohr’s circle.
the circle. Now through O, draw a line OE
Problem 3.23. Solve problem 3.5 by using Mohr’s circle method.
marking an angle 20 with OB.

From E, draw ED perpendicular on AB. Sol. The data is given in problem 3.5, is

Join AE. Then the normal and tangential stresses o, = 120 N/mm2 (tensile)

on the oblique plane are given by AD and ED o = 60 N/mm2 (tensile)
2
respectively. The resultant stress on the oblique
plane is given by AE. 6 = 30°.

From Fig. 3.22, we have Scale. Let 1 cm = 10 N/mm2

Length AD = Normal stress on oblique plane —o, = 120— = 12 cm
Length ED = Tangential stress on oblique plane

Length AE = Resultant stress on oblique plane.
—
Radius of Mohr’s circle = Mohr’s circle is drawn as : (See Fig. 3.23).

Angle <j> = obliquity. Take any point A and draw a horizontal

Proof. (See Fig. 3.22) line through A. Take AB - o = 12 cm and AC =

x

CO = OB - OE - Radius of Mohr’s circle - qi ~°2 o = 6 cm. With BC as diameter (t.e. , BC = 12-6
2 2

O= 6 cm) describe a circle. Let is the centre of

the circle. Through O, draw a line OE making an Fig. 3.23

.

STRENGTH OF MATERIALS PRINCIPAL STRESSES AND STRAINS
130

angle 20 (i.e., 2 x 30 = 60°) with OB. From E, draw ED perpendicular to CB. Join AE. Measure AO = OC-AC
lengths AD, ED and AE.
Oi+o 2 c0! + a, - 2o2 0! -0
2
22
By measurements :
AD = AO + OD
Length AD = 10.50 cm
= AO + OB cos 20 OD OE('.' = cos 20)
Length ED = 2.60 cm

Length AE = 10.82 cm = Ol-O^ + 01+0, cog29 OE = Radius =

Then normal stress = Length AD x Scale AA
= 10.50 x 10 = 105 N/mm2. Ans.
= an or Normal stress
Tangential or shear stress = Length ED x Scale
ED = OE sin 20

= 2.60 x 10 = 26 N/mm 2 Ans. ^pL= sin2e (y OE = ^kp
.
= a or Tangential (or shear) stress. Lj
Resultant stress = Length AE x Scale t

= 10.82 x 10 = 108.2 N/mm 2 Ans. Problem 3.24. Solve problem 3.6 by using Mohr’s circle method.
.

3.6.2. Mohr’s Circle when a Body is Subjected to two Mutually Perpendicular Sol. Given : The data given in problem 3.6, is
Principal Stresses which are Unequal and Unlike (i.e., one is Tensile and other is
Oj = 200 N/mm2
Compressive). Consider a rectangular body subjected to two mutually perpendicular princi-
pal stresses which are unequal and one of them is tensile and the other is compressive. It is o = - 100 N/mm2 (compressive)
2
required to find the resultant stress on tin oblique plane. 0 = 30°.

Let a = Major principal tensile stress, It is required to determine the resultant stress and the maximum shear stress by Mohr’s
1

o = Minor principal compressive stress, and circle method. First choose a suitable scale.
2

6 = Angle made by the oblique plane with the axis Let 1 cm represents 20 N/mm2 .

of minor principal stress. 200 = 10 cm / ~"\E
\
Mohr’s circle is drawn as : (See Fig. 3.24) Then Oj =

ATake any point A and draw a horizontal line through o -100 = - 5 cm \/ // / \
2
Aon both sides of as shown in Fig. 3.24. Take AS = <:;,_(+) towards and = -gjj-

A AACright of and Mohr’s circle is drawn as given in Fig. 3.25. c \~
= o (~) towards left of to some suitable scale. o°° d )B
2 M— 1 00 — 4 J
Bisect BC at O. With O as centre and radius equal to CO or OB, 200
Take any point A and draw a horizontal line through \

draw a circle. Through O draw a line OE making an angle 20 Aon both sides ofA. Take AB = 0 = 10 cm towards right of \. /
j
/
with OB. A and AC = 0 = - 5 cm towards left of A. Bisect BC at O.
2
With O as centre and radius equal to CO or OB, draw a
From E, draw ED perpendicular to AB. Join AE and CE.

Then normal and shear stress (i.e., tangential stress) on the circle. Through O draw a line OE making an angle 20 (i.e.,

oblique plane are given by AD and ED. Length AE represents 2 x 30° = 60°) with OB. From E, draw ED perpendicular to Fig 3 25

the resultant stress on the oblique plane. Fig. 3.24 AB. Join AE and CE. Then AE represents the resultant

.-. From Fig. 3.24, we have stress and angle <j> represents the obliquity.

Length AD = Normal stress on oblique plane, By measurement from Fig. 3.25, we have

Length ED = Shear stress on oblique plane. Length AE = 9.0 cm

Length AE = Resultant stress on oblique plane, and Length AD = 6.25 cm and length ED = 6.5 cm

Angle <j) = 46°

Angle cj> = Obliquity. AE.-. Resultant stress = Length x Scale

—Radius of Mohr’s circle = CO or OB - = 9.0 x 20 = 180 N/mm2. Ans.
1 —

^

Angle made by the resultant stress with the normal of the inclined plane = <p = 46°. Ans.

Proof. (See Fig. 3.24). Normal stress = Length AD x 20

CO = OB = OE = Radius of Mohr’s circle = 6.25 x 20 = 125 N/mm2

a_ cr t + 2 Shear stress = Length ED x 20

~2 = 6.5 x 20 = 1^0 N/mm2.

—

132 STRENGTH OF MATERIALS PRINCIPAL STRESSES AND STRAINS

Maximum shear stress. Shear stress is along a line which is perpendicular to the line 2a 2 + o - o2 0i + a 2
AB. Hence maximum shear stress will be when perpendicular to line AB is drawn from point O. x
Then maximum shear stress will be equal to the radius of Mohr’s circle. 2
2

AD=AO + OD

Maximum shear stress = Radius of Mohr’s circle 01+02 + OE cos (20 - a) OD OE[-.- - cos (20 • a)]

N/mm= ct, + o2 = 200 + 100 = 160 , „ —~—= ~ + OE [cos 20 cos a + sin 20 sin a]
2 Ans.
.

3.6.3. Mohr’s Circle when a Body is Sub- +**.'***** _ q * °2 + OE cos 20 cos a + OE sin 20 sin a
i
jected to two Mutually Perpendicular Principal
Tensile Stresses Accomp anied by a Simple Shear T T
Stress. Consider a rectangular body subjected to two
D Ml II C _ + OE cos a . cos 20 + OE sin a . sin 20
mutually perpendicular principal tensile stresses of I
unequal intensities accompanied by a simple shear oblique I 2
stress. It is required to find the resultant stress on an *
oblique plane as shown in Pig. 3.26. /_

Let Oj = Major tensile stress g plarle ~~~ T 1 P ~~ + CF cos a , cos 20 + OF sin a . sin 20
1 t =l
* Z / _/
v 2
*. —/ -F , >-
( v OE = OF ~ Radius)

—A I- — — >j- B 01+02 + OB cos 20 + BF sin 20

o = Minor tensile stress
2
°?
ADx = Shear stress across face JSC and ( v OF cos a = OB, OF sin a = BF)

0 = Angle made by the oblique plane with Fig. 3.26 —= — + oo cog 20 + t sin 20 (v OB = CO, BF = x)
2
the plane of major tensile stress.

ABAccording to the principal of shear stress, the faces and CD will also be subjected to 0, +C 9 0i-O?2 _. —CO =°1_ a )
- 2
= + ^ cos 20 + x sin 20
a shear stress of x. r-‘

j

Mohr’s circle is drawn as given in Fig. 3.27. = o or Normal stress
71
Take any point A and draw a horizontal line through A.
ED = OE sin (20 - a) = OE (sin 20 cos a - cos 20 sin a)

Take AB = a and AC = o2 towards = OE sin 20 cos a - OE cos 20 sin a
1
Aright of to some suitable scale. Draw per-
= OE cos a . sin 20 - OE sin a . cos 20
pendiculars at B and C and cut off BF and
= OE cos a . sin 20 - OE sin a . cos 20 ( v OE = OF = Radius)
CG equal to shear stress x to the same scale.
= OB . sin 20 - BF cos 20 ( OF cos a = OB, OF sin a = BF)
Bisect BC at O. Now with O as centre and
(v OB = CO, BF^x)
radius equal to OG or OF draw a circle. = CO . sin 20 - x cos 20

'Through O, draw a line OF, making an an- ^ _ ^2. sjn 20 - T cos 20 v CO =

gle of 26 with OF as shown in Fig. 3.27. ^ lT? 2

From E, draw ED perpendicular to CB. = a or Tangential stress.
t
AJoin E. Then length AS represents the

resultant stress on the given oblique plane. N Maximum and minimum value of normal stress. In Fig. 3.27, the normal stress is

And lengths AD and ED represents the Mgiven by AD. Hence the maximum value ofAD will be when D coincides with and minimum

normal stress and tangential stress respec- Dvalue of AD will be when coincides with L.

tlvely Fig. 3.27 Maximum value of normal stress.
‘
Hence from Fig. 3.27, we have MAM(oa)max =
Length AE = Resultant stress on the oblique plane = A0 + 0
Length AD = Normal stress on the oblique plane
Length ED = Shear stress on the oblique plane. = °i + q2 + OF v .40 - - OM = OF - Radius

.

2

Proof. (See Fig. 3.27). Ql + ° 2 + yjOB2 + BF2 (v In triangle OBF, OF = JOB 2 + BF2 )

2

“[°i cr2] CB = a. °1 t£g. 4- If 2l~ a +T 2 —q ,BF = x

?.| OD = - 1
2

AO = AC + CO = o +-[0i- a2 ]
2

134 STRENGTH OF MATERIALS : 135

Minimum value of normal stress, (/ LO = OF = Radius) principal stresses and strains

(a) - = AL = AO - LO

' n'min.

_ Oj+Os _ nw

Pi +02 gi ~ °2 Oblique / 2
plane 15 .
65 N/mm
2 l2

( i ) For maximum normal stress, the point D coincides with M. But when the point D
coincides with M, the point E also coincides with M. Hence for maximum value of normal

stress, 20 = a (v Line OE coincides with line OM)

Angle

q ~ g2 Fig. 3.28
l
tan 20 - tan a = OB BF ~ x, OB =
Sol. Given
Qt -a2
Major principal stress, = 65 N/mm2
2

0i~02 Minor principal stress, a - 35 N/mm2
Shear stress, 2
% = 25 N/mm2

(ii) For maximum and minimum normal stresses, the shear stress is zero and hence the Angle of oblique plane, 0 = 45°.
planes, on which maximum and minimum normal stresses act, are known as principal planes
and the stresses are known as principal stresses. Mohr’s circle method

Let 1cm =10 N/mm2

D(Hi) For minimum normal stress, the point coincides with point L. But when the point Then 65
D coincides with L, the point E also coincides with L. Then ^°i = = 6.5 cm,

Angle 20 = it + a (v Line OE coincides with line OL) —cm and x = = 2.5 cm

a q = ^5" =
2
jt +
0= - (I1)
•• Mohr’s circle is drawn as given in Fig. 3.29.

22

From equations (i) and (ii), it is clear that the plane of minimum normal stress is in-
clined at an angle 90° to the plane of maximum normal stress.

Maximum value of shear stress. Shear stress is given by ED. Hence maximum value
of ED will be when E coincides with G, and D coincides with O.

Maximum shear stress,

(o ) max = OH = OF (v OH = OF = radius)

(

= Job 2 + BF 2 (•• In triangle OBF, OF = JOB 2 + BF 2 )

=f ~ °2 0B = o1 ~a2 bf
in 2

AProblem 3,25. point in a strained material is subjected to stresses shown in Fig. 3.28.

Using Mohr’s circle method, determine the normal and tangential stresses across the oblique
plane. Check the answer analytically.

Fig. 3.29

AC = Take any point A and draw a horizontal line through A. Take AB = Oj = 6.5 cm and
Draw perpendicular at B and C and cut off BF and CG
o2 = 3.5 cm towards right of A.

:,

136 STRENGTH OF MATERIALS PRINCIPAL STRESSES AND STRAINS 137

equal to shear stress x = 2.5 cm. Bisect BC at O. Now with O as centre and radius equal to OF Take any point A and draw a horizontal line through A. Take AB - o = 10 cm and AC =
L
(or OG ) draw a circle. Through O, draw a line OE making an angle of 26 (i.e. 2 x 45° = 90°) with
az ~ 5 cm towards right side of A. Draw perpendiculars at B and C and cut off BF = CG = x = 5
OF as shown in Fig. 3.29. From E, draw ED perpendicular to AS produced. Join AE. Then cm. Bisect BC at O. Now with O as centre and radius equal to OG (or OF), draw a circle cutting

ADlength represents the normal stress and length ED represents the shear stress. AMthe horizontal line through A, atL and Mas shown in Fig. 3.30. Then and AL represent the

By measurements, length AD = 7.5 cm and major principal and minor principal stresses.

length ED = 1.5 cm. By measurements, we have

.-. Normal stress (on ) = Length AD x Scale = 7.5 x 10 = 75 N/mm 2 Ans. AMLength = 13.1 cm and Length AL = 1.91 cm

.

(v 1 cm = 10 N/mm 2 LFOB (or 29) = 63.7°.

)

mmEDAnd = 2
tangential stress (a ) Length x Scale = 1.5 x 10 = 15 N/ Ans. AM= Length
t .

Analytical Answers .-. Major principal stress x Scale

Normal stress (on ) is given by equation (3.12). = 13.1x2 N/mm2 (v 1 cm = 2 N/mm2)
.'. Using equation (3.12), = 26.2 N/mm2. Ans.

—°n ~ — On Minor principal stress = Length AL x Scale
Of + On
+’ (Ji 2~ = 1.91 x 2 = 3.82 N/mm2. Ans.

2 °0S + X S*n 2®

— — —= 65 + 35 + 65--35 cos (2 x 45°) + 25 sin (2 x 45°) Location of principal planes

20 = 63.7°

= 50+15 cos 90° + 25 sin 90“ 63 7° Ans.
0 = = 31.85°.
= 50 + 15 x 0 + 25 x 1 (v cos 90° = 0, sin 90° = 1)
A
= 50 + 0 + 25 = 75 N/nun2. Ans.
The second principal plane is given by

Tangential stress is given by equation (3.13) 6 + 90° or 31.85° + 90° or 121.85°. Ans.

Using equation (3.13), Problem 3.27. An elemental cube is subjected to tensile stresses of 30 N/mm 2 and
10 N/mm2 acting on two mutually perpendicular planes and a shear stress of 10 N/mm2 on
_ gi ~ °2 sin 20 - x cos 29
these planes. Draw the Mohr’s circle ofstresses and hence or otherwise determine the magnitudes
2 and directions of principal stresses and also the greatest shear stress.

65-35 Sol. Given : = 30 N/mm2
Major tensile stress,
sin (2 x 45) - 25 cos (2 x 45)

= 15 sin 90° - 25 cos 90° = 15 x 1 - 25 x 0 = 15 - 0 Minor tensile stress, a = 10 N/mra2
Shear stress, 2

= 15 N/mm2. Ans. x = 10 N/mm2

Problem 3.26. At a certain point in a strained material, the intensities of stresses on two Scale. Take 1 cm = 2 N/mm 2

planes at right angles to each other are 20 N/mm2 and 10 N/mm2 both tensile. They are accom- —= = 15 cm

panied by a shear stress of magnitude 10 N/mm2 Find graphically or otherwise, the location of Then
.

principal planes and evaluate the principal stresses. — —10

Sol. Given a = = c5 cm andJ x= = 5 cm
2

Major tensile stress, = 20 N/mm2 Mohr’s circle of stresses is drawn as given in

Minor tensile stress, a = 10 N/mm2 Fig. 3.31.
Shear stress, 2
Take any point A and draw a horizontal line
x = 10 N/mm2

This problem may be solved analytically or graphi- through A.
cally. Here we shall solve it graphically (i.e., by Mohr’s
Take AB = Oj = 15 cm and AC = o = 5 cm to-
2
circle method). wards right side ofA. Draw perpendiculars at B and C

Scale. Take 1 cm = 2 N/mm2 and cut off BF - CG = x = 5 cm. Bisect BC at O. Now

T™hen Oj = —20 = 10 cm, ~10 with O as centre and radius equal to OG (or OF), draw

and, A as = A = 5 cm a circle cutting the horizontal line through A at L and
M AMas shown in Fig. 3.31. Then
—10 and AL represents

x = = 5 cm. the major and minor principal stresses respectively.

Mohr’s circle is drawn as given in Fig. 3.30. Fig. 3.30 OHAnd represents the maximum shear stress.

'. 1

138 STRENGTH OF MATERIALS PRINCIPAL STRESSES AND STRAINS 139

By measurements, we have 7, When a member is subjected to a simple shear stress (t), then the stresses on an oblique plane

Length AM = 17.1 cm are given as :

Length AL = 2.93 cm Normal stress, an = T sin 26

OHLength = Radius of Mohr’s circle Tangential stress, o = — x cos 26.
t

= 7.05 cm 8. When a member is subjected to two direct stresses (Oj , a2) in two mutually perpendicular directions

/.FOB (or 29) = 45°. accompanied by a simple shear stress (x), then the stresses, on an oblique plane inclined at an

angle 0 with the axis of minor stress, are given by :

Major principal stress ———a = Cl + °2 +

AM= Length x Scale Normal stress, "2 cos 29 + sin 20.

= 17.1x2 (v 1 cm = 2 N/mm 2 2
(v
= 34.2 N/mm2. Ans. )
Minor principal stress = Length AL x Scale
Tangential stress, o = 0l — °2 sin 20 - t cos 29
= 2.93x2 t

= 5.86 N/mm2. Ans. 1 cm = 2 N/mm 2 2t
AFOB or 20 = 45° (a) Position of principal planes is given by tan 20 =
)
°1~°2

—45 9.

9 = = 22.5°. Ans. (b) Major principal stress

a (c) Minor principal stress

The second principal plane is given by 0 + 90°.

Second principal plane = 22.5 + 90 = 112.5°. Ans. (d) Maximum shear stress
The greatest shear stress
= Length OH x Scale
= 7.05 x 20 = 14.1 N/mm2. Ans.

~——(e) Condition for maximum shear stress, tan 20 = .

2t

1. The planes, which have no shear stress, are known as principal planes. Mohr’s circle of stresses is a graphical method of finding normal, tangential and resultant stresses
on an oblique plane.

2. The stresses, acting on principal planes, are known as principal stresses. 10. Maximum shear stress by Mohr’s circle method, is equal to the radius of the Mohr’s circle.
11. The planes of maximum and minimum normal stresses are at an angle of 90° to each other.
3. Analytical and graphical methods are used for finding the stresses on an oblique section.

4. When a member is subjected to a direct stress (a) in one plane, then the stresses on an oblique EXERCISE 3

plane (which is inclined at an angle 0 with the normal cross-section) are given by :

Normal stress, o = a cos2 6 :‘J
n
|
Tangential stress. o. = sin 20 (A) Theoretical Questions
Max. normal stress *2 ’i
1. Define the terms : Principal planes and principal stresses.
=0 %
A2. rectangular bar is subjected to a direct stress (a) in one plane only. Prove that the norma! and
Max. shear stress a 1
~ 9. shear stresses on an oblique plane are given by
;
—cr = cr cos2 0
*
When a member is subjected to two like direct stresses in two mutually perpendicular direc- and =or, sin 20

‘2

tions, then the stresses on an oblique plane inclined at an angle 0 with the axis of the minor where 0 = Angle made by oblique plane with the normal cross-section of the bar,

stress (or with the plane of major stress) are given by : a - Normal stress, and
n

Normal stress, Ri±52 + £LZ£2 2e a = Tangential or shear stress,
Tangential stress. t

"2 2 A3. rectangular bar is subjected to two direct stresses (o' and o2 ) in two mutually perpendicular

———a = L

sin 20 directions. Prove that the normal stress (o ) and shear stress (a,) on an oblique plane which is
n

12 inclined at an angle 0 with the axis of minor stress are given by

Resultant stress. A,= 2 + 2 a = CN + ag Q1-Q2
a,
. cos 20

6. The angle made by the resultant stress with the normal of the oblique plane, is known as obliquity. 22

—It is denoted by <j>. Mathematically, tan <t> = ———and
I a, =

(2 sin 20.

4. Define the term ‘obliquity’ and how it is determined.

^ ^7 a point within a h j
‘
~—-saSSfesa:
, = 2ii^ +a -^ ^r«• the problem 7 d t ^" -Stre %*0,1 30 ob
2 2| cos 20 + T sjn 20
=5i2l22 tii) magnitmjg and ^^tudeofeariTofUm^p^’ 152-07 ^oua2]
( sin 26 - x cos 2 0 the
--*•"*“* [Ans- 154.057, - 4 057 'x, "PuJ stress and
9- Direct stresses of i fin Nr/
2

Z~r ** **»>

are unequal and like.

L ^ Numerical Problems ; S^':r ^**&S5£*
^ rectangular bar of cross
i

“'“I" *«'• whi.h i« .ol.i MW ““nw!“H"» > 25 m~ a strainer! a tresses.
alW»M, .
W» *1M m, Pianf
W J *“” »CWT “*• * ~to l„ li .,
” »*”*’ 7a
“ .J to 1

if a ^3-33-

! • <Ans- 92.378 kN] t50 N/mm 2 ““"» MW,
15

40 N/m m 2

®0 N/mm 2 [SO N/mm 2
40 N/mm2
B

<• The principal tensile stress %• 3.32

.

NW«»3 168 a.« *’ An elemental cube is sub' ^50 N/mm 2
m N/mm N/mmn• Fig, 3 33
^ cl P aI stresses at a noiHntf N/mmCAns. 0 875 W/rr% a^ P^ane inclined at 30°
Det. ermi.ne the resultant strec * , 2 21 -65 2 90 1301

3 bar are 160 2 it ’

- .,

T

aiaa. = + <* — a2 *"*»major principal plane by

I

strength of materials

4142

C>»;» Fnerav and Iwpoc* loading

INTRODUCTION *%£££&£m* abeorted the

4.1.

«^JSStt'ZS&XZ STs“h. ,.* dee. hr t»« *P>“ “

impact. The strain energy store

stretching the body.

4.2. SOME DEFINITIONS , t t„ a body due to gradually

1. Resilience body jg coinmonly known as resil-

2 Proof resilience, and

3 Modulus of resilience.

4.2.1. Besillenee. strain^tods^^^^^^
i *»«*defined .. *.
fence. of .

doing work, tie „ .. straining force. . . j : s known as

a.. P~f "•

Math-naPca®,
»f»“m,^.u as
.,. „ d,f,«d

4.2.3. Modwins -« prop .

unit volume. It is an important P wnf resilience

Modulus of residence - Volume of the body

APPLIED GRADUALLY j that the strain enerrggyy stored in a body is equal to the

In Art. 4.1, we .^T^Tdln stretching the body. “ “K”“tegt upt0 elastic limit. The
onhc 1
e dthe

S£KiSSSEi- s‘ 143

:

144 145

The load P performs work in stretching the body. 4.4. EXPRESSION FOR STRAIN ENERGY STORED IN A BODY WHEN THE LOAD IS

This work will be stored in the body as strain energy which APPLIED SUDDENLY

is recoverable after the load P is removed. When the load is applied suddenly to a body, the load is constant throughout the proc-
Let P = Gradually applied load,
ess of the deformation of the body.
x = Extension of the body,
Consider a bar subjected to a sudden load.
A = Cross-sectional area,
L = Length of the body, Let P = Load applied suddenly,
V = Volume of the body,
E = Young’s modulus, L = Length of the bar,
U = Strain energy stored in the body, and
A = Area of the cross-section,
a - Stress induced in the body. V = Volume of the bar =AxL,

Now work done by the load = Area of load ex- E = Young’s modulus,

tension curve (Shaded area in Fig. 4.1) Fig. 4.1 x = Extension of the bar,

ONM= Area of triangle a = Stress induced by the suddenly applied load, and

U - Strain energy stored.

As the load is applied suddenly, the load P is constant when the extension of the bar

= — X P X X. takes place.

2 Work done by the load = Load x Extension = P x x.

load, P = Stress x Area = a x A The maximum strain energy stored (i.e., energy stored upto elastic limit) in a body is

Extension L.-. Extension = Strain x given by

and extension, x - Strain x Length Length —U = x Volume of the body
2E
Strain = —= xAAxLT. (v Volume = AxL)

= ~ xL. Equating the strain energy stored in the body to the work done, we get

E ^ ~x A X L = P X X = P x X L. ^xt|[••• Frora equation (4.1), * =

Substituting the values of P and x is equation (i ), we get

—=
2
Work done by the load xoxAx— xL=— E xAxL Cancelling to both sides, we get

E^ 2

—= x V (v Volume V = Ax L) ^A=P or 0 = 2x4A- ...(4.5)
2E 2

But the work done by the load in stretching the body is equal to the strain energy stored From the above equation it is clear that the maximum stress induced due to suddenly

in the body. applied load is twice the stress induced when the same load is applied gradually.

.'. Energy stored in the body. After obtaining the value of stress (o), the values of extension Gc) and the strain energy

stored in the body may be calculated easily.

Problem 4.1. A tensile load of 60 kN is gradually applied to a circular bar of 4 cm

mdiameter an d 5 long. If the value ofE = 2.0 x 10s N/mm2 determine .
,
Proof resilience. The maximum energy stored in the body without permanent defor-
mation (i.e., upto elastic limit) is known as proof resilience. Hence if in equation (4.2), the ( i) stretch in the rod,
stress a is taken at the elastic limit, we will get proof resilience.
(it) stress in the rod
,

(iii ) strain energy absorbed by the rod.

Proof resilience x Volume Sol. Given

where a* = Stress at the elastic limit. Gradually applied load,
Modulus of resilience - Strain energy per unit volume
P = 60 kN = 60 x 1000 N

Dia. of rod, d - 4 cm = 40 mm

ETotal strain energy _ 2 mmA = — x 402 = 400 re 2

Volume V 4
L = 5 m = 500 cm = 5000 mm
Length of rod,

N; :

STRAIN ENERGY AND IMPACT LOADING

Volume of rod, V = A xL = 400jtx5000 = 2 x 10« ji mm3 Young’s modulus, E = 2 x 10 5 N/mm 2
Young’s modulus,
E = 2 x 10s N/mml ,

Let * = stretch or extension in the rod, Let ° = Instantaneous stress due to sudden load, and

o' = stress in the rod, and P = Suddenly applied load.

U = strain energy absorbed by the rod. The extension x is given by equation (4.1),
me ^ ^A N/mmNowT.
stress, a = Area = - 60000 =“_ 4,7.746 2 Ans. *=§xL or 1-5=^—8 x 3000
given by
. 400jt .
equation (4.1),
st, ret. c,h or extension is 1.5 x 2 x 10 5

B WT mm.Lv„ _ ~a. OUUU = 100 N/mm2. Ans.
4•*7I.. 7I 46
X Tr = 2~ x 5000 = 1.19 Sriuddi e1 nljy applied load

x 10 Ans.

The strain energy absorbed by the rod is given by equation (4.2), 2x- ^The
instantaneous stress produced by a sudden load is given by
i10nn0 = 29 xv
cCT = „P pi— equation (4.5) as

_° 2 ' 47. 476 2 2x10 * = 36810 -mm = 35.81 N-m. 2 X or

77 vT/ Ans. 1000 x 100

2E = 50000 N = 50 kN. Ans.

kNProblem 4.2. If in problem 4.1, the tensile load of 60 is applied suddenly determine • m mmProblem 4.4. A steel rod is 2

(i) maximum instantaneous stress induced suddenly applied to the rod. Calculate
neous elongation produced in the rod.
long and 50 in diameter. An axial pull of 100 kN is

(“') ^tantaneous elongation in the rod, and the instantaneous stress induced and also t"heimnssttaannttaa-

.(“*). strain energy absorbed in the rod. Take E = 200 GN/m2 .

Sol. Given : Sol. Given :

™ m “™' 2 x 10 Length, L = 2 m = 2 x 1000 = 2000 mm
? P=, Diameter, d = 50 mm
~ 2x 10 d ' Area = 400 * 2 L = 5000 mm. Volume =
N/mm2 and suddenly applied load,
>

(i) Maximum instantaneous stress induced 60000 N.

Using equation (4.5), " Area mm' A = | x 502 = 625 x 2
’

P 60000 Suddenly applied load,
A
W^ 493NW°o-~t2>xv
=2o x =95’ Ans. P = 100 kN = 100 x 1000 N

-

am\ilili)) Value of E = 200 GN/m2 = 200 x 109 N/m2
lIfnlSsift.arfn.nttan rn> eou/so elongation in .7 rod,
(v G
the = Giga = 9
)
Let x = Instantaneous elongation 10

_ 200 x 10 9 • m mm(••• 1 = 1000 mmm2 = 106

Then „ _ SL v r 95.493 „„„„ 6 N/mm 2

10 )

[see equation (4.1)] = 200 x 10a N/mm2

= 2.38 mm. Ans. Using equation (4.5) for suddenly applied load,

ini) Strain energy is given by, _9 P =2,, x 100 x 1000 N/mm2
A
~625x = 101.86 N/mm2. Ans.

ETr _ o* X ,, 95.493 2 143238 N-mm Let dL = Elongation
2
2x2xl05 X 2 x 106 ^Xt

_ = 143.238 N-m. Ans. Then 101.86

20oIl0J X 2000 = 1 -°186
££££££ W ^" “ mlong by ^ mmrtr ‘ A*®-

i dWed
7al^ZTo?unk ^bar due to suddenly «•**T» “*Ze.beLbIem A umf°rm
area and 3 f^? 4’5‘ met°l bar has
™™mm fE =2x10* N/mm* “ mm7 n , a 2 and a length of
k n°Wn ma the extension of the cross-sectional area of 700
nit ude, applied load. Take
applied load 5 MsA! if “ wba, will be
°f
is 1 ° d,7et.er the suddenly be suddenly npplied wilheuNZZZ

Sol. Given
:

Area of bar, mmA = 10 cm2 = 1000 2 Take E = 2x10* N/mm2.
Length of bar,
L = 3 m = 3000 mm ' Sol. Given

Extension due to suddenly applied load, Area, A = 700 mm2
Length,
x = 1.5 mm L = 1.5 m = 1500 mm
" Volume of bar,
mmV = A x L = 700 x 1500 = 1050000 2



150 STRENGTH OF MATERIALS STRAIN ENERGY AND IMPACT LOADING

The section x-x will be acted upon by the weight of the bar

of length ac. Wx = Weight of the bar of length x
Let
= (Volume of the bar of length x)
2 xA I?
=
x Weight of unit Volume —1 x -p X
2E
= (Axi)xp= pAx 3

W*L3
As a result of this weight, the portion dx will experience a = . ah,
small elongation db. Then 6E

Elongation in dx mProblem 4.8. The maximum stress produced by a pull in a bar of length 1 is
Length of dx
Strain in portion dx N!mm150 2 The area of cross-sections and length are shown in Fig. 4.3. Calculate the strain

.

Eenergy stored in the bar if = 2 x 10s N/mm2.

Weight acting on section x-x 2
Stress in portion dx -
c D
Area of section
’
pxAxr
Fig. 4.2 (a)

: 4 475 mm mm50 4 475 mm *

W<- - -k.

ipN\

Strain Fig. 4.3
p x .t p x x x dx
Sol. Given :
Length of bar, f=lm = 1000 mm

Max. stress. a = 150 N/mm2

p x x x dx Part AB : Length, Lj = 475 mm

Area, A = 200 mm2
t
Part BC : Length, mm
Now the strain energy stored in portion dx is given by, L = 50
dU = Average Weight x Elongation of dx Area, 2

Part CD : Length, Ag = 100 mm2

Area, L - 475 mm
3
Value of mmA
= x Ifej X 3 2
= 200
(jj-

E = 2 x 10® N/mm2

=(rH p x x x dx Wx(••• = pAr) In this problem, maximum stress is given. Axial pull P is not known. But stress in equal
?
to load/area. As load (or axial pull) for the bar is same, hence stress will be maximum, when
x^= |xp 2Ax 2
area will be minimum. Part BC is having less area and hence stress in part BC will be maxi-
WTotal strain energy stored within the bar due to its own weight is obtained by inte- mum. As parts AB and CD are having same areas, hence stresses in them will be equal.

grating the above equation from 0 to L. Let o = Stress in part BC = 150 N/mm2
2
rL
= Stress in part AB or in part CD
£/= dU
Now load = Stress x Area
JO
load = o xA = o A
1 1 2 2

cTo An 150 x 100 = 75 N/mm2

Now strain energy stored in part AB,

'"rr-' U, = -^xV,
1 2E 1

—

STRAIN ENERGY AND IMPACT LOADING

e Vj = Volume of part AB The strain in the bar is given by,

= A xL = 200x475 —Strain - Str-ess
1 1
th
= 95000 mm3

Substituting this value in equation (i), we get bL a

a2 i.e., L "E

Ul= * 95000 ~bL = x L
2E

- x 95000 Work done by the load = Load x Distance moved
2 x 2 x 10 6
= P(Ji + bL)

= 1335.938 N-mm The strain energy stored by the rod,

Strain energy stored in part BC, —U V= -grr x =

&U E= 2E 2E x AL
2
2 x V,2 Equating the work done by the load to the strain energy stored, we get

150 2 „2

A Lg- x 2 x 2 P(h + bl)=^.AL

2 x 2 xlO 1

p(h + = .AL “-I- 1

)

x 100 x 50 = 281.25 N-mm ~hPk - p i- L AL

2 x 2 x 10 5
Energy stored in part CD,

2 —SL~

2E
U= * V = 1335 ‘ 938 N-m V UVa = PAL - E . L- Ph = 0
s 3
..

2? lt o = Gl 3 = Uf)
3

Total strain energy stored, Multiplying by to both sides, we get

U uU = C/j + 2 + 3 = 1335.938 + 281.25 + 1335.938 N-mm AL

= 2953.126 N-mm. Ans. d^-P.- .Lx-^---Ph.~ = 0
E A. L AL

4.5. EXPRESSION FOR STRAIN ENERGY STORED a2 -=j- .a- A.L = 0.

IN A BODY WHEN THE LOAD IS APPLIED A

WITH IMPACT The above equation is a quadratic equation in ‘o’,

The load dropped from a certain height before the 2P If 2pV „ 2PEh b 2 ± -J4ac
load commences to stretch the bar is a case of a load ap- (Neglecting - ve root)
plied with impact. Consider a vertical rod fixed at the up- A ± vLJ A.i
per end and having a collar at the lower end as shown in
Kg. 4.4. Let the load be dropped from a height on the col- 2x1
lar, Due to this impact load, there will be some extension
P [4P2 8 .PEh P
in the rod. ± 2+
“A I 4A 4.A.L A
LetP = Load dropped (i.e., load applied with impact)
L - Length of the rod, V

A = Cross-sectional area of the rod, P 2 2PEh
V = Volume of rod = A x L, ffP"|
+
h = Height through which load is dropped, UJ+ |
A.L
bL = Extension of the rod due to load P, ai
Ht 2PEh A2 P P
E = Modulus of elasticity of the material of rod, A.L * P 2 ~ A* A 1

a = Stress induced in the rod due to impact load. = ^fi+Ji7 2AEh)

Fig. 4.4 After knowing the value of ‘o’, the strain energy can be obtained.

)

STRAIN ENERGY AND IMPACT LOADING

Important Conclusions NProblem 4.10. A load of 100 falls through a height of 2 cm on to a collar rigidly
(i) If 5L is very small in comparison with h.
The work done by load = P.h mattached to the lower end of a vertical bar 1.5 long and of 1.5 cm.2 cross-sectional area. The
Equating the work done hy the load to the strain energy stored in the rod, we get
upper end of the vertical bar is fixed.

P . h = —— . AL Determine :
2E
(i ) maximum instantaneous stress induced in the vertical bar,
(z£) maximum instantaneous elongation, and

2E.P.h (iii) strain energy stored in the vertical rod.

, = and =cr . Take E = 2 x 10s N/mm2.

cr AL

.

(n In equation (4.7), if h = 0, we get Sol. Given: P = 100 N
Impact load,
a=-P(l+^ITo)=p~a op
+ l)=-j- Height through which load falls,

which is the case of suddenly applied load. h = 2 cm = 20 mm

Once the stress p is known, the corresponding instantaneous extension (6L) and the Length of bar, L = l,5m = 1500 mm

strain energy (U) can be obtained. mm mmA = 1.5 cm2 = 1.5 x 100
Area of bar, 2 = 150 2
Volume,
A mmProblem 4.9. weight of 10 kN falls by 30 on a collar rigidly attached to a vertical mmV = A x L = 150 x 1500 = 225000 3

aw and mm2 *n sect^on- ^ind the instantaneous expansion of the bar. Take EModulus of elasticity, = 2 x 105 N/mm2
(Bhavnagar University, Feb. 1992)
- 210 GPa. Derive the formula you use. Let a = Maximum instantaneous stress induced in the vertical bar,

Sol. Given : bL = Maximum elongation, and

Falling weight, P = 10 kN = 10,000 N U - Strain energy stored.
Falling height,
h ~ 30 mm (i) Using equation (4.7),
Length of bar,
Area of bar, L = 4 m = 4000 mm --
Value of
A = 1000 mm2 a=-PLi + 2 x 150 x 2 x 10s x 20
1 100 x 1500
E = 210 GPa = 210 x 109 N/m2

G( '•' = Giga = 109 and Pa = Pascal = 1 N/m 2) N/mm• (1 + a/ 1 + 8000 ) = 60.23

N210 x 109 2 Ans.
.

mm m mm' I hi - 1000
mm106 2 and 2 = 10 6 2 («) Using equation (4.6),.
)
h_L-—o xLT
= 210 x 103 N/mm2 = 2.1 x 105 N/mm2 60.23x1500 Ans.
E
Let dL = Instantaneous elongation due to falling weight 2 x10s = 0.452 mm.
cr = Instantaneous stress produced due to falling weight
(iii) Strain energy is given by,
Using equation (4.7), we get
—U = o2 fiO 23 2 s
°^m) 2E x V= 2 x 2 x 10,5
x 225000 = 2045 N-mm

= 2.045 N-m. Ans.

_ MOOO ( L 2 x 2.1 x IQ5 X 1000x30 Problem 4.11. The maximum instantaneous extension, produced by an unknown fall-
lt +t
ming weight through a height of 4 cm in a vertical bar of length 3 and of cross-sectional area
1000 V 10000x4000
5 cm2 is 2.1 mm.
v) ,

= 10 ( 14-717315) = io(i + V3le) Determine :

= 10 X 18.77 = 187.7 N/mm2 (i) the instantaneous stress induced in the vertical bar, and

N/mmE(iii ) the value of unknown weight. Take = 2 x 10s 2

.

6L a_ Sol. Given :

Strain L ~E mmInstantaneous extension, bL = 2.1

Length of bar, L = 3 m = 3000 mm

—6L = x L = 187.7 x 4000 = 3.575 mm. Ans. Area of bar, A = 5 cm2 = 500 mm2
fcj
Volume of bar, mmV = 500 x 3000 = 1500000 3

—J '

STRAIN ENERGY AND IMPACT LOADING

mmHeight through which weight falls, h = 4 cm = 40 =—or 80 1+ hi+ 2 x 600 x 20 * X° 5 * 10 )
— —or 600 P x 5000
EModulus of elasticity, = 2 x 106 N/mm2 \|

^et o = Instantaneous stress produced, and j
P = Unknown weight,
48000 £ 480000.
—Hri. Stress r^ +
We ki now Strain Enor S0„4t.r„.e.s~.s. = mlh + 1

x.. nS.train ror ~
48000 _ l _
480000

E E —Instantaneous stress = x Instantaneous strain = x (— > — -f-Squaring both sides,
L

= 2 x 105 * 21 N/mm2 = 140 N/nun2. Ans. ^f 48000 2x48000

J , f 480000
1+

Equating the work done by the falling weight to the strain energy stored, we get — — —o.r.
2304000000 96—000 = 480—000 ,
—P(h + bL)= xV
9.W. “"2 (cancelling 1 to both sides)

WO ——or or 2304000000 _ 480000 + 96000 _ 576000
p2
+ 2.1) = g- x 1500000 = 73500 p p p

2x 2x 10 5 —23040—00000 = 576000

—or f 1 ,, \

P=or „ 73500 ^I cancelling to both sides I

~^T=, 1745.8 N. Ans. or P„ = 2304000000 ~= 4000 N = 4 kN. Ans.

Note. The value of P can also be obtained by using equation (4.7). 57600 O

4‘ 12- An unknown weight falls through a height of 10
mma„t„tac,hed on a collar rigidly mm mmProblem 4.13. A bar 12
to the lower end, ofa vertical bar 500 cm long and 600 mm? in section, //the maximum diameter gets stretched by 3 under a steady load of

ZZ fkL X mmunknown 8000 N. What stress would be produced in the same bar by a weight of 800 N, which falls
°r lt be 2 what is the corresponding and magnitude of the
weight ? = 2.0x stress (AMIETwinter 1984) vertically through a distance of 8 cm on to a rigid collar attached at its end ? The bar is initially
Tak?e El 10* N/mm2
. Eunstressed. Take = 2.0 x 105 N/mm 2
(AMIE, Winter 1986)
.

Sol. Given : Sol. Given :

mmHeight through which the weight falls, h = 10 Dia. of bar, d = 12 mm

Length of the bar, L = 500 cm = 5000 mm

Area of the bar, A = 600 mm2 v Area of bar. mmA - — (12)2 = 113.1 2

mmMaximum extension, SZ, = 2 Increase in length. 4
Steady load.
Young’s modulus, E = 2.0 x 10s N/mm2 Falling weight, SL = 3 mm
Vertical distance.
Young’s modulus, W = 8000 N
Let
a = Instantaneous stress produced in the bar, and P = 800 N
With steady load
P = Weight falling on the collar. h = 8 cm = 80 mm

We know E _ Stress E = 2.0 x 105 N/mm*

Strain L = Length of the bar, and

EStress = x Strain = E x Strain = - a = Stress produced by the falling weight.

Substituting the known values, we get ( Steady load

——a = 2.0 x 105 x oUUU = go N/mm2. Ans. We know Stress i, Area
Strain
Value of weight falling on the collar or 6L

Using equation (4.7), L

Zcr=P— L1+11 + 2A.E.h f 8000^1 8000 L
A V P.L 2.0 x 105 = [,1131

5

STRENGTH OF MATERIALS, STRAIN ENERGY AND IMPACT LOADING 159

mm2.0 x 10 5 x 1131 x 3 mm2.1 X 10 5 X 122.72 X 3.2
8482.5
L = " 8246 ' 7
Now using equation (4.7), we get 2AEh 1A0M

PL , Now using equation (4.7), we get

P(, L 2AEh

o -Trtr ptt

2 x 113.1 x 2.0 x 105 x 80 N/mm2 700 2 x 122.72 x 2.1 x 10 3 x 75
1 + Jl + “ 1+ +
700 x 8246.7
8.0 x 8482.5 122.72 i V

= 7.0734(1 + 7l + 533.33 ) = 7.0734 x 24.1155 = 153.74 N/mm2. Ans.

= 170.578 N/mm2. Ans. AProblem 4.15. vertical round steel rod 1.82 metre long is securely held at its upper

mm mmProblem 4.14. A rod 12.5 Aend. weight can slide freely on the rod and its fall is arrested by a stop provided at the lower
mmend ofthe rod. When the weight falls from a height of30
in diameter is stretched 3.2 under a steady load above the stop the maximum stress

of 10 kN. What stress would be produced in the bar by a weight of 700 N, falling through reached in the rod is estimated to be 157 N/mm2. Determine the stress in the rod if the load had

mm75 before commencing to stretch, the rod being initially unstressed ? The value ofE may been applied gradually and also the minimum stress if the load had fallen from a height of

be taken as 2.1 x 10s N/mm2. (AMIE, Winter 1988) 47.5 mm.

Sol. Given : Take E =2.1 x 10s N/mm2 .
Dia. of rod,
d = 12.5 mm Sol. Given :
Length of rod, mmmL — 1.82 = 1.82 x 1000 = 1820

.•. Area of rod. mmA -= x 12. 2 = 122.72 2 Height through which load falls, h = 30 mm

Increase in length, 4 Maximum stress induced in the rod, a = 157 N/mm2
Steady load,
Falling load, bL = 3.2 mm Modulus of elasticity, E- 2.1 x 105 N/mm2
Falling height,
W = 10 kN = 10,000 N Let o = Stress induced in the rod if the load is applied gradually and
s
P = 700 N 0, = Maximum stress if the load had fallen from a height of 47.5 mm.

h = 75 mm Strain energy stored in the rod when load falls through a height of 30 mm,

Young’s modulus, E - 2.1 x 10s N/mm2

Let L = Length of the rod, —U = x Volume V2 x
2E r5
o - Stress produced by the falling weight. x 2.1 x
10

We know E = |^i = 0.05868 x V N-m

The extension of the rod is given by equation (4.6),

'Steady load' 5L = |xL
Area j
— mm1 ^7
UJ
= x 1820 = 1.36
' 2.1 x 10 s

10 000 '! .-. Total distance through which load falls

, mm= h + 6L = 30 + 1.36 = 31.36

2.1 x 105 = 122.72 J .. Work done by the falling load = Load x Total distance

m ' =Px 31.36

10,OOP (L_ Equating the work done by the falling load to the strain energy stored, we get ,
'!
X P x 31.36 = 0.05868 x V

122.72 J i 3.2

h

160 STRENGTH OF MATERIALS STRAIN ENERGY AND IMPACT LOADING
;

Weight, P = 10 kN = 10,000 N

Height through which weight falls,

h = 3 mm

Young’s modulus for steel, E. = 2 x 105 N/mm 2

Young’s modulus for brass, E-. = 1.0 x 10 5 N/mm2

Let a = Stress in steel tube, and
s

a = Stress in brass tube.
b

As both the ends are fixed together,

Strain in steel rod = Strain in brass tube

ue - ZL = ?tL Strain = -
E
E. b

^-xEa ab X 2 x10s =2Xo
s 6b
= Eb s lx 10 s
*

Now volume of steel rod, V = Area x Length
s

= A xL- 100 it x 2500
s

mm= 250000 it 3

mmV A L - = 3
b b
Volume of brass tube, = x 114.75 n x 2500 286875 tc

.•. Strain energy stored in steel rod,

y17 = -2L X s = (2x ° 6) X 250000 It
E8 2 2X 2
= 3.4 (l + V1 + 3219.24) X 10 s.
s

= 196.64 N/mm2. Ans. = 7.854 06 .

AProblem 4.16. vertical compound tie member fixed rigidly at its upper end, consists of and strain energy stored in brass tube,

m mm mma steel rod 2.5 long and 200 22
in diameter, placed within an equally long brass tube 21
JT = x 2866887755H71
mmin internal diameter and 30 external diameter. The rod and the tube are fixed together at b
6 2Eb 2 x 1 x 10 5

the ends. The compound member is then suddenly loaded in tension by a weight of 10 kN = 4.506 o2
A.
mmfalling through a height of 3 on to a flange fixed to its lower end. Calculate the maximum
Total strain energy stored in the compound bar;
and Assume E -2 x 10s N/mm2 and E = 10s N/mm2
stresses in steel brass. s b 1.0 x . C7=C7 + G
s6
Sol. Given :
= 7.854 a 2 + 4.506 o 2
m mmLength of steel rod, L = 2.5 = 2500 6

mmDia. of steel rod, = 12.36 a2
.
d
s = 20 Work done by the falling weight = Weight ( + 6L)

.-. Area of steel rod, A =7 x 202 = 10000 (3 + bh)

4 As both the ends are fixed,
.

mm= 100 it 2 The strain in steel rod = Strain in brass rod

mmInternal dia. of brass tube = 21

mmExternal dia. of brass tube = 30 But strain in brass rod Strain = -

.'. Area of brass tube, A ^& = (30 2 ~ 21 2
Length of brass tube,
)

mm= 114.75 it 2

= 250 cm (v L = 2500 mm)

= 2500 mm Fig. 4.5 = 0.025 a .
6

162 STRENGTH OF MATERIALS STRAIN ENERGY AND IMPACT LOADING 163

Substituting this value of 6L in equation (Hi), we get

Work done by falling weight = 10000 (3.0 + 0.025 a ) —(iv)
6

Now equating the work done by the falling weight to the total strain energy stored

[i.e., equating equations (iv) and («)], we get 21 x 2 x 10° x 3000x200

10000 [ 3 + 0.025 a ] = 12.36 or =
6 \ 2000 x 4000

or 30000 + 250 a = 12.36 of ( A = 2000 mm2 L = 4000 mm)
b ,
= 173.2 N/mm2. Ans.
or 12.36 a 2 - 250 of - 30000 = 0
b
N2nd
250 30000 Case. Falling weight, P = 30 kN = 30000
06 2
2 „
°r ° b cm = mm
Height, h = 2 20
12.36 12.36 2

o\ - 20.226 o - 2427.18 = 020.226 Let o = Maximum stress induced.
b 2
or

The above equation is a quadratic equation. In this case falling weight is having a large value. Hence the extension produced by a

20.226 ± J20.226 2 + 4 x 242718 large weight will be large. Moreover the height through which this weight falls is 2 cm only.

a. = Hence the extension in the bar, in comparison to the height through which weight falls, is not

- negligible.

20.226 2

± ^409.09 + 9708.72 .-. Using equation (4.7), we get
20.226
=

2 P L [ 2AEh

± 100.587 °~a 1 *r*~plT

2 1 L 2AEh2

= + 100.587 (/%NTegl,ectxi. ng - ve root) A

N= 60.4 /mm2. Ans. P .L
2

From equation (z), we get o = 2 x o = 2 x 60.4 i 2 x 2000 x 2 x 10 s x 20
s b
30000 f
= 120.8 N/mm2. Ans. 2000
V 30000x4000
A mmProblem 4.17. vertical bar 4 metre long and of2000 2 cross-sectional area is fixed

at the upper end and has a collar at the lower end. Determine the maximum stress induced = 15 (1 + 11.590) = 188.85 N/mm2 Ans.

when a weight of : AProblem 4.18. crane-chain whose sectional area is 6.25 cm2 carries a load of 10 kN.
mAs it is being lowered at a uniform rate of 40 per minute, the chain gets jammed suddenly, at
N(i) 3000 falls through a height of 20 cm on the collar,
which time the length of the chain unwound is 10 m. Estimate the stress induced in the chain
kN(ii) 30 falls through a height of 2 cm on the collar.
due to the sudden stoppage. Neglect the weight of the chain. Take E =2.1 x 10s N/mm2.
Take E = 2.0 x 10s N/mm2.

Sol. Given : (AMIE, Summer 1989)

Length of bar, L = 4 m = 4000 mm Sol. Given :

Area of bar, A = 2000 mm2 Area, mmA = 6.25 cm2 = 625 2
Load,
mm.-. Volume of bar, V= A xL = 2000 x 4000 = 8000,000 3 W = 10 kN = 10,000 N

N1st Case. Falling weight, Pj = 3000

Height, h - 20 cm = 200 mm Velocity, V = 40 m/min ;
Let l
60 3
(Tj = Maximum stress induced.
mmmLength of chain unwound = 10 = 10 x 1000

In this case the falling weight is small as compared to second case. The small weight will L = 10,000 mm
produce a small extension of the bar. Hence the extension in the bar will be negligible as
compared to the height of 20 cm through which the weight falls. Value of E = 2.1 x 10s N/mm2

.-. Using equation (4.8), we get Let 0 = Stress induced in the chain due sudden stoppage.

K.E. of the crane —-mV= 1 tt2i = -1(W) x Vt , 2
UJ2 2

1

.

164 STRAIN ENERGY AND IMPACT LOADING 165

But the maximum energy stored

1 f 10000 N m = 226.5 N m — xA xL= x 2500 x 15000 N mm
2 I 9.81
2E 2x2 x 10s ...'(«);

= 226.6 x 1000 N mm = 226500 N mm ...(£) But K.E. of the cage = Energy stored in the rope

When the chain gets jammed suddenly, the whole of the KE. of the crane is absorbed in

the chain. But the enei'gy stored or absorbed in the chain 30000 x 1000 2x2x10 s x 2500 x 15000

9.81 30000 x 1000 x 2 x 2 x 10 5

= oz xAxL 9.81 x 2500 x 15000

-rzr 30000 x 1000 x2x2x 10s

mmNr- x 625 x 10,000 1 180.61 N/mm2. Ans.

2 x 2.1 x 10 V 9.81x2500x15000

Now K.E. of crane - Energy stored in the chain

226500 : x 625 x 10,000 4.6. EXPRESSION FOR STRAIN ENERGY STORED IN A BODY DUE TO SHEAR

STRESS

2 x 2. lx 10 s Fig 4.6 shows a rectangular block of length /, D

height h and breadth b, fixed at the bottom face AB. "T" D, C C, P

226500 x2x2.1xl0 5 Let a shear force P is applied on the top face CD and h ; 7
°9 ~ hence the top face moves a distance equal to CC r
/ I
625x10,000 Let x = Shear stress produced, /
/
226500 x2x2.1xl05 <j> = Shear strain, and ty /

°~ C = Modulus of rigidity. / /
V 625 x 10,000 Now shear stress,
Vr? v yr j
= 123.37 N/mni*. Ans. _ Shear force
Area / / / / / / // ry
AProblem 4.19. kNcage weighing 60 is attached to the end of a steel wire rope. ft. is
(v Area of top face = l x b) 1
lowered down a mine shaft with a constant velocity of 1 mis. What is the maximum stress
P Fig. 4 6

produced in the rope when its supporting drum is suddenly jammed ? The free length of the

~ope at the moment of jamming is 15 m, its net cross-sectional area is 25 cm 2 and E

= 2 x 10s N/mm 2 The self-weight of the wire rope may be neglected. (AMIE, Winter 1990) lx b
.
P=xx lx b
Sol. Given :
Weight, W = 60 kN = 60,000 N CC,

and shear strain, „ n=<j>
L,r>
Velocity, V = 1 m/s
CC = CB.
L = 15 m = 15,000 mm 1
Free length,
—If the shear force P is applied gradually, then average load will be equal to p
Area, mmA = 25 cm2 = 25 x 100 2

Value of E = 2 x 10s N/mm2 Work done by gradually applied shear force

= Average load x Distance

K.E. of the cage

2 2{g ^ CC |=
x = (t x l x b). (CB. 4.)

l

(v p = xxlxb and CC = CB. <h)
r

= — .x .Ixbxh. (|> ( v CB = h)

30000 x 1000 -= — ,xx$xlxbxh = — ,tx x Volume of block

This energy is to be absorbed (or stored) by the rope. v = Shear strain Shear strain

Let a = Maximum stress produced in the rope when its supporting drum is suddenly —= —1 x x V
2C
jammed.

(y V= Ixbxh)

L C,

166 STRENGTH OF MATERIALS strain energy and impact loading

But the work done is equal to the strain energy stored. where P = Impact load,

h = Height through which load falls.

Strain energy stored = 7^7 x V ...(4.9) 8. To find the expression for the stress induced in a body either by suddenly applied load or by an
impact load, the strain energy stored in a body is equated to the work done by the load.
Problem N/mm 2
4.20. The shear stress in a material at a point is given as 50 Deter- 9. The energy stored in a body due to shear stress (x) is given by
.

mine the local strain energy per unit volume stored in the material due to shear stress. Take x—2 xV

C = 8xl04 N/mm2. U=

Sol. Given : 2

Shear stress, t = 50 N/mm2 where V = Volume of the body, and

Modulus of rigidity, C = 8 x 10 4 N/mm2 C = Modulus of rigidity.

Using equation (4.9), EXERCISE 4

Strain energy —= x Volume = x Volume
2C 2 x 8 x 10 4
(A) Theoretical Questions

= 0.015625 x Volume 1. Define the following terms :
(i) Resilience
Strain energy per unit volume Strain energy

= 0.015625 x Volume = 0.015625 N.../mm2„. (t'ii) Impact loading, and (>«) Spring.

. Ans. 2. Define resilience, proof resilience and modulus of resilience.

Volume

3. Find an expression for the strain energy stored in a body when

HIGHLIGHTS (i) the load is applied gradually («) the load is applied suddenly and

1. The energy stored in a body due to straining effect is known as strain energy. (iii) the load is applied with an impact.
4. Prove that the maximum stress induced in a body due to suddenly applied load is twice the
2. Resilience is the total strain energy stored in a body. Resilience is also defined as the capacity of
a strained body for doing work on the removal of the straining force. stress induced when the same load is applied gradually.
5. Derive an expression for the stress induced in a body due to suddenly applied load and hence
3. The maximum strain energy stored in a body is known as proof resilience. The proof resilience is
find the value of extension produced in the body.
given by,
6. Prove that the maximum strain energy stored in a body is given by,
a
—U = a2 x Volume
Proof resilience - — x Volume 2E
2E

where a = Stress at the elastic limit. where a = Stress at the elastic limit.
7. Explain the terms : Gradually applied load, suddenly applied load, and load applied with an
4. The proof resilience of a body per unit volume is known as modulus of resilience.
5. The maximum stress induced in a body is given by impact.

o = —P 8. Prove that the stress induced in a body when the load is applied with impact is given by,

if the load P is applied gradually

P Pif the load is applied suddenly A = Cross-sectional area of the body,
=2 ^
where P = Load applied with impact,

. if the load P is applied with impact h = height through which load falls, L = Length of the body, and

Awhere = Cross-sectional area of the body, E = Modulus of elasticity.

h = Height through which load falls, If the extension produced in a rod due to impact load is very small in comparison with the height
through which the load falls, prove that stress induced in the body will be given by
E = Modulus of rigidity,
2EPh
l
L = Length of the body. a= Im-
yA.

6. The maximum stress induced in a body due to suddenly applied load is twice the stress induced prove that the strain energy stored in a body due to shear stress is given by
when the same load is applied gradually.

7. If the extension produced in a rod due to impact load is very small in comparison with the height where x = Shear stress,

through which the load falls, then the maximum stress induced in body is given by C = Modulus of rigidity, and
V = Volume of the body.
liE.P.h
a= ,

\ A.

168 . STRENGTH OF MATERIALS STRAIN ENERGY AND IMPACT LOADING 169

11 Explain the following terms : (t) Proof stress, (ii) Proof resilience, and (iii) Modulus of resilience. mm5 on to a flange fixed to its lowe'r end. Calculate the maximum stresses in steel and brass.
(Bhavnagar University, Feb. 1992) E 10c kgfcm2. [Ans. 1173.5 kgfcm2 , 586.76 kgfcm2]
Assume £ =2x 10s kgfcm2 and b = 1.0 x

(B) Numerical Problems A12 . circular rod 5 cm in diameter and 3 metre long hangs vertically and has a collar securely

attached to the lower end. Find the maximum stress induced : (i) when a weight of 250 kgf falls
through 15 cm on the collar, (ii) when a weight of 2500 kgf falls 1.5 cm on the collar. Take
A E m1.
tensile load of 50 kN is gradually applied to a circular bar of 5 cm diameter and 4 long. If the E = 21 x 106 kgfcm2 [Ans. (i) 1635 kgfcm 2 (ii) 1767 kgfcm2 ]
value of 2 .0x stretch in the the rod, and . ,
= 105 N/mm2, determine: (i) rod, (ii) stress in
The shear stress in a material at a point is given as 45 N/mm2. Determine the local strain energy
(iu) strain energy absorbed by the rod. [Ans. (i) 0.0509 cm, (ii) 25.465 N/mm2 (iii) 12.73 Nm] 13 . per unit volume stored in the material due to shear stress. Take C = 8 x 10 N/mm .

2. question
kNIf in
1 the tensile load of 50 is applied suddenly, determine : (i) maximum instanta- [Ans. 0.01265 N/mm2]
, in the rod,
and (Hi) strain energy absorbed in
neous stress induced, («) instantaneous elongation

r°d W[Ads- 50.93 N/mm2 (ii) 0.1018 cm (iii) 50.93 Nm]
' ,

m3. Calculate instantaneous stress produced in a bar 10 cm2 in area and 4 long by the sudden

application of a tensile load of unknown magnitude, if the extension of the bar due to suddenly

Eapplied load is 1.35 mm. Also determine the suddenly applied load. Take = 2 x 105 N/mm2 .

A [Ans. 67.5 N/mm2 33.75 kN]
,

4 . uniform metal bar has a cross-sectional area of 6 cm2 and a length of 1.4 m. If the stress at the
elastic limit is 15 tonne/cm , find the proof resilience of the bar. Determine also the maximum
value of an apphed load, which may be suddenly applied without exceeding the elastic limit.
Calculate the value of the gradually applied load which will produced the same extension as that
produced by the suddenly applied load above. Take = 2000 tonnes/cm2.
E.

A m m5.
tension bar 6 long is made up of two parts, 4 metre of its length has a cross-sectional area
12.5 cm while the remaining 2 length has a cross-sectional area of 25
the total strain energy produced in the cm2 . An axial load of
5 tonnes is gradually applied. Find
bar and compare this
value with that obtained in a uniform bar of the same length and having the same volume when
Eunder the same load. Take = 2 x 10 6 kgf/cm 2
. [Ans. 242 kgfcm, 1.054]

A N m6.
a load of 200 falls through a height of 2.5 cm on to a collar rigidly attached to the lower end of
vertical bar area. The upper end of the vertical bar
2 long and of 3 cm2 cross-sectional is

fixed. Determine : (i) maximum instantaneous stress induced in the vertical bar, (ii) maximum
instantaneous elongation, and

E“ 2 X 10 kg£tan
•
(iii) strain energy stored in the vertical rod. Take

[Ans. (i) 58.4 N/mm 2 (ii) 0.0584 cm (iii) 511.5 Nm]
m m7. The maxi. mum instantaneous, produced by an unknown falling weight through a height of 4 cm
a vertical bar of length 5 and of cross-sectional area 5 cm 2 is 1.80 mm. Determine (i) the
instantaneous stress induced ,
:

in the vertical bar and (ii) the values of unknown weight.
Take E = 2 x 106 kgfcm2
. [Ans. (i) 72 N/mm 2 and (ii) 775.1 N]

mm8. An unknown weight falls through a height of 20 on a collar rigidly attached to the lower end
f <mW mm mm9.
of a vertical bar 5 long and 800 2 in section. If the maximum extension of the rod is to be

.5 mm, what is the corresponding stress and magnitude of the unknown weight ? Take
= 2-° x 10 kS
- [Ans. 1000 kgf'cm 2 444.44 kgf]
,
A bar 1.5 cm diameter gets stretched by 2.5
under a steady load of 100 kgf. What stress
would be produced in the same bar by a weight
of 120 kgf, which falls vertically through a
distance 5 cm on to a rigid collar attached at its
E-2.0x 10- Kgf/cm2 end ? The bar is initially unstressed. Take
kgW. [Ans. 130iU4
A m A10 vertical round steel rod 2 long is securely held at its upper end. weight can slide freely on
the rod and its fall is arrested by a stop provided at the lower end of the rod. When the weight
gh of 2-5 cm above the st°P’ the maximum stress reached
k"gfS/cm»2 in the rod is estimated
. ibfe 1450 . Determine the stress in the rod if the load had been
stress if the load had fallen from a height of 4.5. Take applied gradually and
to
E = 2.0 x 106 kgf'cm2.
also the maximum

A [Ans. 39.743 kgf/cm 2 193.42 kgfcm 2 ]
,

vertical compound tie member fixed

and 20 diameter, placed within
mm m.
rigidly at its upper end, consists of a steel rod 3 long
mm20 external an equally long brass tube 20 ram internal diameter and

member is then diameter. The rod and the tube are fixed together at the ends. The compound
suddenly loaded in tension by a weight of 1200 kgf falling through a height of

-

CENTRE OF GRAVITY AND MOMENT OF INERTIA 171

.

A OFGLet is the centre of gravity of the total area whose distance from the axis is x.

5 Then moment of total area about OF = Ax ...(it)

Centre of Gravity and Momenf of Inertia The moments of all small areas about the axis OF must be equal to the moment of total

5.1. CENTRE OF GRAVITY area about the same axis. Hence equating equations (i) and ( ii ), we get

Axa X =
l1
+ 02*2 + OgX3 + a4x 4 + ...

_ = a-,*-, + a?x9 + a8x3 + a4x4 + ...
*
or ... 1.0.11

Awhere = aj + a + a + a ...
3
2 4
If we take the moments of the small areas about the axis OX and also the moment of

total area about the axis OX, we will get

Ws- * °474 +

y -
_- a iyi * °272 + ...(5.2)

5.2. CENTROID A

where y = The distance of G from axis OX

OXy 4
= The distance of C.G. of the area a from axis
l

OXyv y
y4 = The distance of C.G. of area a a a from axis respectively.
2, 3, 4

P Gby C'°' °r SlmpIy - Thc centr°id and centre of gravity are at 5.4.1. Centre of Gravity of Areas of Plane Figures by Integration Method. The

the same point equations (5.1) and (5.2) can be written as

5.3. CENTROID OR CENTRE OF GRAVITY OF SIMPLE PLANE FIGURES
« S5" ““ “““The
centre of gravity (C.G.) of a uniform rod
•*»
8""‘y °f * Sk lies at its middle point. where i = 1, 2, 3, 4

•»•»“ '*«• *>• ttee medians* of*. x. = Distance of C.G. of area a from axis OF and
l

(iu) The centre of gravity of a circle is at its centre. y. = Distance of C.G. of area a from axis OX.
i
THEMOTODOPM^ots GRAVITY) OF AREAS OF PLANE FIGURES BY
The value of i depends upon the number of small areas. If the small areas are large in
Fig. 5 1 shows a plane figure of total area A whose centre of Y number (mathematically speaking infinite in number), then the summations in the above equa-
tions can be replaced by integration. Let the small areas are represented by dA instead of ‘a’,

then the above equations are written as :

_ Jx*dA ...(5.2 A)

* = T5a”

_ \y*dA -(6 -2B)

and >-7dT

A= Oi + a + a + a + where { x* dA = lxfL
3 i
2 4 ... JdA =

% OFLet = The distance of the C.G. of the areaa, from axis fy*dA = 2y a
* = The ii
2 ~ The distance of the C.G. of the area a2 x* = Distance of C.G. of area dA from axis OF
= distance of the C.G. of the area a
x y* = Distance of C.G. of area dA from axis OX.
° f the C' ' 3
3
°f the areaa
4 4
^ G OOOFFFand
on.* from axis Also
from axis
so from axis 5.4.2. Centroid (or Centre of Gravity) of a Line. The centre of gravity of a line
which may be straight or curve, is obtained by dividing the given line, into a large number of
The moments of ail small areas about the axis OF small lengths as shown in Fig. b.i (a).

= ax + a rjx,, + a^:3 + a 4x 4 + The centre of gravity is obtained by replacing dA by dL in equations (5.2 A) and (5.2 B).
11
- (j) Fig. 5.1

—\x * dL ...(5.2 C)

Then these equations become x% = :
J dL

170

5

172 STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA 173

a2 = Area of rectangle EFGH = 10 x 3 = 30 cm2 —10

y2 = Distance of C.G. of area a from bottom line GF = = 5 cm.
2

Fig. 5.1 (a)

and - \y*dL Fig. 5.2 (a) Fig. 5.2 ( b)
y~
J dL ..:(5.2 D)

where x* = Distance of C.G. of length dL from y-axis, and ...(5.2 E) Using equation (5.2), we have
...(5.2 F)
y* = Distance of C.G. of length dL from x-axis. gi3>i + a2y2 _ aiYi + 02^2

If the lines are straight, then the above equations are written as : Aa + a2
1

£ _ jnfp + L2x2 + L3X3 + 36 x 11.5 + 30 x 5 414 + 150 Ans.
= 8.545 cm.

+ L3 +

y _ -^1^1 + -^2^2 + ^3^3 + Problem 5.2. Find the centre ofgravity of the I-section shown in Fig. 5.3 (a).
+ ^2 + £3 +
Sol. The /-section is split up into three rectangles ABCD, EFGH and JKLM as shown in
5.5. IMPORTANT POINTS
Fig. 5.3 (5). The given /-section is symmetrical about Y-Y axis. Hence the C.G. of the section
(l) The ®xls about will lie on this axis. The lowest line of the figure line is ML. Hence the moment of areas are
>
taken about this line, which is the axis of reference.
ab, ove article, axis
which moments of areas are taken, is known as axis of reference. In
and
OX OFthe are called axis of reference.

(“) The ax s of reference, of plane Figures, is generally taken as the lowest line of the

^
figure for determining y , and left line of the figure for calculating x.

(..«) If the given section is symmetrical about X-X axis or Y-Y axis, then the C.G. of the
section will he on the axis is symmetry.

5.5.1. Centre of Gravity of Composite Bodies. The centre of gravity of composite
bodies or sections like T-section, /-section, L-sections etc. are obtained by splitting them into
rectangular components. Then equations (5.1) and (5.2) are used.

Problem 5.1. Find the centre ofgravity of the T-section shown in Fig. 5.2 fa).

given ^-section is split up into two rectangles ABCD and EFGH as shown in
tig 5.2 (b) The given T-section is symmetrical about Y-Y axis. Hence the C.G. of the section
will he on this axis. The lowest line of the figure is fine GF. (a)
taken about this line GF, which is the axis of reference in Hence the moments
of the areas are
this case.
Fig. 5.3

Let y = The distance of the C.G. of the T-section from the bottom line GF MLLet y - Distance of the C.G. of the /-section from the bottom line

(which is axis of reference) a = Area of rectangle ABCD = 10 x 2 = 20 cm2

= Area of rectangle ABCD - 12 x 3 = 36 cm2 MLyi - Distance of C.G. of rectangle ABCD from bottom line = 2 + 15 + -2 = 18 cm

y x = Distance of C.G. of area a, from bottom line GF = 10 + — = 11. cm

2

—

174 STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA

a = Area of rectangle EFGH = 15 x 2 = 30 cm2 2'o Find x

2 AGLet x = Distance of the C.G. of the L-section from left line
AGx = Distance of the rectangle ABCD from left line
™MLy2 = Distance of C.G. of rectangle EFGH from bottom line = 2 + 7.5 = 9.5 cm
=2+ 1

a = Area of rectangle JKLM = 20 x 2 = 40 cm 2 = —2 = 1.0 cm
3
A
ML>'3 = Distance of C.G. of rectangle JKLM from bottom line - — = 1.0 cm
2 x„ = Distance of the rectangle DEFG from left line AG

^1Now using equation (5.2), we have y = + a2y2 + ‘Mg.

A

_ a l^l + ^2^2 + g 3J3 ^A(, '' = a Using equation (5.1), we get
1
a + a2 "* ^3 + a + ag)
l
2 a2x2

_ a,*, + Awh, ere , = a. + a„
x_ 12
_ 20 x 18 + 30 x 9,5 + 40 x 1 A
20 + 30 + 40 20x 1+ 16x4
_ a,x, + a2x2
_ 360 + 285 + 40 _ 685 a1 + a2 20 + 16 (v a = 20 and a 2, == 16)
90 ~~90 1

= 7.611 cm. Ans. 20 + 64 84 7 ~: 2.' 333 cCmm.'

36 ~ 36 ~
3

Problem 5.3. Find the centre ofgravity of the L-section shown in Fig. 5.4. Hence the C.G. of the L-section is at a distance of 4.33 cm from the bottom line GF and

Sol. The given L-section is not symmetrical about any —+)2 emu 2.33 cm from the left line AG. Ans.

section. Hence in this case, there will be two axis of references. Problem 5.4. Using the analytical method, determine the centre of gravity of the plane
The lowest line of the figure (i.e., line OF) will be taken as axis uniform lamina shown in Fig. 5.5. (U.P. Tech. University, 2001-2002 AMIE, Summer 1975)

of reference for calculating y. And the left line of the L-section ;

(i.e., line AG) will be taken as axis of reference for calculating Sol. Let y be the distance between c.g. of the lamina and the bottom line AS.
Area 1

ABCDThe given L-section is split up into two rectangles D

and DEFG, as shown in Fig. 5.4. Area 2
To Find y

Let y = Distance of the C.G. of the L-section from bottom G

line GF

a - Area of rectangle ABCD = 10 x 2 = 20 cm2 Fig. 5.4
i

>i = Distance of C.G. of rectangle ABCD from bottom line GF

—= 2 + = 2 + 5 = 7 cm Area 3

a = Area of rectangle DEFG = 8 x 2 = 16 cm2 a„ = = 12.5 cm2 Fig. 5.5

2 22

y2 = Distance of C.G. of rectangle DEFG from bottom line GF 5 + —5 = cm.

2 y2 = O 6.67

= - 1.0 cm.

Using the relation, --

Using equation (5.2), we have Wia Dh * q272 +

_ a\yx + a2 y2 where A = a + a2 y
1 a 1 + a2 + a3
*A
——=
a iyi + «2yg 20x7 + 16 x 1 140 + 16 50x2.5+9.82x2.5 + 12.5x6.67 cm = 232.9 = „„ cm.
__ 3.22
20 + 16 " 36 ““rr
50 + 9.82+12.5
72.32

Similarly, let x be the distance between c.g. of the lamina and the left line CD.

156 _ 13 =433 Area 1
.
li""T
cm ' a = 50 cm2
1

x = 2.5 + —10 = 7.5 cm
tA

—/

STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA 177

Area 2 ft f~1 \t
_ 11
iLJ1 (_ ve gigi! is taken due to cut-out hole)

a0 = 9.82 cm2 4. 120 x 6 - 12 x 4 720 - 48 = 6.22 cm.

4r cm = 1.44 cm

120-12 108

Area 3 To Find x

a = 12.5 cm2 ADLet x = Distance between the C.G. of the section with a cut hole from the left line
3
x - Distance of the C.G. of the rectangle ABCD from the left line AD
*3 = 2.5 + 5 + 2.5 = 10 < l

Now using the relation, 10

- ^ alXl + a2 x2 + a3xs 50 x 7,5 + 9.82 x 1.44 + 12.5 x 10 Y= = 5c cm
aq + a2 + a3 5o0u + 9H.8bA2 + I1A2..5o ADx
2
= Distance of the C.G. of the cut-out hole from the left line

514.14 3
= = 7.11 cm.
= 5 + 1 + = 7.5 cm.
72.32
Using equation (5.1) and taking area (a2 ) of the cut hole as negative, we get
Hence the C.G. of the uniform lamina is at a distance of 3.22 cm from the bottom line AB
a lx l — a2x2 A(•.• = dj - a 2)
and 7.11 cm from the left line CD. Ans.
®1 - a2
Problem 5.5. From a rectangular lamina ABCD ^ 10 cm— ] „600 - 90 = 510 = 4,.72 cm.
= 120 x 5 - 12 x 7.5
10 cm x 12 cm a rectangular hole of 3 cm x 4 cm is cut as Ai B 108 108
120-12
shown in Fig. 5.6. '

Find the c.g. of the remainder lamina. ' //////f//////// Hence the C.G. of the section with a cut hole will be at a distance of 6.22 cm from bottom

Sol. The section shown in Fig. 5.6, is having a cut <//////'/./////// r'l/T A *70 A D Anc
hole. The centre of gravity of a section with a cut hole is
determined by considering the main section first as a com- 12 '////////////// Problem 5.5 (A). Determine the co-ordinates
plete one, and then subtracting the area of the cut-out hole, CITI '//////A \ , 3 /
i.e., by taking the area of the cut-out hole as negative. X mmc and Yc of the centre of a 100 diameter circular
..
Let y is the distance between the C.G. of the section
with a cut hole from the bottom line DC. ///////// Y// ' hole cut in a thin plate so that this point will be the

Oj = Area of rectangle ABCD = 10 x 12 = 120 cm2 ///////// &// l centroid of the remaining shaded area shown in Fig.
y. = Distance of C.G. of the rectangle ABCD from hot-
tom line DC 5.6 (a). (U.P. Tech. University, 2001-2002)

////////////1// t~ Sol. The given shaded area is equal to area of a

2m

^ //////'' //f/'/'///'// i mm mmthin rectangular plate of size 200 minus
\ x 150

mm mmthe area of a triangle of length 100
and height 75

‘ mmminus the area of circular hole of dia. 100 as shown

„
Figg.' 5.6

in Fig. 5.6 (5).

Let A = Area of rectangular plate
1
mm= 200 x 150 = 30000 Fig. 5.6 (a)
2

a = Area of cut-out hole, i.e., rectangle EFGH, mm—- (1002 ) = 2500a;
2
= 4 x 3 = 12 cm2

DCy2 = Distance of C.G. of cut-out hole from bottom line

= 2+ —4 = 2 + 2 = 4 cm.

2

Now using equation (5.2 ) and taking the area (a2) of the cut-out hole as negative, we get 4

Wi where A = a - a2 The centre of hole is the centroid of the shaded
3
y_ X Yarea. Hence c and c is the co-ordinates of the cen-

( tre of the hole and also the co-ordinates of the centroid

of the shaded area.

*y = — but for cut-hole area a„ is taken - ve. Hence

01 + 02 Fig. 5.6 (b )

_ Pij; - a2y2
01-02

178 STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA

For area A,, -y mm40U
AjCj = Distance of C.G. of area from y-axis = = 125

t

- 100 + - x 100 = 166.67, — ™ mmyl = Distance of C.G. of area Aj from x-axis = —100 = 50

mm-: 75 + x 75 : : 126 mm100
o Ax = Distance of C.G. of area from y-axis = 150 + —r~ = 200
2
For area A Xx - c and y3 = Yc 2
3, 3
— ^y2
A ANow using equation (5.1) and taking areas 3 as negative, we get 4r 4 x 50 200
and A =
= Distance of C.G. of area 2 from x-axis
2

- =xc _ A lXl - AqXq - A3 x 3 X30000 x 100 - 3750 x 166.67 - 2500a: x c —2 500
-g-
A -A2 -A (30000 - 3750 - 2500*0 x = Distance of C.G. of area A from y-axis = 250 x -
1 3 a
3
X Xc (30000 - 3750 - 2500n) = 30000 x 100 - 3750 x 166.67 - 2500jt x c
X xl Xc (30000 - 3750) - 2500jt c = 30000 x 100 - 3750 x 166.67 - 2500« x c y3 = Distance of C.G. of area A from x-axis = 100 + — mm50 350=
s
Xc (30000 - 3750) = 30000 x 100 - 3750 x 166.67
x, y = Distance of C.G. of the shaded area from y and x-axis.
X(Cancelling 2500 x x x c on both :
X26250 c = 300000 - 625012.5 = 2374987.5 ANow using equation (5,1) and taking area as negative, we get

2

__ Ax - A2x2 + A3X3
11
„ A2J3U7I 4987<.. 5KJ
47mm= 90- - A2 A
3
- Ans. ~Ai +

26250

= rYc = A-iyi ~ A~ 3 y3 25000 x 125 - 1250n x 200 + 6250 x
Aj - -A
Similarly, y
3
A 26000 - 1250jc + 6250
a

30000 x 75 - 3750 x 125 - 250Qjt x Yc 3125000 - 785398 + 1041666 =_ 123.75 mm. Ans.

(30000 - 3750 - 2500.tr) 27323

Yc (30000 - 3750 - 2500ic) = 30000 x 75 - 3750 x 125 - 2500it x Yc Similarly, — —25000 x 50 - 1250it x - + 6250 x
Yc (30000 - 3750) = 30000 x 75 - 3750 x 125

A A Ay yy-x x
(Cancelling 2500jt x Yc on both i „. 3k - 3

26250FC = 30000 x 75 - 3750 x 125 2 2+ 3 3

' = 225000 - 468750 = 1781250 ^ ^^A - Ag + A3 27323
x
250000 - 83333,729166 Ang_

1781250 A 1 OAjU

= ^6250“ = 67-85mm- A03' .-. Centroid of the given section = (x, y ) = (123.75 mm, 69.38 mm).

AProblem 5.5 (B). semi-circular area is 5.5.2. Problems of Finding Centroid or Centre of Gravity ofAreas by Integration

removed from the trapezoid as shown in Fig. 5.6 (c). V Method.
Determine the centroid of the remaining area.
Problem 5.6. Determine the co-ordinates of the C.G. of the area OAB shown in Fig. 5.7,

(U.P. Tech. University, 2000-2001) ^ OBif the curve represents the equation of a parabola, given by

Sol. The given shaded area is equal to the E y = fcc2

mmarea of a thin rectangular plate of size 100 x 8 in which OA = 6 units

mm(150 + 100) plus the area of the triangle of ±7 and AB = 4 units.

mm mmlength 250
and of height (150 - 100) = 50 Sol. The equation of parabola is y = kx2 —(*)

minus the area of semi-circular area of diameter -150 mm 100 mm

mm100 as shown in Fig. 5.6 (c). First determine the value of constant k. The point B is lying on the curve and having co-

Let Aj = Area of rectangular plate Fig. 5.6 (c) ordinates

mm= 100 x 250 = 25000 2 x = 6 and y = 4
Substituting these values of equation (t), we get
A2 = Area of semi-circle - nr KX 50 mm1250jc 2
4 = k x 62 = 36 k
2 2

250 x 50
A

mm3
= Area of the triangle = 6250 2

1

STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA

Substituting the value of k in equation (i), we get Fig. 5.7 y* dA = J fxdA-Jl%xydx = 6 £dx

1 (v dA = ydx, x* = x) J l0

y=Q*2 ...(«> I -2>\ 2

or x 2 = 9y 1 (-6 2 1 6 jc ,
2 Jo r{
j

or x = 3-Jy 2 Jo ^ 9 )

...(Hi) 1 x4 , 1 1 [x 5
2 Jo 81
Consider a strip of height- y and width dx as shown in 1 If6 *. ]
Fig. 5.7. The area dA of the strip is given by
2 81 Jo 2 8l|_5J0
dA=y x dx
61
_ 1 * J_ Xx
” 81 5
The co-ordinates of the C.G. of this area dA are x and ~ 2

.'. Distance of C.G. of area dA from y-axis = x r r6 x6 2 1 x3 1

and distance of C.G. of area dA from x-axis - - f,

2 M- _27_ x ef
810 63
x* = x and y* - — fdA
J 07
2
— —6-1 „ 36 = —6 Ans.
Cet x = Distance of C.G. of total area OAB from axis OY .
y = Distance of C.G. of total area OAB from axis OX. =x
30 30 5
Using equation (5.2 A), we get

x*dA x x y dx Problem 5.7. Determine the co-ordinates of the C.G.

j J" —x 2

dA ydx of the shaded area between the parabola y = and the
j jo

—But y = from equation (it). straight line y = x as shown in Fig. 5.8.
Sol. The equations of parabola and straight line are

6 x2 xdJx If®),
-JQ X dx
f Xx
T_
\o

yJ = X ...(U)

j%dx |fx 2 dx AThe point is lying on the straight line as well as on Q x ! H dx y
H
Jo 9 9 Jo

—r 4 6 the given parabola. Hence both the above equations holds pjg. 5.8

s 4 -1 x6 4 Agood for point A. Let the co-ordinates of point are x, y.

f X 2q dx L Jo _ 4 Substituting the value of y from equation (it) in equation (i), we get
_ JIo
5-3[-x3 6 -x 63 — —x2 -xv
fx* dx 1
x=
Jo or a4 =
4x
8

L Jo

13 Substituting the value of x = 4, in equation ( ii ),

= T4 x T * 6 = 4.5. Ans. y=4

1 AHence the co-ordinates of point are 4, 4.

Using equation (5.2 B), we get Now divide the shaded area into large small areas each of height y and width dx as

\y*dA shown in Fig. 5.8. Then area dA of the strip is given by m

i-

dA = ydx ~(y - y2) dx
l
Dy, = Co-ordinate of point which lies on the straight line OA
where y* — Distance of C.G. of area dA from x-axis where

= ^ (here) y 2 = Co-ordinate of the point E which lies on the parabola OA.

dA = ydx horizontal co-ordinates of the. points and E are same.
DThe

) 1j

STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA

The values of y andy can be obtained in terms of* from —y — —= - 16 x 6 = 2« Ans.
x 2 16 3 16
equations (it) and
(i), \y*dA
Now using equation (5.2 B), y =
• >i = * and ^2=7-
[From equation (0)]
Substituting these values in equation (Hi),
dA = \x-
dA = \*-^jdx
Jy*dA = Jo -^ + ^|x-^
The distance of the C.G. for the area dA from y-axis is given by,
x* = x

And the distance of the C.G. of the area dA from x-axis is given by

_ 2^2 + J'x - y2 yi+ y2
2
2

Vi-x and y2 = 2 3 5x16

= —If X + x 2 = lf£ 4s ^ 1[64 _64"
2 3 5x16 ~2L3 5.
24

Now let x = Distance of C.G. of shaded area of Pig. 5.8 from y-axis

y = Distance of C.G. of shaded area of Fig. 5.8 from x-axis 2 {.3 5
using equation (5.2 A),
Now

X- —lx* dA " „„ X 2 ~ 64
15 15
wwnh, oerroe xv** —= xv

[<“- [From equation (oi)l

—dA = x - 1 dx [See equation (iu)] ,J Ans.
( V x varies from 0 to 4) 16 15 16 5
4 f dA
fi
w*K 1

5.5.3. Problems of Finding Centroid or Centre of Gravity of Line-Segment by
Integration Method

3 4x4 Problem 5.8. Determine the centre of gravity of a y,.

quadrant AB of the arc of a circle of radius R as shown in

Fig. 5.9 (a). B

3 4x4 3 Sol. The centre of gravity of the line AB, which is an y*
64-48 16
arc of a circle radius R, is obtained by dividing the curved |, , X\ dL
3 ”3 O
line AB into a large number of elements of length dL as d>/—
k
shown in Fig. 5.9 (a). Si

tJ* ABThe equation of curve is the equation of circle of \\

radius R. j

4 ~!l
l
_ X2 X3 42 43 The equation of curve AB is given by — ——1 r A x
x2 + y2 = R2
2 3x4 2 3x4 • -*l

- Jo

= Jj* _ l6 _ 48 ~ 32 -16 Differentiating the above equation, ^

23 ~~6 —( vi 2x dx + 2y dy = 0 [v R is constant]
6

,.

184 STRENGTH OF MATERIALS CENTRE OF GRAVITY AND MOMENT OF INERTIA 185

or 2y dy = - 2x dx
—or
=, -2x dx = - x dx

<iy ...W

2y y

Consider an element of length dL as shown in Pig. 5.9 (a). The C.G. of the length dL is at

a distance x* from y-axis and y* from x-axis.

Now using equation (5.2 D) for y we get

—f y*dL .Mi)

. y= f
dL

J

Let us express dL in terms of dx and dy.

But dL = yjdx 2 + dy 2

1. I / , \2 ( From (£), dy _ - xdix \^
y.

dx 2 + ^jdx 2

[¥ Similarly,
= dx JV—r«
(v x 2 + y2 = i?2) (.R cos 9) x (R dQ) cos 0 d6
(v y* = y)
—= .dx.

y

Substituting the value of dL in equation (ii),

R R. R sin 0

y x dx
y
— —f > 2R

I i?[sin90‘
y** x dx -sin0°]
iJ y -= A.ns.
.

jdL jdL

R Ax Problem 5.9. Determine the centre of gravity of the area of the circular sector OAB of

Rdx r L Jo radius R and central angle a as shown in Fig. 5.10.

| *j| dx
[
o

dL dL '72jiR') Sol. The given area is symmetrical about x-axis. Hence Y
J j l4J the C.G. of the area will lie on rc-axis. This means y = 0. To

( v S dL is total length of arc of one quadrant of a circle) find x > the moment of small areas are to be taken along y-axis.

RxR 2R Divide the area OAB into a large number of triangular ele- -
(
“isar“* A,ns- Rments each of altitude and base RdO as shown in Fig. 5.10.
Such triangular element is shown by OCD in which altitude
4
OC = R and base CD = RdO. The area dA of this triangular
Similarly, the value of x can be calculated. Due to symmetry this value will also be
element is given by,
2R
' equal, to OC x CD R x RdO
. dA=

jt 22

—_=_ = 2R . Fig. 5.10
y
.... . x . Ans.

Jt