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MADE
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Preface
I am thankful to Mr. B. Singh, CMD of MADE EASY Group, who is ever ready to help the Student
Community by providing them newest type of books, as in the present book with typical/thought
provoking/mind racking questions asked in IFS and IAS. Prelims and Mains of UPSC, for the
last 30 years for both Civil and Mechanical engineering, in the subject of Strength of Materials.
For the solution of each question a student must be equipped with strong concepts in the
subject, and the students are the beneficiaries of the latest and comprehensive knowledge
of the subject of the qualified and dedicated faculty of MADE EASY.
Further improvements in the text of the book will be made after getting the feedback from the
students.
Any error in printing or calculations pointed out by the reader will be acknowledged with thanks
by the author.
Dr. U.C. Jindal
Author
Contents
Contents
IAS
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& IFS
IFS(Objective
(Objective &
& Conventional)
Conventional) Previous
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Questions
1
Strength
Strength of
of Materials
Materials 1
Sl.
SI.
Chapter
Chapter
Pages
Pages
1.
1.
Simple Stresses
Stresses in
Bars............................................................. 1-19
Simple
inUniform
Uniform and
and Compound
Compound Bars
1-19
2.
2.
Principal
Stresses........................................................................................................................20-38
Principal Stresses
20-38
3.
3.
Thin
Thin and
and Thick
Thick Shells
Shells.................................................................................................................39-51
39-51
4.
4.
Shear Force
Diagrams..................................................................52-69
Shear
Forceand
and Bending
Bending Moment
Moment Diagrams
52-69
5.
5.
Theory
Simple Bending.
Bending......................................................................................................70-88
Theory of
of Simple
70-88
6.
6.
Deflection
of Beams
Beams............................................................................................................... 88-107
Deflection of
88-107
7.
7.
Torsion . ....................................................................................................................................108-123
Torsion
108-123
8.
8.
Springs
124-130
Springs . ....................................................................................................................................124-130
9.
9.
131-137
Struts and
Struts
and Columns..............................................................................................................131-137
Columns
10.
10.
Theories of
of Failure
Failure.................................................................................................................138-148
Theories
138-148
11.
11.
149-157
Strain Energy
Strain
Energy Methods........................................................................................................149-157
Methods
12.
12.
Miscellaneous Questions
Questions....................................................................................................158-165
Miscellaneous
158-165
13.
13.
Rotational Stresses
Stresses................................................................................................................166-169
Rotational
166-169
14.
14.
Unsymmetrical Bending.....................................................................................................170-173
Bending
170-173
Unsymmetrical
15.
15.
174-179
General Objective
Objective Type
Type Questions
Questions..................................................................................174-179
General
01
01
CHAPTER
Simple Stresses in Uniform
and Compound Bars
Bars
Q.1.1 AAsteel
steelrod
rodofoflength
length300
300mm
mmand
anddiameter
diameter30
30mm
mm is
is subjected
subjected to
to a pull
pull P, and the temperature
Q.1.1
rise is
extension of the
is 100°C. IfIf the total extension
the rod
rod is
is 0.40
0.40 mm,
mm, calculate the magnitude of P. Take a for
steel=
1Cr61°C and E
E = 0.215 x 106
106 N/mm2.
N/mm2 .
steel = 12 xx 10-6/°C
[CSE-Mains, 2011, CE : 12 Marks]
[CSE-Mains,
Solution:
P = Pull in N
P=
%
2
2
A=
of cross-section
cross-section == 4 x 30 2 = 706.86 mm
mm2
= Area of
Extension due to pull,
pull, (assuming pull axial)
&11 =
P
x 300
~XL
Px300
_
x -6
xL _
1 974 x 10-6P mm
6 -= 1.974
·
AE
- 706.86
706.86x0.215x10
Pmm
x0.215 x10-
612 extension due to
8/2
to temperature
temperature change
change
a.UT
= aLAT
= 12 x 10-6 x 300 x 100
100 =
=0.36
0.36 + 1.974 x 10-6 P = 0.4
1.974 xx 10-6 P = 0.04
1 6
4 x106
P = 0.04
0.0 x 0 = 20263 N = 20.263 kN
P
1.974
metallicbar
bar250
250mm
mmxx100
100mm
mmx x5050
mmisisloaded
loadedasasshown
shownininthe
thefigure
figure 1.1.
1.1.Work
Work out
out the
the
Q.1.2 AAmetallic
mm
change
in volume. What should
in the 4 MN load in
in order that
change in
should be the change that should be made in
should be no change in
in the
the volume of the bar.
there should
4MN
4 MN
z
,..x
i
50 _ _ _ _ _ _ __
50
400 kN
mm
T ,___ __,,,,,_____.,
250 mm
2MN
2 MN
Fig. 1.1
1.1
Fig.
=
5
2
ratio = 0.25.
0.25.
E = 2 x 106
10 N/mm2,
N/mm , Poisson's
Poisson's ratio
Assume E=
Marks]
[IFS 2011, CE : 15 Marks]
22
~=
/AS &
Previous Solved
Solved Questions
IAS
& IFS
IFS (Objective
(Objective &
& Conventional)
Conventional) Previous
MADE EASY
EASY
Solution:
Stresses
0
x
= +
+ 400,000
= +80
+80 MPa
MPa
400,000 =
5000
2x10
2
x1066
a
= +160 MPa
CJy _
= +
+ 50x250
50x250 = +160 MPa
6
CJz =
Volumetric strain,
Et, -
4x10
4x1066
160MPa
= 160
250x100
MPa
250 x100 = a -Fa +a
1a, +ay +az )
Y
z 2v
80 = 40
40
= 80
80 -2
=
- 2 x0.25x
x 0.25 80
x- =
E
E
E E
E
3
Volume
250 xx 100
100 x 50 mm3
mm
Volume == 250
40
xV =
— x 250 x100 x 50
in volume=
volume . s
6V = change in
evv xV
= Ex250x100x50
81/
6
50 x1066 50
x106
50x10
= sax
10 =
= 250 mm3
mm 3
E
2x10
2x1055
For
ev =
;;;;; 0
0
Ey
+
ay
+
az
2v(a,
+
ay
+
az)
=
0
(Jx (Jy + (Jz - 2v(ax (Jy az> = 0
+ ay)(1
ay) (1-2v}
- 2v) == 2v(az)
(ax+
2v(az>-az
(80 + 160)(1-2v}
160) (1 - 2v) == (2v
- 1) CJz
az
(2v-1}
240
CJz
240Xx0.5
0.5== (2v-1)
(2v -1) = -0.5 az
120
=
120 ;;;;; -<Jz Xx 0.5
a
az==-240
-240MPa
MPa
Py'' =; ; ; 240 x 250 x 100
100;;;;;
where in load,
P
= 6 MN
4 MN load should be increased to 6 MN load in same
same direction
direction
N/mm 2 .
So that az becomes-240
becomes -240 N/mm2.
(b)
cranechain
chainhaving
having an
an area 7.25 cm2
cm2 carries a load of 15
15 kN.
kN. ItIt is
is being
being lowered
lowered at
at aa uniform
uniform
Q.1.3 AAcrane
speed
of 50
50 m/minute, the
the chain gets
gets jammed
jammed suddenly,
suddenly, at
at that
that time
time the length of chain unwound
speed of
is 12 m. Estimate the stress
stress induced
induced in
in the
the chain
chain due
due to
to sudden
sudden stoppage.
stoppage. Neglect
Neglect weight of the
chain. Assume E.
E = 2.1
Hf N/mm2.
N/mm2 .
2.1 x 105
Marks]
[IFS 2012, CE : 10 Marks]
Solution:
(·: self weight of chain
W = 15 kN = 15000 N
(-;
chain negligible)
negligible)
2
725 mm2
mm
A, chain area
area of
ofcross-section
cross-section == 725
Length,
L=
;;;;; 12000 mm
Volumeof
ofchain
chain == 12000
Volume
12000 x 725
725 = 87 x 105
105 mm3
mm3
Say, stress
Say,
stressdeveloped
developed== ai, instantaneous in N/mm
N/mm22
Speed,
50
V=
= 0.833 m/s
60
= 60
mV
mV 22 15000
15000 0.8332=
0.833 2
Kinetic energy
;;;;; 530.5 Nm;;;;;
energy absorbed
absorbed by
bychain
chain ;;;;;
= - - = 9.81 xx
Nm = 530500 Nmm
9.81
2
2
Stresses in
and Compound
Compound Bars
Bars
Simple Stresses
in Uniform
Uniform and
Strength of Materials
Change
in length,
Change in
length,
3
4 3
<1111
0
12
a.x
12000
6L = a
a; xi= 1; x ooo
SL=
E
E
ltVlil
Change in
in potential
potentialenergy
energy == WSl
x a; x12000 N
= 15000
15000XO';X12000
-857143
N
Nmm
aiNmm
mm =
- 857.143
.
a;
mm
2.1x10s
2.1x 105
2
Gi
Total
strainenergy
energyabsorbed
absorbedby
bychain
chain == 530500
Total strain
530500 + 857.143 a.=
cri = —
;~ volume
2E
=
530500 + 857.14a;
857.14 a, Or
or
ai2
O'i
x87x
10 5 == 20.7143
20.7143$52
crf
x €3 7 x105
2x2.1x10
2 x2.1x10'5
cr~-41.38cr;-25610.3
O
aF
- 41.38a1 - 25610.3 == 0
-
41.38+.J41.38
+ 44x25610.3
41.38 +V41.3822 +
x 25610.3
2
41.38+J1712.30+102441.2
41.38
+ ../1712.30+ 102441.2
2
41.38
322. 728
41.38++ 322.728
2
182.054 N/mm2
N/mm2
= 182.054
Provethat
thatPoisson's
Poisson's ratio
ratio cannot
cannot be greater than
Q.1.4 Prove
than 0.5.
[CSE-Mains, 1988, ME
ME:: 15 Marks]
[CSE-Mains,
Solution:
Figure shows a sphere
sphere under uniform
uniform hydrostatic
hydrostatic pressure p.
E = Young's
Young's modulus
If E=
modulus and vis
v is Poisson's ratio,
ratio, v, then
Volumetric strain,
Ev
3p
(1-2v)=0
or
v, Poisson's
Poisson's ratio
ratio == 0.5
0.5, then in place of decrease in
in volume
volume there will
If vis
v is greater than 0.5,
will be
increase in volume,
volume, which is not possible.
possible.
p
Fig. 1.2
Q.1.5 A A
1000
mm
long
bar
subjectedtotoan
anaxial
axialpull
pullPwhich
Pwhichinduces
inducesaamaximum
maximum stress
stress of 1500 kg/crn2.
Q.1.5
1000
mm
long
bar
isissubjected
kg/cm2.
The area of cross-section of the
the bar is 2 cm
cm22 over a length of 950 mm and for the central 50 mm
length, the sectional
cm2 .
sectional area
area is equal to 1 cm2.
2
2cm
2
cm2
P
P
P~-------------------~--------------------~P
2
1 an
cm2
1- 47.5 cm -P-1l---47.5cm---l
k 47.5 cm -id
l---47.5cm---l
5.0cm
5.0 cm
Fig. 1.3
Fig.
Efor
kgf/cm2 , calculate strain energy stored in
Assuming that E
for bar material
material is
is 20
20 xx 1a5
105 kgf/cm2,
in bar
[CSE-Mains, 1990, ME : 20 Marks]
[CSE-Mains,
44
~=
/AS &
Previous Solved
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IAS
& IFS
IFS (Objective
(Objective &
& Conventional)
Conventional) Previous
MADE EASY
EASY
Solution:
2
,
Maximum stress will occur in central portion of area 1 cm
cm2,
2
= 1500
1500 kgf/cm2
kgf/cm
1500 xx 1
1=
-- 750
750 kgf/cm
kgf/cm 22
Stress
inother
otherportion
portion=- 1500
Stress in
2
2
2
1
5002
7502
1500
x 47.5 x 2
x 5 x1+
strainenergy
energy =- ~x5x1+2Ex47.5x2
U, strain
U,
2E
2E
x 103
5625x
103 26718.75
26718.75x
103
5625 x103
=
E
+
E
E
+
E
32343750
32343750
32343750 32343750
=
20x10 5
20x105
E
= 16.17
16.17kgf-cm
kgf-cm (Strain
(Strain energy stored)
Q.1.6 AAsteel
steelrod
rodofofsquare
squarecross-section
cross-sectionisisloaded
loadedas
asshown
shown in
in the figure
figure 1.4.
B
A
T
T
E
~ 150
150 kN
kN
C
75kN
75 kN
4--------- 10 mm
100 kN
-4-
II)
U)
10 mm
k-
D
E
E
25kN
25 kN
N
N
N
!k
l----2m----1---2m----1---2m---l
[4-2 m
2m
>I<
2m>1
Fig. 1.4
Fig.
Find the section which is subjected
subjected to maximum stress, its
its magnitude
magnitude and nature. What will
will be the
change
in its
its length?
length? Take E
E == 200 GPa.
change in
Solution:
A
B
B
....-----.---,
150 kN
150kN
►
E
E
~
150 kN
4—
c
B
50kN
50 kN
—110.
cC
C
E
E
0
~
N
50kN
50 kN
4—
25 kN
4-
D
25 kN
I10 mm
Hollow
Fig.1.5
Considering compressive stress as (-ve)
(-ve) and vice-versa
dAB =
150 000
.
150,000
;
=
240 MPa
suare
= -240
MPa (q
(squaresection
section))
6
5
625
aBc =
50
,000 = -125 MPa
5
0,000
125 MPa
400
25
25000
Oen
= +
000 = +83.33 MPa
Gap —
400-100
400 -100
MPa (in
amax
(inportion
portion AB)
°max =
= -240 MPa
240
125 x 2000 + 83.33 x 2000
240 xx2000
Change
2000 _ 125 x 2000 + 83.33 x 2000
Changeininlength
length= = 200,000
200,000
200,000
200, 000
200, 000
200,
000
= -2.4
-1.25 ++0.833
-2.817 mm
-2.4-1.25
0.833==-2.817
mm (contraction)
(contraction)
►
Simple Stresses in Uniform and Compound Bars
Strength of Materials
4
Q.1.7 A rigid bar AD is pinned at A and attached to the bars BC
and ED as shown in figure 1.5. The entire system is initially
stress free and weight of all bars are negligible. The
temperature of the bar BC is lowered by 25°C and that of bar
ED is raised by 25°C. Neglecting any possibility of lateral
buckling, find the normal stress in bars BC and ED.
For BC, of brass, E = 90 GPa, a = 20 x 10-6/°C
For ED, of steel, E = 200 GPa and a = 12 x 10-6/°C
Cross-sectional area of BC is 500 mm2 and that of ED is
250 mm2.
[IFS 2011, CE : 10 Marks]
I
E
E
0
cv
mm
Solution:
Free expansion in steel bar (if there were no resultant)
8/s = 250x 12x 10-6 x 25 = 0.075 mm
Free contraction in brass bar,
SlB = 300 x 20 x 10-6 x 25 = 0.15 mm
If brass bar is free to move down, then end D would move by
0.15 x
5
350 mm-0-I
Fig. 1.5
600
= 0.36 mm
250
(... ABD is rigid, vertical deflection in ABD will be linearly proportional to distance from A)
But free expansion in steel bar is only 0.075 mm.
Therefore steel bar will be pulled down and brass bar will be pulled up, or stretched to prevent, free
contraction
Say
as = Stress in steel bar
6B = Stress in brass bar
Ps = Force in steel bar = as x 250 N
PB = Force in brass bar = 08 x 500 N
Taking moments about B
600 x Ps = 250 PB
600 x as x 250 = 250 x 500 x as
150000 as = 1250000'B
6B = 0.833 GB
or
aB = 1.2 as
Brass bar
aB
- x 300 = 81g'
Final contraction = 0.15 - —
Eg
Steel bar
as
Final extension = 0.075+ Es x 250 = Sls'
But
Putting the value
,
=
600 RI
X (N g'
250
R7
2.4 viB '
x 300
0.075 + Es x 250 = 2.4[0.15 EB
]
s
1.2a
x 720
x 250 = 0.36
0.075 + Gs
90,000
20Q 000
6 0-
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
as x 1.25 x 10-3 + as x 9.6 x 10-3 = 0.285
as [10.85] = 285
= 26.26 N/mm2 (tensile)
aB = 31.52 Nimm2 (tensile)
Checking
Steel bar, total extension = 0.075 + 0.032825 = 0.107825 = 8/s
31.52x
300 = 0.15 — 0.105066
90,000
Brass bar, final contraction = 0.15
8/B = 0.04493
8/B x 2.4 = 0.10784 ^ 8/s
Q.1.8 A plate is riveted to a channel section in a
structure as shown in figure 1.6. An eccentric
load of 12.5 kN acts as shown on the plate.
Determine the rivet diameter so that the
maximum shear stress in any rivet is not to
exceed 40 MPa. Diameter of the rivet should
be chosen from preferred series diameter
(in mm)
r = 141.4 mm, /2 = 20000 mm2
12, 14, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39,
42, 48 rivet diameter
[CSE-Mains, 2009, ME : 40 Marks]
12.5 kN
'V
Channel
E
E
O
c\I
I1
II
200 mm 50 mm
Solution:
400 mm
Fig. 1.6
Number of rivets = 5
Load = 12.5 kN
12.5
= 2.5 kN
5
A = Area of rivet
Direct shear load on each rivet —
If
i
d—
2500
A N/mm2
Torsional shear stress, is « r
Torsional shear force = krA
4 kA (r2 ) = P.e. = 12500 x (100 + 50 + 400)
4 kA (20000) = 12500 x 550 Nmm
kA =
k—
12500 x 550
80000
= 85.9375
85.9375
A
Torsional shear stress
"Cs ,
in outer rivets = kr —
'Cd =
tir
85.9375 x 141.4
12151.5625
A
A
2500
A
= \('rs x sin45° + td)2 + (Ts x cos4512
Fig. 1.7
Simple Stresses in Uniform and Compound Bars
Strength of Materials
I(12151.5625 x 0.707+ 2500)
\
A
2
+
0 2151.5625
A
A
7
2
x0.707
)
8591.15)2
(11091.15)2
A )
A
1000 x 14.03
1000 v
-I23.01+ 73.80 =
A
A
14030
A
= 40 MPa (permissible shear stress)
A = 350.073 mm2 = 4 d
2
d = 21.11 mm
Rivet diameter from preferred series = 22 mm.
Q.1.9
A compound tube is made by shrinking a thin steel tube on a thin brass tube. The areas of crosssection of these tubes are As and Ab, while the Young's moduli are Es and Eb respectively. Show
that for any tensile load, the extension of the compound tube is equal to that of a single tube of
same length and total cross-section area, but having a Young's modulus of
E s As + E b Ab
AS + Ab
[CSE-Mains, 2005, ME : 20 Marks]
Solution:
Say
Steel tube
P = axial load
As, Es
= asAs + abAb
But
5Is = 81b
a
S1
Es
Brass tube
Ab, Eb
= ab x L
Es
a,
Es
b
Eb
Fig.1.8
Es
Es
P = ab x — x As x ab x Ab = ab[Eb As + At) ]
Eb
[EsAs+AbEb]
P = ob
Eb
1 [Es As + AbEb ]
Eb
sib = sis
ab xi
klb =
b
81b = a b
1
Eb
P
fit' = EsAs+EbAb
(For single bar having modulus of elasticity E)
P
1
x
e=
(As +Ab ) E
Strain,
(From eq. (i)
8 0.
lAS & IFS (Objective & Conventional) Previous Solved Questions
or
Es As + Eb Ab
E(A, + Ab )
E=
or
MADE EASY
E s As + E b Ab ,
for a single bar
As + Ab
Q.1.10 A copper tube 22 mm internal diameter, 30 mm outer diameter and 150 mm long is compressed
by a nut tightening over a steel bolt, 20 mm diameter and 1 mm pitch. (i) If the nut is tightened by
a quarter of a turn beyond the just touching position, determine the stress in the bolt. (ii) what
would be the final stress in the bolt if the temperature of the assembly is to increase by 10°C.
= 18 x 10-6/°C.
Assume E, = 2 x 106 kg/cm2, Ecu = 6 x 106 kg/cm2, as = 12 x 10-6/°C,
[CSE-Mains, 1999, ME : 30 Marks]
Solution:
2
As area of cross-section of bolt = 4 x 20 = 1 007t mm2
Acu, area of cross-section of copper tube
= .(302 _222 )=1O47c
(i) For equilibrium, compressive force in tube = Tensile force in bolt
ac„ x ,4„ = as x As
acu x 104n = as x 1007c
as = 1.04o
Due to tightening of nut as bolt
Pitch = 1 mm
1
0.25 mm = 4 mm = Extension of bolt + Contraction in tube
0.25 mm =
6
sx 150 + a0U x 150
Es
E„
0.25 as a„ 1.04a„ a
„ +
, (here as and a are in kg/mm2)
150
E, E„
2x10" 6 x10'
1.04a„ u„
0.25 x103
150 — 20 6
1.6667 = 0.052 a„ + 0.1666 acu= 0.21866 a„
= —7.622 kg/mm2 (comp.)
as = +7.927 kg/mm2 (tensile)
= —762.2 kg/cm2 (comp.)
=
as +792.7 kg/cm2 (comp.)
(ii) Increase in temperature
a >
as
OT= 10°C
Compressive stress will be developed in copper tube due to temperature rate and tensile stress will
be developed in stress bolt
Strain in Cu tube = Strain in steel bolt
Hence,
oc„,AT — E
r = a. AT GsT
s
E
Simple Stresses in Uniform and Compound Bars
Strength of Materials
4
9
acur ▪ asr _ (a
cu - as) AT
E„ E,
or
acur
asT
2x106 6x105
(here asT and Guff are in kg/cm2)
- (18 -12) x 10-6 x 10
acur ▪
a6sT
6 x 10-6 x 105 x 10 = 6
20
acuT x 104n = sT x 10071
asT = 1.04 acuT
But
1:5,,,T
1.040,,T
" - 6
20
6
am[0.05 + 0.1733] = 6
-26.866 kg/cm2 (comp.)
acuT
a sT = +27.94 kg/cm2 (tensile)
Final stress in steel bolt
asT = 792.7 + 27.94 = 820.64 kg/cm2 (tensile)
Objective Questions
Q.1
In a semi-infinite plate shown in the figure 1.9.
The theoretical stress concentration factor kt,
for an elliptical hole of major axis 2a and minor
axis 2b is given by
(a)
t
S
k
A
2a
(b)
1
Fig. 1.9
(a) kt =
= 2b
(c)
kt
Q.3
S
2a
(d) kt = 1+ —
[CSE-Prelims, ME : 1998 ]
Q.2
0
(b) kt = 1+ -a-
In a simple tensile test, Hooke's law is valid
upto the
(a) elastic limit
(b) limit of proportionality
(c) ultimate stress
(d) breaking point
[CSE-Prelims, ME: 1998]
B
(c)
S
(d)
The stress strain curve for an ideal strain
hardening material will be as in
[CSE-Prelims, ME : 1998]
10
Po. lAS & IFS (Objective & Conventional) Previous Solved Questions
Q.4 The percentage elongation of a material as
obtained from static tension test, depends on
(a) diameter of the test specimen
(b) gauge length of the specimen
(c) nature of end grips of the testing machine
(d) geometry of the test specimen
[CSE-Prelims, ME : 1998]
Q.5
(c) Potential energy of strain : Body is in a state
of elastic deformation
(d) Hooke's law : Relation between stress and
strain
[CSE-Prelims, ME : 1999]
0.8
Match the List-I (material properties) with List-II
(technical definition represents) and select the
correct answer.
List-I
A. Hardness
B. Toughness
C. Malleability
D. Ductility
List-I I
1. Percentage elongation
2. Resistance to indentation
3. Ability to absorb energy during plastic
deformation
4. Ability to be rolled into plates
Codes:
A
BCD
1
4
(a) 3
2
3
1
(b) 2
4
(c) 2
3
4
1
4
2
(d) 1
3
[CSE-Prelims, ME : 1999]
Q.9
A measure of Rockwell hardness is
(a) depth of penetration of indentor
(b) surface area of indentation
(c) projected area of indentation
(d) height of rebound
[CSE-Prelims, ME : 1999]
A horizontal force of 200 N is applied at 'A' to
lift the load W at C, as shown in figure 1.10.
Value of weight W is
200 N 4
A
E
0.075 m
(a) 200 N
(c) 600 N
Q.6
Fig. 1.10
(b) 400 N
(d) 800 N
[CSE-Prelims, ME : 1998]
Which one of the following pairs is not correctly
matched?
If E. Young's modulus
a = coefficient of linear expansion
T= Temperature rise
A= Area of cross-section
L= Original length
(a) Temperature strain with permitted expansion
8 is raTL-Sl
L )
(b) Temperature stress — aTE
(c) Temperature thrust — aTEA
(d) Temperature stress with permitted
expansion 8 — E(aTL - 8)
Q.7
Which one of the following pairs is not correctly
matched?
(a) Uniformly distributed stress: Force passes
through centroid of the cross-section
(b) Elastic deformation : work done by external
forces during elastic deformation is fully
dissipated as heat
MADE EASY
Q.10 If a block of material of length 25 cm, breadth
10 cm and height 5 cm undergoes volumetric
strain of
1
, then change in volume will be
5000
(a) 0.50 cm3
(c) 0.20 cm3
(b) 0.25 cm3
(d) 0.75 cm3
[CSE-Prelims, ME : 2000]
Q.11 For an isotropic homogeneous and linearly
elastic material, which obeys Hooke's law, The
number of independent elastic constants are
(a) 1
(b) 2
(c) 3
(d) 6
Simple Stresses in Uniform and Compound Bars
Strength of Materials
1 1 1
Q.12 Assuming E= 180 GPa and G=100 GPa for a
material a strain tensor is
(0.002
0.004 0.006`
0.004 0.003
0
0.006
0
0
Shear stress xy is
(a) 400 MPa
(c) 800 MPa
Dm-
(b) 500 MPa
(d) 1000 MPa
[CSE-Prelims, ME : 2001]
Q.13 With the increase of percentage of carbon In
steel, which one of the following properties does
increase
(a) modulus of elasticity
(b) ductility
(c) toughness
(d) hardness
[CSE-Prelims, ME : 2001]
0.14 Match List-I (material) and List-II (Stress-strain
curve) and select the correct answers.
List-I
A. Mild steel
B. Pure copper
C. Cast iron
D. Aluminium
List-II
1.
a
2.
a
3.
4.
Codes:
A
(a) 3
(b) 3
(c) 2
(d) 4
BCD
1
4
1
2
4
1
1
4
3
1
3
2
[CSE-Prelims, ME : 2001]
Q.15 Match List-I with List-I I and select the correct
answers.
List-I
A. Ultimate strength
B. Natural strain
C. Conventional strain
D. Stress
List-II
1. Internal structure
2. Change in length per unit instantaneous
length
3. Change in length per unit gauge length
4. Load per unit area
Codes:
A
BCD
(a) 1
2
3
4
(b) 1
3
2
4
2
1
(c) 4
3
2
3
1
(d) 4
[CSE-Prelims, ME : 2002]
Q.16 A steel rod of diameter 1 cm and 1 m long is
heated from 20°C to 120°C, its a = 12 x 10-6/K
and E= 200 GN/m2. If the rod is free to expand,
thermal stress developed in it is
(a) 12 x 103 N/m2
(b) 240 kN/m2
(c) zero
(d) infinite
[CSE-Prelims, ME : 2002]
Q.17 Consider the following statements:
1. There are only two independent elastic
constants
2. Elastic constants are different in orthogonal
directions
12
.
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
3. Material properties are same everywhere
4. Elastic constants are same in all loading
conditions
5. The material has ability to withstand shock
loading
Which of the above statements are true for a
linearly elastic homogeneous and isotropic
material?
(b) 2, 3 and 4
(a) 1, 3, 4 and 5
(d) 2 and 5
(c) 1, 3 and 4
[CSE-Prelims, ME : 2002]
Q.18 A cast iron specimen in a torsion test gives a
(a) cup and cone fracture
(b) fracture along a plane normal to the axis of
the specimen
(c) fracture along a helix of approximately 45°
(d) fracture along a plane inclined at 60° to the
axis
[CSE-Prelims, ME : 2002]
Q.19 The reactions at the rigid support A and B for
the bar loaded as shown in figure 1.11 are
respectively
Fig. 1.11
(a)
0 kN 10 kN
3
3
(c) 5 kN, 5 kN
(b)
10 kN 20 kN
3
3
(d) 6 kN, 4 kN
Q.20 A steel rod 10 mm in diameter and 1 m long is
heated from 20°C to 120°C, E = 200 GPa and
a =12 x 10-6/°C. If the rod is not free to expand,
the thermal stress developed is
(a) 120 MPa (tensile) (b) 240 MPa (tensile)
(c) 120 MPa (comp.) (d) 240 MPa (comp.)
[CSE-Prelims, ME : 2003]
Q.21 The force F is such that the bars AC and BC
(AC and BC are equal to length) as shown in
the figure are identically loaded is
Fig. 1.12
(a) 70.7 N
(c) 141.4 N
(b) 100 N
(d) 108 N
[CSE-Prelims, ME : 2003]
Q.22 A mild steel specimen is tested in tension up
to fracture in a Universal Testing Machine. Which
of the following mechanical properties of the
material can be evaluated from such a test?
1. Modulus of elasticity
2. Yield stress
3. Ductility
4. Tensile strength
5. Hardness
6. Modulus of rigidly
Select the correct answer using the code given
below:
(a) 1, 3, 5 and 6
(b) 2, 3, 4 and 6
(c) 1, 2, 5 and 6
(d) 1, 2, 3 and 4
[CSE-Prelims, ME : 2007]
Q.23 A heavy uniform rod of length 'L' and material
density '8' is hung vertically with its top end
rigidly fixed. How is the total elongation of the
bar under its own weight expressed?
(a) 28L2g
E
8L2g
(c)
(b)
8L2g
E
(d)
61_2g
2E
[CSE-Prelims, ME : 2007]
Q.24 A round bar length 1, elastic modulus E and
Poisson's ratio ji is subjected to an axial pull
'P. What would be the change in volume of the
bar?
Simple Stresses in Uniform and Compound Bars
Strength of Materials
(a)
(c)
Pl
(1— 21.0E
P1µ
(b)
P1(1— 2µ)
E
(d)
—
P/
ptE
[CSE-Prelims, ME : 2007]
Q.25 Which one of the following material has the
highest ultimate tensile strength
(a) mild steel
(b) cast iron
(c) spring steel
(d) wrought iron
[CSE-Prelims, ME : 2007]
Q.26 A straight uniform rod is subjected to axial load.
Which one of the following is the correct
statement?
(a) It induces maximum shearing stress on a
transverse plane
(b) It induces maximum normal stress on a
plane inclined at 45° to the axis of the rod
(c) It induces maximum shear stress on a plane
inclined at 45° to the axis of the rod
(d) It induces zero shear stress on any inclined
plane to the axis of the rod
[CSE-Prelims, ME : 2006]
Q.27 Which one of the following is rupture stress?
(a) Breaking stress
(b) Maximum load/original cross-sectional
area (A)
(c) Load at breaking point/A
(d) Load at braking point/neck area
[CSE-Prelims, ME : 2006]
Q.28 A 2 m long rod of diameter 2 mm is subjected
to an axial pull of 1 kN. The rod is extended by
0.5 cm. What is the approximate value of the
modulus of elasticity of the material of the rod?
(a) 127 G Pa
(b) 180 G Pa
(c) 125 G Pa
(d) 200 G Pa
Q.29 Which one of the following information cannot
be obtained from the static tensile test of a mild
steel specimen?
(a) Modulus of elasticity
(b) Qualitative determination of toughness
(c) Ductility
(d) Weldability
[CSE-Prelims, ME : 2006]
41 13
Q.30 Consider the following statements relating to
Rockwell Hardness Testing Method:
1 It is a quick method to determine hardness
of industrial components.
2. Polishing of test sample is necessary.
3. Test load is applied in two stages.
4. Hardness of relatively softer metals cannot
be determined by this method.
5. Inverted pyramid type of indenter is used.
Which of the statements given above are
correct?
(a) 1, 2, 3 and 4
(b) 3, 4 and 5
(c) 1, 2 and 3
(d) 2, 4 and 5
[CSE-Prelims, ME : 2006]
0.31 If a material has numerically the same value for
its modulus of rigidity and bulk modulus, then
what is its Poisson's ratio?
(a) 0.25
(b) 0.2
(c) 0.15
(d) 0.125
[CSE-Prelims, ME : 2009]
Q.32 A free bar of length 1 is heated uniformly from
0°C to a temperature T°C. a is the coefficient of
linear expansion and E is the modulus of
elasticity. Which one of the following is the stress
induced in the bar?
(a)
aTE
4
(c) aTE
(b)
aTE
2
(d) Zero
[CSE-Prelims, ME : 2009]
Q.33 What property of a material enables it to be
drawn into wires with the application of tensile
force?
(a) Plasticity
(b) Elasticity
(c) Ductility
(d) Malleability
[CSE-Prelims, ME : 2009]
Q.34 ABC is a rigid bar. It is hinged at A and
suspended at B and C by two wires, BD and
CE made of copper and steel respectively, as
shown in the given figure. The bar carries a load
of 10 kN at F, midway between Band C. Given
that
A,-= 4 cm2
Aa = 2 cm2
= 1 x 105 N/mm2
ES = 2 x 105 N/mm2
14 0.
MADE EASY
IAS & IFS (Objective & Conventional) Previous Solved Questions
Subscript c and s stands for copper and steel.
If the extensions in the steel and copper wires
are As and As respectively, the ratio As/Ac would
be
I
I
Q.38 A bar of uniform cross-section of 400 mm2 is
loaded as shown in figure. The stress at section
1-1 is
71
40 kN
-0.
20 kN
4E--
10 kN
----0,-
30 kN
4-
1
400 mm 300 mm 400 mm
m
,-1
Fig. 1.13
(a) 1/4
(c) 2
(b) 4
(d) 1/2
[CSE-Prelims, CE : 2000]
Q.35 Match List-I (Material) with List-II (Modulus of
elasticity N/mm2) and select the correct answer:
List-II
List-I
A. Steel
1. 0.6 x 105
2. 1 x 105
B. Cast iron
C. Aluminium
3. 2 x 105
D. Timber
4. 0.1 x 105
Codes:
A
B
C
D
2
1
4
(a) 3
4
(b) 2
3
1
(c) 3
2
4
1
(d) 2
3
4
1
[CSE-Prelims, CE : 2001]
Q.36 The phenomenon of decreased resistance of a
material due to reversal of stress is called
(b) elasticity
(a) resilience
(d) fatigue
(c) creep
[CSE-Prelims, CE : 2002]
Q.37 Total elongation of a prismatic bar of length L,
weight W, cross-sectional area A and modulus
of elasticity E, under its own weight while
hanging vertically, is
(a)
WL
AE
(c)
WL
2AE
Fig. 1.14
1
. m---1
--
(b)
2WL
AE
WL
3AE
[CSE-Prelims, CE : 2002]
(d)
(a) 50 N/mm2
(c) 25 N/mm2
(b) 100 N/mm2
(d) 200 N/mm2
[CSE-Prelims, CE : 2003]
Q.39 As soon as the external forces causing
deformation in a perfectly elastic body are
withdrawn, the elastic deformation disappears
(a) only partially
(b) completely over a prolonged period of time
(c) completely and instantaneously
(d) completely after an initial period of rest
[CSE-Prelims, CE : 2003]
Q.40 Match List-I (Mechanical property) with List-II
(Feature) and select the correct answer.
List-I
A. Creep
B. Tenacity
C. Ductility
D. Brittleness
List-II
1. Amenability to go through changes of shape
without rupture
2. Susceptibility to deform with time under
sustained loading
3. Ability to be drawn into wire
4. Susceptibility to fail suddenly without
warning
Codes:
A
BCD
(a) 4
1
3
2
(b) 2
3
1
4
3
1
2
(c) 4
(d) 2
1
3
4
[CSE-Prelims, CE : 2003]
Simple Stresses in Uniform and Compound Bars
Strength of Materials
4
15
Q.41 If P is the direct load, L is the length of the
member, A is the uniform area of cross-section
and E is the Young's modulus; then strain energy
due to direct stress caused by a gradually
applied load is
P 2L
P 2L
(a)
2AE
PL
(c) 2AE
(b)
2 A2E
PL
(d) AE
[CSE-Prelims, CE : 2003]
Q.42 Consider the following statements regarding
tensile test diagrams for carbon steels with
varying carbon contents;
As the carbon content increases
1. the ultimate strength of steel decreases
2. the elongation before fracture increases
3. the ductility of the metal decreases
4. the ultimate strength of steel increases
Which of the statements given above are
correct?
(a) 3 and 4
(b) 1 and 3
(c) 1, 2 and 3
(d) 1 and 2
[CSE-Prelims, CE : 2005]
Q.43 What is the maximum possible value of
Poisson's ratio for a non-dilatant material?
(a) 0.67
(b) 0.50
(c) 0.33
(d) 0.25
[CSE-Prelims, CE : 2005]
Q.44 The shear modulus of a material is half of its
Young's modulus. What is the value of its
Poisson's ratio?
(a) -1
(b) -0.5
(c) zero
(d) 0.5
[CSE-Prelims, CE : 2006]
Fig. 1.15
(a) 16.67 N/mm2
(c) 26.67 N/mm2
(b) 13.33 N/mm2
(d) 30 N/mm2
[CSE-Prelims, CE : 2007]
Q.47 In a pure tensile member, the normal stress on
a plane at right angles to the direction of load
is 100 N/mrn2. What is the normal stress at a
plane whose normal is inclined at 60° to the
direction of the load?
(a) 75 N/mm2
(b) 100 N/mrn2
(c) 125 N/mm2
(d) 150 N/mm2
[CSE-Prelims, CE : 2008]
Q.48 How is the maximum strain energy stored per
unit volume in a body without permanent
distortion, termed as?
(a) Modulus of resilience
(b) Modulus of tenacity
(c) Modulus of toughness
(d) Proof resilience
[CSE-Prelims, CE : 2008]
Q.49 A cable is supported at P and Q in which P is
higher than Q. At what point(s) among the
following is the tension in the cable maximum?
Q.45 What is range of values of Poisson's ratio for
ductile materials?
(a) 0.10 to 0.15
(b) 0.16 to 0.20
(c) 0.21 to 0.24
(d) 0.25 to 0.33
[CSE-Prelims, CE : 2006]
Q.46 What is value of stress at the base CC for the
bar shown in the figure given below?
Fig. 1.16
16 0.
lAS & IFS (Objective & Conventional) Previous Solved Questions
(a) Ponly
(c) R only
(b) Q only
(d) P and 0
[CSE-Prelims, CE : 2008]
Q.50 In the following pin jointed truss, what is the
displacement of support B due to the given
load?
5005 kN
MADE EASY
Q.52 At a certain stage under elastic loading, the
elongation observed was 0.03 mm, the gauge
length was 150 mm and the modulus of
elasticity was 2 x 105 N/mm2. What was the
stress at that location?
(a) 4 N/mm2
(b) 40 N/mm2
(c) 80 N/mm2
(d) 60 N/mm2
[CSE-Prelims, CE : 2009]
Q.53 Modulus of elasticity and Poisson's ratio of a
material are 2.1 x 105 N/mm2 and 0.25
respectively. What is the value of modulus of
rigidity of the same material in 105 N/mm2?
(a) 0.84
(b) 0.70
(c) 1.40
(d) 0.50
[CSE-Prelims, CE : 2010]
Fig. 1.17
Answers
(Cross-sectional area of each member
= 500 mm2, modulus of elasticity E = 2 x 105
N/mm2)
(a) 3.25 mm
(b) 2.50 mm
(c) 1.50 mm
(d) 0.50 mm
[CSE-Prelims, CE : 2009]
Q.51 What is the normal stress on a plane inclined at
45° to the axis of a square rod of side a
subjected to an axial tensile force of T?
(a)
(c)
T
a
T
(b) 2a
2
T
(d)
4a2
2
1. (d)
2. (b)
3. (d)
4. (b)
6. (d)
7. (b)
8. (c)
9. (a) 10. (b)
11. (b)
12. (c)
13. (d)
14. (b) 15. (a)
16. (c)
17. (c)
18. (c)
19. (a) 20. (d)
21. (a)
22. (d)
23. (d)
24. (b) 25. (c)
26. (c)
27. (d)
28. (a)
29. (d) 30. (c)
31.(d)
32. (d)
33. (c)
34. (c) 35. (b)
36. (d)
37. (c)
38. (b)
39. (b) 40. (d)
41.(a)
42. (a)
43. (b)
44. (c) 45. (d)
46. (d)
47. (a)
48. (a)
49. (a) 50. (b)
51.(b)
52. (b)
53. (a)
5. (d)
2
8a
[CSE-Prelims, CE : 2009]
Explanations
1.
without having holes, acts, shoulders or narrow
passes.
(d)
2a
kt = 1+--b
2a is major axis perpendicular to axis of loading.
Stress contraction factor (kt) is a dimensionless
factor which is used to quantity how concentrated
the stress is in a material. It is defined as the
ratio of the highest stress in the element to the
reference stress.
2.
(b)
but
ae
aP
abreaking
ame
°max
k=
r aref
Reference stress is the stress in the part within
an element under the same loading conditions
Simple tensile stress
Fig. 1.18
Simple Stresses in Uniform and Compound Bars
Strength of Materials
Hooke's law valid only upto
proportionality.
3.
'
limit of
(d)
Eand [t are two independent elastic constants.
12. (c)
shear strain = 0.004
0.008
G= 100 x 1000 N/mm2
ti~ = yxy x 4
= 0.008 x 100 x 1000 N/mm2
= 800 MPa
a
y =
13. (d)
With the increase of carbon percentage in steel.
Hardness of steel does increase.
Fig. 1.19
OA = elastic
AB= strain hardening portion
(b)
% elongation -
6L
gauge length
change in length
gauge length
5.
(d)
200 x 0.3 = 0.075 W
W= 800 N
6.
17
11. (b)
2
4.
44
(moment about B)
(d)
is not correctly matched.
Temperature stress with permitted explain is not
E(aTL- 6).
14. (b)
Mild steel - 3, Pure copper - 2, Cast iron - 4,
Aluminium - 1.
15. (a)
1. A. internal structure
2. B. 6//instantaneous length
3. C. 6L/L
4. D. load per unit area
16. (c)
Rod is free to expand, no stress will be
developed.
17.
(c)
1, 3 and 4 are correct statements.
7.
(b)
Elastic deformation work done by external forces
during elastic deformation is not dissipated fully
as heat.
18. (c)
A cast iron specimen in a torsion test fails along
a helix approx at 45° to the axis due to tensile
principal stress p1 = +T(shear)
8.
(c)
2. Resistance to inductation
A. Hardness
Toughness
3.
Ability to absorb energy
B.
during plastic
C. Malleability 4. Ability to be rolled into plates
1. Percentage of elongation
D. Ductility
19. (a)
9.
(a)
A measure of Rockwell hardness is a depth of
penetration of indentor.
10 kN
1
4- P2
2m
k- 1m
Fig. 1.20
P1 x1 =P2 x 2
AE
AE
P1 = 2P2
P1 + P2 = 1 0
10. (b)
6V-
25 x10x 5
- 0.25 cm3
5000
20
- 3
18 ••
MADE EASY
IAS & IFS (Objective & Conventional) Previous Solved Questions
20. (d)
TE
= 12 x 10-6 x 100 x 200 x 103
= 240 MPa (comp.)
Because expansion is prevented.
21. (a)
Bars AC and BC are identically loaded.
FAC = FBC
Horizontal component of FAc and FBc are cancelled.
2FAB cos 30° = 100 cos45°
Horizontal component of 100 N = 70.7 N is
balanced by F
So F= 70.7 N
22. (d)
1, 2, 3, 4 are evaluated by tension test on MS.
23. (d)
8L due to self weight =
(pg) 2
2E
29. (d)
Weldability cannot be determined by static tensile
test.
30. (c)
1, 2, 3 are correct statements.
31. (d)
2(1+v) 3(1-2v)
2 + 2v= 3 - v
1 =8v
v=0.125
32. (d)
Bar is free to expand no stress is developed.
33. (c)
Ductility enables the wire to the drawn.
34. (c)
Since ABC is rigid, deflection will be linear.
L
A
8,
24. (b)
8V= evV =
1-1 m
2µ) x IA
1m-►I
I
Fig. 1.21
8s
-=2, Similar triangles
Sc
25. (c)
Spring steel is a high carbon steel, it has highest
out amongst the material gives.
26. (c)
Axial load on uniform straight bar induces
maximum shear on a plane inclined at 45° to the
axis of the rod.
35. (a)
A. Steel
B. Cast iron
C. Aluminium
D. Timber
3. 2 x 105 N/mm2
2. 1 x 105 N/mm2
1. 0.6 x 105 N/mm2
4. 0.1 x 105 N/mm2
36. (d)
Fatigue
37. (c)
27. (d)
Rupture load at breaking point/neck area.
28. (a)
E=
1000 2000
x
TC
0.5x10
=1.27 x 105 N/mm2
=127 GPa
Fig. 1.22
Extension due to own weight
38. (b)
611 =
Ebb = craa x (a)2 = 100 x (0.861)2
= 75 N/mm2
wL
2AE
8-
19
Simple Stresses in Uniform and Compound Bars
Strength of Materials
48. (a)
Maximum strain energy per unit volume without
permanent distortion = Modulus of resilience.
40000
= 100 N/mm2
400
39. (b)
Elastic deformation disappears completely over
a prolonged period of time.
49. (a)
P only, maximum slope of cable at P.
50. (b)
500 x,FakN
40. (d)
A. Creep
2. Susceptibility to deform
with time under
sustained loading
1. Amenability to go
through changes of
shape without rupture
3. Ability to be drawn into
wire
4. Susceptibility to fail
without
suddenly
warning
B. Tenacity
C. Ductility
D. Brittleness
A
250 xsffkN
250 x5- kN
Fig. 1.24
Extension in AB - Components of contraction
in CA and CB
8B =
41. (a)
250 x 103 1000
500 x 103 x 1000
x
2 x
x cos 60°
500
2 x 105
500 x2 x105
= +2.5 - 5 = -2.5 mm or 2.5 mm
PL P2L
Strain energy = -P x
=
2
AE 2AE
1
51. (b)
n
\<
7 45°
42. (a)
(3) and (4).
n
43. (b)
Maximum possible value of Poisson's ratio for a
non-dilatant material is 0.5.
Fig. 1.25
T
6a= a
T
2
abb- -7---X(COS45) .
a2
2a 2
44. (c)
1
E
G= -E =
, so, v = 0
2
2(1+v)
52. (b)
8L= 0.03 mm
L= 150 mm
E= 2 x 105 N/mm2
45. (d)
v = 0.25 - 0.33, for ductile materials.
46. (d)
0cc -
a=
(20+25)1000
1500
= 30 N/mm2
LL xE _
0.03
x2x105
150
= 40 N/mm2
53. (a)
47. (a)
E= 2.1x 105 N/mm2, v = 0.25
b
60°
E
3
100
b
Fig. 1.23
2.1x105
2.5
= 0.84 x 105 N/mm2
G= 2(1+0.25) -
11111111•
02
Principal Stresses
CHAPTER
Q.2.1 At a section in a beam the tensile stress due to bending is 50 N/mm2 and there is shear stress of
20 N/mm2. Determine from first principles, the magnitude and direction of principal stress and
calculate the maximum shear stress
[IFS, 2012, ME : 10 Marks]
Solution:
50
20
(F1 + Q2)
F1
Q2
(b)
(a)
(c)
Fig. 2.1
Figure shows an element ABC unit of thickness subjected to normal and shear stresses as given. On
plane BC there is complementary shear stress of 20 N/mm2 as shown in figure 2.1(a).
F1 = 50 x AC
Normal force,
Shear force,
Q1 = 20 x AC
Shear force,
Q2 = 20 x BC
Normal force on inclined plane AB
Fr,_ (F1 + Q2 ) cos0 + Q1 sin°
On x AB = (F1 + Q2) cos() + Q1 sine
or
on x AB = 50 AC cos0 + 20 BC cos° + 20 AC sine
or
an = 50 cos2O + 20 sine cos() + 20 cos() sin()
50 cos20 + 20 sin 20
tangential force on plane AB
Ft
x AB = (F1 + Q2 ) sin° - 1 cos°
or
To =
sine 02 sine Q1 cos()
AB AB AB
Note from figures, that internal resistance is equal and opposite to the applied forces on inclined plane.
Principal Stresses
Strength of Materials
4
21
Putting the values of F1, Qi, 02
tie -
50 x ACsin0 20 xl3Csin0 20 x ACcos0
AB
AB
AB
= 50 sin0 cos° + 20 sin20 - 20 cos20
50
= — x sin 20 - 20 cos 20 = 25 sin 20 - 20 cos 20
2
Principal planes
On principal planes, to, shear stress is zero
25 sin20 - 20 cos20 = 0
So,
tan20 = 0.8
201 = tan-10.8 = 38.66°
01 = 19.33°
Another plane,
02 = 90 + 19.33 = 109.33°
Principal stresses
01 = 19.33, cos01 = 0.943.6
cos201 = 0.890
Principal stress,
sin201 = 0.6247
p1 = 50 x 0.8904 + 20 x 0.6247
44.52 + 12.494 = 57.016 MPa
02 = 109.33°, cos02 = -0.3310, cos202 = 0.10956
sin202 = -0.6247
Principal stress,
p1 = 50 x 0.10956 + 20 x (-0.6247)
= 5.478 - 12.494 = -7.016 MPa
Principal stresses are + 57.016, - 7.016 MPa
Principal directions are 19.33°, 109.33°
For maximum shear stress, 0,
de = 0
25 x 2 cos 20 + 20 x 2 sin 20 = 0
0 = -25.67° or 64.33°
To = 25 x sin (-2 x 25.67°) - 20 x cos (-2 x 25.67°)
= -19.521 - 12.495 = - 32.016 Mpa
It9I = 32.016 MPa
Q.2.2 Draw Mohr's circle for a 2-dimensional stress field subjected to
(ii) Pure biaxial tension
(i)
Pure shear
(iv) Pure uniaxial tension
(iii) Pure uniaxial compression
[CSE-Mains, 1997, ME : 15 Marks]
Solution:
= +T
OB = -T
ti
B - ti
22 ►
lAS & IFS (Objective & Conventional) Previous Solved Questions
Air
a
MADE EASY
2
- Radius
6
0
OA =
Pure Bi-axial tension
OB= (72
oc =
Cf + 6
'
'
2
R=
Pure uniaxial compression
OA =
OC = --cv/2
(iv)
Pure uniaxial tension
Fig. 2.2
Q.2.3 For the following state of stress, show the stresses on two given planes at right angle of an
element. Find the magnitude and directions of principal stress and the maximum shear stresses
in each case.
(i) Simple uniaxial tension
(ii) Pure equal normal stresses on given planes
(iii) Pure shear stresses on given plane
[CSE-Mains, 2002, ME : 20 Marks]
Solution:
(a)
Fig. 2.3 (a)
(i) Simple uniaxial tension.
Principal stress,
p1 =
p2 = p3 = 0
p1 along x-axis.
Maximum shear stress =
/91 —P2
2
—P1
2
Principal Stresses i
Strength of Materials
23
y
x
(b)
Fig. 2.3 (b)
(along x and y direction)
(along z direction)
Pure equal normal stresses , p1 = p2 = a
p3 = 0
Maximum shear stress -
P1
P2
2
=0
(iii) Mohr's circle diagram will be a point on +x axis with coordiantes (a, 0)
(c)
Fig. 2.3 (c)
Pure shear stresses on given planes
p1 = +T, P2 =
01 = 45°, 02 = —45°
p3 = 0
Maximum shear stress = ti
Note: In case (ii) Mohr circle diagram will be a point on (+ve) x-axis, having coordinate (a, 0)
Q.2.4 At a point in an elastic material the stresses on three mutually perpendicular planes are
: 50 MN/m2 tensile and 40 MN/m2 shear
First Plane
Second Plane : 30 MN/m2 compressive and 40 MN/m2 complementary shear
Third Plane
: No stress
Find: (i) The position of principal planes and magnitude of principal stresses
(ii) The position of the planes on which maximum shear stress act, calculate normal and
shear stress on them.
[CSE-Mains, 1992, ME : 20 Marks]
Solution:
Fig. 2.4
24
P.
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
BC and AC are two mutually perpendicular planes. Reference plane BC
1 = +50 N/mm2
(i)
c = +40 N/mm2
Plane AC
62 = -30 N/mm2
ti = +40 N/mm2
61
2 62 -
10, al 262
-
- 40
Principal stresses
P1 = 10-4402 +402 = 10 + 56.56 = 66.56 N/mm2
P2 = 10_J402 + 402 = 10 - 56.56 = -46.56 N/mm2
12 x 40
20 = tan-1 2T = tan= tan-1(1) = 45°
1
80
al a2
61 = 22.5°, 02 = 112.5°
Tmax = V402 + 402 = 56.56 N/mm2
(ii) Angle of plane of Tam
03 = 0l+ 45° = 67.5° w.r.t plane BC
04 = 67.5 + 90° = 157.5°
Normal stresses on planes of maximum shear stress, 'um,
-
61 + az
2
(wrt plane BC)
- 10 N/mm2
On both shear planes with +Tricia, and -Tma,
Q.2.5 At a point in a material under stress, the intensity of the resultant stress on a certain plane is
50 MN/m2 (tensile) and inclined at 30° to the normal of that plane (Fig. 2.5). The stress on a plane
at right angle to this plane has a normal tensile component of intensity 30 MN/m2.
Find: (i) resultant stress on second plane
(ii) principal planes and stresses
(iii) maximum shear stress and the plane on which it occurs
Solution:
Shear stress on plane BC = 50 sin 30° = 25 MPa
Normal stress on plane BC = 50 cos 30°
a1= 43.3 MPa
Complementary shear stress on plane AB
ti = 25 N/mm2 (as shown)
50 MPa
Resultant stress on plane AB = V302 + 252
= 39.05 N/mm2 = 39.05 MPa
62, normal stress on plane AB = 30 MPa
(ii)
al +62
2
433+ 3° = 36.65 MPa
2
Fig. 2.5
Principal Stresses
Strength of Materials
al — a2
ai — G2 )2 T_2 _
652 252 25.87 MPa
2
Principal stresses
25
43.3 — 30
= 6.65 MPa
2
2
(
4
(31
P1' 1-'2
2
+ G2 + (G1 G2) +T2
2
2
p1 = 36.65 + 25.87 = 62.52 MPa
p2 = 36.65 — 25.87 = 10.78 MPa
Principal angle (wrt plane BC)
1
0 = — tan
1
2
2T
cy1—a2
1
=
2
tan
_12 x 25
13.3
= 37°33'
02 = 90 + 61 = 127°33'
(iii) Maximum shear stress
max
= ±25.86 MPa
03 = 01 + 45° = 37°33' + 45° = 82°33'
04 = 03 + 90° = 172°33'
2
20 N/mm
Q.2.6 The figure given below shows the state of stress at a
point namely 30 N/mm2 tensile in x-direction, 20 N/mm2
tensile in y-direction and shear stress of 50 N/mm2.
Find the position of principal planes, principal stresses
and maximum shear stress graphically or other wise.
[CSE-Mains, 2000, ME : 20 Marks]
50
2
30 N/mm
50
30
2
30 N/mm
50
Solution:
Let us take plane AB as reference plane on which shear
stress is negative shear stress
cr / = 30 N/mm2, a2 = 20 N/mm2, ti = 50 N/mm2
D
50 4
20
— 25
Fig. 2.6
— 5 N/mm2
Principal stresses
P1 ' P3 =
+ G2
(01 — 02
2
2
2
) +ti2
= 25± V52 + 502 = 25± 50.25
p1 = +75.25 N/mm2
p2 = —25.25 N/mm2
Angle of principal plane w.r.t plane AB (in anti-clockwise direction)
_1 2T = 1
1 2 x 50
2 tan= 42°8'41"
0 = -tan _1
1
10
0 G2
02 = 01 + 90° = 132°8'41"
2
l'Crinaxl maximum shear stress = ( al — (32 )
2
Tina, will occur at angle,
and
T 2 = 50.25 N/mm2
03 = 01 + 45° = 87° 8' 41"
04 = 03 + 90° = 177° 8' 41"
26 . lAS & IFS (Objective & Conventional) Previous Solved Questions
Q.2.7 A circle of 10 cm diameter is inscribed on a steel plate
before if is stressed. Then the plate is loaded so as to
produce stresses as shown in the figure 2.7 and circle
is deformed into an ellipse. Determine the length of
major and minor axes of the ellipse and their directions.
Take modulus of elasticity of steel as 2.1 x 107 N/cm2
and Poisson's ratio = 0.3.
[CSE-Mains, 2004, ME : 20 Marks]
MADE EASY
2
12 kN/cm
17
6
P2
Solution:
E = 21000 kN/cm2
Taking AB as reference plane
2
12 kN/cm
= +12 kN/cm2
a2 = -6 kN/cm2
Fig. 2.7
ai +432 - 3 kN/cm2
2
al - 02 - 9 kN/cm2
2
ti = 4 kN/cm2
2
a2
2
2
2
+tie - V9 + 4 2 --9.849 kN/cm
p1 = 3 + 9.849 = 12.849 kN/cm2
p2 = 3 - 9.849 = -6.849 kN/cm2
Principal angle (w.r.t. in plane CD in anti-clockwise direction)
1
1
= 12°
= 1
01 = --tan2
ai --a2 2 tan-12184
02 = 102°
Strains,
E
vp2
E
12.849 0.3 x 0.849 _ 14.9037
+
21000
21000
21000
10.7037
-6.849 0.3 x 12.849
21000
E2 - 21000
21000
= 0.70797 x 10-3
e2 = -0.5097 x 10-3
(In the direction of major principal axis)
Maximum axis = 100 + 0.70797 x 10-3 x 100 = 100.0709 mm
(In the direction of minor principal)
Minimum axis = -100 - 5097 x 10-3 + 100 = 99.949 mm
Q.2.8 At a point in a strained material, planes AB and AC pass through the point A as shown in the
figure 2.8(a). Normal and shear stresses on plane AB are 30 MPa and 50 MPa respectively as
shown. Normal and shear stress on plane AC are 60 MPa and 20 MPa respectively as shown. By
graphical or analytical method, determine angle between the planes AB and AC.
[CSE-Mains, 2001, ME : 30 Marks]
Solution:
Using sign convention
and
Principal Stresses
Strength of Materials
1
27
D
60
(30, +50)
(a)
E
OC = +30
CD = +50
OA = +60
AB = —20
Refer
(60, — 20)
Mohr's stress circle
29 = 90°, 0 = 45°
(b)
Fig. 2.8
Note: In shear stress produce clockwise couple at the center of element then it will be plotted above ax
(+ve) and if shear stress causes anticlockwise couple at the center then it will be plotted below a axis.
Q.2.9 (a) Under which condition of the state of stress at a point in the two dimensions, the Mohr's
circle will be reduced to a point.
(b) How much change in volume would a 100 mm side cube of steel will have when it is kept at
a depth of 2 km in sea water?
Assume specific gravity of sea water equal to 1.02, modulus of elasticity equal to 2.08 GPa and
Poisson's ratio equal to 0.29.
[CSE-Mains, 2007, ME : 20 Marks]
Solution:
(a)
Mohr's stress circle will be reduced to a point
(T = ay =
If
x
or
ax = 0y =
Cr
as shown
fa
Y
X
, 0)
(—a.
•
rX
(a, 0)
Fig. 2.9
28
(b)
0.
MADE EASY
IAS & IFS (Objective & Conventional) Previous Solved Questions
h= 2 km
y= 1.02 x 1 x 10-5 N/mm2
p, pressure = yh = 1.02 x 10-5 x 2000 x 1000 N/mm2 = 20.4 N/mm2
Depth,
Weight density,
K, bulk density =
E
3(1-2v) 3(1-0.29 x 2)
p
Volumetric strain,
Ev
2.08 x105
K=
-1.65 x105 N/mm2
20.4
1.65x105 12.357x 10-5
V = 1003 = 106 mm3
8V = 12.357 x 10-5 x 106 = 123.57 mm3
Change in volume,
Q.2.10 The normal stresses at a point in a strained material across two planes at right angles of each
other are 120 N/mm2 tensile and 60 N/mm2 compressive. The shear stress on these planes is
40 N/mm2. Find
Principal stresses and principal planes
(i)
The direct and shear stresses on a plane inclined at 30° to the vertical
(ii)
[CSE-Mains, 1998, ME : 30 Marks]
Solution:
Let us take BC as reference plane
al = +120 MPa, a2 = -60 MPa, t = -40 MPa on BC
a' 2 a2
(31 CF2
- 90 MPa
- 30 MPa
2
'
2
Pi, P2 = 30 ± V902 + 402 = 30 ± 98.49
p1 = 128.49 MPa
p2 = -68.49 MPa
1 12T
9 ,_ -tan
1
2
al 0,2
(w.r.t palne BC in anticlockwise direction)
12 x 40 1
= 1tan= x 24° = 12°
2 180 2
02 = 90 + 12° = 102°
Reference plane BC
Inclined plane AB
0 = 60° (angle from plane BC)
Normal,
a o
al 1- a2 a1 - a2 cos120°+ tsin120°
2 + 2
co
0_
m
0
co
1
= 30 + 90 cos120° + 40 sin120°
= 30 - 45 + 34.64 = 19.64 MPa
Shear stress,
II
T
40 MPa
T = (a1 - a2 )sin120°-tcos120°
o
2
IT'
= 90 sin 120°- 40 x cos 120°
= 90 x 0.866 - 40(-0.5)
= 77.94 + 20 = 97.94 MPa
= 97.94 MPa
0"
T
C
al= + 120 MPa
Fig. 2.10
Principal Stresses
Strength of Materials
29
Q.2.11 Two planes AB and AC make an angle of 50° at point
A. Plane AB is subjected to tensile stress of 3 kN/cm2
and a shear stress 3 kN/cm2 from B toward A. Plane
AC neglected to a shear stress of magnitude
AB — Ref. Plane
2 kN/mm2 (from C towards A) and an normal stress
of unknown magnitude. Determine (i) Normal stress
on plane AC (ii) Principal stresses.
[IFS 2011, ME : 15 Marks]
Solution:
Stress on plane AB
a = +30 N/mm2 = 3 kN/cm2
ti = -30 N/mm2 = -3 kN/cm2
AB is reference plane shear stress on reference plane is
D
negative. Consider a plane AD perpendicular to plane
Fig. 2.11
AB, say normal stress on this plane is a2. Complementary
shear stress on this plane is +30 N/mm2.
Plane AC
= 61 - 62 sin20 --tcos20
2
0 = -50°
20
- alsin(-100°)-30cos(-100°)
2
(30 2
a2)( 0 9848)- 30 (-0.17365)
= -14.772 + 0.4924 a2 + 5.20945
62 -
20 +14.772 - 5.20945
0.4924
= 60.038 N/mm2
(i) Normal stress on plane BC
a -
61
2
62 4. al
a2 cos(-100°) + Tsin(-100°)
2
30 + 60.038 30 - 60.038
2
2
x cos(-100°)+ 30 x sin(-100°)
= 45.019 - 15.019 (-0.173648) + 30 (-0.9848)
= 45.019 + 2.6080 - 29.544
= 18.083 N/mm2 (Normal stress on plane AC)
(ii) Principal stresses
a1 = 30 N/mm2
a2 = 60.038 N/mm2
01 + 02 —
2
+45.019 N/mm2
61 2G2 - -15.019 N/mm2
T = 30 N/mm2
30 0.
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
2
(ai - az ) +,r2
= V(-15.019)2 + 302 = ±33.55 N/mm2
2
P1 —
al + az +33.55 = 45.019 + 33.55 = 78.519 N/mm 2
2
P2 =
01+ 02 + 33.55 = 45.019 - 33.55 = 11.519 N/mm 2
2
Q.2.12 A1 m x 1m mild steel shell of 1 mm thickness is stretched in its own plane by stresses, ; = 20 MPa,
ay = 30 MPa as shown in figure 2.12. Point 0 is centre of the plane. OC and OD are two mutually
perpendicular lines inclined at 45° to x and y-direction respectively before application of stresses.
Determine the change in angle (in degrees) between OC and OD after application of stresses.
Take modulus of rigidity of plate material = 80 GPa.
ax = 20 MPa
0
30 MPa
ay = 30 MPa
Fig. 2.12
[CSE-Mains, 2011 , ME : 20 Marks]
Solution:
t, shear stress on plane OC - 30 - 20 sin 90° = +5 MPa
2
Shear stress on plane OD = -5 MPa
Change in angle of 90° between OD and OC
5
(I)
xy - =
G
5
rad=0.0036°
80000
30 kN 1r 500 mm
30 kN
;150
Q.2.13 A steel bracket is bolted a column by three similar
bolts as shown in the figure 2.13. It is subjected to
an eccentric load of 30 kN as indicated. Determine
the suitable minimum diameter for the bolts using
a factor of safety of 5.0. Assume the bracket and
the attached plate to be rigid. Take the allowable
stresses for the bolt material in tension as
600 MN/m2 and in shear as 350 MN/m2.
[CSE-Mains, 1996, ME : 30 Marks]
300 mm
1
100 mm
Fig. 2.13
Principal Stresses
Strength of Materials
4
31
Solution:
Load = 30 kN
Load is eccentric about two axes, x and y
Direct shear stress on each bolt —
10000
A
Nimm2 (.0
Shear stress due to 30,000 x 150 Nmm moment will be maximum at bolt E, but ab due to moment
30,000 x 500 Nmm will be maximum at both D.
Shear strain in bolts
r1 = 180 mm
G of the system bar as shown.
r2 = 200 mm
A = area of cross-section of bolt
Shear stresses in bolts
N/mm2
i
d = direct shear stress — 10000
A
'Cs oc r
G of system lies at G', 100 mm for upper bolts
ri = 180 mm, r2 = 200 mm
Ts = kr
Force,
fs = krA
Torque,
T = EWA
k[2r12 A + r22A] = kA[2 x 1002 + 2002]
kA x 10.40 [ 10.48] = 30,000 x 150
Ak —
k—
To = kri —
3 x15 x105
104 x 10.48
42.939
A
42.939 x 180 7729
A
=
A
150
10000
TA
,. —
A
100
tanO = 0.667
0 = 33.7°
i
s cos 33.7 = 6430
i
s sin 33.7 = 4288
Td + Ts cos 33.7 —
Tr —
T r' resultant shear stress —
16430
A
4288 12 11 6 430)2
10A00
270 16980
=
+
—
18.386 +
A
A
9
16980
A
Nimm2
Bending stress
KA [211 2 + 122] = Re
KA [2 x 4002 + 1002] = P.e = 30000 x 500 = 15 x 106 Nmm
KA [320,000 + 100,00] = 15 x 106
3 2 P.
MADE EASY
IAS & IFS (Objective & Conventional) Previous Solved Questions
k'A —
15 x 106 1500
.
33 x 104
33
k'A = 45.45,
45.45
a bi = 0,, —
A
A
x 400 =
18180
A
9090
abi
2
A
For upper bolt (max. principal stress) =
(Gbi )2
9090
A
—
2
+ Tr
2
= 9090 +
Pmax
45.45
K—
+
(9090)2 (16980 )2
A )
A )
1000
A
V82.628 + 288.32 =
9090 + 19260
A
28350
= 120 (since FOS = 5)
A
A= 236.25 mm2 = Ed 2
4
d = 17.344 mm -=-' 18 mm
Bolt dia.,
For upper bolt (max. shear strain),FOS = 5
19260
'CMaX —
A
= 70
A = 275.143 = Ed2
4
d = 18.72 mm
Bolt dia.,
Objective Questions
Q.1
The complementary shear stresses of intensity
T are induced at a point in the material, as shown
in the figure 2.14. Which of the following is the
correct set of orientation of principal planes with
respect to AB.
T
D
C
T
e
I
TI
(a) 30° and 120°
(c) 60° and 150°
Q.2
A material element is subjected to a plane state
of stress such that the maximum shear stress
is equal to the maximum tensile stress would
compound to
T
D
C
(a) 1 0
1
T
A
T<
Flg. 2.14
B
(b) 45° and 135°
(d) 75° and 105°
[CSE-Prelims, ME : 1998]
A
T<---
B
Principal Stresses
Strength of Materials
41
33
(d)
(b)
[CSE-Prelims, ME : 1999]
Q.4
Biaxial stress system is correctly shown in
30
20
a
(c)
(a) 20...
►10
1
20
30
10
(d)
30
6
20
(b) 10
- 10
1
20
[CSE-Prelims, ME : 1998]
Q.3
30
y
For the given stress condition ax = 2 N/mm2,
ay = 0, Txy =0, the correct Mohr's circle is
10
40
(a)
30
(C) 3
40
40
(b)
20
(d) 30
30
1
(c)
20 t
40
[CSE-Prelims, ME : 1999]
34 0.
Q.5
Principal stresses at a point in a stressed solid
are 400 MPa and 300 MPa respectively. The
normal stress on plane inclined at ±45° to the
principal planes will be
(a) 200 MPa and 500 MPa
(b) 350 MPa on both planes
(c) 100 MPa and 600 MPa
(d) 150 MPa and 550 MPa
[CSE-Prelims, ME : 2000]
(b)
Q.6 A Mohr's stress circle is drawn for a body
subjected to tensile stresses f, and fy in two
mutually perpendicular directions such that
fx > fy. Which one of the following statements in
this regard is not correct?
(a) Normal stress on a plane at 45° to fx is equal
to
fx
tO
fy
2
(d)
C
0
2
(d) Maximum shear stress is equal to
(0 = origin and C = centre of circle, OA = al ,
OB= 02 )
fy
fx
2
[CSE-Prelims, ME : 2000]
The resultant stress on a certain plane makes
an angle of 20° with the normal to the plane. On
the plane perpendicular to the above plane. The
resultant stress makes an angle of 0 with the
normal. The value of 0 can be
(a) 0° to 20°
(b) any value other than 0° to 90°
(c) any value between 0° to 20°
(d) 20° only
[CSE-Prelims, ME : 2001]
The correct Mohr's stress circle drawn for a point
in a solid shaft compressed by a shrink fit hub
is as
(a)
B
A
fy
fx
(c) Maximum normal stress is fx
Q.8
0
(c) 0
(b) Shear stress an a plane at 45° to f is equal
Q.7
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
Q.9
When the two principal stresses are equal and
like, the resultant stress on any plane is
(a) equal to principal stress
(b) zero
(c) one half of the principal stress
(d) one third of the principal stress
[CSE-Prelims, ME : 2002]
Q.10 On a plane, resultant stress is inclined at an
angle of 45° to the plane. If the normal stress is
100 N/mm2, shear stress on plane will be
(a) 71.5 N/mm2
(b) 100 N/mm2
(c) 86.6 N/mm2
(d) 120.8 N/mm2
[CSE-Prelims, ME : 2003]
Q.11 The moduli of elasticity and rigidity of a material
are 200 GPa and 80 GPa respectively. What is
the value of Poisson's ratio of the material?
(a) 0.30
(b) 0.26
(c) 0.25
(d) 0.24
[CSE-Prelims, ME : 2007]
Q.12 State of stress at a point of a loaded component
is ax = 30 MPa, ay = 18 MPa,
8 MPa. If the
Txy =
Principal Stresses
Strength of Materials
4
35
larger principal stress at the point is 34 MPa,
what is the value of smaller principal stress?
(a) 12 MPa
(b) 14 MPa
(c) 16 MPa
(d) 18 MPa
[CSE-Prelims, ME : 2008]
Select the correct answer using the codes given
below:
(a) 1 and 3
(b) 2 and 3
(c) 1 and 4
(d) 2 and 4
[CSE-Prelims, CE : 2004]
Q.13 The maximum shear stress occurs on
(a) principal planes
(b) plane at 45° to the principal planes
(c) planes at 90° to the principal planes
(d) planes independent of the inclination to the
principal planes
[CSE-Prelims, CE : 2003]
Q.17 If Ex and Ey are the maximum and minimum
strains, respectively, in the neighbourhood of a
point in a stressed material, then what is the
expression for the maximum principal stress?
(a) EE,
(b) E(E, + µEy)
Q.14 The ratio of the maximum shear stress to the
difference of the two principal stresses is
(a) 1/2
(b) 1/3
(c) 1/4
(d) 1/6
[CSE-Prelims, CE : 2002]
[CSE-Prelims, CE : 2007]
(c)
E(E, + µEy )
1— µ 2
E(Ey +µE,)
(d)
1-112
Q.18 The principal stresses in N/mm2 on a square
element are shown in the figure below. What is
the intensity of tangential stress (pi) on the
plane BD?
100
Q.15 For the two dimensional stresses shown in the
figure, what is the normal stress on the 45°
plane?
oll
llll
100
100
ctfffttB
100
2
10 N/mm
Fig. 2.20
(a) 50 N/mm2
(c) 100 N/mm2
Fig. 2.19
(a) 20 N/mm2
(c) 4 N/mm2
(b) 12 N/mm2
(d) 8 N/mm2
[CSE-Prelims, CE : 2004]
Q.16 A rectangular block is subjected to shear stress
intensity of T. The stresses induced on a plane
inclined at 135° to the horizontal axis shall be
(b) 0
(d) 200 N/mm2
[CSE-Prelims, CE : 2006]
Q.19 A point in an element is stressed as shown in
the figure given below.
100 N/mm2
t
E
t
100 N/mm2
c„,
O.,
1. ax = 4-T (tensile)
2. crx = —T (compressive)
3.
4.
atangential = -FT
atangential =
A
1
1
100 N/mm2
Fig. 2.21
—.-100 N/mm2
36
►
IAS & IFS (Objective & Conventional) Previous Solved Questions
What is the value of normal stress on the oblique
plane BE?
(b) 175 N/mm2
(a) 200 N/mm2
(d) 100 N/mm2
(c) 150 N/mm2
[CSE-Prelims, CE : 2007]
400 N/mm2
400 N/mm2
Q.20 A specimen is subjected to pure shear, the shear
stress being q. Tensile and compressive
stresses of intensity a, occur on planes inclined
at 45° to the shear stress. What is the value of
the ratio a/q ?
(b) 1.5
(a) 2
(d) 1
(c) 1.25
[CSE-Prelims, CE : 2008]
Q.21 The square element in the figure is subjected
to a biaxial stress of 400 N/mm2 as shown. What
is the intensity of normal stress pn on the plane
BD?
MADE EASY
400 N/mm2
Fig. 2.22
(b) 400,
(a) 200
(c) 400/V-
1. (b)
6. (d)
11. (c)
16. (d)
21. (d)
2.
7.
12.
17.
(d)
(b)
(b)
(c)
(d) 0
[CSE-Prelims, CE : 2009]
Answers
3. (d)
8. (d)
13. (b)
18. (b)
4. (d) 5. (b)
9. (a) 10. (b)
14. (a) 15. (d)
19. (d) 20. (d)
Explanations
1.
3.
(b)
45° and 135°
(d)
p1 = +r, P2 =
Fig. 2.23
Cannot Mohr's stress circle.
5.
(d)
40
20
2
Fig. 2.22
2.
30
30
20
(d)
al -(-G1)
max
2
= a = maximum tensile stress
40
Fig. 2.24
Principal Stresses
Strength of Materials
Complementary shear stresses are balanced
Normal stresses is balanced in x and y-direction
5.
4
37
10. (b)
100 mpa
(b)
an
—
D1 + 10
,
,2
2
± P1 — P-2 cos 90°
45°
2
400 + 300
= 350 MPa
2
On both planes.
6.
(d)
fx
Amax
7.
fy
2
Fig. 2.26
=
(b)
100 MPa
11. (c)
E = 2G(1 +v)
200 = 2 x 80 (1 + v)
1.25 = 1 + v
v = 0.25
ar cos 20°
12. (b)
ax + ay = p1 + P2
30 + 18 = 34 + p2
p2 = 14 MPa
13. (b)
The maximum shear stress occurs on planes at
45° to the principal planes.
Fig. 2.25
14. (a)
4) can be other than 0 or 90°
where a = 0
8.
9.
P1 — P2
(d)
For a point in a solid shaft, compressed by a shrink
fit hub is
= ao = p' (junction pressure)
OA = OB = OC = p',
circle will be point
(a)
2
P2
ti max
— max
=2
ti max
— P2
=
0.5
15. (d)
_ Pi
20
= +45°, sin29 = 1
ti
0
P2 sin
2
a
Pi= P2 = P
2—
— P2
2
= 10 N/mm2, p2= 6 N/mm2
P1 + P2
a on any plane = p.
Pn —
2
+
P1 P2
2
cos 20
38 ►
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
cos 20 = cos 90° = 0
18. (b)
= p2 = —100 N/mm2
10+6
2
p_
= 8 N/mm
n
2
pt _ Pi — P2 cos 90° = 0
2
16. (d)
Mohr's stress circle is a point.
19. (d)
P1 = P2 = +100 N/mm2
+ P2
— 100
2
45°
— P2
"
—u
2
. 913°
a=
sin (-2 x 45°)
= 100
= —t (compressive)
0 = 450
atangential =
+ P2 P1 P2 cos 60°
2
2
a _
e
20. (d)
't COS 90° = 0
—=1
Shear
—t
Fig. 2.27
Fig. 2.29
17. (c)
21. (d)
p1 = +400 N/mm2
e _ Pi — 1-IP2
p2 = —400 N/mm2
E
x
v.
Eex =
P2 — 11,131
Pi + P2
E
2
— 1-1P2
ay = P2 — PP1
•• .(i)
—
P1 — P2
— 400 N/mm2
2
0 _ 450
EVEy = µP2—µp1
(1 —1,1,2) P1 = E(ex + 1.tev)
o
p —
n
p1 + P2
+Pi — P2
cos90°
2
2
= 0+400 x 0=0
P1
E
— (1— µ2 ) [ex +"Y]
03
Thin and Thick Shells
CHAPTER
0.3.1 Derive a formula for increase in volume of a thin metallic sphere when it is subjected to an
internal pressure p. A thin spherical shell of copper has a diameter of 400 mm and a wall thickness
of 2 mm and just full of water at atmospheric pressure. Calculate the volume of water pumped in
to raise the inside pressure to 1.5 N/mm2. The modulus of elasticity of copper is 1 x 105 N/mm2.
K (Bulk modulus) is 2.5 x 103 N/mm2 and Poisson's ratio v is 0.25)
[CSE-Mains, 2000, ME : 20 Marks]
Solution:
(a) When a thin spherical shell is subjected to internal pressure p, circumferential stress, sac is developed in
wall of the shell as shown in figure 3.1.
pD
a=
4t
e
where p = pressure, D = diameter, t = wall thickness, Ec = circumferential strain
a, va,
pD
_ 4tE(1- v)
ec - E
E
Volumetric strain,
Ev = 3E, =
3pD
(1 v)
4tE
(Neglecting the effect of p)
V, volume of shell -
703
6
3pD
703
8V, change in volume = evV = — (1 v) x
4tE
6
(b)
D = 400 mm
t = 2 mm
p = 1.5 N/mm2
a_
EV
pD 1.5 x 400
=
= 75 N/mm 2
4x2
4t
pD
3 x 75
(1 0.25)
=
(1-v)
E
= 3 -4
Fig. 3.1
40
0.
IAS & IFS (Objective & Conventional) Previous Solved Questions
225 x 0.75
MADE EASY
= 1.6875 x 10-3
1x105
V = Volume of shell -
it133
TE
X 40 03 = 33.51x 106 mm3
6
6
Expansion in volume of shell = ev x V= 1.6875 x 10-3 x 33.51 x 106
= 56.548 x 103 mm3
Decreases in volume of water =
k
xV =
1.5 , (33.51 x 106 )
2.5 x 10'
= 20.106 x 103 mm3
Volume of water pumped in = 56.548 x 103 + 20.106 x 103 mm3
= 76.654 x 103 mm3
= 76.654 cc
Q.3.2 A thin cylinder 150 mm internal diameter and 2.5 mm thick has its ends closed by rigid plates and
is then filled with water under pressure. When an axial pull of 37 kN is applied to the ends, water
pressure is observed to fall by 0.1 N/mm2. Determine the value of Poisson's ratio. Assume
E. 140000 N/mm2, K for water = 2200 N/mm2.
[CSE-Mains, 2008, ME : 20 Marks]
Solution:
Op = 0.1 N/mm2
Reduction in volumetric strain of cylinder =
Reduction in volumetric strain of water =
x
4tE
(5 4v)
Op
0.1
K
2200
Axial force = 37000 N
37000
Axial stress - 37000
701-
x 150 x 2.5
a (1- 2v) =
Increase in volumetric strain due to a = T
31.406
E
= 31.406 N/mm2
x (1- 2v)
0.1x
or
or
0.1
31.406
150(5-4v)+
(1 2v)
4 x 2.5E
2200 =
E
1.5(5 - 4v) +
0.1x 140000
2200
31.406 (1 - 2v)
7.5 - 6v + 6.3636 = 31.406 - 62.812v
56.812v = 31.406 - 6.3636 - 7.5 = 17.5424
or
Poisson's ratio, v -
17.5424
56.812
= 0.308
Q.3.3 A tyre is shrunk on a wheel of 12 meter diameter. Assuming the wheel to be rigid, calculate the
internal diameter of the tyre if after shrinking, hoop stress in the tyre is 1200 kgf/cm2, a for the tyre
= 11.7 x 10-6/°C and E
=
2 x 106 kgf/cm2. Find the least temperature to which the tyre must be
heated above that of the wheel before it could be fitted.
[CSE-Mains, 1986, ME : 30 Marks]
Thin and Thick Shells
Strength of Materials
Solution:
Hoop stress,
csc = 1200 kgf/cm2,
E = 2 x 106 kgf/cm2,
Hoop strain,
E =
41
a = 11.7 x 10-6 /°C
D= 12m
1200 _ a,
= 600 x 10-6 = 0.6 x 10-3
E
2 x 106
= A Ta = AT x 11.7x 10-6 = 0.6 x 10-3
AT -
0.6 x 1000
= 51.28°C
11.7
Q.3.4 Distinguish between thin and thick cylinders. A thin cylindrical shell having hemispherical ends is
subjected to internal pressure p. The internal diameter is 'd and thickness of cylinder and
hemisphere are t1 and t2 respectively. Assuming Poisson's ratio, v = 0.3, prove that.
1.
2.
t2
7
For no distortion at junction, -6- = 17
For equal maximum hoop stress in cylinder and hemisphere, ratio it = 0.5 .
ti
[CSE-Mains, 2002, ME : 20 marks]
Solution:
D
In a thick shell or a thin shell, D is internal diameter and t is wall thickness, if — > 20 , then it is a thin
shell. When the shell is subjected to internal pressure, the hoop stress developed in the shell does not
vary much across the thickness. If — < 20 , i.e., thickness of shell is considerable in comparison to
f
diameter, then there is variation of hoop and radial stresses
across the thickness of thick shell.
Figure 3.2 shows a thin cylindrical shell with hemispherical ends,
subjected to internal pressure p.
In cylindrical portion:
pD
a hoop stress = 2t1
Fig. 3.2
pD
Axial stress, as = 4t
1
c
hoop strain =
pD
v pD
(2 v)
1 E ptDE
In hemispherical portion at junction
pD
a = hoop stress = 4t
2
E,
c / hoop
strain =
PD (1 v)
4t 2E
For no distortion
Sc' = EC"
4 2 0. lAS & IFS (Objective & Conventional) Previous Solved Questions
4t0 (2
or
v) =
4t2
(1— v)
E
t2
1—V
1— 0.3
t1
2—v
2 — .07
pD
Maximum stress in cylindrical portion
0.7 = 7
1.7 17
=
2t1
Maximum stress in hemispherical portion =
a
MADE EASY,
pD a
=
4t2
= ac
pD = pD
2t1
4t2
t2 =
0.5
tl
Q.3.5 A steel sleeve is pressed onto a solid steel shaft which has 5 cm diameter. The radial pressure
between shaft and sleeve is 1800 N/cm2 and hoop stress at the inner surface of sleeve is
4500 N/cm2. If an axial compressive load of 50 kN is applied to the shaft, determine change in
radial pressure at the interface of shaft and sleeve. Assume, v = 0.3.
[IFS 2011, ME : 15 Marks]
Solution:
Shaft diameter,
Radial stress,
Hoop stress,
Poisson's ratio,
Compressive load,
Axial compressive stress,
Say
d = 50 mm
a = 18 N/mm2 = p' at inner radius
a = 45 N/mm2
v = 0.3
Pc = 50,000 N
50000 4
x , 25.46 N/mm2
lc
50'
= additional radial stress developed in shaft due to axial
compressive stress of 25.46 N/mm2
CYa =
Additional stresses in shaft
p,, = radial compressive stress
ID" = hoop stress compressive
aa" = axial compressive stress
additional circumferential strain in shaft
cs
—p" vp" va,"
+
+
E E
—p" + 0.3p" + 0.3 x 25.46
—0.7p" + 7.638
Sleeve
Additional radial stress = p" at inner surface
45
additional hoop stress = — p" = 2.5pItensile)
18
Thin and Thick Shells
Strength of Materials
esi, strain (hoop direction) -
2.5p" + 0.3p"
E E
4
43
2.8p"
E
Strain compatibility
Es/ = Ecs
7.638 - 0.7p"
2.8p"
E - E
7.638
= 2.18 N/mm2
3.5
= Increase in radial pressure
p" =
Q.3.6 (a)
(b)
(c)
Solution:
(a)
(b)
How will you distinguish between a thin walled and a thick walled pressure vessel.
What advantages you obtain by wire winding a thin cylinder?
What largest internal pressure can be applied to a cylindrical tank 1.8 meter in diameter
and 14 mm wall thickness, if the ultimate tensile strength of steel is 467 MPa and factor of
safety of 7 is desired?
In a pressure vessel if D is diameter and t is wall thickness and internal pressure is p, then if D/t
ratio is greater than 20 then it is termed as thin walled pressure vessel. Hoop stress and axial
stresses developed in thin pressure vessel are assumed to be constant along radial thickness
of pressure vessel. If Dlt ratio is much less then 20, then it is termed a thick walled pressure
vessel. Hoop stress and radial stress developed in thick pressure vessel vary along its thickness.
When a thin pressure vessel is subjected to internal pressure, then axial and hoop stresses
developed in thin shell are
as = - and ac=
pD
2t
ac = aya
Material of the shell is not gainfully used in axial direction. By wire winding, a thin wire under
tensile stress crw is wound tightly along the outer circumference of shell introducing initial
compressive stress ac' in the hoop direction in shell.
When the thin cylinder with wire wound on it is subjected to internal pressure then hoop stress
developed in shell due to internal pressure is ac" resultant hoop stress will be ad, = 0:- ac'.
Pressure bearing capacity of the cylinder is increased by wire winding.
(c)
aut = 467 MPa
FOS = 7
Cr a I = allowable stress =
=
pD
2t
467
7
= 66.71 N/mm 2
, hoop stress
D= 1800 mm, t = 14 mm
P-
2taa 2x14 x66.71
.
D
1800
= 1.038 MPa, maximum internal pressure
44 1.
MADE EASY
IAS & IFS (Objective & Conventional) Previous Solved Questions
Q.3.7 A compound cylinder is formed by shrinking a tube 16 cm external diameter and 12 cm internal
diameter on to another which has an internal diameter of 8 cm. If after shrinking the radial
compressive stress at common surface is 300 kg/cm2, find the circumferential stress at the inner
and outer surfaces and at the common surface.
[CSE-Mains, 1987, ME : 30 Marks]
Solution:
Inner cylinder
p' = 300 kg/cm2
R2 = 8 cm
Ri = 4 cm (junction pressure)
R3 = 6 cm (junction)
2pRE
acRi
-Rf
= —2 x 300
acR3 =
6
1080 kgf/cm2
x36 -16=
28 cm
Compass cylinder
RE +R-f
P x R2
3 — R21
. -300 x
Fig. 3.3
36 +16
= 780 kg/cm2
36 -16
+ 1071.43 kgf/cm2
Outer cylinder
+ 771.43 kgf/cm2
RE +RE
a cR3
+13' X 2
2
R2 - R3
. +300 x
64 + 36
= +1071.43 kgf/cm2
64 - 36
—1080 kgf/cm2
2R
6 cR2 =
x De
,2
Dp2
112 - 13
Hoop Stress distribution
Fig. 3.4
= +300 x 2 x 36 = 771.43 kgf/cm2
64 - 36
Q.3.8 On the outer surface of a closed thick cylinder of diameter ratio 2.5, were fixed strain gauges to
measure the longitudinal and circumferential strains. At an internal pressure of 230 MN/m2 the
strains were recorded as 91.8 x 10-6, and 369 x 10-6 respectively. Determine the value of Young's
modulus and, modulus of rigidity and Poisson's ratio.
[CSE-Mains, 1993, ME : 20 Marks]
Solution:
R2
Ri
= 2.5
p = 230 N/mm2
Outer surface
Hoop stress,
ac
2/312
2 x 6.25
P X 2 - 2 = 230 x 6.25 -1 = +547.62 N/mm2
Thin and Thick Shells
Strength of Materials
Axial stress,
aa = +PX
4
45
2 = 273.81 N/mm2
2
R2 —
= 0, radial stress is zero at outer surface
E strain =
c axial strain —
547.62 v x 273.81 273.81
=
(2 v)
E
E
273.81 v x 547.62 273.81
(1 2v)
E
E
E
273.81
(2 v) = 369 x 10-6
E
273.81
(1 2v) = 91.8 x 10-6
E
or
2—v
369
= 4.0196
1— 2v
91.8
4.0196-4.0196 x 2v = 2 —v
2.0196 = 7.0392v
v = 0.287
E—
Modulus of rigidity,
273.81(2 — 0.287)
369
E
G= 2(1+v)
x 10
6
= 1.271 x 106 N/mm2
1.271 X 1 06
4.94 x 105 N/mm2
2 x1.287 =
Q.3.9 Find the ratio of thickness of internal diameter of a thick tube subjected to internal pressure when
the pressure is 5/8 of the value of maximum permissible circumferential stress.
Find the increase in internal diameter of such a tube of 100 mm internal diameter, when the
internal pressure is 100 MN/m2, E. 200 x 109 N/m2, Poisson's ratio = 0.286.
[CSE-Mains, 1992, ME : 20 Marks]
Solution:
Say, R1 = Internal radius, R2 = External radius, p= Internal pressure
R2
R 22 +
1
ac max = P — 2
132 —1312
But,
5
P = 8 6 cmax
R 2 +R12
5
—XpX 22
— p
R2 —131 2
or
5F2 + 513 = 8/3 — 8R1
3F2 = 13131
R2 = 2.08 R1
R1 + t = 2.08 R1
t = 1.08 R1 =
Di
= 0.54D1
46 0.
MADE EASY
IAS & IFS (Objective & Conventional) Previous Solved Questions
= 0.54
= 50 mm
Now,
p = 100 N/mm2
R2 = 104 mm
ac max = 50 x
13316
1042 + 502 _ 50 x 10816+2500
= 50x
10816 - 2500
8316
1042 - 50'
= 1.6 x 50 = 80 N/mm2
Axial stress,
as = p x
R2
502
- 50 x
2
R2
1042 - 502
"2 "1
= 50 x
502
= 15.03 N/mm2
8.316
Principal stresses at inner radius
ac max = +80, a,. = 15.03 N/mm2
p = -50 N/mm2
Strain,
80
Ec
0.286 x15.03
0.286 x 50
E
80 +10
E
90
= 4.5 x10-4
200,000
Change in internal diameter
SDI = 4.5 x 10-4 x 100 = 0.045 mm
Q.3.10 A compound cylinder is made by shrinking a jacket with outer diameter of 20 cm on a hollow
cylinder with diameters 10 and 15 cm. When the compound cylinder is subjected to an internal
pressure of 350 kgf/cm2, the maximum circumferential stress in both the cylinder is the same.
Calculate the maximum stress developed and internal diameter of the jacket. Take the value of
E = 2 x 106 kg/cm2.
[CSE-Mains, 1988, ME : 30 Marks]
Solution:
Inner radius,
Junction radius,
Outer radius,
Internal pressure,
Inner cylinder,
R1 = 5 cm
R3 = 7.5 cm
R2 = 10 CM
p = 350 kgf/cm2
+
acRi = 350 x R2
1=6 -13-F
100+25
100 - 25
= 583.33 kgf/cm2
= 350x
Outer cylinder,
acR3 = Px
Fig. 3.5
FifF? +RfFIE .=
52 X 102 ± 52 X 7.52
350 x
02 (R 2 R2)
7.5 2 (10 2 -5 2 )
/13 2 - 1 /
Thin and Thick Shells
Strength of Materials
4
47
2500 +1406.25
3906.25
= 350 x
56.25(100 - 25)
75 x 56.25
= 324.074 kgf/cm2
= 350 x
Junction pressure p',
aCRi (inner cylinder) = -p' x
2R2
, 2 x7.5 2
3 = p x
56.25 - 25
RE -Fq
acR, = -p' x 3.6
+Fq
, 102 + 7.52
-+P x 2
acR3 = +13' X R22-R32
10 - 7.5 2
6CR3 = p' X
156.25
= +p' x 3.5714
43.75
Finally
583.33 - 3.6p' = 324.074 + 3.5714p'
_ 259.256
amax
= 36.15 kgf/cm2
7.1714
= 583.33 - 3.6 x 36.15
= 583.33 - 130.145 = 453.185 kgf/cm2
Shrinkage allowance
D,
-=[3.6p'+ 3.5714p1
8D3
=
E
7.1714x 36.15
7.5x
2 x 10'
= 1.944 x 10-3 cm
D3, inner diameter of jacket = 15 -1.944 x 10-3 = 14.998 cm
0.3.11 A compound cylinder is made by shrinking an outer tube of outside diameter 200 mm and inside
diameter 150 mm onto an inner tube, internal diameter 100 mm with a radial interference of
0.2 mm.
Both the tubes are made of steel with elastic modulus E = 2 x 106 kg/cm2 and Poisson's ratio
v = 0.3.
Calculate the value of pressure at the interface and values of hoop stress in the two tubes at the
interface.
Work from the first principle assuming the basic Lame's equation.
ar = A-
r2
ae A+ 2
r
[CSE-Mains, 1999, ME 30 Marks]
Solution:
= 5 cm, R2 = 10 CM, R3 = 7.5 cm, 8R3 = 0.02 cm
8R3 -
p'R3 [R5 + R3 R3 +Rf]
E
+
R2 -RE 14 -13?
48 ►
IAS & IFS (Objective & Conventional) Previous Solved Questions
MADE EASY
102 +7.52 7.52 +52
0.02 - P' x 7.5
+
52
2 x 106 102 - 7.52 7.52
5333.3 _
,[156.25 81.25]
+
P
43.75 31.25
p'[3.571+ 2.6]= 6.171 p'
p' = 864.2 kg/cm2
= 8.64 kg/mm2, Pressure at interface
Fig. 3.6
Hoop stress at interface
Inner tube
Outer tube
R3 +Ri2
a' CR3 =
,
'IcR3
3
7 52 4. 52
= 864.2 x ' 2 '
7.5 - 5 2
864 2 x
+P X Fq+R
=
2246.9 kg/cm2
10 2 +7.52
„
10` -7.5'
= 864.2 x 3.571 = 3086.06 kg/cm2
Objective Questions
Q.1
Match List-I (Terms used in thin cylinder stress
analysis) with List-II (Mathematical expression)
and select the correct answer.
A.
B.
C.
D.
Q.2
List-I
Hoop stress, a
Maximum in plane shear stress
Longitudinal stress
Cylinder thickness
List-II
(a)
pd 2
pd
2t
3.
pd
2o
4.
(d)
pd2
2tE
(1+µ)
pd2
(2+ it)
4tE
Q.3 The percentage change in volume of a thin
cylinder under internal pressure having hoop
stress = 200 MPa, E. 200 GPa and Poisson's
ratio = 0.25
(a) 0.40
(b) 0.30
(c) 0.25
(d) 0.20
[CSE-Prelims, ME : 2002]
pd
8t
Codes:
A
(a) 2
(b) 2
(c) 2
(d) 2
(b)
[CSE-Prelims, ME :1998]
4t
2.
pd2
(2 µ)
4tE
(c) 7E-(2+11)
pd
1.
A thin cylinder of diameter d, thickness t is
subjected to an internal pressure' p'. Change in
diameter is (where E is the modulus of elasticity
andµ is the Poisson's ratio.)
BCD
3
1
4
4
3
1
4
1
3
4
1
3
[CSE-Prelims, ME : 1998]
Q.4
A thin cylinder shell of mean diameter 750 mm
and wall thickness 10 mm has its ends rigidly
closed by flat steel plates. The shell is subjected
to internal fluid pressure of 10 N/mm2 and an
external pressure p1 . If the longitudinal stress
in the shell is to be zero, what should be the
approximate value of p1?
Thin and Thick Shells
Strength of Materials
(a) 8 N/mm2
(c) 10 N/mm2
Q.5
Q.6
(b) 9 N/mm2
(d) 12 N/rnm2
[CSE-Prelims, ME : 2007]
A thin walled water pipe carries water under a
pressure of 2 N/mm2 and discharges water into
a tank. Diameter of the pipe is 25 mm and
thickness is 2.5 mm. What is the longitudinal
stress induced in the pipe?
(a) 0
(b) 2 N/mm2
(c) 5 N/mm2
(d) 10 N/mm2
[CSE-Prelims, ME : 2007]
What is the value of volumetric strain in a thin
walled cylindrical vessel with modulus of
elasticity E, Poisson's ratio µ and of diameter
'd, wall thickness 't' and subjected to internal
fluid pressure p?
(a)
pd
(5 - 4µ)
4tE
(c)
pd
(3 20
2tE
(b)
pd
3tE
pd
- 3µ)
(d) — (2 tE
)
[CSE-Prelims, ME : 2006]
Q.7
Q.8
A thin cylindrical shell 500 mm in diameter and
25 mm thick having E. 2.0 x 105 N/mm2 and
= 0.25, is subjected to internal fluid pressure
of 20 N/mm2. What is unit change in volume of
the shell?
(a) 4 x 10-3
(b) 2.5 x 10-3
(c) 2 x 10-3
(d) 1.5 x 10-3
[CSE-Prelims, ME : 2008]
Which one of the following is the correct
expression for hoop stress, if p is internal
pressure in a thin walled cylinder of diameter
'd and thickness t?
(a)
pd
(c) pd
—
4t
Q.9
(b)
49
2. Circumferential stress is twice the
magnitude of longitudinal stress.
3. Hoop stress and longitudinal stress remain
constant along the vessel thickness.
Which of these statements are correct?
(a) 1 and 2 only
(b) 2 and 3 only
(c) 1, 2 and 3
(d) 2, 3 and 4
[CSE-Prelims, ME : 2009]
Q.10 In a hollow thick cylinder the radial stress or
under an internal pressure 'p'
(a) increase from a minimum at the innermost
surface to a maximum value at the
outermost surface
(b) decreases from a maximum at the
innermost surface to a minimum value at
the outermost surface
(c) increase from zero at the innermost surface
to a value or = p at the outermost surface
(d) decreases from a value ar. = p at the
innermost surface to zero at the outermost
surface
[CSE-Prelims, ME : 2001]
Q.11 A seamless pipe with 80 cm diameter carries a
fluid under a pressure of 2 N/mm2. If the
permissible tensile stress is 100 N/mm2, the
minimum required thickness of the pipe is
(a) 2 mm
(b) 4 mm
(c) 8 mm
(d) 10 mm
[CSE-Prelims, ME : 2003]
Q.12 Note: 'id is the internal fluid pressure
A thin cylindrical shell is subjected to the loads
as shown in figure. The element marked 'A' will
be subjected to
pd
2t
pd
8t
[CSE-Prelims, ME : 2009]
(d)
Consider the following statements in respect of
thin cylindrical pressure vessel subjected to
internal fluid pressure.
1. State of stress along the vessel thickness
is biaxial except on the inside surface.
4
Fig. 3.7
(a) biaxial compressive stresses
(b) biaxial tensile stresses
50
0. IAS & IFS (Objective & Conventional) Previous Solved Questions
(c) uniaxial compressive stresses
(d) tensile stress along the longitudinal
direction only
[CSE-Prelims, ME : 2004]
Q.13 The maximum hoop stress in a thick cylinder
under internal pressure would occur at which of
the following locations?
(a) At the inner surface
(b) At the outer surface
(c) At mid thickness
(d) Between the inner surface and mid
thickness
[CSE-Prelims, ME : 2008]
Q.14 A thin cylindrical pressure pipe with both ends
closed has diameter 1000 mm. The pipe is
subjected to an internal pressure of 4 N/mm2.
The permissible tensile stress in the material is
100 N/mm2. What is the minimum required
thickness of the pipe?
(a) 5 mm
(b) 10 mm
(c) 40 mm
(d) 20 mm
[CSE-Prelims, ME : 2008]
MADE EASY
pD
(b)
tE-(1- ")
pD2
(d)
(c)
4tE
(1 21.0
[CSE-Prelims, ME : 2010]
Q.16 A thin cylindrical shell is subjected to internal
pressure, such that hoop strain is approximately
five times the axial strain, what is the material
of the shell (depending on Poisson's ratio)
(b) Cast iron
(a) Steel
(c) Aluminium
(d) Wrought iron
Q.17 A thin cylindrical shell made of steel is subjected
to internal pressure such that hoop stress
developed in shell is 120 MPa. This shell is
subjected to axial compressive stress such that
there is no change in the length of the shell.
(There are end plates on the shell). If E= 200
GPa, Poisson's ratio is 0.3. What is axial
compressive stress applied on cylinder
(a) 90 MPa
(b) 60 MPa
(c) 36 MPa
(d) 24 MPa
Answers
Q.15 What is the change in diameter D of a thin
spherical shell of wall thickness t when subjected
to an internal fluid pressure p? (E = Young's
modulus of la = Poisson's ratio)
2. (a)
3. (d)
4. (c)
5. (c)
7. (c)
8. (b)
9. (c)
10. (d)
12. (b)
13. (a)
14. (d)
15. (c)
17. (d)
Explanations
1.
(d)
od =
pD
al, Hoop stress -- 2
2t
3.
Maximum shear stress, 13) - 4
8t
Longitudinal Stress, ±
15 1 - 1
4t
Cylinder thickness t , PD -3
2a
2.
c
xd =
d
(2 v)
4E
(d)
c = 200
as = 100
200 - 0.25 x 100
175
100 - 0.25 x 200
50
E°
Ea
=
(a)
pd - vpd _ pd
E= (2-v)
2tE 4tE 4tE
2ec + ea =
e=2x
400
E
0.2%
400
200 x 1000
Thin and Thick Shells
Strength of Materials
4.
(c)
10 x 750 pi x750
-0
4 x 10
4 x 10
p1 = 10
5.
14. (d)
(c)
p = 4 N/mm2
D = 1000 mm
= 100 =
(a)
pd
Volumetric strain = — (5-4g)
4tE
7.
t
(c)
E - PD (5 - 4v) , as v = 0.025
v 4tE
pD x 4 pD
4tE
tE
20 x 500
9.
pD
11)
4tE (
G
ED=Ec xD
pd
pD 2 (1-v)
4tE
(c)
1, 2, 3 are correct statements.
16. (c)
= Poisson's ratio
10. (d)
ar decreases from maximum value of ar = pat the
inner most surface to zero at the outermost surface.
2hoop strain
axial strain - 1- 2g
or
D = 800 m, p = 2 N/mm2
= 100 N/mm2
pD
µ= 1 = 0.333
is for aluminium i.e., 0.33
2t
pD
2 x 800
= 8 mm
=
t = 2a, 2 x 100
17. (d)
a = 120 MPa
as = 60 MPa
12. (b)
as
Via c
E
as
1' a,
Fig. 3.8
5
2-g= 5- 10 p.
9g=3
11. (c)
=
- 4 x 1000
= .20 mm
2 x 100
6
(b)
Hoop stress =
D
2t
15. (c)
Thin spherical shell
t = wall thickness
= Poisson's ratio
p = internal pressure
D = diameter
25 x 2 x105
= 200 x
= 2 x 10-3
8.
51
13. (a)
Maximum hoop stress in a thick cylinder under
internal pressure would occur at the inner surface.
pD 2 x 25
a_
=
= 5 N/mm2
a
4t 2.5 x 4
6.
4
as'
E
60 - 0.3 x 120 = cr;
as = 24 MPa
Axial compressive stress.
04
Shear Force and Bending
Moment Diagrams
CHAPTER
Q.1
A beam ABCD, 10 m long is supported at B, 1 m from end A and at C, x is from end D. The beam
carries a point load of 10 kN at end A and a UDL of 4 kN/m throughout its length fig. 4.1. Determine
the value of x if centre of the beam is the point of contraflexure. Draw the BM diagram.
10 kN
4 kN/m
Al
B
Rc
R8
-'---1m
(9—x)
x
-°1
Fig. 4.1
[CSE-Mains, 2004, ME : 30 Marks]
Solution:
Total load on beam = 10 + 40 = 50 kN
Reaction
Take moments about B
(9 - x)Rc + 10 x 1 = 40 x 4
Rc -
150
9 -x
RB = 50
150
450 - 50x -150
300 - 50x
9-x
9-x
9-x
Moments about centre of the beam
-10 x 5 + R8 x 4 - 4 x 5 x 2.5 = 0
-50+
4 (300 - 50x)
9 -x
50 = 0
4 (300 - 50x)
9 -x
--
100
300 - 50x = 225 - 25x
75 = 25x
x = 3m
RB
300 - 50x 150
= 25 kN
9 -x
6
Shear Force and Bending Moment Diagrams
Strength of Materials
11
53
BMD
MA = 0
MB = -10 x 1 - 4 x 0.5 = -12 kNm
M3 =-10 x 3 + 25 x 2-4 x 3 x 1.5
= -30 + 50 - 18 = +2 kNm
+ 2kN/m
M5 =-10 x 5 +25 x 4-5 x 4 x 2.5
= -50 + 100 - 50 = 0
—12 kN/m
Mc = M7 = -3 x 4 x 1.5 = -18 kNm
— 18 kN/m
BMD
MD= 0
Fig. 4.2
Q.4.2 On a simply supported beam (10 m span), a concentrated load (10 kN) and a moment of 40 kNm
act a section 7 m from one of the ends. Draw the shear force and bending moment diagrams.
Indicate the points of contraflexure, if any
10 kN
[ICSE-Mains, 2012, ME : 10 Marks]
Solution:
Fig. shows a beam AB, 10 m long, simply supported at
ends carrying a load of 10 kN at Cand moment 40 kNm at
RA = —1 kN
RB = +11 kN
C, 7 m from A
Reaction
3 —0Taking moment about A
70 + 40 = 10 RB
Reactions,
RB = 11 kN
— 11 kN
(b)
Reaction,
Ftc = 10 - 11 = -1 kN
SF diagrams
1 kN
FAC
FCB = 1 - 10 = -11 kN
Fig. (b) shows SFD
BM diagram
A
MA
Mc = -1 x 7 = -7 kNm
Mc = -7 + 40 = +33 kNm
MB = 0, Fig. (C) shows BMD
— 7 kNm
BMD
Fig. 4.3
Q.4.3 A beam ABCD is loaded as shown in figure 4.4. The beam is of rectangular section 50 mm x
100 mm.
600 N
60 N/m
RA = 393.33 N t
Rc = 386.67 N
5000 Nmm
1-*-3 m
3m
5m
Fig. 4.4
(i)
(ii)
Sketch the SF and BM diagrams of the beam
Determine the maximum bending stress at section B of the beam
[CSE-Mains, 2011, ME : 25 Marks]
54 1.
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
Solution:
Reaction
Moment about A
60 x 6
x 4 + 600 x 11 = 5000 + 6/3c (Note that CG of triangular load lies at 4 m from A)
2
720 + 6600 = 5000 + 6/3c
2320 = 6/qc
= 386.67 N
Total load on beam = 600 + 180 = 780 N
RA = 780 - 388.67 = 393.33 N
Reaction,
SFD
FA = +393.33 N
F3 = 393.33
30 x 3
= 348.33 N
2
A
386.67 N
600N
393.33
213.33 N
(Just left to C) F6 = 393.33-180=213.33N
(Just right to C) F6 = 213.33 + 386.67 = 600.00 N
•
D
A
Fig. 4.5
BMD
MA = 0
(Just left to B) MB = 393.33 x 3
30 x 3
x 1= 1180 - 45 = 1135 Nm 1135 Nm
(Just right to B)M3'= 1135 - 5000 = -3865 Nm
x
M6 = 393.33x6-5000 6026 x2
= 2360 - 5000 - 360 = -3000 Nm
MD =
Jump moment at B, 5000 Nm
bxd2 50x1002
= 8,33 x104 mm3
6
6
M 5000 x 10 3
= 60 N/mm2
=
z
8.33 x 104
za
max
— 3000 Nm
— 3865 Nm
Fig. 4.6
Q.4.4 Draw the bending moment and shear force diagram of the beam as shown in fig. 4.7(a).
0= Constant.
[CSE-Mains, 2012, CE : 15 Marks]
Solution:
BM diagram (Let us consider this as a combination of to cantilevers AEB and CFD and a simply supported
beam BC)
Portion BC
Carries UDL = 4 kN/m
Total load = 4 x 4 = 16 kN
Reaction at end = 8 kN
x 42
Maximum BM = w/2 = 4 8 = 8 kNm
8
Shear Force and Bending Moment Diagrams
Strength of Materials
2 kN
A
1
4
55
2 kN
4 kN/m
Mt-VW\
1
C
F D
--1-1 1 I 2 .(-- 4 m—•-1 2 I 1 [4-(a)
8kN
E B
(b)
BMD
—26
—26 kNm
10 kN
— 10 kN
Fig. 4.7
Cantilever AB
MB =
ME = -16 kNm
MA = -8 x 3 - 2 = -26 kNm
BM diagram for CD cantilever will be similar
SF diagram,
BC,
FB = +8 kN, Fcenter = 0
FBE = -8 kN
Fa4 = -10 kN
SFD for portion CFD portion will be similar to that of portion BEA
Figure shows BMD and SFD for the beam AB, BC, CD.
Q.4.5 The bending moment diagram of a simply supported beam AB is shown in figure 4.8(a). Sketch
the loading, diagram and SF diagram of the beam.
[CSE-Mains, 2013, ME : 10 Marks]
Solution:
BM diagram is of a beam subjected to a building couple Mat C, at a distance of 3 m from A. Length of
beam is 5 m.
M = 12 + 8 = 20 kNm (anticlockwise)
Moment at B,
RB x5 = 20
56
P. lAS
MADE EASY
& IFS (Objective & Conventional) Previous Solved Questions
RB = 4 kN1..
RA = 4 kNT (to balance RB)
Mc = 4 x 3 = 12 kNm
Mc = 12 - 20 = -8 kNm
SFD
Fig 4.8
SF diagram
FAB = +4 kN (constant)
Q.4.6 The shear force diagram of a beam is shown in the figure. Draw bending moment and load
diagrams.
2P
B
+
P/3
D
E
2P/3
2a
.1- a•-•-1
Fig. 4.9
[CSE-Mains, 2002, CE : 12 Marks]
Solution:
Beam is supported at B and D
Reaction
P
RB = 2P —
+ 7P
3 3
2P 8P
RD = 2P + — =
3
3
Shear Force and Bending Moment Diagrams
Strength of Materials
1
57
7P 8P
Total load on beam = — + — = 5P
3
3
Rate of loading between A and B = 2P sl,
a
P 2P ,
Vertical load at C = - + — = P 4,
3
3
Vertical load at E = 2P1
2P
1". a
.
A.,,rem.„, B
H'— a
BMD
RB
P
2P
1, C
11 1 E
a -044- 2 a
RD
(a)
A4A = 0
MB=
2P
a
-- x ax- = -P.a
2
a
— Pa
— 2Pa
3
— 2Pa
ME= °
MD = -2Pa
BMD
Fig. 4.10
8P
2Pa
Mc= -2P x 3a + — x 2a = 3
3
Verification of reaction
Moment about B
2P
a
ax3RD+- xax- = Pa + 8Pa
2
a
3RD = 8P
8
RD = iP
Q.4.7 A beam ABCD, 8 m long hinged at end A and roller
supported at end D carries transverse loads as
shown in the figure 4.11(a). Determine support
reactions and draw BM diagram for the beam.
Solution:
A load at end of the lever equal to 20 kN can be
replaced by a load of 20 kN and a CWmoment of
40 kNm at point C.
Reaction
Taking moments at A
2 x 20 + 20 x 4 + 40 = 8 RD
RD = 20 kN
RA = 20 + 20 - 20 = 20 kN
BMD
MA =O
MB = +20 x 2 = +40 kN
Mc = 20 x 4 - 20 x 2
= 40 kNm
Mc = 40 + 40 = 80 kNm
20 kN
2m
(a)
20 kN
20 kN
Fig. 4.11
58
0.
IAS & IFS (Objective & Conventional) Previous Solved Questions
MADE EASY
Q.4.8 Construct the SF and BM diagrams for the beam shown in the figure 4.12.
100 kNm
20 kNm
A
50 kN
D
-id 0.5 1-4-- 3 m
40 kN
E
F G
1-1.4 1.5 m--1.-1 0.5 1 0.5
RB =45kN
RE =
55 kN
Fig. 4.12
[CSE-Mains, 2006, ME : 20 Marks]
Solution:
Taking moments about B
20 x0.5 x 0.25+100+4.5 xRE = 50 x 3 +40 x5
2.5+ 100+4.5 RE = 150+200
0,
350 -102.5
= 55 kN
RE 4.5
RB = 20 x 0.5 + 50 + 40 - 55 = 45 kN
SF diagram
AB Portion
= x x20
= 0 at 0
= -10 kN, at x = 0.5
BCD Portion
DE Portion
EF Portion
Fx = -10 + 45 = +35 kN (remains constant)
= -10 + 45 - 50 = -15 kN (remains constant are DE)
= -10 + 45 - 50 + 55 = +40 kNm
+ 40 kN
Fig. 4.13
+ 32.5 kNm
BMD
MA
MB = -20 x 0.5 x 0.25 = -2.5 kNm
Mc = -20 x 0.5 x 1.25 + 45 x 1 = +32.5 kNm
Mc = 32.5 - 100 = - 67.5 kNm
MF =
ME = -4 x 0.5 = -20 kNm
MD =-40x 2+55 x 1.5
= -80 + 82.5 = +2.5 kNm
- 67.5 kNm
BMD
Fig. 4.14
Shear Force and Bending Moment Diagrams
Strength of Materials
Q.4.9 A beam ABCD, 6 m long hinged at both the ends A and
D, is subjected to moment 6 kNm (CW) at B and 3 kNm
(CCW) at C as: shown in figure 4.15(a). Sketch the
bending moment diagram of the beam. How many point
of contraflexure are then in the beam)
[CSE-Mains, 2009, ME : 5 Marks]
Solution:
Reaction: Moment about A,
6 kNm
4
59
3 kNm
AD
0.5 kN t
ho— 2m
2m
RA = 0.5 kN
RD x 6 + 3 = 6 kNm
2m —0-1
(a)
+5 kNm
RD = 0.5 kN T
+4 kNm
RA = 0.5 kN
BMD
MA
=
+ 1 kNm
MB = -0.5 x 2 = -1 kNm
MB = -1 + 6 = 5 kNm
Mc = -0.5 x 4 + 6 = +4 kNm
— 1 kNm
Mc = 4 - 3 = +1 kNm
Mo =
There is only one point of contraflexure in the beam.
BMD
(b)
Fig. 4.15
Q.4.10 For the semicircular simply supported member shown
in the figure 4.16, find the value of bending moment at
a function of 0 and also draw its bending moment
diagram.
[CSE-Mains, 2013, CE : 15 Marks]
Solution:
Consider a section xx at angular
displacement 8 as shown
Taking moment about A, for reaction
Px 1.5R = 2R x RB
RB = 0.75P and so RA = 0.25P
Bending moment, Me =
(R - Rcos0)
= 0 at x = 0
= 0.0335 PR, at 0 = 30°
= 0.07325 PR, at 8 = 45°
= 0.125 PR, at 0= 60°
= 0.25 PR, at 0 = 90°
Portion BC (0 = 0 - 60°)
Fig. 4.16
0.375 PR
3P
Me = — (R - R cos0)= 0.75PR (1- cos())
At 90°
= 0 at 0 = 0
= 0.1005 PR, at 0= 30°
= 0.21975 PR, at 0 = 45°
= 0.375 PR (at point C), at 0 = 60°
M90 = 0.75 P(R)- Px 0.5 R
= 0.25 PR
1.5 R
0.5 R-0-1
BMD
Fig. 4.17
60 0.
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
Objective Questions
Q.1
Which one of the following figure represents the
correct shear force diagram for the loaded beam
shown in the figure 4.18.
a
I.- 3a
a
Q.3
The beam is loaded as shown in the fig. 4.19.
Select the correct BM diagram
1.4
Fig. 4.19
Fig. 4.18
(a)
(a)
(b)
(b)
(c)
(d)
(c)
[CSE-Prelims, ME : 1998]
Q.2
A cantilever carrying an uniformly distributed
load throughout its length. Select the correct
bending moment diagram of the cantilever
(d)
[CSE-Prelims, ME : 1999]
(a)
Q.4
If the beam shown in the fig. 4.20 is to have
zero bending moment at its middle point, the
overhang x should be
r
(b)
w
(c)
Fig. 4.20
(a)
wL2
4P
(c)
wL2
8P
(d)
[CSE-Prelims, ME : 1999]
(b)
wL2
6P
wL2
12P
[CSE-Prelims, ME : 2000]
(d)
Shear Force and Bending Moment Diagrams
Strength of Materials
Q.5
Match List-I with List-II and select the correct
answer.
List-I
A.
f
Q
If the SF diagram for a beam is a triangle with
length of the beam as its base, the beam is
(a) A cantilever with a concentrated load at its
free end
(b) A cantilever with UDL over its whole span
(c) Simply supported with a concentrated load
at its mid point
(d) Simply supported with a UDL over its whole
span
[CSE-Prelims, ME : 2000]
Q.7
The bending moment diagram for a simply
supported beam is a rectangle over a larger
portion of the span except near the supports.
What type of load does the beam carry?
(a) A uniformly distributed symmetrical load
over a larger portion of the span except near
the supports
(b) A concentrated load at mid span
(c) Two identical concentrated loads equidistant
from the supports and close to midpoint of
the beam
(d) Two identical concentrated loads equidistant
from the mid span and close to supports
[CSE-Prelims, ME : 2007]
Q.8
Which one of the following is the correct
statement?
B.
P
Q
Q
t
D.
F
F
F
F
List-II
P
T
1.
dM
If for a beam, —
dx
R
4.
p
S
T
T
R
5.
P
Codes:
A
(a) 4
(b) 1
(c) 1
(d) 4
Q
BCD
2
5
3
4
5
3
4
3
5
2
3
5
[CSE-Prelims, ME : 2001]
0 for its whole length, the
beam is a cantilever
(a) free from any load
(b) subjected to a concentrated load at its free
end
(c) subjected to an end moment
(d) subjected to a UDL over its whole span
[CSE-Prelims, ME : 2007]
Q
S
61
Q.6
tT
SI
41
Q.9
Which one of the following is the correct answer?
The point of contra flexure in a beam is a point
on its length where
(a) the shear force is zero
(b) the bending moment is maximum
(c) the bending moment changes its algebraic
sign and is zero at that point
(d) the shear force changes its algebraic sign
[CSE-Prelims, ME : 2006]
62
I,. lAS & IFS (Objective & Conventional) Previous Solved Questions
MADE EASY
Q.10 A overhanging beam AB has simple supports
at D and E as shown in the figure below. It
carries concentrated loads P at its free ends
and a uniformly distributed load of intensity
w (N/m) over its length DE. If bending moment
at the middle point C of the beam is to be zero,
then what is the value of P?
P (N)
P (N)
Fig. 4.22
(b) 4(2W+ w)
(a) 8(W + 2
J
(c) 212W + —
w
2 11
[CSE-Prelims, ME : 2008]
Fig. 4.21
(b)
(d) 4(W + ;)
w/
Q.14 A cantilever beam is subjected to moments as
shown in the given figure. The BM diagram for
the beam will be
w/
(d) 16
30 kNm
50 kNm
[CSE-Prelims, ME : 2006]
Q.11 For a cantilever
dM
= a , constant for its whole
dx
length. What is the shape of the SF diagram for
the beam
(a) rectangle
(b) triangle
(c) a parabola
(d) a hyperbola
[CSE-Prelims, ME : 2008]
Q.12 SF diagram for a simply supported beam is a
rectangle with its longer side equal to beam
length. What type of load is acting on the beam?
(a) Concentrated load at its mid span
(b) UDL over its whole span
(c) Concentrated load alongwith a couple at a
point on beam length
(d) Couple at a point on the beam length
[CSE-Prelims, ME : 2008]
Q.13 A beam AB of length 1 and weight wsupports a
load Wat its free end Band is hinged at its end
A. A wire CD supports the beam at the mid
point C making an angle of 30° with the
horizontal. What is the value of tension T in the
wire CD?
2m
2m
Fig. 4.23
A
(a)
20 kNm
50 kNm
50 kNm
(b) 80 kNm
A
(c)
(d)
20 kNm
50 kNm
50 kNm
80 kNm
[CSE-Prelims, CE : 2001]
Q.15 A cantilever is loaded as shown in the below
figure. The bending moment along the length is
p
Q.18 Consider the following statements in regard to
the shear force diagram for an overhanging
beam supported at A and C
a
P
Fig. 4.24
(a)
(b)
(c)
(d)
6 kN
2 kN
uniform
uniformly varying
zero
concentrated at the free end
A 1-4-- 2m
Q.17 A cantilever AB carries loading as shown in the
figure. The BM diagram for the beam is
4 kNm
1m
M
Fig. 4.25
6kN-m
(a)
4kN-rn
6kN-m
(-)
(b)
B
2 kN
2m
I.- .1
C
2 kN
m
3 kN
Q.16 Consider the following statements:
1. Point of contraflexure is the point where the
bending moment is maximum
2. Point of contraflexure is the point where the
bending moment changes sign
3. Point of contraflexure is the point where the
shear force is zero
Which of these statements is/are correct?
(a) 1, 2 and 3
(b) 2 and 3
(c) 2 only
(d) 1 only
[CSE-Prelims, CE : 2002]
F-
63
Shear Force and Bending Moment Diagrams
Strength of Materials
4kN-m
7 kN
Fig. 4.26
1. The beam is carrying a uniformly distributed
load of 2 kN/m throughout
2. The beam is carrying a uniformly distributed
load of 2 kN/m over the supported span
AC and concentrated load of 2 kN at the
free end B
3. The beam is carrying a uniformly distributed
load of 2 kN/m over the supported span
AC, and concentrated load of 5 kN at the
centre of supported span BC and also a
point load of 2 kN at the free end B.
4. The points of contraflexure occurs between
the supported region AC and nearer to
support C.
Which of the above statements is/are correct?
(a) 1, 2, 3 and 4
(b) 4 only
(c) 2 and 3
(d) 3 and 4
[CSE-Prelims, CE : 2004]
Q.19 A propped cantilever beam shown in the figure
given above is having internal hinge at its mid
span. Which one of the following is the shape
of bending moment diagram for the given
loading?
eWslY=NPWY-1e-
6kN-m
l
(c)
Fig. 4.27
4kN-m
(d) 4kN-m
6kN-m
4kN-m
[CSE-Prelims, CE : 2003]
Parabolic
(a)
A
B
64 ►
IAS & IFS (Objective & Conventional) Previous Solved Questions
Q.21 A cantilever beam of span 1 carries a uniformly
varying load of zero intensity at the free end
and w per metre length at the fixed end. What
does the integration of the ordinate of the load
diagram between the limits of free and fixed
ends of the beam give?
(a) Bending moment at the fixed end
(b) Shear force at the fixed end
(c) Bending moment at the free end
(d) Shear force at the free end
[CSE-Prelims, CE : 2006]
Straight line
(b) A
Parabolic
(c)
—
A
(d)
MADE EASY
A
[CSE-Prelims, CE : 2005]
Q.20 A simply supported beam AB is loaded as
shown in the figure given below. CDE is a rigid
member. A load 4 kN is applied at E. Which
one of the following is the SFD for the beam?
Q.22 In the cantilever beam shown below, what is
the percentage of bending moment at 1/2 with
respect to the maximum bending moment at
the fixed support?
w/unit length
C1.4— 1/2
4 kN
D
.1
E 0.5m
C
(a) 15%
(c) 25%
t
k0.5 m
2m
—141
2m
Fig. 4.28
2 kN
(a)
2 kN
(b)
1.5 kN
2.5 kN
(c)
2.5 kN
1.5 kN
1 kN
(d)
3 kN
[CSE-Prelims, CE : 2005]
Fig. 4.29
(b) 20%
(d) 30%
[CSE-Prelims, CE : 2006]
Q.23 Match List-I with List-II and select the correct
answer using the code given below the lists:
List-I
A. Cantilever beam
B. Overhanging beam
C. Fixed beam
D. Simply supported beam
List-II
1. At the supports, shear force exists, BM is
zero
2. Deflection is zero, but BM exists at the
supports
3. Shear force and BM exist at the supports
4. BM exists, deflection is zero at the supports
Codes:
A
BCD
(a) 4
1
2
3
(b) 2
3
4
1
(c) 4
3
2
1
(d) 2
1
4
3
[CSE-Prelims, CE : 2006]
Shear Force and Bending Moment Diagrams
Strength of Materials
Q.24 What is the maximum bending moment at mid
span of the beam given below?
r
W (Rate of loading)
Codes:
A
(a) 3
(b) 4
(c) 3
(d) 4
B
2
1
1
2
65
C
4
3
4
3
[CSE-Prelims, CE : 2007]
Q.26 The figure shows the shear force diagram for
an overhanging beam ACDB.
55 kN
Fig. 4.30
(a)
wi2
2
(b)
w/2
(c)
(d)
12
1/4/12
25 kN
3 m —0,40— 3 m
6
Wi t
Fig. 4.31
[CSE-Prelims, CE : 2007]
Q.25 Match List-I (Loaded Beam) with List-II
(Maximum bending moment) and select the
correct answer using the code given below the
lists:
List-I
r
w/m
A.
L
H
w/m
Consider the following statements with respect
to the above beam:
1. The beam has supports at A and D
2. The beam carries a concentrated load at C
of 25 kN
3. Bending moment at D is 15 kNm
4. The beam carries a uniformly distributed
load of 80 kN over the portion AC.
Which of the statements given above is/are are
correct?
(a) 1, 2 and 4
(b) 1 only
(c) 2, 3 and 4
(d) 1 and 4 only
[CSE-Prelims, CE : 2009]
Q.27 A simple supported beam AB of span 2 m is
loaded as shown in the figure. Which one of the
following pairs corresponds to SFD and BMD
for the beam?
B. f
L
1 kN-m
3 kN-m
G
C.
E)
2m
'Alm
Fig. 4.32
List-II
1.
wL2
8
2.
wL2
12
3.
wL2
2
4.
wL2
6
2 kN
(+)
2 kN
2 kN-m
(—)
3 kN-m
(a)
66 0.
MADE EASY
1AS & IFS (Objective & Conventional) Previous Solved Questions
1 kN
(-)
1 kN
1 kN-m
(-)
2 kN-m
3 kN-m
2 kN
(b)
(d)
2 kN-m
(-)
1 kN-m
[CSE-Prelims, CE : 2010]
(-)
1 kN
1 kN
Answers
1. (a)
(c)
3 kN-m
(-)
1 kN-rn
2. (c)
3. (d)
4. (c)
5. (d)
6. (b)
7. (d)
8. (c)
9. (c)
10. (c)
11. (a)
12. (d)
13. (d)
14. (a)
15. (a)
16. (c)
17. (b)
18. (d)
19. (d)
20. (a)
21. (b)
22. (c)
23. (c)
24. (c)
25. (c)
26. (d)
27. (c)
Explanations
1.
(a)
Reactions = Weach.
4.
(c)
C IW
wt
H.- a -1.
L/2-0-1-4- LJ2
Fig. 4.37
tw
3a
-I- a
wic -
SFD
-
Fig. 4.34
2.
(c)
x=
3.
2M
x
2
wL2
) (p 4.
2 )2 2x4
PL wL2 wL2
- Px +-+ 2
4 8
-Px +
-w
Fig. 4.35
-PI
wL2
8
0
wL2
8P
5.
(d)
ABCD
4 2 3 5
6.
(b)
(d)
BMD
Fig. 4.36
SFD
Fig. 4.38
Shear Force and Bending Moment Diagrams
Strength of Materials
7.
41
67
12. (d)
(d)
M
a
.1.
(L — 2a)
M
L
SFD
Fig. 4.39
8.
Fig. 4.43
(c)
13. (d)
dM
=0
dx
Fig. 4.40
dM
0 end moment M
,
dx
9.
(c)
At point of contraflexure, bending moment
changes its algebraic sign and is zero at that point.
14. (a)
10. (c)
30kNm
cvmeme_v_w,
1
50 kNm
L
L
2
Fig. 4.41
wL) L
wL2
- P x 1.5L +(P+— - 2 2 8
- 1.5PL+
L
+
wL2
-
wL2
2 4 8
PL =
A
0
20
50 kNm
30
= 0
wL2
8
WL
P=
2
Fig. 4.45
15. (a)
Uniform
2 kN
8
4 kNm
11. (a)
dM
dx
= a constant = SF is constant
Ii
M
.1.
1m
(-)
4 kNm
BMD
SFD
Fig. 4.42
6 kNm
Fig. 4.46
68
0.
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
16. (c)
2 only
2. Point of contraflexure is the point where
bending moment changes sign.
20. (a)
4 kN
DrIE
0.5 m
A
17. (b)
11114111
0.5 m
RA = 2 kNt
RB = 2 kN
2m
Fig. 4.50
(-)
6 kNm
R8=
4 kNm
10
=
2.5 kA
RA = 1.5 kN
BMD
z
Fig. 4.47
18. (d)
3 and 4 are correct statements
(+)
ko-- 1.5 m
SFD
5 kN
2 kN
Fig. 4.51
21. (b)
SF at fixed end.
A
6 kN
}-0— 2 m --144— 2 m —1±1- 1 m-0-1
-2 kNm
BMD
SFD
Fig. 4.48
19. (d)
Parabolic
Fig. 4.55
22. (c)
25%
I-0— 2 m
2m
Fig. 4.56
MA —
BMD
Fig. 4.49
M, -
w/
2
2
wit
8
Strength of Materials
23. (c)
Shear Force and Bending Moment Diagrams
4
69
25. (c)
List-I
A. Cantilever beam
B. Overhanging beam
C. Fixed beam
D. Simply supported beam
List-II
4. BM exists, deflection is zero at the supports
3. Shear force and BM exist at the supports
2. Deflection is zero, but BM exists at the
supports
1. At the supports, shear force exists, BM is
zero
A.
3.
B.
1.
C.
4.
w/ 2
2
w/
2
8
w/
2
6
26. (d)
1. Beam has supports at A and D.
4. Beam caries a UDL of 80 kN over portion AC.
27. (c)
3 kNm
1 kNm
24. (c)
RA =
1 kN /4
1 kNm
wit
= 12
(-)
,
4 RB
= 1 kN
1 kNm
SFD
Fig. 4.57
wi i wi i wit wg
c 4 X 2 4 x 6 8 24
2m
1 kNm
(-)
3 kNm
BMD
Fig. 4.58
NM MI
05
Theory of Simple Bending
CHAPTER
Q.5.1 State the restrictions or assumptions made in deriving the formula for theory of bending
M a
I
y
E
R
[CSE-Mains, 1998, ME : 15 Marks]
Solution:
Following assumptions are taken while deriving the flexure formula:
1. The material of the beam is homogeneous and isotropic.
2. Young's modulus of the material in tension = Young's modulus in compression.
3. Material is considered as consisting of layers.
4. Each layer is independent to extend or contract irrespective of the layers above or below it.
5. Beam is initially straight.
6. Elastic limit of the material i.e., ae is not exceeded.
7. Beam is symmetrical about plane of bending.
8. Transverse sections which are plane before bending remain plane after bending.
Q.5.2 A wire of diameter d is wound round a cylinder of diameter D. Determine the bending stress
produced in the cross-section of the wire. Hence or other wise find the minimum radius to which
a 1 cm diameter circular rod of high tensile steel can be bent without undergoing permanent
deformation. Take yield stress = 17000 kg/cm2 and E. 2 x 106 kg/cm2. What is the magnitude of
BM necessary for this?
[CSE-Mains,1991, ME : 20 Marks]
Solution:
Wire diameter = d
R=
Drum radius,
2
Bending stress, ab
_ E _ 2E
d/2 R D
b
20 b
2E
d
Gb
E-5
Bending stress in wire
71
Theory of Simple Bending
Strength of Materials
E= 2 x 106
ab = 17000 kg/cm2
d= 1 cm
Now,
=
Ed
ab
2x106 x1
=117.647 cm
17000
It X13
1°3- =17000 x
M = ab Z = b X
32
32
M = 1668.97 kg-cm
Q.5.3 Compare the moments of resistance for a given maximum bending stress of a beam of square
section placed (i) with two sides vertical and (ii) with a diagonal vertical. The bending in each
case is parallel to vertical plane.
[CSE-Mains, 2012, CE : 12 Marks]
Solution:
M = Zab
Section-I (with two sides vertical)
a3
-1
z1 =
6
Section-II (with a diagonal vertical)
a 4 2 a3
,___
Z2 12 X ,/2a 6J
Say
ab = Maximum bending stress
a3
M1 =
bX 6
a3
M2 = ab x 6,//2
a
—h
II
M2
=
=1.414
Fig. 5.1
Q.5.4 Compare the bending strengths of three beams of same material, same weight and same depth,
if one of them has solid rectangular area 6 x 20 cm, second beam is a hollow rectangular section
having a wall thickness of 2 cm. The third beam has I-section of equal flanges have web and
flange thickness equal to 2 cm.
[IFS 2011, ME : 10 Marks]
Solution:
For solid rectangular area 6 x 20 cm
6 x 202
= 400 cm3
6
Hollow rectangular section (assume width 8)
208 - (8- 4) x 16 = 6 x 20 = 120 cm2 (area same)
20B- 16B + 64 = 120
4B = 56 = 14 cm
Z1 =
4-
[14x203 10x163 ]
10
12
12
1
1
= — x — [112000 - 40960] = 592 cm3
10 12
[-4-- 6 mm
- 20 mm
Fig. 5.2
...(ii)
7 2 a.
MADE EASY
1AS & IFS (Objective & Conventional) Previous Solved Questions
I
B
2 cm
I
I
20 cm
16 cm
20 cm
2 cm
B
TT
Fig. 5.4
Fig. 5.3
I-section
28 + 2B + 16 x 2 = 120 cm2
4B= 88
= 22 cm
-
=
63]
203
=
[176000 - 81920] = 784
120
10 [2212
2012
Beam strength is proportional to section modulus
Zi : Z2 : Z3 = 400: 592: 784
= 1 : 1.48 : 1.96
Q.5.5 Compute second moment of area of the plane lamina shown
in the figure 5.5 about an axis parallel to the base and
passing through the centroid.
[CSE-Mains, 2012, ME : 20 Marks]
Solution:
Area,
Al = 22 x 16 = 352 cm2
A2 —
22 x 9
= 99 cm2
2
A3 =
TcR2 n x 82
=
= 100.531 cm2
2
2
[4-16 cm --+-- 9 cm
Fig. 5.5
4R 4 x 8
=
= 3.395 cm (from diameter)
3n 3 x7c
Location of centroid (from bottom)
7
for semi circle _
99 x22
+100.531(22+3.395)
3
352+99+100.531
16 x 22 x11+
111
3872 + 726 + 25529.85
551.531
7150.985
- 12.966 cm
551.531
Moment of inertia of semi circular section about its own centroidal axis.
irR 4
1x/x
=
itR2 (412
8 2 37r
Theory of Simple Bending
Strength of Materials
73
nR 4 nR 2 16R2 ER 2 8R 4
x
8
2
9112
8
911
= R4 [0.3927 - 0.2829]
= 0.1098 R4
= 0.1098 x 84 = 449.741 cm4
1182
2
- 100.531 cm2
22 2
16 x 223
x 223
2
+352(12.966 11) + 36 + 99112.966 - -3-/xx = 12
)
+449.741 + 25.1327 (22 + 3.395 - 12.966)2
= 14197.333 + 1360.535 + 2662 + 3141.338 + 449.741+ 15530.03
= 37340.98 cm4
= 3.7341 x 10 cm4
Q.5.6 The shear force diagram for rectangular cross-section beam
is shown in the figure 5.6. Width of the beam in 100 mm and
depth is 250 mm, determine the maximum bending stress in
the beam.
[CSE-Mains, 1995, ME : 30 Marks]
Solution:
SF diagram is only on one side. This is the case of a cantilever
3 m long AB, fixed at A and free at end B carrying UDL of 5 kN/m
over BC= 1 m, a point load of 5 kN at Cand a point load of 5 kN
at D. Loading diagram is shown in the figure as above.
Maximum bending moment occurs at fixed end A
MA = 5(2.5) + 5 x 2 + 5 x 1
= 12.5 + 10 + 5 = 27.5 kNm
= 27.5 x 106 Nmm = ob x Z
Z = Section modulus
5 kN
5 kN
T
5 kN
5 kN
LW-4/
1-0-1 m -1+-1 m -44-1 m
Fig. 5.6
2 100 x 2502
- bd
= 1.04166 x106 mm3
6
6
27.5 x106
= 26.4 N/mm 2 (Max. bending strtess in beam)
as 1.04166 x 10'
Q.5.7 A cantilever beam of cross-section 50 mm x 100 mm in subjected to a compressive force parallel
to the longitudinal axis of the beam at the free end. The point of application of the load is at 10 mm
above the bottom surface of the vertical centroidal axis. Span of the beam is 4 m. Calculate the
maximum stress in the beam.
[CSE-Mains, 2012, CE : 15 Marks]
Solution:
b = 50 mm, d = 100 mm
Eccentricity of loading,
bd 3 = 50 x 1003
= 4.1667 x106 mm4
12
12
e= 50 - 10 = 40 mm
74 ►
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
Pin Newton,
A
—0-150
50
Bending moment
M = P x 40 Nmm
8
P 10 mm
4°
4m
Maximum bending stress,
My
/
=
+ 50 x 40P
Fig. 5.7
4.1667 x 106
= +480 x 106 P = ± 4.8 x 10-4 P NImm2
Direct shear,
= 2 x 10-4 P N/mm 2 (Comp.)
a =
area 50 x 100
d
amax at edge B = —(4.8 + 2) x 10-4 P. —6.8 x 10-4 P N/mm2 (Comp.)
6max at edge A = +4.8 x 10-4 — 2 x 10-4 = 2.8 x 10-4 P N/mm2 (Tensile)
Q.5.8 Compare the flexure strength of the following three beams of equal weight per unit length.
I-section 300 mm x 150 mm with flanges 20 mm thick and web 12 mm thick.
1.
Rectangular section having depth twice the width
2.
3.
Solid circular section
[CSE-Mains, 2012, ME : 15 Marks]
Solution:
Having same material and equal weight per unit length means same area
of cross-section for all three beams.
8
20
I-section
Area of cross-sections = 2 x 150 x 20 + 260 x 12
= 6000 + 3120 = 9120 mm2
260
150 x 3003 138 x 2603
zr
12
20
12
337.5 x 106 — 202.124 x 106
= 135.376 x 106 mm4
135.376 106
150
—150 --xd
Fig. 5.8
= 0.9025 x 106 mm3
Rectangular section
Area = 282 = 9120 (same area of cross-section)
B = 67.53 mm
D = 135.055 m
4
BD2
67.53 x 135.0552
6
6 =
= 0.20529 x 106 mm3
Circular section
Fig. 5.9
Tcd2
4
Diameter,
D = 28
— 9120
d = 107.758 mm
nd3
Z3
x (107.7583 )
—
—
32
0.122845 x 106 mm4
32
: Z2 : Z3 = 0.9025: 0.20529: 0.122845
= 7.346 : 1.671 : 1
Theory of Simple Bending
Strength of Materials
4
75
Q.5.9 A beam is of square section with diagonals 60 mm long, vertical
and horizontal, as shown in figure 5.10. Shear force at a particular
section is 5 kN. What is the shear stress at layer AB.
[CSE-Mains, 2009, ME : 15 Marks]
Solution:
Square section
Diagonal,
d = 60 mm
60 mm
d4 60 4
= 27 x 104 mm4
= 48 48
'-J
Fig. 5.10
Layer, AB, breadth = 30 mm = b = d/2
1
OAB = - x 30 x 15 = 225 mm2
2
Area,
y
15
= 15+ — = 20 mm
3
ay = 225 x 20 = 4500 mm3
F= 5 kN = 5000 N
TAB =
Fay
lb
5000 x 4500
27 x104 x 30
= 2.778 N/mm2
Q.5.10 Two long wooden planks form a T-section of a beam as shown in
figure 5.11. If this beam transmits a constant vertical shear of 3000 N,
find the necessary spacing of the nails between the two planks to
make the beam act as a unit. Assume that allowable shear force per
nail is 700 N.
Solution:
Location of G of section
200
50
r•-•
#
a
a
co
4G
x 250
= 162.5
200 x 50 x100+200 x50 x 225
2 x200 x 50
= 50 + 112.5 = 162.5 mm
y2 = 250 - 162.5 = 87.5 mm
I
Fig. 5.11
Moment of inertia, I
2 200 x 503
50 x 2003
+ 50 x 200(162.5 100) +
+200 x 50 (87.5 - 25)2
12
12
= 33.33x 106 + 78.125 x 106 + 2.083 x 106 = 113.538x 106 mm4
Shear stress along a-a, nail section
=
ti
Fay
aa = Ib
b= 50 mm, I =113.538 x 106 mm4, F= 3000 N
ay about neutral axis = 200 x 50 x (87.5 - 25) = 200 x 50 x 62.5 = 6.25 x 105 mm3
3000 x6.25 x10 5
x 50 - 0.33 N/mm2
as
113.538 x106
Allowable shear force = 700 N
76 ►
MADE EASY
1AS & IFS (Objective & Conventional) Previous Solved Questions
Spacing between nails = S mm (say)
700 = S x 0.33 x 50
S, spacing = 42.42 mm
Nails are fixed at spacing of 40 mm
Q.5.11 An I beam with the following dimensions is subjected to a shearing force
of 20 kN.
Flange: breadth = 50 mm, web thickness = 3.5 mm
Thickness = 5.5 mm, web depth = 189 mm
Area of cross-section = 9.4 x 102 mm2
Ad/ = = 220 x 104 mm4
Calculate the value of the transverse shear stress at the neutral axis x-x
and at the top of the web.
[CSE-Mains, 2012, ME : 12 Marks]
Solution:
F= 20 kN
Shear force,
= 220 x 104 mm6
b = breadth = 3.5 mm
Shear stress at neutral axis
b= 3.5 mm
I
5.5 mm
I
I
189 mm
x
5.5 mm
1-4— 50
I
Fig. 5.12
a)7 = 50x 5.5 x(100 - 2.25)+ 3.5 x 189(189)
= 26743.75 + 1562.9375 = 42371.69 mm3
20000 x 42371.69
= 110.056 N/mm2
220 x104 x3.5
Shear stress at top edge of web=
20000)26743.75
= 69.46 N/mm2
220x104 x3.5
Q.5.12 A horizontal beam of square cross-section is so placed that the loading in the transverse plane is
along one of the diagonals of length d. If the shear
A
force at a section of the beam is S. Draw shear
2s
2
stress distribution diagram for the section and
T a
a
d
2.25s
indicate the position and magnitude of maximum T
2
d
d -t
shear stress on it.
ir
[CSE-Mains, 1998, ME : 30 Marks]
I.
j M8X
Solution:
Square section
Shear stress
Diagonal = d
distribation
d
Fig. 5.13
Side =
4
(d )
=
Shear force = S
Take a section at a distance y. from NA, xx
Area Aaa (distance of centroid) from xx
x
1 d4
12 - 48
Theory of Simple Bending
Strength of Materials
Height of the triangle
Aaa =
4
77
(d -y )
2
Base of the triangle = 21d -y)
2
Area =
Ay =
Breadth,
b=
Transverse shear stress,
T=
id
2
)1
(d
) (d
)2
y 2 x2 2 y
2 y
2 2y d
2 - Y ) (V d
(
d
2(
2-y )
Say
Tu b
s(d _ )(2yd ) 48
x
+
y
3 6)
d4 x2(2-y)
=
d
At- = Y
4
'
At — = y Outer edge,
2
'
At y = 0,
T=
[ d 2 4_ dY 2 v 2
d 4 ' 2 - '
8S[ d2
d d
d2]
— +— x— -2 x
16
d4 4 2 4
8S [d2 d2 d2]
TA =
d4 4 4 2
8S
TNA =
8S [d2 d 2 d 2
d4 4 8 8
2S
= d2
=o
d 2 2S
d 4 X 4 —d2
tmax
For Tma,„
d'r _
0
—
d - 4y) = 0
dy = (2
Y =
Tom _
Refer equation (i)
d
8
8S [d2 d d
d2]
9S
s
= 2.25x—
,
— +— x— -2x— =
64
4d2
4
2
8
d'
d"
Q.5.13 An I-beam has flanges 10 cm wide and 1 cm thick, and the web 12 cm high and 1 cm thick. At a
section of this beam act a bending moment of 1000 kg-m and a shear force of 10,000 kg, find the
normal and shear stresses at the following points on the vertical center line.
(i)
Top of the flange
(ii) In the web at the junction with the flange
(iii) At the neutral axis
[CSE-Mains, 1999, ME : 20 Marks]
78 0.
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
Solution:
I-section shown in the figure 5.14
I
1cm
I
I
= 10 x 143 9 x 123
12
12
About neutral axis,
12 cm
= 2286.667 - 1296 = 990.667 cm4
M = 1000 kg-m = 1000,00 kg-cm
(i) Bending normal stress
1 cm
GA =
•
1000,00 x 7
= 706.6 kg/cm2'
990.667
1000,00 x 6
= 605.65 kg/cm2
=
990.667
I-4— 10 —0-1
I
Fig. 5.14
ac = 6NA=Q
(ii) Shear stresses
For
Breadth,
TA = 0
TB, ay = 10 x 1 x 6.5 = 65 cm3
b = 1 cm
F = 10,000 kg
Fay _ 10,000 x 65
Ib 990.667 x 1
tic
656.12 kg/cm2
ay = 10 x 1 x 6.5 + 1 x 6 x 3=65 + 18 = 8.3 cm3
b= 1 cm
10000 x 8.3
= 837.82 kg/cm2
T C
990.667 x 1
Q.5.14 Simply supported beam of 3 m span is subjected to loads
as shown in figure 5.15(a) the beam is of I-section and all
of the dimensions are in mm. Determine the principal
stresses at point D in the web. The section is located at a
2m
1m
distance of 1 m for right hand support.
-4
5.667 kN
4.333 kN
[CSE-Mains, 2013, ME : 25 Marks]
(a)
Solution:
Fig. 5.15 (a)
Reactions,
2 x 3x 1.5 +4 x 2= 3RB
17 = 3RB
Reaction,
RB = 5.667 kN
Reaction,
RA = 2 x 3 + 4 - 5.667
10 mm
= 10 kN - 5.667 = 4.333 kN
40 mm
Mc = Bending moment at C
50 mm
180 mm
= 4.333 x 2-2 x 2 x 1
= 8.666 - 4 = 4.666 kN-m
15m
= 4.666 x 106 Nmm (Sagging)
10 mm
Shear force at C
F°—
Fc = 4.333 - 2 x 2 = 0.333 (just left to C)
(b)
Fc = 0.333 - 4 = -3.667 kN (just right to C)
Fig. 5.15(b)
Jump shear force at C = 4 kNI
I
I
Theory of Simple Bending 4
Strength of Materials
c x Y E)
ab, bending stress - M
l x„
Mc x 50
75 x 2003 60 x 1803
12
12
= 20.84 x 106 mm4
xr =
(ado -
79
50 x106 -29.16 x106
4.666 x 106 x 50
20.84 x 106
= -11.20 N/mm2 (sagging moment) compressive
Shear stress at D
'co =
Fay
lb
when,
b = 15 mm
ay (about neutral axis) = 15 x 40 x 70 + 75 x 10 x 95 = 113250 mm2
F = 4000 N
113250 x 4000
= 1.45 N/mm2
- 20.84x106 x15
Principal stresses
P1 , P2 =
11.20 44
11.20)2
l
2
+0.45)2
2
11.385 N/mm
2
1.45 N/mm
= -5.6 ± V31.56 + 2.1025
= -5.6 ± 5.79
p1 = -11.385 N/mm2
p2 = +0.19 N/mm2
2
1.45 N/mm
Objective Questions
Q.1
Q.2
A cantilever beam of rectangular cross-section
is 1 m deep and 0.6 m thick. If the beam were
to be 0.6 m deep and 1 m thick then the beam
would
(a) be weakened 0.5 time
(b) be weakened 0.6 time
(c) be strengthed 0.6 time
(d) have the same strength on the original beam
because the cross-sectional area remains
the same
[CSE-Prelims, ME : 1999]
If the T-beam cross-section shown in the given
figure 5.16 has bending stress of 30 MPa in the
top fibre, then the stress in the bottom fibre
would be (G is centroid)
(a) zero
(c) -80 MPa
Q.3
Fig. 5.16
(b) -30 MPa
(d) 0.50 MPa
The distribution of shear stress of a beam in
shown in the figure 5.17. The cross-section of
the beam is
80
0.
Fig. 5.17
(b) T
(d)
Q.4
py 0 0
(a) 0 qy 0
By conjugate beam method, the slope at any
section of actual beam is equal to
(a) El times the SF of the conjugate beam
(b) EI times the BM of the conjugate beam
(c) SF of the conjugate beam
(d) BM of the conjugate beam
[CSE-Prelims, ME : 2002]
Q.8
Select the correct shear stress distribution
diagram for a square beam with a diagonal in a
vertical position
/\
(b) 0
(b)
(c)
(d)
qy 0
ry
py 0
0
0
0 (d) 0 qy 0
0 0 qy
py,
py
(a)
py 0 0)
0 0
0 0
0 0 0,
py
0 0
[CSE-Prelims, ME : 2001]
Q.5
Q.7
A uniform bar lying in x-direction is subjected
to pure bending. Which one of the following
tensors represents the strain variation when
bending moment is about z-axis (p, q, and x
constant)
(c)
Two identical planks of wood are connected by
bolts at a pitch distance of 20 mm. The beam
is subjected to a bending moment of 12 kNm
and the shear force in bolts will be
I
[CSE-Prelims, ME : 2002]
Q.9
A pipe of external diameter 3 cm and internal
diameter 2 cm and of length 4 m supported at
its ends. If carries a point load of 65 N at its
center-section, modulus of pipe will be
657E
(a)
10
10
I
(a) zero
(c) 0.2 kN
Q.6
MADE EASY
IAS & IFS (Objective & Conventional) Previous Solved Questions
Fig. 5.18
(b) 0.1 kN
(d) 4 kN
[CSE-Prelims, ME : 2001]
A straight bimetallic strip of copper and steel
is heated. It is free at ends. The strip will
(a) expand and remain straight
(b) not expand but bend
(c) expand and bend also
(d) first only
[CSE Prelims, ME : 2002]
(c)
3
64 cm
657t
96
cm
3
65/r
3
(b) 32 cm
(d)
657t
124
cm
3
[CSE-Prelims, ME : 2002]
Q.10 In a loaded beam under bending
(a) both the maximum normal and maximum
shear stresses occur at the skin fibres
(b) both the maximum normal and maximum
shear stresses occur at the neutral axis
(c) the maximum normal stress occurs at the
skin fibres and the maximum shear stress
occurs at the neutral axis
(d) the maximum normal stress occurs at the
neutral axis while the maximum shear stress
occurs at the stress fibres
[CSE-Prelims, ME : 2003]
Theory of Simple Bending
Strength of Materials
Q.11 Consider the following statements:
Two beams of identical cross-section but of
different materials carry same bending moment
at a particular section, then
1. the maximum bending stress at that section
in the two beams will be same
2. the maximum shearing stress at that section
in the two beams will be same
3. maximum bending stress at that section
will depend upon the elastic modulus of the
beam material
4. curvature of the beam having greater value
of E will be larger
Which of the statements given above are
correct?
(a) 1 and 2 only
(b) 1, 3 and 4
(c) 1, 2 and 3
(d) 2, 3 and 4
[CSE-Prelims, ME : 2002]
Q.12 For the shear force to be uniform throughout the
span of a simply supported beam, it should
carry which one of the following loadings?
(a) A concentrated load at mid span
(b) UDL over the entire span
(c) A couple anywhere within its span
(d) Two concentrated loads equal in magnitude
and placed at equal distance from each
support
[CSE-Prelims, ME :2007]
Q.13 Which one of the following is the preferable
cross-section of a beam for bending loads?
(a) Circular
(b) Annular circular
(d) I-section
(c) Rectangular
[CSE-Prelims, ME : 2006]
Q.14 Consider the following statements for simple
bending of beams:
1. Neutral axis always passes through the
centroid of the beam cross-section
2. Bending stress in a fibre is a longitudinal
stress
3. Bending stress is zero at the neutral axis
4. Shearing stress is always zero on the neutral
axis
Which of the statements given above are
correct?
(a) 1, 2 and 3 only
(b) 1, 2, 3 and 4
(d) 1 and 4 only
(c) 2, 3 and 4 only
[CSE-Prelims, ME : 2006]
81
Q.15 Three beams of 100 mm x 100 mm section are
made from RCC, aluminium and timber
respectively. These three beams are loaded
identically with similar supports. Maximum
bending stress will occur in which beam?
(a) Timber beam
(b) Aluminium beam
(c) RCC beam
(d) The three beams will have same maximum
bending stress
[CSE-Prelims, ME : 2008]
Q.16 A simply supported beam of span Land flexural
rigidity EI carries a unit point load at its centre.
What is the strain energy in the bema due to
bending?
(a)
L3
16E/
L3
(c) 96E1
(b)
L3
48 E1
(d)
L3
192 E1
[CSE-Prelims, ME : 2009]
Q.17 For a triangular section with base 'b' and height
' h' , the ratio of the moment of inertia about an
axis passing through its vertex and parallel to
its base to the moment of inertia about an axis
passing though its centre of gravity and also
parallel to its base would be
(a) twelve to one
(b) nine to one
(c) six to one
(d) four to one
[CSE-Prelims, CE : 2001]
0.18 A beam has rectangular section 100 mm x 200
mm. If it is subjected to a maximum BM of
4 x 107 Nmm, then the maximum bending stress
developed would be
(a) 30 N/mm2
(b) 60 N/mm2
(c) 90 N/mm2
(d) 120 N/mm2
[CSE-Prelims, CE : 2001]
Q.19 Consider the following statements:
For each component in a flitched beam under
the action of a transverse load.
1. the radius of curvature will be different.
2. the radius of curvature will be the same.
3. the maximum bending stress will be the
same.
82 ►
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
4. the maximum bending stress will be
dependent upon the modulus of elasticity
of the material of the component.
Which of these statements are correct?
(a) 1 and 3
(b) 1 and 4
(c) 2 and 3
(d) 2 and 4
[CSE-Prelims, CE : 2001]
Q.20 A timber beam is simply supported at the ends
and carries a concentrated load at mid span.
The maximum longitudinal stress 'f is 12 N/mm2
and the maximum shear stress 'q' is 1.2 N/mm2.
The ratio of span to depth would be
(a) 10
(b) 6
(c) 5
(d) 4
[CSE-Prelims, CE : 2001]
0.21 The moment of inertia of a given rectangular
area is minimum about
(a) its longer centroidal axis
(b) its polar axis
(c) its axis along the diagonal
(d) its shorter centroidal axis
[CSE-Prelims, CE : 2003]
Q.22 For a beam of rectangular section under bending
the shear stress across the depth varies
(a) Linearly
(b) Exponentially
(c) Hyperbolically
(d) Parabolically
[CSE-Prelims, CE : 2003]
Q.23 A cantilever beam is loaded with a uniformly
distributed load of intensity w along its entire
length. The span of the beam is L. Which of the
following Mohr's circle diagram correctly
represent(s) the state of stress above the
neutral axis of the beam?
Shearing stress
Shearing stress
(i)
Normal
Tensile
stress
Normal
Tensile
stress
(ii)
Shearing stress
Shearing stress
Normal
Tensile
stress
Fig. 5.20
Normal
Tensile
stress
( v)
Select the correct answer from the codes given
below:
(a) Fig. (i)
(b) Fig. (iii)
(c) Figl. (ii) and (iv) (d) Fig. (iii) and (iv)
[CSE-Prelims, CE : 2014]
Q.24 In a thin uniform lamina having symmetrical
central axis as shown below, the distance of
centre of gravity from AD is
A B
oco
1.5 m
K
D
T
C
—►] 2 cm I-4— 8 cm
(a) 3 cm
(c) 23/7 cm
►H
Fig. 5.21
(b) 22/7 cm
(d) 24/7 cm
[CSE-Prelims, CE : 2004]
Q.25 In the cross-section of a rectangular beam, what
is the ratio of the average shear stress to the
maximum shear stress?
(a) 3/2
(b) 2/3
(c) 4/3
(d) 3/4
[CSE-Prelims, CE : 2006]
Q.26 Which of the following are implied in the
assumption of plane sections remaining plane
(in simple bending)?
1. Stress is proportional to the distance from
neutral axis.
2. Displacement is proportional to the distance
from neutral axis.
3. Strain is zero across the cross-section.
4. Strain is directly proportional to the distance
from neutral axis.
Select the correct answer using the codes given
below:
(a) 1 and 4 only
(b) 2 and 3 only
(c) 3 and 4 only
(d) 1, 2 and 4
Q.27 A beam of rectangular cross-section is to be
cut from a circular beam of diameter D. What is
Theory of Simple Bending
Strength of Materials
the ratio of the depth of the beam to its width
for maximum moment of resistance?
(b)
3
(d)
[CSE-Prelims, CE : 2009]
Q.28 A cantilever beam as shown in figure is
subjected to end moment M0. What is the
deflection at the free end?
1
21
4m -IA
(a)
(b)
(c)
(d)
-41 83
top edge of the flange
centre of the web
junction of the flange and web
bottom edge of the flange
[CSE-Prelims, CE : 2010]
Q.32 Two tubes of outer diameter 20 mm and inner
diameter 16 mm each are combined with a solid
bar of diameter 20 mm. Their centres make an
equilateral triangle of side 20 mm. The section
is of a cantilever, with solid tube on top,
subjected to load at free end. If the maximum
tensile stress developed in section is 100 MPa.
What is the maximum compressive stress
developed.
4
Fig. 5.22
10 M0
EI
(b)
15 M0
El
20 M0
(c) El
(d)
30 M0
El
(a)
[CSE-Prelims, CE : 2010]
Q.29 A mild steel beam is simply supported. It has
constant moment of inertia =106 mm4. The entire
length of the beam is subjected to a constant
BM of 107 Nmm, E = 2 x 105 N/mm2. What is
the radius curvature of the bent beam in meters?
(a) 200
(b) 2
(c) 20
(d) 0.2
[CSE-Prelims, CE : 2010]
Q.30 In a beam of solid circular cross-section, what
is the ratio of maximum shear stress to the
average shear stress?
3
(a) 7
1
(c)
3
(b)
4
2
(d) 3
[CSE-Prelims, CE : 2010]
Q.31 In case of a beam of I-section subjected to
transverse shear force 'F, the maximum shear
stress occurs a the
(a) 136 MPa
(c) 86 MPa
Fig. 5.23
(b) 116 MPa
(d) 73.2 MPa
Q.33 A beam of uniform strength refers which one of
the following?
(a) A beam in which extreme fibre stresses are
same at all cross-section along the length
of the beam
(b) A beam in which the moment of inertia about
the axis of bending is constant at all crosssection of the beam
(c) A beam in which the distribution of bending
stress across the depth of cross-section is
uniform at all cross-sections of the beam
(d) A beam in which the bending stress is
uniform at the maximum bending moment
cross-section.
Q.34 The cross-section of a beam in bending is as
shown in the figure given below. It is subjected
to a shear force acting in the plane of crosssection. Which among the following figures
shows the correct shear stress distribution
84
0. IAS & IFS (Objective & Conventional) Previous Solved Questions
(a) 3.2
(c) 3.4
across the depth of the cross-section of the
beam?
T
2
Fig. 5.24
MADE EASY
(b) 3.3
(d) 3.5
[CSE-Prelims, CE : 2006]
Q.37 The below diagrams show the details of two
simply supported beams B1 and 82. EI is
constant throughout the length and same for
both the beams. The beams have the same area
of cross-section and the same depth. What is
the ratio of maximum bending stress in 82 to
that in Bi?
4 kN
kN/m X3
A1y-v-v-v111100.11
(b)
m -+- 2 m-od
B1
(a) 4
(c) 2
(d)
[CSE-Prelims, CE : 2005]
Q.35 A mild steel plate is subjected to a moment M
each at its ends such that it bends into an arc
of a circle of radius 10 m. The plate has width
60 mm and thickness 10 mm. E = 2 x 105
N/mm2.
What is the maximum bending stress produced
in the plate?
(a) 100 MPa
(b) 200 MPa
(d) 400 MPa
(c) 300 MPa
[CSE-Prelims, CE : 2005]
1-4— 4 m
B2
Fig. 5.26
(b) 1
(d) 1/2
[CSE-Prelims, CE : 2009]
Q.38 Cross-sections of two beams A (600 mm x 200
mm) and 8(800 mm x 60 mm) are shown in the
figure given below. Both the beams have the
same material. By how many times is the beam
A stronger than the beam B in resisting
bending?
200 mm-1-1
i
T
1
E
E
0
co
800 mm
A
Q.36 What is the y-coordinate of the centroid of the
area ABCDEshown in the figure given below?
Fig. 5.27
(a) 80
(c) 50
Fig. 5.25
(b) 60
(d) 25
[CSE-Prelims, CE : 2007]
Q.39 A mild steel structural beam has cross-section,
which is an unsymmetrical I-section. The overall
depth of the beam is 250 mm. The flange
Theory of Simple Bending
Strength of Materials
stresses at the top and bottom are 200 N/mm2
and 50 N/mm2, respectively. What is the depth
of the neutral axis from the top of the beam?
(a) 50 mm
(b) 100 mm
(c) 150 mm
(d) 200 mm
[CSE-Prelims, CE : 2008]
1 85
Answers
1. (b)
6. (c)
11. (a)
16. (c)
21. (a)
26. (d)
31. (b)
36. (c)
2.
7.
12.
17.
22.
27.
32.
37.
3. (b)
8. (d)
13. (d)
18. (b)
23. (a)
28. (c)
33. (a)
38. (d)
(c)
(c)
(c)
(b)
(d)
(b)
(b)
(d)
4. (d)
9. (c)
14. (a)
19. (d)
24. (b)
29. (c)
34. (c)
39. (d)
5.
10.
15.
20.
25.
30.
35.
Explanations
6. (c)
1. (b)
Z
1-
0.6 x13
6
Z2
1 x 0.63
= 0.036 m3
6
Z2
Z
2.
0.1 m3
= 0.36 weakened by 0.64 times
Fig. 5.29
Strip will bend and expand.
(c)
Y2 = 3°
Y1 = 80
f2 = 30
7.
(c)
Slope = SF of conjugate beam.
8.
(d)
Square with diagonal vertical
(d) is correct shear stress distribution
= -80 MPa
linearly proportional to distance y
3.
4.
(b)
Shear stress distributed is of T-section.
d
max
(d)
9.
at 8 from NA.
(c)
x
z
Fig. 5.28
G oc y
Stress
a
Ey=
5.
Fig. 5.30
= PY = E x
=
VG
= _ vpy = qy
(a)
Zero, as BM is constant and shear force is zero.
TC
65n
/ = 6 £31- 16) =
4
64
Z-
657C
1 657C 3
X = - cm
64 1.5 96
(a)
(c)
(d)
(c)
(b)
(b)
(a)
86 ►
MADE EASY
IAS & IFS (Objective & Conventional) Previous Solved Questions
10. (c)
=
Ixx
bh3 2bh3 bh3
36 9 4
=9
IGG
- ;Mx
I2h
3
1
Fig. 5.31
11. (a)
1 and 2 are correct statements.
5
b
12. (c)
Fig. 5.33
18. (b)
b = 100, h = 200 mm
bh 2 100 x 2002 4
= x 106
=
6
6
6
M= 4 x 107 Nmm
Z=
RA
a —
Fig. 5.32
13. (d)
I-section has maximum moment of inertia for same
area of cross-section.
w
at = —
Lx
12
4 bd` —
tia = 1.5
L
d
W 2 L3
96EI
L3
96E1
1
1
—
=1.2
2
2 bd
10 = 1.5wL 2bd
12
x
1.5W
bd 2
1.2 =
or
16. (c)
Strain energy,
48E1
6
= 60 MPa
106
19. (d)
Flitched beam (Each component)
5. Radius of curvature will be the same
4. Maximum crb dependent on Eof component
15. (d)
Stress depends on BM and section modulus
independent of material properties.
3
1
wL
=—WX
x
20. (c)
14. (a)
1, 2, 3 are correct statements.
U=
4
b
SFD
2
W= 1
4 x 107
5
21. (a)
MI minimum about the longer centroidal axis.
D
17. (b)
x
bh 3
GG
I
=
36
bh3 bh 14h2 )
Ixx = 36 ± 2 9
2L
d
Fig. 5.34
BD 3
x
12
Theory of Simple Bending
Strength of Materials
22. (a)
Parabolically
av
"c max -
1
$7
2
3
26. (d)
1, 2 and 4 are correct statements.
27. (b)
Fig 5.35
23. (a)
Fig. 5.36
At a +ve normal stress
shear stress
Fig. 5.39
b2 +d2 =
b = D sing, d =D cos()
Z —
dz
=
O
ti
Fig. 5.37
bd 2
6
1-1— 7 —""i
=
sin() =
8 cm —01
16x1+12x6
28
16 + 72 88 22
= = —cm
28 28 7
'ray
—
1.5
3 0))
( 0
sin — sin
k
3
[cos() — 3 sin2 cos 0] = 0
6
1
core =
28. (c)
Fig. 5.38
max
6
d_
b
T
25. (b)
Rectangular beam section
3
= dt
24. (b)
D3 sinOcos2 0
6
D3
= — sinO(1—sin2
6
d_1
x
—
L
L
• G
• G
21
Fig. 5.40
8—
MnL L M L 3L
x
x —+
D. 2 2E1 2
M0L2 3M0L2 5 M0 42 M0 x 20
2E1
4E1 • 4 El
El
88 P.
IAS & IFS (Objective & Conventional) Previous Solved Questions
35. (a)
R = 10 mm, b = 60 mm, t = 10 mm
E = 2 x 105 N/mm2
29. (c)
M
I
107
106
MADE EASY
E
R
2 x 105
_20m
R
R
a E
—
Y
R
2 x 104 mm
30. (b)
Beam of circular section
4
'c max
3
Tav
31. (b)
I-section beam, Tmax occurs at centre of the web.
y=
_
2x105 x5
10000
=100 MPa
36. (c)
Y=
32. (a)
Ai .7c x 100, solid
A2 = n(100 — 82) = 367c (hollow)
2 =5mm
24 x 3+6 x 5 102
=
28
30
3.4
37. (d)
M1 = 4 kNm
IV12 —
1x16
8
= 2 kNm
Same area, same depth
as M
02
al
= 0.5
38. (d)
Fig. 5.41
1007cx17.32+36nx0 x2
10.07
172n
y1 = 17.32 — 10.07 + 10 = 17.25 mm
y2 = 10.07 + 10 = 20.07
Y-
6
20.07
x 100 =116 MPa
"p 17.25
33. (a)
Beam of uniform strength — in which extreme fibre
stress are same at all cross-section along the
length of the beam.
34. (c)
For cross
ZA -
200 x 6002
=12x106 mm3
6
ZB
800 x 602
= 48 x 104 mm3
6
ZA
ZB = 25
39. (d)
200 MPa
I
200 mm
N
A
50 mm
500 MPa
T
Fig. 5.44
Depth of neutral axis from top.
•• • •
Shear stress distribution
Fig. 5.42
06
Deflection of Beams
CHAPTER
Q.6.1 A simply supported beam of length L carries a concentrated load Wat a distance 'a' from one end
and 'b' from the other end (a > b). Find the position and magnitudes of the maximum deflection
L
and show that the position is always within — approx from the center of the beam.
13
[CSE-Mains, 2003, ME : 30 Marks]
Solution:
Wa = RB X L
Wa
L
Reaction,
RB =
Reaction,
RA = W
x
Wa Wb
L
L
Take a section x-x at a distance x from A, in portion CB
El
2y
dx 2
Wb
b
RA = y
R _ Wa
B
a>b
Fig. 6.1
= RAx /- W(x - a)
Wbx
L
W(x - a)
,r dy
/ W(x-a)2
2
Li — = Wbx ±ci
dx
2L
2
Wbx 3
+ Ci x +C2 / W (x-a)3
6
6L
y = 0 at x = 0, therefore constant, C2 = 0 (second term not used)
Ely =
Moreover,
,
y = 0, x = L (second term -- v - a)3also used)
6
0-
0=
0=
WbL3 W
- (L - a)3 + C1 x L
6L
6
WL2
b
6
6
6
(L-a)3 + Ci L
bL2 -W x b3 + Ci L
6
90 ►
lAS & IFS (Objective & Conventional) Previous Solved Questions
WbL
Wb3 Wb r h2
6 ± 6L = 6L
Constant,
MADE EASY
,21
= 6b [b2 - a2 - b 2 -2ab]= Wb
— [-a 2 - 2a b]
6L
6L
Wab
_
El
dy
dx
=
6L
[a + 2b]
Wb x2
W
2L
2
(x a)
2
Wa b
6L
(a + 2b)
dy
ymax will occur, where — = 0
dx
n = Wbx2 W (
x2
2L
=
2
2 k +a
b 2
L
(
2
-zax)
Wab
—
6L (a + 2b)
ab
+ a 2 -2ax)-- (a + 2b)
3L
0 = 3bx2 -3L(x2 + a 2 -2ax)-ab(a+2b)
0 = 3bx2 - 3Lx2 -3La 2 + 6axL-a2b-2ab2
0 = -3x2 (L -b)-3La2 +6axL-a 2b-2ab 2
or
0 = -3x2 -3La+6xL-ab-2b 2
3x 2 -6xL+3La+ ab+2b 2 = 0
3x2 -6xL+3a(a + b)+ab+2b 2 = 0
3x2 - 6xL+3a2 +3ab+ ab+2b 2 = 0
3x2 -6xL+3a 2 +4ab+2b 2 = 0
x=
6L + \1(602 - 4 x 3(3a2 +4ab+2b 2 )
6
x=
6L + 6 1L2 -1(3a2 +4ab+2b2 )
3
1,
6
L±/ L2 -(3a2 +4ab+2b 2 )
3
X
x
X
,
= L±,\IL2 - - [a3
,
L
(
IL2 2L2 + a 2
3
i3L2 - 2L2 - a 2
= L+ \
3
L±
L2 - a 2
\1
3
2
+ 2ab + b2 )]
Deflection of Beams
Strength of Materials
x=
,
91
2ab + b 2
3
L±
Let us take a= 0.6L, b = 0.4L
x= L±
.12x.6x0.4L2 +(0.4) L
3
= L±LV0.16+0.0533
= L±LV0.21333
x = L±Lx 0.4619
x = -L x 0.4619
= 0.538 L= 0.5 L+ 0.038 L
x = 0.5L + 0.038L -_-_ — from the center
26 '
L
26
or
L
-_) _ L , total
26 )
13
II 2 _ a2
I-
x=
3
Q.6.2 A cantilever beam of length L, is subjected to two concentrated loads of 2P and Pat its mid length
L
and free end respectively. If the deflection at its free end is limited to — . What should be the
500
value of P? Take flexural rigidity of the beam as El.
[CSE-Mains, 2011, ME : 15 Marks]
Solution:
Flexural rigidity = El
Deflection at free end,
3 2P (
L
PL3 2P ( 1
x
x
5 - —+- x - +
3E/ 3E/ 2
2E/ U) 2
Fig. 6.2
PL3 PL3 P L3
+
+ x3E112E1 El 8
=
PL3 [1 1 + 1]
+
EI 3 12 8
PL3 r8+ 2 +3]
3E1 1_ 24
13 PL3 L
x— =
24 E/ 500
P=
El
L2
X
24 1
24E1
X =
500 13 6500 L2
value of P
Q.6.3 A cantilever 3 m long, and of symmetrical section 250 mm deep carries a uniformly distributed
load of 30 kN/m run throughout together with a point of 80 kN at a section 1.2 m from fixed end.
Find the deflection at the free end. E = 200 GPa, I = 54000 cm4
[CSE-Mains, 2012, ME : 15 Marks]
92 P
IAS & IFS (Objective & Conventional) Previous Solved Questions
MADE EASY
Solution:
Fig. 6.3 shows a cantilever,
AB fixed at end B,
Free at end A carrying a point load W at L1 from B end UDL of w
throughout to length
= flexural rigidity
8 = deflection at free end
80 kN = W
w = 30 kN/m
A
L
Fig. 6.3
WO1
wL4
WL3 +
x L2 +
3E/ 2E1
8E1
where, L1 = 1.2 m, L2 = 1.8 m, L = 3 m, W= 80 kN and w = 30 kN/m
EI = 200 x 106 kN/m2 x 54000 x 10-8 1114
= 108000 kNm2
80 x 1 22
30 x 34
80 x 1.23
+
1.8 +
2
x
108000
x
8
x 108000
3 x 108000
= 0.42666 x 10-3 + 0.96 x 10-3 + 2.8125 x 10-3
= 4.2 x 10-3 m = 4.2 mm
8-
Q.6.4 A cantilever beam of length L and uniform flexural rigidity El is subjected to continuously distributed
externally applied moment m kg-cm per cm of length of the beam. Using the area moment method
show that deflection of the free end of the beam is 8 = mL3/3 El and explain why this is the same
as that obtained for the case of a concentrated force P = is applied at the end of the beam.
[CSE-Mains, 1991, ME : 20 Marks]
Solution:
kg-nn/m
m
Area of BM diagram = mL x
Ditance of CG of BMD from A =
L mL2
= 2
A
2L
2L mL3
=
First moment of BM area from A = mL2 X
2
3
3
Deflection at the free end =
BMD
mL3
3E1
Fig. 6.4
(since deflection at fixed end is zero)
Because in BM diagrams = mL = PL
m= P
Therefore same deflection.
Q.6.5 Use Macaulay's method or area moment method to show
that the deflection of a simply supported beam at point B
with an off centre load at point A as shown in fig. 6.5 is
given by
pL2 x (L2 _ x2 _ L2
2
YB =
1.
R
x'
*1.
L,
L2
R2
L
Fig. 6.5
6 EL
[CSE-Mains, 1989, ME : 30 Marks]
93
Deflection of Beams
Strength of Materials1
Solution:
Reactions,
R1 =
PL2
L
R2 =
PLi
L
Consider a section yy at a distance x' from point 1
Mx' =
P(x - Li )
Bending moment,
EI
d y
dx 2
=
P(x' - L ) 1
PL2
L
x' P(x' - Li )
dy PL2 'x 2 P
2
=
x—-(x' - L1) + C1
dx L
2
2
PL2 x' 3 P
3
Ely = — x- --6(x'-Li) +
L
6
C2 = O,x = 0, y = 0
x' = L, y = 0
El x0 =
C1
=
Ely =
So,
Deflection at B, (at distance x)
PL2
2
L xL
+ C2
PL2 L2
P (1--1-1)3±C11- -
P (1_32 )+4
6 L
PL2L
P
6xL
6
IL2 x
x'3
px
(x' 1-1)3 + 16
L
6
6
PL21-1,c,
6
x < x'
ElyB =
P
al
6
x-- omitted term+ x —
L
6
6 L
2 x3
PL
PL2 x 3 PL32 x PL2Lx
6L
Y
6L
PL 2 x 112
s
6
Px1-2r 2
Lx +L22 -L2 ]
6L
x 2 _ L21
6 EI L L
Q.6.6 A beam ABCD, 5 m long is supported at A and C as
shown in figure 6.6. It carries a point load of 2 kN at
end D, and a moment of 2 kNm (CW) at B. What is the
flexural rigidity of the beam (El), if the deflection at D
is not to exceed 1 mm.
[CSE-Mains, 2009, ME : 20 Marks]
2
2 kN-m
:1410
1•-- 2 m
RA =
Solution:
Reaction
Taking moment about A
x
H
x J, 2kN
x
B/
2 m--4-1 m-R
Rc = + 3kN
1kN
Fig. 6.6
5 x 2 + 2 = 4Rc
Rc = 3 kNT
RA = 2- 3 = -1 kNL
94
0.
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
Take a section x-x at a distance x from A, in portion CD (Fig. 6.6)
d 2y
El — = -1x x + 2(x - 2)° + 3(x - 4))
dx 2
Ely=
Integrating two times,
x3
/
nN2
/
+
A
y = 0, x = 0, C2 = 0, constant
x = 4 m, support C, y = 0
64
0 = -- +4+0+4C1
6
0 = -10.6667 + 4 + 4C1
C1 = 1.666
3
Ely =
6
2
+ (X — 2) -I-
3
2
(x — 4) +1.666x
At end D, x = 5 m, Y = YD
125
1
E/YD = - — + 9 + - + 1.666 x 5
6
2
= -20.833 + 9.5 + 8.33 = -3.003
3.003
El m
YD =
3.003 = 0.001 m
EI
3.003
= 3003 kNm2 (Flexural rigidity)
0.001
El
Q.6.7 A cantilever of length L, is loaded with uniformly increasing load, starting from zero at the free
end and to a maximum of Wo at the fixed end. The free end is propped to the level of the fixed end.
Determine the reaction at the prop and equation of elastic curve along with the slope at the
propped end. EI is assumed constant.
[CSE-Mains, 2006, ME : 30 Marks]
Solution:
Say prop reaction = R
Rate of loading at a distance,
Wx = woL
x x
L 2
Load uptox, = wo- X —
wo
x
( x2 x)
BM at section x-x = WO — x 2L 3
X3
BM at section x-x = Rx - wo -E
EI
d2y
(12
x3
= Rx -wo -6-E
w
Fig. 6.7
Deflection of Beams
Strength: Of Materials
El
x 2 ,.,
vv ox4
- Rx
+c
dx 1
2 24L
dy
dx
0=
woL3
24
95
41
0, x = L
RL2
2
woL3
24 +1
RL2
Cl
2 Ely =
Rx 3
6
wox 5 ( w oL3
120E + 24
RL21 r
2 )x + -2
x = 0, y = 0, C2 = 0
So,
Ely =
Rx 3
6
wox 5 ( w oL3 RL2 )
120L + 24
2 )x
x = L, y = 0
woo woo RL3
120 + 24
2
0=
RL3
6
0 =
_RL3 4.woL4
3
30
R= wol10
Ely = w°1-x3 w°x5 1w°1-3 w°L x L2 x
60
120L + 24
20
woi-x3 wox5 wo0
60 120L 120
Slope at free end
x= 0
woL3 w01-3 w0L-3
20
120
EliA - Cl = 24
Slope at A,
iA =
woL3
120E1
0.3 m 10 kN
20 kN/m
.._
Q.6.8 A simply supported beam is subjected to the
loading as shown in the figure 6.8(a). Calculate
the deflection at a section 2 m from end A.
Assume E= 70 GN/m2, I= 830 cm4)
[CSE-Mains, 1996, ME : 30 Marks]
11i1:3
101(N
141---- 1.5 m --144-- 0.9 m -414-0.6 m -44
20 kN/m 16
3 kNm
Solution:
Reaction
3xRB + 3 = 16 x 2.4 + 20 x 0.9 x(1.5 + 0.45)
3 + 3/38 = 38.4 + 18 x 1.95 =38.4 +35.1
RB = 23.5 kN T
Total load = 20 x 0.9 + 16 = 34 kN
16 kN
A
D
t
4-- 1.5 m
'14
RA = 10.5 kN
Fig. 6.8
x
0.9.m --4-}4- 0.6 m
AB
9 6 0. lAS & IFS (Objective & Conventional) Previous Solved Questions
Reaction,
MADE EASY
RA = 34 - 23.5 = 10.5 kN
Figure shows the simplified loading diagram UDL extended to portion is portion DB upto x and negative
UDL applied beyond D.
10.5x - 3(x -1.5)° - 2 (x - 1.5)2 +14-(x
/ - 2.4)2 -16(x - 2.4)
2
w = 20 kN/m
El
d2y
dx 2
EI
_ 10.5x - 3(x - 1.5)° -10(x - 1.5)2 + 10(x - 2.4)2 - 16(x - 2.4)
dy
5.25x2 - 3(x -1.5)-
dx
10
3
1 .5)3 +
(x _
3
(x - 2.4)3 - 8(x - 2.4)2 +
10
— (x -1.5)4 + 10 (x 2.4)4 8 (x 2.4.)3
12
12
3
Ely = 1.75x3 -1.5(x -1.5)2
CiX + C2
x = 0, y = 0, C2 = 0, Omitting II, Ill, IV and V terms
x = 3 m, y = 0
0 = 1.75 x 27 - 1.5 x 2.25 -
10
2
4
(1.5) +
\ 4
8
3
(0.6) - -(0.6) + 3C1
2
3
10
0 = 47.25 - 3.375 - 4.21875 + 0.108 - 0.576 + 3C1
= 39.188 + 3C1
C1 = -13.063
Ely = 1.75
-1.5(x -1.5)2
+ 10
2 (x 1
1
4
5) ;(x - 2.4)4 - 3 (X - 2.4)3 - 1 3.063x
x = 2m
Sy2 . 1.75 x; - 1.5 x 0.25 + ;(0.5)4 - Omitted - Omitted + 13.063 x 2
= 2.333 - 0.375 - 0.05208 - 26.126 = -24.22
y2EI = 24.22
=
70 x 106 kN/m2 x 830 x 10-8 kNm 2 = 581 kNm2
24.22
Y2 =
581
= 0.04168 m = -41.08 mm
Q.6.9 A simply supported beam carries a uniformly varying load with zero intensity at left support and
an intensity of w at the right support. Calculate the maximum deflection and maximum slope and
mention the position on the beam when these occur.
[CSE-Mains, 1988, ME : 30 Marks]
Solution:
Figure 6.9 shows a beam AB of span length L, carrying uniformly varying load with maximum intensity
w/m at right hand B.
wL
Total load on beam = 2
Deflection of Beams
Strength of Materials
Moment about A,
97
wL 2L
-T x 3 = RB x L
wL
3
Reaction,
R8 =
Reaction,
wL wL wL
RA = 2 — 3 6
RA =
Consider a section xx at a distance x from A
wL
6
wx
= rate of loading = L
x=
Total load upto
F2,-=
Fig. 6.9
wx x = wx2
x
L 2 2L
CG of load ties at — from x-x
3
x
2
x
x
2L 3
Wx
= RAx
Bending moment,
wL
Wx 3
-x x
6L
6
ci — =
dx
or
Ely =
wLx2 wx4
12
24L
wLx3
36
+ C1 (constant of integration)
wx 5
120L + Ci x+ C2 (constant of integration)
y= 0, x = 0, C2 = 0
y= 0 at x = L
0=
wLx 3 wx 5 r ,
+
+ C2
36
120E
0 — 3) 4
wL A- CiL = 7 x wL4 +Ci L
— (1
360
360
Content,
3
wL
360 x
C1
o
r.,dy
wLx 2 wx 4 7
", — =
w
dx
12 24L 360
dy
Maximum deflection occurs, where — = 0
dx
,,
4
7
vv X
24L 360 w1-3
4
3)
x
y2
2L — 30 "'
WL 2
0 = 12
W
0
=
4
or
41
LX2 X
2L
12
7
30
=0
wL
3
98 0.
IAS & IFS (Objective & Conventional) Previous Solved Questions
4
7
I-2X2 - X 2 30
4
MADE
=0
30L2x2 -15x4 - 7L4 = 0
15x4 - 30L2x2 + 7L4 = 0
2 + V900L4 - 420L4
x2 - 30L
30
2=
x
300 ± NFINTIL71 30L2 ± L2 x 4.156
30
30
4 r301:
2 - , x- N/30
30
= L2 - 0.730L2 = 0.27L2
x = 0.5196L
x2
ymax at x = 0.5196L
EIy
syma,
=
wLx3
36
7 ,3
WL x +1.1x
360
wL
3
-k(0.5196L)
5
7
3
x (0.51960 - — wL x 0.5196L
-12
0L
360
= wL4 [0.003897 - 10003156 -0.0101] (Putting the value of C1)
= 6.522 x 10-3 wL4
Ymax =
6.522 x10-3wL4
at x = 0.5196 from A
EI
7wL3 1
360 xt aend A
EI
Slope at A
Slope at B
EBB =
wL3 wL3 7
wL3
12 - 24 - 360 x
(30 -15 - 7)
wL3 = +
360
360 wL3
wL3
ja = 45E1 , (Hence, maximum slope at B)
Q.6.10 Draw SF and BM diagram of the beam ACDB, where two couples are acting. Find the ratio of
deflection at location C and D where couples are acting.
[CSE-Mains, 2002, ME : 30 Marks]
Solution:
Reaction:
Moment about A,
10 + 20 = 3RB
RB = 10 kN
RA = -10 kN
Deflection of Beams
Strength of Materials
X
= 10 kNm
-10 kNm
M2 = 20 kNm
-10 kNm
(c)
BM diagram
Fig. 6.10
SF diagram
A vertically, SF is -10 kN constant
BM diagram
MA =
Mc = -10 x 1 = -10 kNm (Hogging) (Just left of C)
Mc = -10 + 10 = 0 kNm (Just right of C)
MD = -10 x 2 + 10 = -10 kNm (Hogging) (Just left of D)
Mc = -10 + 20 = +10 kNm (Sagging) (Just right of D)
Me =
Deflection
Taking a section in the last portion DB of the beam
Mx = -10x + 10(x -1)° + 20(x - 2)°
El
d2
Y = -10x + 10(x-1)0 + 20 (x - 2)°
dx 2
dy
EI—
dx
-5x2 +10(x-1)+20(x-2)+C1
Ely = -
5x 3
+5(x-1)2 +10(x-2)2 + Cpc + C2
3
y = 0, x = 0, constant, C2 = 0
y= 0, x = 3
= — — x3
3
3
+5x4+10x1+3Ci
0 = -45 + 20 +10 + 3C1
C1 = +5
4
99
100 10-
lAS & IFS (Objective & Conventional) Previous Solved Questions
MADE EASY
Ely = -5c+ 5(x -1)2 +10(x-2)2 +5x
5
Elyc =
10
0 + Omitted+ 5 x 1= 3 = +3.33
- 3 x 8+ 5(1)+100)+5 x 2
= -13.33+5+0+ 10 = +1.667
Yc
3.33
=2
1.667
Yo
Q.6.11 A beam of flexural rigidity 20 MN m2 is simply supported over a span of 6 m as shown in figure
6.11. It carries a concentrated load of 24 kN, 2m from the left hand support and a UDL of 6 kN/m
on central 2 m part. The beam has vertical member welded onto it 2 m from right hand support
which carry two horizontal loads of 24 kN as shown in the figure. Distance between these loads
is 1 m.
24 kN
24 kN
6 kN/m
WyTheThrY\
m
C
(a)
2m
► 1.1
2m
D
I18 kN
24 kN
2m
'14
x
A
(b)
Fig. 6.11
(i)
(ii)
Draw BM diagram of the beam.
Calculate the vertical deflection at the central points of the beam.
Solution:
Reaction:
Moment about A
24 x 2 + 24 + 6 x 2 x 3 =
48 + 24 + 36 =
Reaction,
130 =
Reaction,
RA =
BM diagram
Just left to C,
6130
6R0
18 kN
24 + 12 - 18 = 18 kN
MA = °
MB = 18 x 2 = 36 kNm
M3 = 18 x 3 - 24 x 1- 6 x 1 x 0.5 = 54 - 24 - 3 = 27 kNm
M4 = 18 x 4 - 24 x 2 - 6 x 2 x 1 = 72 - 48- 12 = 12 kNm
Deflection of Beams
Strength of Materials
4
101
M4' = 12 + 24 = 36 kNm (Just right to C)
MD = °
36 kNm
A
27
36 kNm
B
BMD
Fig. 6.12
Deflection at central point
Extend UDL to portion CD and apply -w from C to Das shown.
MC = 18x - 24(x - 2)- --t4 (x - 2)2 + 1.;1-(x - 4)2 + 24(x - 4)°
2
W = 6 kN/m
El
d y
x2 = 18x - 24(x - 2)- 3(x- 2)2 + 3(x - 4)2 + 24 (x - 4)°
d
dy
EI — = 9x2 - 12(x - 2)2 -(x- 2)2 + (x - 4)3 + 24(x - 4) +
dx
3
Ely = 3x3 - 4(x -2)
(x - 2) 4
4
+
(x-44)4
+12(x -4)2 +Cix+ C2
x = 0, y = 0 Constant C2 = 0
y= Oatx= 6 m
24 +12x22 +6Ci
0 = 3 x 63 - 4(4)3 --•-• -+4 4
Constant,
0 = 648 - 256 - 64 + 4 + 48 + 6C1
0 = 380 + 6C1
C1 = -63.33
3
Ely = 3x3 - 4(x -2)
0y3 = 3 x 33 - 4 -
4
4
(x - 2) 4
4
+
(x - 4) 4
+12(x -4)2 - 63.33x
4
+ omitted + omitted term - 63.33x
= 81-4-O.25-190=-113.25
81 -4El = 20000 kNm2
113.25
Y3 - 20000
= 5.66x10-3 m= - 5.66 mm
Q.6.12 A cantilever beam AC of span '3a' and carrying a concentrated load 'P at distance '2a' from the
fixed end A is strengthened by supporting its free end on the free end of another cantilever BC as
shown in the figure 6.13. Determine percentage reduction in the moment at the support A.
10 2
IP. IAS
& IFS (Objective & Conventional) Previous Solved Questions
MADE EASY
IC
AI
2a
—.4.-- 2a
--id
Fig. 6.13
[CSE-Mains, 2002, CE : 20 Marks]
Solution:
MA = -2Pa (without strengthening)
Say reaction at C = RT
Take a section at a distances of x from C
d2y
2c1— = R xx- P(x- a)
dx 2
dy
Rx 2 P(x 2E1— =
2
dx
2
+Ci
Fig, 6.14
dy
= Oatx= 3a
dx
R9a2 P
(4a2)+Ci
0=
Ci 2 2
Constant,
= 2Pa2 - 4.5Ra2
2
dy
2E1— = Rx
dx
2
P (x -a)2 .1- 2Pa2 - 4.5a2R
2
Rx3 P
2Ely = — --(x - a)3 +2Pa 2 x - 4.5Ra2x + C2
6 6
y= Oatx= 3a
0 = 4.5Ra
C2 = 9Ra
3
3
3
3
-- Pa +6Pa -13.5Re + C2
6
14 3
- 3Pa
x3
2Ely =
But
or
or
P
3 14
2
--(x - a) 3 + 2Pa x — 4.5Ra2x +9Ra - — Pa
3
6
6
y= 0 at x = 0, from C
14
= 9Ra3 - -5-Pa- , Omitting Ilnd term
R=
14P
27
Deflection of Beams
Strength of Materials
1 03
14
14
4
- —
MA =
27 x P x 3a - 2Pa = + 9 Pa - 2Pa = -- Pa
Bending moment,
Pa
4
-Reduction in BM = 12Pal H--Pa
9
149
14
7
1
% reduction = — Pax — x 100 = - x100 = 77.78%
9
9
2Pa
0.6.13 Compare the elastic deformations at the points of loading produced in a simply supported beam
of length L., carrying a concentrated load Pet its mid-point to that of a cantilever of the length L,
carrying a load P at its free end. The moment of inertia of the section of the simply supported
beam is J in the left half and 0.5 J in the right half, For the cantilever, the moment of inertia of the
section Is 2J upto a distance of 112 from the fixed end and J for the remaining portion. Assume
that both the beams are of the same material.
[CSE-Mains, 1990, ME : 20 Marks]
Solution:
BMD
Fig, 0,15
For S,S, Beam
Fig. 6.16
E
d2y
P
=
2J
dx2
dy
c— =
dx
2
4j
P
0.5J
p(
L.) Px _2P( L
= 2J J
2
- -j- x -
+C
213 Cix + C2
3J k
Y= 0, x = 0, C2 = 0 (Omitting lind term)
Y= 0, X = L
Ey =
0=
• =
Ey =
3
J
P2
PL3 PL
P L3 P L3 r,
3 r,
X — - — X + = — - — + iL
12 J 3J 8
12J 24J
PL3 1 _
- 24J x — 24J
Px 3 P
L)3 PL2
- 24J
104 10. IAS & IFS (Objective & Conventional) Previous Solved Questions
Deflection at centre (x= L/2)
PL3 PL3 PL3
EYc
96J 48J = 96J
PL3
96EJ
Y0
For Cantilever Beam
d2 y
E — xdx
dx 2
Px 2dx
px3 L/2
d - yi B E[xJ
dx
A
E[(L.iB - y B L) - (0 x i A - y
+Ey A =
3J o
px3 L
6J
PL3
PL3
24J
6J
+
L/2
PL3
48J
3
:J+ l a =
,
where iB= 0, YB = 0
PL3
2 -4
3
YA =
i L Px2dx
JI-I2 2J
16
x
x J
PL3
EJ
Q.6.14 For the beam shown in the figure 6.17 compute
the deflection of end B by force method.
[CSE-Mains, 2013, CE : 15 Marks]
Fixed
end
Solution:
Consider a section at a distance x from A
wx 2
- Rx
M = bending moment = Rx2
where W= wL, total load,
El
d2y
dx 2
w x2
L
2L
Fig. 6.17
W 2
_ Rx-- x
2L
dy = Rx 2 Wx 3 r ,
El —
dx 2 6L
dy =
0 at x = L
dx
0=
Constant,
C1 dy
El — =
dx
EIy =
RL2 WL3 c
2
6L
1
WL2 RL2
6
2
Rx2
Wx3
RL2
2 6L 6 2
Rx3 Wx4 WL2
6
24L + 6
Y= 0, x = L
RL2
,-,
x 2 x+u2
Deflection of Beams
Strength of Materials
0=
RL3 WL3 WL3 RL3
6 24 6 2
RL3
3
Constant,
+
WL3
8
+L.2
+ C2
_ WL3 RL3
8
3
C2
Rx3 Wx 4 WL2 RL2 WL3
+
x
6 24L 6 2 8
y = —8, x = 0
So,
EIy
At end A,
10 5
RL3
3
WL3 RL3
+
8
3
2(-5)
R = k8
—81E/ +
—E18 =
WL3 k8L3
8 4- 3
k
WL3
8
3
=—
WL3
8=
81E/ + 1(12
3
Objective Questions
Q.1
Q.2
The ratio of area under the bending moment
diagram to the flexural rigidity between two
points along a beam gives the change in
(a) deflection
(b) slope
(c) shear force
(d) bending moment
[CSE-Prelims, ME : 1998]
(a)
(c) 3M/2
2E1
Q.3
For cantilever beam shown in the given figure
6.18, the deflection at C due to a couple M
applied at B is equal to
M
A
(b)
EI
(d) 2M/2
El
[CSE-Prelims, CE : 2003]
In a simply supported beam AB of span L, the
mid point is C. In case-1, the beam is loaded
by a concentrated load W. In case-2, the beam
is subjected to a UDL of intensity w such the
wL = W. The ratio of central deflection in case 1
to that in case-2 is
5
(a) —
3
(b) 3
5
(c) —
8
(d) 8
[CSE-Prelims, CE : 2003]
El = Constant
Q.4
Fig. 6.18
M/ 2
2E/
A simply supported beam AB of span L is
subjected to a concentrated load W at the
centre C of the span. According to Mohr's
10 6 0.
IAS & IFS (Objective & Conventional) Previous Solved Questions
moment area method, which of the following
gives the deflection under the load?
(a) Moment of the area of M/E1 diagram
between A and C taken about C
(b) Moment of the area of M1E1 diagram
between A and B taken about B
(c) Moment of the area of M/E1 diagram
between A and B taken about A
(d) Moment of the area of M1E1 diagram
between A and C taken about A
[CSE-Prelims, CE : 2005]
Q.5
Q.6
(a)
12EIA
6EIA
(b)
/3
4E1A
(C)
3
3E/A
3
/
[CSE-Prelims, CE : 2006]
(d)
3
A propped cantilever is acted upon by a
moment M0 at the propped end. What is the
prop reaction?
Q.7
A cantilever has rectangular cross-section and
supports concentrated load at its free end
initially. If depth and width of the beam section
are doubled, the deflection at free end of the
cantilever will reduce to what percentage of the
initial deflection?
(a) 20%
(b) 15.72%
(c) 9.57%
(d) 6.25%
[CSE-Prelims, CE : 2005]
Due to settlement of support at A of propped
cantilever shown in the figure 6.19 given below,
what is the vertical reaction at B?
MADE EASY
(a)
0.5M0
L
(b) Mo
(c) 1.5M0
(d) 2.0Mo
L
[CSE-Prelims, CE : 2007]
A simply supported beam AB of span L carries
a concentrated load Wat the centre of the beam
with constant El throughout. What does the
moment of the area of the M/Eldiagram between
A and C, taken about A represent?
[CSE-Prelims, CE : 2010]
Q.8
B
1
El = Constant
Answers
Fig. 6.19
1. (b)
6. (d)
3. (d)
4. (d)
8. Deflection
2. (c)
7. (c)
Explanations
1.
(b)
3.
El
dx
=
xi
Mdx
(d)
Case-1:
Yc
f" Mdx
(L2' — L1') =
2.
xl El
= change of slope
(c)
+ ML
xL
YOc 2E1 El
3 ML2
— 2 E/
Case-2:
_ WL3
•
48E1
_ 5 wL4
Yc 3
• 84 E/
wL = W
, 5 WL3
Yc —•384 x EI
Yc
8
Yc' — 5
5. (d)
Deflection of Beams
Strength of Materials
107
7. (c)
4. (d)
Moment of area of
Mo
diagram between A and C
taken about A.
--t)
x
Fig. 6.23
L
2
2
EI
y
2
dx
= Px + Mo
dy
El —
dx
Px 2
—+Mox+Ci
2
a
y = 0 at x = L
Fig. 6.21
dx
r,
PL2 A A
0 = — ivio...+L.,1
2
1 WL L(L) WL3
Moment about C= - — x- - =
2 4 2 6 96E/
1 WL L( L) WL3
Moment about A= x— x- - =
48
2 4 2 3
5.
PL2
C1 = ---MoL
2
Px 2 A A
dy
El — =
+ ivi o x
dx
2
(d)
81 =
WL3
3E1
P63
82 =
6.
2
AA
I
ivi oL
p 2
L.
T x-M°Lx+C2
ElY =
I, = (2b)(2d)3 16bd3
=
16/
12
12 =
2
PC
y= 0,x=0, C2 =0
y = 0 at x = L
81 = 6.25%
0=
(d)
PL3
3
P L3 M L2 P L3
2
6
2
Moe
-
M01-2
2
Mo
P = 1.5—
Fig. 6.22
ARB =
RBL3
8. (Deflection)
Deflection at the centre C of the beam = moment
3E/
3EIA
of area of — between A and C taken about A.
E/
M
07
Torsion
CHAPTER
Q.7.1 Design two solid circular shafts to transmit 200 HP each without exceeding a shear stress of
70 MPa at 20 rpm and other at 20,000 rpm. Give your inference about the final results from the
view point of economy. Do you have any other suggestion to improve the economy further?
[CSE-Mains, 2013, CE: 10 Marks]
Solution:
HP = 200
Shaft-1
RPM = 20
2.TE x 20
60
Torque,
Ti _
= 2.0944 rad/sec
200 x746
2.094 = 71251.194 Nm = 71251.194 x 103 Nmm
ic
3
it
71251194 = — xdi X T = — xdi3 x70
16
16
d13 = 5183990.724 mm3
d1 = 172.98 mm, shaft diameter
Shaft-2
RPM = 20,000
27c x 20000
60
Torque,
T2 -
= 2094.4 rad/sec
200 x 746
2094.4 = 71.251 Nm = 71.25 x 103 Nmm
4
4 = 5183.0
X 70,
16
d2 = 17.3 mm, shaft diameter
71251 = 16 cll x
Shaft 1 at 20 rpm
Shaft diameter is too large. There is hardly any machine which runs at 20 rpm for power transmission.
Shaft 2 at 20,000 rpm
Shaft diameter is reduced to one tenth from 173 mm to 17.3 mm. There is saving in material by 99%. But
at high speed of rotation, large centrifugal forces are developed if there is any eccentric mass on shaft,
which will produce large deflection and slope in the shaft, rigidity is drastically reduced. Even to instal
any gear, pulley for power transmission, keyways cannot be made on a shaft of 17.3 mm.
Torsion
Strength of Materials
1 109
For ordinary power transmission in simple machine speeds are 200-500 rpm, say at 400 rpm, shaft
diameter will be approx. 47 mm. To install any gear, pulley on the shaft, keyways can be easily cut.
Q.7.2 Find the diameter of a solid cylindrical shaft subjected to 100 rpm and transmitting 350 kW power,
when is the shear stress not to exceed 90 N/mm2.
What percent saving in weight would be obtained if this shaft is replaced by a hollow one, whose
internal diameter equals to 0.65 of the external diameter, the length, the material and the maximum
shear stress being the same.
[IFS 2012, CE : 15 Marks]
Solution:
Power = 350 kW = 350 x 1000 Nm/s
N, speed = 100 rpm
Cylinder velocity,
Torque transmitted,
T
2rc x 100
= 10.472 rad/sec
60
350 x 1000
10.472 = 33422.54 Nm = 33422.54 x 103 Nm
ti = 90 N/mm2, maximum shear stress
T=
16
d
3
X d 3 x 90
16
d3 = 1.891298 x 106 mm3
d = 123.66 mm, shaft diameter
(b) For same power transmission, same shear stress, polar modulus should be the same
33.422 x 106 =
Zps for solid -
rcd3
16
4 2
[ 4
Zph, for hollow = — D -(0.650 ]32
D
—[0.82149]E)3
16
d3
=1,2173d3
0.82149
D = 1.06774 d
Inter dia. = 0.694 d
Saving in material depends on area of cross-section of solid and hollow shaft
D3 -
Area of solid shaft,
As -
Area of hollow shaft,
Ah =
itd 2
4
4
[(1.06774d)2 - (0.694d)2 ] = Ed2 [1.14 - 0.4816]
4
Ed2 x0.658
4
Ah
AS = 0.658
Weight of hollow shaft is 0.658 x weight of solid shaft
Saving in material = 34.2%.
MADE EASY
110 10- IAS & IFS (Objective & Conventional) Previous Solved Questions
Q.7.3 Compare the weights of equal lengths of hollow and solid shaft to transmit a given torque for the
same maximum shear stress if the inside diameter is 2/3 of the outside diameter.
[IFS 2012, ME : 10 Marks]
Solution:
Say,
d = Diameter of solid shaft
D = External diameter of hollow shaft
2D
= Internal diameter of hollow shaft
3
T = torque
= Maximum shear stress
Solid shaft
=
16T
3
70
...(1)
Hollow shaft
0.5D
(D4
n
=
16T
TCD3
32T 81
32T
=
x
16 D.4) 704 65
81 )
81
65
x
...(ii)
From eq. (i) and eq. (ii)
16T 81
x
tD3 65
16T
7cd3
81
65
...(iii)
D = 1.0761 d
Since length is same, material is the same (not given in problem), weight of shaft depends on area of
cross-section.
3
u x
Ah =
.0761d)2 -(; x 1.0761d)2 ]
= E[1.1580 - 0.51466]c/2 = -7-c10.643341d2
4
4
Ah
W
= 0.64334
=
AS
Ws
Weight of hollow shaft is only 64.334% weight of solid shaft.
0.7.4 A stepped shaft ABC, is 0.8 m long. For a length AB = 0.4 m, shaft diameter is 40 mm and the
length BC = 0.4 m, shaft diameter is 20 mm. Shaft is fixed at both the ends A and C. At the section
B, a torque T is applied which causes a maximum shear stress of 100 MPa in stepped shaft.
Determine the magnitude of torque T.
Torsion
Strength of Materials
41
111
Fixed
A
Fig. 7.1
[CSE-Mains, 2001, ME : 20 Marks]
Solution:
Torque Tapplied at section B is shared by shaft AB and BC, as shown
± T2 = T
...(i)
(since
eAc
=
0)
eAB = eBc
T2 x0.4
T1 x0.4
GJ2
GJ1
J1 = 16 J2 as di = 40 mm, d2 = 20 mm
T1
16J2
(d1 = 2d2 )
T2
J2
= 16T2
T1 = 0.94117 T
T2 = 0.05883 T
Assume that max. shear stress occurs in AB
So,
3
—
TC X 100 x 403 = 1256637 Nmm = 0.94117 T
T = — T x 40
1
16
= 16
Torque,
T = 1335186 Nmm = 1.335 kNm
T2 = 78540.35 Nmm = 16 x r' x 20 3
Then,
= 50 N/mm2 in portion BC. (Less than permissible hence, OK)
T = 1.335 kNm
Hence,
Q.7.5 A shaft of 12 cm external diameter and 8 cm internal diameter is subjected to a bending moment
of 300 kgf-m, twisting moment of 100 kgf-m and a direct thrust of 10,000 kgf. Determine the
maximum principal stress and diameter in which it acts with reference to the axes of the shaft at
the end points P and Q of diameter PQ as shown.
[CSE-Mains, 1987, ME : 30 Marks]
Solution:
D = 12cm
d = 8 cm
Area of cross-section = 5- (D2 - d2 ) = 'It-(144 - 64) = 20 TG cm2
4
4
= 207c cm2 (Comp.)
10,000
= 159.155 kgf/cm2
2rc
O
Bending stress (sagging moment)
Direct stress, ad =
64
(D4
-d 4 )=
(12 4 -84
4 -d
(D
) -1
)
6 cm
Q
Fig. 7.2
112
MADE EASY
► IAS & IFS (Objective & Conventional) Previous Solved Questions
-(20736 - 4096) = 260n cm4
= 1
64
ae = +
My
+ 300 x 100 kgf-cmx 6 cm
I - 2607c
cm4
= ±220.37 kgf/cm2
Shear stress
"C
=
x6
J = Polar moment of inertia = 21= 520n cm4
—
100x100 x6
520n
= 36.73 kg/cm
2
Point P (normal stress = 22037 + 59.155 = 379.525 N/mm2 (comp.) and shear stress = 36.73 kgf/cm2)
Principal stresses
379.525
+
2
/31, p2
379.125
2
2
)
(36.73)2
379.525
+ ab
36.73
= -189.76 ± J36009.8 +1349.1
= -189.76 ± 193.285
ti
= -383.04 kgf/cm2, + 3.525 kgf/cm2
36.73
Point Q (Normal stress = 220.37 - 159.155 = 61.215 kgf/cm2)(tensile) and sehar stress = 36.73 kgf/cm2)
Principal stress
Pi, P2 =
61.215 \I(61.215)2+ (36.73)2
2 ±
= 30.605 ± V936.82 +1349.1
= 30.605 ± 47.81
= +78.416 kgf/cm2, - 17.205 kgf/cm2
36.73 kgf/cm2
Q.7.6 A brass tube fits closely over a steel shaft of 100 mm diameter. Find the thickness of the brass
tube which would ensure that the torque applied to the assembly is shared equally by the two
materials. Find the maximum shear stress in each material and the angle of twist in a length of
3 m. The torque applied is equal to 20 x 104 kgf-cm. Assume modulus of rigidity of steel = 8 x 105
kgf-cm2, modulus of rigidity of brass = 4 x 105 kgf/cm2.
[CSE-Mains, 1990, ME : 20 Marks]
Solution:
GS = 2 GB
Say outer radius of bras tube = R2 cm
es e9
(Since its is parallel combination)
But,
TSLS
TB LB
GSJS
GB JB
Brass
R
Ts = TB as given
Ls = TB
Fig. 7.3
Torsion
Strength of Materials
GsJs =
LIS
GBJE3
GO
-=
4
JE1
JD
X Gs =2
or
JI3 = 2Js
Js' polar moment of inertia of steel shaft
Js = EX
54 = 312.57c cm4
2
II (F? - 5 4 ) = 625 x TC
JB = 2 x 312.5 p= 625x 7C CM 4 = 2
R24 - 625 = 1250
R24 = 1875
R2 = 6.58 cm
Thickness of brass tube = 6.58 - 5 = 1.58 cm
Steel shaft
It x c x d3
Ts = 10x104 kg-cm= —
Torque,
16
10x104 = EXTXr 3 =EXts X 53
2
Ts
Shear stress,
2
- 10 x104 x 2
= 509.3 kgf/cm2
Icx125
Brass shaft
TB = 10 X 104 kg-cm
4
_ 7c R -R/ x T B
2
R2
10 x 104 =
n[6.58 4 -5 41
TC [1875 -625]
625ic
x TB - 2
TB =
XTB
2
6.58
6.58
6.58
Maximum shear stress in brass tube
TB
= 10 x104 x6.58
= 335.116 kg/cm2
625n
0 =
TsL
GsJs
=
10x104 x300
- 0.0382 radian = 2.99°
8x105 x 312.5n
Q.7.7 A shaft circular in section and of length L is subjected to
a variable torque given by kx 2 , where x is the distance
L
measured from one end of the shaft and k is a constant.
Find the angle of twist for the shaft by use of Castigliano's
theorem. Torsional rigidity of the shaft is GJ, see fig. 7.4.
[CSE-Mains, 1994, ME : 20 Marks]
Solution:
Take length dx along the length
Strain energy due to twisting
Fig. 7.4
113
114 ►
MADE EASY,
IAS & IFS (Objective & Conventional) Previous Solved Questions
LT2dx
u= IJO
2GJ
au =
9
2Tdx 1
L kx2
1
GJ f0 L
But T= k x2
L
f L Tdx
GJ 0
ar Jo 2a/
dx -
kL2
kL3
3GJL 3GJ
Q.7.8 Obtain the angle of twist of one end relative to the other end of a tapered shaft having radii ri and
r2 at its ends and of length L. The shaft is subjected to equal and opposite torque Tat it ends.
Modulus of rigidity of shaft is G.
[CSE-Mains, 1995, ME : 30 Marks]
Solution:
r1 = radius at one end
r2 = radius at other end
I
= radius at distance x (from r1 )
=
+ kx where
k- ro
I
L
=
Polar moment of inertia,
x
angular fracture length, dx,
T
T
L(
dx
kx) 4
Torque
Fig. 7.5 [
2Tdx
dO =
G1C(ri
L
0
x
`4
r4 = 7
2 x
2
T
Here, dO =
+ kx)4
2Tdx A =
Tdx]
GJ
2T L
S (ri + kx)-4 dx
° TtG(r1 + kx)- nG °
L
=
2T [(ri + kx) 3
TcG
=
-3k
=
o
2T 1
37ckG[ ( ri + k x\3 1
1 0
1
11
2T [ 1 1 ]
2T [
=+
ankG (ri + kL)3 r13
3TckG r13 q
2T [r? - r1
3nkG ri3r?
_
=
L
(r2 - ri )(ri2 + ri r2 + r22 )
2T
x
anG (r2 -r1)
r13r2
2TL [ rig + rir2 +
anG
3 3
ri r2
d1
]
Q.7.9 A solid circular uniformly tapered shaft of length L with a small angle of taper is subjected to a
torque T. The diameter at the two ends of the shaft are d and 1.2d. Determine the error introduced
if the angular twist for a given length is determined on the basis of the uniform mean diameter of
the shaft.
[CSE-Mains, 1996, ME : 20 Marks]
Torsion
Strength of Materials
115
Solution:
Tapered shaft 1.2d to d (di = d, d2 = 1.2d)
+d1d2
3271
x
3nG
0-
d2 +1.2d2 +1.44d2 ]
d1 d2
d3 x(1.2d)3
32TL [ 3.64 d
x
3itG 1.728
7.152 7±
GJ
0', on the basis of mean diameter 1.1d
32TL
TL
0' = rc(1.1 A G = 6.9571GJd 7.152 - 6.9531
x 100 = 2.78%
7.152
% Error _
Q.7.10 For the shaft loaded as shown in figure 7.6, calculate the maximum shear stress induced and the
angle of twist for cross-section at A value of modulus of rigidity is G.
4M
M
C
t
2t
3d
t
B
O(A
2L
0-14
2L
L
Fig. 7.6
[CSE-Mains, 1994, ME : 20 Marks]
Solution:
Polar moment of inertia
nd 4
J
1
it(2d)4
J,32 - 32
= 16,11,
7C (3d)4
J, =
- 32
= 81J1
4M - T2
JZ
A
M2L
Fig. 7.7
Let torque at 0 is Ti and at D is T2 as shown,
Therefore in portion AB
M- T1 = 4 M- T2
T2 - T1 = 3 M
°Ao OBA + °DB = 0
x 2L (114-7-1)2L + T2 L
L
1 J2 J3
Ji
J2
=0
116
P.
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
T1 x2L
(M-Ti )x2L
J1
16,11
Lx2Ti
Lx2Ti -
+T2
on- Ti)2L
16
M x2L
16
17
x 2LTi
16
+
+
IL
16J1
T2 x L
16
Tix2L
16
2ML
4.
16
17
8
+
+
T2L
16
L )
81J1
T2 X L
81
+
T2L
81
(81+16)T2L
16x81
M
T1
+
8
+
= 0
= 0
= 0
= 0
97T2
16 x 81
=
M
97T2
17.,
/1 +
8
16 x 81
Also
or
0
...(ii)
8
M 8
97 1
8
—
T1+- x- x-T2 = 8
x1
16 81 17
M
Ti + 0.03522 T2
17
0.0588 M
Ti + 0.03522 T2
T2 - Ti
3M
T2
2.9547 M
-0.0453 M
7-1
From eq. (ii)
Ti x 2L = -0.0453M x 2L
e AO
GJ
0.0453M x 2Lx 32 = 0.9228ML
Gxrcd 4
Gd 4
16T2 _ 16 x 2.9547M 1.881M
tin portion BC
n(2d)3
8icd3
d3
16T2 = 0.557 M
ti in portion CD
Hence, tmax is in portion on BC
d3
TC(3d)3
1.881M
-
d3
Q.7.11 A cantilever ABC, 200 mm long and 20 mm diameter is fixed at end A as shown in the below figure
7.8. A horizontal axial load of 10 kN, a vertical load of 1 kN acts at end C. Torque T applied at
section C, produces a maximum shear stress of intensity 50 MPa in cantilever. Determine principal
stresses at point B of cantilever.
10 kN
H
100 mm-01
100 mm
Fig. 7.8
[CSE-Mains, 2001, ME : 20 Marks]
Torsion
Strength of Materials
41
117
Solution:
Axial compressive force = 10,000 N
7C X 202 = 100 nmm2
A= —
4
as = compressive stress, uniform
Area,
10,000
1007c
31.83 MPa
(-ve sign indicate compressive nature)
BM at B = 1 kN x 100 = 1000,00 Nmm
Z =
ird 3
IC
3
=
x 20 = 0.7854 x 103 mm3
32
32
1000,00
= 127.324 MPa (at point B)
CS,
" - 0.7854 x 10°
(Tensile at upper fibre)
Net direct stress at B, (upper fibre) aB = 127.324-31.830 = 95.494 MPa
Shear stress due to torque,
ti = 50 MPa
Bending stress,
Principal stress at B, !Du p2 =
\ (GB)2
2 - 2
= 47.747 ± V47.7472 + 502
= 47.747+ V2279.8 + 2500 = 47.747 ± 69.136
pi = 116.883 MPa
1)2 = -21.389 MPa
Q.7.12 A solid circular shaft is encased in a hollow copper shaft so as to make a compound shaft. The
diameter of steel shaft is 8 cm and outside diameter of the copper shaft in 11 cm. The compound
shaft is 2 m long and is subjected to a torque of 8 kNm. Determine maximum shear stress in steel
and copper.
Gsteel = 2 Gcopper = 80 kN/rne
[CSE-Mains, 2007, ME : 20 Marks]
Solution:
Solid steel shaft
d = 8 cm
rcd 4 Tcx8 4
J S — 32 - 32
128ncm4
Hollow copper shaft
d= 8, D = 11 cm
(114 - 84 ) =
(14641- 4096) = 329.53n cm4
32
32
Gs = Gcu x 2
Angle of twist will be equal in both shafts
JCL/ =
°s = esu
TsLs
JsGs
T„L„
J„G„
118 0. lAS & IFS (Objective & Conventional) Previous Solved Questions
Ls
Ts
Tcu
= Lcu
2
128ir
Js x Gs
x = 0.778
Gcu
329.1341
1
Jcu
Ts = 0.778 Tcu
Ts +Tcu = 8 kNm
1.778 Tcu = 8
Tc„ = 4.50 kNm
Ts = 3.50 kNm
Also,
Stress in solid steel shaft
s
3.5 x106
7C
x
, = 43.51 N/mm2
- 16 128 x10- Tc
Maximum shear stress in solid steel shaft
Copper shaft
Ta, = JCL, X
55 = 329.5371 x 104
tcu
55
X
CU
- 329.537c x104 Tcu
55
tai = 23.90 N/mm2, Maximum shear stress in hollow copper shaft.
4.5 x 106
Q.7.13 A mild steel shaft of 200 mm diameter is to be replaced by a hollow shaft of alloy steel for which
the allowable shear stress is 25% greater. If the power to be transmitted is to be increased by
20% and speed of rotation increased by 5%, determine the maximum internal diameter of the
hollow shaft taking its external diameter to be limited to 200 mm.
[CSE-Mains, 1998, ME : 30 Marks]
Solution:
Tc
Solid shaft
3
—(200) X
= 16
P= coT
=T
PH = 1.2 P
(4-/ = 1.05 w
Hollow shaft
T=
1.2P
1.05w
it
1 .05
it [200 4 - d 4 ] x 1.25T
P
x
= (200)3 ti [Since Ts = TH]
— =
w
16 1.2
16
200
or
or
...(ii)
1.05
1.2
(2004 -d4 )
- 2003
200
1.09375[2004 - c/4] = 2004
(1.09375-1)2004 = 1.09375 d4
(0.09375)2004 = 1.09375 d4
x1.25
_
4
0.09375
x 200
1.09375
d = 0.541 x 200 = 108.2 mm, internal diameter of hollow shaft.
Torsion
Strength of Materials
1 119
Q.7.14 The pulley A exerts a torque on shaft B as shown in figure 7.9. The total vertical tension on both
sides of the belt on each pulley is 400 kg-f. Diameter of the shaft is 6 cm. If the tensile and shear
stress intensities are not to exceed 3200 kgf/cm2 and 1600 kgf/cm2 respectively, what is the
maximum power that can be transmitted by the shaft when running at 150 rpm. Shaft may be
assumed to the simply supported at the bearings C and D.
400 kgf
Rc
400 kgf
RD = 66.667 kgf
Fig. 7.9
[CSE-Mains, 1989, ME : 30 Marks]
Solution:
Reaction
Moment about C
RD x 0.9 + 400 x 0.3 = 400 x 0.45
60
RD = —
0 . 9 = 66.667 kgf
Rc = 733.33 kgf
BM diagram
MA
Mc = 400 x 0.3 = - 120 kgf-m (Hogging)
MB = +66.67 x 0.45 = 30 kfg-m
Mmax = 120 kgf-m = 120,00 kg-cm
crmax = 3200 kgf/cm2
Amax = 1600 kgf/cm2
Me =
Equivalent BM = 32 X d 3 X a
ic
3
= --X 6 x 3200 = 67858.4 kgf/cm
32
Te = Equivalent TM =
16 x
d 3 xTmax
It
3
= — x 6 x 1600 = 67858.4 kgf-cm=VT2 + Mmax 2
16
(Putting the value of Mmax)
= 46047626629 = 144000000 + T 2
T2 = 4460762629
T = 66789 kgf-cm = 667.89 kf-m
Mi4T2 I- M 2
Me =
2
=
67858.4
120 N.
MADE EASY
IAS & IFS (Objective & Conventional) Previous Solved Questions
2 x 67858.4 = 120,00 + VM2 +T2
123716.8 = Vm2 + T2 = V12000 2 + T2
T = 123133.4 kgf-cm = 1231.33 kgf-m
or
Thus value of Twill for greater than 667.89 kgf-m (hence permissible value for shear stress will reach first
than permissib;e value of tensile stress)
2ic x 150
= 15.708 rad/sec
60
Power = 667.89 x 15.708 kgf-m/s = 10491.2 kfg-m/s
= 140 HP, as 1 HP = 75 kfg-m
=
Objective Questions
Q.1
Two steel shafts, one solid of diameter D and
other hollow of outside diameter D and inside
diameter DI2 are twisted to the same angle of
twist per unit length. Ratio of maximum shear
stress in solid shaft to that in the hollow shaft is
4
(a) 4
(c)
16
15
Q.3
A stepped solid circular shaft shown in the figure
7.11 is built in at its ends and is subjected to a
torque To at the shoulder section. The ratio of
reactive torque Ti and T2 at ends is (J1 and J2
are polarmoment of inertia)
T,
(b) 8
(d) 1
[CSE-Prelims, ME : 1998]
Q.2
F
A load perpendicular to the plane of the handle
is applied at the free end as shown in the figure
7.10. The value of (shear force), bending moment
(BM) and torque (T) have been determined
respectively on 400 N, 340 Nm and 100 Nm by
a student. Among the values these are
(a)
(b)
(c)
(d)
Fig. 7.10
SF, BM and torque are correct
SF and BM are correct
BM and torque are correct
SF and torque are correct
[CSE-Prelims, ME : 1994]
L2 -1.1
Fig. 7.11
J 2L2
(a)
jiLi
J1L2
(c) J2L1
Q.4
400 N
1"
J2L1
(b) J1L2
J1L1
(d)
J21-2
[CSE-Prelims, ME : 2001]
Steel shaft and brass shaft of same length and
diameter are connected by a flange coupling.
The assembly is rigidity held at its ends and is
twisted by a torque through the coupling.
Modulus of rigidity of steel is twice that of
brass. If torque of the steel shaft is 500 Nm,
then the value of the torque in brass shaft will be
(a) 250 Nm
(b) 354 Nm
(c) 500 Nm
(d) 708 Nm
[CSE-Prelims, ME : 2001]
Q.5 A shaft is subjected to a bending moment of
M = 400 Nm and a torque = 300 Nm. The
equivalent bending moment is
Torsion
Strength of Materials
A hollow shaft of length L is fixed at its both
ends. It is subjected to torque Tat a distance
of L/3 from one end. What is the reaction torque
at the other end of the shaft?
2T
(a) —3--
(b)
(c) 3
(d) 4
2
[CSE-Prelims, ME : 2007]
Q.7
121
(b) 700 Nm
(d) 450 Nm
[CSE-Prelims, ME : 2002]
(a) 900 Nm
(c) 500 Nm
Q.6
4
Consider the following statements, in connection
with a metallic rod of a circular section being
subjected to equal and opposite torque Twithin
elastic limit:
1. The transverse section of the rod does not
experience warping.
2. The diameter of rod does not alter.
3. Angle of relative twist between two sections
is proportional to the lengths between these
sections.
4. A surface element of the rod is under pure
shear state of stress.
Which of these statements are correct?
(a) 1, 2, 3 and 4
(b) 1, 2 and 3 only
(d) 2, 3 and 4 only
(c) 2 and 3 only
[CSE-Prelims, ME : 2009]
Q.8
Which one of the following is true for torsional
shear stress at the axis of a circular shaft?
(a) Minimum
(b) Maximum
(c) Negative
(d) Zero
Q.9
A circular shaft of length 'L' is held at two ends
without rotation. A twisting moment 'T is applied
at a distance l'/3 from left support as shown in
the given figure. The twisting moment in the
portion AB will be
Fig. 7.12
(b) T/3
(d) 2T/3
(a) T
(c) T/2
Q. 10 A circular shaft of length 'L' a uniform crosssectional area 'A' and modulus of rigidity '6' is
subjected to a twisting moment that produces
maximum shear stress
in the shaft. Strain
energy in the shaft is given by the expression
T2AL/kG, where k is equal to
(a) 2
(b) 4
(c) 8
(d) 16
[CSE-Prelims, CE : 2001]
Q. 11 When subjected to a torque, a circular shaft
undergoes a twist of 1° in a length of 1200 mm,
and the maximum shear stress induced is.
80 N/mm2. The modulus of rigidity of the material
of the shaft is 0.8 x 105 N/mm2. What is the
radius of the shaft?
(a) 90/7r mm
(b) 108/Tc mm
(c) 180/Tt mm
(d) 216/7cmm
[CSE-Prelims, CE : 2008]
Q. 12 Power is transmitted through a shaft, rotating at
2.5 Hz (150 rpm). The mean torque on the shaft
is 20 x 103 Nm. What magnitude of power in
kW is transmitted by the shaft?
(a) 501c
(b) 120 TC
(c) 100 Tc
(d) 1507c
[CSE-Prelims, CE : 2009]
Answers
1. (d)
6 . (c)
11. (d)
2. (d)
7. (a)
12. (c)
3. (c)
8. (d)
4. (a)
5. (d)
9. (d) 10. (b)
Explanations
1.
(d)
Ti
J1
T22
G6 =
_. T
=
L
J2
Ti . Ji = TcD4 x
T2 J2 32
32
16
04 04)= 15
It (
16
12 2 0.
lAS & IFS (Objective & Conventional) Previous Solved Questions
MADE EASY
15
T,
16 '
T
2
Ti =
T2 =
16
3
D T
• ( -15 D4 )x_
32 06
2 xi'
) D
Fig. 7.14
• (15 D3 ), ,
t
16 x 16 —3 :7 16 16)
15 7C n
Because,
Ts Ls
—
TO LB
LB
TB
GBJ B
Ts _Gs
TB GB
= e, Ratio = 1
2. (d)
SF = 400 N
BM = 100 Nm
5.
(d)
m = 400 Nm
T= 300 Nm
Mc =
400+,4002 + 3002
2
= 450 Nm
400 N
6.
(c)
T= 100 Nm
80 Nm
400 N
400 N
T= 100 Nm
Fig. 7.15
Fig. 7.13
°BA = °BC
3.
(c)
0
TA
T1 L1 = T2 L 2
GJ1 GL/2
GJ
x
L
3
—
TB X 2L
GJ x 3
TA = 2TB
T1 = J 1 L 2
T2
J2 L1
TA + TB = T
TB = T/3
4.
(a)
T =G
Ts
s =2
TB GB
Ts = 2TB
TB = 0.5 Ts = 250 Nm
7.
(a)
All statements 1, 2, 3 and 4 are correct.
8.
(d)
At the axis of the bar, torsional shear stress is
zero.
Torsion
Strength of Materials
4
12 3
11. (d)
9. (d)
TAB =
L = 1200 mm
2T
0=
180
ti = 80 N/mm2
G = 0.8 x 105 N/mm2
T
Bc = 3
Because
TAB x L
GJ
TBc 2L
3 — GJ 3
TAB = 2 TBc
T
GO
R
L
T1._
R= GO
TAB 4- TBC = T
TAB =
=
80 x 1200
x180
0.8 x105 xic
= 216/ir mm
2T
12. (c)
10. (b)
Strain energy, U =
7C
T2AL
4G
co = 2.5 x 27c = 57r rad/sec
T= 20 x 103 Nm
Power = Tco Nm/s
= 100 x 7C X 103 W
= 100 7c kW
11•IM•
08
Springs
CHAPTER
Q.8.1 A close coiled helical spring is of 80 mm mean coil diameter. The spring extends by 37.55 mm
when loaded axially by a weight of 500 N. There is angular rotation of 45° when the spring is
subjected to an axial couple of 20.0 Nm. Determine the Poisson's ratio of the material of the
spring.
[CSE-Mains, 1996, ME : 20 Marks]
Solution:
Axial load,
Stiffness,
W = 500N
Extended load = 37.75 mm
k_
Axial couple,
500
= 13.265 N/mm
37 .75
Gd 4
8nD3
M = 20 x 103 Nmm
0 = Angular rotation =
ML _ 7C
El - 4
L = 27EnR =TcnD
nd4
64
D = 80 mm
/
13.245 -
G=
G x d4
8n x 803
13.245 x 8n x 803 n
x 54251520
=
d4
d4
ML
El
th
1.
E=
ML 20x103 xrcx nx80x4x64
=
(t)./
it xlcd4
n
= —, x130379724.14
cr
Springs •
Strength of Materials
G=
d" x
12 5
54251520
E
130379729.4
= 2.4032 = 2(1+v)
G - 54251520
v = 0.2016, Poisson's ratio
Q.8.2 A close coiled spring has coil diameter to wire diameter ratio of 6. Spring deflects 30 mm under a
load of 500 N and maximum shear stress is not to exceed 350 MPa. Find diameter and length of
wire required. Modulus of rigidity of wire material = 80 GPa.
[CSE-Mains, 2012, ME : 10 Marks]
Solution:
G = 80,000 N/mm2
k = Stiffness = 300 = 16.6667 N/mm
Gd 4
8nD3
D = coil diameter
d = number of coil
Where,
16.6667 =
80000
(d)3
xdx
8n
D
1000 d
d xi 1 )3 =
10000x —
n
6 216 x n
d
— = 16.6667 x 216 = 0.36
n
10000
D
3
500x — - —
7c x d
2
16
Torque
X "C
3
7E X d x 350
250 D = —
16
250 xr) = 6 d2 x350
116
d
16
1
x 350
— =21.82696
It
d = 4.67 mm, wire diameter
d2 = 250 x 6 x
nLength of the wire required,
d = 13 , number of turns
0.36
L = icDn = It x 6 x 4.67 x 13= 1144.36 mm
Q.8.3 Design a suitable helical spring for a balance which is used to measure 0 - 100 kg over a scale
of 80 mm. Spring is to be enclosed in a casing of 25 mm diameter. Approximate number of turns
is 30. Also calculate the maximum shear stress induced, G = 0.85 x 105 N/mm2.
[IFS 2011, ME : 10 Marks]
126
0.
MADE, EA
1AS & IFS (Objective & Conventional) Previous Solved Questions
Solution:
Load = 0-100x9.8=0-980N
Range = 80 mm
980
k = — = 12.25 N/mm
80
Stiffness of spring,
k -
Gd 4
8nD3
G = Shear modulus = 85000 N/mm2
Where,
d = wire diameter
n = number of coils = 30
Casing diameter = 25 mm
D = 25 - d
So, mean coil diameter
Putting the values in k expression
12.25 -
85000 x d4
8 x30 xD3
D3 = 28.91156 d4
D = 3.069 C/4/3
25 - d = 3.069 d4/3
25 = d + 3.069 d4/3
Say,
d= 4 mm
RHS = 4 + 19.486 = 23.486
d = 4.1 mm
RHS = 4.1 + 20.138 = 24.138
d = 4.2 mm
RHS = 4.2 + 20.795 = 24.995
So, wire diameter,
d= 4.2 mm
Coil diameter,
D= 25 - 4.2 = 20.8 mm
980 =
16
d3T
980 x 20.8 x 16 = 700 MPa
2xnx 4.23
For a closed coil helical spring under compression, illustrate the stress distribution across
wire diameter.
(ii) Differentiate between compound and composite helical spring.
(iii) The spring load against which a valve is opened is provided by an inner helical spring
arranged within and concentric with an outer helical spring. Free length of inner spring is
6 mm larger than that of outer helical spring. Outer spring has 12 coils of mean diameter
25 mm, wire diameter 3 mm and initial compression 5 mm when the valve is closed. Find the
stiffness of the outer spring, if the greatest force required to open the valve of 10 mm is
150 N.
If the radial clearance between the springs is 1.6 mm. Find the wire diameter of the inner spring,
if it has 10 coils. For both springs, G = 82000 N/mm2.
[CSE-Mains, 2002, ME : 30 Marks]
Q.8.4 (i)
Springs
Strength of Materials
12 7
Solution:
(i)
On wire diameter, Ts = torsional shear stress =
16WR
703
R = mean coil radius
4W
Td = Direct shear stress — icd2
At inner coil radius, resultant shear stress
'Cr = 'Cs + "Cd
At outer coil radius, resultant shear stress,
Tr = is — Td
(ii) Compound spring
Springs in parallel are called compound spring
keq = k1 + k2
Composite spring
Fig. 8.1 Shear stress distributions
kik2
kequivalent = k1 + k2
Where k1 and k2 are stiffnesses of individual springs
(iii)
Initial compression in outer spring = 5 mm
Initial compression in inner spring = 5 + 6 = 11 mm
(As the free length of the inner spring is 6 mm more than the free length of the outer spring)
Say
k1 = Stiffness of inner spring in N/mm
k2 = Stiffness of center spring in N/mm
Initial load on valve = 11k1 + 5k2
Outer spring
n= 12, R = 12.5, d = 3 mm
k2 =
Gd4
82000 x 34
64nR3 64 x12 x 12.53
= 4.428 N/mm
The valve is opened by 10 mm, additional force required to open the value
F0 = 10 ki + 10 k2
Ft = Total force = 10 + 10 k2 + 11
= 21 k1 + 15 k2 = 150 N
k2 = 4.428 N/mm
15 k2 = 66.42
21 k1 = 150 — 66.41 = 83.58 N
k1 —
—
83.58
= 3.98 N/mm
21
Gd4 = 82000 x
16n1F3 16 x10 xl:q
Radial clearance between the springs = 1.6 mm
Outer spring,
R2 = 12.5 mm
Radial clearance = 1.6 mm
Outer coil radius of inner spring = 12.5 — 1.6 = 10.9 mm
mean coil radius of inner spring -= (10.9 — 0.5d1) mm
+ 5 k2
128 ••
MADE EASY
IAS & IFS (Objective & Conventional) Previous Solved Questions
82000 x dj
3.98 —
160 x (10.9 — 0.5d)3
82000 d4
x 1 = 128.770
160
3.98
10.9 — 0.5d1 = 5.049 d11.33
5.049 d11.33 + 0.5 di = 10.9
(10.9 — 0.5d1)3 —
say
d1 = 2 mm
LHS = 13.72
d1 = 1.8 mm
LHS = 11.95
d1 = 1.7 mm
LHS = 11.095
d1 = 1.68 mm
LHS = 10.923
Wire diameter of inner coil spring is 1.68 mm
Objective Questions
Q.1
Two identical springs, each of stiffness k are
assembled as shown in the figure 7.2.
Combined stiffness of assembly is
1
2 1
2
(a) — kx2 - - kX1
2
2
(b) 2 k(xi — x2 )2
1
2, + xi)2
(c) —
(d) k
(x1
X2
2
[CSE-Prelims, ME : 1999]
Q.4
A close coiled helical spring has wire diameter
10 mm and spring index 5. If the spring contains
10 turns, then the length of the spring wire
would be
(a) 100 mm
(b) 157 mm
(c) 500 m
(d) 1570 mm
Q.5
Match List-I (Type of spring) with List-II
(Application) and select the correct answer
List-I
A. Leaf/helical spring
B. Spiral springs
C. Beleville springs
List-II
1. Automobile/Railway coaching
2. Shearing machine
3. Watches
Codes:
A
B
C
2
3
(a) 1
2
(b) 1
3
1
2
(c) 3
3
1
(d) 2
[CSE-Prelims, ME : 2000]
k2
Fig. 7.2
(a) k2
(c) k
Q.2
(b) 2k
(d) k/2
[CSE-Prelims, ME : 1998]
Two close coiled springs are subjected to same
axial force. If the second spring has 4 times
the coil diameter, double the wire diameter and
double the number of coils of the first spring,
then the ratio of deflection of the second spring
to that of the first will be
(b) 2
(a) 8
(d) 1/16
(c) 1/2
[CSE-Prelims, ME :1998]
0.3 A spring of stiffness k is extended from a
displacement x1 to a displacement x2 . Work
done by the spring is
Springs
Strength of Materials
Q.6
The equivalent spring stiffness for the system
shown in figure 8.3. CS is the spring stiffness
of each of the three springs.
(a) S/2
(c) 2S/3
Q.7
Q.8
Fig. 8.3
(b) S/3
(d) S
[CSE-Prelims, ME : 2001]
If the number of turns in a spring are halved, its
stiffness is
(a) halved
(b) doubled
(c) increased from turns
(d) not clayed
[CSE-Prelims, ME : 2007]
A closed coiled helical tension spring having
20 coils is cut in two parts such that part A has
8 coils and part B has 12 coils. What is the
ratio of stiffness of the spring A to the stiffness
of the spring B
(a) 1
(b) 1/2
(c) 3/2
(d) 3/4
[CSE-Prelims, ME : 2006]
Q.9
A close-coiled helical spring has 100 mm mean
diameter and is made of 20 turns of 10 mm
diameter steel wire. The spring carries an axial
load of 100 N. Modulus of rigidity is 84 GPa.
The shearing stress developed in the spring in
N/mm2 is
(b) 160//n
(a) 120/it
(c) 100/it
(d) 80/7c
[CSE-Prelims, CE : 2003]
Q.10 For a laminated spring, 1= length, n = number
of plates, width of each plate = b, thickness of
each plate = t, central point load = W What is
the expression for the maximum bending stress?
(a) f =
(c) f
3WI
,
4nkt
(b) f =
2W1
(d) f =
(b)
Same deflection 8 for both, k = k1 + k2 = 2k
3W1
Q.11 A close coiled helical spring is subjected to an
axial moment M, producing an angle of rotation
of 90° at free end with respect to the fixed end.
Strain energy absorbed in spring is 1007t Nmm,
what is M
(a) 100 Nm
(b) 200 Nmm
(c) 300 Nmm
(d) 400 Nmm
Answers
1. (b)
6. (c)
11. (d)
2. (a)
7. (b)
3. (a)
8. (c)
2. (a)
k,
Gd 4
64nR3
W
k1 =
W = 88
k2
1
Fig. 8.4
3W1
2nb2t
2nbt 2
[CSE-Prelims, CE : 2007]
3nb2t
Explanations
1.
129
4. (d) 5. (b)
9. (d) 10. (d)
130 0-
k, =
Gx16d4
7.
64 x 2n x 64R3
(b)
Doubled
Gd 4
Gd 4
16
64nR3 x 128
Gd 4 [11
64nR3 L8
3.
MADE EASY
IAS & IFS (Objective & Conventional) Previous Solved Questions
=x
64 Ran
n is in denominator.
8.
(c)
(a)
A = 8 coil
B =12 coils
kA oc
1
8
,
r‘B c'c
k A 12 , 3
= — = 1.0 =
kB
8
-4'i
[4
X2
Fig. 8.5
9.
(d)
D= 100 mm
R 50 mm
d= 10 mm
n = 20
W= 100N
G = 84 GPa
— kx2 2 --kxi2 = Work done
2
2
4.
(d)
L= irDn.nx5dn
= nx5x10x10
= 1570 mm
5.
6.
(b)
A. Leaf/helical spring
B. Spiral Spring
C. Belleville spring
1. Automobiles/Railway coach
3. Watches
2. Shearing machine
12
16WR
rcd3
16 x100 x 50
Tcx103
L = Length
n = number of beam
W = Load
t = thickness
b = width
WL
n6 x
4 = 6 x'
f_
3 WL
x
,
2 nbt`
11. (d)
7r
1
2Mx2 = 10076
S8
`je
2S x S _ 2 s
2S+ S 3
80
10. (d)
(c)
Fig. 8.5
=
M= 400 Nmm
N/mm2
09
Struts and Columns
CHAPTER
Q.9.1 The critical buckling load of a cast iron hollow cylindrical column 3 m in length, when hinged at
both the ends is equal to PkN. When the column is fixed at both the ends, its critical load increases
to (P+ 300) kN. If the ratio of external diameter to internal diameter is 1.25, E= 100 GPa, determine
external diameter of column.
[CSE-Mains, 2012, ME : 12 Marks]
Solution:
Critical load, when both ends hinged = P
Critical load when both ends fixed = 4P = P + 300 kN
So
P = 100 kN
E = 100000 N/mm2
L = 3 m = 3000 mm
P-
100000 -
_
(Since effective length is half)
n2EI
12
ic 2 X100000 XI
30002
30002
Tc
2 —
1
-t--(D4
64
d= 0.8D
911890.653 mm4
-d4 )
[ 4 - 0 4096D4 ] = 0.02898D4
= 64 D
0.02898 D4 = 911890.653
D4 = 3146.49 x 104 mm4
D = 74.89 mm, external diameter of CI column.
I
Q.9.2 A hollow column, 400 mm external diameter and 300 mm internal diameter, is hinged at both
ends. If the length of the column is 5 m, E = 0.75 x 105 N /mm2, factor of safety 5. Rankine's
constants 1/1600 and crushing stress 587 N/mm2, find the safe load, the column can carry
without buckling. Use Euler and Rankine, formulae.
[CSE-Mains, 2005, ME : 30 Marks]
13 2 P.
IAS & IFS (Objective & Conventional) Previous Solved Questions
MADE EASY
Solution:
E = 75 x 103 N/mm2
D = 40 cm
d = 30 m
700
A = 4 (1600 — 900) = -iL = 175 5c cm2
/ = 7 (402 — 302 ) = 8.59 x 104 cm4 = 8.59 x 108 mm4
64
1= 5m
n2E1 5c2 x 75000 x 8.59 x108
P = Euler buckling load — 2 =
50002
= 25433970.54 N = 25433.97 kN
FOS = 5
Pe safe = 5086.8 kN
_ I _ 8.59 x108 = 1.5624 x 104 mm2
— A — 1755tx100
12
50002
k2
k2
PR—
=1600
.A
/2 =
1+ax--g—
k2
587x1757cx100
1
x 1600
1+
1600
= 161360.0 x 100 N = 16136 kN
PR safe = 3227.2 kN
Safe load without buckling = 3227.2 kN
Q.9.3 A vertical column 6 m length is fixed at the base and a clockwise moment of 1.4 kNm is applied at
the top of the column, a horizontal force of P is applied to the column at a height of 3 m above the
base so as to give a CCW moment.
Determine the value of force P so that horizontal deflection at the top of the column and at the
point of application of P shall be equal (i) when the deflections are on the same side (ii) when the
deflections are on the opposite sides of the vertical line through the foot of the column
[CSE-Mains, 1993, ME : 20 Marks]
Solution:
Take a section yy at distance x from A
Mx = 1.4 — P(x — 3)
El
d2y
, = 1.4 — P(x — 3)
Az'
El
dy
= 1.4x --(x— 3)2 +ci
2
dx
dy
0, at x = 6
dx =
0 = 1.4 x 6- 2 (3)2 +ci
Constant,
= 8.4 — 4.5 P+
c1 = 4.5 P— 8.4
Fig. 9.1
Struts and Columns
Strength of Materials
4
133
dy
El- = 1.4x - -(x - 3)2 + (4.5P - 8.4)
dx
2
El y = 0.7x 2 - f(x - 3)2 + (4.5P - 8.4)x+ C2
y = 0, at x = 6, from top
= 0.7 x 36-i x 27 +(4.5 P-8.4)x 6 +C2
0 = 25.2 - 4.5 P+ 27 P- 50.4 + C2
= 22.5 P- 25.2 + C2
C2 = 25.2 - 22.5 P
Deflection at, x = 0, yA =
25.2 - 22.5P
El
Elyc = 0.7 x3 2 -0 + (4.5 P- 8.4) x 3 + 25.2 - 22.5 P
= 6.3+ 13.5 P- 24.6 + 25.2 - 22.5 P
At x = 3 m,
= - 9P+ 6.9
6.9 - 9P
Y,
Deflections on same side
25.2 - 22.5P
6.9 - 9P
El
El
18.3P = 13.5P
P = 1.335 kN
Deflections on opposite sides
9P - 6.9
25.2 - 22.5P
El
El
31.5P = 32.1
= 1.019 kN
(For deflection on opposite side)
Q.9.4 A cast iron column of circular section 20.0 cm external diameter and 2.0 cm thickness and of
height 4.0 meters is required to carry 15.0 kN at an eccentricity of 2.5 cm. If both the ends are
fixed, find the extreme stresses In the column section. E. 100 GPa.
[CSE-Mains, 2011, CE : 15 Marks]
Solution:
P= 15 kN,
D= 20 cm,
d= 16 cm
A, area of cross-section, A = -71(202 -162 ) = = 113.1 cm2
4
15000
ad , direct shear
- 132.626 N/cm2 = -1.32626 N/mm2 (Compressive)
113.1
eccentricity, e = 2.5 cm = 25 mm
4 .1p
2 E/-
Pe
sec = P-+MaX
A Z
le = equivalent length =
I=
64
(D4-
=
64
= 2 m (fixed ends)
(204 -164 )
134 ►
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
= it-[-I 60000 - 65536] = 4637 cm4 = 4637 x 104mm4
64
E I = 100000 x 4637 x 10+4 Nmm2 = 4637 x 109 N/mm2
1/EPI
_
I 15000 = 1 I 15 _
1
\ 4637 x 109 1000 4637
\
1000 x 17.582
1
le X,
=
1
\IP = 1000x
17.582
2
El
1000 x 17.582
sec 3.258° = 1.0016
Z=
100
0.05688 rad = 3.258°
= 4637 x 104 mm4 = 4637 x 100 mm3
p
15000 x 25
x 1.0016 = +0.810 N/mm3
e sec3.258° =
463700
Amax = 1.32626 + 0.810 = 2.13626 N/mm2 (comp.)
Amin = 1.32626-0.810 = 0.51626 N/mm2 (comp.)
Q.9.5 A built up column consists of three ISWB 450 CD
0.794 kN per m connected effectively to act as one
column (below figure).
Determine the safe load carrying capacity of built
up section, if the unsupported length of column is
4.25 m. Properties of ISWB 450 10 0.794 kN/m.
Area = 101.15 cm2
I = 35057.6 cm4
/YY= 17067.7 cm4
tW = Thickness of web = 9.2 mm
f = 250 MPa, yield stress.
Slenderness Ratio
-x
450 mm
10
20
30
40
50
60
70
550
148
145
139
132
122
112
Permissible stress in
axial compression in MPa
fr = 25 MPa.
[CSE-Mains, 2013, CE : 10 Marks]
Solution:
Area of cross section = 3 x 101.15 cm2 = 303.45 cm2
I = 2 x 35057.6 + 1706.7 = 71821.9 cm4
/wPassing through G
= 1 + 2Iyy + 2 x 101.15 (22.5 + 0.46)2
= 35057.6 + 2 x 176.7 + 202.3 x 527.1616
= 35057.6 + 3413.4 + 106644.8 = 145115.8 cm4
<
I yy
71821.9
,,
- 236.684 cm2
K- 303.45
k = 15.38 cm
L = 4.25 m = 425 cm
L
- = 27.63
k
Struts and Columns
Strength of Materials
4
13
5
From the table
f = 145 +
30 - 27.63
10
Safe load =
=
=
=
x 3 = 145 + 0.711 = 145.71 N/mm2
145.71 x 303.45 x 100 N
4421570 N
4421.57 kN
4.42157 MN
Objective Questions
Q.1
Which one of the following statements is
correct?
(a) Eulers formula holds good only for short
columns
(b) A short column is one which has the ratio of
its length to least radius of gyration greater
than 100
A
(c) column with both ends fixed has minimum
equivalent length or effective length
The
equivalent length of column with one
(d)
end fixed and other end hinged is half of
its actual length
[CSE-Prelims, ME : 2000]
Q.2
A strut's cross-sectional area A is subjected to
a load Pat point S(h, k) as shown in the figure.
Stress at the point Q(x, y) is
Q.3
(a) S
(c) R
Which one of the pairs is not correctly matched
(a) Slenderness ratio 1. Ratio of length of
the column/least
radius of gyration.
(b) Buckling Factor 2. Ratio of maximum
load to the
permissible axial
load on the
column.
(c) Short Column
3. A column for which
slenderness ratio
< 32.
(d) Strut
4. A member of a
structure in any
position
and
carrying an axial
compressive load.
[CSE-Prelims, ME : 2003]
Q.5
What is the cause of failure of a short MS strut
under an axial load?
x
y
P + Phy Pkx
+
(a) —
A
Ix I y
(c)
(d)
P Phx Pky
A Iy
P _Phx Pky
A Ix I y
P + Phx Pky
—
A ly I x
[CSE-Prelims, ME : 2000]
(b) Q
(d) P
[CSE-Prelims, ME : 2002]
Q.4
Q(x, y)
S(h, k)
(b)
A short vertical column having a square cross
section is subjected to an axial compressive
force, centre of pressure of which passes
through the point R, as shown in figure.
Maximum compressive stress occurs at point.
13 6 0.
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
(a) Fracture stress
(c) Buckling
(b) 1/2
(d) 1/4
[CSE-Prelims, CE : 2007]
(a) 2
(c) 4
(b) Shear stress
(d) Yielding
[CSE-Prelims, ME : 2007]
Q.6
Which one of the following columns has effective
length twice the value of actual length?
(a) Hinged-Hinged column
(b) Fixed-Fixed column
(c) Fixed-Hinged column
(d) Fixed-Free column
[CSE-Prelims, ME : 2009]
Q.11 In an axially loaded compressive member with
a circular cross-section of radius r, what is the
radius of the core section which is proof against
tensile stress?
(a) r /2
(b) r /3
(d) r /6
(c) r /4
(CSE-Prelims, CE 2009]
Q.7
Critical Euler buckling load for a long column of
diameter D was evaluated as P. If the diameter
of the section is reduced to D/2, what is the
load carrying capacity of the modified column?
(b) P/4
(a) P/2
(d) P/16
(c) P/8
[CSE-Prelims, ME : 2009]
Q.12 A Short column of external diameter D and
internal diameter 01 is subjected to an eccentric
load P at an eccentricity of e thereby causing
tensile stress at the extreme fiber. What is the
magnitude of eccentricity?
D2 +02
D2 +
(a)
(b)
8nD
8D
Q.8
The axial load which just produces the condition
of elastic instability in a column is
(b) Euler load
(a) Rankine load
(c) Yield load
(d) Crushing load
[CSE-Prelims, CE : 2002]
Q.9
If the stress on the cross-section of a circular
short column of diameter D is to be wholly
compressive, the load should be applied within
a concentric circle of diameter
(b) D/8
(a) D/2
(d) D/6
(c) D/4
[CSE-Prelims, CE : 2003]
Q.10 Column C1 has both the ends hinged while the
column C2 has one end hinged and other end
fixed. What is the ratio of the critical load for C1
to that of C2 according to the Euler's formula?
(c)
D2 — D1
(d)
870
80
[CSE-Prelims, CE : 2010]
Q.13 A column of length 4 m, an area of cross section
2000 mm2, moments of inertia, /xx = 720 cm4,
l
ye = 80 cm4, is subjected to a buckling load.
Both the ends of the column are fixed. What is
the slenderness ratio of the column
(b) 120
(a) 200
(d) 80
(c) 100
Answers
1. (c)
6. (d)
11. (c)
2. (b)
7, (d)
12. (b)
3. (a)
8. (c)
13, (c)
Explanations
1.
(p)
Only statement (c) is correct.
In column if the both the ends fixed, has minimum
equivalent length or effective length i.e. 2 .
D2 —D-f
2.
(D)
All compressive in I-quadrant
P Ph x_Pyk
A
1),
1,,
e =h
4. (b) 5. (d)
9. (0) 10. (b)
Struts and Columns 4
Strength of Materials
13 7
10. (b)
(x, y)Q
(h, k)S
---
C1,both ends hinged P1 =
C2,one end fixed, other hinged P2 =
=
C2
e=k
P Px
----xe -PY-xe
A Iy
x
Y
x
1 1 . (c)
Column of circular section of radius r
(Compressive)
(a)
p P(QR)
x (QS)
+
°max = + ab =A
I YY
4.
(b)
(b) is not correctly matched
5.
External diameter D,
2
2
Internal diameter 01, A= 7C (D - D )
4
= 1`— (D4 —D41 )
64
(d)
4P
a
(d)
n(D 2 - Of)
c
for a fixed, free column, effective length is twice
the actual length.
7.
4
12. (b)
ms strut short in length, fails by yielding
6.
P = 0.5
P2
radius of cone =
3.
n2E1
ab —
(d)
P.e x x 64
2 x n(D4 -D4)
32 Pe D _
m(D4 -D11) n(D2+EX)(02 -DD
D is reduced to 2
I is reduced to —
32 Pe D
32 Pe D
4P
n(D2+ DD(D2 -
n(D 2 - Di
)
16
Pis reduced to
8.
9.
e-
P
iF
6
D2 +
8D
13. (b)
Yield load produces elastic instability in a column
Imin = 80 cm4
A = 20 cm2
(c)
kmin
(c)
D , core of section.
Concentric circle of diameter _
4
= 180 = 2 cm
,\ 20
L
le = - (Fixed ends) = 200 cm
2
le =
200 =100
kmin
2
II • III •
10
Theories of Failure
CHAPTER
Q.10.1 Explain what do you understand by theories of failure. Compare any three failure theories
graphically for an element subjected to two mutually perpendicular direct stresses.
[CSE-Mains, 1989, ME : 20 Marks]
Solution:
A body may be subjected to various combinations of M (bending moment), T (twisting moment) and
F(axial force) and may fail. At the critical section of the body principal stresses p1 > p2 > p3 can be
worked out. Then three principal stresses individually or in combination are compared with the uniaxial
tensile stress for a bar, under which the bar fails. For a brittle material it is no ultimate tensile strength but
for ductile material it is aYP' yield point stress.
Generally for any plane stress problem, two principal stresses p1 > p2 and
p3 = 0 are determined as shown. There are five theories of failure but we let
us take following 3 theories of failure
1. Maximum principal stress theory
2. Maximum shear stress theory
3. Strain energy theory
P2
P2
YP
YP
0J
0
P1
—= x
aYP
(c)
(b)
(a)
Fig. 10.1
Maximum principal stress theory (Rankine's theory)
P11
Gyp
P2 5' Gyp
p1
< ±1
YP
P2 <
YP
e.A
as shown in figure (a).
Theories of Failure
Strength of Materials
4
13 9
Maximum shear stress theory (Tresca theory)
P1 P2
YP
2 - 2
1P1 - P21 5- cryp
If p1 and p2 both are positive
P2
Pi or —
i
t=—
2
2
Then
If p1 > 0 and p2 < 0, then
(P1 P2) < cryp
Fig. (b) shows graphical representation of maximum shear stress theory.
Strain energy theory
[p12 p22 - 2vp1p2]
0. 2
yp
2
2
P1 + P2
2v P1P22
2
2
a YP
a
aYP
YP
or
or
< 1
(x2 + y2 - 2vxy) 5_ 1, ellipse
where,
x-
P1
a YP
=
P2
a YP
Fig. (c) shows graphical representation of strain energy theory.
Q.10.2 What is failure theory? Discuss its importance. Enumerate various failure theories and mention
their fields of application.
[CSE-Mains, 1997, ME : 15 Marks]
Solution:
There are various theories of failure depending upon various types of stresses on a body and different
types of strain energies absorbed by a body. These theories of failure are based on three dimensional
response of any component due to external load, and at a particular section of interest, these principal
stresses pi> p2 > p3 are determined. Analysis is based on the parameters of a simple tensile test on the
material. In case of ductile material, yield point stress Gyp and in case of brittle material but are determined.
Principal stresses or strain energies are compared with stresses in simple tensile test and strain energies
in simple tensile test.
There are 5 theories of failure
1. Maximum principal stress theory
<
used for brittle materials
2. Maximum shear stress theory
G
YP
, used for ductile materials
2
P1 P3
2
3. Maximum principal strain theory
(pi - vp2 - vp3
)
or
[P1 - v(P2
P3)]
<
YP
E
ayp
theory is not acceptable to designers
140 0. lAS & IFS (Objective & Conventional) Previous Solved Questions
MADE EASY
4. Maximum strain energy theory
2E
2
3
[Pi + P2 + P3 - 2 v(PiP2
1 a via
P2P3 + P3P1)] - '-
2
2E
v, Poisson's ratio
This theory is also not acceptable to designers
5. Shear strain energy theory
2ayp2
[(P1 - P2)2 + (P2 - P3)2 + (P3 For ductile materials, the theory is in good agreement with the experimental results and is used for the
design of shafts.
Q.10.3 Explain the salient features of maximum distortion energy theory of failure and discuss how it
compares with maximum strain energy theory and maximum shear stress theory.
Solution:
Maximum distortion energy theory
It principal stresses are p1 > p2 > p3 at a point of a body, then as per distortion energy theory
aryl!
[(P1 - P2)2 + (P2 - P3)2 + (P3 - P1)2]
Where aYP = yield point stress of the material in simple tensile test.
This theory is used for ductile materials and specially in the design of shafts.
Maximum strain energy theory
[P12 + p22 + p32 2V(P1 P2 + P2P3 P3P9] 5 ayp2
where, v = Poisson's ratio
However this theory is not used by designers.
Maximum shear stress theory of p1 > p2 > p3
Pi
-P3
<
a YP
2 - 2
This theory is used for the design of components of ductile materials. This theory provides more factor of
safety and a conservative design of a component is obtained.
Q.10.4 State and deduce the strain energy of distortion theory of elastic failure and compare it with
maximum shear stress theory with respect to the field of application and suitability for optimization.
[CSE-Mains, 1998, ME : 15 Marks]
Solution:
u = total strain energy per unit volume
= —1 [Pi +14 + - 2 v(P 1P2 P2P3 P3P1)]
2E
= uv + vs
= Volumetric strain energy + Shear strain energy
= due to volumetric stress, here pv = Pi + P2 + P3 = Pm
3
uv
2E
3,2
-2v(3p,i.1 )] = t'm (1 2v)
2E
Theories of Failure
Strength of Materials
3
= —X
2E
4I
141
(Pi + P2 +P3 )2 (1— 2v)
u = u— uv, taking (1 + v) = G , taking v = 0.5, volumetric strain energy becomes zero.
2
us = (1+ v) 3E
+ p2 + /313 — (PiP2 P2P3 P3P1)]
1 [
2 + P32 /
kP1P2 + P2P3 + P3P1)]
= — Pi+ P2
6G
1
= 12G
[4
0 D,2
2p2
'
[(P1 — P2 )2 + (P2 + P3)2 + (P3
a_ 0 2
,p3 — 0
— 2p2p3.— 2p3pi]
a yp
2
6G
2ayp2
Maximum shear stress theory is used for ductile material and gives the maximum size of the component.
Distortion or shear strain energy is also used for ductile material, but gives slightly lower value of dimensions
of a component, as diameter of a shaft. Using this theory, shaft diameter can be optimized.
Q.10.5 (a) What is the drawback of maximum principal stress theory?
(b) A body is under action of two principal stresses of +40 N/mm2 and —70 N/mm2 and the third
principal stress is zero. If the elastic limit in simple tension as well as in compression is
200 N/mm2, find the factor of safety based on the elastic limit according to
(i) maximum shear stress theory
(ii) maximum strain energy theory
(iii) maximum shear strain energy theory
Take v = 0.3
[CSE-Mains, 2008, ME : 20 Marks]
Solution:
(a) Maximum principal stress theory is used for brittle materials. It is not advisable to use this theory
for ductile materials
If principal stresses
p1 > 0, P2 >
Then this theory coincides with maximum shear stress theory
p1 > 0, p2 < 0, then
If
Pi + p2
"Cmax
GYP
2 — 2xFOS
av„
, (Factor of safety is reduced)
+ P2)
p1 = +40 N/mm2
(b)
p2 = —70 N/mm2
cse = 200 N/mm2
(i) Maximum shear stress theory
FOS —
40 — (-70)
2
FOS
200
2 xFOS
200
=1.82
= 110
14 2 0.
(ii)
IAS & IFS (Objective & Conventional) Previous Solved Questions
MAINEEA
Maximum strain energy theory
2
63q3
FOS
[(p12 + p22-2v(p1p2)] =
G
2
yp
[702 + 402 - 2 x 0.3(-70)(40)] = 1—
FOS
2
(4900 + 1600 + 1680) =
GY
FOS)
a „
90.44 -
FOS
FOS - 200 = 2.211
90.44
(iii) Maximum shear strain energy theory
Factor of safety,
)2
( a
[ p
1
2
p
2
2
p
p
1
]
FOYPS
2
(
[(-70)2 + (40)2 + 70 x 40]
)2
a
YP
FOS
2
YP
(4900 + 1600 + 2800) <
FOS 2
96.43
6 YP
FOS
200
FOS -
96.43
= 2.074
Q.10.6 A round member is subjected to a direct tensile load of 20 kN and shear load of 12 kN. The yield
stress in tension is 25 kN/cm2, and Poisson's ratio 0.3. Determine the diameter of the member,
using a factor of safety of 2 according to
(i)
Maximum principal stress theory
(ii)
Maximum shear stress theory
(iii)
Maximum distortion energy theory
Solution:
Direct stress,
61 =
Shear stress,
T -
Principal stress,
p1 =
20 kN
A
12 kN
A
10 kN + (10 kN)2 (12k12
A
A )
15.62 kN
-5.62 kN
P2 =
A
A
' tmax -
A
25.
62kN
A
Theories of Failure
Strength of Materials
-4I
143
25 kN/cm2
Gyp
FOS = 2
Gyp all = 12.5 kN/m2
(i) Maximum principal stress theory
=
25.62 kN
- 12.5 kN/cm2
A
A = 2.0496 cm2 = 4 d
2
d= 1.6 cm = 16 mm
(ii) Maximum shear stress theory
15.62
A
12.5 kN
2 cm 2
A = 2.4992 cm2 F
d= 1.78 cm = 17.8 mm
(iii) Maximum distortion energy theory
(pi 2
2_
p2
2
2
F25.621 +1-5.621 + [25.62 5.62]
- (12.5)2
L A
1_ A
L
A2
p
p
i
)
2
156.25 A2 =
=
A=
d=
a
2
yp
656.3844 + 31.5844 + 143.9844
831.9532
2.3075
1.714 cm = 17.14 mm
Q.10.7 A hollow shaft 30 mm inner diameter and 50 mm outer diameter is subjected to a twisting moment
of 800 Nm and an axial compressive force of 40 kN. Determine the factor of safety according to
theories of feature based on normal stress theory, shear stress theory and distortion energy
theory. The tensile and compressive yield strength of material is 280 N/mm2 and Poisson's ratio is 0.3.
[CSE-Mains, 2006, ME : 20 Marks]
Solution:
Shaft,
D. 50 mm,
II (502
A
Area,
d = 30 mm
3U
-2) _ 400 7t mm2
4
Torque,
504 3-4)
U = 53.40 x104 mm4
32 (
T= 800 Nm = 800 x 1000 Nmm
T
J
Shear stress,
25
T
800 x 103x 25
ti = — x 25 =
53.4 x 104
= 37.45 N/mm2
a, axial compressive stress -
P
=
A
= 33.45
40,000
400/1
= -31.83 N/mm2
31.83
144 ►
IAS & IFS (Objective & Conventional) Previous Solved Questions
MAD
EAU
Principal stresses
P1' P2 =
(31.83)2
31.83
+37.452
2 ±
2
= -15.915 ± V253.28 +1402.5
= -15.915 ±,1655.7825
= -15.985 ± 40.691
p1 = +27.776 N/mm2 (tensile)
p2 = -56.606 N/mm2 (comp.)
Itmaxi = ±40.691 N/mm2
yp
=
280 N/mm2
Maximum normal stress theory
4:3 YP
P2 —
FOS -
FOS
280
56.606
= 4.946
Maximum shear stress theory
—
max
40.691 -
FOS -
CT YP
2 x FOS
280
2 x FOS
140
40.691
= 3.44
Distortion energy theory
a )2
YP
[(27.776)2 + (-56.606)2 - (27.776X-56.606)] =
IFOS
(771.506 + 3204.24 + 1572.29) =
or
p
FOS
yp
15548.036 - FOS
FOS -
280
= 3.759
74.485
Q.10.8 A hollow circular steel shaft is subjected to a torque of 800 Nm and a bending moment of 1200 Nm.
The internal diameter of the shaft is 60% of the external diameter. Determine the external diameter
of the shaft according to
(i) maximum principal stress theory
(ii) maximum shear stress theory
(iii) shear strain energy theory
Take factor of safety as 2 and the yield strength of the material as 27 kN/cm2.
Theories of Failure
Strength of Materials
Solution:
aYP = 27000 N/cm2 = 270 N/mm2
T= 800 Nm
M = 1200 Nm
FOS = 2
ayp = allowable = 135 N/mm2
External diameter = D, internal diameter = 0.6 D
1 = 6 [D4 —(0.604 ]= 0.042726D4
4
J = 21= 0.08545 D4
ab, maximum bending stress =
M
D
x
T, maximum shear stress =
1200x103 xD
14043 x 103
2 — 0.042776 x 2D4
D3
T D 800 x 10 3 D
1
4681x 10 3
=
x =
x x
0.08545
J 2
2 D4
D3
Principal stresses
ab
2
(5b2
2
'T
2
7021.5 8438.8 15460.3
D3
D3
D3
7021.5 8438.8 —1417.3
P2
(I)
—
D3
D'
=
,
x 10
X
3
103
Maximum principal stress theory
103° x 15460.3 < 135
D3
D3 = 114.5153 x 103
D = 48.55 mm
(ii)
Maximum shear stress theory
(iii)
8438.8 x 103 _ 135
—2
D3
D3 = 125.02 x 103
D = 50 mm
Shear strain energy theory
(p
1
115460.3 x10D3
2 + p
2
2
p
p
1
)
2
2
1-141,3 x101 (15460.3 x1417.3 x106
= 1352
)
D3
D3 x D3
)
2
1352 x D6 = 106[239020.8761 + 2008739.3 + 21911883.2]
1352 x D6 = 106[262941498.6]
06 = 106 x 14427.517
D = 49.34 mm
4
145
146
0. lAS & IFS (Objective & Conventional) Previous Solved Questions
MADE EASY
Q.10.9 A hollow shaft of outside diameter 50 mm and inside diameter 20 mm is subjected to a torque of
T Nm and a bending moment of 0.5 T Nm. If the tensile yield stress of the shaft material is
250 N/mm2, what is maximum permissible value of Tto avoid failure according to Tresca's failure
theory? Take a factor of safety of 2.0 for a given application.
[CSE-Mains, 2011, ME : 15 Marks]
Solution:
aYP= 250 N/mm2
FOS = 2
a = 125 N/mm2
Allowable,
Allowable shear stress,
ti = 2 = 62.5 N/mm2
Shaft,
D= 50 mm
d= 20 mm
Torque,
T-
it
16
x
(D4 -d 4 )
D
it (50 4 - 20 4 )
x 62.5 - 16
x 62.5
x
50
71 609
x
x 62.5 x 104 Nmm = 149.47 x 104 Nmm = 1494.7 Nm
16 50
T = T Nm
M' = 0.5 T Nm
_
Torque,
BM,
Tv equivalent torque = VT2 +(0.5T)3 = 1.118 T Nm = 1494.7 Nm
T = 1336.9 Nm
Q.10.10 The load on a rod consists of an axial pull of 10 kN along with a transverse shear force of 5 kN.
Determine the diameter of the rod by using the following theories of failure
Strain energy theory
(i)
Shear strain energy theory
(ii)
Elastic limit in tension is 270 N/mm2 and a factor of safety of 3 is to be used.
Poisson's ratio = 0.3
[CSE-Mains, 2011, CE: 15 Marks]
Solution:
6-
10,000
A
5000
A
A = area of cross-section
Principal stresses are
2
5000 +
2
a 11(1
—
= — + 2 +i =
A
2
P2 =
2
2J
± T2
= 5000
A
(i) Strain energy theory
ae = 270 N/mm2
x 5000
A
.4 x 5000
A
2.414 x 5000
A
12070
A
0.414 x 5000
=
A
2070
A
Theories of Failure
Strength of Materials
147
FOS = 3
270
= 90 Nimm2
3
Poisson's ratio = 0.3
[p12 + p22 - 2v pi p2 ] 5. al
aAllowable
[(
=
12070)2 +( 2070)2 2 x 0.3 x 12070 x 2070] =902
) +
A
A )
A2
[14.5684900 + 4284900 - 14990940] = 8100A2
A2 = 20365.523
—d
A = 142.708 mm2 = 4
Rod diameter,
(ii) Shear strain energy theory
2
d = 13.48 mm
[P12- P1P2 + ,022] = aa2
12070)2 112070 X 2070) ( 2070)2 ] = 902
+
+
[1
A )
A2
A
1
--2 [145684900 + 24984900 + 4284900] = 902
A
A2 = 18823.23
4
A = 137.2 mm2 = 1-t-d2
Rod diameter,
d = 13.22 mm
Objective Questions
Q.1
0.2
A certain steel has proportionality limit of
300 N/mm2 in simple tension. Under three
dimensional stress system, the principal
stresses are 150 N/mm2 (Tensile), 75 N/mm2
(Tensile) and 30 N/mm2 (Compressive), p. = 0.3.
The factor of safety according to maximum shear
stress theory would be
(a) 1.4
(b) 1.5
(c) 1.3
(d) 1.66
[CSE-Prelims, CE : 2001]
Which one of the following theories of failure is
widely used in the design of a machine element
made of a ductile material?
(a) Maximum normal stress theory
(b) Maximum strain theory
(c) Strain energy theory
(d) Maximum shear stress theory
[CSE-Prelims, CE : 2003]
Q.3
Which of the following is the most appropriate
theory of failure for mild steel?
(a) Maximum principal stress theory
(b) Maximum principal strain theory
(c) Maximum shear stress theory
(d) Maximum strain energy theory
[CSE, Prelims-CE : 2009]
Q.4
A shaft is subjected to a torque and an axial
compressive force. Shear stress due to torque
is 30 MPa and compressive stress due to axial
force is 80 MPa. If yield strength of material is
255 MPa, what is factor of safety as per the
maximum principal stress theory?
(b) 3.01
(a) 3.19
(c) 2.83
(d) 3.64
Q.5
The principal stresses developed at a point are
+80, -80 MPa, 0.0. Using shear strain energy
148 0
theory factor of safety is 2. What is yield strength
of the material in MPa.
Q.6
MADE EASY
1AS & IFS (Objective & Conventional) Previous Solved Questions
(a) 150 MPa
(b) 80fd MPa
(c) 240 MPa
(d) 277 MPa
The principal stresses at a point are 70, 60 and
-80 MPa. If
is the yield point stress of the
material, using maximum principal stress theory
factor of safety is 3.6. What is FOS if the
maximum principal strain theory is used.
Poisson's ratio of material is 1/3
(a) 3.29
(b) 3.9
(c) 4.5
(d) 5.5
Q.7
A thick cylinder of an internal diameter 100 mm
and external diameter 200 mm is subjected to
internal pressure p. Yield strength of the material
is 240 MPa. Using maximum shear stress theory
FOS is 2. What is the maximum value of internal
pressure p?
(a) 120 N/mm2
(b) 90 N/mm2
(c) 72 N/mm2
(d) 60 N/mm2
Gyp
Answers
1. (d)
6. (a)
2. (d)
7. (b)
3. (c)
4. (c)
5. (d)
Explanations
5.
1. (d)
150+30)
Gyp
(P12 - P1P2 + P22) = FOS
-Maximum shear stress = 90
2
ae
2
802 + 802 + 802= 3 x 802 =
= i8 = 150
ayp
150
=1.66
FOS 90
2.
3.
4.
(d)
Shear strain energy theory
p1 = 80, p2 = -80, p3 = 0
a = 300 N/mm2
Principal stress are
+150, +75, -30 N/mm2
Poisson's ratio = 0.3
6.
(d)
For ductile material
(d) Maximum shear stress theory is widely used.
=
2
138.56 x 2 = 277 MPa
(a)
Maximum principal stress = 70
FOS = 3.6
ayp 70 x 3.6 = 252 MPa
Maximum principal strain
=
1
1
70 - 3(60)+ 3(80)
(c)
Mild steel — a ductile material
(c) Maximum shear stress theory
E
FOS =
(c)
7.
80
252
76.66
= 3.287
(b)
R22 + Ri2
5
ac ax = P x 2
2
=
R2 - Ri
3P
5
3 P+P
tmax
pm = -40 -,402 +302 = -90 N/mm2
a = 255
255
FOS = — = 2.83
90
=
2
4
-p = 120
3
p = 90 N/mm2
-1-A
p
3
=
76.66
E
11
Strain Energy Methods
CHAPTER
Q.11.1 The L-shaped bar shown in figure 11.1 is of uniform cross-section 60 mm x 120 mm. Calculate
the total strain energy. Take E = 2 x 105 MPa, G = 0.8 x 105 MPa.
b= 60 mm
d= 120 mm
Fig. 11.1
[IFS 2011, CE : 10 Marks]
Solution:
Total strain energy
U= UB + Us + UA
= Strain energy due to bending of BC + Strain energy due to shear of BC +
Strain energy due to axial load W
w20 3w2L w2L _
1+
2
6EI
5Gbd 2AE
Where, W = 10,000 N, L1 = 2000 mm, L2 = 1000 MM, E = 2 x 105 MPa, G = 0.8 x 105 MPa
/-
bd3
12
60 x1203
= 8.64 x106 mm4
12
b = 60 mm, d = 120 mm
A=bxd= 60 x 120 = 7200 mm2
Putting the values
Strain energy,
(10000)2 x1000
+
6 x2 x105 x8.64 x106
0.8x105 x60 x120 2 x7200 x 2 x105
= 77160.5 + 208.33 + 357.143 Nmm
= 77725.967 Nmm = 77.725 Nm
U-
(10000)2 x(2000)3 + (10000)2 x 0.6 x 2000
150 N.
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
Q.11.2 The cross-sectional area of each member of the truss
shown in figure 11.2 is A = 400 mm2 and E= 200 GPa.
Determine the vertical displacement of joint C if a 4 kN
force is applied.
Solution:
Reactions
4 kN
RBV=
RAV
RAH
12
8 = 1.5 kN T
1.5 kN T
4 kN
Fig. 11.2
Force in members
2
FAB -
cosO
2
2
= — = -2.5 kN
0.8
FcA
+
FAB
+2 kN (tensile)
cos()
= +2.5 kN(tensile)
4 kN
Unit load diagram (Fig. 11.3)
1 kN
1.5 kN
2 kN
0.5 kN
0.666 kN
0.5 kN
0.5 kN
Members
n
N
nNL
CB
-0.833
-2.5 kN
10.4125
FCA FCB -0.833
CA
-0.833
FAB = +0.666
AB
+0.666
1nNL = 10.656 kNm
A = 400 mm2
= 400 x 10-6 m2
= 4 x 10-4 m2
E = 200 kN/mm2 = 200 x 106 kN/m2
AE = 4 x 10-4 x 200 x 106 = 800 x 100 kN
= 80000 kN
+2.5 kN
-10.4125
+2 kN
10.656
Fig. 11.3
Vertical deflection at C,
Z nNL = 10.656 kNm
EnNL
10.656
0.1332 x10-3 m
80000 = 0.1332 mm
AE
Q.11.3 A shaft is supported by two anti-friction bearings with loads of 140 Newton each acting at points
B and F as shown in Fig.11.4. Portion of the shaft between B and C has diameter of 2D as
Strain Energy Methods 4
Strength of Materials
compared to a diameter D for the portion of the
shaft between A and B and between C and F. Using
Castigliano's theorem, determine the deflection of
the shaft at points B and F.
[CSE-Mains, 1992, ME : 20 Marks]
151
= —P
Solution:
Let us take length L, with AB = BC = CF=
RA =P
Moments about A
PL
1.-150 cm-44-150 cm-44-150 cm
L/3
L/3
L/3
= Re xL
3
2PL
3
Fig. 11.4
2L
Re x —
3
Rc =
RA = +PT
E
BMD (Fig. 10.5)
MA = 0
PL
3
PL
MB +3
MC -
C
B
2PL PL PL
3
3 - 3
Fig. 10.5
MF =
704
1AB =
18c =
64
CF
7c(204 1670 4
64
64
18c = 16 /AB
f L/3 (Px)2
Strain energy,
UAB — UCF = JO 2EI AB
dx
L/3 P2x2
2EIAB
p 2x3 L/3
xdx _
P 2 L3
6EIAB 0 -162E/ AB
2
(11)
UBc
U—
x3 =
P2L3
'
27 x 2 x E I Eic
2EIBc
2P2L3
162E/AB
+
P2L3
54 E/Bc
=
P2L3
E I AB 81
+
P2L3 1 1 11,667P2L3
DAB 1_81 864 j
/AB -
(:a
704
64
DAB x 864
P2L3
54 x16 E BAB
F
15 2 0. IAS & IFS (Objective & Conventional) Previous Solved Questions
U—
MADE EASY
11.667P2L3 64 746.667P2L3 237.67P2L3 0.275P2L3
x
=
E x 704
864 ED47c x 864
ED4 x864
ED4
Say deflection at B and F is 8, then
8—
au 0.275 x 2PL3
=
ED4
ap
8 — 0.550 PL3
ED4
P= 140N
L= 4500 mm, D in mm
8—
0.550 x140 x 45003 7.02 x1012
ED4
ED4
mm
Q.11.4 A steel tube having outside and inside diameters of 10 cm and 6 cm
respectively is bent into the form of a quadrant of 2 m radius as shown
in fig. 10.6 One end is rigidity attached to a horizontal base plate to
which a tangent to that end is perpendicular, and the free end supports
a load of 1000 N. Determine the vertical and horizontal deflection of
the free end under this load using Castigliano's theorem. E. 2 x 1011
N/mm2.
[CSE-Mains, 1993, ME : 20 Marks]
1000N = P
Solution:
Vertical deflection
Take a small element Rd 0 at an angle 0 from vertical
Bending moment at small element = P.R sin()
rt/2(M9)2Rdo
U, strain energy = So
2E/
—
Ht--- 2 m
Fig. 10.6
2
fir/2 p 2R 2 sin
ORA
2EI
au
= sv , vertical deflection
ap
=
=
PR 3
fola EI
sine Ode =
n12 1 PR3
fo 2
So
E1 X de n/2
ic/2 PR3 (1— cos 20) d0
E,
PR3 cos 20
E/ x 2
7EPR3
PrcR 3
.
0=
4EI
4EI
=
64
0 0 4 _ 6 4) = 427.256 cm4
E= 2 x 107 N/cm2
= 854.513 x 107 Ncm2
1000 x icx 2003
2
- 4x 854.513x10',= 0.1838 cm
d0
Strain Energy Methods
Strength of Materials
Horizontal deflection
Put frictitious horizontal load,
41
153
H = 0 at A
M0 = P(R sine) + H(R - R dose)
l
2
(M ) Rd()
171/2
J.
2E1e
U
cir/2 1
au
OH = .10
aH
2E1[PRsine][R-Rcos0]
x Rd()
/r/2 2
§-1 = So 2EI [PR3][sin0-sin0cose]d0
3 r it/2
PR 3 f rE/2 sin20
= PR
c/0
sin0d0
2
EI Jo
El Jo
_
PR 3
cos0
El
=+
n/2
0
PR 3 cos20
+ EI
4
nI2
0
PR3 PR [cosic cos01
+
4 j
El
El
4
PR3 PR3 [ 1 1] PR3
.
+— - - =
El
4 4
2E1
El
2
§4 = it- x8 =0.117 cm
Q.11.5 Using the Castigliano's theorem, calculate the vertical deflection 8 at the middle of a simply
supported beam which carries a uniformly distributed load of intensity w over the full span. The
flexural rigidity El of the beam is constant and only strain energy of bending is to be considered.
[CSE-Mains, 1991, ME : 20 Marks]
Solution:
Let us apply a fractious load P. 0 and the centre of the beam reactions,
wL P
RA = Rg = —+ —
2
BM at section x,
MX=
P=
2
0
l'x —1-I
w
wL P wx 2
Tx+ V 2
Beam is symmetrically loaded about center
P
Strain energy,
U= 2
U2 (Mx) 2
dx
.10 2E1
2 j
am
u2 2m x x—x dx
2E/ 0
DP
=
El
0
[ wL
P
wx 2 ]x
.dx
2 x+ 2 x 2 2
But P = 0
8=
2 ulwLx 2 wx3 d
i
El o
4
4 )"
Flg. 10.7
154 0. lAS & IFS (Objective & Conventional) Previous
L/2
2 wLx3 wx4
EI 12
16 0
=
MADE EASY
Solved Questions
2 wL4 wL4
96 256
El
2 x 8-3
xwL4 = 5wL4
El 768
384E/
Q.11.6 A circular rod of diameter d is bent at right angle. It is fixed
at one end and a load W is applied at the other end as shown
in the fig. 10.8. Determine the deflection under the load W, if
E and G are elastic and shear modulii of the material.
[CSE-Mains, 1996, ME : 20 Marks]
Solution:
w 2b3
Strain energy due to bending in BC 6E/
2a
Strain energy due to twisting in member AB - (Wb)
2GJ
W2a3
Strain energy due to bending in member AB -
6E1
IA/ 4,3
Total strain energy,
au
aw
Li
2,3
a + vv
m/a
2GJ
6EI
8 - Wb3 Wb2a Wa3
+
3E1
3E1 GJ
J= 21
8-
Q.11.7 (a)
(b)
vv
U = "
6E1
Wb3
3E/
+
Wa3
3E/
+
Wb2a
2GI
I -
ird 4
64
8-
W [ b3 a3 b 2al 64W
+
+
=
I 3E 3E 2G
704
r
+ a3
3E
b 2a
2G ]
State the theorem of Castigliano's.
Using the above theorem, find the horizontal displacement along the load line of the frame
shown in fig. 10.7. Considering the deflection due to bending only. The moment of inertia
is the same for all sections.
B
Pa)
B '
10-P P44
P
'
Pa c)
P8
Pe
1-4-- b
a
D
Fig.10.7
[CSE-Mains, 1996, ME : 30 Marks]
Strain Energy Methods
Strength of Materials
4
15 5
Solution:
Castigliano's Theorem
If a body is acted upon by forces F1 , F2, F3 .. F and U is the strain energy stored in the body due to these
forces (causing, axial stresses, bending moments or twisting moments etc. then partial derivative of the
strain energy with respect to force F., gives the displacement of the body in the direction of
au
aF;
8i, deflection along force F;
Free body diagram of the figure is given
P 2a3
Uco 6 E1
Member CD
Member BC
p2 a 2 x b
UBC
2E1
Member AB
UAB
P2a3
6E/
P2a3 P 2 a 2 b
2E1
3E./
Total strain energy,
au
ap
= 8 = deflection along P
P
2Pa3 Pa 2 b Pa2
_
[2a + 3b]
3E1 EI 3E1
Q.11.8 Determine the vertical deflection of the joint C of the frame shown in fig. 10.10. Area of crosssection of each member is 850 mm2. E = 200 kN/mm2.
5 kN
10 kN
RDH = 1 kN
Unit load at C
3
RAV =15kN
RDV = 10 kN
Fig. 10.10
[IFS 2013, CE : 10 Marks]
Solution:
Reaction: Taking moments about D
10 x 3 + 5 x 3 = RAvx3
RAC = 15 kN
156
1.
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
RDH
= 10 kN
Roy = 10kN
J.
Forces in the members
FED = -10 kN (comp.)
FDD = 0
FAB = -15 kN (comp.)
= +14.14 kN(Tensile)
FBD = +1
Unit load at C
FDD = -1 (comp.)
FBD = o
o
=0
FBD
FAB
Members
N
n
L
NnL
AB
—15
0
3
0
BC
—10
0
3
0
CD
0
1
3
0
BD
+14.14 kN
0
3'I2
0
E nNL = 0
Vertical deflection at joint C is zero
Vertical deflection at point C is zero.
Q.11.9 The steel truss shown in fig. 11.11 is anchored at
A and supported on roller at B. If the truss is so
designed that under the given loading all tension
members are stressed to 100 N/mm2, and all
compression members to 80 N/mm2, find the
vertical deflection of the point C. E= 2 x 105 N/mm2.
Find also the lateral displacement of the end B.
[CSE-Mains, 1998, CE : 40 Marks]
6 kN
V
E
cq
Fig .11.11
0.444
2
F
0.888
0-)
0
D
RB = 5
6 kN
[4— 2.4 m —44-- 2.4 m —44— 2.4 m
Solution:
There are 9 members in truss.
Zero force member is FD
Compression member AE, EF, FB, ED, FC
Tension member AD, DC, CB
Draw unit load diagram, unit load at Cas shown below:
= tan-1 0.75
= 36.87°
sine = 0.6
cos() = 0.8
For unit vertical load at C,
Reactions at A and B
RA =
6 kN
0.888
co.
-1
2.4 m —.4.-- 2.4 m
Fig. 10.12
2.4 m
Strain Energy Methods
Strength of Materials
Joint A
Joint B
0.444
0.888
Joint E
FED = 0
FEF = 0.444
Joint F
Joint C
0.444 (Net)
0.444
0.888
Members
Type of force
AE
C
EF
n
nN/A
L(mm)
-80
-0.444
+35.52
2400
85248
C
-80
-0.444
+35.52
2400
85248
FB
C
-80
-0.888
+71.04
2400
170496
AD
T
+100
+0.555
+55.5
3000
166500
ED
C
-80
0
0
1800
0
DF
0
0
-0.555
0
3000
0
FC
C
-80
+0.333
-26.64
1800
-47952
CB
T
+100
+1.111
+111.1
3000
333300
DC
T
+100
+0.888
+88.8
2400
213120
En
A
N/A
n(N/A) L
L = 340992 + 499500-47952 + 213120 =1006160
1006160
NL
= 5.03 mm
Zn-- =
AE
2 x 105
MUM
41
15 7
12
Miscellaneous Questions
CHAPTER
Q.12.1 Determine the bar forces in the members CB, BE and EF
of the truss as shown.
[IFS 2013, CE : 15 Marks]
10 kN
10 kN
Solution:
Force in various members can be determined by method of
sections.
(i) Let us break the frame by section x-x as shown
Taking moments about B (force in BF passes through
point B so no moment caused by it)
[4-4 m -+- 4 m
10 x 4 + 20 x 4 = FEF X= FEF x distance B P)
vL
4m
Fig. 12.1
120 = FEF x 2.828
FEF= —42.443 (Comp.), Foo = +20 kN (Tensile),
FOE = —10 kN (comp.)
10 kN
(ii) Cut off by yy-section
Taking moments about E
FcB x 4 = 20 x 4 — 10 x 4 = 40
10 kN (tension)
FCB
Taking cut along section z-z as shown
C
y
10 kN
10 kN
m —444-- 4 m
Member BC, BE, BFare cut
10 kN (tensile)
FCB
Taking moments about point F
20 x 8 — 10 x 4 — 20 x 8 + FBE x 4 = 0
FBE = 10 kN (comp.)
20
Miscelaneous Questions
Strength of Materials
FBE
10 kN (comp.)
FCB
+10 kN (tension)
FEF
-42.443 kN (Comp.)
4
15 9
Objective Questions
Q.1
Figure 12.2 shows the loading pattern on a
truss. The force in the member AC is
2t
4t
4t
4t
(a) 1
(c) 3
2t
Q.4
8t
(b) 2
(d) 4
[CSE-Prelims, CE : 2001]
A pin-jointed lower truss is loaded as shown in
the below figure. The force induced in the
member DFis
A
8t
Fig. 12.2
(a)
(b)
(c)
(d)
Q.2
Zero
2t
8t
Statically indeterminate
[CSE-Prelims, ME : 2001]
What is the vertical deflection of joint 'N' of the
pin-jointed frame as shown in the figure below?
4m
Fig. 12.5
(a) 1.5 kN (Tension) (b) 4.5 kN (Tension)
(c) 1.5 kN (Comp.)
(d) 4.5 kN (Comp.)
[CSE-Prelims, CE : 2002]
0.5
(a) PL/6AE
(c) PL/AE
Q.3
FIg. 12.3
(b) PL/3AE
(d) 2PL/AE
[CSE-Prelims, CE : 2008]
For the truss shown in the figure, which one of
the following members has zero force induced
in it?
20 kN
The degree of static indeterminacy of the pinjointed plane frame shown in figure is
w2
(a) BC
(c) DE
Fig. 12.4
Fig. 12.6
(b) AD
(d) BD
[CSE-Prelims, CE : 2003]
160 ►
Q.6
MADE EASY
IAS & IFS (Objective & Conventional) Previous Solved Questions
(b) 40 kN
(d) 100 kN
[CSE-Prelims, CE : 2006]
(a) 0
(c) 80 kN
The force induced in the vertical member CDof
the symmetrical plane truss shown in the
figure is
50 kN
Q.9 In the plane truss shown above, how many
members have zero force?
V
50 kN
A
Fig. 12.7
(a) 50 kN (tension)
(c) 50 kN (Comp.)
Q.7
A
(b) 100 kN (tension)
(d) Zero
[CSE-Prelims, CE : 2004]
(a) 3
(c) 7
In a truss work as shown in the figure given
below, what is the force induced in the member
Fig. 12.10
(b) 5
(d) 9
[CSE-Prelims, CE : 2006]
DE?
Q.10 What is the force in the member EH for a pinjointed tower truss as shown in the figure?
50 kN
2 kN
A
T
3m
2 kN
3m
Fig. 12.8
(a) 50 kN (tensile)
(c) 50 kN (Comp.)
Q.8
2 kN —110.E
(b) Zero
(d) 25 kN (Comp.)
[CSE-Prelims, CE : 2005]
3m
M
A
.4
Due to horizontal pull of 60 kN at C, what is the
force induced in the member AB?
C
I
~-4 m H
60 kN
Fig. 12.11
(a) 6.0 kN (Tension) (b) 6.0 kN (Comp.)
(c) 7.5 kN (Comp.)
(d) None of the above
[CSE-Prelims, CE : 2008]
Answers
3m
Fig. 12.9
1. (a)
6. (a)
2. (c)
7. (b)
3. (d)
8. (d)
4. (c)
5. (a)
9. (d) 10. (c)
Miscelaneous Questions
Strength of Materials
Explanations
1.
FDF = 4.5 kN (T)
FFE = 4 kN (T)
(a)
2t
At joint F
8t
A
8t
Zero force member at joint A.
2.
(c)
FDF = 5 kN
5 cos() = 4 kN
Balance FEF
5 kN sine = 3 kN
3 + FDF = 4.5 kN
FDF = 1.5 kN (Comp.)
5.
(a)
In members BC and CD, no force.
Vertical displacement of N
1 [P
3.
x
FAB =
FED = 0
Fa), = 0
L 1
L 1 P
x +
xx
A E
A
AE
6.
(d)
Members = 21 = m
Joints = 11 = j
Ds = m + Re - 2j
= 21 +(2 x 2 +1)- 2 x 11 = 4
Static indeterminacy 2.
4.
(a)
At joint D
Fop = 50 kN (T) Tension
7.
(b)
50 kN
(c)
RE-
12 + 6
4 = 4.5 kN
4 kN
= 4.5 kN
4.5 kN
At the joint E,
FDE -= 0
161
16 2 0. lAS & IFS (Objective & Conventional) Previous Solved Questions
8.
MADE EASY
FEH cos° = 6
(d)
At joint C
6
= 7.5 kN(Comp.)
F =
84
0.8
Q.12.2 Analyse the continuous beam as shown in
fig. 12.8 by moment distribution method.
Draw the bending moment diagram.
Assume El = constant.
20 kN
60 kN
A'
4
sin° = - = 0.8
5
cos° = 0.6
2m
60
60
=
= 100 kN
F cos() 0.6
Bc
FEic = FAB = 100 kN
FBD = 0
9.
20 kN/m
4m
3m
-40 IN/m
+24 kN/m
-20.488
(d)
Fig.12.8
[CSE-Mains, 2012, CE : 15 Marks]
Solution:
Mp = 0
Mc = -40 kNm
ME =
None zero force members.
a1.71
10. (c)
3 x 42
144_2
= + 8 = +6 kNm
8
24x
4 24x3x(
x
2+1)
2 2 3 +2
= 32 + 108 = 140 kNm2
=
a1.771' ( Origin at B)
24x3
2
x 2+
24 x2
2
2
x3+3
11
72 + 24-) = 160 kNm2
3
a272 (origin at B or C)
9 kN
9 kN
Considering span AA' (imaginary) of span
zero.
Clapeyron's theorem for spans A'A, AB
sin° = 0.6
cos0 = 0.8
RH -
18+12+6
4
= 2
-x 6x 4x 2 = 32 kNm2
3
= 9 kN
M A' X01-2MA(5-1-0)+5MB
Miscelaneous Questions
Strength of Materials
=
6 x ai.71
0
5
-96
+40
+2.58 kNm
MB
=
-5618MB = -56 - 18 x 2.58
5MA
-56 - 46.44 = -102.44
-20.488 kNm
MA
+2.58 kNm
MB
-40
kNm
MC
-CO
10MA + 5MB = -192
Or
5MA + 2.5MB = -96
Spans AB and BC
5MA + 2MB(4 + 5)+ 4Mc
6 xaqi 6 xa 2x2
5
4
-6 x 140 6
5
4
163
5MA + 2.5MB
15.5%
(origin at B)
10MA + 5MB= -5 x 160 —192
41
(origin at A)
zE
0
x 32
= -168-48 =-216
5MA + 18MB - 4 x 40 = -216
5MA + 18MB = -56
A
+2.58
, P2
points of contraflexure
BM Diagram
Fixed and Continuous Beams
Q.1
A fixed beam as shown in the figure 12.15 has
a span L and uniform flexural rigidity El. It is
subjected to a concentrated clockwise moment
at the centre. The deflection at the centre of the
beam is
(a) 0.5
(c) 1.5
(b) 1
(d) 2
[CSE-Prelims, CE : 2006]
Q.3
A beam shown in fig. 12.16 of span 10 m and
having El = 10000 kNm2 is subjected to a
rotation of 0.001 radian at end B. What is the
fixed end moment at A?
14e-- L/2
)-I4
L/2 —H
Fig. 12.15
10m
ML2
, upwards
(a)
8 EI
(a) 1.5 kNm
(c) 3.0 kNm
(b) ML2, downwards
(c) Zero
ML2
, downwards
(d) 384E/
[CSE-Prelims, CE : 2002]
Q.2
A fixed beam and a simply supported beam
having same span and develop same maximum
bending moment due to uniformly distributed
load on entire span, what is the ratio of uniformly
distributed load on fixed beam to that on simply
supported beam
Q.4
Fig. 12.16
(b) 2.0 kNm
(d) 4.0 kNm
A heavy weight W attached to a rod can slide
in a grooved support as shown in fig. 12.17.
What is the equilibrium sliding distance A?
w
L---4
Fig. 12.17
1 64
►
MADE EASY
lAS & IFS (Objective & Conventional) Previous Solved Questions
WL3
(a) 3E/
clockwise. What is the distance of the point of
contraflexure from the left support?
WL3
(b)
12 EI
3L
(b) TC)
WL3
(d) 48E1
WL3
(c) 24E1
(d)
[CSE-Prelims, CE : 2010]
Q.5
What is the bending moment at A for the beam
shown in fig. 12.18
E= 200 GPa
L
2L
[CSE-Prelims, CE : 2006]
Q.7 What is the fixed end moment for the beam
shown in fig. 12.19
E= 100 GPa
-I-
L
Fig. 12.18
PL
(a) -3-(c)
(b)
PL
(d)
3PL
Fig. 12.19
2
(a) M0
2PL
3
[CSE-Prelims, CE : 2010]
Q.6 A horizontal fixed beam AB of length L has
uniform flexural rigidity El. During loading the
right support A rotates through an angle 0
(c)
(b)
Mo
3
2M0
(d)
1. (c)
2. (c)
6. (a)
7. (b)
3. (b)
(c)
M
L/2
—
L/2 —al
Zero deflection at centre.
2.
1-2
W2 L2
(c)
w21-2
— = 1.5
W2
L/2
5.
(b)
Wi
M -L2
24
wi
MA = M1 = 12 = Mmax
AA
8 ""max
12 - 8
Hi— L/2
60
Answers
Explanations
1.
3
10m
4. (c)
5. (d)
Miscelaneous Questions
Strength of Materials
Pi + P2 = P
2P
P1= 3
Say equivalent load is W
WL2
=0.001
0=
° 2E1
2P
MA _ 3
Wx102
- 0.001
2 x 10000
W 0.2 kN
MA = 0.2 x 10 = 2 kNm
4.
6.
(a)
(c)
He— L/3
A fixed beam of length 2L, central load W,
Ac
4E/0
P
2E10
L
5.
16 5
4
L
I-1-4
2L/3
L
Point of contraflexure lies at - from left support.
3
W (203 WL3
- 192E/ 24E/
7.
(b)
(d)
C
L
L
-14
L-3
81
M
P2 1-3
82 = 3E21
['*-. 4
2/
►
ML 1
M0L
x
21 - 4 21
= 82
P1
1
P2
21
E
I= 2
E2
M= -Mo
3
P2
t
C P2
FBD
B
4
ML 1 ML 1
x
x
+
4 21 2 1
ML ML _ 3 MI_
+
- 4/ 21 - 4 I
P1 = 2P2
P
2
13
Rotational Stresses
CHAPTER
Q.13.1 A circular disc 50 cm outside diameter has a central hole and rotates at a uniform speed about
an axis through its center. The diameter of the hole is such that the maximum stress due to
rotation is 85% of that in the thin ring whose mean diameter is also 50 cm. If both are of the same
material and rotate at the same speed, determine the diameter of the central hole and speed of
the disc for the datas given
Allowable stress = 900 kg/cm2
Specific weight = 7.8 gm/cm3
Poisson's ratio = 0.3
[CSE-Mains, 2006, ME : 30 Marks]
Solution:
In a thin disc, oc max occurs at inner radius, R1
If R2 is outer radius = 250 mm
C max =
k1 k2 -
,,„2
" [ki (2Fq +
3+v
8
=
.3
8
k 2R1
= 0.4125
1+ 3v 1.9
=
= 0.2375
8
8
2
PC° [0.4125(2 x 2502
= 0.85
x 2502
+Rn- 0.2375Rn
xp
[stress in ring =
(02r2p1
g
0.85 x 2502 =
0.025 x 2502 =
R1 =
Allowable stress =
2 x 2502
7.8 x10-6x 0.85 x (0
9810
0.825 x 2502 + 0.175 R12
0.175 R12
94.5 mm
900 kg/cm2 = 900 x 9.81/cm2 = 88.29 N/mm2
88.29 N/mm2
(02 = 2090195.84
= 1445.175 rad/sec
N = 13806 rpm
(:4
Rotational Stresses
Strength of Materials
1
16 7
Q.13.2 A steel rotor disc of uniform thickness 50 mm has an outside diameter 800 mm and a central hole
of diameter 150 mm. There are 200 blades each of weight 2N at an effective radius of 420 mm
pitched evenly around the outer periphery of the disc. Determine the maximum rotational speed
such that maximum shearing stress in the disc does not exceed 375 MN/m2. Take (p) of steel as
7470 kg/m3. Following basic relations for radial stress (ar) and Hoop stress (a0) at radius of
rotating disc at w rad/sec can be used with usual notations
= A — —(3+v)
pco 2 r 2
po) 2r 2
, Go = A+ 2 (i + 3V)
8
8
[CSE-Mains, 1995, ME : 30 Marks]
Solution:
t= 50 mm, R2 = 400 mm, R1 = 75 mm, timax= 375 N/mm2 and p = 7470 kg/m3
Number of blades = 200
Effective radius = 420 mm
Weight of each blade = 2N
Centrifugal force on the periphery due to blades
200 x 2
400 x co2 x 420
=17.1254t w2N
9810
Resulting area = it x 800 x 50 = 12.566 x 104 mm2
Radial stress at the periphery of the disc
17.1254(1)2
=1.363 x10-4 w2Nimm2
12.566x104
Radial stress at the inner radius 75 mm is zero
pco 2r 2
=
A——(3+v)
6r
8
x co2 xr
Taking v = 0.3, Poisson's ratio for steel
2
2
or = 0 = A
E3 (3 + 0.3)Pc° r
8
752
p = 7470 kg/m3 = 7470 x 10-9 kg/mm3 = 7.47 x 10-6 kg/mm3
3.3
8 = 0.4125
0 = A
=A
B
75“r,
B
5625
0.4125x 7.47 x10-6 x(02 x752
1
1.7332 x10-2 co2
A
B — 1.7332 x 10-20 Nimm2
5625
Radial stress at periphery of the disc = 1.363 x 10-4 (02 Nmm2
1.363 co2 x 10-4 = A —
1.363 co2 x 10-4 = A
B
4202
176400
3+0.3 7.47 x10-6 w2 x 4202
1
8
0.543 w2
168 ►
IAS & IFS (Objective & Conventional) Previous Solved Questions
MADE 'EASY'
B
176400
0 0.5431363 = A
B
5625
1.7332 x 10-2 w2 = A
...from equations (i)
B x 170775
0.525800)2 =
5625 176400 5625 x 176400
—
0.525800)2 x 5625 x 176400
— 3055 0)2
170775
B
176400
= 0.5431363 0 + 0.017318 0)2
= 0.560450
Maximum Hoop stress occurs at inner radius 75 mm.
A
0.5431363a)2 +
ao. = A+ 2
(1+ 3v)
8
x 30)2r 2
2 305542 1.9
6 x 032 x 752
0.560450) + 75,
,
8 x 7.4 x 10= 0.56045 0)2 + 0.5431 0)2 — 0.998 x 10-2 (02
GA
'Cmax
-max
2
= 375 mN/m2
Goma, = 750 N/mm2 = 1.09337 0
w = 685.827 rad/sec
27EN
60
N = 6550 rpm
Q.13.3 (a) What do you understand by rotating disc of uniform strength?
(b) A turbine disc is required to have a uniform stress of 150 MPa at a speed of 3200 rpm. The
disc is to be 30 mm thick at the centre. What will be its thickness at a radius of 40 mm?
Assume density of disc material = 7800 kg/m3.
[IFS 2012, ME : 10 Marks]
Solution:
(a) Rotors of a steam or gas turbine, on the periphery of which
turbine blades are attached are designed as disc of uniform
strength, in which the circumferential and radial stresses,
developed due to centrifugal forces on account of high angular
speed ware equal and constant and independent of radius.
Both radial and circumferential stresses are maximum at the
centre of the disc, while radial stress becomes zero at outer
radius but circumferential stress reduces to a minimum at the
outer radius. So as, the stresses increase towards the centre
of the disc, thickness of the disc also increases to maintain
constant stress ar = Go = a as shown in figure. Relation between
Fig. 13.1. Disc of uniform strength
Rotational Stresses 4
Strength of Materials
thickness t at any radius r and thickness to at the centre is t = to.e
pa 2 r 2
2ga
Where,
r= radius, at which thickness is t
co = angular velocity of rotation of disc in rad/sec
p = density of the material
g= acceleration due to gravity
a = constant stress
r = 40 mm
p = 7800 kg/m3
(b)
7800 x 9.8
109
= 0.07644 x10-3 N/mm3
27c x 3200
= 335.10 rad/sec
60
a = 150 N/mm2
g = 9810 min/sect
—
p(.02r2
0.07644 x 10-3 x 335.12 x 402
= 0.004666
2 x 9810+150
2ga
pt 2r2
e
296
= e-0.004666 = 0.99534
t = to x 0.99534 = 30 x 0.99534 = 29.86 mm
Note that radius given is too small
Let us take 40 cm radius or 400 mm radius
pco 2r 2
2ga —
43-0.4666
0.07644 x10-3 x 335.12 x 4002
= 0.4666
2 x9810 x150
= 0.6270
t = to x 0.6270 = 30 x 0.6270 = 18.8 mm
169
14
Unsymmetrical Bending
CHAPTER
Q.14.1 A cantilever beam of span 3 m is subjected to a vertical load of 1.0 kN at free end. The crosssection of the beam consists of equal angle 100 mm x 100 mm x 12 mm with one of its legs placed
vertically. Find the magnitude and direction of the resultant deflection. Given /uu = 329.3 cm4.
= 84.7 cm4, E= 2 x 105 N/mm2, centroidal distance = 29.2 mm.
[IFS 2013, CE : 15 Marks]
Solution:
/„ = 329.3 cm4,
/w =
84.7 cm4
Position of neutral axis
/
tam = tan()
100
0 =45°
‘,
x
I„
0 = 45° and tang = 1
tang —
329.3
84.7
= 3.888
r
a = 75.57°
S = deflection at free end
29.2-1
00
Fig. 14.1
2
wL3
3E1„
sin2 0(iruu + cos2 0
W = 1 kN
/„ = 329.3 x 10-8 m4
E = 200 kN/mm2 = 200 x 106 kN/m2
3EI„= 3 x 200 x 106 x 329.3 x 10-8
= 1975.8 kNm2
L = 3 m, = 45°
2
sin2 01W + cos2 0 = \10.5 x1329'312
+0.5
84.3
l'w )
Fig. 14.2
= V7.55765 + 0.5 = 2.8386
S—
3
1975 .8
x 2.8386 = 0.0388 m = 38.8 mm
Resultant deflection in a direction perpendicular to neutral axis.
Unsymmetrical Bending
Strength of Materials
4
171
Q.14.2 Stress concentration factor is not considered harmful for ductile materials
in static loading but for brittle materials, it has damaging effect in both
static and dynamic loading. Justify the above statement giving
illustrations.
[CSE-Mains, 2011, ME : 15 Marks]
Solution:
In a brittle material, due to stress concentration, if ama, = SCF x aav is more
than nut, ultimate stress, and the crack is developed in sample then it is
carried through and specimen breaks. Figure shows a flat sample with a central
hole, amax = 3 (Yaw a SCF = 3, crack may develop at inner edge of the hole and
carried through axes aa. But in a ductile material, the plastic deformation at
the tip of the fine crack blunts the edge of crack and further propagation of
crack becomes difficult in a region where SCF is less. In fatigue load, brittle
material should never be used. Sometimes in the presence of a notch in a
machine member, ductile material behaves as a brittle material and machine
member breaks.
Fig. 14.3
Q.14.3 A cast iron sample when tested in compression fails along
approximately 45° plane from its axis while when tested in torsion
also fails along a 45° (approx) helical plane from its axis. Explain
the reason for such failure and mention about the dominating stress
causing failure.
[CSE-Mains, 2011, ME : 10 Marks]
Shear failure
Solution:
Cast iron is weak in tension and shear, but strong in compression. Its
compressive strength is about three times its tensile strength.
When CI sample is tested in compression, ac is the axial compressive
stress, but shear stress "E is maximum on planes at ±45° to the axes of
the sample. So sample breaks under shear on inclined plane at approx
45° to the axis as shown.
Fig. 14.4
\
- ti
\p 2
Failure under torsion
Fig. 14.5
When a specimen is subjected to twisting moment, angular twist 0 and shear stress "C act on the sample.
Shear stress is maximum at outer radius and gradually reduces to zero at the centre. Principal stress
p1 = +T acts on a plane about 45° to the axis of the sample. Due to angular twist, fractured surface
becomes helical.
Q.14.4 How ductility of structural steel in evaluated from static tension? What is the effect of gauge
length on the value of ductility?
[CSE-Mains, 2011, ME : 10 Marks]
MADE EASY
17 2 0- IAS & IFS (Objective & Conventional) Previous Solved Questions
Solution:
A standard sample of structural steel is tested under tension using Universal testing machine. Original
diameter and original gauge length of the sample are noted down before applying tensile force. Tensile
force on specimen is slowly and gradually increased till the specimen breaks into two pieces. Final
gauge length is measured using an internal callipers.
Find gauge length = L'
Initial gauge length = L
% elongation —
L' —L
L x 100
Ductility is measured by percentage elongation of the specimen
d
A
4
L
Gage length
Fig. 14.5
As per Barba's law
Change in length
SL = L'— L
= bL+c,174
Where b and care Barba's constants
L = gauge length, A= area of cross-section
As the gauge length increases, bL increases but cJ remains constant. Therefore as the gauge length
of the specimen increases, percentage elongation also increases.
Q.14.5 (a) What is fatigue? Define fatigue limit and fatigue strength.
(b) Explain how various factor of design effect the fatigue strength.
(c) What is cumulative fatigue damage? Explain Miner's hypothesis and bring out the drawbacks
of this hypothesis.
[CSE-Mains, 2008, ME : 5 + 5 + 10 = 20 Marks]
Solution:
Behaviour of any material under fluctuating loads or stress cycle is
termed as fatigue. For different stress levels, fatigue tests are
performed on samples and cycles required for fracture of sample
at a particular stress level S1, noted down say it is N1. Then S1 is
known as fatigue strength of material at stress cycles N1.
If a graph is plotted between S,. and
then the stress Se at which
the graph becomes asymptotic is known as stress limit as shown
in the figure.
Various factors affecting the fatigue life are (i) surface roughness
(ii) size factor (iii) stress concentration factor (iv) factor of safety
(v) order of reliability.
s'
S.
I
_
SB
Fig. 14.5
Depending upon the surface roughness fatigue life of member is reduced. As the size increases life of
the member decreases. As the stress concentration factor increases life of member decreases under
fatigue loading. Finally depending upon the reliability factor more reliability factor less is the life.
Unsymmetrical Bending 41
Strength of Materials
17 3
Cumulative fatigue Damage: There Is cumulative effect of stress levels and number of cycles put in at
that level, that forms a fraction of total life upto failure
k n.
N
1.1 i=C
ni = number of cycles accumulated at stress S,
N= total number of cycles upto failure at stress level S,
+ n2 n, nk
Nk = 1
N1 N3 ...
or
+ C2 ... Ci + Ck =
C is the fraction of life consumed by exposure to the cycles at different stress levels.
This is known as Miner's rule or hypothesis.
Drawbacks: Miner's rule becomes invalid if any stress level is more than
than Se (endurance limit stress).
11.11111111
Cry p
(yield point stress) or less
15
General Objective Type
Questions
CHAPTER
Directions: The following 4 questions, items consist
of two statements one labelled as 'Assertion A' and
other labelled as 'Reason R'. You are to examine these
two statements carefully and decide if Assertion A and
Reason R are individually true and so weather the
Reason R is a correct explanation of the Assertion.
Select your answers to these items using the codes
given below and mark your answer sheet accordingly.
(a) Both A and R are true and R is the correct
explanation of A
(b) Both A and R are true but R is not the correct
explanation of A
(c) A is true but R is false
(d) A is false but R is true
Q.1
Q.2
Assertion (A): I, T are channel sections are
preferred for beams.
Reason (R): A beam cross-section should be
such that the greatest possible amount of area
is as far away from the neutral axis as possible.
A channel cross-section of the beam shown in
the figure 15.1 carries a uniformly distributed
load.
w N/m
w N/m
•
CG
Fig. 15.1
Assertion (A): The line of action of the load
passes through the centroid of the crosssection. The beam twists besides bending.
Reason (R): Twisting occurs hence the line of
action of the load does not pass through the
web of the beam.
Q.3
Assertion (A): A thin cylindrical shell is
subjected to internal fluid pressure that induces
2-D stress state in the material along longitudinal
and circumferential directions.
Reason (R): The circumferential stress in the
cylindrical shell is two times the magnitude of
the longitudinal stress.
Q.4
Assertion (A): Brittle material such as gray cast
iron cannot be extruded by hydrostatic pressure.
Reason (R): In hydrostatic extrusion billet is
uniformly compressed from all sides by the
liquid.
[CSE-Prelims, ME : 2000]
Q.5
Match List-I with List-II and select the correct
answer.
List-I
A. Proof stress
B. Endurance limit
C. Leaf spring
D. Modulus of rigidity
List-I I
1. Torsion test
2. Tensile test
3. Fatigue test
4. Beam of uniform strength
Codes:
A
(a) 2
3
4
1
3
1
4
(b) 2
2
4
1
(c) 3
2
1
4
(d) 3
Strength of Materials
Directions: The following 4 (four) items consist of the
statement, one labelled as Assertion 'A' and other
labelled as Reason 'R'. You are to examine these two
statements carefully and decide if the Assertion 'A'
and Reason 'R' are individually true and of so whether
the Reason is a correct explanation of the Assertion.
Select your answers to these items using the codes
given below and mark your answer sheet accordingly.
(a) Both A and R are true and R is the correct
explanation of A
(b) Both A and R are true but R is not the correct
explanation of A
(c) A is true but R is false
(d) A is false but R is true
Q.6
Assertion (A): When an isotropic, linearly elastic
material is loaded biaxially, the direction of
principal stresses are different from those of
principal strains.
Reason (R): For an isotropic linearly elastic
material, Hooke's law gives only two
independent material properties.
0.7
Assertion (A): In theory of torsion, shearing
strain increases radially away from the
longitudinal axis of the bar.
Reason (R): Plane transverse sections before
loading remain plane after the torque is applied.
Q.8
Assertion (A): For a thin cylinder under internal
pressure, a minimum three strain gauges are
needed to know the stress state completely at
any point on the shell.
Reason (R): If the principal stress directions
are not known, the minimum number of strain
gauges needed is three in a biaxial field.
Q.9
Assertion (A): Buckling of long column causes
plastic deformation.
Reason (R): In a buckled column, the stresses
do not exceed the yield stress.
[CSE-Prelims, ME : 2001]
Q.10 Match List-I (stress induced) with List-II
(situation/location) and select the correct answer
using the code given below the lists:
List-I
A. Membrane stress
B. Torsional shear stress
General Objective Type Questions
1 17
5
C. Double shear stress
D. Maximum shear stress
1. Neutral axis of beam
2. Closed coil helical spring under axial load
3. Cylindrical shell subject to fluid pressure
4. Rivets of double strap butt joint
Codes:
A B C D
(a) 3
1
4
2
(b) 4
2
3
1
(c) 3
2
4
1
(d) 4
1
3
2
[CSE-Prelims, ME : 2007]
0.11 Match List-I (Type of thread) with List-II (Use)
and select the correct answer using the code
given below the lists:
List-I
A. Square thread
B. Acme thread
C. Buttress thread
D. Trapezoidal thread
List-II
1. Used in vice
2. Used in lead screw
3. Used in screw jack
4. Used in power transmission devices in
machine tool
Codes:
A
BCD
(a) 2
3
4
1
(b) 2
3
1
4
(c) 3
2
1
4
(d) 3
2
4
1
[CSE-Prelims, ME : 2007]
0.12 Match List-I (Natural of failure) with List-II
(Nature of member) and select the correct
answer using the code given below the lists:
List-I
A. Structural member fails in axial compression
B. A member fails along a 45° helical plane
subjected to torsion
C. A structural member bends and collapse
under axial compression load
D. A member fails in double shear for a joint
176 ►
lAS & IFS (Objective & Conventional) Previous Solved Questions
List-II
1. Knucide joint
2. Long column
3. Strut
4. Cast iron round bar subjected to torsion
Codes:
A
BCD
(a) 3
1
2
4
(b) 2
4
3
1
(c) 3
4
2
1
(d) 2
1
3
4
[CSE-Prelims, ME : 2006]
Q.13 What type of fracture occurs when a brittle
material is under torsion?
(a) Cup and cone
(b) Granular transverse
(c) Granular helicoidal
(d) Smooth transverse
[CSE-Prelims, ME : 2006]
Q.14 Which one of the following is not correctly
matched?
(a) Torsional rigidity : Torque per unit angle of
twist
(b) Modulus of rigidity : Strain energy per unit
volume
(c) Creeping effect : Loss of mechanical energy
[CSE-Prelims, ME : 2008]
Directions: The following items consist of the
statement, one labelled as the 'Assertion (A)' and the
other as 'Reason (R)'. You are to examine these two
statements carefully and select the correct answers to
these items using the codes given below:
(a) Both A and R are true and R is the correct
explanation of A
(b) Both A and R are true but R is not the correct
explanation of A
(c) A is true but R is false
(d) A is false but R is true
Q.15 Assertion (A): Plane state of stress does not
necessarily result in a plane state of strain.
Reason (R): A uniaxial state of stress in general
produces a 3-dimensional strain state.
[CSE-Prelims, ME : 2008]
MADE EASY
Q.16 Which one of the following is true regarding the
fatigue life of a set of identical ball bearings?
(a) Directly proportional to load
(b) Inversely proportional to load
(c) Inversely proportional to the square of load
(d) Inversely proportional to the cube of load
[CSE-Prellms, ME : 2009]
Q.17 Match List-I with List-II and select the correct
answer:
List-I
A. Partial derivative of strain energy w.r.t. load
B. Derivative of deflection
C. Derivative of slope
D. Derivative of moment
List-II
1. Equation for shear force
2. Equation for slope
3. Equation for bending moment
4. Deflection under the load
Codes:
A BCD
3
2
(a) 4
1
(b) 3
2
4
1
(c) 4
2
3
1
1
4
2
(d) 3
[CSE-Prelims, ME : 2003]
Directions: The following item consists of two
statements, one labelled as the 'Assertion (A)' and the
other as 'Reason (R)'. You are to examine these two
statements carefully and select the correct answers to
these items using the codes given below:
(a) Both A and R are true and R is the correct
explanation of A
(b) Both A and R are true but R is not the correct
explanation of A
(c) A is true but R is false
(d) A is false but R is true
Q.18 Assertion (A): To a cantilever beam of circular
cross-section, if a moment is applied with its
axis perpendicular to the axis of beam, no shear
stress will be induced in the beam.
Reason (R): To the above beam, if moment is
applied with its axis along the axis of the beam,
no bending stress will be induced in the beam.
[CSE-Prelims, CE : 2004]
General Objective Type Questions
Q.19 Match List-I with List-II and select the correct
answer using the code given below the lists:
List-I
A. Beam
B. Column
C. Circular section of diameter d of a shaft
D. Close-coiled helical spring
List-II
1. Member subjected to twisting
2. Member used to store strain energy
3. Member subjected to buckling
4. Member subjected to bending
Codes:
A BCD
(a) 4
3
1
2
(b) 1
2
4
3
(c) 4
2
1
3
(d) 1
3
4
2
[CSE-Prelims, CE : 2009]
Q.20 Given that F
+(3.12 ) where F denotes
force and t time; how is r3 described
dimensionally?
(a) MLT-3
(b) MLT-2
(c) LT-4
(d) MLT-4
[CSE-Prelims, CE : 2009]
List-II
1. T-1
2. M LT-2
3. mL.2-r2
4. ML-1T-2
Codes:
A BCD
(a) 1
3
2
4
3
2
1
(b) 4
2
3
4
(c) 1
1
(d) 4
2
3
[CSE-Prelims, CE : 2010]
Q.23 Match List-I with List-I I and select the correct
answer using the codes given below the lists:
List-I
A. Moment of inertia about diametral axis
B. Moment of inertia about an axis tangent to
the perimeter
C. Moment of inertia about an axis through
centroidal axis
D. Polar moment of inertia
List-I I
Q.21 What is the unit vector of the resultant of the
following two forces?
= 21+31+4k;
(a) 61 + 61+ 6k
(c) -21+ 2k
j
+ +
(d) 21- 2k
[CSE-Prelims, CE : 2009]
Q.22 Match List-I with List-II and select the correct
answer using the codes given below the lists:
List-I
A. Modulus of elasticity
B. Work
C. Force
D. Frequency
1.
ico4
32
2.
17704
64
3.
5704
64
4.
704
64
Codes:
A
(a) 4
(b) 1
(c) 4
(d) 1
F2 = 41+31+2k
(b)
41 177
BCD
3
2
1
3
2
4
2
3
1
2
3
4
[CSE-Prelims, CE : 2010]
Answers
1. (a)
2. (c)
3. (b)
4. (b)
5. (a)
6. (d)
7. (b)
8. (d)
9. (c)
10. (c)
11. (c)
12. (c)
13. (c)
14. (b)
15. (b)
16. (d)
17. (c)
18. (b)
19. (a)
20. (d)
21. (b)
22. (b)
23. (a)
178 N.
MADE EASY
IAS & IFS (Objective & Conventional) Previous Solved Questions
Explanations
1. (a)
Both A and R are true, R is the correct explanation
of A.
2.
(c)
A is true but R is false,
Line of action of the load must pass through the
shear centre.
3.
(b)
Both, A and R are true but R is not the correct
explanation of A.
4.
(b)
Both A and R are true but R is not the correct
explanation of A.
5.
(a)
A. Proof stress
B. Endurance limit
C. Leaf spring
D. Modulus of rigidity
2. Tensile test
3. Fatigue test
4. Beam of uniform
strength
1. Torsion test
6.
(d)
A is false, R is true.
7.
(b)
Both A and R are true but R is not the correct
explanation of A.
8.
(d)
A is false but R is true.
9.
(c)
A is true but R is false.
10. (c)
A. Membrane stress
3. Cylindrical shell
subjected to fluid
pressure
coiled
B. Torsional shear stress 2. Close
helical spring
under axial load
4. Rivets of double
C. Double shear stress
strap butt joint
D. Maximum shear stress 1. Neutral axis of
beam
11. (c)
A. Square thread
B. Acme thread
C. Buttress thread
D. Trapezoidal thread
3. Used is screw jack
2. Used is lead screw
1. Used in vice
4. Used in power
transmission
devices in machine
tool
12. (c)
A. Structural member falls in axial compression
B. A member falls along a 45° helical plane
subjected to torsion
C. A structural member bends and collapse under
axial compression load
D. A member falls in double shear for a joint
3. Strut
4. Cast iron round bar subjected to torsion
2. Long column
1. Knuckle joint
13. (c)
Granular helicoid - Type of fracture in brittle
material under torsion.
14. (b)
Modulus of rigidity strain energy per unit volume.
15. (b)
Both A and R are individually correct but R is not
the correct explanation of A.
16. (d)
Fatigue life of a ball bearing is inversely
proportional to cube of load.
17. (c)
dU
A. dW
dW
B.
C.
D.
dy
dx
d2y
dx 2
dM
dx
4. Deflection
2. Slope
3. Equation of BM
1. Shear force
General Objective Type Questions
Strength of Materials
18. (b)
Both A and R are individually true but R is not the
correct explanation of A.
19. (a)
A. Beam
B. Column
C. Shaft
D. Spring
4.
3.
1.
2.
Bending
Buckling
Twisting
Strain energy storage
22. (b)
A. Modulus of elasticity
B. Work
C. Force
D. Frequency
4.
3.
2.
1.
17
41
9
ML-1T-2
mL2T-2
MLT-2
T-1
23. (a)
A. MI about diametral axis
4.
B. AN about an axis
3.
itD4
64
20. (d)
MLT-2, = MLT-4
21. (b)
6i+61+6k
C. Micentroidal axis
2.
D. Polar moment of inertia
1.
-11)E3
61+61+6k
6i
I
64
tangent to perimeter
+ F2 = 61+61+6k
Unit vector —
5704
k
= - + - + -
17704
64
TcD4
32
MINIM