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Theory with Exercises
o
Practice Question Bank
Recruitment Exam Guide
Handbook to
sse
Junior Engineer
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disha
o
Theory with Exercises
o
Practice Question Bank
Recruitment Exam Guide
Handbook to
sse
Junior Engineer
1. Engineering Mechanics and Strength of Materials A-I - A-40
2. Theory of Machines & Machine Desig A-4I - A-82
3. Thermal Engineering A-83 - A-135
4. Fluid Mechanics and Machinery A-136 - A-173
5. Production Engineering A-174 - A-220
(x-yplane)
F3 .'
F2 ./
Fl •.•
»" -:
/
(Complete classification of force system)
1. Coplanar collinear : In this case, all the forces act in the same plane and also have a common line of action. Non-concurrent Non-parallel Parallel Concurrent Non-concurrent Non-parallel
l
Parallel ConcurrentL
CollinearNon-coplanar
J
Coplanar I
Force
l
systemStatics deals with forces in terms of their distribution and effect on a body at absolute or relative rest.
Dynamics deals with the study of bodies in motion. Dynamics is further divided into kinematics and kinetics.Kinematics is concerned with the bodies in motion without taking into account the forces which are responsible for the motion.
kinematics deals with the bodies in motion and its causes.
Force System: A force system may be coplanar/non-coplanar. In a coplanar force system,all the forces act in the same plane. In a non-coplanar force system,all the forces act in different planar. Classification of force system: (For coplanar forces)
Kinetics Kinematics
Dynamics
I
Statics
Itis the branch of Engineering Science which deals with the principles of mechanics along with their applications to the field problems.
Engineering Mechanics can be divided into its sub-groups as below
Engineering Mechanics
I
ENGINEERING MECHANICSI~NfJINI~I~IIINfJ)11~(~II1INI(~S
lINI) srl'111~NfJrl'II
f)lf
)11Irl'I~III1II
..S
SECTION
A : MECHANICAL
ENGINEERING
-P-=_g_=~
sma sin B sin y
Moment of a force : It is defined or the product of the magnitude of the force and the perpendicular distance of the point from the line of action of the force.
a, ~, y=Angles included between three forces P, Q and R then,
Q
Lami's theorem:
According to Lam is theorem, if three forces are acting at a point and the forces are in equilibrium, then the each ofthe three forces is directly proportional to the sine ofthe angle between the other two forces.
Let, P, Q, R=Three forces in equilibrium
( P sin a
J
-1 ( P sin aJ
tan a = :::::>a=tanQ+pcosa Q+pcosa
or,
Let a = Angle between the two forces 'PI and 'Q' a= Angle between resultant 'R' and one of the force ('Q' in this case)
=direction of the resultant then,
Resultant 'R'
=
~p2 + Q2 + 2PQ cos a Psina Angle made by resultant 'R'=
Q + P cos aL?7
Qp
Law of parallelogram :
According to law of parallelogram, if two forces are acting at a point and may be showed in magnitude and direction by two adjacent sides of the parallelogram, then the resultant ofthe two forces will be shown by the diagonal of the parallelogram in megnitude and direction.
Let 'PI and 'Q' are two forces acting at the point '0' Here 'PI and' a' shows the sides ofthe parallelogram and 'R' is the resultant.
z
F}
(Non-coplanar concurrent forces)
(xy plane)
y
4. Coplanar non-concurrent, non-parallel: In this case, the lines of action of these forces act in the same plane but they are neither parallel nor meet intersect at a common point.
---~F2
---~Fl
(xyplane)
Coplanar parallel force: All the forces act in a plane and parallel with each other irrespective of direction. 3.
2. Coplanar concurrent : Inthis case all the forces act in
the same plane and meet or intersect at a common point.
(xy plane)
Engineering Mechanics and Strength of Materials
A-2
Methods of reducing friction :
There are many ways to reduce friction some of them are given as follows:
1. Surfaces ofthe mating parts or contacting surfaces should be smooth
2. Lubrication is also implemented for reducing friction by making surfaces smooth
3. Streamlined shapes should be used because these shapes offers least resistance against air flow or water flow. 4. If the forces are reduced on contacting surfaces, the value
of friction is reduced
5. Lesser contact between the mating surfaces also reduces the friction.
Cone
of friction
Cone of friction : It is defined as the right circular cone with vertex at point of contact of two surfaces and axis in the direction of normal reactions.
Inclined plane
with horizontal w=mg
From fig:tan <I>= : = Il =><I>= tan-1(FIR)
Angle of repose (<X) : When a body rests on an inclined plane, the angle by which the body is at the verge (just) to start moving in terms as angle of repose.
by the resultant of normal reaction with the limiting force of friction with the normal reaction.
21R
F I
w=mg From the fig : R = w= mg
P=F
If, P is less than F, the body will not move.
But, if P is increased aftera stages achieved by limiting force of friction, the bodywill start moving.
Co-efficient of friction (J..t) : It is defined as the ratio of limiting force of friction (F) to the normal reaction (R) between two rigid bodies.
F u> -=>F=IlR
R
Angle of friction (<1»: It is defined as the angle subtended
Horizontal surface F (Frictional for~
~
Some conceptor/terms of friction:
Friction: Friction may be defined as the resistive force acting at the surface of contact between two bodies that resist motion of one body relative to another.
=> Based on the nature of two surfaces in contact, friction in categorised in the following two kinds/types. (a) Static friction: When two contact surfaces are at
rest, then the force experienced by one surfuce is termed or static friction.
(b) Dynamic friction : When one of the two contact bodies starts moving and the other in at rest, the force experienced by the body in motion is called dynamic friction.
R
Surface/ Support R
Action and Reaction: From Newton's third law, for every action there isa equal and opposite reaction.
d
Moment (M) = F x r
Couple : Two parallel forces equal in magnitude and opposite in direction and separated by a definite distance are said to form a couple.
Gravitational law is also known as universal law of gravitation. According to this law, Every substance or body has an attractive force with another substance or body and this attractive force is directly proportional to the product oftheir masses and inversely proportional to the square of distance between their centers. This attrative force is directed along the line which joins the centers of bodies.
Let MJ and M2be the masses of two bodies and 'R' be the
distance between the centers of two bodies. 'F' be the attractive force or force of attraction between those bodies.
Now, According to law, Fo:MJ X M2
GRAVITATIONAL LAW
where
I
F
=m~=m
a
l
In =mass ofthe body v =velocity of the body F =Force acting on the bodya
=acceleration produced in the body.3. Newton's third law of motion: This law states that there is always an equal and opposite reaction to every action.
d(mv)
---=F dt
There are three laws of motion known as Newton's laws of motion.
1. Newton's first law of motion: This law statesthat if a bodyis in the state of rest it remains in the state of rest and ifit is in motion it remains in the state of motion until the body is acted upon by some external force.
2. Newton's second law of motion: It statesthattherateofchange of momentum is directly proportional to the impressed force, and take place in the same direction, in which the force acts.
Momentum=mv
NEWTON'S LAWS OF MOTION Methods of analysis:
(i) Method joints (ii) Method of sections (iii) Caraphical method
Displacement, Speed, Velocity and Acceleration
Displacement: Change of position of a body with respect to a certain fixed reference point is termed as displacement. Displacement is a vector quantity.
Speed: Rate of change of displacement with respect to its surrounding is called as speed of the body.Since the speed of a bodyis irrespectiveof its direction, therefore it is a scalar quantity. Velocity: The rate of change of position of a bodywith respect to time is called velocity.Velocityis a vector quantity. In other way we can say velocityis the speed of a bodyin a particular direction. Acceleration: The rate of change of velocity of a body with respect to time is called acceleration.A negative acceleration is called retardation.
Beam is subjected to following set of forces after the beam is detached from the supports.
(a) Weightofthe beam W acting vertically downwardsthrough mass centre of the beam.
(b) Reaction Rt, normal to the beam at its smooth contact with the corner.
(c) Horizontal applied force P and couple M
(d) Vertical and horizontal reactions (Ravand Rah)extented at the pin connection at B.
=>
Principle of equilibrium/ Equilibrium conditions : According to the principle of equilibrium,A body, either in co-pl-anar or concurrent or parallel system, will be in equilibrium if the algebric sum of all the external forces is zero and also algebric sum of moments of all the external forces about any point in their plane is zero.So, LF =0, LM =0=>
Equilibrium equations for non-concurrent forces LFx = 0' LFy = 0 LM, = 0=>
Equilibrium equations for concurrent forces LFx = 0, LFy = 0 (only two conditions are required)w
~M
The free body diagram (FBD) of the above system can be drawn as in Fig.
FREE BODY DIAGRAMS
A free body diagram (FBD) is a simplified representation of particle or rigid body that is isolated from it's surroundings, and all applied forces and reactions on the body are put together in a diagram. These diagrams are the simplest abstraction of the external forces and moments acting on a physical object. Creating a free body diagram involves mentally separating the system (the portion of the world you are interested in) from its surroundings (the rest ofthe world) and then drawing a simplified representation of the system.
All forces acting on a particle,original body must be considered and equally important. Any force not directly applied on the body must be excluded.
Let us consider a system of a beam loaded and supported as shown in Fig.
Engineering Mechanics and Strength of Materials A-4
Z =Z
xy
yx'Z =Zxz zx'Z =Z~ yzPlane Stress problems are those in which the stress acting in one of the mutual perpendicular directions is assumed to be zero
.. cr =0z Z =0xz Z =0yz
[cr]=[crx
crxy]
cryx cryFor a given stress tensor Z face
o , o .o are normal stressesx y z Remaining are shear stress. STRESS TENSOR
Torsional shear stress
I
I
Axi I B di Direct shear stress ~mg Tensile Compressive Shear Stress (acting parallel to corresponding plane)
I
Normal Stress (acting perpendicular to corresponding plane)I
Type of StressesI
When deformation or strain occurs freely in a direction, stress developed in that direction is zero.
When deformationis restrictedcompletely,or partially stress is developed.Hence strain is the cause of stress.
• NOIE MKS : kgf/cm? 1 MPa = 106N/m2 1GPa = 109N/m2 1 Pa = 1N/m2 1kg;:::::0.1 MPa cm SI : Pa, MPa, GPa
Stress developed in one direction ~ uniaxial state of stress Stress developed in two direction ~ biaxial state of stress Stress developed in three direction ~ triaxial state of stress Units of Stress L ,
·
, •••·
p ---- --- --- ---_ •• _•• _ ... _.-,·
.
, , ,.
.
.
,.
.
, / \ F d cr=-A p STRENGTH OF MATERIALSLoad : It is defined as external force or couple to which a component is subjected during its functionality.
Stress: It is defined as the intensity of internal resisting force developed at a point against the deformation cuased due to the load acting at the member.
Centripetal and Centrifugal Force
Essential force for a circular motion acting radially inwards is called centripetal force which is given by
Fe = mv?r
where m is the mass of the body w = angular velocity
r = radius ofthe circular path
As per Newton's third law of motion, the body must exert a force radially outwards of equal.
ANGULAR ACCELERATION
The rate of change of angular velocity is called angular acceleration.Itis expressed in radls2
dO)
a=cit
ANGULAR VELOCITYThe rate of change of angular displacement of a bodywith respect to time is called angular velocity.
de
0)=-dt
if a bodyis rotating at N r.p.m. then correspondingangular velocity 21tN
0)=
60
rad/sIf the body is rotating 0)rad/s along a circular path of radius r, then its linear velocity (v) is given by
v =(l}r
1 and Foc2
R
On combining the above two expressions, Foc MIM2
R2 F=G MIM2
R2
where, G = universal gravitational constant
=6.67x 10-11NM21kg? ANGULAR DISPLACEMENT
The displacement of a body in rotation is called angular • displacement.Angular displacement is a vector quantity.Angular displacementO can be measuredin radians, degreesor revolutions.
1 revolution = 21tradians = 360 degrees
Strain tensor is used to define the state of a strain at a point
c : normal strain y: shear strain
•
STRAIN TENSOR
b
Shear strain=Shear angle (<l»
BJ
. .. L B B' pIt is defined as the change in initial right angle between two line elements which are parallel toxand y axes respectively.
V=/bt 'bV 0/ ob ot Cv
=--:Y-=T+-';+t
For a sphere,oD
Cv= -
D:diameter of sphere D SHEAR STRESS 'bV Cv=v :
s, +cy +czAnother example of rectangular block is considered p+-~~~----~-~---¥---~-!--+p I I I I : La : I I : L, :
~'
---+'
Compressive Tensile Strain StrainConsider a rod oflength La subjected to load P Volumetric Strain (s.)
I
Lateral Strain (Slateral)I
I
Longitudinal Strain (Slong) Shear Strain Normal Strain StrainI
Strain is defined as the ratio of change in dimension to original dimension.
STRAIN
L2
Elongation of a conical bar under its self weight = ~E
DI
L
: : 8
'1. 1
y=selfweight per unit volume
2 Elongation ofa prismatic bar under its self weight = yL
2E
p
Elongation of a tapered bar subjected to axial load P
'b=~
AE
Ol
l
Zxz=10(i.e., shear stress acting onxface along Zdirection) Zzy = 0
ox = 100'yo =50'zo=25
Elongation of a bar Subjected to axial load P
Units: MPa
[
100 120 10]
rr= 20 50 0
10 0 25
Engineering Mechanics and Strength of Materials A-6
oct
IfJ.! = 0 => - = 0 do J.!= I ---- f--- ---~dr
do"
1..- Lo ...,..
Lr ... " , => Ev= 0v= 0The material neither expands in volume nor contracts in volume. Thus, it is called as incompressible material and for that J.!= 0.5.
Poisson's Ratio
~ used to determine lateral strain theoretically.
I -lateral strain I J.!= longitudinal strain K=oo For HYDROSTATICSTATEOF STRESS (NO DISTORTION,ONLY
VOLUME CHANGES) (J I 1 1 1 ... ",...",..,; -
---~".",..
--+-~(J ---,. .. " 1 ~_+--____,,_,,"''',... 1 1 1 1 1 1 1 (J K= Bulk Modulus (K) Normal stress o 1for a given 't, G ex. -.
Y
Shear Modulus or Modulus of Rigidity
As per Hooke's law,
Shear stress ex. shear strain
1't=Gyl
Engineering Stress(o)= 0" I .
ngma xsection area Instantaneous load
True stress = .
Instantaneous x section Load
'E' is the slope of ovis ~E diagram EL --- B PL--- ~ Yxy=Yyx Strain Tensor in 3D [ Ex Yxy/2 YXZ/2] [EhD
=
YyxJ2 Ey Yyz/2 Y2xJ2 Yzy/2 EzRelationship Between Elastic Constants
E=2G(1 +J.!) E=3K(1-2J.!) 9KG E= 3K+G E 1 G=-x--2 1+J.! E 1 K=-x--3 1-2J.! Value of any Ee ~ 0 Note:J.!cork=0
Young's Modulus or Modulus of Elasticity
As per Hooke's law upto proportional limit normal stress is directly propotional to longitudinal strain
o ex. E)ong
o = E = young's modulus E10ng E
t
=> E10ng.J,=>0IJ..
A material having higher E value is chosen
EMS=200 GPa ECI= lOOGPa
E = 200 GPa
AI 3
.. (oI)MS < (ol)CI < (0)Ai
Elastic Limit: Maximum value of stress upto which a material can be completely elastic.
ProportionalLimit
It is the maximum value of stress upto which materials obey Hooke's Law.
Shear strain like complementary shear stress are equal in magnitude but opposite in direction.
[EhD
=
[Ex YXY/2] Yyx/2 Ey•
Let, L = Length of the bar
F
where,
E = Young's modulus of elasticity A= Area of cross - section L = length of wire
I = increase in length of wire
Extension of a tapered bar:
Let us consider a circular bar whose diameters are DJ and D2as
shownin figure. Let 'F' be the tensile load which is applied axially. Uy =_1_(Stress)2
2E or
EAI where F=---, L
Energy stored/unit volume of wire
u,
=..!.E (Strain)2 21 Energy stored in wire, (U) =
"2
FI_ h
+J.l E2) _ (E2 +J.l Edal - E 2' a2 - E 2·
1-u 1-J.l
Work done (stretching wire) :
When a wire is stretched, the work is done against internal restoring forces. This work is stored in wire as strain energy.
Now, 1
El = E [al - J.l(a2 +
(3)]
E2= ~ [a2 - J.l(a3 +ad]
1
E3= E [a3 - J.l(a1+a2 )]
.. for biaxial state of stress/plane stress problemsa3 = 0but
E3*- 0
al =E(El) +J.la2
a2 = E E2+J.la1
or
Relationship Between Principal Stress and
Principal Strain
(a) Elasticity: Itis the property of the material due to which it regains its original shape after the external load is removed after applied.Perfectly elastic bodies are those bodies which returns to their original shape completely. (b) Plasticity: Itis the property ofthe material due to which it does not regain its original shape after the removed of external load. Plasticity is the opposite of elasticity of external load. Plasticity is the opposite of elasticity. (c) Ductility: Itis the property of the material due to which
if can be drawn into thin wires.The length of deformation is very large in a ductile materiaL
(d) Brittleness: material is said to be brittle if the length of
deformation is very little in tension.A brittle material has lack of ductility. A brittle material tails at a very small deformation.
(e) Malleability: It is the property of the material due to which it can be converted into thin sheets in compression. This property is used in forging, rolling etc.
(f) Toughness: Itis the property ofthe material due to which a maximum amount of energy stored in a material upto fractors. This property is utilized under the action of shock or impact loading.
(g) Hardness: Itis the property of the material due to which it resists cutting, scractehing, pinetration or inditation.
PROPERTIES OF MATERIALS
A ~ Proportional Limit B ~ Elastic Limit C ~ Upper yield point D ~ Lower yield point F ~ Ultimate point G ~ Fracture point DE ~ Yielding region
EF ~ Strain Hardening region FG ~ Necking region
~ Sudden fall of stress occurs from C to D due to slipping of carbon atoms in molecular structure of mild steeL ~ Increase in carbon content increases strength, cast surface
hardness and modulus of resilience.
~ Increase in carbon content decreases ductility. ~ For the most metals, its value is between 0.25 to 0.33.
Eng. stress visEng. strain curve MS under tension test
~---~
~
Engineering Mechanics and Strength of Materials
F
o A-8
Pressure vessel is defined as a closed cylindrical or spherical container designed to store gases or liquids at a pressure substantially different from ambient pressure.
THIN CYLINDERS
(e) Continuous beams: In continuous beams, there are more than two supports.
I
I
(d) Fixed beam: In fixed beam, both of its ends are rigidly fixed into the supporting walls.
I
(c ) Overhanging beam:
In overhanging beam, the supports are not placed at the end of the beam and also one or both ends are entended over the supports.
Simply supported beam: A simply supported beam has both of its ends are supported.
(b)
L ~
~
Types of plans:
Various types of beams are given as follows:
(a) Cantiliver beam: A cantiliver beam has one of its end is fixed and the other end is free.
8L
=
FILl + F2L2 + F3L3Al EI A2E2 A3E3
Ifthe bars are of same material, then EI = E2= E3=E, then,
8L
=
!.[.s_+.!2_+~]
E Al A2 A3
Composite bars:
Let us consider a composite bar which is attached at the top and force F is applied.
Now, F= FI+F2 =o.A,+cr2~
As strains in the bars are equal, then 8L =crILl
+
cr2L2+
cr3L3Now, E E E
1 2 3
Let, LI' L2, L3=Lengths of bars
AI' A2, A3=Area of cross - section of bars EI' E2, E3= Young's moduls of elasticity Here, F=F I = F2= F3
Let, 8L = total change in length
IE
F~ A" E,
I
A2, E, where,8L = Elongation
w = specific weight of bar material L = Length of bar
E=Young is modulus of elasticity P=Mass density of bar material Stresses in bars of variable cross-sections:
Let us consider a stepped bar of different lengths and different cross - sections.
where,
8L = Elongation
w=Area ofweight of bar A=Area of cross - section E = Young's modulus of elasticity L =Length of bar
Case (ii)For coxical bar
2
8L
=
wL =Pg L2 crE 6E 8L=
wL2AE
Elongation of a bar due to self weight: Case (i)For uniform cross - section:
=>
Ifbar is of uniform diameter 'D', then,~
E = Young's modulus of elasticity Extension of tapered bar (8L),
8L= 4FL
1tE DID2
Beams
I
Statically Statically Determinate Beam Indeterminate Beam
I
~
J
J
t
J
l
Cantilever Simply Over Fixed Propped Continuous Supported Hanging Beam Cantilever Beam
Beam Beam Beam
X
I I
Pil
I
ILp
i1
+ve shear force0
X
I I I
Ii P
1
i-ve shear forcepLI
G
II I
Bending Moment Sign Convention X I
C]
ID
Ci)
I I I concave IX upwards +ve bending
(+ve) moment SAGGING BENDING X I
CJ
ID
Ci)
I I I X HOGGINGconvex -ve bending
upwards -ve
HOGGING
BENDING
SHEAR FORCE AND BENDING MOMENT DIAGRAMS
~ SFD and BMD play an important role in design of beams.
~ To design a beam, maximum value of shear force and
bending moment are required which are determined from SFD and BMD.
Shear Force Sign Convention
Moreover,it can be seen from expressionsofEhoopandElongthat
Elong < Ehoop
:. The chances offailure ofthin cylinder is more longitudinally.
pd pd along = -4--' ahoop = t 11eJ 2t 11u pd pd .. a1=
21'
a2=4t
al pd Absolute ~max =2
=4t
_ 8d _ pd (2 _ ) Ehoop - d - 4tE ~ - 6L _ ~(1 - 2 ) Elong - L - 4tE ~ 6V pd Ev= -y=
4tE(5-4~)STATEOF STRESS ATA POINT IN THIN CYLINDER
pd pd
along = 4t' ahoop = 2t
Sometimes11 of circumferentialjoint and longitudinaljoint are given.
In that case, ---If---~ (jlong z ,1 __ -y
)-
x
Example of Thin Cylinder:
Hydraulic Cylinder.
Example of Thick Cylinder:
LPG Cylinder, Steam Pipes.
Assumptions for Thin Cylindrical Vessels
~ Stresses are assumed to be uniformly distributed as
thickness 't' is small.
~ Radial stresses are neglected.
Spherical Pressure Vessel Cylindrical Pressure Vessel on the basis of shape of shell Thin d : diameter d t: thickness -> 20 t Thick d -~ 20 t
Engineering Mechanics and Strength of Materials
Pressure vessel
A-tO
PAy
't=---INA' b
Shear Stresses in Beams
D ~ diameter of log (given)
·. final dimensions of strongest rectangular cross-section are
d
b
x=
b[~]
width should be varied linearly.
Consider a log, out of which a rectangle is to be cut such that it is strongest in bending.
band d ~ arbitrary dimensions of rectangle Zl1 = Z22= Zxx
Ml1 = M22 = Mxx (crb)ll = (crbb = (crb)xx
· . (crb) is independent of 'x' .
If beam is subjected to transverse shear load, the bending moment varies.
· . (crb)varies.
To make beam a beam of uniform strength:-(i) depth is varied.
d = d
Ix
x
~L
depth should be varied parabolically. (ii) width 'b' is varied
(+ve)
11
bx
Beams of Uniform Strength
A beam is said to be a beam of uniform strength when bending stress developed at each and every cross-section is same.
(crb)max Y
(crb)= y .
max
A beam offering higher moment of resistance is stronger. I-section beams are strongest as they have high section modulus.
Fora giveneross-sectionalareaandmaterialsquarecross-section beam is strongerthan circular cross-sectionbeam as
Zgquare > Zcircle This fibre is subjected to tension NeutralAxis is neither in tension norin compression This fibre is subjected tocompression M (crb)max= ±--ZNA
. . ~A
t
=>(crb)max "l.. =>chances of failure "l.. For a given beam, (crb)ex yMR : moment of resistance offered by plane of cross-section of beam.
(crb) :bending stress at a distance 'y' from Neutral Axis.
R : Radius of curvature. E :Young's modulus.
INA : area moment of inertia of plane of cross-section about Neutral Axis.
From bending eq",
crb=
i ~
to be used when 'R' is known. As BM =constabove beam is under pure bending.
Bending Equation
I
-ve ML..---lP
-0
~~
I I MA=:P(CD) M ~ P(CD) I I I Bending StressNormal stresses introduced due to the bending of a shaft / member.
Pure Bending: If the magnitude of bending moment remains constant throughout the length of beam, the beam is said to be under pure bending.
---)-e ~
+ve ---)- Deflection upwards (+ve)e
= dy dxJ
Mxx
+C,=
EI(:)
--> slope equation is obtianedIf
Mxx + C1x + C2=
EI(y) ---)-deflection eq"Sign Convention
Deflection of beams plays an important role in design of beams for rigidity criterion.
The expressions of deflections are further used for determination of natural frequencies of shaft under transverse vibrations.
For a cantilever beam under any loading condition deflection is maximum at free end.
In simply supported beam, deflection is maximum at mid-span (when beam is subjected to symmetric loading only).
Relationship between R, q and Y
e :
slope Y : deflection R : radius of curvature d2y Mxx -- - --dx2 Rsr.,
DEFLECTION OF BEAMS 9 I----~ rmax="8
tavg~ Square with DiagonalsVertical: ~ T section:
Shear stress Distribution
1'max=
"2
1'avg' 1'NA="3
1'avg 1'max 9 --=- =1.125. 1'NA 8 4 3 4 in a circular cross-section1'max="3
1'avg·---)- For square, circle, rectangle, 1'NAis the maximum shear stress.But in triangular cross-section, it isn't so. In triangular cross-section, 4 K=-3 For circle, A= bd 3 K= -2 A= a2 3 K= -2 P
where,1'avg= A·
Expression for Maximum Shear Stress Across Various Cross-Sections .. ra: y2(parabolicvariation) Asr o: f(y2) :. As
'
s
'
t
t ~ at extreme fibresl'=0 1'= b By using the above formulae, we getP shear force on plane of cross-section.
A area.
y distance of hatched portion from neutral axis.
INA: moment of inertia of entire cross-section about neutral
axis.
b width.
Consider a Beam of Rectangular Cross-Section
~ I section:
Engineering Mechanics and Strength of Materials A-12
WL WL3 Mmax =
4;
Ymax = Yc
= 48EIWL2
9 =9
=-max A, B 16EI
Case III: Simply supported beam subjected to uniformly distributed load
L
ML ML2
9max=9BA=, 2EI; Ymax=YC=--· 8ill
Case II: Simply supported beam subjected to concentrated point load 'W' at mid-span.
W 1
I
1A
rl
~*~
C
---~1
B4$
Ll2~
G
AMr---: ---'ltF---C ---::I~ ~4$
:G
_)
1 Ll2 >1 Ll2 1 IE '" WL2 5 WL39B = 9
c
= 9max = 8EI' Ymax = Yc
= 48EI
CaseVI: Cantilever beam subjected to uniformly distributed load over half its length from fixed end.
Aro~o~WN/m
DC7 WL4 WL3
Yc = Ymax = 384
ill'
9max = 9B = 9c
= 48EIExpressions for Deflections in Simply Supported Beam CaseI: Simply supported beam subjected to pure bending.
Alr(---
L
-
/
2
--~!_
B
---~OC
ML ML2
9max = 9B = EI' Ymax = YB = 2EI
CaseV: Cantilever beam of length 'L' subjected to point load 'W' at its mid-span.
J.I---"
-L-DB
)
~~.~---~>M
CaseIV: Cantilever beam subjected to concentrated moment 'M' at free end.
WL4 Ymax = YB = 30EI
WN/m
A L
IB
CaseIII: Cantilever beam subjected to uniformly varying load WL4
Ymax=YB=
8EI
W N/m
A
For cantilever,y = Ymaxat x = 0
_WL3 .. Ymax=~.
CaseII: Cantilever beam subjected to uniformly distributed load
L
EId4y __,.4times integration to / Wxx = dx4----r
J? obtain deflection 'y'
load intensity
Expression for Deflection in Cantilever Beams CaseI: Cantilever beam subjected to point load W at free end
X W
EI d3Y ~ 3 times integration to
JCFxx = dx3 obtain deflection 'y'
shear force
9 (-ve Deflection downwards (-ve)
Also,
TL 9 = GJ
GJ
i
=>9 ,j.. =><P,j.. =>r ,j.. =>chances of failure ,j.. ~ A shaft offering higher value of Tr' has more strength.Shafts with high value of polar section modulus are preferred. ~ Torsional Rigidity GJ : Torsional rigidity d K= -D 9A' = 9B, = 9C' = 9A = 9B =ge <PA= <PB= <Pe= <P <PA'= <PB'= <PC'« <p) T 'tmax= -Zp
Zp : polar section modulus
. 7t 3
For sohd shafts, Zp = 16 d
For hollow shafts, Zp = 1: D3 (1 - K4) D : outer diameter
d : inner diameter tA = 'tB = 'te = 'tmax
'tA' = 'tB' = 'te' = 'tmax
I C
Cross-section of a shaft at free end
-B
Moreover, <pex rand 9 exL
A
I
9 : maximum angle of twist. <P:maximum shear angle. J : polar moment of inertia. Tr :Twisting moment R9 Now,<p=
L
R L J Torsion Equatione
:
angle of twist <1>: shear angleL:distance of cross-section from fixed end
--
-
~~
::~
_
~
_
~
~~/:
~(
~
~~----~
L
---~
Cm~P Shear Stress Distribution
Pure Torsion
A member of a shaft is subjected to pure torsion when the magnitude oftwisting moment remains constant throughout the length of shaft. TORSION 5 WL3 Y =---W=W Be 48 EI' e A 5 WL3 YeB= 48
ill'
WB=W We YeB = WB YBe A Deflection atC due to load at B Deflection at B due to load atC ,?ILoad at C = We YeB ~ ~ Stiffness of beam = Max. deflectionHigher flexural rigidity is an indicative of higher stiffness
of beam but lower deflection and slope.
~ Maxwell's Reciprocal Theorem
(valid for beams under point load and having same L, E and I) Load
/Y
c doesn't give max. deflectiono
= Wb (a2 - ab)s/
c 3EIL doesn't give max. slope b 5 WL4 WI}Ymax= Yc= 384
ill ;
9max= 9B = 9A = 24EI Case IV: Simply supported beam subjected to a concentratedpoint load acting not at mid-span
W
1
Engineering Mechanics and Strength of Materials A-14
1
f=
a2~ (end fixity coefficient)
Assumptions
~ The self weight of column is neglected. ~ Crushing effect is neglected.
~ Flexural rigidity is uniform. ~ Load applied is truly axial.
~ Length is very large compared to cross-section. . . Pe o: f [E, Imin' end conditions, L2]
2
1t E Imin
.. Pe = 2
Le
Pe : Euler's buckling load. Imin : min [Ixx and Iyy].
L, : effective length of column. L : actual length of column.
L =aL
e
4
length fixity coefficient are more.Euler's Formulae
Short Columns (fail due to crushing) Long Columns
(fail due to buckling)
Medium Columns (fail due to buckling as well as crushing)
~ As the length of structure, chances of it failing by buckling
Columnis defined as a vertical structural member which is fixed at both ends and is subjected to an axial compressive load.
Strutis defined as a structural member subjected to an axial compressive load.
All columns are struts but vice-versa isn't true. THEORY OF COLUMNS = -3. 6. T1= TA' T2= TA - T. TA+Tc=T. 91 + 92 = <1>0
=>
91 = (-92) 3T=>
TA=4
·
G1J1=G2J2 1. Net TM = T (anti-clock) Rxn = T (clock)1'.
CD
T=T1+T2 9=>
T= (G1J1 + G2J2) L 91 = 92 = 9Shafts with Both Ends Fixed
CD
T Shafts in Parallel Shafts in Series ~ Torsional Stiffness (q) 2. T GJ 3. q=-=-9 L 4.~ Torsion of a Tapered Shaft
2 2
SE=_!_T9=_!_ T L=2._(AL).
2 2 GJ 4G
where
T :twisting moment.
Zp : polar section modulus for circular x section.
~ =
C~)d3
T
t=-Z 'p
SE of bar = work done by load P
1 P2L cr2 crEAL
Strain energy of bar= -Po=-- =- x AL=--.
2 2AE 2E 2
~ Strain energy of solid circular shaft subjected to torsion
P
Strain energy is defined as energy absorbtion capacity of the component during its functionality.
Resilience is energy absorbtion capacity of the component
within elastic region.
Energy absorbtion capacity of a component just before fracture is known as toughness.
STRAIN ENERGY METHODS
oe : buckling stress
n2E
cr
=--e s2
S
t
=>
Pe .,l..=>
buckling tendency is increased (S)sc < (SMC)< (SLC) SC:Short Column MC : Medium Column LC : Long Column For steels, if S ~ 30=>
short column S > 100=>
long column 30 < S ~ 100=>
medium column I I 1 -=-+-RR PE Pc where, PR= Rankine's LoadPE=Crippling load by Euler's formula Pc=crushing load
h cry.A
were PR
=
_--=- __l+a(~r
whereK=radius of gyrotion (minimum) a = Rankini's constant
A=Area of cross - section of column
Slenderness Ratio
~ Used to compare buckling loads of various columns having same material and same cross-section.
a4 nr4 na4 a4 I = - I = -=--=-1 12' 2 4 n2(4) 4n (Pe
)l
=
4n 072=
4n(0.72) _ (Pe)2
12· 12 -0.513 . . (2) is stronger. Rankines formula:Itis a combination of Euler and crushing load.Itis also known as Rankine Gordon formula.
(2) (1)
If remaining all other parameters are same, (Pe)BF > (Pe)FH > (Pe)BH > (Pe)FF Which of the following column is stronger?
Engineering Mechanics and Strength of Materials A-16
~
Both Ends Both Ends Fixed and Fixed and Hinged Fixed Hinged Free
(BH) (BF) (F &H) (FF) 1 1 a 1
-J2
2 2 1 1 11=- 1 4 2 -0.2 4Badboys2
Now,Let o. (maximum principal stress)and o, ora3(minimun principal stress) and a by the yield stress.y
For design creterion, maximum principal stress must not exceed the working stress (aw)
al,2 ~away
For considering yield creterion, a1=±ayora2=±ay
This theory is utilized for brittle meterials.
(b) Maximum principle strain theory (St venant's theory)
According to this theory, material failure will take place during tensile testing under a three dimensional complex stress starts system when maximum strain value reaches the value of strain due to yielding.
0"3
J 0"2
~,.
Various thories of failure are given as follows:
(a) Maximum principal stress theory or Rankine's theory
According to this theory, material failure will take place when the maximum principal stress exceeds the value of yield stress under a state of complex stress system during of yield stress under a state of complex stress system during a tensile test.
Wb Wa
e
e
In this case using the above relation, we get
Wa2b2 U=---6EIf Theories of failure: ( b A Mxx : moment at section x-x X I I r- ~;
B
~
~
~1(---7)M X x ~x =-M L M2 M2L U ~!
2Eldx~ 2EI W 2 U=.!_
P8= 2P L A 2 nd2E UB = UI +U2 2p2 (~) p2 (~) nd2 E + 2nd2 E = 0.5 UA" STRAIN ENERGY DUE TO BENDINGb
(M
)2
u=J
xx dx 2EIxx a U : strain energy P (B) P (A) Modulus ofResilience~ Two bars A and B are as
shown:-Modulus of Toughness EL PL ~2
T
SE = -- (AL) (1 +K2), where t = 4G ZpProof Resilience: It is the maximum strain energy stored up to elastic limit.
Modulus of Resilience is proof resilience per unit volume. Modulus of Resilience is the property of material. Proof Resilience is function of volume of component.
0" d K= -D K =0 for solid K<l
~ Strain energy of hollow circular x section shaft. d:Inner diameter.
D : Outer diameter.
1 61= E [al -Y (a2 + (3)] 1 62= E [a2 - Y(al +(3)] 1 63= E[a3-y (al +(2)]
Now,According to failure creterion,
10"1 - r0"2 - r0"3 = 0"Y
I
(c) Maximum shear stress theroy or Tresca's theory:
According to this theory, material failure will take place.
Itmaximum shear stress in complex stress state will be equal to the value of maximum shear stress in simple tension.
Ifal=maximum principal stress
a2= minimum principal stress a =y yield stress
then, ay=al -a2
(d) Maixmum strain energy theory :
According to this theory, material failure will take place under complex stress state, when total strain on the body or specimen reaches the value of strain energy at elastic limit in simple tension.
[(at +a~ +(J~ )-2y(ala2 +a2 a3 +a3aI)] ~ a~
Let, 6)' 62, 63=three principle strains 6 = strain at yielding
y
61,2,3::;6y
Now,
A-IS Engineering Mechanics and Strength of Materials
PL PL
(a) -- (b)
-2AE AE
(c) --PL2 (d) --PL2
AE 2AE
13. In the above question, if w be the total weight of the bar hanging fixed at one end,then elongation (8L) will be equal to: (a) 8L = wL (b) 8L = 2wL AE AE 2 8L = wL (c) 8L = wL (d) AE 2AE
12. IfP = axial load applied,A = cross- sectional are ofuniform circular bar, L = length of the bar, E = Young's modulus of elasticity, then elongation of the bar will be equal to :
(b) Uu =..!..E(strain) 2 1 (d) Uu = - E(stress) 2 (a) U; = ..!..E(strain)2 2 (c) U,= ..!..E(stress)2 2 (c)
--=--=--cos2a cos2J3 cos2 y (d) None of these
5. Streamlined shapes offers resistance against air flow or water flow, the magnitude ofthe resistance is :
(a) Least (b) Maximum
(c) Negative (d) Positive
6. Ifthe forces are reduced on contacting surfaces, the value offriction:
(a) increases (b) decreases
(c) remains constant (d) None of these
7. The property due to which the material can be drawn into thin wires is knows as :
(a) Malleability (b) brittleness (c) Ductility (d) Elasticity
8. The property due to which the material can be converted into thin sheets is known as :
(a) Ductility (b) Malleability
(c) Hardness (d) Resilience
9. The property of the material due to which the maximum amount of energy stored in a material upto fracture limit is called as:
(a) Hardness (b) Resilience
(c) Plasticity (d) Toughness
10. The property of the material due to which it resists against indentation is known as:
(a) Hardness (b) Toughness
(c) Elasticity (d) None of these
11. The work stored in a stretched wire in the form of strain energy per unit volume of wire is given by:
C
A
B
A B C
(a) -- =-- =
--cosu cosJ3 cosy
A B C (b) --=--=--sm u sinJ3 sin y C A 4. 3.
Iftwo co-planar forces 'PI and 'Q' are acting at a point and '9' being the angle between them and also resultant 'R' is making an angle awith force Q, then the magnitude of resultant will be equal to:
(a) R =
Jp
2 +Q2 +2PQsinQ (b) R =Jp
2 +Q2 -2PQsin9 (c) R=~p2 +Q2 +2PQcos9(d) R =
Jp
2 +Q2 -2PQcos9In the question number 2, the direction or angle mode by the resultant will be equal to:
(a) u
=
tan-I(Q:~::se)
(b) a= tan-1 ( Pcos9 ) Q+Psin9 (c) a= tan-1 ( Psin9 ) Q-Pcos9 (d) a= tan-I ( P cos9 ) Q-Psin9Which of the following expressions represents Lami's theorem, ifA,B, C are three are in equilibrium and as shown in figure.
2.
(a) F=G MIM2 (b) F=G (MIM2)2
R2 R2
(c) F=G--MIM2 (d) F=G (MIM2)2
R R
If M1 and M2 are two masses of two bodies, 'R' is the distance between their centers then which ofthe following expression represents gravitational law 'G' is universal gravitational constant. B 1.
...,
EXERCISE
I···..
Badboys2
wL4 wL4 (d) 8max = 8B= --, Ymax = YB
=--2EI 3EI
28. Ift= shear stress, G = shear modulus and v = volumn ofthe body then the expression of strain energy stored withing (a) 8max = 8B= --,wL2 Ymax =YB =--wL2
3EI 5EI
(b) 8max = 8B= --,wL3 Ymax =YB =--wL4
6EI 8EI
(c) 8max = 8B= --,wL6 Ymax =YB =--wL6
6EI 4EI
27. Ifcantilever beam is subjected to uniformly distributed load (UdI), then expressions for deflection are given by:
l=
G8=!_
(d) "["max L R T, =G8= "["max L J R (c) "["max L J G8 R Tr =G8= "["max (a) J L R r (c) v = - (d) y= 0)2r (J)24. According to Hook's Law, stress is directly proportional to strain within :
(a) Plastic limit
(b) yield point
(c) elastic limit ofproportionality (d) None of these
25. The value of slenderness ratio (s) for short column ofsteels is in the range of:
(a) s~30 (b) s~30
(c) s<20 (d) None
26. Torsion equation is given as: if8 = maximum angle of twist, J =Polar moment of inertia, T r = Twisting moment,
R8
<I>= maximum shear angle =
L
23. If the body is rotating atoi radls along a circular path of radius 'r', then its linear velocity (y) is given by:
(a) y= r20) (b) y= rro 21tN (b) -rad/s 360 21tN (d) -rad/s 180 21tN (a) --rad/s 120 21tN (c) --rad/s 60
(b) Rate of change of momentum is inversely proportional to impressed force and takes place in the direction to opposite of force acting
(c) To every action there is always in equal and opposite reaction
(d) None of these
22. If a body is rotating at N rpm, the corresponding angular velocity will be equal to :
20. Equilibrium equations given for non - concurrent forces are given as :
(a) EFx = 0, EFy = 0, EM =
°
(b) EFx = 0, EFy =
°
(c) EFx=O,EFy=O,EM:;tO (d) None of tliese21. Newton's second law of motion states that:
(a) Rate of change of momentum is directly proportional to the impressed force and takes place in the direction of force acting
(b)
"v"
(')1-(')22 2
(d) o y
=
o1 - o219. If(')1'(')2and (')yare maximum principal stress, minimum principal stress and yield stress, then according to maximum shear stress theory, which of the following expression satisfies: (a) (')y=(')1+(')2 2 2 (c) (')y
=
(')1+(')2 1 1 1 1 1 1 (a) -=-+-PE PR (b) -=-+-Pc Pc PE PR 1 1 1 (c) - =--- (d) -=-+-PR PE Pc PR PE Pc P (Ll L2 L31 (a) 8L = E Al + A2 + A3) 8L = PE(.!1_+~+ L31) (b) Al A2 A3 P(c) 8L = p(LIAI + L2A2 + L3A3)
(d) 8L = PE (LIAI + L2A2 + L3A3)
15. A beam whose one of its ends is fixed is known as : (a) simply supported beam
(b) continuous beam (c) cantilever beam (d) overhanging beam
16. A beam whose both ends are fixed rigidly into the supporting walls is called as :
(a) continuous beam (b) fixed beam (c) cantileverbeam (d) None of these
17. A beam whose both ends are supported is known as : (a) simply supported beam
(b) fixed beam (c) overhanging beam (d) continuous beam
18. IfPR = Rankin's Load, PE= crippling load by Euler's formula ~nd.P c = crushing load, then Rankin's formula for columns
ISgiven as :
P~ A"E
I
A"E A3,E ~PIE ~IE ~IE ~I
L1 L2 L3
14. Ifin case ofa stepped bars of same material i.e., E =E, = E2 = E3 as shown in figure, then elongation ofthe bar will be
Engineering Mechanics and Strength of Materials A-20
42. In case of curved beams, the location of the neutral axis does:
(a) coincide with the geometric axis (b) lie at the top of the beam (c) lie at the middle of the beam (d) coincide with normal axis
(c) (b) Lx Al. 2L (d) ~L L
~L
~L
L (a)40. Range of Poisson's ratio for steel is given by:
(a) 0.21-0.22 (b) 0.23-0.27
(c) 0.37-0.43 (d) 0.57-0.63
41. IfL = original length of specimen,&= Increase in length, then the strain (E) will be equal to :
Dr-D~
(d) 4Dl
(c) (a)
IfDI = external diameter of short column D2 = internal diameter of short column F = external load applied
The highest value of eccentricity will be equal to :
Df-D~
(b) 8D1
39.
(c)
PP (l-!':.)
(d)PP(I-!':. )
2tE 2 3tE 2
38. If the both the ends of the column made of mild steel are hinged, then Rankin's constant value will be equal to :
1 1 (a) -- (b) --7500 6500 1 1 (c) -- (d) --5500 4500 (c)
~('L~)2
+('LFv )2
+2('LFH )('LFv)
(d)~('LFH)2
+('LFy )2 -2('L~)('LFv)
36. In the above question, the direction or angled of the resultant will be give by:
LFy
(a)tana=--Lftl
(b)tana=Lftl
LFy
(c) tan a=2)H x2)v
(d) tana=2x
LFH
x~:)v
37. IfD = diameter of thin cylindrical shell L=length
t = thickness
~ = internal pressure, J..l= Poisson's ratio then Hoop strain will be given by:
(a) PP(2-!) (b) PiD(1_!:)
2m 3J..l 3m 3
~
(b)
~
YF:
33. A cyelindrical elastic body subjected to pure forsion about its axis develops:
(a) compressive stress in a direction 45° to the axis (b) shear stress in a direction 45° to the axis (c) tensile stress in a direction 45° to the axis (d) None of these
34. The forces whose line of action lie on the same plane and also must at a point is known as :
(a) co-planar non concurrent forces (b) co-planar concurrent forces (c) Non - coplanar concurrent forces (d) Non - coplanar-Non concurrent forces
35. When number of forces are acting on the body and
L
FHand
L
Fv be the algebric sum of all the horizontal forces and the algebric sum of all the vertical forces, then the resultant will be given by;Longitudinal strain Lateral strain (a) Lateral strain (b)
Longitudinal strain
stress strain
(c) strain (d) stress
32. Poisson's ratio is described as the ratio of:
29. IfoI' cr2,cr3are three principal stresses, J..l= Poisson's ratio and E = Young's modulus of elasticity, then the expression for strain energy/volume is given by
(a)
!
[crt + cr~+ cr~+ J..l(crlcr2+ cr2cr3+ cr3crd] (b)_!_[
crt + cr~+ cr~- J..l(crlcr2+ cr2cr3+ cr3crt)] E (c) 2~ [ crt + cr~+ cr~- 2J..l(crlcr2+ cr2cr3+ cr3crd] (d) _1_[crt +cr~ +crj +2J..l(crlcr2+cr2cr3+cr3crd] 2E30. Impact strength of a material represents:
(a) Hardness (b) Resilience
(c) Ductility (d) Toughness
31. If'i' is the actual length of column andIEis the effective length of column, then ifboth ends of a column are fixed, then the effective length (IE) will be equal to:
I
(a) IE
="2
(b) IE=21I
(d) IE
="4
the body is given as :
'[2 '[ (a) -xV (b) -xV 2G 2G '[2 r (c) -xV (d) -xV G G
Badboys2
(a) 1600x 10--6 (b) 400x 10-6 (c) 800x 10-6 (d) 300x 10-6
65. In terms of Poisson's ratio (u), the ratio of young's modulas of elasticity(E) to shear modulus (C) of elastic material is :
(a) 2 (1-u) (b) 2 (1+ u)
(c)
(1-
Jl)
(d)(1
+Jl)
2 2
(a) 300MPa (b) 200MPa
(c) 100MPa (d) 400 MPa
62. A rod of length I and diameter 'd' is subjected to a tensile load P which of the following do we need to calculate the change in diameter.
(a) Young's modulus of elasticity (b) Poisson's ratio
(c) Shear modulus
(d) Young's modulus of elasticity and shear modulus 63. An elasti~ body is subjected to a tensile stress O"tand a
compressive str~ss O"ei~ its perpendicular direction.O"tand 0"eare not equalInmagnitude,then on the plane ofmaximum shear in the body, there will be:
(a) normal stress only (b) shear stress only
(c) normal and shear stress both (d) Maximum shear stress only
64. Ittwo principal strains at a point are 1000 x 10-6 and - 600x 10-6,then the maximum shear strain will be equalto
20emO 10 em 55. Poisson's ratio generally depends on:
(a) Material of specimen (b) Area of cross section (c) Magnitude ofload (d) None of these
56. Which of the following has the largest value of Poisson's ratio?
(a) Mild steel (b) Rubber
(c) ceramics (d) stainless steel
57. A wire is stretched by a load, ifits radius is doubled the young's modulus of elasticity of material of wire will become:
(a) trippled (b) doubled
(c) No change (d) one fourth
58. The value of Poisson'sratio for aluminium material is equal to:
(a) 0.33 (b) 0.43
(c) 0.53 (d) 0.63
59. If o"w= working stress, O"u= ultimate stress then the which ofthe following relation is free?
(a) O"w=O"u (b) O"w<O"u (c) O"w>O"u (d) None of these
60. The point of contraflexure is found to be in which of the following beam?
(a) cantilever beam (b) Simplesupportedbeam (c) overhanging beam (d) None of these
61. Alarge plate (uniform) consistingof a rivet hole is subjected to uniform uniaxial tensilestressof 100MPa. The maximum stress in the plate will be equal to :
1 (c) 0" ocz2 (d)
O"boc-b z2
51. Neutral plane ofa beam is defined as the plane: (a) whose length changes during deformation
(b) whose length does not change during deformation (c) which lies at top most layer
(d) None of these
52. In case of a continous beam, which of the following statement is true?
(a) Ithas two supports at ends only (b) Ithas less than two supports (c) Ithas more than two supports (d) None of these
53. Stiffness is measured in which of the following: (a) Modulus of elasticity (b) Toughness (c) density (d) ultimate strength 54. Percentage elongation is associated with which of the
following terms during tensile test? (a) Malleability (b) creep
(c) Hardness (d) ductility 1 (a) O"boc-z 50. 49. 48. 47. 46. 45. 44.
In case of curved beams, the bending stresses are distributed in the shape of:
(a) Parabola (b) ellipse
(c) circle (d) Hyperbola
Which ofthe followinghas given maximum principal stress theory
(a) Rankins (b) Tresca
(c) ST. venant (d) Mohr
Which of the following has given maximum shear stress theory:
(a) Rankins (b) Tresca
(c) Mohr (d) ST. venant
Maximum shear stress theory is utilized for which of the followingmaterials.
(a) brittle material
(b) ductilematerial
(c) brittle and ductile materials (d) None of these
Maximum principal stress theory is used for which of the followingmaterials:
(a) ductile and brittle materials (b) ductile materials
(c) brittle materials (d) None of these
If llwefficiency of weldedjoint, llR = efficiency of riveted joint, then which of the following relation is true:
(a) llw>llR (b) llw<llR
(c) llw=llR (d) llw~llR
When the cyclic or repeated stresses are applied to the material, then its behaviour is termed as :
(a) creep (b) fatigue
(c) stiffness (d) endurance
I( O"b= stress in a beam, z = section modulus, then, which of the following expression represents the relation between them:
43.
Engineering Mechanics and Strength of Materials A-22
(d) Y "C aL (c) Y (a) ac (c) -= E aT
Free Body diagram shows:
(a) No forces are acting of the body
(b) All the internal forces acting on the body
(c) All the internal and external forces acting on the body (d) None of these
During tensile testing for cast iron specimen, the stress-strain curve shows:
(a) No yield point
(b) upper yield point only
(c) lower yield point only
(d) Both upper and lower yield points
In a stress strain curve, the area under stress strain curve upto fracture shows which of the following property:
(a) Hardness (b) Ductility
(c) Toughness (d) Brittleness
IfG = Modulus of rigidity,"C= shear stress, Y= shear strain,
aL = Longitudinal stress,EYL= Longitudinal strain, then
the expression for 'G' will be given by:
75. True stress is associated with:
(a) Instantaneous cross - sectional area
(b) Average cross - sectional area
(c) Original cross - sectional area (d) Final cross - sectional area
76. The unit of stress in SI system is given as :
(a) N!mm2 (b) N/m2
(c) Kgim2 (d) None of these
77. IfaT= True stress, ac = conventional stress, then their
relationship is represented by : where E= strain n2EI n2EI (a) 4L2 (b) 2L2 n2EI n2EI (c) 8L2 (d) 16L2
74. For the case of the slender column of length L, flexural
rigidity EI built in at its base and free at the top, Euler's critical bucking load will be equal to :
1 .. 1 ..
(a) - xongmal value (b) - xongmal value
2 8
1 .. 1 .. 1
(c) - xongmal value (d) - xongmal va ue
4 16
73. Ifthe length of the column is doubled, the value of critical
load becomes: -8F (d) nd2 (c) Fi 2FI (a) (b) -4 9 78. FI FI (c) - (d) -9 3
71. The second moment of a circular area about the diameter is
given by if'd'is the diameter: 79.
(a) nd4 (b) nd4
64 32
(c) --nd4 (d) --nd4
16 8 80.
72. A circular rod of dimeter 'd' and length 3d is subjected to a
compressive force F acting at the point as shown in figure. Then the stress value at bottom most support at point A.
3d 81.
1
~)
(~
F6F -12F
(a)
nd2 (b) nd2
(a) Shear only (b) bending only (c) twisting only
(d) Shear and bending both
70. A concentrated load F acts on a simply supported beam of
I
span l at a distance of"3 from the left - end.The bending
moment at the point of application ofload is given by :
66. If the principal stresses in a plane stress systems are 100 MPa and 40 MPa, then maximum shear stress will be equal to:
(a) 30 (b) 40 (c) 100 (d) 50
67. A thin cylinder of 100 mm internal diameter and 5mm
thickness is subjected to an internal pressure of 10 MPa
and a torque of 2000 Nm. The magnitude of principal
stresses will be equal to:
(a) a1=1098MPa,a2=41MPa
(b) 0'1= 502 MPa,0'2= 62 MPa
(c) 0'1=2018,MPa,a2=46MPa
(d) 0'1= 702 MPa,0'2= 88 MPa
68. In case of simply supported beam on two end support, the
value of bending moment is maximum will be : (a) On the supports
(b) at mid - span
(c) where there is no shear force (d) where the deflection is maximum
69. A steel cube is subjected to tangential force on its top
surface and its bottom is fixed rigidly as shown in figure:
then the deformation of the cube will be due to:
)P
nnlnmllLm
(b) S< 32 (d) S~32
97. Under the action of torsion, the shear stress at the centre of a circular shaft is equal to :
(a) maximmn (b) minimum
(c) zero (d) None of these
98. If two shafts are connected in paralled position, then (a) angle of twist of both shafts are equal
(b) angle of twist of both shafts are unequal (c) torque of both shafts are equal
(d) None of these
99. If two shafts are connected in series, then (a) torque of both shafts are equal
(b) angle of both shafts are equal (c) shear stress of both shafts are equal (d) torsional stiffness of both shafts are equal
100. Ifs = slenderness ratio, then the value of's' for short column should be in the range of :
(a) S=32 (c) S>32 (c) (b) TxJ (d) Tx
o
T J Te
(a)95. Ifac= radial stress, ah = hoop stress, then the radial stress value in a thin spherical vessel will be equal to:
(a) Zero (b) 2ah
(c)
a;
(d) None of these96. If T = Torque transmitted, O = angle of twist,J = Polar moment of inertia, then torsional rigidity ofthe shaft will be equal to: (d) 16 (c) 2 1 8 (b) 4 (a) (c) 3ae (l-v) (d) 3ae (l+v) 2E 2E
Ifwe round a thin cylinder with a wire under the application of tensile stresses, then the hoop stress will be of nature:
(a) twisting (b) compressive
(c) shear (d) tensile
The ratio of maximum shear ("Cmax)to hoop stress (aH) in
case of thin cylinderical pressure vessel is equal to
PR PR (a) - (b) 2H H PR PR (c) (d) -4H 8H
92. In a thin spherical pressure vessel, the volumetric strain is given by:
(a) 3ae (l-v) (b) 3ae (1+ v)
E E
91. IfH = wall thickness, P = pressure, R = mean radius, then maximum shear stress in case of thin cylindrical pressure vessel will be: (a) ah = 2 (b) ~=4 at at (c) ah = 8 (d) ah = 16 at at 90.
(a) Hyperbolic (b) linear
(c) Parabolic (d) None of these
When the concentrated load is applied, then the nature of variation of bending moment will be :
(a) Linear (b) Parabolic
(c) Hyperbolic (d) Uniform
If ah = hoop stress, at = longitudinal stress, then the ratio of ah to at in case of thin cylindrical pressure vessels is equal to
89. 88. 87.
Area under shear force diagram represents the (a) Shear force at a point
(b) Bending moment at a point (c) load at a point
(d) None of these
In shear force and bending moment diagram, if the bending moment is maximum then the shear force at that location will be equal to :
(a) Zero (b) maximmn
(c) minimum (d) None of these
When the uniformly distributed load is applied, then the nature of variation ofthe bending moment diagram will be 86.
(c)
Wl2 (b) 3 (a) wl2
84. In case of a simply supported beam, whose span is 'L' and carries a UDL at wi unit length, the value of maximum
bending moment will be equal to : 93.
wL3 (b) wL2 (a) 8 4 wL2 wL2 94. (c) 8 (d) 16
85. In case of a cantiliver beam, whose span is 'L' and carries a UDL of intensity wlunit length then maximum bending moment will be equal to :
wL 2 (d) wL 8 (c) wL 4 (b) wL 3 (a)
82. Three plans on which the principal strains occurs are: (a) Mutually perpendicular to each other
(b) Mutually inclined to each other than 90° (c) Inclined at 45° only
(d) None of these
83. In case of simply supported beam, the maximum bending moment ofa beam having span 'L' and a concentrated load wat mid-span will be equal to :
Engineering Mechanics and Strength of Materials A-24
F (d) 2 (c) F (b) FL FL 2 (a) (a) Jlk (b) &_
R
R
(c) Jl~ (d) Jl~121. During calculation of shear force, the upward forced to the left the of the section are taken as :
(a) Negative (b) Positive
(c) Zero (d) None of these
122. In a shear force and bending moment diagrams, Area of load diagrams provides:
(a) shear force change (b) bending moment (c) shear force (d) Point of contra flexure 123. There is a cantilever beam whose length is L and it carried
a point load F at its true end. Shear force at the center of the beam will be equal to:
1 (a) <I>=a (b)
¢=-a
(c) <1>=a2 (d) a= 2<1>
118. If <I>= angle of friction, u= coefficient offriction, then which ofthe following relation is true?
(a) <I>=COC1(Jl) (b) <I>=tan-1(Jl) (c) <I>=sin-1(Jl) (d) <I>=cos-1(Jl)
119. When a block of weight w is resting on a rough inclined plane with angle of inclination being 'a', the force offriction will be equal to :
(a) wsin
e
(b) w cose
(c) wtan
e
(d) w cote
120. If Jls= coefficient of static friction, Jlk= coefficient ofkinefic friction, R = Normal reaction, then frictional force of a moving body with constant velocity will be equal to :
116. If <I>= angle of friction, Jl = coefficient of friction, then
which of the following relation is true?
(a) Jl = cot <I> (b) Jl = sin <I>
(c) Jl = tan <I> (d) Jl = cos <I>
117. If <I>= angle of friction, a = angle of repose, then which of the following relation is true?
R (a) Jl=- (b) Jl=RxF F F Jl=F2R (c) Jl=- (d) R
113. The value of frictional force depends on (a) weight of the body
(b) area of contact (c) Normal reaction (d) roughness of surface
114. The value of maximum force of friction when the body begins to slide over another body/contacting surface is known to be:
(a) limiting friction (b) rolling friction (c) sliding friction (d) None of these
115. IfF = limiting friction, R = normal reaction, then coefficient offriction (u) is given as:
(b) Normal reaction and frictional force (c) Force on the body and normal reaction (d) None of these
112. The maximum inclination of the plane at which the body just starts to move is termed as :
(a) Cone of friction (b) Angle of repose (c) friction angle (d) None of these
and normal reaction
101. IfS = slenderness ratio, then the value of's' for long column should be in range of:
(a) S> 120 (b) S< 120
(c) S= 120 (d) S= 60
102. Euler's buckling formula is associated with: (a) Short column (b) long column (c) medium column (d) None of these
103. If'D' is the diameter of a circular column, then radius of gyration (K) will be given by:
D
D
(a) -
(b)-2 4
(c) 2D (d) 4D
104. A beam column is described as a column which carries: (a) axialloads only
(b) transverse loads
(c) axial and transverse loads (d) None of these
105. When two forces are in equilibrium, then which of the following conditions is true.
(a) Magnitudes are equal (b) opposite directions (c) collinear in action (d) All of the above 106. In case of co-planar non-concurrent forces, when EH= 0,EV
= 0, then the resultant may be:
(a) moment (b) couple
(c) force (d) None of these
107. When a sphere is placed on a smooth surface, then the reaction will act:
(a) inclined to contact plane (b) perpendicular to contact plane (c) horizontal to contact plane (d) All of the above
108. For aquiring equilibrium condition, How many are minimum number of coplaner and non - collinear forces required?
(a) 1 (b) 5
(c) 3 (d) 4
109. Ifthree co-planar and concurrent forces are acting on a rigid body at different points then the body will be in : (a) equilibrium
(b) not in equilibrium
(c) mayor may notbein equilibrium (d) None of these
110. A body having a weight of 50 N is resting on a rough horizontal floor, then the force of friction acting on the body will be equal to:
(a) 50N (b) lOON
(c) zero (d) None of these
111. Angle offrictiion is defined as the angle between
(a) normal reaction and the resultant of frictional force