SOLUTION OF PRESEMESTER EXAMINATION: ENGINEERING MECHANICS SOLUTION OF PRESEMESTER EXAMINATION: ENGINEERING MECHANICS©subhankar_karmakar

GROUP-A Q.1 Answer the following questions as per the instructions 2x20=20 Choose the correct answer of the following questions: (i) A truss hinged at one end, supported on rollers at the other, is subjected to horizontals load only. Its reaction at the hinged end will be (a) Horizontal;                   (b) Vertical (c) Both horizontal & vertical (d) None of the above. Ans: (c) both horizontal & vertical (ii) The moment of inertia of a circular section of diameter D about its centroidal axis is given by the expression (a) π(D)4/16           (b) π(D)4/32 (c) π(D)4/64           (d) π(D)4/4 Ans: (c) π(D)4/64 Fill in the blanks in the following questions: (iii)The distance of the centroid of an equilateral triangle with each side(a) is …………. from any of the three sides. Ans a /(2√3) (iv)Poisson’s ratio is defined as the ratio between ……. and ………… . Ans: Lateral Strain, Longitudinal Strain (v)If two forces of equal magnitudes P having an angle 2Ө between them, then their resultant force will be equal to ________ . Ans: 2P CosӨ Choose the correct word/s. (vi) Two equal and opposite force acting at different points of a rigid body is termed as (Bending Moment/ Torque/ Couple). Ans: Couple

Choose correct answer for the following parts: (vii) Statement 1: In stress strain graph of a ductile material, yield point starts at the end of the elastic limit. Statement 2: At yielding point, the deformation becomes plastic by nature. (a) Statement 1 is true, Statement 2 is true. (b) Statement 1 is true, Statement 2 is true and they are unrelated with each other (c) Statement 1 is true, statement 2 is false. (d) Statement 1 is false, Statement 2 is false. Ans: (a) Statement 1 is true, Statement 2 is true. (viii) Statement 1: It is easier to pull a body on a rough surface than to push the body on the same surface. Statement 2: Frictional force always depends upon the magnitude of the normal force. (a) Statement 1 is true, Statement 2 is true. (b) Statement 1 is true, Statement 2 is true and they are unrelated (c) Statement 1 is true, statement 2 is false. (d) Statement 1 is false, Statement 2 is false. Ans: (a) Statement 1 is true, Statement 2 is true. (ix) In a cantilever bending moment is maximum at (a) free end            (b) fixed end (c) at the mid span (d) none of these Ans: (b) fixed end (x) The relationship between linear velocity and angular velocity of a cycle (a) exists under all conditions (b) does not exist under all conditions (c) exists only when it moves on horizontal plane. (d) none of these Ans: (a) exists under all conditions

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SOLUTION OF PRESEMESTER EXAMINATION: ENGINEERING MECHANICS PART-II ©subhankar_karmakar

QUESTION PAPER: ENGINEERING MECHANICS

Vivekanand Institute of Technology & Science; Ghaziabad
PRE-SEMESTER EXAMINATION (odd SEMESTER 2009-10)
B.Tech…first Semester

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Sub Name: Engineering Mechanics Max. Marks: 100
Sub Code: EME-102 Max. Time: 3: 00 Hr

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(i) This paper is in three sections, section A carries 20 marks, section B carries 30 marks and section C carries 50 marks.
(ii) Attempt all the questions. Marks are indicated against each question
(iii) Assume missing data suitably if any.

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Group A

Q.1 Answer the following questions as per the instructions 2x20=20
Choose the correct answer of the following questions:

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(i) The magnitudes of the force of friction between two bodies, one lying above the another depends upon the roughness of the
(a) Upper body;                 (b) Lower body
(c) Both the bodies               (d) The body having more roughness

(ii)The moment of inertia of a circular section of diameter D about its centroidal axis is given by the expression
(a) π(D)⁴/16               (b) π(D)⁴/32
(c) π(D)⁴/64               (d) π(D)⁴/4

Fill in the blanks in the following questions:

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(iii)The distance of the centroid of an equilateral triangle with each side(a) is …………. From any of the three sides.
(iv)Poisson’s ratio is defined as the ratio between ……………………. and
………………………… .
(v)If two forces of equal magnitudes P having an angle 2Ө between them,
then their resultant force will be equal to ________ .

Match the following columns for the following two parts:

(vi) Match the column I to an entry from the column II:

COLUMN – I	COLUMN - II
(i) BMD of an UDL	(a) stored strain energy per unit volume
(ii) Resilience is	(b) brittle materials
(iii) Bulk Modulus	(c) parabolic in nature
(iv) Yield Point	(d) volumetric stress & strain
	(e) Ductile materials
	(f) Shear stress

(vii) Match the Following columns:

COLUMN – I	COLUMN - II
(i) Square of side (b)	(p) π b4 / 64
(ii) Equilateral Triangle of side (b)	(q) b4 / 12
(iii) Circle of diameter (b)	(r) b4/ 36
(iv) Isosceles right angle triangle of base (b)	(s) b4/(32√3)
	(e) Ductile materials
	(f) Shear stress

Column II gives the value of Moment of Inertia Ixx about a centroidal axis.

Choose correct answer for the following parts:

(viii) Statement 1:
In stress strain graph of a ductile material, yield point starts at the end of
the elastic limit.

Statement 2:
At yielding point, the deformation becomes plastic by nature.

(i) Statement 1 is true, Statement 2 is true.
(ii) Statement 1 is true, Statement 2 is true and they are unrelated with
each other
(iii) Statement 1 is true, statement 2 is false.
(iv) Statement 1 is false, Statement 2 is false.

(ix) Statement 1:
It is easier to pull a body on a rough surface than to push the body on
the same surface.

Statement 2:
Frictional force always depends upon the magnitude of the normal force.

(i) Statement 1 is true, Statement 2 is true.
(ii) Statement 1 is true, Statement 2 is true and they are unrelated with
each other
(iii) Statement 1 is true, statement 2 is false.
(iv) Statement 1 is false, Statement 2 is false.

Choose the correct word/s.
(x) Two equal and opposite force acting at different points of a rigid body
is termed as (Bending Moment/ Torque/ Couple).

SECTION-B

Q.2: Answer any three parts of the followings: 10X3=30

(a)	Find the shear force and moment equation for the beam as shown in the figure. Also sketch SFD (shear force diagram) and BMD (bending moment diagram)
(b) Explain and prove “the parallel axis theorem of moment of inertia.” Also find the centroid of the following composite area.
(c) Find the centroidal Moment of Inertia of the following shaded area.
	(d) Two cylinders P and Q rest in a channel as shown in fig. below. The cylinder P has a diameter of 100 mm and weighs 200 kN, where as the cylinder Q has a diameter of 180 mm and weighs 500 kN. Find the support reactions at all the point of contact.
(e)	Two blocks A and B of weights 1 kN and 2 kN respectively are in equilibrium as Shown in the figure. If the co-efficient of friction everywhere is 0.3, find the force P required to move the block B.

SECTION C:

(3) Answer any two parts of the following 5X2=10

(a) A simply supported beam of 2 cm wide and 4 cm high having a length 2 m long
in concentrated load of 3 kN (acting perpendicular to the axis of beam) at a point
0.5 m from one of the supports. Determine
(i) the maximum fiber stress (σb max); (ii) the stress in a fiber located at a
distance of 1 cm from the top of the beam at mid-span.

(b) Explain and justify the assumptions taken during analysis of a perfect truss.

(c)A flywheel is making 180 rpm and after 20 second it is running at 120 rpm.
How many revolutions will it make and what time will elapse before it stops, if the retardation is constant.


(4) Answer any one part of the following: 1X10=10

(a) Explain the terms angular momentum and radius of gyration.
A wheel rotates for 5 seconds with a constant angular acceleration and
describes during this time 100 rotations. It then rotates with a constant angular
velocity and during the next five seconds describes 80 radians. Find the initial
angular velocity and the angular acceleration.

(b) Analyse the following truss:



(5) Answer any three question;                           3X10=30

(a) Compare the stress – strain diagrams of a ductile material to that of a brittle Material. Also explain the term bulk modulus and modulus of elasticity.

(b) Explain and prove the “Bending Equation” (M/I)=(б/y)=(E/ρ). What is
called as Section Modulus?

(c) A cylinder of radius R, length L and total mass M is suspended vertically
from the floor, if the modulus of elasticity of the cylinder be E, find the total
deflection and maximum stress induced in the cylinder.

(d) A cylinder of 200 mm diameter is subjected to a twisting moment of
250 kN-m, the length of the cylinders is 1 m, if the modulus of rigidity of the
cylinder material be 150 GPa, find the maximum shear stress induced in the
cylinder. Also find the total angular deformation.

MULTIPLE CHOICES : ENGG MECHANICS Sub: Engineering Mechanics, Sub Code: EME-202, Semester: 2nd Sem, Course: B.Tech,

MULTIPLE CHOICES:
sub: ENGINEERING MECHANICS

Sub Code: EME - 102 /201
B.Tech First Semester (all branch)
University: UPTU (GBTU), Lucknow, Uttar Pradesh
OBJECTIVE TYPE QUESTIONS

PROBLEM SET - 1

Q1) If the coefficients of friction of an inclined plane be (1/√3), then the angle of repose of the plane will be
a) 90° b) 60° c) 45° d)30°

Q2) Three forces of equal magnitude acts along the side of an equilateral triangle, then the body will be in
a) static equilibrium                                 b) dynamic equilibrium
c) translational motion                             d) rotational motion.

Q3) (1/2).E.e² is called
a) total strain energy                                  b) resilience
c) dynamic loading energy                          d) none of this

Q4) The ratio of shear stress and shear strain is known as,
a) modulus of elasticity                                      b) poissons' ratio
c) modulus of rigidity                                         d) bulk modulus

Q5) The bending moment curves generated by UVL is,
a) parabolic                                          b) straight line
c) cubic                                                d) constant

Q6) The centroidal moment of inertia of a quarter circle lamina is
a) 0.11r⁴                                              b) 0.05r⁴
c) 0.4242r⁴                                          d) 0.5868r⁴

Q7) The Point of Contraflexure is a point in the beam where
a) shear force is zero                        
b) bending moment is zero
c) bending moment is zero and it changes sign
d) none of these

Q8) If a body of mass M is moving with an acceleration (a) then the Inertial Force on the body is equals to
a) -Ma          b) Ma        c) |Ma|     d) none of these

Q9) A truss is made of seven linkages and five joints, then the truss is
a) deficient                                              b) redundant
c) perfect                                                d) none of these

Q10) A beam has 3 no.s of supports is known as
a) cantilever beam
b) continuous beam
c) overhanging beam
d) simply supported beam


PROBLEM SET - 2:

Q1) The maximum value of frictional force that comes into play when a body tends to move on a surface called:
a) sliding friction,                                   b) limiting friction,
c) milling friction,                                   d) none of these.

2) The ratio of static friction to dynamic friction is:
a) less than 1,                                      b) equal to 1,
c) greater than 1,                                 d) none of these.

3) The angle of friction is equal to the:
a) ratio of frictional force to the normal reaction.
b) angle of inclined plane when a body tends to slide down.
c) angle of an inclined plane when a body is sliding.
d) none of these.

Q.4) A particle is moving along a circle with constant speed. The acceleration of the particle is:
a. along the circumference                      b. along the tangent
c. along the radius                                 d. zero

Q.5) The area moment of inertia of a quarter circle of radius (r) about the centroidal
axes is
a. (0.055r²)                                        b. (0.11r²)/3
c. (πr²)/4                                           d. π.r²

Q.6) The centroid of a semi circular area of radius r is
a. (r, 3r/4π )                                      b. ( r, 2r/π)
c. (r/π, r)                                           d. (r, r/2π)

Q.7) In the truss analysis, if the no. of linkages is 12 and the no. of joints is 7, then
a. perfect truss                            b. redundant truss
c. deficient truss                          d. indeterminate truss

Q.8) When shear force is zero at a point in a beam, bending moment at a certain point is,
a. zero                                                 b. maximum
c. minimum                                          d. increasing

Q.9) The mass moment of inertia of a solid sphere of mass M and radius R about an diameter of the sphere will be
a. (1/5).M.R²              b. (2/5).M.R⁴           c. (2/5).M.R²       d. (1/5).M.R⁴

Q.10) In a simply supported beam of length L, a concentrated load W acts at the mid span, the maximum bending moment would be
a. W.L/4                                                   b. W.L²
c. W.L/2                                                   d. 0

PROBLEM SET - 3

Q1) A couple can be balanced by
a) a direct Force                                    b) a moment
c) a torque                                            d) an equal and opposite couple.

Q2) Opening a Cold drinks bottle, one has to apply
a) a moment                                            b) a torque
c) parallel forces                                      d) a couple

Q3) A truss is said to be plane truss if
a) all the members lies in one plane
b) any two members lies in one plane
c) all the members are perpendicular to one plane
d) none of these

Q4) The ratio of frictional force normal reaction is called
a) angle of repose
b) angle of friction
c) limiting friction
d) coefficient of friction

Q5) Yielding means
a) elastic deformation
b) plastic deformation
c) fatigue failure
d) shear failure

Q6) The force required to move a body up an inclined plane will be least when the angle of inclination is:
a) equal to friction angle,
b) greater than friction angle,
c) less than friction angle,
d) none of these.

Q7) In a cantilever beam loaded with a point load at the free end, has maximum bending moment,

(i) free end,                                            (ii) fixed end,
(iii) at the mid span,                                (iv) none of the above.

Q8) An idealized truss can only be loaded at
a) only at the mid point of the linkages
b) only at the joints
c) both at the mid points and joints
d) none of the above

Q9) UTM can be used to find
a) Modulus of Elasticity
b) compressive breaking strength
c) ultimate tensile strength
d) all of the above

||| the question paper is made by subhankar karmakar, 2011 |||

Q10) Bulk Modulus is defined as the ratio between
(i) linear stress and strain,                 (ii) shear stress and strain,
(iii) volumetric stress and strain,        (iv) none of the above.

PROBLEM SET - 4:

Q.1 Answer the following questions as per the instructions 2x20=20

Choose the correct answer of the following Questions:

(i) The magnitudes of the force of friction between two bodies, one lying above the another depends upon the roughness of the
(a) Upper body                               (b) Lower body
(c) Both the body                           (d) The body having more roughness

(ii) The moment of inertia of a circular section of diameter D about its centroidal axis is given by the expression
(a) π(D)⁴/16
(b) π(D)⁴/32
(c) π(D)⁴/64
(d) π(D)⁴/4

Fill in the blanks in the following questions:

(iii) The distance of the centroid of an equilateral triangle with each side (a) is …………. From any of the three sides.

(iv) Poisson’s ratio is defined as the ratio between ……………. and…………………………

(v) If two forces of equal magnitudes P having an angle 2Ө between them, then their resultant force will be equal to ________

Match the following columns for the following two parts:

(vi) Match the column I to an entry from the column II:

COLUMN I----                            ---COLUMN II

(i) BMD of an UDL                      (a) stored strain energy per unit volume
(ii)Resilience is                            (b) brittle materials
(iii) Bulk Modulus                        (c) parabolic in nature
(iv) YieldPoint                             (d) volumetric stress and strain
                                                  (e) Ductile materials
                                                  (f) Shear stress

(vii) Match the Following columns:

COLUMN I  ------------------       COLUMN II

(i) Square of side b                                              (p) π b⁴ / 64

(ii) Equilateral Triangle of side b                           (q) b⁴ / 12

(iii) Circle of diameter b                                       (r) b⁴/ 36

(iv) Isosceles right angle triangle of base b            (s) b⁴/(32√3)

Column II gives the value of Moment of Inertia I_xx about a centroidal axis.

Choose correct answer for the following parts:

(viii) Statement 1:
In stress strain graph of a ductile material, yield point starts at the end of the elastic limit.
Statement 2:
At yielding point, the deformation becomes plastic by nature.

(i) Statement 1 is true, Statement 2 is true.
(ii) Statement 1 is true, Statement 2 is true and they are unrelated with each other
(iii) Statement 1 is true, statement 2 is false.
(iv) Statement 1 is false, Statement 2 is false.

(ix) Statement 1:
It is easier to pull a body on a rough surface than to push the body on the same surface.
Statement 2:
Frictional force always depends upon the magnitude of the normal force.
(i)Statement 1 is true, Statement 2 is true.
(ii)Statement 1 is true, Statement 2 is true and they are unrelated with each other
(iii) Statement 1 is true, statement 2 is false.
(iv) Statement 1 is false, Statement 2 is false.

Choose the correct word/s.
(x) Two equal and opposite force acting at different points of a rigid body
is termed as (Bending Moment/ Torque/ Couple).

PROBLEM SET - 5

Q.1) The example of Statically indeterminate structures are,

a. continuous beam,
b. cantilever beam,
c. over-hanging beam,
d. both cantilever and fixed beam.

Q.2) A redundant truss is defined by the truss satisfying the equation,
a. m = 2j - 3,
b. m < 2j + 3,
c. m > 2j - 3,
d. m > 2j + 3

Q.3) The property of a material to withstand a sudden impact or shock is called,
a. hardness                                           b. ductility,
c. toughness,                                        d. elasticity  of the material

Q.4) The ratio of the stress generated by a dynamic loading to the stress developed by the gradually applying the same load is
a) 1           b) 2               c) 3                    d) none of the above

Q.5) The ratio between the volumetric stress to the volumetric strain is called as
a. young's modulus
b. modulus of elasticity
c. rigidity modulus,
d. bulk modulus

Q.6) In a Cantilever beam, the maximum bending moment is induced at
a. at the free end
b. at the fixed end
c. at the mid span of the beam
d. none of the above

Q.7) The forces which meet at a point are called
a. collinear forces
b. concurrent forces
c. coplanar forces
d. parallel forces

Q.8) The coefficients of friction depends upon
a. nature of the surface
b. shape of the surface
c. area of the contact surface
d. weight of the body

Q.9) The variation of shear force due to a triangular load on simply supported beam is
a. uniform                                                       b. linear
c. parabolic                                                     d. cubic

Q.10) A body is on the point of sliding down an inclined plane under its own weight. If the inclination of the plane is 30 degree, then the coefficient of friction between the planes will be
a. (1/√3)        b. √3            c. 1            d. 0

PROBLEM SET - 6:

Q.1) Moment of Inertia of a body means
a) The resistance of the body against any movement or the tendency of movement of the body
b) The resistance of the body against any rotation of the body about a certain point
c) The force multiplied by the distance of the force from the body
d) None of these

Q.2) A truss is a
a) load bearing structure
b) framed mechanical structure
c) structure made of linkages and joints
d) all of the above

Q.3) The type of joints that can resist a moment is called as
a. roller joint                                            b. pin joint
c. hinged joint                                          d. fixed joint

Q.4) The slope of the normal stress-strain graph is equal to the
a. modulus of rigidity
b. bulk modulus
c. modulus of elasticity
d. stiffness constant.

Q.5) The shear stress in a circular shaft under torsion varies
a. linearly                                                  b. parabolically
c. hyperbolically                                        d. uniformly

Q.6) If for a material poissons ratio is 0.2 and modulus of elasticity is 200 GPa, then the value of modulus of rigidity would be
a) 100.33 GPa                                          b) 93.33 GPa
c) 83.33 GPa                                            d) 100 GPa

Q.7) If the section modulus of a beam decreases, then bending stress will
a. decrease                                   b. increase
c. remain same                             d. bending stress is independent of section modulus

Q.8) Which one of the following statements is correct?
a. Energy and work are scalars
b. Force and work are vectors
c. Energy, momentum and velocity are vectors
d. Force, momentum and velocity are scalars

Q.9) Stress may be defined as:
a. load per unit area
b. external force per unit area
c. internal resistance per unit area
d. same as pressure

Q.10) Hooke's law is valid up to the
a. yield point                                           b. elastic limit
c. proportional limit                                 d. ultimate point.

PROBLEM SET - 7:

Q.1) The first law of motion provides the definition of
a. momentum                                         b. force
c. energy                                               d. acceleration

Q.2) A man in a lift will weigh more when the lift is
a. accelerated upwards
b. accelerated downwards
c. descends freely
d. lift going up is slowing down

Q.3) The motion of a bicycle wheel is
a. linear                                                   b. rotary
c. translatory                                           d. rotary as well as translatory

Q.4) A particle is moving along a circle with constant speed. The acceleration of the particle is :
a. along the circumference
b. along the tangent
c. along the radius
d. zero

Q.5) The mass moment of inertia of a thin disc of mass (m) and radius (r) about the centroidal axes is
a. (m.r²)/2                                                  b. (m.r²)/3
c. (m.r²)/4                                                  d. m.r²

Q.6) The centroid of a semi circular arc of radius r is
a. (3r)/π                                                     b. 2r/π
c. r/π                                                        d. r/(2π)

Q.7) In the method of sections for trusses, the section must be passed so as to cut not more than
a. two members
b. three members
c. four members
d. five members

Q.8) When bending moment at a certain point is maximum, shear force is
a. zero b. maximum
c. minimum d. increasing

Q.9) The mass moment of inertia of a solid sphere of mass M and radius R about an diameter of the sphere will be
a. (1/5).M.R²
b. (2/5).M.R⁴
c. (2/5).M.R²
d. (1/5).M.R⁴

Q.10) In a cantilever beam of length L, a concentrated load W acts at the free end, the bending moment at the free end would be
a. W.L                              b. W.L²
c. (W.L)/2                         d. 0

Multiple Choice type Question:

TOPIC : FRICTION

1) The maximum value of frictional force that comes into play when a body tends to move on a surface called :
a) sliding friction,
b) limiting friction,
c) milling friction,
d) none of these.

2) The ratio of static friction to dynamic friction is :
a) less than 1,
b) equal to 1,
c) greater than 1,
d) none of these.

3) The angle of friction is equal to the :
a) ratio of frictional force to the normal reaction.
b) angle of inclined plane when a body tends to slide down.
c) angle of an inclined plane when a body is sliding.
d) none of these.

4) The coefficient of friction depends upon :
a) area of contact,
b) shape of the body,
c) nature of contact surfaces,
d) none of these.

5) Kinetic friction is :
a) limiting friction,
b) friction when a body is moving,
c) friction when a body is stationary,
d) none of these.

©subhankar_karmakar

PROBLEM SET - 8

Q1) The enclosed area in the shear force diagram for a beam is the magnitude of
a) shear force
b) bending moment
c) applied load
d) bending stress.

Q2) If a particle is moving in a circular path at constant velocity on a plane, then the direction of the angular velocity will be
a) along the tangent to the circular path
b) along the radial direction
c) normal to the plane
d) none of this.

There will be 2 statements, which may be right or may be wrong, read those sentences and then choose the right choice

Q3) statement 1: Two forces act on a particle are collinear forces.
statement 2: The ratio of longitudinal strain and lateral strain is the value of poisson's ratio.

a) both of them are true
b) statement 1 is true, but statement 2 is false
c) statemen 2 is true while stateme 1 is false
d) both of them are false.

ENGINEERING MECHANICS

CONCEPTS OF FORCE AND FORCE SYSTEM                                 ©subhankar_karmakar

CONCEPTS OF FORCE AND FORCE SYSTEM

               To understand force one has to understand Energy. Energy is the transferable physical quantity of an object which has the capacity to do work. Energy can be transferred from one place to another place, or from one object to another object.



               Force is the physical quantity which when acts on a body, may do some work on the body by displacing from a certain point in the space to another point in the space. When a body gets displaced due to the application of force, some work has been done on the object. So work done on a body is basically nothing but putting some more energy on the body.



               Any kind of displacement of an object is called a change in position and it is a universal truth that to go through any kind of change in position, an object needs some time interval, the process of changing position involves time, hence we can say an amount of ∆x change in position it needs ∆t amount of time. Hence, we can get a rate of change of position with respect to time.



               If in (∆t) time the body travels a distance (∆x) than,
in unit time the body travels a distance (∆x/∆t) and it is known as average velocity.



               When the time interval ∆t is very small, then the ratio of (∆x/∆t) is represented by differential operator and then it is called instantaneous velocity, V=dx/dt



               An heavier object needs more energy to attain the same velocity of that of a lighter object. It has been observed that body at a higher motion needs more energy to change velocity incrementally. Suppose we have two identical object A and B, while the velocity of A is (u), the velocity of B is larger at (v), now if we try to increase the velocity of both the body by a amount ∆v, then the body with the higher motion would need more energy per unit change in velocity. 

MOMENTUM : AN IMPORTANT CONCEPT

NEWTON'S LAW OF MECHANICS:

Although we know there are three laws of motion proposed by Issac Newton, but it can be shown that the 2nd law of motion is the fundamental laws of motion, and the other two laws are nothing but special cases of second law.

The second law states that the rate of change of momentum is equal to force, which is another physical quantity and it is a vector.

SO WHAT DOES MOMENTUM MEAN?

MOMENTUM

Tuesday, 16. November, 02:41

Objects in motion are said to have a momentum. This momentum is a vector. It has a size and a direction. The size of the momentum is equal to the mass of the object multiplied by the size of the object's velocity. The direction of the momentum is the same as the direction of the object's velocity.

Momentum is a conserved quantity in physics. This means that if you have several objects in a system, perhaps interacting with each other, but not being influenced by forces from outside of the system, then the total momentum of the system does not change over time. However, the separate momenta of each object within the system may change. One object might change momentum, say losing some momentum, as another object changes momentum in an opposite manner, picking up the momentum that was lost by the first.

IMPORTANCE OF MOMENTUM

Momentum is a corner stone concept in Physics. It is a conserved quantity. That is, within a closed system of interacting objects, the total momentum of that system does not change value. This allows one to calculate and predict the outcomes when objects bounce into one another. Or, by knowing the outcome of a collision, one can reason what was the initial state of the system.

MOMENTUM IS MASS TIMES VELOCITY

When an object is moving, it has a non-zero momentum. If an object is standing still, then its momentum is zero. To calculate the momentum of a moving object multiply the mass of the object times its velocity. The symbol for momentum is a small p.

MOMENTUM IS A VECTOR QUANTITY

Momentum is a vector. That means, of course, that momentum is a quantity that has a magnitude, or size, and a direction. The above problem is a one dimensional problem. That is, the object is moving along a straight line. In situations like this the momentum is usually stated to be positive, i.e., to the right, or negative, i.e., to the left.

MOMENTUM IS NOT VELOCITY

.

Sometimes the concept of momentum is confused with the concept of velocity. Do not do this. Momentum is related to velocity. In fact, they both have the same direction. That is, if an object has a velocity that is aimed toward the right, then its momentum will also be directed to the right. However, momentum is made up of both mass and velocity. One must take the mass and multiply it by the velocity to get the momentum.

MOMENTUM IS DIRECTLY PROPORTIONAL TO VELOCITY

If the mass is kept constant, then the momentum of an object is directly proportional to its velocity. In the example at the left, the mass is kept constant at a value of 2.0 kg. The velocity changes from 0 m/s to 10 m/s while the momentum changes from 0 kg-m/s to 20 kg-m/s. This creates a straight line graph when momentum is plotted as a function of velocity. (The symbol for momentum as a function of velocity would be p(v).) The straight line graph demonstrates the direct proportion between momentum and velocity.


That is, if one were to double the velocity of an object, then the momentum of that object would also double. And, if one were to change the velocity of an object by a factor of 1/4, then the momentum of that object would also change by a factor of 1/4.

MOMENTUM IS DIRECTLY PROPORTIONAL TO MASS

If the velocity is kept constant, then the momentum of an object is directly proportional to its mass. In the example at the left, the velocity is kept constant at a value of 3.0 m/s. The mass changes from 0 kg to 10 kg while the momentum changes from 0 kg-m/s to 30 kg-m/s. This creates a straight line graph when momentum is plotted as a function of mass. (The symbol for momentum as a function of mass would be p(m).) The straight line graph demonstrates the direct proportion between momentum and mass.

That is, if one were to triple the mass of an object, then the momentum of that object would also triple. And, if one were to change the mass of an object by a factor of 1/2, then the momentum of that object would also change by a factor of 1/2.

COURSE FILE: VITS GHAZIABAD

COURSE FILE: ©subhankar_karmakar

COURSE FILE:

INSTITUTE NAME : VIVEKANAND INSTITUTE OF TECHNOLOGY AND SCIENCE
SUBJECT NAME : ENGINEERING MECHANICS
SUBJECT CODE : EME-102
FACULTY NAME : SUBHANKAR KARMAKAR
DEPARTMENT : MECHANICAL ENGINEERING DEPARTMENT
YEAR  : FIRST YEAR

INDEX:

1 : LESSON PLAN
2 : TIMETABLE
3 : COURSE PLAN
4 : ASSIGNMENT
5 : LECTURE NOTES

LESSON PLAN : EME-102 / 202 : ENGINEERING MECHANICS

UNIT	TOPICS SYNOPSIS	No. of Lectures	DATE
UNIT I	Two Dimensional Force Systems: Basic concepts, Laws of motion, Principle of Transmissibility of forces, Transfer of a force to parallel position , Resultant of a force system, Simplest Resultant of Two dimensional concurrent and Non-concurrent Force systems, Distributed force system, Free body diagrams, Equilibrium and Equations of Equilibrium, Applications.	5	date
UNIT I	Friction: Introduction, Laws of Coulomb Friction, Equilibrium of Bodies involving Dry-friction, Belt friction, Application.	3	date
UNIT II	Beam: Introduction, Shear force and Bending Moment, Differential Equations for Equilibrium, Shear force and Bending Moment Diagrams for Statically Determinate Beams.	5	date
UNIT II	Trusses: Introduction, Simple Truss and Solution of Simple truss, Method f Joints and Method of Sections.	3	date
UNIT III	Centroid and Moment of Inertia: Centroid of plane, curve, area, volume and composite bodies, Moment of inertia of plane area, Parallel Axes Theorem, Perpendicular axes theorems, Principal Moment Inertia, Mass Moment of Inertia of Circular Ring, Disc, Cylinder, Sphere and Cone about their Axis of Symmetry.	6	date
UNIT IV	Kinematics of Rigid Body: Introduction, Plane Motion of Rigid Body, Velocity and Acceleration under Translation and Rotational Motion, Relative Velocity.	4	date
UNIT IV	Kinetics of Rigid Body: Introduction, Force, Mass and Acceleration, Work and Energy, Impulse and Momentum, D’Alembert’s Principles and Dynamic Equilibrium.	4	date
UNIT V	Simple Stress and Strain: Introduction, Normal and Shear stresses, Stress- Strain Diagrams for ductile and brittle material, Elastic Constants, One Dimensional Loading of members of varying cross-sections, Strain energy.	3	date
UNIT V	Pure Bending of Beams: Introduction, Simple Bending Theory, Stress in beams of different cross sections.	3	date
UNIT V	Torsion: Introduction, Torsion of shafts of circular section, torque and twist, shear stress due to torque.	3	date

Text books:

1. Engineering Mechanics by Irving H. Shames, Prentice-Hall

2. Mechanics of Solids by Abdul Mubeen, Pearson Education Asia.

3. Mechanics of Materials by E.P.Popov, Prentice Hall of India Private Limited.

TIME TABLE:

DAY	9.30	10.20	11.10	12.00	LUNCH	1.40	2.30	3.20	4.10
MON	9.30	ME-D	11.10	ME-F	LUNCH	1.40	ME-D	3.20	ME-D
TUE	9.30	ME-D	11.10	12.00	LUNCH	ME-F	2.30	ME-D	4.10
WED	ME-F	10.20	ME-D	12.00	LUNCH	1.40	2.30	3.20	4.10
THU	ME-D	10.20	11.10	ME-F	LUNCH	1.40	2.30	3.20	4.10
FRI	9.30	ME-F	11.10	ME-D	LUNCH	1.40	2.30	3.20	4.10
SAT	ME-D	10.20	ME-F	12.00	LUNCH	ME-F	2.30	3.20	4.10

COURSE PLAN:

SUBJECT NAME: ENGINEERING MECHANICS

SUBJECT CODE : EME-102

SCOPE : The course aims to provide deeper knowledge, a wider scope and improved understanding of the study of motion and the basic principles of mechanics and strength of materials. It is a concept based subject and it needs the application capabilities of the concepts on the part of the students.

SESSIONAL EVALUATION SCHEME:

PARTICULAR	WEIGHTAGE	MARKS
TWO SESSIONALS	60%	30
ATTENDANCE	20%	10
TEACHER'S ASSESSMENT(TA)*	WEIGHTAGE	MARKS

*TA will be based on the Assignments given, Unit test Performances and Attendance in the class for a particular student.

Lecture Schedule of Unit – 1

Total Number of Lectures: 8

• Lecture Details & Synopsis :



• Lecture- 1: Introduction, mass, particle, rigid body, position vector, change of position, velocity, momentum, change of momentum, force acceleration, Newton’s law of motion, conservation of momentum, conservation of energy.



• Lecture- 2: Definition of force, characteristics of force, Force as a vector, Force addition, triangle’s & parallelogram laws of force addition, magnitudes & direction of resultant force, negative force, resolution of force, oblique and orthogonal resolutions, component of a vector along a line, classification of a force system, force system in one dimension, like & unlike forces, two dimensional force system, co-planar force system, non coplanar force system, concurrent force system, coplanar concurrent force system, coplanar parallel force system



• Lecture- 3: The concepts of rigid body, principle of transmissibility of forces, resultant of coplanar concurrent force system, equilibrium of forces, conditions of static equilibrium for concurrent force system, actions & reactions in case of equilibrium in (i) spherical balls in a channel, (ii) blocks of mass in an inclined plane, (iii) reactions in strings, wires & ropes. Types of force (i) tension (ii) compression. Concepts of free body diagrams. Lami’s theorem.



• Lecture- 4: Applications of the conditions of static equilibrium in case of concurrent forces in the analysis of a concurrent force system & numericals based on this. Numericals based on the resultant of a force system. Numericals based on Lami’s theorem.



•Lecture- 5: Normal reactions, concepts of friction, angle of friction, coefficient of friction, angle of repose, laws of coulomb friction, limiting friction, coefficient of static friction & kinematics friction, Equilibrium of bodies involving dry friction. Use of Friction, Friction as a necessary EVIL.




• Lecture- 6: Numericals based on static friction, ladder friction, friction in inclined plane, numericals on ladder friction & friction in inclined plane. Objective type questions in friction.




• Lecture- 7: Theory of Belt Friction, Slack & tight side of a belt, Concepts of Included angle, power delivered by belt drive, Numericals on Belt friction & objective type Questions.



• Lecture- 8: Doubt clearing Sessions on Unit- 1, (Static Equilibrium Analysis, Resultant Forces, Resolution of Forces, Lami’s Theorem, Concepts of Dry & belt Friction.)





•Reference books:
•(i)Engineering Mechanics by Timoshenko & Young
(ii)Engineering Mechanics by R. K. Rajput
(iii) Engineering Mechanics by Irving H. Shames

Lecture Schedule of Unit – 2

Total Number of Lectures: 8

• Lecture Details & Synopsis:



• Lecture- 9: Concepts of Beam, Classification of Beams, simply supported beam, cantilever beam, over hanging beam, continuous beam,Types of Support Reactions, Pin/hinged joints, Roller joints, fixed joints, determination of support reactions in beam, types of loading in beams, concentrated load, distributed load on the beam, uniformly distributed load (UDL), uniformly varying load (UVL), pure moment loading.



• Lecture- 10: Concepts of Shear Force, sign convention for shear force, determination of shear force at each point of the beam over the complete length of the beam, shear force diagrams (SFD), differential equations for equilibrium, concepts of bending moments, sign conventions for bending moments, determination of bending moments at each point of the beam over the complete length of the beam, bending moment diagrams (BMD), maximum bending moment, point of contra-flexure and its importance.



• Lecture- 11: SFD & BMD in case of (i) simply supported beam, (ii) cantilever beam, (iii) over-hanging beam with (a) concentrated loading, (b) uniformly distributed loading, (c) uniformly varying loading.



• Lecture- 12: Numericals on SFD & BMD for all types of beam.



• Lecture- 13: Numericals on SFD & BMD for all types of beam and to find point of contra-flexure.



• Lecture- 14: Concepts of Truss, Linkages, and Joints, Classification of Trusses, Perfect Truss, Deficient Truss, Redundant Truss, Simple Truss, Analysis of a Truss by (i) Method of Joints (ii) Method of Sections.



• Lecture- 15: Numericals on Truss analysis by method of joints & method of Sections.



• Lecture- 16: Numericals on Truss analysis by method of sections.

Lecture Schedule of Unit – 3

Total Number of Lectures: 6

• Lecture Details & Synopsis:



• Lecture- 17: Concepts of geometrical Centroid, Center of Mass & Center of Gravity, Centroid of Plane, Curve, Area, & Volume, determination of centroid of composite bodies.



• Lecture- 18: Numericals on determination of Centroid of composite bodies.



• Lecture- 19: Concepts of Rotation & Moment of Inertia, concepts of area moment of inertia & mass moment of inertia, Determination of moment of Inertia with the help of calculus, Parallel axis theorem & Perpendicular axis theorem of Moment of Inertia.



• Lecture- 20: Concepts of Principal Moment of Inertia, determination of Mass Moment of Inertia of (i) Circular Ring, (ii) Disc, (iii) Cylinder, (iv) Sphere & (v) Cone about their axis of symmetry



• Lecture- 21: Numericals on determination of Moment of Inertia of different objects.



• Lecture- 22: Numericals on determination of M.O.I of different objects.

Lecture Schedule of Unit – 4

Total Number of Lectures: 8

• Lecture Details & Synopsis:



• Lecture- 23: Introduction of rigid body, Motion of Rigid Body, Velocity & Acceleration under Translational Motion, Equation of motion due to gravity, concepts of Relative Velocity, Problems on Projectile Motion.



• Lecture- 24: Concepts of Rotational Motion, Angular Displacement, Angular Velocity, Laws of Motion for Rotation, Concepts of Moment, Torque & Couple, Angular Acceleration, Relations between angular velocity & linear velocity, Relation between angular acceleration & linear acceleration, concepts of centripetal acceleration, concepts of Pseudo Force ie. Centrifugal acceleration.



• Lecture- 25: Motion on Level road, Banking of road & Super elevation of rails, Analysis of Slider-crank mechanism (Four bar mechanism) & numericals on rotational motion.



• Lecture- 26: Numericals on Rotational motion & its application.



• Lecture- 27: Concepts of Force, Newton’s Laws of Motion, Definition of Mass, Gravitational Mass & Inertial Mass, Concepts of Work & Energy, Conservation of Mass Principle, Principle of Conservation of Momentum.



• Lecture- 28: Principle of Conservation of Energy, Work- Energy Theorem, Concepts of Conservative Force & Potential Energy. Collision of two bodies, Elastic & Inelastic Collision, Impulse & Impulsive Force, Impulse & change of Momentum. Power.



• Lecture- 29: Concepts of Dynamic Equilibrium, Inertial Mass & D’ Alembert’s Principle of Dynamic Equilibrium, Motion on an Inclined Plane, Analysis of Lift Motion, Analysis of Motion of Connected Bodies (i) System of Pulleys (ii) Two Bodies connected by a string.



• Lecture- 30: Numericals on Dynamic Equilibrium & System of Pulleys.

Lecture Schedule of Unit – 5

Total Number of Lectures: 10

• Lecture Details & Synopsis:



• Lecture- 31: Deformation of Rigid Bodies under the action of External Force, Resistance against deformation & induction of internal resistive force, Unit deformation & strain, internal force & stress, linear deformation and normal stress, Hooke’s Law & Modulus of Elasticity ( E, Young’s modulus), angular deformation, Shear Strain, & shear Stress, Modulus of Rigidity ( G ), Complimentary Shear Stress.




• Lecture-32: Simple Stress-Strain Diagrams for (i) Ductile Materials, (ii) Brittle Materials, One Dimensional Loading of members of Varying Cross Sections (i) Circular Bar of Uniform Taper, (ii) Bar of Uniform Strength (iii) Bar of (a) Uniform & (b) Taper Cross section due to Self Weight, (iv) Composite Bar, Impact loading (i) Gradually Applied load, (ii) Suddenly Applied Load.



• Lecture- 33: Concepts of Strain Energy & Resilience, Concepts of (i) Longitudinal & (ii) Lateral Strain, Poisson’s Ratio, Hydro-static Compression & Volumetric Strain, Bulk Modulus (K), relation between (i) E, G, & K, (ii) E, K, m (iii) E, G, m. simple numericals on stress & strain.



• Lecture- 34: Numericals on Simple Stress & Simple Strain, Shear Strain & Shear Stress, numericals on composite bars.



• Lecture- 35: Concepts of Pure Bending, Assumptions in simple theory of bending, Concepts of Bending Stress, Neutral Layer & Neutral Axis, Bending Stress Diagrams, Difference between Simple Stress & Bending Stress, Derivation of Bending Equation, Section Modulus (Z), Relation between max. Tensile & max. Compressive Stress,



• Lecture- 36: Stress in Beams of different cross sections, Numericals on Bending Stresses.



• Lecture- 37: Doubt clearance class on (i) stress, strain (ii) pure bending



• Lecture- 38: Introduction of Shaft & Torsion, concept of pure torsion, Polar Moment of Inertia ( J ), Section Modulus (Z), Polar Modulus ( Zp), Assumption for Deriving the Torsional Formulas, Torsional Equation,



• Lecture- 39: Torsional Rigidity or Torsional Stiffness ( K ), Comparison of strength of (i) Solid & (ii) Hollow Circular Shaft (Tmax), Power Transmission by a Shaft, Importance of Angle of Twist, numerical based on Torsion in Shaft.



• Lecture- 40: Doubt clearing classes on Torsion.





• Reference Books:

• Engineering Mechanics by R. S. Khurmi

• Engineering Mechanics by Bhavikatti

• Engineering Mechanics by D. S. Kumar.

• Engineering Mechanics by Timoshenko & Young.

ENGINEERING MECHANICS: ENGINEERING MECHANICS: THEORY OF FRICTION & FRICTIONAL FORCES

HOW TO FIND THE RESULTANT OF A FORCE SYSTEM?

For a force system i.e. a system of several forces acting on an object, it is possible to get the same effect on the object by the force system replacing it by a single force, that will be equivalent to the summation of the component forces acting on the object. The single force that will produce exactly the same effect on the object in stead of the force system is called Resultant of the force system.

We know that two forces acting on an object lying on a plane can be added together by

• (i) Triangle's Law or
• (ii) Parallelogram Law.
  For more than two vectors we use
• (iii) Polygon Law of Force Addition.
• (iv) Force Resolution Method.

The resultant of a force system is the Force which produces same effect as the combined forces of the force system would do. So if we replace all components of the force by the resultant force, then there will be no change in effect.

The Resultant of a force system is a vector addition of all the components of the force system. The magnitude as well as direction of a resultant can be measured through analytical method.

THE STEPS TO FIND A RESULTANT OF A CON-CURRENT FORCE SYSTEM:

STEP 1:

RESOLVE ALL THE COMPONENT FORCES ALONG X-AXIS AND Y-AXIS.

If a force F acts on an object at an angle ß with the positive X-axis, then its component along X-axis is F cosß, and that along Y-axis is F sinß.


STEP 2:

ADD ALL THE X-COMPONENTS OR HORIZONTAL COMPONENTS AND IT IS DENOTED BY ΣF_x AND

ADD ALL THE Y-COMPONENTS OR VERTICAL COMPONENTS AND IT IS DENOTED BY ΣF_y.

STEP 3:

FIND THE MAGNITUDE OF THE RESULTANT R

We know from Geometry that

R = √{(ΣF_x)² + (ΣF_y)²}

STEP 4:

FIND THE DIRECTION (α) OF THE RESULTANT FORCE (R)

We know that

tan α = (ΣF_y/ΣF_x)

hence,

α = tan^-1(ΣF_y/ΣF_x)

TWO DIMENSIONAL FORCE SYSTEM

Q: WHAT DO YOU UNDERSTAND BY THE TERM "FORCE"? WHAT IS THE EFFECT OF FORCE ON A PARTICLE AND A RIGID BODY? EXPLAIN WITH SUITABLE EXAMPLES.

Answer:

FORMAL DEFINITION:

A FORCE is that which can cause an object with mass to ACCELERATE. Force has both MAGNITUDE and DIRECTION, making it a vector quantity. According to Newton's second law, an object with constant mass will accelerate in proportion to the net force acting upon it and in INVERSE PROPORTION TO ITS MASS (M). An equivalent formulation is that the net force on an object is equal to the RATE OF CHANGE OF MOMENTUM it experiences. Forces acting on three-dimensional objects may also cause them to rotate or deform, or result in a change in pressure. The tendency of a force to cause angular acceleration about an axis is called TORQUE. Deformation and pressure are the result of stress forces within an object.


EXPLANATION OF MECHANICAL FORCE AND IT'S EFFECT ON A PARTICLE:

CHANGE IN POSITION:

To know force well, first we have to understand what do we mean by Change. What does it mean when we say the position of the body has been changed? Whenever we find the state of object becomes different than that of the same object before some time say Δt, then we say that there exists a change in the state of the object. Suppose the change occurs in the position of the body. But to find the initial position of a body, we need a co-ordinate system.

THE CAUSE OF CHANGE:

It has been seen that to induced a change or to make a change in the position of an object we must have to change the energy possess by the body. To transfer energy into the object we shall have to apply FORCE on the body. Therefore Force is the agency that makes a change in position of a body.

THE CONCLUSION: GALILEO'S LAW OF INERTIA OR NEWTON'S FIRST LAW OF MOTION.

So, if there is no force on an object the position of the object won't change with respect to time. It means if a body at rest would remain at rest and a body at uniform motion would remain in a steady motion. This law is known as Galileo's Law of Inertia or Newton's first law of motion.

• 2 DIMENSIONAL FORCE

In physics, force is a vector quantity that is used to describe the interaction between two objects. In a two-dimensional system, forces can act in two different directions, which are typically labeled as the x-axis and the y-axis.

When dealing with two-dimensional force, it is essential to use vector addition to determine the net force acting on an object. The net force is the vector sum of all the forces acting on the object. The direction of the net force is determined by the angle of the resultant force vector.

To calculate the net force in two dimensions, we must first break down each force into its x and y components. The x-component of a force is the amount of force acting in the x-direction, and the y-component is the amount of force acting in the y-direction. Once we have the x and y components for each force, we can add them together to find the net force.

The magnitude of the net force can be calculated using the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. In this case, the hypotenuse is the magnitude of the net force, and the other two sides are the x and y components of the net force.

In summary, when dealing with two-dimensional force, it is essential to use vector addition to determine the net force acting on an object. To calculate the net force, we must first break down each force into its x and y components and then add them together. The magnitude and direction of the net force can be determined using trigonometry.

• ORTHOGONAL RESOLUTION OF A FORCE

Orthogonal resolution of a force is a technique used in physics to break down a force vector into its components along two orthogonal axes, typically the x and y axes. This technique is useful in analyzing the motion of an object under the influence of a force and can help determine the net force acting on an object.

To perform orthogonal resolution of a force, we first need to identify the angle that the force vector makes with respect to one of the axes, usually the x-axis. We can then use trigonometry to determine the components of the force vector along the x and y axes.

If the angle between the force vector and the x-axis is θ, the x-component of the force can be found using the equation F_x = F cos(θ), where F is the magnitude of the force. Similarly, the y-component of the force can be found using the equation F_y = F sin(θ).

Once we have the x and y components of the force, we can use vector addition to determine the net force acting on an object. The net force is the vector sum of all the forces acting on the object and can be found by adding the x and y components of each force separately.

Orthogonal resolution of a force is a powerful technique that is used in many areas of physics, including mechanics, electromagnetism, and fluid dynamics. By breaking down a force vector into its components, we can better understand the forces acting on an object and predict its motion under different conditions.

WHAT IS A FORCE SYSTEM? CAN WE CLASSIFY FORCE SYSTEMS?

ANSWER:
                         A force system may be defined as a system where more than one force act on the body. It means that whenever multiple forces act on a body, we term the forces as a force system. We can further classify force system into different sub-categories depending upon the nature of forces and the point of application of the forces. Almost any system of known forces can be resolved into a single force called a resultant force or simply a Resultant. The resultant is a representative force which has the same effect on the body as the group of forces it replaces. (A couple is an exception to this) It, as one single force, can represent any number of forces and is very useful when resolving multiple groups of forces. It is important to note that for any given system of forces, there is only one resultant.


Different types of force system:

• (i) COPLANAR FORCES:

If two or more forces rest on a plane, then they are called coplanar forces. There are many ways in which forces can be manipulated. It is often easier to work with a large, complicated system of forces by reducing it an ever decreasing number of smaller problems. This is called the "resolution" of forces or force systems. This is one way to simplify what may otherwise seem to be an impossible system of forces acting on a body. Certain systems of forces are easier to resolve than others. Coplanar force systems have all the forces acting in in one plane. They may be concurrent, parallel, non-concurrent or non-parallel. All of these systems can be resolved by using graphic statics or algebra.

• (ii) CONCURRENT FORCES:

A concurrent coplanar force system is a system of two or more forces whose lines of action ALL intersect at a common point. However, all of the individual vectors might not actually be in contact with the common point. These are the most simple force systems to resolve with any one of many graphical or algebraic options. If the line of actions of two or more forces passes through a certain point simultaneously then they are called concurrent forces. concurrent forces may or may not be coplanar.

• (iii) LIKE FORCES:

A parallel coplanar force system consists of two or more forces whose lines of action are ALL parallel. This is commonly the situation when simple beams are analyzed under gravity loads. These can be solved graphically, but are combined most easily using algebraic methods. If the lines of action of two or more forces are parallel to each other, they are called parallel forces and if their directions are same, then they are called LIKE FORCES.

• (iv) UNLIKE FORCES:

If the parallel forces are such that their directions are opposite to each other, then they are termed as "UNLIKE FORCE".

• (v) NON COPLANAR FORCES:

The last illustration is of a "non-concurrent and non-parallel system". This consists of a number of vectors that do not meet at a single point and none of them are parallel. These systems are essentially a jumble of forces and take considerable care to resolve.

on 20th November, 2010: ©subhankar

CENTROID OF COMPLEX GEOMETRIC FIGURES:

So in this articles, we are going to discuss the concepts of centroid for one dimensional as well as two dimensional objects. 

Let's first discuss about 1D and 2D objects, one by one, an 1D object is a line, practically a metallic rod will be considered as a linear, 1D object. Where as any thin plate of negligible thickness can be considered a 2D body. Suppose we have a thin metallic rectangular plate. If it is considered as a 2D rectangular area of b X h.

The concept of centroid has been developed on the basis of resultant of several areas. We know that an area can be represented as the cross product of two vectors, hence it is also an vector. Suppose we have an area A, in a cartesian 2D coordinate system. We just divide the area into n parts, and represent them as a₁, a₂, a₃, ..... a_n.

Let the elemental areas are at a distance x₁, x₂, x₃, ..... x_n, from Y axis and y₁, y₂, y₃, ...y_n from X axis.

The total moments produced about Y axis will be equal to the summation of all the individual moments produced by n elemental areas. 

Now moment is a vector quantity and we know vectors of same kind can be added together, therefore, all the n moment vectors can be added to get a single value of Resultant Moment. 

We also know this resultant moment's position vector. Let the resultant moment passes through a point G. The point through which resultant moment passes through is called Center of the Area or Centroid.

How can we find out the point G, whose coordinates are (X_g,Y_g)?

As moment of an area also obeys VARIGNON'S THEOREM OF MOMENT, then sum of all the moments produced by individual elemental areas will exactly be equal to the moment produced by the total area, i.e. the resultant of all those elemental areas. Now if all the areas are added to have the resultant area which will pass through the centroid G such that it produces a moment of X_gA about Y axis and Y_gA about X axis.

But Varignon's theorem states us that, for a vector system, resultant vector produces the moment about a point, is exactly equal to the sum of all the moments produced by all elemental areas about the same point and in the same plane. Hence, we can write now that,

Sum(a₁x₁+ a₂x₂+ + +a_nx_n) = AX_g
we can use summation sign ∑ to represent these equations,
∑a_ix_i = (∑a_i)X_g
=> X_g = (∑a_ix_i)/((∑a_i)

Sum(a₁y₁+ a₂y₂+ + +a_ny_n) = AY_g
∑a_iy_i = (∑a_i)Y_g
=> Y_g = (∑a_iy_i)/((∑a_i)

Algorithm to find out the Centroid G(X_g, Y_g) of a Complex Geometric Figure.

Step1:
Take a complex 2D figure like an Area or Lamina.

Step2:
Try to identify the basic figures whose algebraic combination produces our problem figure, whose centroid we shall find out.

Step3:
Choose a coordinate system, and make it as our frame of reference. All the distances and coordinate must be define with respect to our frame of reference.

Step4:
Compute the area (a_i), coordinates of their own centroid G_i (x_i, y_i) for each and every elemental areas. While measuring the centroids, all the measurements will be based on according to our chosen Axes.

Step5:
If any particular area has to subtracted to get the complex figure, the area will be negative, where as any area addition will be positive area.

Step6:
If the Centroid of the complex figure be G(X_g,Y_g)then,

=> X_g = (∑a_ix_i)/((∑a_i)

=> Y_g = (∑a_iy_i)/((∑a_i)

Here G₁ is the centroid of the part one where G₂ is the centroid of the circular area that has to be removed where as G₃ is the centroid of the triangular area that has to be removed also.

If we are asked to find moment of inertia of an area, which is nothing but the "second moment of area" then we shall have to find the centroidal moment of inertia first. Then we shall transfer the Moment of Inertia to another axis ie we shall apply parallel axis theorem to transfer moment of inertia from one axis (here centroidal axis) to another parallel axis.

INTELLIGENT OBJECTIVE QUESTIONS IN MECHANICS

1) A cantilever beam of square cross-section (100 mm X 100 mm) and length 2 m carries a concentrated load of 5 kN at its free end. What is the maximum normal bending stress at its mid-length cross-section?

(a) 10 N/mm²
(b) 20 N/mm²
(c) 30 N/mm²
(d) 40 N/mm²

2) A hollow shaft of outside diameter 40 mm and inside diameter 20 mm is to replaced by a solid shaft of 30 mm diameter. If the maximum shear stresses induced in the two shafts are to be equal, what is the ratio of the maximum resistible torque in the hollow to that of solid shaft?

(a) 10/9
(b) 20/9
(c) 30/9
(d) 40/9

(3) A cannonball is fired from a tower 80 m above the ground with a horizontal velocity of 100 m/s. Determine the horizontal distance at which the ball will hit the ground. (take g=10 m/s²)

(a) 400 m,
(b) 280 m,
(c) 200 m,
(d) 100 m.

(4) Water drops from a tap at the rate of four droplets per second. Determine the vertical separation between two consecutive drops after the lower drop attained a velocity of 4 m/s. Take g=10 m/s².

(a) 0.49 m
(b) 0.31 m
(c) 0.50 m
(d) 0.30 m