Engineering Mechanics
I Year B.Tech
By
N.SRINIVASA REDDY., M.Tech.
Sr. Assistant Professor
Department of Mechanical Engineering
Vardhaman College of Engineering
Basic concepts of Mathematics & Physics for
Engineering Mechanics
Measurement of angles: There are two common systems of measuring angles
1. sexagesimal measure : According to this system, divide a right angle 90 into equal parts called
degrees. Each degree is divided into sixty equal parts called minutes and each minute is divided into sixty
equal parts called seconds. Denoting one degree, one minute and one second by the symbols 10 1’1”.
2. Circular measure: The unit of measurement is called radian . ∏ is an irrational number. Its
approximate value is 22/7. Its value correct to 5 decimal places is 3.14159. Sometimes1/∏ value also
required & is given by 0.31831.
Relation between degree & radian {1 radian = 1800 x 1/∏ = 57017’44.8”};
10 = ∏ /180 radian
Formulas from Trigonometry:
1. sin2A + cos2A = 1 ; sin2A = 1- cos2A ; cos2A = 1- sin2A
2. 1+tan2A = sec2A ; tan2A = sec2A -1 ; tan2A - sec2A = 1
3. sin(A+B) = sin A . cos B + cos A . sin B
4. sin(A - B) = sin A . cos B - cos A . sin B
5. cos(A+B) = cos A . cos B – sin A . sin B
6. cos(A - B) = cos A . cos B + sin A . sin B
7. sin(A+B) . sin(A - B) = sin2A-sin2B = cos2B- cos2A
8. cos(A+B) . cos(A - B) = cos2A – sin2B = cos2B-sin2A
9. sin2θ = 2 sinθ cosθ = 2tan θ / 1+tan2θ
10. cos2θ = cos2θ – sin2θ =1- tan2θ / 1+tan2θ = 1-2 Sin2θ =2cos2θ -1
Formulas from Differentiation:
1. y = xn,
= nxn-1
2. y = ex,
= ex
3. y = log x,
4. y = ax,
Formulas from Integration
1.
=xn+1/ n+1
2.
= x1-n / 1- n
3.
= log x
4.
=
=
= ax log a
5. y = sinx ,
= cos x
6. y = cosx,
= - sinx
5.
= sinx
6.
= - cosx
= F(b) – F(a)
Scalars & Vectors
Those quantities which have only magnitude and are not related to any direction in space are called
scalars. Whereas those which have both magnitude and direction are called vector quantities.
1. Vectors are said to be like vectors when they have the same sense of direction, unlike vectors
when they are in opposite direction.
2. When two or more vectors are said to be collinear vectors when they act along the same line of
action & parallel vectors when they are in parallel lines
3. Three or more vectors are said to be coplanar when they are parallel to the same plane or lie in
the same plane whatever their magnitudes be (**two vectors are always coplanar)
4. Two vectors are said to be equal when they have the same length(magnitude &are parallel having
the same sense of direction)are called as equal vectors.
5. If the origin and terminal points of a vector coincide, then it is said to be a zero vector
6. A vector whose magnitude is of unit length is called a unit vector. If
magnitude is , then unit vector in the direction
vector
by its magnitude
. Thus
=
is denoted by
is a vector whose
and is obtained by dividing the
.{ Unit vector = Force vector / Magnitude of vector}
7. The vectors having the same initial point are call co –initial vectors.
8. A vector drawn parallel to a given vector through a specified point in space is called a localized
vector. But if the origin of vectors is not specified the vectors are called free vectors.
Formulas from Vectors
1. a.b = b.a = ab cosθ
if the scalar product is of two vectors is zero, then at least one of the vectors is a zero vector or
they are perpendicular.
2. a.(b+c) = a.b + a.c
3. i2=j2=k2=1
4. i.j=j.k=k.i=0
5. a x b ≠ b x a but a x b = - b x a
6. i x i = j x j = k x k =0
7. i x j = k = - j x i ; j x k = i = - k x j; k x i = - i x k
8. if a = a1i+a2j+a3j ; b = b1i+b2j+b3k then a x b is given by
= (a2b3 – a3b2) i +(a3b1-a1b3) j + (a1b2 – a2b1) k
axb=
Conversion Equivalents:
Length
1in = 2.540cm ; 1ft =12in =30.48cm
1mile=5280ft=1.609km
Force
1lb= 0.4536kg = 4.448N
Velocity
88fps=60mph=96.54km/hr
Acceleration
g = 32.2fps2 = 9.81m/s2
Pressure
1atm = 14.7psi=760mmof Hg= 1.013 x 105 N/m2
Volume
1cuft=7.481gallons=28.32litres
Sign Conventions:
I Quadrant
II Quadrant
III Quadrant
IV Quadrant
X -Axis
+
-
-
+
y-Axis
+
+
-
-
INTRODUCTION TO ENGINEERING MECHANICS
It is the branch of science which deals with the behavior of a body and predicts the
conditions of rest or motion of bodies under the action of forces. Under the category of Rigid
bodies which is a branch of engineering mechanics, the subcategories are statics & dynamics.
Rigid body is defined as a definite amount of matter that parts of which are fixed in
position relative tone another. Actually solid bodies are never rigid; they deform under the
action of applied forces. In many cases, this deformation is negligible compared to the size of
the body and the body may assumed to be rigid.
Statics is the branch of science which study the forces and their effects, when the body
is at rest.
Dynamics is the branch of science which study the forces and their effects, when the
body is in motion. It is further sub classified as kinematics & kinetics
Kinematics which deals with study of the motion, but not considering the forces which
are responsible for the motion.
kinetics which deals with study of the motion, considering the application of forces
which are responsible for the motion.
A Particle is a body of infinitely small volume and the mass of the particle is considered
to be concentrated at a point. Hence a particle is assumed to a point and the mass of the
particle is concentrated at this point
Mass it is that invariant property of a body which measures it resistance to a change of
motion.
Force is defined as an agent which produces or tends to produce, destroys or tends to
destroy motion, in statics by consideration ”tends to produce motion”.
Engineering Mecahnics
Dynamics
[Branch of mechanics which
deals with forces of body
which are at motion]
Kinetics
Kinematics
System of Units:
Quantity
Plane Angle
Force
Frequency
Energy, work
Power
Stress, Pressure
Velocity
Acceleration
Angular Velocity
Units
Radians
Newton
hertz
Joule
Watt
Pascal
metre/second
metre/(second)2
radian/second
Multiplication factors:
tera
T
Giga
G
Mega M
Kilo
K
milli
m
micro μ
nano
n
pice
p
-
Symbol
rad
N
Hz
J
W
Pa
m/s
m/s2
rad/second
1012
109
103
103
10 –3
10 – 6
10 – 9
10 – 12
Statics
[Branch of mechanics to study
the forces which are at rest]
UNIT - I
Fundamentals
Theory:
- Engg Mechanics is a branch of science which deals with the behavious of a body when the body is at rest or
in motion.
- It is divided into statics and Dynamics
1. Statics (Which deals with the study of a body when body is at rest)
2. Dynamics (Study of a body when the body is in motion)
a. Kinetics (When the body is in motion if the forces causing the motion also considered is called as kinetics)
b. Kinematics (Study of a body in motion the forces which cause motion are not considered)
Vector Quantity:
- A quantity which is completely specified by magnitude and direction.
- A vector quantity is represented by means of a straight line with an arrow.
Eg: Velocity, accleration, force and momentum.
Scolar Quantity:
- A quantity which is completely specified by magnitude only is known as scolar quantity.
Eg: Mass, Length, time and temparature.
Particle:
- A particle is a body of infinitely small volume and the mass of the body particle is considered to be
concentrated at a point.
- Hence a particle is assumed to be a point and the mass of the particle is concentrated at this point.
B
Porallelogran Law:
- If two forces acting at a point be represented in magnitude and direction by the two
adjacent sides of a parallelogram then there resultant is represented in magnitude
and direction by the diagonal of the parallelogram passing through that point.
Q
α
0
➤
Q
➤
➤
0
α
➤
➤
Magnitude of the Resultant (R):
- From 'C' draw CD ..... to OA Produced.
- Let α equal to angle b/w two forces P and Q = ∠AOB
α = ∠AOB
- Now ∠DAC = ∠AOB (corresponding angles) = α
- In parallelogram OACB, Ac is parallel to OB.
∴ AC = Q
- In ΔACD, AD = AC cosα = Qcosα
CD = ACsinα = Qsinα
- In ΔOCD, OC2 = OD2 + DC2
But OC = R, OD = OA + AD = P + Q cosα
B
θ
C
R
➤
P
➤
➤
Proof:
- Let two forces P and Q act at a point '0' as shown in fig.
- The force 'P' is represented in magnitude and direction by 0A.
- Where as the force Q is represented in magnitude and direction by 0B.
- Let the angle between the two forces P and Q is α
- The Resultant of these two forces will be obtained in magnitude and direction
by the diagonal (passing through '0') of the parallelogram of which 0A and 0B
are two adjacent sides.
- Resultant is represented by 'R' makes an angle θ with OA.
A
P
A
α
D
DC = Qsinα
∴ R2 = (P + Q cosα)2 + Q2Sin2α
⇒ R2 = P2 + Q2cos2α +2PQcosα + Q2sin2α
⇒ R2 = P2 + Q2 + 2PQ cosα
∴ R = P 2 + Q 2 + 2PQ Cos α
Directions of Resultant:
- Let θ equal to angle made by Resultant with OA.
CD
- from ΔOCD, Tanθ = OD
⇒ Tanθ =
Qsin α
P + Qcos α
⎛ Qsin α ⎞
∴θ = Tan −1 ⎜
⎟
⎝ P + Q cos α ⎠
Case (i):
- If the two forces P and Q at right angles i.e. α = 90°
∴ R = P2 + Q2
⎛Q⎞
θ = Tan −1 ⎜ ⎟
⎝P⎠
case (ii):
- If the two forces P and Q are equal and are acting at an angle a b/w then.
R = P 2 (1 + cos α ) = 2P cos α / 2
⎛ sin α ⎞
−1 ⎛ 2sin α / 2 cos α / 2
θ = Tan −1 ⎜
⎟ = Tan ⎜⎜
2cos 2 α / 2
⎝ 1 + cos α ⎠
⎝
⎞
⎟⎟
⎠
⇒ θ = α /2
Law of Traingle Forces:
- If three forces acting at a point be represented in magnitude and direction by 3 sides of triangle taken in order
they will be in equlibrium.
Lami's theorem (sine Law):
- It states that if three forces acting at a point are in equlibrium each force will be proportional to the sine of
angle b/w the other two froces.
i.e.
Proof:
- Suppose the three forces P, Q and R and they are in
P
equilibrium as shown.
- Let α = Angle b/w Q and R,
β = Angle b/w P and R,
γ = Angle b/w P and Q
β
- According to Lani's theorem:
P α Sine of angle b/w Q and R
P α Sin α
⇒
0
R
α
P
= cons tan t
Sinα
Similarly, Q α sine of angle b/w R and P
Q α sin β
⇒
γ
Q
Q
= cons tan t
Sinβ
Similarly, R α sine of angle b/w P and Q
R α sin γ
⇒
P
P
R
=
=
cons tan t
Sinα Sin β Sin γ
Case (i):
➤
➤
∴
R
= cons tan t
Sinγ
θ2
•
θ1
(When an object is hanged by two chains/ropes from the same plane)
W
T1
T2
=
=
sin 90° Sin (180 − θ1 ) Sin (180 − θ2 )
➤
- Between T1 and T2 always 90° (θ1+ θ2 = 90)
- ∠AOC =180° - θ2
BOC = 180 - θ1
C
•
T3
W
case (ii):
T3
T1
T2
=
=
sin (180 − (90 − α )) sin (180 − θ1 ) sin (90 − α + θ1 )
i.e.
T3
T1
T2
=
=
sin (90 ° + α ) sin (180 − θ 1 ) sin (90 − α + θ 1 )
Resolution of a force:
- It means finding the component of a given force in given two directions.
- Let a given force (P) which makes as angle 'θ' with x - axis as shown in fig.
- Component of P along x - axis is
Px = Pcos θ
- Component of P along y-axis is
Py = Psin θ
Problem:
Two forces of magnitude 10N and 8N are acting at a print. If the angle b/w 2 forces is 60°. Find magnitude
of resultant forces
Sol:
R = P 2 + Q 2 + 2PQ cos θ
= 100 + 64 + 160 ×
P = 10N, 2 = 8N
1
= 164 + 180 = 244N
2
= 15.62N
- Two equal forces are acting at a point with a angel of 60°. If the R = 20x3N then find magnitude of each force.
Sol:
R2 = (2pcos α/2)2
⇒ 20 3 = 2P
3
⇒ P = 20N
2
- The resultant of two forces when they act at an angle of 60° is 14N if same forces are acting at right angles
their resultant is √136 N Determine magnitude of 2 forces.
Sol:
R = P 2 + Q 2 + 2PQ Cos60 °
14 = P 2 + Q 2 + PQ
136 = P 2 + Q 2
...............................(1)
...............................(2)
196 = P 2 + Q 2 + PQ
from (2)
196 = 136 + PQ ⇒ PQ = 60 ⇒ Q = 60/P
⇒ P2 + Q2 = 14 - 60 = - 46
⇒ P2 +
3600
P2
= − 46 ⇒ P 4 + 46P 2 + 3600 = 0
PQ = 60 2PQ = 120
196 = (P + Q)2 – PQ
(P + Q)2 = 196 - 120 = 76
(P - Q)2 = 196 - 120 = 76
(P − Q )2 = P 2 + Q 2 − 2PQ = 136 −120 =16 ⇒ P − Q = 4
⇒P−
60
=4
P
⇒ P2 - 4P - 60 = 0
⇒ P2 - 10P + 6P - 60 = 0
⇒ P (P - 10) + 6 (P - 10) = 0
⇒ P = 6N (or) P = 10N
- Two forces are acting at a point 'o' as shown in fig. Determine the resultant in magnitude and direction.
sol:
Given P = 50N, Q = 100N
α = 30°
R = P 2 + Q 2 + 2PQ cos α
(
= 2500 + 10000 + 100 × 100
50
= 2500 + 500
3
+ 10,000
2
= 10 25 + 100 +
5 3
2
) 23
= 10 125 + 5 3 2
⎛ Qsin α ⎞
Tanθ = Tan −1 ⎜
⎟=
⎝ P + Q cos α ⎠
100 / 2
50 + 100
3
2
=
50
(
50 1 + 3
=
)
3 −1
= 0.366
2
θ = 20.1°
i.e. direction
- Angle subtended by Resultant with x - axis is (20.1 + 15)° = 35.1°
- Determine the magnitude of resultant of two forces of 12N and 9N acting at a point and if the angle b/w them
is 30°
Sol:
R = P 2 + Q 2 + 2PQ cos α
144 + 81 + 18 ×12 ×
(
3
2
)
= 144 1 + 3 + 81
=20.3 N
- Find the magnitude of 2 equal forces with an angle of 60° b/w them if R = 30 √3 N.
Sol:
R = 2P cos α/2
⇒30√3=2P cos 30°
⇒ P = 30 N
- The Resultant of two forces when they act at right angles is 10 N whereas they act at an angle of 60° is
R = 148
determine magnitude of two forces.
Sol:
100 = P2 + Q2
148 = P 2 + Q 2 + 2 PQ
1
2
⇒ PQ = 48
⇒ If P = 6N ⇒ Q = 8N
If P = 8N ⇒ Q = 6N
- The weight of 1000N is supported by two chains from the same roof. Chain (1) makes angle 30° and chain
(2) 60°. Draw the diagram and determine the tension in the chains.
Sol:
T2
T1
1000
=
=
1
sin 30 ° sin 60 °
⇒ 1000 = 2T2 =
2T 1
3
⇒ T2 = 500N, T1 =
1000 × 3
= 866N
2
- A weight of 900 N is supported by two chains of lengths 4m and 3m as shown. Determine the tension in each
chain
Sol:
AB2 = AC2 + BC2
= 16 + 9
= 25
⇒ AB = 5m
3
4
⇒ sinα = ,sin β =
5
5
T1
T2
W
=
=
sin (180 ° − α ) sin (180 ° − β ) sin 90 °
⇒
5T1 5T1
=
= 900
3
4
⇒ T1 =
900 × 3
= 540N
5
⇒ T2 =
900 × 4
= 720N
5
- An electrical light fixer 15N hangs from a point 'c' by two sittings AC and BC. AC is inclined at 60° to the
horizontal and B at 45° to the vertical. Using Lamis theoremm or otherwise determine the forces in the strings
AC and BC.
Sol:
T1
T2
15
=
=
sin (180 ° − 30 ° ) som (180 ° − 45 ° ) sin (90 ° −60 ° +45 ° )
⇒
T1
T2
15
=
=
sin 30 ° sin 45 ° sin 75 °
⇒ P1 =
T2 =
15
15
1
× sin 30 ° = ×
= 7.81N
sin 75 °
12 0.96
15
1
×
= 10.98 N
0.96
2
Method 2:
T1 sin 45° = T2 cos 60°
⇒ T2 = √2T1
T1cos45° + T2 sin60° = 15
⎡ 1
2× 3⎤
⇒ T1 ⎢
+
⎥ = 15
2 ⎦
⎣ 2
⇒ T1 =
15 2
1+ 3
T1 = 7.76 N
T2 = 2T1 = 10.98 N
Resolution of number of coplanar forces:
- Let a number of coplanr forces (forces acting in one plane are coplanar forces) R1, R2, R3 are acting at a point
'o'.
- Let θ1, θ2, θ3 the angles made by R1, R2, R3 respectively with x - axis
- Then H = Resultanat component of all forces along x - axis
component of R1 along x - axis = R1 cosθ1
component of R2 along x - axis = R2 cosθ2
component of R3 along x - axis = R3 cosθ3
H = ΣR1 cosθ1
- V = Resultant component of all forces along y - axis
component of R1 along y - axis = R1 sinθ1
component of R2 along y - axis = R2 sinθ2
component of R3 along y - axis = R3 sinθ3
V = ΣR1 sinθ1
R = H2 + V2
- Resultant force,
- Angle made by 'R' with x - axis is given by
Tanθ = V / H
Problem:
Sol:
H = – P1 cos 30° – P2 cos 120° – P3 cos 190°
V = – P1 sin 30° – P2 sin 120° – P3 sin 190°
R = H2 + V2 =
Moment of a force:
- The product of a force under perpendicular distance of the line of action of the force from a point known as
moment of foce.
Types of Beams:
1. Candidates Beam (One end fixed and other end is free)
2. Simply supported Beam (both ends are supported) Eg: Benches
3. Over hanging Beam (Additional beam is existing other than support prints). Eg: Diver in swimming
4. Continuous Beam (no. of supports are more as well as beam length is also long) Eg: Railway Bridge.
5. Fixed beam (both end are fixes).
Types of Loads:
1. Point Load (concentration at one point) (W)
2. Uniformly distributed load (w × l) [w - weight per unit run, l - length of the beam]
- Length of beam
concentrating at perpendicular midpt length.
Types of Supports:
1. Rollen support (Only Vertical reaction)
2. Hinged support (Both Vertical and horizontal Reaction)
Problem:
Find RA and RB
Sol: RA + RB = 240 N ----------- (1)
a) case (i):
- Taking moments about 'A'.
Note: The force existing on 'A' is treated as zero.
RA = 0
∴ Clockwise moments = Anticlock moments
Clockwise moment: The wt and the distance towards A.
Anti - Clock wise moment: (upward force (or) reaction) × (the distance up to A)
RB × 10 = 60 × 2 + 80 × 6 + 100 × 8.
RB = 140N RA = 100N (RA + RB = 240 N)
Case (ii):
Considering moments about B.
RA × 10 = 100 × 2 + 80 × 4 + 60 × 8
⇒ RA = 100 N
∴
⇒ RB = 140 N
RA + RB = 240 N
Action and Reaction:- Using Newton's Third Law of motion, "For every action there is equal and opposite reaction."
- Any face on a support causes equal and opposite force from the support so that action and reaction are two
equal and opposite forces.
Note:
- If weight acting on the surface.
- Always reaction is perpendicular to the contact surface either it is vertical, horozontal (or) inclined.
Con
- A Stasionary body which is subjected to coplanar forces (concurrent (or) parallel) will be in equilibrium if
the algebraic sum of all the external forces is zero and also the algebraic sum of moments of all the external
forces about any point in their plane is zero.
Mathematically,
ΣM = 0, - Moment law of Equilibrium
Σfx = 0, (Foce law of
Σfy = 0 Equilibrium)