engineering mechanics ( course content as per csvtu) reference A.K.TAYAL

1. GENERAL SYSTEM OF FORCES
 Equations of equilibrium for a system of
concurrent forces in a plane.
 Constraint, Action and Reaction. Types of
support and support reactions.
 Free Body Diagram – Body subjected to two
forces & Body subjected to three forces.
 Moment of a force. Theorem of Varignon,
Equations of equilibrium.
BASIC CIVIL ENGINEERING & MECHANICS :
UNIT- 4
THIS SLIDE IS MADE BY PARIMAL JHA, H.O.D- CIVIL ENGG, C.E.C DURG, C.S.V.T.U ,CHATTISGARH

2. FORCE
Force Action of one body on the other (push or pull)
Point ofApplication
Direction
Magnitude

3. What is the need of knowing
MECHANICS?
Mechanics Deals with forces

4. TYPES OF APPLIED FORCES
▪ TENSION
▪ COMPRESSION
▪ BENDING / FLEXURE
▪ SHEAR
▪ TORSION
▪ COMBINED

5. CONCEPT OF ENGINEERING MECHANICS
▪ ENGINEERING MECHANICS is the science, which deals with the physical
state of rest or motion of bodies under the action of forces .
▪ Depending upon the nature of the body involved it can be further divided
into mechanics of Rigid body( deformable) or mechanics of solids and the
mechanics of fluids
MECHANICS
FLUID
MECHANICS
SOLID
MECHANICS

6. ENGINEERING MECHANICS

7. Mechanics
Mechanics of Rigid
Bodies
Mechanics of
Deformable Bodies
Mechanics of
Fluids
Statics
kinematics
kinetics
Dynamics

8. BRANCHES OF MECHANICS IN ENGINEERING

9. Statics
 Deals with forces and its effects
when the body is at rest
Dynamics
 Deals with forces and its effects when
the body is in moving condition
TrussBridge IC Engine

10. Studying
forces on
External effect of
a body such as
acceleration,velocity,
displacement etc.
Studying Internal effect of
asforces on a body such
stresses (internal resistance),
change in shape etc.
Rigid body mechanics
Deformable body mechanics

11. Rigid body mechanics
Negligible deformation (no deformation) under the action of forces.
Assuming 100% strength in the materials. Large number of particles occupying
fixed positions with each other.
Actual structures and machines are never rigid under the action of external loads
or forces.
But the deformations induced are usually very small which does not affect the
condition of equilibrium.

12. SOLID MECHANICS
▪ In this Subject we shall deal with the
mechanics of rigid bodies, which do not
deform under the action of applied
forces
▪ A Rigid body is defined as a body on
which the distance between two points
never changes whatever be the force
applied on it
▪ The mechanics of rigid bodies are
studied in two parts statics (deals with
bodies at rest) and dynamics (deals
with bodies at motion)
SOLID
MECHANICS
STATIC DYNAMIC

13. STUDY OF MECHANICS
▪ The study of mechanics involves concepts of space, time, mass and force.
1. Concept of SPACE is essential to fix the position of a point. To fully define position of a point in
space we shall need to define some frame of reference and coordinate system.
2. Concept of TIME is essential to relate sequence of events, for example, starting and stopping
of a motion of a body.
3. Concept of MASS is essential to distinguish between the behavior of two bodies under the
action of identical force.
4. Concept of FORCE is essential as an agency which changes or tends to change the state of rest
or of uniform motion of a body.
A body can be said to be at rest or in motion only with respect to some reference frame. This
reference should be fixed in space, so the earth surface is employed as reference frame.

14. FUNDAMENTAL PRINCIPLES
▪ The elementary mechanics rests
on a few fundamental principles
based on experimental
observations, these are :-
1. Newton's three laws of motion
2. Newton's law of gravitation
3. The parallelogram law for the
addition of forces
4. The principle of transmissibility
of a force.

15. SYSTEMS OF UNITS
▪ FORCE = Mass x Acceleration
▪ F = m x a
▪ F= { mass x length/ (time)2 }
▪ Different system of units are :
▪ Centimetre gram sec system (C.G.S)
▪ Foot pound second system (F.P.S)
▪ Metre kilogram second system (M.K.S)
▪ International system of units (S.I)
Force is a vector quantity thus rules of vector addition and subtraction are applicable to the addition and subtraction of forces

16. SYSTEMS OF UNITS

17. USEFUL FORMULAS IN MECHANICS

21. Types of forces
Concurrent coplanar forces
Non Concurrent coplanar
(Parallel)
Collinear forces
Concurrent non-coplanar

22. ▪ A concurrent force system
contains force whose line of
action meets at one point
CONCURRENT FORCES IN A PLANE
force
Scalar
Vector
▪ SCALAR QUANTITY : Quantities like Time, Mass, Volume & Energy can be completely defined by stating
their magnitudes.These quantities can be added and subtracted according to law of Algebra.
▪ VECTOR QUANTITY : Quantities like displacement ,Velocity ,Acceleration ,Momentum & force posses both
magnitude as well as direction. To define these quantities we have to specify their magnitude, direction &
point of action such quantities can be added according to the Parallelogram law are termed as vector
quantities.

23. Components of a Force
Plane Force
Space Force

24. ADDITION OF TWO FORCES: PARALLELOGRAM LAW
▪ In Mechanics ,most of the time, we are concerned
with the forces having an equivalent effect on the
rigid body rather than equal forces
▪ The resultant of the two forces acting on a body
,in this sense, is equivalent force

25. Parallelogram law:
Two forces acting on a particle can be replaced by the single
component of a force (RESULTANT) by drawing diagonal of the
parallelogram which has the sides equal to the givenforces.
Parallelogram law cannot be proved mathematically . It is an
experimental finding.
A = point of application , AB= Magnitude , AC= Reference line
B
C
B
C
30o
R = P + Q

26. Law of Triangle of forces
▪ Instead of constructing a parallelogram the sum of the resultant of two forces can be
determined by the triangle law.
▪ Triangle law can be stated as :
▪ “If two forces acting at a point are represented by two sides of a triangle taken in order, then
their sum or resultant is represented by the third side taken in an opposite order.”
R = P+Q = Q+P

27. Law of Triangle of forces
▪ The remaining angles can be computed by
using the law of sines as :
α β
γ
R= √ P2+Q2 - 2PQ Cos β
R= magnitude of the resultant R
Angle β should be known for finding the magnitude
P Q R
sinγ sin α sinβ
SUBTRACTION
Here forces are working in
opposite direction
ADDITION

28. ▪ Q.1 – Two forces are acting at a point as shown in figure. Determine the
magnitude and direction of the resultant.
Trigonometric solution :
Instead of drawing triangle to the scale and measuring resultant “R” we can determine it
by applying law of cosines to the ▷ ABC
R= √ P2+Q2 - 2PQ Cos β = √ 502 + 1002 – 2*50*100* cos 1500 = √21160 = 145.46 N

29. The two vectors can also be added by head to tail by using triangle law.
Triangle law states that if three concurrent coplanar forces are acting at a
point be represented in magnitude and direction by the sides of a triangle,
then they are in static equilibrium.

30. Lami’s Theorem states that if three concurrent
coplanar forces are acting at a point, then
each force is directly proportional to the
sine of the angle between the other two
forces.

31. Lami’s theorem considering only
the equilibrium of three forces
acting on a point not the stress
acting through a ropes or strings
The principle of transmissibility is
applicable only for rigid bodies not
for deformable bodies

32. ▪ Point force
▪ Coplanar forces
▪ Non coplanar forces
▪ Concurrent forces
▪ Non concurrent forces
▪ Parallel forces
▪ General system of
forces
CONCEPT OF FORCES:
Point force

33. CONCEPT OF SYSTEM OF FORCES:

34. ▪ The rectangular components Fx & Fy are also
called the scalar components of force F
F = √ FX
2 + FY
2
Tan θ = Fy
Fx
θ = angle between force F
and X axis
Concurrent force-Two or more forces
are said to be concurrent at a point if their
lines of action intersect at that point.
The rectangular or Scalar
Components of a Force in a plane

35. RESULTANT OF SEVERAL CONCURRENT COPLANAR FORCES BY
SUMMING RECTANGULAR COMPONENTS (METHOD OF PROJECTIONS)
Here forces P,Q & S are coplanar concurrent forces. Each forces acting at A can be replaced by its rectangular
components in corresponding x & y direction as Rxi & Ryj ,these forces produce same effect on the particle as
the forces themselves. Now algebraic sum of forces can be determined by adding in particular directions x &
y as Σ Fx & Σ Fy .

36. F1
F2
F5
F4
F3
A B
E
D
C
Polygon Law of Forces
“If many number of forces acting at a
point can be represented as a sides of
a polygon, then they are in
equilibrium”
F1
F2
F3F4
F5

37. Q.1 - Four forces act on bolt A as shown. Determine the resultant of the force
on the bolt.

38. Q.1 - Four forces act on bolt A as shown. Determine the resultant of the force
on the bolt.
SOLUTION:
• Resolve each force into rectangular components.
• Determine the components of the resultant by adding
the corresponding force components.
• Calculate the magnitude and direction of the resultant.

39. Q.1 - Four forces act on bolt A as shown. Determine the resultant of the force
on the bolt.
Resolve each force into rectangular components.

40. Q.1 - Four forces act on bolt A as shown. Determine the resultant of the force
on the bolt.
Resolve each force into rectangular components.
force Magnitude
in N
X component
in N
Y component
in N
F1 150 +129 +75
F2 80 -27.4 +75.2
F3 110 0 -110.0
F4 100 +96.6 -25.9

41. Q.1 - Four forces act on bolt A as shown. Determine the resultant of the force
on the bolt.
force Magnitude
in N
X component
in N
Y component
in N
F1 150 +129 +75
F2 80 -27.4 +75.2
F3 110 0 -110.0
F4 100 +96.6 -25.9
Fx= +199.1 Fy= +14.3
R = √ 199.12 +14.32
,R = 199.6N
Tanα = 14.3 N
199.1N
, α = 4.1

42. Q.2 - Find the magnitude and direction of resultant R of 4 concurrent forces acting as
shown in figure below ?
60o
90o
150o
p
2p
4p
3 √3 p

43. Q.2 - Find the magnitude and direction of resultant R of 4 concurrent forces acting as
shown in figure below ?
ΣFx = -P/2
ΣFy = √3/2 x P
60o
90o
150o
p
2p
4p
3 √3 p
ΣFx = p cos 0+ 2p cos 60 + 3 √3 P cos 150 + 4p cos 300
ΣFy = p sin 0+ 2p sin 60 + 3 √3 P sin 150 + 4p sin 300

44. Q.2 - Find the magnitude and direction of resultant R of 4 concurrent forces acting as
shown in figure below ?
ΣFx = -P/2
ΣFy = √3/2 x P
Resultant of forces R = √ RX
2 + RY
2 = √ (-P/2)2 + (√3/2 x P)2 = P
60o
90o
150o
p
2p
4p
3 √3 p
ΣFx = p cos 0+ 2p cos 60 + 3 √3 P cos 150 + 4p cos 300
ΣFy = p sin 0+ 2p sin 60 + 3 √3 P sin 150 + 4p sin 300
Tanα = (-1/2 P/√3/2 P) = -√3
α = tan-1 ( -√3 )

45. EQUATIONS OF EQUILIBRIUM FOR A SYSTEM OF
CONCURRENT FORCES IN A PLANE
▪ Equations below are called the
equilibrium equations, if a
number of concurrent forces
lying in a plane are in
equilibrium, these equations
are to be satisfied
ΣFx = 0
ΣFy = 0
▪ Equilibrium state means
restriction of movement of
body in any of the direction in
the plane under application of
forces.

48. TYPES OF SUPPORT AND SUPPORT REACTIONS

49. CONSTRAINT ACTION AND REACTION

52. ▪ A free body diagram consists primarily of a sketch of the body in question and arrows representing
the forces applied to it (TO the body) Another way is to say the force of something ON the body.
How to find out forces and reactions of a body subjected to applied
forces :
1. Isolate Body
▪ There must be only 1 body. Make sure you isolate the body exactly – as if you cut it out in a
silhouette or outline. It is important to be very clear about the boundary you have made around
the body.
2. Find Force locations
▪ Forces are applied by contact, gravity or inertia. Identify all the points where forces are applied to
the body. Gravity always acts through the centre of mass, pressure acts through the centre of
pressure.
▪ The only forces to consider are those that CROSSTHE BOUNDARY. Ignore all other forces.
FREE BODY DIAGRAM

53. 3. Line of Action of Forces
▪ Point contact: ‘Smooth’: No friction. Force can only be applied perpendicular to smooth surface.
Wheels: Perpendicular to surface and acting through Centre of axle.
▪ Cable. Force can only be tensile (pulling). Force must be along direction of cable.
▪ Pin Joint: Force can be any direction but no moment around pin.
▪ Other Joints: Solid: Force in any direction, moment in any direction. Slider: Moment any direction.
Force perpendicular to slider
▪ Gravitational. Centre of gravity
▪ Inertial. Centre of inertia (not always the same as C.O.G).These are due to acceleration.
▪ Fluid pressures. Usually taken as a single resultant force in statics calculations, acting through the
Centre of pressure.
▪ Other. Magnetic, electrostatic (charge) etc.
These are pretty rare outside of electric equipment design.
4. Direction of the Forces
▪ Include all the forces actingTOTHE BODY.The direction is determined by thinking TO the body,
or ON the body. Eg Gravity acts down ON the body, floor pushes up ON the body, what does the
road doTO the wheel…etc.

54. FREE BODY DIAGRAM

57. Couple Two equal and opposite forces are acting at
some distance forming a couple

58. How Rotational Effect
will change with
distance?

59. Free body diagram
Isolated body from the structure of machinery which shows all the forces and reaction forces
acting on it.

60. Examples for free body diagram

61. Examples of Free Body Diagrams :