Concept of Particles and Free Body Diagram Why FBD diagrams are used during the analysis? It enables us to check the body for equilibrium. By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body. It helps to identify the forces and ensures the correct use of equation of equilibrium. Note: Reactions on two contacting bodies are equal and opposite on account of Newton's III Law. The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces. Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable. Physical Meaning of Equilibrium and its essence in Structural Application The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium. A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium. The rigid body in equilibrium means the body is stable. Equilibrium means net force and net moment acting on the body is zero. Essence in Structural Engineering To find the unknown parameters such as reaction forces and moments induced by the body. In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters. For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.

1. Engineering Mechanics: Statics
Force System Resultants

2. Chapter Objectives
 To discuss the concept of the moment of a force and
show how to calculate it in 2-D and 3-D systems.
 Definition of the moment of a couple.

3. Chapter Objectives
• To present methods for determining the resultants of
non-concurrent force systems in 2D and 3D systems.
• Reducing the given system of forces and couple
moments into an equivalent force and couple
moment at any point.

4. Chapter Outline
Moment of a Force
Force Couple
Principles of Moments

5. Moment of a Force
Carpenters often use a hammer in this way to pull a stubborn nail.
Through what sort of action does the force FH at the handle pull the
nail? How can you mathematically model the effect of force FH at
point O?

6. MomentMoment of a force about a point (or an axis) is
a measure of the tendency of the force to
cause a body to rotate about the point or axis.
Moment of a Force
2D System

7. Case 1:
Fx horizontal and acts perpendicular to
the handle of the wrench and
is located dy from the point O
Fx tendstends to turn the pipe about the z axisto turn the pipe about the z axis
 The larger the force or the distance dy the greater the
Moment of a ForceMoment of a Force
2D System2D System

8. Moment of a ForceMoment of a Force
2D System2D System
The larger the force ‘W’ or the distance ‘D’ the greater the turning
effect at point ‘P’.

9. Note that:
Moment axis (z) is perpendicular to shaded
plane (x-y). i.e. The remaining third axis: ‘z’
Fx and dy lies on the shaded plane (x-y)
Moment axis (z) intersects
the plane at point O
Moment of a Force
2D System

10. Case 2:
Apply force Fz to the wrench
 Pipe does not rotate about z axis
 The pipe not actually rotates but
Fz creates a tendency for rotation
so causing (producing)moment
along (Mo)x.
 Moment axis (x) is perpendicular to
the shaded plane (y-z)
Moment of a Force
2D System

11. Case 3:
Apply force Fy to the wrench
 NoNo momentmoment is produced about point O
 Lack of tendency to rotateLack of tendency to rotate
as line of action passesas line of action passes
through Othrough O
Note that, Fy and dy both
lies on the same line (and not forming any
Moment of a Force
2D System

12.  Moment of a force does not always cause rotation
 Force F;
 tends to rotate the beam clockwise about A with
moment MA = dAF
 tends to rotate the beam counterclockwise about B
with moment MB = dBF
Moment of a Force
2D System

13. In General
 Consider the force F and the point O which lies in the
shaded plane
 The moment MO about point O,
or about an axis passing
through O and which is
perpendicular to the plane, is a vector quantityvector quantity..
Moment of a Force
2D System

14. Moment MO is a vector having
specified magnitudemagnitude and
directiondirection.
Moment of a Force
2D System

15. M = F x d
M = Magnitude of the moment
about point or axis [N.m]
F= Magnitude of the force [N]
d = perpendicular distance [m]
Direction is determined by
using the right hand rule
Moment of a Force
2D System

16. PositivePositive direction of the moment :direction of the moment :
Anti clockwise
Moment of a Force
2D System

17. PositivePositive momentmoment
NegativeNegative momentmoment
Moment of a Force
2D System

18. Magnitude:
Use simple multiplication;
For magnitude of MO: MO = FF .. dd
d = moment armmoment arm or perpendicularperpendicular
distancedistance
from the axis at point O to its line of action
of the force.
F = Magnitude of the force

Moment of a Force
2D System
Scalar Formulation

19. Moment of a Force
2D System
Scalar Formulation
Direction:
Direction of MO is specified by
using “right hand ruleright hand rule”
Fingers of the right hand are
curled to follow the sense of
rotation when force rotates about
point O.

20. Direction:
Thumb points along the moment
axis to give the direction and
sense of the moment vector
Moment vector is upwards and
perpendicular to the shaded
plane
Moment of a Force
2D System
Scalar Formulation

21. Direction
MMOO is shown by a vector arrow with a curl tois shown by a vector arrow with a curl to
distinguish it from force vectordistinguish it from force vector Fig b.
 MO is represented by the counterclockwise
curl, which indicates the action of F.
 Arrowhead shows the sense of rotation caused
by F.
 Using the right hand rule, the direction and
sense of the moment vector points out of the
page.
Moment of a Force
2D System
Scalar Formulation

22. FrM

×=
=r
 Position vector which runs from the
moment reference point to any point
on the line of action of the force.
In some two dimensional problems and
most of the three dimensional problems,
it is convenient to use a vector approach
for moment calculations.
The MOMENT of a force about point A
may be represented by the cross productcross product
expression.
Moment of a ForceMoment of a Force
2D2D System
VectorVector FormulatiFormulationon

23. Without using the right hand rule
directly apply the equation through
cross product of vectors.
MO = d XX F
The cross product directly gives the
magnitude and the direction.
Moment of a ForceMoment of a Force
2D2D System
VectorVector FormulatiFormulationon

24. FrMo

×=
Sarrus’ RuleSarrus’ Rule
+k- k
)kFr)kFrM xyyxo

(( −=
Moment of a ForceMoment of a Force
2D2D System
VectorVector FormulatiFormulationon

25. Example:
For each case, determine the moment of the
force about point O
Moment of a ForceMoment of a Force
2D2D System

26. Solution
 Line of action is extended as a dashed line to
establish moment arm d
 Tendency to rotate is indicated and the orbit is
shown as a colored curl
o
o
(a)M (2m)(100N) 200.000 N.m (CW)
(b)M (0.75m)(50N) 37.500 N.m (CW)
= =
= =
Moment of a ForceMoment of a Force
2D2D System

27. Solution
o
o
o
(c) M (4m 2cos30 m)(40N) 229.282 N.m (CW)
(d) M (1sin45 m)(60N) 42.426 N.m (CCW)
(e) M (4m 1m)(7kN) 21.000 kN.m (CCW)
= + =
= =
= − =
o
o
Moment of a ForceMoment of a Force
2D2D System

28. Example:
Determine the moments of
the 800 N force acting on the
frame about points A, B, C
and D.
Moment of a Force
2D System

29. Solution
(Scalar Analysis)
Line of action of F passes through C
A
B
C
D
M = (2.5m)(800N) = 2000 N.m (CW)
M = (1.5m)(800N) = 1200 N.m (CW)
M = (0m)(800N)= 0 N.m
M = (0.5m)(800N) = 400 N.m (CCW)
Moment of a ForceMoment of a Force
2D2D System

30. Moment of a ForceMoment of a Force
2D2D System

31. Principles of Moments
 Principles of MomentsPrinciples of Moments
Also known as Varignon’s Theorem
This principle states that the
moment of a force about a point is
equal to the sum of moments of the
force’s components about the
point.

32. Principles of Moments
“Moment of a force about a point is equal to
the sum of the moments of the forces’
components about the point”

33. Principles of Moments

34. Principles of Moments
Solution
Method 1:
From trigonometry using triangle BCD,
CB = d = 100cos45° = 70.7107mm
Thus,
MA =dF= (0.07071m) 200N
= 14142.136 N.mm (CCW)
= 14.142 N.m (CCW)
As a Cartesian vector,
MA = {14.142 k} N.m

35. Principles of Moments
Solution
Method 2:
 Resolve 200 N force into x and y components
 Principle of Moments
MA = ∑dF
MA =(200)(200sin45°) – (100)(200cos45°)
= 14142.136 N.mm (CCW)
= 14.142 N.m (CCW)

36. Principles of Moments
“Moment of a force about a point is equal to the sum of the
moments of the forces’ components about the point
( ) m898.275sin3 =°=d
( )( )
mkN489.14
898.25
⋅−=
−=
−= FdMO
( )( )[ ] ( )( )[ ]
mkN489.14
30cos345sin530sin345cos5
⋅−=
°−°−=
−−= xyyxO dFdFM
Example:

37. Example:
Determine the moment of the 600 N force with respect to
point O in both scalar and vector product approaches.
Moment of a ForceMoment of a Force
2D2D System

38. Example:
The force F acts at the end of the angle
bracket. Determine the moment of the force
about point O.
Moment of a ForceMoment of a Force
2D2D System

39. Solution
Method 1:
Resolve the given force into components and
than apply the moment equation.
MO = 400sin30°N(0.2m)-400cos30°N(0.4m)
= -98.5641 N.m
As a Cartesian vector,
MO = {-98. 5641k} N.m
Moment of a ForceMoment of a Force
2D2D System

40. Solution
Method 2:
 Express as Cartesian vector
r = {0.4i – 0.2j} m
F = {400sin30°i – 400cos30°j} N
= {200.000i – 346.410j}N
For moment,
{ }
O
i j k
M rXF 0.4 0.2 0
200.000 346.410 0
-98.564k N.m
= = −
−
=
r r r
r rr
r
Moment of a ForceMoment of a Force
2D2D System

41. Resultant Moment of
System of Coplanar Forces
Resultant moment MRo = addition of the moments
produced by all the forces
algebraically since all
moment forces are
collinear (for 2D case).
MRo = ∑F.d
taking counterclockwise (CCW), to be positive.

42. Resultant moment, MRo = addition of the moments
produced by all the forces algebraically, since all moment
forces are collinear (for 2D case).
MRo =M1 – M2 + M3
=∑dF= d1 F1 – d2 F2 + d3 F3
taking counterclockwise (CCW)
to be positive.
Resultant Moment of
System of Coplanar Forces

43. Counterclockwise is positive
ORM Fd+ = ∑
Resultant Moment of
System of Coplanar Forces
+

44. Example:
Determine the resultant moment of the four
forces acting on the rod about point O.
Resultant Moment of
System of Coplanar Forces

45. Solution: (by scalar analysis)
Note that always positive moments acts in the +k
direction, CCW
)CW(m.N923.333
m.N923.333
)m30cos3m4)(N40(
)m30sin3)(N20()m0)(N60()m2)(N50(M
d.FM
Ro
Ro
=
−=
+−
++−=
∑=


Resultant Moment of
System of Coplanar Forces

46. Solution: (by vector analysis)
(CW)or(-k)N.m333.923
(-k)][263.923(k)][30[0]k)]([100
(-j)]40X)(i)3cos30(4[
(i)]20X(-j)[3sin30(i)]60X[0(-j)]50X(i)[2MRo
=
+++−=
++
++=
∑=


dXFMRo
Resultant Moment of
System of Coplanar Forces

47. Moment of a CoupleMoment of a Couple

48. Moment
Moment force F about point O can be
expressed using cross product
MO = r X F
where r represents position vector
from O to any point lying
on the line of action of F.

49. • Remember M = r (vector) X F (vector)
• Find the length ‘r’ vectorially for each
force ‘F’
• if not given, find the vectorial
representation of ‘F’ also.
Moment

50. In some two dimensional problems and
many three dimensional problems, it is
convenient to use a vector approach for
moment calculations. The MOMENT of
a force about point A may be
represented by the cross product
expression
FrM
rrr
×=
Moment

51. FrM
rrr
×=
=r
r Position vector which runs from the
moment reference point to any point
on the line of action of the force
MA
Due to the principle of
transmissibility, can act at any point
along its line of action and still create
the same moment about point A.
=F
r
FrFrFrM A
rrrrrrr
×=×=×= 321
Moment

52. The Moment Vector
The result obtained from r X F doesn’t depend on
where the vector r intersects the line of action of F:
r = r’ + u
r × F = (r’ + u) × F = r’ × F
because the cross product of the parallel vectors u
and F is zero.

53. Moment of a CoupleMoment of a CoupleMoment of a CoupleMoment of a Couple
CoupleCouple
- two parallel forces
- same magnitude but opposite direction
- separated by perpendicular distance dseparated by perpendicular distance d
 Resultant force = 0
 Tendency to rotate in specified direction
 Couple moment = sum of
moments of both couple
forces about any arbitrary point

54. Moment of a CoupleMoment of a Couple
A couple is defined as twoA couple is defined as two
parallel forces withparallel forces with the samethe same
magnitudemagnitude butbut opposite inopposite in
directiondirection separated by aseparated by a
perpendicular distance d.perpendicular distance d.
The moment of a couple is defined asThe moment of a couple is defined as::
MO = F . d (using a scalar analysis; right hand rule for direction),
MO = d X F (using vector analysis).

55. Moment of a CoupleMoment of a Couple
The net external effect of a couple is zero since
the net force equals zero and the magnitude of
the net moment equals F.d
Moments due to couples can be added using
the same rules as adding any vectors.
The moment of a couple is a free vector.
It can be moved anywhere on the body
and have the same external effect on the
body.

56. O
a
d
F
FA
B
C
MO= F (a+d) – F a = F d
MMOO==MMAA==MMBB==MMCC
Moment of a couple has the same value for allMoment of a couple has the same value for all
moment centers.moment centers.
Moment of a Couple ‘2D’

57. M
M
M
M
2D CCW couple
2D CW couple
Moment of a Couple ‘2D’

58. The moment of a couple is a free vector. It can be moved
anywhere on the body and have the same external effectexternal effect on
the body.
==
Moment of a Couple ‘2D’

59. Moment of a Couple ‘2D’
APPLICATIONSAPPLICATIONS

60. Moment of a Couple ‘2D’
APPLICATIONS
(continued)

61. Moment of a Couple ‘2D’
Scalar Formulation
Magnitude of couple moment
M = F.d
 Direction and sense areDirection and sense are
determined by right hand ruledetermined by right hand rule
In all cases, M acts perpendicular
to plane containing the forces.

62. Moment of a Couple ‘2D’
Vectorial Formulation
M = d X F
In all cases, M acts perpendicular
to plane containing the forces.

63. Moment of a Couple ‘2D’
Example:
A couple acts on the gear teeth. Replace it
by an equivalent couple having a pair of
forces that act through points A and B.
==

64. Moment of a Couple ‘2D’
Solution
 Magnitude of couple
M = 24 N.m
 Direction out of the page since
forces tend to rotate CCW
 M is a free vector and can
be placed anywhere.

65. Moment of a Couple ‘2D’
Solution
 To preserve CCW motion,
vertical forces acting through
points A and B must be directed
as shown
 For magnitude of each force,
M = F.d
24 N.m = F (0.2m)
F = 120.000 N

66. Moment of a Couple ‘2D’
Example:

67. Two different couples are equivalent if they
produce the same moment, (magnitude as
well as direction).
Equivalent Couples
Moment of a Couple ‘2D’

68. Two couples are equivalent if they produce the same moment
with magnitude and direction.
== ======
Moment of a Couple ‘2D’

69. Example:
Moment of a Couple ‘2D’

70. Due principle of
transmissibility, the
distance r that can
act at any point
along the line of
action of the Force
creates the same
moment about point
A.
MA
FrFrFrMA

×=×=×= 321
Moment of a Couple ‘3D’

71. • select one of the couple forces to start with,
• along the line of action of this selected
force, select any point over it,
• determine a point also along the line of
action of the other force,
• get the position vector between those two
selected points,
• using cross-product obtain the couple
moment, since the principle of
transmissibility applies M = rXF.
Moment of a Couple ‘3D’

72. Example:
For the given force-couple system:
1) Determine the resultant couple moment in Cartesian form?
2) Express the magnitude of the resultant couple moment?
3) Find the direction cosines of this resultant couple moment?
Moment of a Couple ‘3D’