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. THREE DIMENSIONAL SYSTEMS
‘3D - CARTESIAN VECTORS’

2. - Cartesian (x, y, z) Coordinate System
- Right-handed coordinate system
- Positive z axis points upwards, this axis helps
us in measuring the height of an object or
finding the altitude of any point.
3-D Cartesian Vectors

3. x
z
y
x y
zx
y
z
Cartesian (x, y, z) Coordinate System
Right-handed coordinate system
Various Orientations

4. 3-D Cartesian Vectors
[Right-Handed Coordinate System]
A rectangular coordinate system is said to be
right-handed provided:
Thumb of right hand points
in the direction of the positive
z axis. When the right-hand
fingers are curled about this
axis, the fingers direct from the
positive x axis towards the positive y axis

5. ESTABLISHING VECTOR IN 3D
• READING THE COORDINATES WITH RESPECT TO
ANY REFERENCE
(THE ORIGIN IN GENERAL)
• A = < xA , yA , zA >
• B = < xB , yB , zB >

6. Example:
Obtain the coordinates of the points A, B and C .
3-D Cartesian Vectors

7. Example:
Obtain the coordinates of the points A, B and C .
Always give in the form of point name (x, y, z)
B (0, 2, 0)
C (6, -1, 4)
A (4, 2, -6)
3-D Cartesian Vectors

8. 3-D Cartesian Vectors
Rectangular Components of a Vector
A vector A may have one, two or three
rectangular components along x, y and z
axes, depending on its orientation.
By two successive application of the
parallelogram law
A = A’ + Az
A’ = Ax + Ay
Combining the equations, A can be
expressed as:
A = Ax + Ay + Az

10. 3-D Cartesian Vectors
• Cartesian Vector Representations
Three components of A act in
the positive i, j and k directions
A = Ax i + Ay j + AZ k
*Note the magnitude and
direction of each components
are separated, for easing the
vectors algebraic operations.

11. 3-D Cartesian Vectors
Cartesian Unit Vectors
Cartesian unit vectors, i, j and k are used to designate
the directions of x, y and z axes.
Sense (or arrowhead) of these vectors
are described by:
plus (+) or minus (-) sign,
depending on pointing towards
the positive or negative axes.

12. 3-D Cartesian Vectors
• Unit Vector
- Direction of A can be specified using a unit vector.
- Unit vector has a magnitude of 1.
- If A is a vector having a magnitude of A ≠ 0,
unit vector having the same direction as A is
expressed by uA = A / A
So that: A = A uA
λA
λA
λA

13. 3-D Cartesian Vectors
• Unit Vector
Magnitude A has the same sets of units,
hence the unit vector λA is dimensionless.
A (a positive scalar) defines magnitude of A
where λA defines: the direction and sense. of A
λA

14. • Right-handed coordinate system:
Express vector λ in terms of vector components λx, λy
& λz parallel to the x, y & z axes respectively:
λ = λx + λy + λz
λz
λy
λx
λ

15. Position vector rOP is defined as:
a fixed vector which locates
a point in space relative to
another point.
In this case the origin O.
Position Vectors

16. rOP extends from the origin, ‘O (xo, yo, zo)’ which is
O (0, 0, 0)’ to point ‘P (xp, yp, zp)’
then, in Cartesian vector form
rOP = (xp – x0 )i + (yp – y0)j + (zp - z0 )k
rOP = xp i + yp j + zp k
Position Vectors
P= xp, yp, zp

17. Note the tail to head vector addition of the
three components
Start at origin O, one travels
x in the +i direction,
y in the +j direction and
z in the +k direction,
arriving at point P (x, y, z)
Position Vectors

18. ESTABLISHING POSITION VECTOR IN 3D
between two interested points
READING THE COORDINATES
WITH RESPECT TO ANY REFERENCE
(THE ORIGIN IN GENERAL)
A = < xA , yA , zA >
B = < xB , yB , zB >

19. Finding POSITION VECTORE ‘r??’:
• Establish an arrow between the two interested points
• Give letter names to the tail and the head of this arrow
• The name of the line is: r??
• Obtain the coordinates of these two ends by
getting the difference of the respective
coordinates from head to tail.
letter name of the head
letter name of the tail
ESTABLISHING POSITION VECTOR IN 3D

20. OBTAINING THE DIRECTION VECTOR COMPONENTS
[rletter name of tail; letter name of head (xHead-Tail, yHead-Tail, zHead-Tail)]
While reading use: tail head
˂arrow head coordinate – arrow tail coordinate ˃
ESTABLISHING VECTOR IN 3D

21. Finding the length ‘rAB’

22. i, j, k components of the positive vector r
may be formed by taking the coordinates of
the tail, A (xA, yA, zA) and subtract them
from the head B (xB, yB, zB). i.e other than origin.
Note that:
the tail to head vector
addition of the three
components
Position Vectors

23. Position Vectors
- Position vector maybe directed from point A to point B
- Designated by rAB
Vector addition gives
rA + rAB = rB
Solving
rAB = rB – rA = (xB – xA)i + (yB – yA)j + (zB –zA)k
or rAB = (xB – xA)i + (yB – yA)j + (zB –zA)k

24. Finding the length ‘r’:
• Read the coordinates of the two ends of the line
• Get the difference of the coordinates from
the tail of the force coordinate to the point
of interest.
OBTAIN THE DIRECTION VECTOR COMPONENTS
[r(xHead-Tail, yHead-Tail, zHead-Tail)]
While reading use: tail head
˂arrow head coordinate– arrow tail coordinate˃

25. The length ‘rAB’
• A = < xA , yA , zA >
• B = < xB , yB , zB >
• rAB = < xB-xA i , yB-yA j , zB-zA k >

26. The length ‘rAB’
Example:
An elastic rubber band is
attached to points A and B.
Determine its length
from A towards B.
1.5 m

27. The length ‘rAB’
Coordinates B (-2, 1.5, 3)
A (1, 0, -3)
1.5 m

28. Solution
Position vector
r = [-2m – 1m]i + [1.5m – 0]j + [3m – (-3m)]k
= {-3.000 i + 1.500 j + 6.000 k} m
Magnitude = length of the rubber band
Unit vector in the director of r
u = r /r
= -3.000/6.874 i + 1.500/6.874 j + 6.000/6.874 k
      m874.6000.6500.1000.3r
222

The length ‘rAB’
λ
1.5 j m

29. 1. By COORDINATES
(with the help of )
2. By ANGLES given
 from any AXIS
(cosine of this angle gives the component of
that axis α (θx), β (θy) and γ (θz))
 from PLANE
((θxy , θxz or θyz) use sine of the angles aswell)
WAYS OF FINDING THE
COMPONENTS OF FORCES IN 3D
λ

30. 1. By COORDINATES
(with the help of )
WAYS OF FINDING THE
COMPONENTS OF FORCES IN 3D
λ

31. Force Vector Directed along a Line
The Lambda ‘λ’ Vector
• Force F acting along the chain can be presented
as a Cartesian vector by establishing x, y, z axes
and forming a position vector r along length of
chain.
λ

32. Force Vector Directed along a Line
The Lambda ‘λ’ Vector
λ
Unit vector, λ = r/r that defines the direction of
both the chain and the force
• We get F = F λ

33. Similarly, obtaining the vectorial components
of the force ‘F’, the unit vector lambda ‘λ’
should be defined.
[In some books lambda is denoted by ‘u’].
F(vector) = F(scalar) . λ(vector)
. 

FF 
Finding the Lambda vector ‘λ’:

34. 222
)-()-()-(
ˆ-,ˆ-,ˆ-
ABABAB
ABABAB
AB
zzyyxx
kzzjyyixx



222
)-()-()-(
ˆ-,ˆ-,ˆ-
..
ABABAB
ABABAB
AB
zzyyxx
kzzjyyixx
FFF

 

222
)-()-()-( ABABAB
AB
AB
zzyyxx
r




35. Force Vector Directed along a Line
The Lambda ‘λ’ Vector
Example:
The man pulls on the cord
with a force of 350 N.
Represent this force as a
Cartesian vector components.

36. Force Vector Directed along a Line
The Lambda ‘λ’ Vector
Example:
The man pulls on the cord
with a force of 350 N.
Represent this force as a
Cartesian vector components.
FAB

37. Solution
End points of the cord are A (0m, 0m, 7.5m)
and B (3m, -2m, 1.5m)
rAB = (3m – 0m)i + (-2m – 0m)j + (1.5m - 7.5m)k
= {3.000 i - 2.000 j - 6.000 k} m
Magnitude = length of cord BA
Unit vector, λ = r /r
= 3.000/7.000i - 2.000/7.000j - 6.000/7.000k
      m000.72m000.62m000.22m000.3r 
Force Vector Directed along a Line
The Lambda ‘λ’ Vector
][ kˆ7/jˆ7/2iˆ7/3350ABF 6

38. Example:
Obtain the components of the cable that
carries tension TAB of magnitude 62 kN.
TAB

39. Example:
A : ˂0, 18, 30˃ B : ˂6, 13, 0˃ TAB = 62 kN
TAB
222AB
)()()6(
)kˆ),jˆ,iˆ6(
-30-5
30-(018-(130)-



31
kˆ,jˆ,iˆ6
AB
30-5-


][ kˆjˆiˆ62TAB
31
30
31
5
31
6

][ kˆ5926.14cosjˆ7182.80cosiˆ8401.78cos62TAB 

40. 2. By ANGLES given
 from any AXIS
(cosine of this angle gives the component of
that axis α (θx), β (θy) and γ (θz))
WAYS OF FINDING THE
COMPONENTS OF FORCES IN 3D

41. Direction Cosines of Cartesian Vectors
• Force, F that the tie down rope exerts on the ground
support at O is directed along the rope
• Angles α, β and γ can be solved with axes x, y and z
λ

42. λ
• Cosines of their values forms a unit vector u that acts
in the direction of the rope
• Force F has a magnitude of F
F = Fu = Fcosαi + Fcosβj + Fcosγk
λ
λ
Direction Cosines of Cartesian Vectors

43. 3-D Cartesian Vectors Directions
• Direction of a Cartesian Vector
i. All angles measured from COORDINATE AXES
Orientation of A is defined as the coordinate
direction angles α (θx), β (θy) and γ (θz)
measured between the tail of A
and the positive x, y and z axes.
0° ≤ α ≤ 180°
0° ≤ β ≤ 180°
0° ≤ γ ≤ 180°
0° ≤ γ ≤ 180°

44. • Direction of a Cartesian Vector
For angles α (θx), (blue colored triangle),
one can calculate the direction cosines of A
A
A
cos x
α
3-D Cartesian Vectors Directions

45. • Direction of a Cartesian Vector
For angle β (θy) (blue colored triangle), one can
calculate the direction cosines of A
A
Ay
cos
3-D Cartesian Vectors Directions

46. • Direction of a Cartesian Vector
For angle γ (θz) (blue colored triangles), one
can calculate the direction cosines of A
A
Az
cos
3-D Cartesian Vectors Directions

47. Components in Three Dimensions
• Direction Cosines:
λx i
λ
λy j
3-D Cartesian Vectors Directions

48. • Direction of a Cartesian Vector
Vector A expressed in Cartesian vector form:
A = A λA
= A cosα i + A cosβ j + A cosγ k
= Ax i + Ay j + AZ k
3-D Cartesian Vectors Directions

49. Direction of a Cartesian Vector
Angles α, β and  can be determined by the inverse
cosines:
A
Ay
cos
A
Az
cos
A
Ax
cos
3-D Cartesian Vectors Directions

50. • Direction of a Cartesian Vector
uA can also be expressed as
uA = cos α i + cos β j + cos γ k
Since and
magnitude of uA = 1,
222
zyx AAAA 
11coscoscos 2222
 γβα
λA
λA
λA=
3-D Cartesian Vectors Directions

51. 2)AzB(z2)AyB(y2)AxB(x
iAx-Bx


ˆ
cosxcos αθ
2
AB
2
AB
2
AB
AB
y
)zz()yy()xx(
jˆyy
coscos


 βθ
2
AB
2
AB
2
AB
AB
z
)zz()yy()xx(
kˆzz
coscos


-
γθ
Direction Cosines of Cartesian Vectors

52. Example:
Determine the components of the force F=200.
Direction Cosines of Cartesian Vectors

53. Since only 2 angles are given, the 3rd angle will be
determined by:
where there are two possibilities;
either
or
  
6515.1113690.0cos 1
 
α
   
  

3485.683690.0cos
3690.05736.07314.01cos
155cos43coscos
1coscoscos
1
22
222
222





α
α
α
γβα
Direction Cosines of Cartesian Vectors

54. By inspection we observed that α = 68.3485° since
component of Fx is at +x axis direction.
Hence F = 200 N
F = F cosα i + F cosβ j + F cosγ k
= (200cos68.3548°N)i + (200cos43°N)j +
(200cos55°N)k
= {73.760 i + 146.271 j + 114.715 k} N
Control:
      N200715.114271.146760.73
FFFF
222
2
z
2
y
2
x


Direction Cosines of Cartesian Vectors

55. Example:
The man pulls on the cord with
a force of 350 N. Determine the
direction cosines of this force.
Direction Cosines of Cartesian Vectors

56. Example:
The man pulls on the cord with
a force of 350 N. Determine the
direction cosines of this force. FAB
Direction Cosines of Cartesian Vectors

57. Solution
End points of the cord are A (0m, 0m, 7.5m)
and B (3m, -2m, 1.5m)
rAB = (3m – 0m)i + (-2m – 0m)j + (1.5m - 7.5m)k
= {3.000 i - 2.000 j - 6.000 k} m
Magnitude = length of cord BA
Unit vector, λ = r /r
= 3.000/7.000i - 2.000/7.000j - 6.000/7.000k
      m000.72m000.62m000.22m000.3r 
Direction Cosines of Cartesian Vectors

58. The magnitude of the force F is 350 N, so with λ unit vector
components
F = F λ
= 350 N (3.000/7.000 i - 2.000/7.000 j - 6.000/7.000 k)
= {150.000 i - 100.000 j - 300.000 k} N
Angles are:
α = cos-1(3/7) = 64.6231°
β = cos-1(-2/7) = 106.6016°
 = cos-1(-6/7) = 148.9973°
y
z
x
β=106.6016°
α=64.6231°
Direction Cosines of Cartesian Vectors

59. Example:
The coordinates of point C of the truss are:
xC = 4 m, yC = 0, zC = 0, and
the coordinates of point D are:
xD = 2 m, yD = 3 m, zD = 1 m.
What are the direction cosines
of the position vector rCD (from point C to point D)?
Direction Cosines of Cartesian Vectors

60. Knowing the coordinates of points C and D, one
can determine rCD in terms of its components.
Then the magnitude of rCD (the distance from C
to D) can be calculate using the direction
cosines.
Direction Cosines of Cartesian Vectors

61. The position vector rCD in terms of its components.
rCD = (xD  xC)i + (yD  yC)j + (zD  zC)k
= (2  4)i + (3  0)j + (1  0) k (m)
= 2i + 3j + k (m)
     
m3.742
2m12m32m2


 2
zCD
2
yCD
2
xCDCD rrrr
Direction Cosines of Cartesian Vectors

62. Solution
Determine the direction cosines.



74.5023
m3.742
m
cos
36.7073
m3.742
m
cos
122.3100
m3.742
m2
cos








γ
2672.0
1
cos
β
8017.0
3
cos
α
5345.0cos
CD
CD
z
CD
CD
y
CD
CD
x
z
y
x
r
r
r
r
r
r
γθ
βθ
αθ
Direction Cosines of Cartesian Vectors

63. β=36.7073
γ=74.5023
α=122.3100
Solution
Determine the
Direction Cosines of Cartesian Vectors
y
z
x

64. Example:
For the given elastic rubber
band AB, determine its
direction cosines.
Direction Cosines of Cartesian Vectors

65. Coordinates B (-2, 2, 3)
A (1, 0, -3)
Direction Cosines of Cartesian Vectors

66. Solution
Position vector
r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k
= {-3.000 i + 2.000 j + 6.000 k} m
Magnitude = length of the rubber band
Unit vector in the director of r
u = r /r
= -3.000/7.000i + 2.000/7.000j + 6.000/7.000k
      m000.7000.6000.2000.3r
222

λ
Direction Cosines of Cartesian Vectors

67. α = cos-1(-3.000/7.000) = 115.3769°
β = cos-1(2.000/7.000) = 73.3985°
 = cos-1(6.000/7.000) = 31.0027°
α=115.3769 β=73.3985
γ=31.0027
r = 7 m
y
z
x
Direction Cosines of Cartesian Vectors

68. 2. By ANGLES given
 from PLANE
((θxy , θxz or θyz) use sine of the angles aswell)
WAYS OF FINDING THE
COMPONENTS OF FORCES IN 3D

69. • Direction of a Cartesian Vector
ii. Angle measured from coordinate PLANE
3-D Cartesian Vectors Directions
γcosAzA 
γ
θxy
γsinA'A 
xysinsinAxysin'AyA θγθ 
xyssinAxys'AxA θγθ coco 

70. Example:
37°
X
Y
AX
AY
AZ
Z
22°
Determine the components of the vector A= 5 units.

71. Example:
636.422cos5yA 
1.49637  cos22sin5zA
.12737 1sin22sin5xA 
37°
X
Y
AX
AY
AZ
Z
22°
Determine the components of the vector A=5 units.
Be careful about x-y-z coordinates!
units
units
units

72. Example
Given: A = Ax i + Ay j + AZ k
and B = Bx i + By j + BZ k
Vector Addition
Resultant R = A + B
= (Ax + Bx) i + (Ay + By ) j + (AZ + BZ)k
Vector Subtraction
Resultant R = A - B
= (Ax - Bx) i + (Ay - By ) j + (AZ - BZ)k
Addition and Subtraction of Cartesian Vectors

73. Addition and Subtraction of Cartesian Vectors
Concurrent ‫متزامن‬ Force resultant is the vector sum of
all the
forces in the system
FR = ∑F = ∑Fx i + ∑Fy j + ∑Fz k
where:
∑Fx , ∑Fy and ∑Fz represent the algebraic sums
of the x, y and z or
i, j and k components of each force in
the system.

74. Addition and Subtraction of Cartesian Vectors
Example:
Determine the magnitude and coordinate
direction angles of resultant force acting on the
ring.

75. Solution
Resultant force
FR = ∑F
= F1 + F2
= {60j + 80k} kN
+ {50i - 100j + 100k} kN
= {50.000i -40.000j + 180.000k} kN
Magnitude of FR is found by
     
kN050.191050.191
000.180000.40000.50F 222
R


Addition and Subtraction of Cartesian Vectors

76. Solution
Unit vector acting in the direction of FR
λFR = FR /FR
= (50.000/191.050)i + (-40.000/191.050)j + (180.000/191.050)k
= 0.2617 i - 0.2094 j + 0.9422 k
So that
cosα = 0.2617 α=74.8290° α= 74.8283°
cos β = - 0.2094 β=102.0872° = 102.08500°
cosγ = 0.9422 γ =19.5756° γ = 19.5820°
Addition and Subtraction of Cartesian Vectors

77. Addition and Subtraction of Cartesian Vectors
α=74.8290°
β=102.0872°
γ =19.5756°

78. Example: (T)
The roof is supported by
cables. If the cables exert
FAB = 100 N and FAC = 120 N
on the wall hook at A,
determine the magnitude of
the resultant force acting at A.
Addition and Subtraction of Cartesian Vectors

79. Solution
rAB = (4m – 0m) i + (0m – 0m) j + (0m – 4m) k
= {4.000 i – 4.000 k} m
FAB = 100N (rAB /r AB)
= 100N {(4/5.657)i - (4/5.657)k}
= {70.711 i - 70.711 k} N
    m657.5m000.4m000.4r
22
AB

Addition and Subtraction of Cartesian Vectors

80. Solution
rAC = (4m – 0m) i + (2m – 0m) j + (0m – 4m) k
= {4.000 i + 2.000 j – 4.000 k} m
FAC = 120N (rAC/r AC)
= 120N {(4.000/6.000) i + (2.000/6.000) j –
(4.000/6.000) k}
= {80.000 i + 40.000 j – 80.000 k} N
      m000.6m000.4m000.2m000.4r
222
AC

Addition and Subtraction of Cartesian Vectors

81. Solution
FR = FAB + FAC
= {70.711 i - 70.711 k} N + {80.000 i + 40.000 j – 80.000 k} N
= {150.711 i + 40.000 j – 150.711 k} N
Magnitude of FR:
     
N859.216
711.150000.40711.150F
222
R


Addition and Subtraction of Cartesian Vectors

82. Example:
Direction Cosines of Cartesian Vectors
49.2°

83. Example:
Determine the magnitude and coordinate direction angles
of resultant force.
x
z
750 N
900 N
57.720
61.250
26.470
32.730
Y
Addition and Subtraction of Cartesian Vectors

84. Example: (T)
a) Determine the resultant force R of the given 3 forces (OA,
OB and OC),
b) Find the direction cosines (θx, θy, θz) of this resultant R.
Addition and Subtraction of Cartesian Vectors
Y
X
137 kN
426 kN
57.120
41.270
31.670
62.310
Z
209 kN
C: ˂ 13, 9, -5 >
A
B
C
O

85. Example:
Find the magnitude and direction of the resultant of
the two forces shown if P= 300 N and Q = 400 N.
Addition and Subtraction of Cartesian Vectors

86. Example: (T)
The end of the coaxial cable AE is attached to the pole AB, which is strengthened
by the guy wire AC and AD. Knowing that the tension in AC and is 150 N and that
the resultant of the forces exerted at A by wires AC and AD must be contained in
the xy- plane, determine
a) the tension in AD
b) the magnitude and direction of the resultant of these two forces (AC and AD).

87. 28°
54°
735 N
If the resultant force acting on the system is R = {800 j} N. Determine:
•the magnitude and
•the coordinate direction angles (α, β,γ) of the force F.
Example: (T)

88. Chapter Summary
Parallelogram Law
• Addition of two vectors
• Components form the side and resultant form
the diagonal of the parallelogram
• To obtain resultant, use tip to tail addition by
triangle rule
• To obtain magnitudes and directions, use Law
of Cosines and Law of Sines

89. Chapter Summary
Cartesian Vectors
• Vector F resolved into Cartesian vector form
F = Fxi + Fyj + Fzk
• Magnitude of F
• Coordinate direction angles α, β and γ are
determined by the formulation of the unit vector
in the direction of F
u = (Fx /F) i + (Fy /F) j + (Fz /F) k
222
zyx FFFF 

90. Chapter Summary
Cartesian Vectors
• Components of λ represent cosα, cosβ and cosγ
• These angles are related by
cos2α + cos2β + cos2γ = 1
Force and Position Vectors
• Position Vector is directed between 2 points
• Formulated by distance and direction moved along
the x, y and z axes from tail to tip

91. Chapter Summary
Force and Position Vectors
• For line of action through the two points, it acts
in the same direction of λ as the position vector
• Force expressed as a Cartesian vector
F = F λ = F (r/r)

92. Chapter Review

93. Chapter Review

94. Chapter Review

95. Chapter Review

96. Chapter Review