Utilizamos seu perfil e dados de atividades no LinkedIn para personalizar e exibir anúncios mais relevantes. Altere suas preferências de anúncios quando desejar.
Próximos SlideShares
Carregando em…5
×

# Lecture 3(95)

326 visualizações

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.

• Full Name
Comment goes here.

Are you sure you want to Yes No
• Seja o primeiro a comentar

### Lecture 3(95)

1. 1. THREE DIMENSIONAL SYSTEMS ‘3D - CARTESIAN VECTORS’
2. 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. 3. x z y x y zx y z Cartesian (x, y, z) Coordinate System Right-handed coordinate system Various Orientations
4. 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. 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. 6. Example: Obtain the coordinates of the points A, B and C . 3-D Cartesian Vectors
7. 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. 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
9. 9. 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.
10. 10. 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.
11. 11. 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
12. 12. 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
13. 13. • 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
14. 14. 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
15. 15. 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
16. 16. 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
17. 17. 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 >
18. 18. 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. 20. Finding the length ‘rAB’
21. 21. 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
22. 22. 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
23. 23. 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˃
24. 24. The length ‘rAB’ • A = < xA , yA , zA > • B = < xB , yB , zB > • rAB = < xB-xA i , yB-yA j , zB-zA k >
25. 25. 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
26. 26. The length ‘rAB’ Coordinates B (-2, 1.5, 3) A (1, 0, -3) 1.5 m
27. 27. 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
28. 28. 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
29. 29. 1. By COORDINATES (with the help of ) WAYS OF FINDING THE COMPONENTS OF FORCES IN 3D λ
30. 30. 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
31. 31. 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
32. 32. 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’:
33. 33. 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   
34. 34. 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.
35. 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. FAB
36. 36. 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
37. 37. Example: Obtain the components of the cable that carries tension TAB of magnitude 62 kN. TAB
38. 38. 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 
39. 39. 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
40. 40. 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
41. 41. 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
42. 42. 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°
43. 43. • 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
44. 44. • 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
45. 45. • 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
46. 46. Components in Three Dimensions • Direction Cosines: λx i λ λy j 3-D Cartesian Vectors Directions
47. 47. • 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
48. 48. 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
49. 49. • 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
50. 50. 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
51. 51. Example: Determine the components of the force F=200. Direction Cosines of Cartesian Vectors
52. 52. 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
53. 53. 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
54. 54. 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
55. 55. 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
56. 56. 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
57. 57. 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
58. 58. 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
59. 59. 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
60. 60. 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
61. 61. 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
62. 62. β=36.7073 γ=74.5023 α=122.3100 Solution Determine the Direction Cosines of Cartesian Vectors y z x
63. 63. Example: For the given elastic rubber band AB, determine its direction cosines. Direction Cosines of Cartesian Vectors
64. 64. Coordinates B (-2, 2, 3) A (1, 0, -3) Direction Cosines of Cartesian Vectors
65. 65. 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
66. 66. 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
67. 67. 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
68. 68. • 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 
69. 69. Example: 37° X Y AX AY AZ Z 22° Determine the components of the vector A= 5 units.
70. 70. 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
71. 71. 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
72. 72. 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.
73. 73. Addition and Subtraction of Cartesian Vectors Example: Determine the magnitude and coordinate direction angles of resultant force acting on the ring.
74. 74. 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
75. 75. 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
76. 76. Addition and Subtraction of Cartesian Vectors α=74.8290° β=102.0872° γ =19.5756°
77. 77. 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
78. 78. 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
79. 79. 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
80. 80. 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
81. 81. Example: Direction Cosines of Cartesian Vectors 49.2°
82. 82. 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
83. 83. 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
84. 84. 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
85. 85. 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).
86. 86. 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)
87. 87. 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
88. 88. 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 
89. 89. 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
90. 90. 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)
91. 91. Chapter Review
92. 92. Chapter Review
93. 93. Chapter Review
94. 94. Chapter Review
95. 95. Chapter Review