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Lecture 4 (27)

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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.

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Lecture 4 (27)

  1. 1. Multiplication of 2 Vectors by: BAC   (read as C equals A cross B)  CROSS PRODUCT  DOT PRODUCT Dot product of vectors A and B is written as: BAC  . (read as C equals A dot B) Cross product of vectors A and B is written as:
  2. 2. DOT PRODUCT Dot product of vectors A and B is written as: BAC  . (read as C equals A dot B)
  3. 3. Dot Product • Laws of Operation 1. Commutative law A·B = B·A 2. Multiplication by a scalar a(A·B) = (aA)·B = A·(aB) = (A·B)a 3. Distribution law A·(B + D) = (A·B) + (A·D)
  4. 4. Angle between 2-Vectors ‘Dot Product of Vectors’ ‘Scalar Product of Vectors’ • First define the magnitudes of A and B • The angle between their tails is θ A·B = |A||B|cosθ where 0° ≤ θ ≤ 180° • The result is a scalar
  5. 5. • Cartesian Vector Formulation - Dot product of Cartesian unit vectors Eg: i·i = (1)(1)cos 0° = 1 and i·j = (1)(1)cos 90° = 0 - Similarly i·i = 1 j·j = 1 k·k = 1 i·j = 0 i·k = 0 j·k = 0
  6. 6. • Cartesian Vector Formulation - Dot product of 2 vectors A and B A·B = (Axi + Ayj + Azk)· (Bxi + Byj + Bzk) = AxBx(i·i) + AxBy(i·j) + AxBz(i·k) + AyBx(j·i) + AyBy(j·j) + AyBz(j·k) + AzBx(k·i) + AzBy(k·j) + AzBz(k·k) = AxBx + AyBy + AzBz Note: since the result is a scalar, be careful of not including any unit vectors in your result.
  7. 7. The angle formed between: two vectors or intersecting lines are: θ = cos-1 [(A·B)/(|A||B|)] 0°≤ θ ≤180° Note 1: if A·B = 0, cos-1 0 = 90°, so A is perpendicular to B Note 2: if A·B = 1, cos-1 1 = 0°, so A is parallel to B
  8. 8. U V Example: The components of two vectors U and V are: U= 6i – 5j – 3k and V= 4i + 2j + 2k. (a) What is the value of U.V (b) What is the angle between U and V when they placed tail to tail.
  9. 9. x x y y z z 2 2 2 2 2 2 -1 U.V U V U V U V (6)(4) ( 5)(2) ( 3)(2) 8 U.V U.V U V cos so cos = U V 8 cos 0.1952 (6) ( 5) ( 3 ). (4) (2) (2 ) =cos 0.1952 78.7436                         r r r r r r
  10. 10. Multiplication of 2 Vectors by Cross Product BAC   read as C equals A cross B
  11. 11. Direction: Right Hand Rule λC Magnitude: C ABsin 
  12. 12. Cross Product • Laws of Operation 1. Commutative law not applicable A X B = B X A 2. Multiplication by a scalar a(AxB) = (aA)xB = Ax(aB) = (AxB)a 3. Distribution law Ax(B + D) = (AxB) + (AxD)
  13. 13. Cross Product of 2 Vectors Not Commutative. ABBA   )AB(BA  
  14. 14. Right Hand Rule
  15. 15. • Cartesian Vector Formulation - Cross product of Cartesian unit vectors Eg: i x i = (1)(1) sin 0° = 0 and i x j = (1)(1) sin 90° = 1 Hence; i x i = 0 j x j = 0 k x k = 0 i x j = k i x k = -j j x k = i j x i = -k k x i = j k x j = -i i k j +
  16. 16. Unit Vectors Cross Product o 90 sin 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆi i 0 i j k i k j ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆj i k j j 0 j k i ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆk i j k j i k k 0                          
  17. 17. Consider cross product of vector A and B A X B = (Ax i + Ay j + Az k) X (Bx i + By j + Bz k) = AxBx (i X i) + AxBy (i X j) + AxBz (i X k) + AyBx (j X i) + AyBy (j X j) + AyBz (j X k) + AzBx (k X i) + AzBy (k X j) + AzBz (k X k) = + (AyBz – AzBy) i – (AxBz - AzBx) j + (AxBy – AyBx) k
  18. 18. zyx zyx BBB AAA kji BA   X In determinant form, Sarrus’ Rule
  19. 19. Cross Product Of even more utility, the cross product can be written as Each component can be determined using 2  2 determinants.
  20. 20. zyx zyx zyx zyx BBB AAA kji BBB AAA kji BA   X In determinant form: + + +- - - = + (AyBz – AzBy) i + (AzBx -AxBz) j + (AxBy – AyBx) k Sarrus’ Rule
  21. 21. U V Example: The components of two vectors U and V are: U= 6i – 5j – 3k and V= 4i + 2j + 2k. (a) Determine the parallelface area formed by these vectors? i.e find [U x V]? (b) determine the smallest angle formed between these vectors?
  22. 22. 224 356 kji    VXU Determinant form: (a) = (-10+6) i + (-12-12) j + (12+20) k = - 4 i -24 j + 32 k = 1616 = 40.200 𝑢𝑛𝑖𝑡2 U = 70 V = 24 θ = sin-1 [(A X B)/(|A||B|)] θ = sin-1 [( 1616)/( 70 24)]=78.7448° (b)
  23. 23. Example: A=(4,0,-1), B=(1,1,2) and C=(-2,1,2) are the corners of the tringle, determine its area? z y x A C B
  24. 24. AB = (1-4) i + (1-0) j + (2-(-1)) k = -3i+1j+3k A=(4,0,-1), B=(1,1,2), C=(-2,1,2) AC = (-2-4) i + (1-0) j + (2-(-1)) k = -6i+1j+3k AB X AC = area of the rectangle So (AB X AC) / 2 area of the triangle 316 313 kji    ABXAC = (3-3) i + (9-18) j + (-3+6) k = -9 j + 3 k = 90 = 9.487 𝑢𝑛𝑖𝑡2 Area of the triangle = 9.487/2= 4.744 𝑢𝑛𝑖𝑡2
  25. 25. Example: Calculate the volume of the parallel surface bounded by lines OP, OQ and OT, where O= (0,0,0), P= (4,0,-1), Q= (1,1,2) and T= (-2,1,2)? Note: To determine the volume; use the mixed triple vector product approach [(OP X OQ ) . OT]. O P Q T
  26. 26. OP = 4 i -1 k O=(0,0,0) P=(4,0,-1), Q=(1,1,2), T=(-2,1,2) OT = -2 i + 1 j + 2 k OP X OQ = area of the rectangle 211 104 kji   OPXOQ = (0-(-1) i + (-8-1) j + (4-0) k = 1i -9 j + 4 k 𝑢𝑛𝑖𝑡2 OQ = 1 i +1j + 2 k So (OP X OQ ) . OT= (1i -9 j + 4 k) . (-2i +1j+2k)= -2-9 +8 =3 𝑢𝑛𝑖𝑡3
  27. 27. Chapter Summary • Dot product between two vectors A and B (vectors expressed as Cartesian form) A · B = AxBx + AyBy + AzBz • Angle between the tails of two vectors θ = cos-1 [(A·B)/(IAI IBI)] • The projected component of A onto an axis defined by its unit vector λ A = A cos θ = A·λ Dot Product • Dot product between two vectors A and B A·B = AB cos θ

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