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Mechanics and Vector Operations - Civil Engineering Lecture

This lecture covers the basics of mechanics, including deformable bodies, rigid bodies, and fluid dynamics, as well as vector operations. Learn about vector addition, subtraction, multiplication, and resolution, and their applications in civil engineering.

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Mechanics and Vector Operations - Civil Engineering Lecture

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  1. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 1 1 1 PHYSICS MECHANICS DEFORMABLE BODIES RIGID BODIES FLUIDS DYNAMICS (ME 201) STATICS (CE 201) • STATICS (CE 201) • DYNAMICS (ME 201) • STRUCTURAL • MECHANICS (CE 203) • STRUCTURAL • ANALYSIS (CE 305) • CONCRETE • DESIGN (CE 315) • STEEL • DESIGN (CE 408) • CONCRETE II (CE 415) • STEEL II (CE 418 )

  2. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering   LECTURE 1 2 2 PARTICLES & RIGID BODIES: BASIC QUANTITIES: 1 - LENGTH : l, (cm, mm, m, in.) 2 - MASS : m, (kg, lb.) 3 - FORCE : f, (N, kN, lbf.) PARTICLES : DIMENSIONLESS RIGID BODIES : HAVE DIMENSIONS SCALERS & VECTORS: VECTOR DEFINITION: 1 - MAGNITUDE 2 - DIRECTION 3 - POINT OF APPLICATION SCALER : MAGNITUDE ONLY (e.g. length) VECTOR : MAGNITUDE & DIRECTION (e.g. Force, velocity). 20 kN force • A FORCE IS FULLY REPRESENTED BY A VECTOR. • A VECTOR IS FULLY REPRESENTED BY AN ARROW.

  3. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 1 3 3 VECTOR OPERATIONS 1 - Multiplication by a Scaler: a ( A ) = aA 2 - Vector Addition: A + B = R a - ( Parallelogram law ): b - ( Triangular law ):  Review “Vector Algebra”  A 2A   3 - Vector Subtraction: A - B = C 4 - Resolution of a Vector: R = A + B  A         R R a  b  B A  B   A B     R

  4. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 2 1 4  b C A A a  c B B VECTOR ADDITION OF FORCES Ex. TRIANGLE LAWS: R = 2 + B - 2  B cos a 2 I - SINE LAW: B R =  sin  sin a  A sin a B C sin c B sin b = =   B sin a  = sin-1 R II - COSINE LAW:   A a C = A2 + B2 - 2AB cos c   R

  5. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 2 2 5 EXAMPLE Determine the magnitude and direction of force P such that the resultant of the two tug boats ( P & T ) is equal to 4.00 kN. SOLUTION: Using Cosine Law:  42 + (2.6)2 - 2 x 4 x 2.6 cos 20o IPI =   P IPI = 1.8 kN 4.0kN  Using Sine Law: 20o  1.8 2.6 P  = T = 2.6 kN 20o  = 30o sin  sin 20 2.6kN  20o 4.0kN

  6. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering    F1 = F1x + F1y F2 = F2x + F2y … = … + … Fn = Fnx + Fny   Fx = F cos  Fy = F sin    y       F F F y x     x    |R| = |Rx|2 + |Ry|2 LECTURE 3 1 6 ADDITION OF 2-D VECTORS    R =  Fi =  Fix +  Fiy     i-1 i-1 i-1  Rx =  Fix   , = F1x + F2x + F3x i=1  Ry =  Fiy   , = F1y + F2y + F3y y i=1  F2y   |Ry| F2  F1y  F1 , tan  =  |Rx|   F2x x F1x  F3x RULE:  Ry  R   F3   Rx F3y

  7. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering    à F = Fz + F' A = |Ã|    F' = Fx + Fy à = |Ã| A      F = Fx + Fy + Fz  Unit vector on the x-direction = i Unit vector on the x-direction = j Unit vector on the x-direction = k    |F| = |Fz|2 + |F'|2      z |F'| = |Fz|2 + |Fy|2  k  j y      |F| = |Fx|2 + |Fy|2 + |Fz|2  x i LECTURE 3 2 7 UNIT VECTOR CARTESIAN VECTORS à z  Fz  A F 1  Fy  Fx y  F' x

  8. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering    à = Axi + Ayj + Azk Ay Az Ax , , cos  = cos  = cos  = |Ã| |Ã| |Ã| LECTURE 3 3 8 DIRECTION OF CARTESIAN VECTOR à = Ãx + Ãy + Ãz The angle any vector makes with +ive x-axis =  The angle any vector makes with +ive y-axis =  The angle any vector makes with +ive z-axis =  |Ã| = à z Ãz à Ax Ay Az à   A  = — = — i + — j + — k |Ã| |A| |Ã| |Ã|    k     = cos i + cos j + cos k y Ãy cos2 + cos2  + cos2  = 1  x j  i    à = |A| cos i + |A| cos j + |A| cos k Ãx

  9. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering Ry Rx 20 30 cos   — = —   = 73.4o cos   — = —  = 64.6o 70 70 |R| |R| LECTURE 4 1 9 Solution: Example 1: n R =  Fi = F1 + F2 + F3      Determine magnitude and direction cosine of resultant (R) of the following force vectors: F1 = 5i + 15 j + 30 k (N) F2 = 25i + 30 j + 40 k (N) F3 = - 25j - 50 k (N) i=1      R = 30 i + 20 j - 60 k      |R| = = 70 (N)        Check cos2 + cos2 + cos2 = 1 0.18 + 0.08 + 0.73 = 1 Rz -60 cos   — = —   = 149.0o 70 |R| OK

  10. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering z F2 = 250N 5 3 y 4 60o 60o 30o 45o x F1 = 350N LECTURE 4 2 10 SOLUTION: Example 2: Express the following two vectors in cartesian form. Find their resultant. F1 = |F1| cos.  i + |F| cos.  j + |F|cos. k = 350 cos. 60i + 350 cos. 60j + 350 x cos. 135k = 175i + 175j - 247.5k F2 = |F2 | cos. 30 i - |F2| sin 30 j + |FzZ| k = 200 x 0.87 i - 200 x 0.5 j + 150 k = 174 i - 100 j + 150 k R = F1 + F2 = 350 i + 75 j - 97.5 k                  1 1            

  11. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering ~ r A LECTURE 5 1 11 POSITION VECTOR z (xB, yB, zB) B (xA, yA, zA) A ~ r y B O x

  12. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering SOLUTION rAB     (m) rA = 3i + 4j - 2k rB = i - 2j + 1k rAB = rB - rA = -2i - 6j + 3k   |rAB|    (m)       (m)  uAB = rAB   |rAB|  |rAB| = 22 + 62 + 32 = 7 (m) -2 6 3     uAB = — i - — j + — k 7 7 7       (N) F = |F| uAB = - 20i - 60j + 30k LECTURE 5 2 12 FORCE ALONG A LINE  F B F = |F| uF  rAB A uF = uAB =   uAB = uF EXAMPLE Represent the force F in Cartesian from its magnitude = 70N and is directed from point A(3,4, -2)m towards B(1, -2, 1)m.

  13. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering ~ ~ ~ ~ ~ ~ i · i = 1 j · j = 1 k · k = 1 i · k = 0 i · j = 0 j · k = 0 ~ ~ ~ ~ ~ ~ LECTURE 6 1 13 DOT PRODUCT ~ A Definition: ~ ~ ~ ~  A · B = |A| |B| cos  ~ B Applications: Features: ~ ~ ~ ~ A · B = B · A A · (B + C) = A · B + A · C a(A·B) = (aA) ·B = A ·(aB) = (A ·B)a 1 - Angle between two vectors: 2 - Component of a Vector Along a Line: ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ | | | |

  14. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering ~ ~ F1·F2 -28 cos  = ——— = ——— = -0.62 ~ ~ |F1||F2| 7.076.4 LECTURE 6 2 14 EXAMPLE: ~ ~ Determine the angle () between the force vectors F1 & F2. ~ ~ ~ ~ F1 = 3i + 5j - 4k F2 = 2i - 6j + 1k F1·F2 = 32 - 56 - 41 = -28 (N) ~ ~ ~ ~ (N) ~ ~ |F1| = 32 + 52 + 42 = 7.07 |F2| = 22 + 62 + 12 = 6.40 (N) (N)  = cos-1 - 0.62 = 128.23o

  15. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering ˜ ˜ F = 0 Fz = 0 Fx = 0 Fy = 0 LECTURE 7 1 15 EQUILIBRIUM OF A PARTICLE ˜ F2 ˜ F1 • ˜ Fi ˜ Fn Springs: Pulleys: F = k Fx = 0 T cos  = T cos   Fy = 0 2T sin  = W T T k = Spring Coefficient = FORCE / LENGTH    W  = Displacement

  16. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 8 1 17 A T T 2m C B   W 10m A G  2m C   10 15 CE BE E  B 15m D CE tan  4.6 1.12 FIG.: 2 EXAMPLE: A soldier wanted to cross a river 10m wide with a 15m rope and a smooth pulley. Will he make it ? From Fig. : 2  BCE   BED  BC = BD GD = (15)2 - (10)2 = 11.2m  CE = ED = ———— = 4.6m  = cos-1 —— = 48.2o / tan  = ———  BE = —— = —— = 4.11m (11.2 -2) 2 short of crossing

  17. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering EXAMPLE: Find (  ) for equilibrium ?  Fy = 0 + 4 sin + 4 sin - 4 = 0  sin =  = 30o  LECTURE 7 2 16 SOLUTION: From Geometry tan  =   = 1 tan 30 = 0.58m From Equilibrium,  Fx = 0 4 cos  = 4 cos     2.0m y 4N 4N 1m   30 x  4N 4kN 4kN 4kN

  18. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 8 2 18 Problems: The gusset plate P is subjected to the forces of three members as shown. Determine the force in member C and its proper orientation  for equilibrium. The forces are concurrent at point O. y 9 kN 8 kN  A SOLUTION: B 3 5 4  x P O  Solve equation  &  for T &  T =  = C T

  19. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering FAB = |FAB| j FAC = |FAC| i rAD = -2i -4j +4k |rAD| = 22 + 42 + 42 = 6m UAD = —— rAD |rAD| 4 6 -2 6 4 6 UAD = — i - — j + — k FAD = |FAD| UAD LECTURE 10 2 EXAMPLE: (Prob. 3·31) 19 z (0, 0, 4)m D (2, 4, 0)m  y  B A C   x 800N     For Equilibrium FA = 0  Fx = 0  |FAC| - |FAD| = 0 ……….  Fy = 0  |FAB| - |FAD| = 0 ……….  Fz = 0  |FAD| - 800 = 0 ..………. From EQU  |FAD| = = 1200N From EQU  |FAC| = x 1200 = 400N From EQU  |FAB| = x 1200 = 800N    2 6  4 6 4 6     800 x 6 4   2 6 4 6 Q ·E·D·

  20. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering  C  A  B 1 - Magnitude: |C| = |A| |B| sin  2 - Direction: Right Hand Rule    LECTURE 11 1 20 CROSS PRODUCT Definition: Properties:     1 - A × B = - ( B × A ) 2 - A × ( B + D ) = A × B + A × D         i × i = o i × j = k i × k = -j j × i = -k j × j = o j × k = i k × i = j k × j = -I k × k = o i j + k

  21. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering Alternatively: LECTURE 11 2 21 + + +

  22. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 11 3 22 MOMENT OF A FORCE  MoF = Moment of force (F) about point (o). r = Position vector from (O) to any point on the line of action of F  O Proof d  O 1 - Magnitude: 2 - Direction: Right Hand Rule  r  O

  23. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering    F F R LECTURE 12 1 23 Principle if Transmissibility: A force can be transmitted along its line of action! Moment of Resultant    MoR = r × R = r × (F1 + F2) = r × F1 + r × F2 MoR = MoF + MoF         F1    2  1  r F2 Moment of Resultant is equal to the sum of moments of its components. O   n Mo =  Mo R Fi i = 1

  24. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering + + 3 - q = = 1 o sin 36 . 8 , 5 6 b = = o sin 1 63 . 4 , - 6 . 32 a = q + - b = o 90 63 . 4 \ = = m o d 6 . 72 sin 63 . 4 6 . 00 + × = = ´ = N m M F d 50 6 300  50N 30N 40N A 3m 6m O LECTURE 12 2 24 EXAMPLE: 30N Find the moment of couple (F) about point (O). 3 4 SOLUTION ~ A r F OA 1 - Scalar Solution (A) 3m C  6m O  A d  6.72m 3m B O 6m 3 - Vector Solution (C) 2 - Scalar Solution (B)

  25. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering ~ ~ ~ ~ = - + - + - r ( 4 1 ) i ( 5 3 ) j ( 0 6 ) k AB ~ ~ ~ = + + 3 i 2 j 6 k ~ = + + = 2 2 2 m r 3 2 6 7 AB ~ ~ r 3 2 6 ~ ~ ~ = = + - AB u i j k AB r 7 7 7 AB ~ ~ ~ ~ ~ = = + - F F u 15 i 10 j 30 k AB ~ ~ ~ ~ = + + r i 3 j 6 k OA i j k i j 1 3 6 1 3 15 10 -30 15 10 ~ ~ ~ = - + - 150 i 120 j 35 k _ _ _ + + + LECTURE 12 3 25 EXAMPLE: Find the moment about the origin of force F (|F| = 25N), this force is acting along the direction A (1,3,6) to B (4,5,0).. SOLUTION OR i j k 4 5 0 15 10 -30

  26. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering  F LECTURE 13 1 26 a Moment of a Force about an Axis:  Maa    MoF = r × F |Maa| = uaa • MoF Maa| = uaa • (r × F)    r     O a This is called Triple Product. uaa = uaxi + uayj + uazk r = rxi + ryj + rzk F = Fxi + Fyj + Fzk Cross product is executed first followed by Dot product, or:         uax uay uaz rx ry rz Fx Fy Fz    

  27. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering ~ ~ ~ ~ = - - + F 20 i 30 j 25 k z C B(-3,2,4) rBC (1,5,2) A y (7,-3,1) x LECTURE 13 2 27 EXAMPLE: Find the force F about axis AB. = - 64.5 + 17.2 - 31.2 + 15.6 + 51.6 - 43.0 = -54.3 N·m +0.43 3 -30 -0.86 4 -20 +0.26 -2 25 -0.86 4 -20 0.43 3 -30 =

  28. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering     -F -F F F d   rA + r = rB  rB - rA = r     LECTURE 15 1 28 MOMENT OF A COUPLE: “COUPLE” is defined as two parallel forces, equal in magnitude, and opposite in direction.  r      Moc = rA × (-F) + rB × F = (rB - rA) × F Moc = r × F  rA  rB   O    Moment of a couple about any point in the space is a constant quantity and is independent of the location of the point. Scaler Solution

  29. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering ~ ~ ~ ~ c c c c M M M M i i z z   |M| = |F| • d = Constant LECTURE 15 1 29 RESULTANT COUPLE: FREE VECTOR: A vector with no fixed point of application. e.g. Mc  EQUIVALENT COUPLES:  Mc    Mc MR Mc 40N 60N 3m 2m 20N 6m 40N 60N 20N

  30. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering = - 480 240 × = N m 240 LECTURE 15 2 30 EXAMPLE: 50N Find the moment of couple (F) about point (O). 4 3 SOLUTION O • D A) Scalar Solution: 6m 6m 5 4 50N 3  B A C) Vector Solution: 6m 40N B) Scalar Solution: 30N A O å B = ´ + - ´ + M 40 12 0 40 6 0 o 30N A B 40N

  31. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering       F F F F -F F A B LECTURE 16 1 31 MOVEMENT OF FORCE: A) TRANSLATION OF FORCE ALONG ITS LINE OF ACTION: B) TRANSLATION OF FORCE OUTSIDE ITS LINE OF ACTION: When F is moved from (A) to (B) it becomes: F + Mc where Mc = rBA × F M is “free” vector B A  Mc   B    

  32. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering O  Mi    F1 F2 Fn  M  O Mn  M      Fn F1 F2 FR MR O LECTURE 16 2 32 RESULTANT OF A FORCE SYSTEM: n  FR =  Fi  i-1 MR =  Mi   n i-1 where    Mi = roi× Fi

  33. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering • • LECTURE 16 3 33 EXAMPLE: Reduce the force system into a single force & single couple at (O). SOLUTION    FR = -25 i - 15 jN 35N 50N   4 MR = 330 kN·m y 3 20N O x A B C 25N 3m 5m 3m 70N 30° 330 N • m 15N

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