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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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