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ENGINEERING MECHANICS STATICS & DYNAMICS

University of Palestine College of Engineering & Urban Planning. ENGINEERING MECHANICS STATICS & DYNAMICS. Instructor: Eng. Eman Al.Swaity. Lecture 6. Chapter 4: Force resultant. Lecture 6. Chapter 4: Force resultant. Today’s Objectives : Students will be able to:

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ENGINEERING MECHANICS STATICS & DYNAMICS

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  1. University of Palestine College of Engineering & Urban Planning ENGINEERING MECHANICS STATICS & DYNAMICS Instructor:Eng. EmanAl.Swaity Lecture 6 Chapter 4:Force resultant

  2. Lecture 6 Chapter 4:Force resultant

  3. Today’s Objectives: Students will be able to: 1) Determine the moment of a couple . 2) Determine the effect of moving a force. 3) Find an equivalent force-couple system for a system of forces and couples. Lecture 6 Chapter 4:Force resultant

  4. APPLICATIONS What is the net effect of the two forces on the wheel? Lecture 6 Chapter 4:Force resultant

  5. APPLICATIONS What is the resultant effect on the person’s hand when the force is applied in four different ways ? Lecture 6 Chapter 4:Force resultant

  6. APPLICATIONS (continued) Several forces and a couple moment are acting on this vertical section of an I-beam. | | ?? Can you replace them with just one force and one couple moment at point O that will have the same external effect? If yes, how will you do that? Lecture 6 Chapter 4:Force resultant

  7. A coupleis defined as two parallel forces that have the same magnitude and have opposite directions, and are separated by a perpendicular distance. • Since the resultant force is zero, the only effect of a couple is to produce a rotation or tendency of rotation in a specified direction. • The moment produced by a couple is called a couple moment. • We can determine its value by finding the sum of the moments of both couple forces about any arbitrary point. Lecture 6 Chapter 4:Force resultant

  8. F1 d d A h F1 • Basic terms • The moment produced by two equal, opposite, and parallel forces is called a couple • Couple can cause rotation only • The couple moment equal the force times the perpendicular distance between the two forces M = Fh • Moment about A MA= (-)F1d + (-)F1d = -2F1.d • Couple moment M= (-)F1h = -2F1.d Lecture 6 Chapter 4:Force resultant

  9. Scalar Formulation. The moment of a couple, M, is defined as having a magnitude of • where F is the magnitude of one of the forces and d is the perpendicular • distance or moment arm between the forces. • The direction and sense of the couple moment are determined by the right-hand rule, where the thumb indicates the direction when the fingers are curled with the sense of rotation caused by the two forces. • In all cases, M acts perpendicular to the plane containing these forces. Vector Formulation. The moment of a couple can also be express (by the vector cross product) using: Lecture 6 Chapter 4:Force resultant

  10. 400N B A 400N 3m 2.5m • Example 1 • Fine the couple moment, moment at A & B The couple moment M = F.h M = (-)4002.5 = -1000 N.m Moment at A MA= (-)4001.25 + (-)4001.25 MA= -1000 N.m Moment at B MB= (-)4005.5 + 4003 MB= -1000 N.m • The couple moment at any point is constant Lecture 6 Chapter 4:Force resultant

  11. Equivalent couples • A couple is not affected if the forces act in a different but parallel plane • The unique effect of a couple is to produce a pure twist or rotation regardless of where the forces are located Lecture 6 Chapter 4:Force resultant

  12. A couple acts on the gear teeth as shown in Fig. 4-29a. Replace it by an equivalent couple having a pair of forces that act through points A and B. • M = Fd = 40(0.6) = 24 N. m • also • M = Fd • 24 N.m = F(0.2 m) • F= 120 N Lecture 6 Chapter 4:Force resultant Lecture 6 Chapter 4:Force resultant

  13. The couple moment can be determined about any point If point D is chosen It is easier to determine the moment about A or B Lecture 6 Chapter 4:Force resultant Lecture 6 Chapter 4:Force resultant

  14. Lecture 6 Chapter 4:Force resultant

  15. Lecture 6 Chapter 4:Force resultant

  16. = When a number of forces and couple moments are acting on a body, it is easier to understand their overall effect on the body if they are combined into a single force and couple moment having the same external effect The two force and couple systems are called equivalent systems since they have the same external effect on the body. Lecture 6 Chapter 4:Force resultant

  17. MOVING A FORCE ON ITS LINE OF ACTION Moving a force from A to O, when both points are on the vectors’ line of action, does not change the external effect. Hence, a force vector is called a sliding vector. (But the internal effect of the force on the body does depend on where the force is applied). Lecture 6 Chapter 4:Force resultant

  18. MOVING A FORCE OFF OF ITS LINE OF ACTION Moving a force from point A to O (as shown above) requires creating an additional couple moment. Since this new couple moment is a “free” vector, it can be applied at any point P on the body. Lecture 6 Chapter 4:Force resultant

  19. When several forces and couple moments act on a body, you can move each force and its associated couple moment to a common point O. Now you can add all the forces and couple moments together and find one resultant force-couple moment pair. Lecture 6 Chapter 4:Force resultant

  20. RESULTANT OF A FORCE AND COUPLE SYSTEM (continued) If the force system lies in the x-y plane (the 2-D case), then the reduced equivalent system can be obtained using the following three scalar equations. Lecture 6 Chapter 4:Force resultant

  21. Lecture 6 Chapter 4:Force resultant

  22. Lecture 6 Chapter 4:Force resultant

  23. Example • Replace the force by an equivalent force & couple at O Technique I M = r  F r = 8j + 5k in F = -40i lb M = (8j + 5k )  (-40i) M = -200j + 320k lb.in (Vector) M = 377 lb-in (magnitude) Technique II M = F.d M = 409.43 = 377 lb-in Lecture 6 Chapter 4:Force resultant

  24. = = If FR and MRO are perpendicular to each other, then the system can be further reduced to a single force, FR, by simply moving FR from O to P. In three special cases, concurrent, coplanar, and parallel systems of forces, the system can always be reduced to a single force. Lecture 6 Chapter 4:Force resultant

  25. Example • Replace the three forces and the moment by a single force & moment RX = 500 sin40 + 700 sin60 = 928 N RY = 600 + 500 cos40 cos45 = 871 N RZ = 700 cos60 + 500 cos40 sin45 RZ = 621N R = 928i + 871j + 621k Moving the 500 N M500 = r  F M500 = (0.08i + 0.12j + 0.05k)  500(i sin40 + j cos40 cos45 + k cos 40 sin 45) M500 = 18.95i – 5.59j – 16.90k N.m Lecture 6 Chapter 4:Force resultant

  26. Moving the 600 N M600 = (600)(0.06)i + (600)(0.04)k M600 = 36.0i + 24.0k N.m Moving the 700 N M700 = (700 cos60)(0.03)i – [(700 sin60)(0.06) +(700 cos60)(0.1)]j – (700 sin60)(0.03)k M700 = 10.5i – 71.4j – 18.19k N.m The given couple MC = 25(-i sin40 – j cos40 cos45 – k cos40 sin45) MC = -16.07i – 13.54j – 13.54k The resultant M M = 49.4i – 90.5j – 24.6k Lecture 6 Chapter 4:Force resultant

  27. EXAMPLE 1 Given: A 2-D force and couple system as shown. Find: The equivalent resultant force and couple moment acting at A and then the equivalent single force location along the beam AB. Plan: 1) Sum all the x and y components of the forces to find FRA. 2) Find and sum all the moments resulting from moving each force to A. 3) Shift the FRA to a distance d such that d = MRA/FRy Lecture 6 Chapter 4:Force resultant

  28. +  FRx = 25 + 35 sin 30° = 42.5 lb +  FRy = 20 + 35 cos 30° = 50.31 lb + MRA = 35 cos30° (2) + 20(6) – 25(3) = 105.6 lb·ft FR = ( 42.52 + 50.312 )1/2 = 65.9 lb  = tan-1 ( 50.31/42.5) = 49.8 ° EXAMPLE (continued) FR The equivalent single force FR can be located on the beam AB at a distance d measured from A. d = MRA/FRy = 105.6/50.31 = 2.10 ft. Lecture 6 Chapter 4:Force resultant

  29. EXAMPLE 2 Given: The building slab has four columns. F1 and F2 = 0. Find: The equivalent resultant force and couple moment at the origin O. Also find the location (x,y) of the single equivalent resultant force. Plan: o • Find FR • 2) Find MRO=  (ri  Fi) = MRxi + MRyj • 3) The location of the single equivalent resultant force is given as x = MRy/FRz and y = MRx/FRz Lecture 6 Chapter 4:Force resultant

  30. EXAMPLE 2 (continued) o FRO= {-50 - 20} = {-70 k} kN MRyO= 10*(-20) + 4*(-50) = -400 MRyO = 3*(-50) = -150 = {-150 i + 400 j } kN·m The location of the single equivalent resultant force is given as, x = -MRyo/FRzo = -400/(-70) = 5.71 m y = MRxo/FRzo = (-150)/(-70) = 2.14 m Lecture 6 Chapter 4:Force resultant

  31. GROUP PROBLEM SOLVING Given: A 2-D force and couple system as shown. Find: The equivalent resultant force and couple moment acting at A. Plan: 1) Sum all the x and y components of the forces to find FRA. 2) Find and sum all the moments resulting from moving each force to A and add them to the 500 lb - ft free moment to find the resultant MRA . Lecture 6 Chapter 4:Force resultant

  32. + MRA = { (4/5)(150)(2) – 50 cos30° (3) + 50 sin30° (6) + 500 } = 760 lb·ft GROUP PROBLEM SOLVING (continued) Summing the force components: +  Fx = (4/5) 150 lb + 50 lb sin 30° = 145 lb +  Fy = (3/5) 150 lb + 50 lb cos 30° = 133.3 lb Now find the magnitude and direction of the resultant. FRA = ( 145 2 + 133.3 2 )1/2 = 197 lb and  = tan-1 ( 133.3/145) = 42.6 ° Lecture 6 Chapter 4:Force resultant

  33. Lecture 6 Chapter 4:Force resultant

  34. Lecture 6 Chapter 4:Force resultant

  35. Lecture 6 Chapter 4:Force resultant

  36. Lecture 6 Chapter 4:Force resultant

  37. Example 2 • Find , if P=400N, to replace the couple at the bottom with an equivalent couple at the top Couple at the bottom M = 1000.1 =10N.m Couple at the top M=4000.04cos  Equating the two couples 10 = 16cos  Lecture 6 Chapter 4:Force resultant

  38. Example 3 • Replace the 80-lb force by an equivalent force and moment (force-couple system) at point O d = 9sin60 = 7.79” couple moment M = 807.79 = 624 lb-in Thus the original force 80-lb equivalent to the 80-lb at O & the moment 624 lb-in Lecture 6 Chapter 4:Force resultant

  39. F1Y 80N 30o F1X F2Y 60N 45o F2X • Example • Determine the resultant of the four forces and the couple • Forces component F1X= 80cos30 = 69.28N F1y= 80sin30 = 40N F2X= 60cos45 = 51.96N F2y= 60sin45 = 30N Lecture 6 Chapter 4:Force resultant

  40. 42.4N 42.4N 40N 69.3N • Example RX= 69.3 + 40 - 42.4 RX= 66.9 N RY= 40 + 50 + 42.4 RY= 132.4 N R = 148.3 N Moment at O MO= 140 + 42.44 – 505 – 42.47 MO= 237 N.m Lecture 6 Chapter 4:Force resultant

  41. d • Example We can replace the moment and the force with a resultant force if we make a couple Rd equal to the moment Mo d Lecture 6 Chapter 4:Force resultant

  42. 42.4N 42.4N 40N 69.3N 42.4 42.4 42.4 50 50 7m 5m 4m O 42.4 42.4 O O Couple=169.6N.m Couple=250N.m Couple=296.8N.m 50 42.4 • Example • Another way Lecture 6 Chapter 4:Force resultant

  43. 140 296.8 250 50 42.4 Ry R=148.3N 40N 169.6 42.4 69.3N 63.2o 40 237 Rx O O • Example RX= 66.9 N RY= 132.4 N R = 148.3 N  = 63.2O MO= 140 + 169.6 – 250 – 296.8 MO= 237 N.m RYd = MO R=148.3N Ry d Rx O 237 Lecture 6 Chapter 4:Force resultant

  44. University of Palestine College of Engineering & Urban Planning ENGINEERING MECHANICS STATICS & DYNAMICS End of the Lecture Let Learning Continue Lecture 6 Chapter 4:Force resultant

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