1 / 45

ENGINEERING MECHANICS STATICS & DYNAMICS

University of Palestine College of Engineering & Urban Planning. ENGINEERING MECHANICS STATICS & DYNAMICS. Instructor: Eng. Eman Al.Swaity. Lecture 5. Chapter 4: Force resultant. Chapter(4) Force System Resultants. Chapter Objectives

lowri
Download Presentation

ENGINEERING MECHANICS STATICS & DYNAMICS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. University of Palestine College of Engineering & Urban Planning ENGINEERING MECHANICS STATICS & DYNAMICS Instructor:Eng. EmanAl.Swaity Lecture 5 Chapter 4:Force resultant

  2. Chapter(4) Force System Resultants • Chapter Objectives • To discuss the concept of the moment of a force and show how to calculate it in two and three dimensions. • To provide a method for finding the moment of a force about a specified axis. • To define the moment of a couple. • To present methods for determining the resultants of nonconcurrent force systems. • To indicate how to reduce a simple distributed loading to • a resultant force having a specified location. Lecture 5 Chapter 4:Force resultant

  3. Today’s Objectives : Students will be able to: a) understand and define moment, and, b) determine moments of a force in 2-D and 3-D cases. The moment (Torque) of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis. For example, consider the horizontal force Fx, which acts perpendicular to the handle of the wrench and is located a distance dy from point 0, is seen that this force tends to cause the pipe to turn about the Z axis. Lecture 5 Chapter 4:Force resultant

  4. Magnitude:The magnitude of Mo is where d is referred to as the moment arm or perpendicular distance from the axis at point 0 to the line of action of the force. Units of moment magnitude consist of force times distance, e.g., N. m or lb. ft. Lecture 5 Chapter 4:Force resultant

  5. Direction: The direction of Mo will be specified by using the "right-hand rule." To do this, the fingers of the right hand are curled such that they follow the sense of rotation, which would occur if the force could rotate about point 0, The thumb then points along the moment axis so that it gives the direction and sense of the moment vector, which is upward and perpendicular to the shaded plane containing F and d. Lecture 5 Chapter 4:Force resultant

  6. READING F = 10 N 1. What is the moment of the 10 N force about point A (MA)? A) 10 N·m B) 30 N·m C) 13 N·m D) (10/3) N·m E) 7 N·m d = 3 m • A Lecture 5 Chapter 4:Force resultant

  7. APPLICATIONS (continued) What is the effect of the 30 N force on the lug nut? Lecture 5 Chapter 4:Force resultant

  8. MOMENT IN 2-D The moment of a force about a point provides a measure of the tendency for rotation (sometimes called a torque). Lecture 5 Chapter 4:Force resultant

  9. MOMENT IN 2-D(continued) In the 2-D case, the magnitude of the moment is Mo = F d As shown, d is theperpendicular distance from point O to the line of action of the force. In 2-D, the direction of MO is either clockwise or counter-clockwise depending on the tendency for rotation. Lecture 5 Chapter 4:Force resultant

  10. F a b O d F F y F x a b O MOMENT IN 2-D(continued) For example, MO = F d and the direction is counter-clockwise. Often it is easier to determine MO by using the components of F as shown. Using this approach, MO = (FY a) – (FX b). Note the different signs on the terms! The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive. We can determine the direction of rotation by imagining the body pinned at O and deciding which way the body would rotate because of the force. Lecture 5 Chapter 4:Force resultant

  11. EXAMPLE1 Given: A 400 N force is applied to the frame and  = 20°. Find: The moment of the force at A. Plan: 1) Resolve the force along x and y axes. 2) Determine MA using scalar analysis. Lecture 5 Chapter 4:Force resultant

  12. EXAMPLE 1(continued) Solution +  Fy = -400 cos 20° N +  Fx = -400 sin 20° N + MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m = 1160 N·m Lecture 5 Chapter 4:Force resultant

  13. 4.5m + EXAMPLE 2 • Find the moment of F1 at point o F2=200kN F1=100kN O M = F.d M = 100  4.5 = 450 kN.m • Find the moment of F2 at point o M = F.d M = 200  0 = 0 kN.m Lecture 5 Chapter 4:Force resultant

  14. - • Example 3 • Fine the moment of all forces at point A 50N 100N 100N 1m 50N 1m 50N A Don’t forget the signs 3m 3m 2m 2m Moment at A = F1d1 + F2d2 + F3d3 +…. = Five forces means five moments MA= (-)500 + (-)1002 + 1003 + (-)506 + 501 MA= -150 N.m Lecture 5 Chapter 4:Force resultant

  15. - • Moment theorem (Varignon’s theorem) • The moment of force F at point O is equal to the sum of moments of the force components at the same point O Example 4 Fined the magnitude of the moment about point O in four deferent ways Technique I the moment arm d d = 4cos40o + 2sin40o d = 4.35m the moment = Fd M = -6004.35 = -2610 N.m Lecture 5 Chapter 4:Force resultant

  16. - - Example 4 Technique II Replace the force by its components F1 = 600cos40o = 460N F2 = 600sin40o = 386N Moment theorem M0= (-)4604 + (-)3862 M0 = -2610N.m Technique III the principle of transmissibility d1 = 4 + 2tan40o = 5.68m M0= (-)4605.68 + 3860 M0 = -2610N.m Lecture 5 Chapter 4:Force resultant

  17. - Example 4 Technique IV the principle of transmissibility d1 = 2 + 4tan50o = 6.77m M0= 4600 + (-)3866.77 = -2610N.m Lecture 5 Chapter 4:Force resultant

  18. Solution: +  Fy = - 40 cos 20° N +  Fx = - 40 sin 20° N + MO = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm = -7107 N·mm = - 7.11 N·m GROUP PROBLEM SOLVING Given: A 40 N force is applied to the wrench. Find: The moment of the force at O. Plan: 1) Resolve the force along x and y axes. 2) Determine MO using scalar analysis. Lecture 5 Chapter 4:Force resultant

  19. For each case illustrated in Fig. 4-4, determine the moment of the force about point O. Lecture 5 Chapter 4:Force resultant

  20. Lecture 5 Chapter 4:Force resultant

  21. In general, the cross product of two vectors Aand Bresults in another vector C , i.e., C = A  B. The magnitude and direction of the resulting vector can be written as C = A  B = A B sin  UC HereUC is the unit vector perpendicular to both A and B vectors as shown (or to the plane containing theA and B vectors). Lecture 5 Chapter 4:Force resultant

  22. CROSS PRODUCT The right hand rule is a useful tool for determining the direction of the vector resulting from a cross product. For example: i  j = k Note that a vector crossed into itself is zero, e.g.,i  i = 0 Lecture 5 Chapter 4:Force resultant

  23. CROSS PRODUCT (continued) Of even more utility, the cross product can be written as Each component can be determined using 2  2 determinants. Lecture 5 Chapter 4:Force resultant

  24. The moment of a force F about point 0, or actually about the moment axis passing through 0 and perpendicular to the plane containing 0 and F, Fig. 4-12a, can be expressed using the vector cross product, namely, Lecture 5 Chapter 4:Force resultant

  25. Lecture 5 Chapter 4:Force resultant

  26. Lecture 5 Chapter 4:Force resultant

  27. Moment M = r  F r = xi + yj + zk F = FXi + FYj + FZk M = (ryFz – rzFy)i + (rzFx – rxFz)j + (rxFy – ryFx)k Mx = ryFz – rzFy My = rzFx – rxFz Mz = rxFy – ryFx Lecture 5 Chapter 4:Force resultant

  28. EXAMPLE 2 Given: a = 3 in, b = 6 in and c = 2 in. Find: Moment of F about point O. Plan: o 1) Find rOA. 2) Determine MO = rOA F. SolutionrOA= {3 i+ 6 j – 0 k} in MO = i j k = [{6(-1) – 0(2)} i– {3(-1) – 0(3)} j + {3(2) – 6(3)} k] lb·in = {-6 i + 3 j– 12 k} lb·in 3 6 0 3 2 -1 Lecture 5 Chapter 4:Force resultant

  29. Lecture 5 Chapter 4:Force resultant

  30. Lecture 5 Chapter 4:Force resultant

  31. Lecture 5 Chapter 4:Force resultant

  32. Lecture 5 Chapter 4:Force resultant

  33. Lecture 5 Chapter 4:Force resultant

  34. Recall that when the moment of a force is computed about a point, the moment and its axis are always perpendicular to the plane containing the force and the moment arm. In some problems it is important to find the component of this moment along a specified axis that passes through the point. To solve this problem either a scalar or vector analysis can be used. Lecture 5 Chapter 4:Force resultant

  35. APPLICATIONS With the force F, a person is creating the moment MA. What portion of MAis used in turning the socket? Lecture 5 Chapter 4:Force resultant

  36. SCALAR ANALYSIS Recall that the moment of a force about any point A is MA= F dA where dA is the perpendicular (or shortest) distance from the point to the force’s line of action. This concept can be extended to find the moment of a force about an axis. In the figure above, the moment about the y-axis would be My= 20 (0.3) = 6 N·m. However this calculation is not always trivial and vector analysis may be preferable. Lecture 5 Chapter 4:Force resultant

  37. VECTOR ANALYSIS Lecture 5 Chapter 4:Force resultant

  38. Lecture 5 Chapter 4:Force resultant

  39. Lecture 5 Chapter 4:Force resultant

  40. Lecture 5 Chapter 4:Force resultant

  41. Example • Determine the moment MZ of T about z-axis passing O • Technique I Minus sign indicate that the Mz is in the negative Z direction Lecture 5 Chapter 4:Force resultant

  42. Technique II The force T resolved into two components Tz & Txy TZ is parallel to the Z-axis. That’s means no moment comes from TZ about Z-axis. MZ is due only to TXY Lecture 5 Chapter 4:Force resultant

  43. EXAMPLE Given: A force is applied to the tool to open a gas valve. Find: The magnitude of the moment of this force about the z axis of the value. Plan: A B 1) We need to use Mz = u • (r  F). 2) Note that u = 1 k. 3) The vector r is the position vector from A to B. 4) Force F is already given in Cartesian vector form.

  44. K EXAMPLE (continued) A rAB = {0.25 sin 30°i + 0.25 cos30°j} m = {0.125 i + 0.2165 j} m F = {-60 i + 20 j + 15 k} N Mz = (rAB F) B Mz = = 1{0.125(20) – 0.2165(-60)} N·m = 15.5 N·m Lecture 5 Chapter 4:Force resultant

  45. University of Palestine College of Engineering & Urban Planning ENGINEERING MECHANICS STATICS & DYNAMICS End of the Lecture Let Learning Continue Lecture 5 Chapter 4:Force resultant

More Related