1 / 37

4.6 Moment due to Force Couples

4.6 Moment due to Force Couples. A Couple is defined as two Forces having the same magnitude, parallel lines of action, and opposite sense separated by a perpendicular distance.

abraham-clay
Download Presentation

4.6 Moment due to Force Couples

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. 4.6 Moment due to Force Couples • A Couple is defined as two Forces having the same magnitude, parallel lines of action, and opposite sense separated by a perpendicular distance. • In this situation, the sum of the forces in each direction is zero, so a couple does not affect the sum of forces equations • A force couple will however tend to rotate the body it is acting on

  2. Moment Due to a Force Couple • By multiplying the magnitude of one Force by the distance between the Forces in the Couple, the moment due to the couple can be calculated. • M = Fdc • The couple will create a moment around an axis perpendicular to the plane that the couple falls in. Pay attention to the sense of the Moment (Right Hand Rule)

  3. Two couples will have equal moments if Moment of a Couple • the two couples lie in parallel planes, and • the two couples have the same sense or the tendency to cause rotation in the same direction. Do example

  4. Why do we use Force Couples? • The reason we use Force Couples to analyze Moments is that the location of the axis the Moment is calculated about does not matter • The Moment of a Couple is constant over the entire body it is acting on • Equivalent couples – two couples that produce the same magnitude and direction (example).

  5. Couples are Free Vectors • The point of action of a Couple does not matter • The plane that the Couple is acting in does not matter • All that matters is the orientation of the plane the Couple is acting in • Therefore, a Force Couple is said to be a Free Vector and can be applied at any point on the body it is acting

  6. Example: Force Couple

  7. 4.7 Simplification of a Force and Couple System (Resolution of Vectors) • The Moment due to the Force Couple is normally placed at the Cartesian Coordinate Origin and resolved into its x, y, and z components (Mx, My, and Mz).

  8. Vector Addition of Couples • By applying Varignon’s Theorem to the Forces in the Couple, it can be proven that couples can be added and resolved as Vectors.

  9. Force Couple System • Two opposing force can be added to a rigid body without affecting the equilibrium of it. • If there is a force acting at a distance from an axis, two forces of equal magnitude and opposite direction can be added at the axis with out affecting the equilibrium of the rigid body. • The original force and its opposing force at the axis make a couple that equates to a moment on the rigid body. • The other force at the axis results in the same force acting on the body

  10. Force Couple Systems • As a result of this it can be stated that any force (F) acting on a rigid body may be moved to any given point on the rigid body as long as a moment equal to moment of (F) about the axis is added to the rigid body.

  11. Resolution of a System of Forces in 3D • Each Force can be Resolved into a Force and Moment at the point of interest using the method just discussed. • The Resultant Force can then be found by Vector Addition. • The Resultant Moment must also be found using Vector Addition.

  12. SIMPLIFICATION OF FORCE AND COUPLE SYSTEMS Objectives: : • Determine the effect of moving a force. b) Find an equivalent force-couple system for a system of forces and couples.

  13. READING QUIZ 1. A general system of forces and couple moments acting on a rigid body can be reduced to a ___ . A) single force B) single moment C) single force and two moments D) single force and a single moment 2. The original force and couple system and an equivalent force-couple system have the same _____ effect on a body. A) internal B) external C) internal and external D) microscopic

  14. APPLICATIONS What are the resultant effects on the person’s hand when the force is applied in these four different ways? Why is understanding these difference important when designing various load-bearing structures?

  15. Several forces and a couple moment are acting on this vertical section of an I-beam. | | ?? For the process of designing the I-beam, it would be very helpful if you could replace the various forces and moment just one force and one couple moment at point O with the same external effect? How will you do that? APPLICATIONS (continued)

  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. SIMPLIFICATION OF FORCE AND COUPLE SYSTEM (Section 4.7)

  17. MOVING A FORCE ON ITS LINE OF ACTION Moving a force from A to B, when both points are on the vector’s 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).

  18. MOVING A FORCE OFF OF ITS LINE OF ACTION B When a force is moved, but not along its line of action, there is a change in its external effect! Essentially, moving a force from point A to B (as shown above) requires creating an additional couple moment. So moving a force means you have to “add” a new couple. Since this new couple moment is a “free” vector, it can be applied at any point on the body.

  19. SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 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.

  20. WR = W1 + W2 (MR)o = W1 d1 + W2 d2 If the force system lies in the x-y plane (a 2-D case), then the reduced equivalent system can be obtained using the following three scalar equations. SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM (continued)

  21. Given: A 2-D force system with geometry as shown. Find: The equivalent resultant force and couple moment acting at A and then the equivalent single force location measured from A. Plan: EXAMPLE #1 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 component to A. 3) Shift FRA to a distance d such that d = MRA/FRy

  22. +  FRx = 150 (3/5) + 50 –100 (4/5) = 60 lb +  FRy = 150 (4/5) + 100 (3/5) = 180 lb + MRA = 100 (4/5) 1 – 100 (3/5) 6 – 150(4/5) 3 = – 640 lb·ft FR = ( 602 + 1802 )1/2 = 190 lb  = tan-1 ( 180/60) = 71.6 ° EXAMPLE #1 (continued) FR The equivalent single force FR can be located at a distance d measured from A. d = MRA/FRy = 640 / 180 = 3.56 ft.

  23. 1. The forces on the pole can be reduced to a single force and a single moment at point ____ . A) P B) Q C) R D) S E) Any of these points. CONCEPT QUIZ • • • • 2. Consider two couples acting on a body. The simplest possible equivalent system at any arbitrary point on the body will have A) One force and one couple moment. B) One force. C) One couple moment. D) Two couple moments.

  24. Given: A 2-D force and couple system as shown. Find: The equivalent resultant force and couple moment acting at A. Plan: GROUP PROBLEM SOLVING 1) Sum all the x and y components of the two forces to find FRA. 2) Find and sum all the moments resulting from moving each force to A and add them to the 45 kN m free moment to find the resultant MRA .

  25. Summing the force components: + MRA = {30 sin 30° (0.3m) – 30 cos 30° (2m) – (5/13) 26 (0.3m) – (12/13) 26 (6m) – 45 } = – 239 kN m GROUP PROBLEM SOLVING (continued) +  Fx = (5/13) 26 – 30 sin 30° = – 5 kN +  Fy = – (12/13) 26 – 30 cos 30° = – 49.98 kN Now find the magnitude and direction of the resultant. FRA = ( 5 2 + 49.98 2 )1/2 = 50.2 kN and  = tan-1 (49.98/5) = 84.3 °

  26. ATTENTION QUIZ 1. For this force system, the equivalent system at P is ___________ . A) FRP = 40 lb (along +x-dir.) and MRP = +60 ft ·lb B) FRP = 0 lb and MRP = +30 ft · lb C) FRP = 30 lb (along +y-dir.) and MRP = -30 ft ·lb D) FRP = 40 lb (along +x-dir.) and MRP = +30 ft ·lb y 30 lb 1' 1' x • 40 lb P 30 lb

  27. ATTENTION QUIZ 2. Consider three couples acting on a body. Equivalent systems will be _______ at different points on the body. A) Different when located B) The same even when located C) Zero when located D) None of the above.

  28. Examples: Simplification of a Force and Couple System:

More Related