**6-Moments Couples and Force Couple Systems_Partb**

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### 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. 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### Moment Due to a Force Couple:

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 = Fd c 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)### Moment of a Couple:

Moment of a Couple Two couples will have equal moments if 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### Why do we use Force Couples?:

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).### Couples are Free Vectors:

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### PowerPoint Presentation:

Example: Force Couple### 4.7 Simplification of a Force and Couple System (Resolution of Vectors):

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 (M x , M y , and M z ).### Vector Addition of Couples:

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.### Force Couple System:

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### Force Couple Systems:

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.### Resolution of a System of Forces in 3D:

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.### PowerPoint Presentation:

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.### PowerPoint Presentation:

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### PowerPoint Presentation:

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?### PowerPoint Presentation:

APPLICATIONS (continued) 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? | | ??### PowerPoint Presentation:

SIMPLIFICATION OF FORCE AND COUPLE SYSTEM (Section 4.7) 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.### PowerPoint Presentation:

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).### PowerPoint Presentation:

MOVING A FORCE OFF OF ITS LINE OF ACTION 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. B### PowerPoint Presentation:

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.### PowerPoint Presentation:

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) W R = W 1 + W 2 (M R ) o = W 1 d 1 + W 2 d 2### PowerPoint Presentation:

EXAMPLE #1 1) Sum all the x and y components of the forces to find F RA . 2) Find and sum all the moments resulting from moving each force component to A. 3) Shift F RA to a distance d such that d = M RA /F Ry 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:### PowerPoint Presentation:

EXAMPLE #1 (continued) + F Rx = 150 (3/5) + 50 – 100 (4/5) = 60 lb + F Ry = 150 (4/5) + 100 (3/5) = 180 lb + M RA = 100 (4/5) 1 – 100 (3/5) 6 – 150(4/5) 3 = – 640 lb·ft F R = ( 60 2 + 180 2 ) 1/2 = 190 lb = tan -1 ( 180/60) = 71.6 ° The equivalent single force F R can be located at a distance d measured from A. d = M RA /F Ry = 640 / 180 = 3.56 ft . F R### PowerPoint Presentation:

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. 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. • • • •### PowerPoint Presentation:

GROUP PROBLEM SOLVING 1) Sum all the x and y components of the two forces to find F RA . 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 M RA . Given: A 2-D force and couple system as shown. Find: The equivalent resultant force and couple moment acting at A. Plan:### PowerPoint Presentation:

GROUP PROBLEM SOLVING (continued) + F x = (5/13) 26 – 30 sin 30 ° = – 5 kN + F y = – (12/13) 26 – 30 cos 30 ° = – 49.98 kN + M RA = {30 sin 30 ° (0.3m) – 30 cos 30 ° (2m) – (5/13) 26 (0.3m) – (12/13) 26 (6m) – 45 } = – 239 kN m Now find the magnitude and direction of the resultant. F RA = ( 5 2 + 49.98 2 ) 1/2 = 50.2 kN and = tan -1 (49.98/5) = 84.3 ° Summing the force components:### PowerPoint Presentation:

ATTENTION QUIZ 1. For this force system, the equivalent system at P is ___________ . A) F RP = 40 lb (along +x-dir.) and M RP = +60 ft · lb B) F RP = 0 lb and M RP = +30 ft · lb C) F RP = 30 lb (along +y-dir.) and M RP = -30 ft · lb D) F RP = 40 lb (along +x-dir.) and M RP = +30 ft · lb P 1' 1' 30 lb 40 lb 30 lb • x y### PowerPoint Presentation:

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.### PowerPoint Presentation:

Examples: Simplification of a Force and Couple System:## View More Presentations

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