SUB:-MOS (2130003) Guided By:- Prof. Megha Patel

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 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)

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? 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.

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

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).

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 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 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 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.

Determine the effect of moving a force. 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. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 4.7-4.9

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? Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 4.7-4.9

APPLICATIONS 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? | | ?? Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 4.7-4.9

SIMPLIFICATION OF FORCE AND COUPLE SYSTEM 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. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 4.7-4.9

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). Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 4.7-4.9

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. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 4.7-4.9

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. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 4.7-4.9

SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 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. WR = W1 + W2 (MR)o = W1 d1 + W2 d2 Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 4.7-4.9

RESULTANT OF COPLANAR NON CONCURRENT FORCE SYSTEM   Coplanar Non-concurrent Force System:   This is the force system in which lines of action of individual forces lie in the same plane but act at different points of application. F2 F2 F1 F1 F5 F3 F3 F4 Fig. 1 Fig. 2

TYPES Parallel Force System – Lines of action of individual forces are parallel to each other. Non-Parallel Force System – Lines of action of the forces are not parallel to each other.

RESULTANT OF COPLANAR NON-CONCURRENT FORCE SYSTEMS The resultant of coplanar, non-concurrent force systems is the one which will produce same rotational and translational effect as that of the given system. It may be a force and a moment or a pure moment. F1 Let F1, F2 and F3 constitute a non concurrent system as shown in the fig. ‘O’ – be any convenient reference point in the plane. F2 O F3

F1 F1 R F2 ∑Mo ∑Mo O O F2 O F3 F3 Fig. C Fig. A Fig. B Each force can be replaced by a force of the same magnitude and acting in the same direction at ‘O’ and a moment about ‘O’ as in Fig B. The non concurrent forces can be combined as in the case of concurrent system to get the resultant force R. Thus, the resultant of the system is equal to a force R at ‘O’ and a moment ‘∑Mo’ as shown in fig.C.

R R d ∑Mo O O Fig. D Fig. C The single force R and the moment ‘∑Mo’ shown in the fig.C can be replaced by a single force R acting at a distance ‘d’ from ‘O’, which gives the same effect. Thus, the resultant can be reduced to a single force acting at a certain distance from ‘O’. Mathematically,

Y-axis A Rx Rx Y d R X-axis B O X Ry Ry X and Y intercepts of Resultant: In some problems, it may be required to determine distances of the resultant R along x-axis and y-axis i.e., X and Y intercepts. Let ‘d’ be the perpendicular distance of the resultant from ‘O’ as shown in the fig. Let Rx=∑Fx and Ry=∑Fy be the components of the resultant in X and Y directions. By Varignon’s theorem, R×d= ∑Mo At B, ∑Mo = Rx×0 + Ry×X Y-axis A Rx Rx Y d R X-axis B O X Ry Ry

R R d ∑Mo O O Fig. D Fig. C The single force R and the moment ‘∑Mo’ shown in the fig.C can be replaced by a single force R acting at a distance ‘d’ from ‘O’, which gives the same effect. Thus, the resultant can be reduced to a single force acting at a certain distance from ‘O’. Mathematically,

Y-axis A Rx Rx Y d R X-axis B O X Ry Ry Therefore, X= ∑Mo/Ry Similarly, at A, Ry×0 + Rx×Y = ∑Mo Y= ∑Mo/Rx Y-axis A Rx Rx Y d R X-axis B O X Ry Ry

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