Engineering Mechanics Statics

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1 Mechanical Systems Engineering_2016 Engineering Mechanics Statics 6. Moment of a Couple Dr. Rami Zakaria

Moment of a Couple We need a moment (or torque) of

A couple is defined as two parallel forces with the

2 Moment of a Couple We need a moment (or torque) of (12 N m) to rotate the wheel. Notice that one of the two grips of the wheel above require less force to rotate the wheel. A couple is defined as two parallel forces with the same magnitude but opposite in direction separated by a perpendicular distance (d).

3 Moment of a Couple The moment of a couple is defined as : M O = F d (using a scalar analysis) or as: M O = r F (using a vector analysis). Here r is any position vector from the line of action of F to the line of action of F. We can add Couple Moments together using the same rules as adding any vectors.

4 Scalar approach Example Given: Two couples act on the beam. The resultant couple is zero. Find: The magnitudes of the forces P and F and the distance d. Solution: From the definition of a couple: P = 2 kn, F = 4 kn Determine the net moment + M = (2)(0.3) (4)(d) It was given that the net moment equals zero. So: + M = (2)(0.3) (4)(d) = 0 Now solve this equation. d = (0.6) N m / (4) N = 0.15 m

6 Classroom Exercise

7 Vector approach Example Given: A 35 N force couple acting on the rod. Find: The couple moment acting on the rod in Cartesian vector notation. Plan: 1) Use M = r F to find the couple moment. 2) Set r = r AB and F = {35 k} N. 3) Calculate the cross product to find M.

8 r AB = { 0 i (0.25) j + (0.25 tan 30 ) k} m r AB = { 0.25 j k} m F = {0 i + 0 j + 35 k} N M = r AB F = i j k N m = {( ) i (0 0) j (0 0) k} N m = { 8.75 i + 0 j + 0 k} N m

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

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.

10 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).

11 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 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.

12 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. W R = W 1 + W 2 (M R ) o = W 1 d 1 + W 2 d 2

13 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 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.

14 Solution: Summing the force components: + F x = (5/13) sin 30 = 5 kn + F y = (12/13) cos 30 = kn

15 Now find the magnitude and direction of the resultant. F RA = ( ) 1/2 = 50.2 kn and = tan -1 (49.98/5) = M RA = {30 sin 30 (0.3m) 30 cos 30 (2m) (5/13) 26 (0.3m) (12/13) 26 (6m) 45 } = 239 kn m

16 Classroom Exercise

FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM If F R and M RO are perpendicular to each other, then the system can be further reduced to a single force, F R,

18 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM If F R and M RO are perpendicular to each other, then the system can be further reduced to a single force, F R, by simply moving F R 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.

19 EXAMPLE 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 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: d x M F RA Ry

20 Solution: F R + F Rx = 150 (3/5) (4/5) = 60 lb + F Ry = 150 (4/5) (3/5) = 180 lb + M RA = 100 (4/5) (3/5) 6 150(4/5) 3 = 640 lb ft F R = ( ) 1/2 = 190 lb = tan -1 ( 180/60) = 71.6 The resultant force F R can be located at a distance d measured from A. d x = M RA /F Ry = 640 / 180 = 3.56 ft.

21 EXAMPLE (simplification of a system) Given: The slab is subjected to three parallel forces. 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: 1) Find F RO = F i = F Rzo k 2) Find M RO = (r i F i ) = M RxO i + M RyO j 3) The location of the single equivalent resultant force is given as M RyO M x, F RZ y F RxO RZ

22 Solution: F RO = {100 k 500 k 400 k} = 800 k N M RO = (3 i) (100 k) + (4 i + 4 j) (-500 k) + (4 j) (-400 k) = { 300 j j 2000 i 1600 i} = { 3600 i j }N m The location of the single equivalent resultant force is given as, x = + M Ryo / F Rzo = (1700) / (800) = 2.13 m y = + M Rxo / F Rzo = (3600) / (800) = 4.5 m

23 Questions: 1. In statics, a couple is defined as separated by a perpendicular distance. A) two forces in the same direction B) two forces of equal magnitude C) two forces of equal magnitude acting in the same direction D) two forces of equal magnitude acting in opposite directions 2. The moment of a couple is called a vector. A) Free B) Spin C) Romantic D) Sliding

24 3. A couple is applied to the beam as shown. Its moment equals N m. A) 50 B) 60 C) 80 D) 100 1m 50 N 2m 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 5. 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

25 DISTRIBUTED LOADING In many situations, a surface area of a body is subjected to a distributed load. Such forces are caused by winds, fluids, or the weight of items on the body s surface. We will analyze the most common case of a distributed pressure loading. This is a uniform load along one axis of a flat rectangular body. In such cases, w(x) is the load intensity distribution function and it has units of force per length.

26 MAGNITUDE OF RESULTANT FORCE Consider an element of length dx. The force magnitude df acting on it is given as df w( x) dx The net force on the beam is given by + F df w( x) dx R L L A Here A is the area under the loading curve w(x).

27 LOCATION OF THE RESULTANT FORCE The force df will produce a moment of (x)(df) about point O. The total moment about point O is given as M xw( x) dx Ro xdf L L Assuming that F R acts at x, it will produce the moment about point O as M ( x) F ( x) w( x) dx Ro R L

28 Comparing the last two equations, we get You will learn more detail later, but F R acts through a point C, which is called the geometric center or centroid of the area under the loading curve w(x).

29 Example: Determine the magnitude and location of the equivalent resultent force (F R ) acting on the shaft

30 Note: for simples shapes we can find the shift easily using the geometric centroid (we will learn more about the centroid later!) Area C x b.h b/2 b.h/2 (a +b)/3 π.a 2 a π.a 2 /2 a

Examples The rectangular load: F R x 400 10 10 2 5 ft

31 Examples The rectangular load: F R x ft 4000lb The triangular loading: F R x m 1800N

32 Classroom Exercise

34 Home Exercise

EQUIVALENT FORCE-COUPLE SYSTEMS

EQUIVALENT FORCE-COUPLE SYSTEMS Today s Objectives: Students will be able to: 1) Determine the effect of moving a force. 2) Find an equivalent force-couple system for a system of forces and couples. APPLICATIONS