SKAA 1213 Engineering Mechanics

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1 SKAA 1213 Engineering Mechanics TPIC 5 Moment and Couple Lecturers: Rosli Anang Dr. Mohd Yunus Ishak Dr. Tan Cher Siang

2 Moment of a Force Moment of a force about a point/axis the tendency of the force to cause the body to rotate about the point/axis. Moment is a vector quantity

3 Moment of a Force Moment axis 1. Scalar Formulation of Moment M o M = Fd M Where d is the perpendicular distance from the axis of point to the action of the force F). F d Direction of force : specified by using the right hand rule.

4 Moment Arm l θ F d l θ F M = Fd d F Note: d = l sin θ θ = 0 o l d = 0 F θ = 90 o l d d = l

5 Example 1 Determine the moment of the 70N force about point A. [ Answer : (a)m A 2800Nmm (b) M A =2704.6Nmm ] (a) (b)

6 Example 2 Determine the moment of the 70N and 60N forces about point A. [ Answer : M A = Nm ] 6

7 Example 3 Determine the moment of each the force about point. [ Answer : M 1 = 200kNm, M 2 =70kNm, M 3 = 50kNm, M 4 = 70kNm, M 5 = [ 1, 2, 3, 4, 5 125kNm ]

8 Example 4 Determine the moments of the 40kN force about points A, B, C and D. [ Answer : M A = 0 M B = 48kNm M C = 20kNm M D = 20kNm ]

9 Resultant Moment of Coplanar Forces determined by total up the moments of all the forces algebraically. + M R = Fd z The counterclockwise curl written along the equation indicates that, the moment of any force d 3 d 1 willbepositive ifitis it is directed along the +z axis. d 2 F 2 F 1 F 3

10 Example 5 Determine the moment of the three forces about point. [Answer : M = -120kNm ] 10

11 Example 6 Determine the moments of the three forces about point B and C. [Answer : M B =-85kNm, M C = 125Nm ]

12 Cross Product cross product of two vectors A and is written as C = A x B Magnitude of C = AB sin θ A The direction of vector C is perpendicular to the plane A & B such that C is specified by the right hand rule. A u C θ C C = A x B C = AB sin θ θ B C = AB sin θ B

13 Cross Product - Laws of perations Commutative law: A x B B x A A xb= B xa Multiplication by a scalar: a (A xb) ) = (aa) xb= A x (ab) = (A xb)a) Distributive law: A x (B + D) = (A x B) + (A x D) 13

14 Cross Product of the Cartesian unit vectors z k = i x j In a similar manner, ixj = k ixk= j ix i = 0 j x k = i j x i = kk j x j = 0 kxi = j kxj= i k x k = 0 x i j y j + i k This diagram is helpful for obtaining the result of cross products of unit vectors Tips : Apply right hand rule 14

15 Cross product of two vectors A and B. Dt Determinant tform: i j k A x B = A x A y A z B x B y B z

16 Moment of a Force 2. Vector Formulation of Moment The moment of F about point discussed earlier, can be expressed using the vector cross product; M = rxf Where r represent the position vector drawn from to any point lying on the line of action of F. F Moment axis r M Magnitude: M = rf sin θ = F(r sin θ) = Fd

17 Direction: Apply right hand hand rule at the intersection point of the tails of extended r and F Moment axis Note that the moment axis is perpendicular to the plane containing r and F r d θ θ r F M r is treated as a sliding vector

18 Example 7 A force F=90N directed from C to D. Determine the magnitude of the moment created about the support at point A, and their coordinate direction angles. [ Answer :M A = 253.4Nm α = o β = 81 o = o ] r C r D r D = 3j + 0.6k

19 Resultant moment of a System of Forces F 3 z F 2 M R r 3 r 2 r 1 y x F 1 M R = (r x F)

20 Example Three forces acting on the rod. Determine the resultant moment about and the coordinate direction i angles. Given F 1 = 20i + 80k, F 2 = 40i + 30j 25k and F 3 = 35i + 50j 15k. [ Answer :M R = 410Nm α = o β = 117 o = 43 o ]

21 Principle of Moments (Varignon s theorem) The moment of a force about a point is equal to the summation of the moments of the force s components about the point. Proof: M = rxf 1 + rxf 2 = rx(f 1 + F 2 ) = r x F F F 2 F = F 1 + F 2 F 1 r Useful to determine the moment arms of the force s components than the moment arm of the force itself.

22 Moment of a Force about a Specified Axis Can be solved by scalar or vector analysis. z M y 0.5 m 0.4 m M b 0.3 m y In some situations we need the component of the moment along a specified axis that passes through the point. Let say component of M about y axis, M y. x F = 20 N

23 1. Scalar Analysis z b M y 05m 0.5 M 0.3 m y x 0.4 m F = 20 N a) M o = 20(0.5) = 10 Nm (direction defined by RH Rule about theb axis) M y = (3/5)(10) = 6 Nm (component method) b) M y = 20(0.3) = 6 Nm (direct method)

24 If the line of action of a force F is perpendicular to any specified axis aa, then; M a = Fd a where d a is the perpendicular distance from the force line of action to the axis. Thedirection is determined from thethumb thumb of the Right Hand when the fingers are curled in accordance with thedirection of rotation. A force will NT contribute a moment about a specified A force will NT contribute a moment about a specified axis if the force line of action is parallel to the axis or its line of action passes through the axis.

25 Example What are the values moment about the x,y, z axes. [Answer : M x =13 Nm M y == 59 Nm M z = -32 Nm] z x y

26 Example Find the Moment about a specified axis using Scalar notation method. [Answer : M x = 48Nm M oy = 0Nm, M oz = 0Nm] (0,0,8) (0,6,0)

27 Example Find the Moment about a specified axis using Scalar notation method. [Answer : M x= =0Nm, M oy =0Nm 0Nm, M oz =50Nm] (0,5,0) 27

28 2. Vector Analysis z b r A M y M u a = j 0.3 m y M = r A xf = (0.3i + 0.4j) X ( 20k) = ( 8i + 6j) Nm x 0.4 m F = (-20k) N The component of this moment along the y axis is then dt determined dfrom the dot product Since the unit vector of this axis is u a =j, then M y = M u a = ( 8i + 6j) j = 6 Nm

29 Vector analysis is advantages to find moment of force about an axis when the force components or the moment arms are difficult to determine. b M a a b θ M = r x F y r How to get M a? 1. Find M = r x F 2. M a = M cos θ = M u a u a M a = (r x F) u a = u a ( r x F) a F In vector algebra, combination of dot and cross product yielding ildi the scalar M a is called the triple scalar product.

30 The triple scalar product : i j k u ax u ay u M az a = (u ax i + u ay j + u az k) r x r y r z = r F x F y F x r y r z z F x F y F z nce M a is determined, M a as a Cartesian vector ; M a = M au a = [u a (rxf)]u a To find the resultant of a series of forces about the axis aa, the moment components of each force are added together algebraically, since the component lies along the same axis. M a = [u a (rxf)] = u a (rxf)

31 Example The force F= 35i + 50j 15k acts at C. Determine the moment of this force about x and a axes. [Answer:M x = -210Nm M a = -161Nm ]

32 Couples and Couple Moments Definition : two parallel forces that have the same magnitude but opposite direction,, and separated by a perpendicular distance, d. The moment produced dis called couple moment. F = 0, the only effect is tendency of rota on. F d -FF 2D F B r B r r A A -F 3D

33 Moment of a Couple Determination of moments of couple forces about any point : F about A: M = rxf about : M = r B X (F) + r A X ( F) F B r B r r A A 3D This indicates that a couple moment is a free vector. It can act at any point since M only depends upon the position vector r, not r A and r B.

34 2D Scalar Formulation: M =Fd d F F Vector Formulation: M = r x F F B r 3D A F

35 Properties of Moment of a Couple 1. The couple moment is unaffected by the pivot location *Couples at the same position for example below. M A =20(0.3)+20(1.7) = 40Nm M A =20(6) 20(4) = 40Nm

36 2. A couple can be shifted and still have the same moment about a given point. Couple at different position & moments calculated at the same point. M A =30(2) = 60Nm

37 Equivalent Couples Two couples which produce the same moment lie either in the same plane or in planes parallel to each other. The direction of the couple moments is the same and is perpendicular to the parallel planes. Resultant Couple Moment Since couple moments are free vectors they can be applied at any point on a body and added vertically.

38 M 1 M 2 Two set of couple forces Two couple moments M M R 1 Moved to any arbitrary M 2 point and added to obtain P resultant couple moment M R = M 1 + M 2

39 Example Replace the forces acting on the structure by an equivalent resultant force and couple moment at A. [ Answer: M RA = = -46.6Nm ( ) ]

40 Example Determine the moment of the couple on the member shown. [ Answer : M= Nm ] d

41 Equivalent System Replacing system of forces and couple moments acting on a body by a single force and couple acting on a specified point that produce the same external effects of translation and rotation. Case 1: Point t Is n the Line of Ati Action F A A F = -F F = F A

42 Case C 2: Point t Is Not tn the Line of Ati Action M = r x F F A F = F A = F P A -F *Note: Since couple is a free vector, it may be applied at any point

43 Resultants of a Force System M R = M + M R C F R = F F M F = r 2 x F 2 M R F R = F 1 + F 2 r r 1 2 F 2 F = 1 M C M 1 = r 1 x F 1 M C = θ

44 NTE: Both the magnitude & direction of F R are independent of the location of, however, M R depends on the location of since the moment M 1 & M 2 are determined dby using the position vectors r 1 & r 2. M R is a free vector and can act at any point on the body.

45 Example Determine themagnitude, direction and location of a resultant force which is equivalent to the given system of forces measured horizontally from A. [ Answer : F R = 272N( ) θ = 68.4 o d = 0.18m ]

46 Example Determine the magnitude and direction of a resultant force equivalent force system and locate its point of application. [ Answer : F R = -1190N ( ),y = 2.84m x = 1.24m]

47 Simplification to a Single Force System Consider a special case for which the system of forces and couple moments reduces at point of the resultant force F R and couple M R which are perpendicular to each other. M F 1 F R 2 F 1 b a b F R a M 2 r 2 r 1 r 3 F 3 = a M R b = a dd P b d = M R F R

48 If the system of forces is either concurrent, coplanar, or parallel, it can be reduced (as in the above case), to a single resultant t force F R. This is because in each of these cases F R and M R will always be perpendicular to each other when the force system is simplified at any point. 48

49 1.Concurrent ConcurrentForce System F 1 F 2 F R = F 2 F 3 =

50 2.Coplanar Force System Couple moments are to plane of forces y F 3 M 2 r 3 r 2 r 1 F 1 M 1 F 2 x y y d= M R /F R M R = M + r x F = d x = F R = F F R = F x

51 3.Parallel l Force System Couple moments are to the forces M 1 x F 1 r 1 z r 2 r3 F 2 F y F 3 M 2 = M R = M + r x F x z F R = F y = F R = F x z y

52 Example Determine the magnitude and location of the equivalent resultant force acting on the beam. _ [ Answer : F R = -1190N ( ), x = 3m ] da = w dx = 50x 2 dx

53 Example Determine the magnitude and location of the equivalent resultant force acting on the beam. _ [ Answer : F R = 140KN, x = 1.86m ]

54 Reduction of a simple Distributed Loading y Uniform pressure along one p x axis on a flat rectangular p=p(x) surface. The load intensity is L of the load represented ese by the arrows form a system of a/2 a/2 parallel forces, infinite in numbers, each acting on a separate differential area.

55 da x df L dx w=w(x) w(x) Load function, p = p(x) [pressure uniform in y axis] Multiply pypp() p=p(x) with the width a, we obtain; w= p(x) a = w x This loading function is a measure of load distribution ib ti along the line y=0 which h is the plane of symmetry of the loading. Note: it is load per unit length. 55

56 A F R C In a system of coplanar parallel forces, the load intensity can be represented by w = w(x) x L This system of forces can be simplified to a single force F R and its location x can be specified.

57 Magnitude of Resultant Force w=w(x) For an elemental l length dx as df shown in the diagram, the force da acting is; df = w(x) dx = da [shaded area] x L dx For entire length; + F R = F: F R = w(x) dx = da = A R R () Hence, the magnitude of the resultant force is equal to the total area A under the loading diagram w = w(x). 57

58 Location of Resultant Force M R = M Equating the moment of the F R and the force distribution about. + M R = SM : x F R = x w(x) dx L Solving for x; x = x w(x) dx L w(x) dx L = x da A da A A df w=w(x) da x dx L df produces a moment of x df =x w(x) dx about. F R C x L

59 Location of Resultant Force df w=w(x) da This eqn represents the x coordinate for the geometric x dx center (centroid) of the area L df produces a moment of under the distributed ib d loading x df =x w(x) dx about. diagram w(x). The resultant force has a line of action which passes through the centroid C fo the area defined by the distributed loading diagram w(x). A F R C x L

Force System Resultants. Engineering Mechanics: Statics

Force System Resultants. Engineering Mechanics: Statics Force System Resultants Engineering Mechanics: Statics Chapter Objectives To discuss the concept of the moment of a force and show how to calculate it in 2-D and 3-D systems. Definition of the moment of