Presentation on theme: "Engineering Mechanics"— Presentation transcript:
1Engineering Mechanics Chapter 3 :Moments and CouplesClick on VIEW and select SLIDE SHOW to view the presentation.
2Objectives Understand the Principle of Transmissibility of a Force; Determine the Moment of a Force System from Basic Definition and/or with the application of Varignon’s Theorem;Understand the Properties of a Couple;Determine a Force-Couple Equivalent to Replace a System of Forces.
3Introduction A force will cause motion along its direction. A force may also cause rotation about a fixed point some distance away.This rotation or turning effect of a force is called MOMENT.
4Principle of Transmissibility If we move (TRANSMIT) the Force P from point A to point B which lies on the Line of Action of Force P, the same effect would be expected.Point of ApplicationLine of actionPABPrinciple of transmissibility states that a force acting on a rigid body at different points along the force’s line of action will produce the same effect on the body.
5Moment of A ForceThe moment M of a force F about a fixed point A is defined as the product of the magnitude of force F and the perpendicular distance d from point A to the line of action of force F.MA = F x dWhere force F is in newtons, NAnd distance d is in meters, mThus moment MA is in newton-meter, Nm.AdF
6Moment Sign Convention Anti-clockwise : + VE (Counter-clockwise)Clockwise : - VE
7Example 3.1Calculate the moment of the 500 N force about the point A as shown in the diagram.500 NA1.5 m
8500 NExample 3.1 Solution1.5 mASince the perpendicular distance from the force to the axis point A is 1.5 m, fromMA = F x dMA = x 1.5= Nm= 750 Nm
9Example 3.2Calculate the moment about point A caused by the 500 N force as shown in the diagram.A1.5 m60o500 N
10Example 3.2 Solution: MA = F x d = -500 x AB = -500 x 1.5 sin 60 Line of action600dBAMA = F x d= x AB= x 1.5 sin 60= Nm= Nm
11Addition of Moments of Coplanar Forces Coplanar Forces refers to forces acting on the Same Plane ( i.e. 2-D ).For a system of several coplanar forces, the combined turning effect of the forces can be determined by adding algebraically the moment caused by each individual force, taking into account their sense of rotation,i.e +ve for anti-clockwise and–ve for clockwise rotation.
12Example 3.3Determine the resulting moment about point A of the system of forces on bar ABC as shown in the diagram.500 N800 N1.5 m0.5m60 0ABC
13Example 3.3 Solution: AD = (1.5 + 0.5) sin 60 = 2.0 sin 60 = 1.732 m 500800 NExample Solution:1.5 m0.5 m60 0A1.5 m0.5 mLine of Action600DC800 N500 NABCAD = ( ) sin 60= 2.0 sin 60= mMA = (-500 x 1.5) + (-800 x 1.732)= –= Nm= Nm
14Varignon’s TheoremVarignon’s Theorem states that “ the moment of a force about any point is equal to the sum of the moments of its components about the same point”.To calculate the moment of any force with a slope or at an angle to the x or y-axis, resolve the force into the Fx and the Fy components, and calculate the sum of the moment of these two force components about the same point.
15Re-Calculate Example 3.3 Above Employing Varignon’s Theorem. 60 0ABC
16Example 3.4 Solution MA = (-500 x 1.5) + (- 800 sin 60 x 2) 60 0ABC1.5 m0.5m600A800 NBCFxFy = 800 sin 600500 NFy= 800 cos 600MA= (-500 x 1.5) + (- 800 sin 60 x 2)= Nm= Nm* Same answer as in example 3.3
17Example 3.5Determine the resulting moment about point A for the system of forces acting on the plate ABCD as shown in the diagram.15 cm20 ND45oC8cm30o35 NAB80o10.38 cm45 N
18Example 3.5 Solution 15 cm 8cm 10.38 cm 45 N 20 N 35 N 80o 45o 30o A B DExample 3.5 SolutionFxFy45 N80045 N force: Fx = 45 cos 80 = NFy = 45 sin 80 = NFxFy35 N30035 N force: Fx = 35 cos 60 = 17.5 NFy = 35 sin 60 = N
1920 N0.15 mExample 3.5 Solution60oD45oC0.08mAB80o35 Nm45 NFyFx20 N45020 N force: Fx = 20 cos 45 = NFy = 20 sin 45 = NMA = (- 45 sin 80 x ) + (- 35 cos 60 x 0.08)+ (20 cos 45 x 0.08) + (20 sin 45 x 0.15)= –= Nm = 2.74 Nm
20- Equal magnitude and opposite in direction CouplesFdA couple consists of a pair of 2 forces which has the following properties :-- Equal magnitude and opposite in direction - Act along parallel lines of action - Separated by a perpendicular distance d.
21What a couple does?A couple causes a body to rotate only without translational motion since the two forces ‘cancels’ out each other giving zero resultant.A couple acting in a system of forces will only contribute to the resulting moment but not to the resulting force.
22Magnitude of a CoupleConsider a light bar acted upon by a couple as shown :-d1dd2FABThe moment of the couple about A isMA = + ( F x d2 ) - ( F x d1 ) MA = F x ( d2 - d1 ) MA = F x dWhat is the total moment of the couple aboutpoint B ?
23The couple’s moment about any pivot point is equal to F x d From above, we see that :The couple’s moment about any pivot point is equal to F x dA couple has the same moment about all points on a body.F
24Example 3.6A light bracket ABC is subjected to two forces and two couples as shown. Determine the moment at (a) point A and (b) point B.120 Nm80 Nm30o2 KN1.5 m3 KN2.5 mBA
25Example 3.6 Solution: 2 KN 80 Nm 30o 120 Nm 1.5 m A 3 KN B 2.5 m a) MA = (2000 cos 60 x 2.5) – (2000 sin 60 x 1.5) + ( 3000 x 0)+ 120 – 80= –= Nm = 58 Nm(b) MB = (2000 cos 60 x 0) – (2000 sin 60 x 1.5) + (3000 x 0)+ 120 – 80= Nm= 2558 Nm
263.7 Force Couple Equivalent The Force-Couple Equivalent concept will enable us to transfer a force to another location outside its line of action.Consider a force F acting at a point B on a rigid body as shown in diagram (a) below. How do we transfer the force F from point B to point A?FdAB
27BFAdABFMA = F x dABFdForce Couple EquivalentThe above shows how a force can be replaced by a force-couple equivalent.
28Example 3.7 (Single Force System) Determine the force-couple equivalent at point A for the single force of 20 kN acting at point C on the bracket ABC.40o20 KN1.2 m2.3 mBAC
29Example 3.7 solution 40o 20 KN 1.2 m 2.3 m B A C Fy Fx MA = (20 sin 40 x 2.3) – (20 cos 40 x 1.2)= kNm Answer:11.18 kNm kN40 0 A
30Example 3.8 (Multiple force system) Another 30 kN horizontal force is added to Example 3.7 at point B. Determine the force-couple equivalent at point A.40o20 KN1.2 m2.3 mBAC30 KN
31Example 3.8 Solution 40o 20 KN 1.2 m 2.3 m B A C Fy Fx 30 KN Fy Fx 400 Rx = Fx = 20 cos= kN Ry = Fy = 20 sin 40= kNTherefore R = ( )= kNAnd tan = Ry = =Rx
32 MA = (20 sin 40 x 2.3) – (20 cos 40 x 1.2) + (30 x 0) 20 KN1.2 m2.3 mBACFyFx30 KNExample 3.8 Solution MA = (20 sin 40 x 2.3) – (20 cos 40 x 1.2) + (30 x 0)= kNmAnswer:47.11 kN11.18 kNmEnd of Chapter 3