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Engineering Mechanics

Engineering Mechanics . Chapter 3 : Moments and Couples. Objectives. 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 ;

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Engineering Mechanics

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  1. Engineering Mechanics Chapter 3 : Moments and Couples

  2. Objectives • 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.

  3. Introduction • 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.

  4. P Point of Application B A P Line of action Principle 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. Principle 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.

  5. F d A Moment of A Force • The moment Mof a force Fabout a fixed point A is defined as the product of themagnitude of force F and the perpendicular distance d from point A to the line of actionof force F. MA = F x d Where force F is in newtons, N And distance d is in meters, m Thus moment MAis in newton-meter, Nm.

  6. Moment Sign Convention Anti-clockwise : + VE(Counter-clockwise) Clockwise : - VE

  7. Example 3.1 • Calculate the moment of the 500 N force about the point A as shown in the diagram. 500 N A 1.5 m

  8. 500 N Example 3.1Solution 1.5 m A Since the perpendicular distance from the force to the axis point A is 1.5 m, from MA = F x d MA = - 500 x 1.5 = - 750 Nm = 750 Nm

  9. 500 N A 1.5 m 60o Example 3.2 • Calculate the moment about point A caused by the 500 N force as shown in the diagram.

  10. 500 N 1.5 m 600 A 600 d Line of action B Example 3.2Solution: A 500 N 1.5 m 60o MA = F x d = -500 x AB = -500 x 1.5 sin 60 = -649.5 Nm = 649.5 Nm

  11. Addition 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.

  12. Example 3.3 Determine the resulting moment about point A of the system of forces on bar ABC as shown in the diagram. 500 N 800 N 1.5 m 0.5m 60 0 A B C

  13. 500 N 1.5 m 0.5 m 800 N A C 600 Line of Action D 500 800 N Example 3.3 Solution: 1.5 m 0.5 m 60 0 A B C AD = (1.5 + 0.5) sin 60 = 2.0 sin 60 = 1.732 m MA = (-500 x 1.5) + (-800 x 1.732) = - 750 – 1385.6 = - 2135.6 Nm = 2135.6 Nm

  14. Varignon’s Theorem • Varignon’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.

  15. Example 3.4 Re-Calculate Example 3.3 Above Employing Varignon’s Theorem. 500 N 800 N 1.5 m 0.5m 60 0 A B C

  16. Fy 500 N 800 N 1.5 m 0.5m 600 A B C Fx Fy = 800 sin 600 600 = 800 cos 600 Fx 500 N 800 N Example 3.4Solution 1.5 m 0.5m 60 0 A B C MA = (-500 x 1.5) + (- 800 sin 60 x 2) = -2135.6 Nm = 2135.6 Nm * Same answer as in example 3.3

  17. Example 3.5 Determine the resulting moment about point A for the system of forces acting on the plate ABCD as shown in the diagram. 15 cm 20 N D 45o C 8cm 30o 35 N A B 80o 10.38 cm 45 N

  18. 15 cm 35 N 45o D 20 N C 8cm 30o A B 80o 10.38 cm 45 N Fx Fy 800 45 N Fx Fy 35 N 300 Example 3.5Solution 45 N force: Fx = 45 cos 80 = 7.814 N Fy = 45 sin 80 = 44.32 N 35 N force: Fx = 35 cos 60 = 17.5 N Fy = 35 sin 60 = 30.31 N

  19. 20 N Fy 450 Fx 20 N 0.15 m Example 3.5Solution 60o 45o D C 0.08m A B 80o 35 N 0.1038 m 45 N 20 N force: Fx = 20 cos 45 = 14.14 N Fy = 20 sin 45 = 14.14 N MA = (- 45 sin 80 x 0.1038) + (- 35 cos 60 x 0.08) + (20 cos 45 x 0.08) + (20 sin 45 x 0.15) = - 4.591 – 1.4 + 1.131 + 2.121 = - 2.74 Nm = 2.74 Nm

  20. d d F F   F F 3.6 Couples A 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.

  21. What 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.

  22. F d1 d A d2 B F Magnitude of a Couple Consider a light bar acted upon by a couple as shown :- The moment of the couple about A is MA = + ( F x d2 ) - ( F x d1 )  MA = F x ( d2 - d1 ) MA = F x d What is the total moment of the couple about point B ?

  23. F F F F From above, we see that : The couple’s moment about any pivot point is equal to F x d A couple has the same moment about all points on a body.

  24. 2 KN 80 Nm 30o 120 Nm 1.5 m A 3 KN B 2.5 m Example 3.6 A light bracket ABC is subjected to two forces and two couples as shown. Determine the moment at (a) point A and (b) point B.

  25. 2 KN 80 Nm 30o 120 Nm 1.5 m A 3 KN B 2.5 m Example 3.6Solution: a)  MA = (2000 cos 60 x 2.5) – (2000 sin 60 x 1.5) + ( 3000 x 0) + 120 – 80 = 2500 – 2598 + 0 +120 - 80 = - 58 Nm = 58 Nm (b)     MB = (2000 cos 60 x 0) – (2000 sin 60 x 1.5) + (3000 x 0) + 120 – 80 = -2558 Nm = 2558 Nm

  26. 3.7 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? F d A B

  27. F F d A B A B MA = F x d F F d A B F The above shows how a force can be replaced by a force-couple equivalent. Force Couple Equivalent

  28. 20 KN 40o C 1.2 m A B 2.3 m Example 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.

  29. Fy 20 KN 40o C Fx 1.2 m A B 2.3 m Example 3.7solution  MA = (20 sin 40 x 2.3) – (20 cos 40 x 1.2) = 11.18 kNm Answer: 11.18 kNm 20 kN 40 0   A

  30. 20 KN 40o C 1.2 m A 30 KN B 2.3 m Example 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.

  31. Fy 20 KN 40o C Fx 1.2 m A 30 KN B 2.3 m Fy 400 Fx Example 3.8Solution Rx =  Fx = 20 cos 40 + 30 = 45.32 kN Ry =  Fy = 20 sin 40 = 12.86 kN Therefore R =  (45.322 + 12.862) = 47.11 kN And tan  = Ry = 12.86 = 15.84 0 Rx 45.32

  32. Fy 20 KN 40o C Fx 1.2 m A 30 KN B 2.3 m Example 3.8Solution  MA = (20 sin 40 x 2.3) – (20 cos 40 x 1.2) + (30 x 0) = 11.18 kNm Answer: 47.11 kN 11.18 kNm 15.84 0 End of Chapter 3

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