1. Ken YoussefiMechanical Engineering Dept. 1 Force Analysis Kinetic analysis has to be performed to design the shape and thickness of the links, the joints and to determine the required input torque/force and the resulting output torque/force. Known: The length of all links; r 1, r 2, r 3, and r 4 The results of position, velocity, and acceleration analysis; θ 3, θ 4,  3,  4, and α 3, α 4 Input information; θ 2,  2, α 2, and T 2 (torque) All external loads acting on the links Unknown: Forces acting on all joints; F O 2, F A, F B, and F O 4 Inertia forces acting on all links; F I 2, F I 3, F I 4, Inertia torques acting on all links; T I 2, T I 3, T I 4, The input torque (or force) needed to obtain the required output torque (or force) O4O4 Force analysis of a 4-Bar A B O2O2 2 3 4 ω2ω2 α2α2

2. Force Analysis Ken YoussefiMechanical Engineering Dept. 2 Assume the input torque is applied in the opposite direction of the motion to stop the mechanism from moving. This results the mechanism to be in equilibrium Graphical method, construct the force polygon Analytical method, matrix method (Draw FBD of each link and use static equilibrium condition)

3. Ken YoussefiMechanical Engineering Dept. 3 Force Analysis – Static Equilibrium Static equilibrium – 2D case ∑ F x = 0, ∑ F y = 0, ∑ M = 0 Two force members – only two forces acting, one at each joint. A rigid link acted on by two forces is in static equilibrium only if the two forces are collinear and have the same magnitude in opposite direction F 1 = - F 2 X X

4. Ken Youssefi+Mechanical Engineering Dept. 4 Force Analysis – Static Equilibrium Three force members – only three forces act on a link. Either three joint forces, or two joint forces and an external or inertia or gravitational force. The link is in static equilibrium only if the force-vectors add up to zero and they pass through the same point (summation of moment about that point is zero).

5. Ken YoussefiMechanical Engineering Dept. 5 Force Analysis – Graphical Method O2O2 O4O4 2 3 4 P = 120 lb A B C O 2 A = 3, AB = O 4 B = 6, O 4 C = 4.5, O 2 O 4 = 2.4 Find the input torque, T 2, to maintain static equilibrium at the position shown or what should be the input torque to exert 120 lb of force for the position shown. Use graphical method, construct force polygon. AB is a two force member, F 34 (force of link3 on link 4) has to be collinear to link 3. M Parallel to link3 F 34 F 14 Direction of F 14 has to pass through point M Free-Body Diagram of link 4 B O4O4 P = 120 lb

6. Ken YoussefiMechanical Engineering Dept. 6 Force Analysis – Graphical Method B O4O4 P = 120 lb F 34 F 14 M Force polygon Parallel to F 34 direction P = 120 lb Draw and scale force P 1 - Parallel to F 14 direction Draw two lines parallel to F 14 and F 34 directions from the two ends of vector P 2 - F 34 = 113.52 lb. Locate the intersection, complete the polygon and scale the force F 34 3 -

7. Ken YoussefiMechanical Engineering Dept. 7 Force Analysis – Graphical Method Free-Body Diagram of link 3 3 A B F 43 F 23 = F 43 = 113.52 T 2 = (F 32 )(d) = 113.52 x 2.428 = 275.6 in-lb (ccw) Free-Body Diagram of link 2 Draw and scale link 2, O 2 A = 3 Scale the perpendicular distance from O 2 to the force F 32 d = 2.428 A O2O2 2 F 32 = 113.52 F 12 = 113.52 lb

8. Ken YoussefiMechanical Engineering Dept. 8 Force Analysis – Graphical Method Principle of Superposition Principle of superposition can be used to solve problems involving linear systems by considering each of the inputs to the system separately. And then, combine the individual results to obtain the total response of the system. In linear systems the output (response) is directly proportional to the input, linear relationship between the output and the input exist. O2O2 O4O4 2 3 4 A B C D P Q O2O2 O4O4 2 3 4 A B C O2O2 O4O4 2 3 4 A B D PQ = + T 2 = (T 2 ) Due to force P + (T 2 ) Due to force Q

9. Ken YoussefiMechanical Engineering Dept. 9 Dynamic Force Analysis Inertia forces and D’Alembert’s Principle An unbalanced set of forces acting on a rigid body = The translation and rotation caused by the unbalanced forces Resultant force R = ∑ F R = (mass)(linear acceleration of the center of gravity), R = (m)a G M G = (area moment of inertia)(angular acceleration) = I G α = (R)( h )

10. Ken YoussefiMechanical Engineering Dept. 10 Dynamic Force Analysis D’Alembert’s Principle R - (m)a G = 0, inertia force = F I = - ma G The inertia force has magnitude of ma G and it acts in the opposite direction of linear acceleration. It stops motion. The inertia torque has a magnitude of (F I )(h) and it acts in the opposite direction of angular acceleration. M G - I G α = 0, inertia torque = T I = - I G α = - R h = F I h FIFI TITI

11. Ken YoussefiMechanical Engineering Dept. 11 Inertia Force and Torque - Example O2O2 2 A Input info: θ 2 = 144 o, ω 2 = 12 rad/sec (ccw), α 2 = 60 rad/sec 2 ( ccw) Geometry: I 2 =.04 lb-ft-sec 2, W 2 = 6 lb. O 2 A = 12 in., O 2 g 2 = 6.6 in. g2g2 Determine the inertia force and the inertia torque of link 2, magnitude and direction. Show the force and the torque on the link. First calculate the linear acceleration of the center of gravity of link 2, a g 2 Velocity of g 2, V g 2 = r g 2 ω 2 i e i θ 2 Acceleration of g 2, a g 2 = r g 2 α 2 i e i θ 2 - r g 2 (ω 2 ) 2 e i θ 2 r g 2 = r g 2 e i θ 2 Position of g 2, θ2θ2

12. Ken YoussefiMechanical Engineering Dept. 12 Inertia Force and Torque - Example a g 2 = r g 2 e i θ 2 [ α 2 i - ( ω 2 ) 2 ] a g 2 = 6.6 e i (144) [ 60 i - (12) 2 ] a g 2 = 6.6 [ cos (144) + i sin (144) ] (60 i – 144) a g 2 = 536.1 – 879 i a g 2 = √ ( 536.1) 2 + (879) 2 = 1029.6 in/sec 2 β 2 = tan -1 (-879/536.1) = -58.62 O2O2 2 A g2g2 ag2ag2 -58.62 Inertia force of link 2 = F I 2 = m 2 a g 2 F I 2 = (6/386)(1029.6) = 16 lb h 2 = I2 α2I2 α2 FI 2FI 2.04x12x60 = 16 = 1.8 “ T I 2 = F I 2 x h 2 = 16 x 1.8 = 28.8 in-lb

13. Ken YoussefiMechanical Engineering Dept. 13 Inertia Force and Torque - Example The inertia force is in the opposite direction of the acceleration of the center of gravity A O2O2 g2g2 ag2ag2 -58.62 h2h2 FI 2FI 2 Inertia force an torque acting on link 2 O2O2 A g2g2 F I 2 = 16 α2α2 180 – 58.62 = 121.38 T I 2 = 28.8 O2O2 A g2g2 ag2ag2 -58.62 h2h2 FI 2FI 2 α2α2 The inertia torque is in the opposite direction of α 2

14. Ken YoussefiMechanical Engineering Dept. 14 Force Analysis – Analytical, Matrix Method g2g2 3 α3α3 g3g3 g4g4 ag3ag3 ag2ag2 A B O2O2 O4O4 2 4 α2α2 α4α4 ag4ag4 Four bar mechanism FI4FI4 FI3FI3 Link 3Link 4 F I 3 = m 3 a g 3 F I 4 = m 4 a g 4 FI2FI2 F I 2 = m 2 a g 2 Link 2 Inertia force Inertia torque T I 2 = F I 2 h 2 T I 3 = F I 3 h 3 T I 4 = F I 4 h 4 ccwcw h2h2 h3h3 h4h4 Torque arm h 2 = ( I 2 α 2 )/F I 2 h 3 = ( I 3 α 3 )/F I 3 h 4 = ( I 4 α 4 )/F I 4

15. Ken YoussefiMechanical Engineering Dept. 15 Force Analysis – Analytical, Matrix Method g2g2 3 α3α3 g3g3 g4g4 A B O2O2 O4O4 2 4 α2α2 α4α4 FI2FI2 FI4FI4 FI3FI3 TI4TI4 TI3TI3 TI2TI2 T2T2 Show all inertia forces and torques on the four bar mechanism Show all external and gravitational forces, location and magnitude of force P is given. P W4W4 W3W3 W2W2

16. Ken YoussefiMechanical Engineering Dept. 16 Force Analysis – Analytical, Matrix Method g2g2 A O2O2 2 FI2FI2 TI2TI2 T2T2 W2W2 (F I 2 ) y (F I 2 ) x AxAx AyAy O 2x O 2y Free body diagram of link 2. Assume the direction of forces at joints A and O 2 O 2 A = r 2, O 2 g 2 = r g 2 θ2θ2 - A x (r 2 – r g 2 )sinθ 2 + A y (r 2 – r g 2 )cosθ 2 + O 2x (r g 2 )sinθ 2 – O 2y (r g 2 )cosθ 2 + T 2 – T I 2 = 0 (3) ∑ M g = 0 A y + O 2y + (F I 2 ) y - W 2 = 0 (2) ∑ F y = 0 A x + O 2x + (F I 2 ) x = 0 (1) ∑ F x = 0 - A x C 1 + A y C 2 + O 2x C 3 – O 2y C 4 + T 2 – T I 2 = 0

17. Ken YoussefiMechanical Engineering Dept. 17 Force Analysis – Analytical, Matrix Method 3 g3g3 A B FI3FI3 TI3TI3 W3W3 (F I 3 ) y (F I 3 ) x θ3θ3 AB = r 3, Ag 3 = r g 3 AxAx AyAy ByBy BxBx Free body diagram of link 3. Direction of the force at A has to be opposite of the direction assumed on link 2. Assume the direction of the force at joint B. - A x + B x - (F I 3 ) x = 0 (4) ∑ F x = 0 - A y + B y - (F I 3 ) y - W 3 = 0 (5) ∑ F y = 0 - B x (r 3 – r g 3 )sinθ 3 + B y (r 3 – r g 3 )cosθ 3 - A x (r g 3 )sinθ 3 + A y (r g 3 )cosθ 3 - T I 3 = 0 (6) ∑ M g = 0

18. Ken YoussefiMechanical Engineering Dept. 18 Force Analysis – Analytical, Matrix Method g4g4 B O4O4 4 FI4FI4 TI4TI4 P W4W4 Free body diagram of link 4. Direction of the force at B has to be opposite of the direction assumed on link 3. Assume the direction of the force at joint O 4. O 4x O 4y ByBy BxBx O 4 B = r 4, O 4 g 4 = r g 4 θ4θ4 (F I 4 ) x (F I 4 ) y + O 4x - B x - (F I 4 ) x + P = 0 (7) ∑ F x = 0 - O 4y - B y + (F I 4 ) y - W 4 = 0 (8) ∑ F y = 0 B x (r 4 – r g 4 )sinθ 4 + B y (r 4 – r g 4 )cosθ 4 + O 4x (r g 4 )sinθ 4 - O 4y (r g 4 )cosθ 4 + T I 4 - pd= 0 (9) ∑ M g = 0 d d = (O 4 C - r g 4 )sin θ 4 C

19. Ken YoussefiMechanical Engineering Dept. 19 Force Analysis – Analytical, Matrix Method Nine equations nine unknowns; A x, A y, B x, B y, O 2x, O 2y, O 4x, O 4y and T 2. AxAx - (F I 2 ) x 100000100 A x + O 2x = - (F I 2 ) x (1) AyAy - (F I 2 ) y + W 2 010001000 A y + O 2y = - (F I 2 ) y + W 2 (2) BxBx (3) - A x C 1 + A y C 2 + O 2x C 3 – O 2y C 4 = - T 2 + T I 2 -C 1 C2C2 100-C 4 C3C3 00 + T I 2 Coefficient matrix Unknown vector Constant vector ByBy O 2x O 4x O 2y O 4y T2T2 - 100100000 (F I 3 ) x 00000010 (F I 3 ) y + W 3 -C 7 C8C8 0-C 5 000C6C6 0 TI3TI3 00000000 0000000 000C9C9 C 11 00C 10 -C 12 (F I 4 ) x - P - (F I 4 ) y + W 4 Pd - T I 4 =