The driver can rotate by 360 but at the dead center

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Unformatted text preview: a double rocker with two fully rotating cranks. Choosing l , p or q as the frame results in a crank rocker or double rocker with one or zero rotating cranks, respectively. (c) l = 4.0 in s = 2.0 in p = 3.5 in q = 2.25 in s + l < p + q Grashof type 1 4.0 + 2.0 > 3.5 + 2.25 6.0 > 5.75 No link can rotate fully. The mechanism is a type 2 double rocker. - 40 - Problem 1.37 If the link lengths of a four-bar linkage are L1 = 1 mm, L2 = 3 mm, L3 = 4 mm, and L4 = 5 mm and link 1 is fixed, what type of four-bar linkage is it? Also, is the linkage a Grashof type 1 or 2 linkage? Answer the same questions if L1 = 2 mm. s + l < p + q Grashof type 1 s + l > p + q nonGrashof type 2 a) s = 1; l = 5; p = 3; q = 4 1 + 5 < 3 + 4 6 < 7 Grashof type 1 Since the shortest member is connected to the frame, the linkage is a crank rocker s + l < p + q Grashof type 1 s + l > p + q nonGrashof type 2 b) s = 2; l = 5; p = 3; q = 4 2 + 5 < 3 + 4 7 = 7 Transition linkage This is a transition linkage. The driver can rotate by 360˚, but at the dead center position, the linkage must be "helped" to continue the rotation. - 41 - Problem 1.38 You are given two sets of links. Select four links from each set such that the coupler can rotate fully with respect to the others. Sketch the linkage and identify the type of four-bar mechanism. a) L1 = 5”, L 2 = 8”, L 3 = 15”, L4 = 19”, and L5 = 28” b) L1 = 5”, L 2 = 2”, L 3 = 4”, L 4 = 3.5”, and L5 = 2.5” (a) Let: l = 28.0 in s = 5.0 in p = 19.0 in q = 15.0 in 19.0" Coupler 15.0" Frame 5.0" s + l < p + q Grashof type 1 5 + 28 < 19 + 15 ⇒ 33 < 34 28.0" Since the shortest link is the coupler, the mechanism is a type 1 double-rocker. (b) Let: l = 5.0 in s = 2.0 in p = 4.0 in q = 3.5 in 5.0" Coupler 2.0" 3.5" Frame 4.0" s + l ≤ p + q Grashof type 1 5 + 2 < 3.5 + 4 ⇒ 7 < 7.5 Since the shortest link is the coupler, the mechanism is a type 1 double-rocker. - 42 - Problem 1.39 The mechanisms shown below are drawn to scale. (a) Sketch kinematic schematics showing the relationships between the members and joints. (b) Determine the Grashof type of each four-bar linkage in each mechanism. (a) (b) (a) l = 0.94 in s = 0.26 in p = 0.56 in q = 0.89 in Driver 2 0.94" 1 C 4 B 0.56" Frame A D 3 0.26" s + l < p + q Grashof type 1 0.26 + 0.94 < 0.56 + 0.89 ⇒ 1.20 <1.45 0.89" (a) The mechanism is a crank rocker if link 1 or link 3 is fixed and link 2 is the driver. (b) Links 1-2-3-4 l = 1.59 in s = 0.31 in p = 1.13 in q = 1.41 in Links 4-5-6-1 a = 0.35 in b = 1.12 in c = 0.88 in - 43 - 0.31 3 B 1.41 1.59 A 1 Frame (b) 0.88 6 F 1.13 D C Driver 4 E 5 0.35 1.12 s +l < p + q Grashof type 1 Links 1-2-3-4 0.31 +1.59 < 1.13 + 1.41 ⇒ 1.90 < 2.54 Links 4-5-6-1 1.12 > 0.35 ⇒ b > a 1.12 − 0.35 < 0.88 ⇒ b − a < c 2 The mechanism containing links 1-2-3-4 is a crank-rocker mechanism because link 4 is the driver. The mechanism containing links 4-5-6-1 is a non Grashof, slider-crank mechanism. The crank cannot make a continuous rotation relative to the other links. - 44 - Problem 1.14 Determine the mobility and the number of idle degrees of freedom for each of the mechanisms shown. Show the equations used to determine your answers. Cam Contact Cam Contact Pin in Slot (a) (b) (c) 1 2 3 4 5 6 7 8 9 11 10 n = 11 j j = 15 M = 3(n − j − 1) + ∑ fi i =1 = 3(11 − 15 − 1) + 17 = − 15 + 17 = 2 Mobility = 2 Idle DOF = 1 n =5 j j =6 M = 6(n − j − 1) + ∑ fi i =1 (a) 2 1 5 4 3 (b) 1 2 3 4 5 6 9 7 11 8 = 6(5 − 6 − 1) + 4(3) + 2 + 1 = − 12 + 15 = 3 Mobility = 3 Idle DOF = 1 n = 11 j j = 14 M = 3(n − j − 1) + ∑ fi i =1 10 (c) = 3( 11 − 14 − 1) + 16 = − 12 + 16 = 4 Mobility = 4 Idle DOF = 1 - 45 - Problem 1.16 Determine the mobility and the number of idle degrees of freedom associated with each mechanism. Show the equations used to determine your answers. Cam joints Pin in Slot Sliders (a) (b) (c) Pin-in-Slot Joint Slider (d) n =5 j =6 Cam joints 2 3 1 4 5 (e) j i =1 M = 3( n − j − 1) + ∑ fi = 3(5 − 6 − 1) + 8 = − 6 + 8 = 2 Mobility = 2 Idle DOF = 1 (a) n =7 5 7 2 3 Pin in slot j =9 j i =1 6 4 M = 3( n − j − 1) + ∑ fi = 3( 7 − 9 − 1) + 10 = − 9 + 10 = 1 Mobility = 1 Idle DOF = 0 1 (b) - 46 - 7 4 5 3 2 1 Sliders n j 8 9 = 11 = 14 j i =1 6 M = 3(n − j − 1) + ∑ fi 11 10 (c) = 3( 11− 14 − 1) + 14 = − 12 + 14 = 2 Mobility = 2 Idle DOF = 0 5 4 3 8 6 7 n =8 j = 10 j i =1 M = 3(n − j − 1) + ∑ fi 1 2 = 3( 8 − 10 − 1) + 10 = − 9 + 10 = 1 Mobility = 1 Idle DOF = 0 (d) 7 6 5 Pin-in-slot joint Slider n =7 j =8 j i =1 M = 3(n − j − 1) + ∑ fi 4 2 1 3 = 3(7 − 8 − 1) + 9 = −6 + 9 = 3 Mobility = 3 Idle DOF = 0 (e) - 47 -...
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