Problem 1 16 Problem 2 24 Problem 3 40 Problem 4 40 RESULT 120

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1 ME 301 THERY F MHINES I SEN MITERM EXMINTIN Section 01 : M. Kemal Özgören Name Last Name Signature Problem 1 16 Problem 2 24 Problem 3 40 Problem 4 40 RESULT 120 Tpe of Eamination: losed ook. uration: 150 minutes RELEVNT FRMULS egree of Freedom Formulas F 2( n 1) j F 3( n 1) (2 j j ) F 6( n 1) (5 j 4 j 3 j 2 j j ) i j F ( n 1 j) f i 1 i j j j j j j ; Number of Independent Loops L j n 1 Necessar and Sufficient Number of Joint Variables V 2L F for 3 Some Trigonometric Formulas di d j dk 2d jdk cos i (sin i)/ di (sin j)/ d j (sin k )/ dk sin( ) sin cos cos sin cos( ) cos cos sin sin atan ( ksin, kcos ) with k M. K. Özgören

2 = b 3 = b 4 = d 1 s 23 = θ 14 PRLEM 1 onsider the mechanism shown in Figure 1. Figure 1. Si-Link Mechanism Eplain wh the revolute joints at the points and have been selected as the closure joints. Eplain also wh the do not violate the connectivit of the open kinematic chain obtained upon their disconnection. In the figure, the joint variables appear as, θ 14,,, and s 23. Verif that the are assigned properl and their number is necessar and sufficient for the kinematic analsis of the mechanism. Justif that their subscripts are also assigned properl. SLUTIN to PRLEM 1 The revolute joints R 35 and R 34 at the points and are floating joints and the do not have an joint variables assigned. These are the reasons wh the have been selected as the closure joints. When these joints are both disconnected, the connectivit of the open kinematic chain is conserved because it is still possible to reach an link from an other link through a path that ma involve the fied link too. For this mechanism, which is planar, λ = 3, n = 6, j 1 = 7 (R 12, R 34, R 35, R 46 ; P 15, P 23, P 56 ), j 2 = 0; j = j 1 + j 2 = 7 Therefore, its degree of freedom is F = 3 (6 1) = 1 The number of independent loops (which is also equal to the number of closure joints) is L = j n + 1 = = 2 eing a planar mechanism, the necessar and sufficient number of joint variables required for its kinematic analsis is V = 2L + F = = 5 s for the subscripts, those of and θ 14 are justified because the angular joint variables are used conventionall to denote the orientation of the relevant link with respect to the base, i.e., link 1. The second subscript of is also justified because it is actuall associated with link 2, which is the swinging block pivoted to the base with R 12. The subscripts of and are justified because the denote the positions of links 5 and 6 with respect to the fied point of the base. Finall, the subscripts of s 23 are also justified because s 23 denotes the position of link 3 with respect to link 2. 2 M. K. Özgören

3 PRLEM 2 onsider again the mechanism shown in Figure 1. Identif the independent loops generated b reconnecting the disconnected closure joints. For each independent loop, write the loop closure equation first as a point-to-point vector equation and then as a comple number equation. c) Show that the corresponding scalar loop closure equations can be written as follows: (b 3 s 23 ) sin = (1) (b 3 s 23 ) cos = d 1 (2) + s 23 sin = (3) b 4 sin θ 14 s 23 cos = d 1 (4) SLUTIN to PRLEM 2 Loop 1 is. It involves the links 1, 2, 3, and 5. Loop 2 is. It involves the links 1, 2, 3, 4, and 6. Loop 1 is generated b reconnecting R 35 and keeping R 34 disconnected. That is, 3 = + 5 Loop 2 is generated b reconnecting R 34 and keeping R 35 disconnected. That is, 3 = The corresponding comple number equations are written as follows: c) Loop 1 (b 3 s 23 )e i(π/2++π) = id 1 + Loop 2 s 23 e i(π/2+) = id 1 + e iπ + b 4 e iθ 14 The preceding comple number equations can be simplified as shown below: Note that Loop 1 (b 3 s 23 )ie i = id 1 + Loop 2 s 23 ie i = id 1 + b 4 e iθ 14 e i = cos + i sin ie i = i cos sin Hence, the following scalar equations are obtained: (b 3 s 23 ) sin = (1) (b 3 s 23 ) cos = d 1 (2) + s 23 sin = (3) b 4 sin θ 14 s 23 cos = d 1 (4) 3 M. K. Özgören

4 PRLEM 3 onsider the scalar loop closure equations given in Problem 2. Let the specified input joint variable be. For this input, derive formulas for the other joint variables in the following order (not in another order): s 23,, θ 14, Identif the values of the closure sign variables (σ 1 and σ 2 ) for the configuration of the mechanism shown in Figure 1. Justif our identifications clearl. c) Sketch the same mechanism in its three other possible closures that correspond to the values of σ 1 and σ 2 different from those in Part b. If necessar, ou ma add hpothetical etensions to the links in drawing our sketches. Justif our sketches clearl. lso, do not forget to indicate the joint variables clearl on each sketch. SLUTIN to PRLEM 3 The squares of Equations (1) and (2) add up to Hence, (b 3 s 23 ) 2 = s d 1 (b 3 s 23 ) = σ 1 s d 1 ; σ 1 = ±1 s 23 = b 3 σ 1 s d 1 With found s 23, if s 23 b 3, Equations (1) and (2) lead to = atan 2 [ /(b 3 s 23 ), d 1 /(b 3 s 23 )] With found s 23 and, Equation (4) gives Hence, sin θ 14 = η 14 = (d 1 + s 23 cos )/b 4 2 cos θ 14 = ξ 14 = σ 2 1 η 14 ; σ 2 = ±1 θ 14 = atan 2 (η 14, ξ 14 ) Finall, with found s 23,, and θ 14, Equation (3) gives directl as = + s 23 sin In Figure 1, 0 < s 23 < b 3 and θ 14 is an acute angle with cos θ 14 > 0. Therefore, c) σ 1 = σ 2 = +1 The other three possible closures are shown in the net page. 4 M. K. Özgören

5 losure with σ 1 = +1 and σ 2 = 1: s 23 = θ 14 losure with σ 1 = 1 and σ 2 = +1: (Possible with a hpothetical etension of Link 3) θ 14 < 0 s 23 = losure with σ 1 = 1 and σ 2 = 1: (Possible with a hpothetical etension of Link 3) s 23 = θ 14 < 0 5 M. K. Özgören

6 PRLEM 4 For our convenience, the derivatives of the scalar loop closure equations given in Problem 2 are taken and arranged as follows: (b 3 s 23 )θ 12 cos s 23 sin = s 15 (5) (b 3 s 23 )θ 12 sin + s 23 cos = 0 (6) s 23 sin + s 23 θ 12 cos b 4 θ 14 sin θ 14 = s 16 (7) s 23 cos = s 23 θ 12 sin + b 4 θ 14 cos θ 14 (8) Let the specified input joint variable be. For this input, obtain the velocit influence coefficients so that the derivatives of the other joint variables are related to s 15 as follows: s 23 = g 23/15 s 15, θ 12 = g 12/15 s 15, θ 14 = g 14/15 s 15, s 16 = g 16/15 s 15 Identif the conditions on the joint variables that ma cause dead center positions for the input. iscuss whether these conditions are phsicall realizable or not. If the are realizable, eplain wh the cause dead center positions in phsical terms and also sketch the mechanism in those dead center positions. SLUTIN to PRLEM 4 Equations (5) and (6) can be combined into the following matri equation. [ (b 3 s 23 ) cos sin ] [ θ (b 3 s 23 ) sin + cos 12 s 23 The determinant of the coefficient matri is = b 3 s 23 ] = [ 1 0 ] s 15 (9) If 0, i.e., if s 23 b 3, Equation (9) can be solved as shown below. Hence, [ θ 12 ] = 1 s 23 [ cos sin ] [ 1 (b 3 s 23 ) sin (b 3 s 23 ) cos 0 ] s 15 θ 12 = cos s 15 g 12/15 = cos (10) s 23 = (sin )s 15 g 23/15 = sin (11) With found θ 12 and s 23, if cos θ 14 0, Equation (8) gives θ 14 = cos s 23 s 23 sin θ 12 θ 14 = cos (sin )s 15 s 23 sin cos s 15 θ 14 = [ sin cos + s 23 sin cos b 4 ( ) cos θ 14 ] s 15 θ 14 = b 3 sin cos b 4 ( ) cos θ 14 s 15 g 14/15 = b 3 sin cos b 4 ( ) cos θ 14 (12) Finall, Equation (7) gives 6 M. K. Özgören

7 s 16 = s 23 sin + s 23 θ 12 cos b 4 θ 14 sin θ 14 s 16 = [g 23/15 sin + s 23 g 12/15 cos b 4 g 14/15 sin θ 14 ]s 15 g 16/15 = g 23/15 sin + s 23 g 12/15 cos b 4 g 14/15 sin θ 14 (13) If desired, the epression of g 16/15 can be worked out further as follows. g 16/15 = [ sin ] sin + s 23 [ cos ] cos b 4 [ b 3 sin cos b 4 ( ) cos θ 14 ] sin θ 14 g 16/15 = (b 3 s 23 ) sin sin + s 23cos cos + b 3 sin cos tan θ 14 g 16/15 = s 23 b 3 sin sin +b 3 sin cos tan θ 14 g 16/15 = s 23 cos θ 14 b 3 sin sin cos θ 14 +b 3 sin cos sin θ 14 ( ) cos θ 14 g 16/15 = s 23 cos θ 14 +b 3 sin (cos sin θ 14 sin cos θ 14 ) ( ) cos θ 14 g 16/15 = s 23 cos θ 14 +b 3 sin sin(θ 14 ) ( ) cos θ 14 (14) s noticed above, for the input, dead center positions occur if s 23 = b 3 and/or cos θ 14 = 0. ccording to Equation (2), the equalit s 23 = b 3 implies that d 1 = 0. Therefore, as long as d 1 0, which is alwas the case for a meaningful operation of the mechanism, the dead center position with s 23 = b 3 cannot occur. n the other hand, the dead center position with cos θ 14 = 0, i.e., with θ 14 = ±π/2, can occur if Link 4 is short enough so that b 4 < b 3. Such a dead center position with θ 14 = π/2 is illustrated in the sketch drawn below. The position illustrated below is indeed a dead center position for the input, because, as the illustration implies, an attempt to change the shown value of will tr to rotate Link 3 and thus tr to stretch or compress Link 4. However, such an attempt will be unsuccessful because Link 4 is practicall rigid. s 23 = θ 14 = π/2 7 M. K. Özgören