Solution manual for mechanisms and machines kinematics dynamics and synthesis 1st edition by stanisic

Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ Chapter Introduction - Solutions Problem 1.1 1111111 0000000 0000000 1111111 0000000 1111111 N =6 P1 = P2 = F = 3(N − 1) − 2P1 − P2 = 3(6 − 1) − 2(7) − = P1 + P1 P1 P1 P1 P 11111111111111111111 00000000000000000000 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 drawing 015.soltn P1 11111111 00000000 00000000 11111111 00000000 11111111 00000000 11111111 If you fix (freeze) the P1 joint between bodies and then cannot roll on 4, so and cannot move and M = F in this mechanism If cannot move, neither or will move Since only one P1 was eliminated to make this mechanism into a structure we have M = M and F in this mechanism Problem 1.2 If you fix (freeze) the P1 joint between bodies and then and cannot move If cannot move, neither or will move Since only one P1 was eliminated to make this mechanism into a structure we have M = M and F in this mechanism Full file at https://TestbankDirect.eu/ c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ P1 N =6 P1 = P2 = F = 3(N − 1) − 2P1 − P2 = 3(6 − 1) − 2(7) − = 11 00 00 11 00 11 00 111 00 11 00 11 00 11 P1 , P1 P1 , P1 P1 P1 111111111 000000000 000000000 111111111 000000000 111111111 fig001.soltn Problem 1.3 0 1 11 1 1111111 0000000 N =5 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 P1 = 0000000 1111111 0000000 1111111 0000000 0000000 1111111 P11111111 P =1 0000000 1111111 0000000 1111111 0000000 1111111 F = 3(N 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 0000000 1111111 P1111111 0000000 1111111 0000000 1111111 P1 + P2 P1 P1 − 1) − 2P1 − P2 = 3(5 − 1) − 2(5) − = fig 005.soltn If you freeze the P1 joint between and 1, cannot rotate without breaking the slipping contact between and If and not move, then and are fixed in place Thus the mechanism has become a structure and M = M = F in this mechanism Problem 1.4 N =3 P1 = P2 = F = 3(N − 1) − 2P1 − P2 = 3(3 − 1) − 2(2) − = 0000 1111 1111 0000 0000 1111 0000 1111 0000 1111 0000 11111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000P1 1111 P2 P1 111 000 000fig008a.soltn 111 If you freeze the P1 joint between and 3, then can no longer move and maintain a rolling contact with So M = and M = F in this mechanism c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ Problem 1.5 P1 P1 11111 00000 00000 11111 11111 00000 P2 N =4 P1 = 11111111111 00000000000 00000000000 11111111111 00000000000 11111111111 P2 = 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 F = 3(N − 1) − 2P1 − P2 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 = 3(4 − 1) − 2(3) − = 00000000000 11111111111 00000000000 11111111111 P 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 fig151.soltn P1 If we freeze the P1 joint between and 2, then cannot move without breaking (violating) one of the two slipping contacts between and Hence this makes the mechanism into a structure and M = M = F in this mechanism Problem 1.6 P1 P1 P1 1111 0000 0000 1111 00001 1111 N =4 + P1 = P2 = P2 F = 3(N − 1) − 2P1 − P2 = 3(4 − 1) − 2(3) − = 11111111111111111 00000000000000000 00000000000000000 11111111111111111 fig093a.soltn 00000000000000000 11111111111111111 00000000000000000 11111111111111111 If we freeze the P1 joint between and 1, can still spin and roll against 2, which slips against If we then freeze the P1 joint between and 4, can no longer roll on and maintain contact with so the system is a structure and M = M = F for this mechanism c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ Problem 1.7 0 1 1 1 1 1 1 1 1 1 1 1 1 1 P1 1 1 1 11 fig107a.soltn P2 + 111 000 000 111 000 111 + P1 N =3 P1 = P2 = F = 3(N − 1) − 2P1 − P2 = 3(3 − 1) − 2(2) − = If you freeze the P1 joint between and 2, can no longer move and the system is a structure so M = and M = F for this mechanism Problem 1.8 P1 P1 P1 1111 0000 0000 1111 0000 1111 111111111111111111111 000000000000000000000 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 P1 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 N =4 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 fig025.soltn 000000000000000000000 111111111111111111111 P1 = 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 P2 = F = 3(N − 1) − 2P1 − P2 = 3(4 − 1) − 2(4) − = If you freeze the P1 joint between and 2, can no longer roll on and we have a structure, so M = and M = F in this case c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ Problem 1.9 P1 P1 P2 111 000 000 111 N =5 P1 = 5 P2 = P1 F = 3(N − 1) − 2P1 − P2 = 3(5 − 1) − 2(5) − = P1 111 000 000 111 000 111 P1 fig273.soltn If we freeze the P1 joint between and 2, then the four bar formed by links 1, 2, and does not move so can no longer move and M = M = F for this mechanism Problem 1.10 P1 111 000 000 111 N =6 P1 , P1 P1 = P1 + P1 , P1 P2 = F = 3(N − 1) − 2P1 − P2 = 3(6 − 1) − 2(7) − = P2 00000000000 11111111111 fig274.soltn 11111111111 00000000000 00000000000 11111111111 P1 1111 0000 0000 1111 0000 1111 This mechanism is already a structure so M = To see this consider the four bar mechanism of links 1, 2, and which moves with degree-of-freedom We see that as this four bar moves, gets get pulled and it slips (drags) along The problem is that as makes this motion cannot maintain a sliding contact with it So there is no relative movement between and Consequently the four bar mechanism does not move and the mechanism does not move M = F for this mechanism c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ Problem 1.11 P1 111111111 000000000 000000000 111111111 000000000 111111111 P1 P1 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 N =4 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 P1 = 000000000 111111111 fig275.soltn P2 = 1111 0000 0000 1111 P1 P1 111 000 000 111 000 111 F = 3(N − 1) − 2P1 − P2 = 3(4 − 1) − 2(5) − = −1 Link cannot be rotated without breaking its contact with 1, therefore 1, and are a structure, so cannot move either In fact, will either be too long or too short, making this system a statically indeterminate structure Since no P1 joint needs to be frozen to make this a structure, M = In this mechanism M = F and M > F because the structure is statically indeterminate Problem 1.12 P1 P1 111 000 000 111 111111111 000000000 000000000 111111111 000000000 111111111 000000000 111111111 P1 P1 000000000 111111111 N =4 000000000 111111111 000000000 111111111 000000000 111111111 P1 = 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 P2 = 000000000 111111111 000000000 111111111 fig276.soltn 000000000 111111111 000000000F = 3(N − 1) − 2P1 − P2 = 3(4 − 1) − 2(4) − = 111111111 Freezing the P1 joint between and makes the mechanism a structure and so M = M = F Problem 1.13 111 000 000 111 000 111 000 111 000 111 000 111 000 111 P2 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 + P1 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 P1 0000000 0000000 1111111 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 P2 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 P1 111 000 000 111 000fig277.soltn 111 N =4 P1 = P2 = F = 3(N − 1) − 2P1 − P2 = 3(4 − 1) − 2(3) − = If you freeze the P1 joint between and 2, can no longer move and cannot roll on and maintain a slipping contact with So M = and M = F c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ Problem 1.14 P1 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 N =4 P1 = P1 P2 = F = 3(N − 1) − 2P1 − P2 = 3(4 − 1) − 2(4) − = P1 P1 1111 0000 0000 1111 0000 1111 fig011a.soltn If you freeze the P1 joint between and 1, then and cannot move without breaking its rolling contact with 4, so the system becomes a structure and M = M = F Problem 1.15 N =4 P1 = P2 = F = 3(N − 1) − 2P1 − P2 = 3(4 − 1) − 2(4) − = P1 P1 P1 1111 0000 0000 1111 0000 1111 P1111 000 000 111 000 111 0001 111 fig011b.soltn If you freeze the P1 joint between and 1, then and cannot move without breaking its rolling contact with 4, so the system becomes a structure and M = M = F c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ Problem 1.16 P1 111 111 000 000 000111 111 000 000111 111 000 000 111 000 111 000 111 000 111 000 111 000 111 000111 111 000 N =3 P1 = P2 = F = 3(N − 1) − 2P1 − P2 = 3(3 − 1) − 2(2) − = P2 P1 1111 0000 0000 1111 0000 1111 fig011c.soltn If you freeze the P1 joint between and 2, can no longer rotate without violating its slipping contact with 2, so the mechanism becomes a structure and M = M = F Problem 1.17 P2 P1 1111 0000 0000 1111 0000 1111 00001 1111 P1 111 000 000 111 000fig011d.soltn 111 N =3 P1 = P2 = F = 3(N − 1) − 2P1 − P2 = 3(3 − 1) − 2(2) − = If you freeze the P1 joint between and 2, can no longer rotate without violating its slipping contact with 2, so the mechanism becomes a structure and M = M = F c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ Problem 1.18 P1 N =5 P1 = P2 = P1 P1 F = 3(N − 1) − 2P1 − P2 = 3(5 − 1) − 2(5) − = P1 P1 111111111111111111 000000000000000000 000000000000000000 111111111111111111 fig528a.soltn If you freeze the P1 joint between and 2, then can still roll against while the ”dyad” formed by and will extend or contract to reach the connection point between and 5, so it is still moveable If you now freeze the P1 joint between and 1, then the system becomes a structure, so M = M = F Problem 1.19 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 P1 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 N =3 P1 = P2 = F = 3(N − 1) − 2P1 − P2 = 3(3 − 1) − 2(3) − = P1 111111111111111111111111 000000000000000000000000 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 P1 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 000000000000000000000000 111111111111111111111111 fig154.soltn 000000000000000000000000 111111111111111111111111 The system is moveable If you push to the left, you see that will move up If you freeze the P1 joint between and 1, then is now longer moveable, so M = M = F , and M > F The explanation for this is as follows The development of Gruebler’s Criterion assumes that every body in the system has the potential of a rotational degree-of-freedom But, when the system consists entirely of sliding joints, as in this case, this rotational degree-of-freedom is not possible Gruebler’s Criterion as developed is not applicable to this system Problem 1.20 The system is moveable, and it moves the same way as the prior problem If you push to the left will slide up If you freeze the P1 joint between and then the system is a structure so M = In this case M = F In this example there are joints besides sliding joints, so rotational degrees-of-freedom are possible and Gruebler’s Criterion as developed is applicable c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ 11111 00000 00000 11111 00000 11111 00000 11111 P1 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 111111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 N =3 P1 = P2 = F = 3(N − 1) − 2P1 − P2 = 3(3 − 1) − 2(2) − = P2 11111111111111111111111 00000000000000000000000 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 P1 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 fig153.soltn 10 c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ Problem 1.21 In the mechanism shown below, what should the questioned joints be in order for the mechanism to have F =1? P 00 11 11 00 00 11 001 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 P1 , P1 P1 P1 ? + + 111111111111111 000000000000000 000000000000000 111111111111111 000000000000000 111111111111111 fig002.soltn ? 1111111111111111 0000000000000000 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 The questioned joints are point contacts, so they can only of rolling (P1 ) or slipping (P2 ) Not counting those two joints we have, N = 6, P1 = and P2 = 0, so F = 3(N − 1) − 2P1 − P2 = 3(6 − 1) − 2(5) − = In order that the mechanism have F = 1, more degrees of freedom must be removed, which means that both of the unknown contacts have to be P1 contacts, i.e rolling contacts 11 c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ Problem 1.22 Give all the kinematic inversions of the mechanism shown in figure 1.31 a.) 12 0 1 11 1 11 00 00 11 00 11 00 11 00 11 00 11 00 11 0 1 21 1 0 1 21 1 11 00 00 11 00 11 004 11 00 11 00 11 00 11 111 000 000 111 1111 0000 0000 1111 0 04 1 1 5 1 279a.soltn The fifth kinematic inversion is the mechanism as shown in figure 1.31 No need to redraw it 12 c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ Problem 1.23 Give the mechanism which results from figure 1.30 c.) when link is grounded Determine the theoretical dof (F) and mobility (M) of that mechanism P1 P1 P1 1111 0000 0000 1111 N =6 P1 1111 0000 0000 1111 0000 1111 P1 = P2 = P1 F = 3(N − 1) − 2P1 − P2 = 3(6 − 1) − 2(6) − = P1 fig278.soltn If we freeze the P1 joint between and 6, is immobilized, but the rest of the mechanism still moves If we continue and freeze the P1 joint between and 2, is immobilized, but 1, and continue to be moveable If we then freeze the P1 joint between and 2, then and are also immobilized and the system becomes a structure Thus M = and M = F Problem 1.24 The figure below shows a pair of gears in mesh The smaller gear is referred to as the “pinion” and the larger gear is referred to as the “gear.” Typically, the pinion is the driver and the gear is driven Since the pinion is smaller than the gear, the gear rotates slower than the pinion, but the torque is increased Generally speaking, that is the purpose of a gear pair, to reduce speed and to increase torque This is because most prime movers (motors, engines etc.) run at high speeds but produce relatively low torques You will see all this later when we study gears What type of joint must exist between these two gear teeth so that the system is movable, with one dof? P1 P1 pinion 1111 0000 0000 1111 0000 1111 gear 111 000 000 111 000 111 fig120a.soltn A B ? If we neglect the contact in question, we have N = 3, P1 = 2, P2 = and then F = 3(N − 1) − 2P1 − P2 = 3(3 − 1) − 2(2) − = so in order that F = 1, the contact between a pair of gear teeth must be a slipping contact 13 c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ 14 c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ .. .Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ P1 N =6... or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ Problem... or in part Full file at https://TestbankDirect.eu/ Solution Manual for Mechanisms and Machines Kinematics Dynamics and Synthesis 1st Edition by Full file at https://TestbankDirect.eu/ Problem