上海交通大学：《Design & Manufacturing II and Project》课程教学资源（作业）07 solutions Homework

Homeworks-7 solution 1.The figure below shows a crimping tool. a.Find all instant-center locations. b.Determine the mechanical advantage for the position shown. Fnand 3/C Fhand 4.26 49° AB=0.80,BC=1.23.CD=1.55.AD=2.4 D all lengths in inches a)C=m(n-D=6 2 b) ll (2,4-1,4) 8 ×4.26=31.16 (2.4-(1,2) 2.Find all the instant centers of the linkages shown in the following figures. A 3 A 0 an 2 002 3 B 2 34 02 02 Assume rolling contact (a) (b)

Homeworks‐7 solution

23 002 24 13到元中一 Assume rolling contact Assume rolling contact (c) (d) (2.4@ > 14@ 4 02 B 23 03 02 A 3 2 Assume roll-slide contact (e) (f)

DESIGN OF MACHINERY SOLUTION MANUAL 7-15a-1 PROBLEM 7-15a Statement: The linkage in Figure P7-5c has the dimensions and coupler angle given below.Find aA and Ac for the position shown for V=10 in/sec and A=15 in/sec2 in the directions shown.Use the acceleration difference graphical method. Given: Link lengths and angles: Link 3(4 to B) b=1.8-in Coupler angle 03=128deg Slider 4 angle 04=59deg Coupler point: Distance A to C p=1.44-im Angle BAC δ3=-49deg Input slider motion VA=-10-imsec-1 A 15im-sec-2 Solution: See Figure P7-5c and Mathcad file P0715a. 1.In order to solve for the accelerations at points B and C,we will need 3.From the layout below and Problem 6-18, 03=128deg 03-13.288.ad sec Direction of AB 0 0.5 1 in Direction of ABAT Direction of ACAt 49.000r° 59.000 28.000 2.The graphical solution for accelerations uses equation 7.4: (Ap'+Ap")=(A+A ")+(Ap+Ap") 3.For point B,this becomes:Ag=A+(AB+AB"),where AA=15.000m-sec2 0:=0.deg ABAn=b-03 ABAn=317.828 insec-2 04BAn=03+180-deg 04B4n=308.000deg 4.Choose a convenient acceleration scale and draw the vectors with known magnitude and direction.From the origin,draw Aat an angle of 4 From the tip of A,draw AB"at an angle of ABAn Now that the vectors with known magnitudes are drawn,from the origin and the tip ofA",draw construction lines in the directions ofAg and AB,respectively.The intersection of these two lines are the tips of AB,and Ag

DESIGN OF MACHINERY SOLUTION MANUAL 7-15a-1 PROBLEM 7-15a Statement: The linkage in Figure P7-5c has the dimensions and coupler angle given below. Find 3 , AB , and AC for the position shown for VA = 10 in/sec and AA = 15 in/sec2 in the directions shown. Use the acceleration difference graphical method. Given: Link lengths and angles: Link 3 (A to B) b 1.8in Coupler angle 128deg Slider 4 angle 59deg Coupler point: Distance A to C p 1.44in Angle BAC 49deg Input slider motion VA 10in sec 1   AA 15in sec 2   Solution: See Figure P7-5c and Mathcad file P0715a. 1. In order to solve for the accelerations at points B and C, we will need 3 . From the layout below and Problem 6-18, 128deg  13.288 rad sec   49.000° Direction of AB 59.000° VA b Direction of ABAt 3 4 B 128.000° A 2 X A A 0 0.5 1 in Direction of ACAt Y C 2. The graphical solution for accelerations uses equation 7.4: (AP t + AP n ) = (AA t + AA n ) + (APA t + APA n ) 3. For point B, this becomes: AB = AA + (ABA t + ABA n ) , where AA 15.000in sec 2   AA 0deg ABAn b  2   ABAn 317.828 in sec 2   ABAn 180deg ABAn 308.000deg 4. Choose a convenient acceleration scale and draw the vectors with known magnitude and direction. From the origin, draw AA at an angle of AA. From the tip of AA , draw ABA n at an angle of ABAn. Now that the vectors with known magnitudes are drawn, from the origin and the tip of ABA n , draw construction lines in the directions of AB and ABA t , respectively. The intersection of these two lines are the tips of ABA t , and AB

DESIGN OF MACHINERY SOLUTION MANUAL 7-15a-2 0100IN/S/S m Acceleration Scale 9.126 8.638 5.From the graphical solution above, Acceleration scale factor ka-100.m dec AB=9.126ka AB=912.6insec-2 at an angle of 239 deg AB4=8.638ka ABA=863.8 insec-2 6.Calculate a using equation 7.6. ABAL a3= cw b a3=479.9rad-sec2 7.For point C,equation 7.4 becomes:Ac=A+(Ac+Ac"),where 2 ACAn P.03 4Ch=254.262im-sec2 04PAn=03+δ3+180deg 04PAn=259.000 deg AcAr:=p-a3 AcA=691.040 imsec-2 04p4=(03+δ3)+90-dg 04p4=169.000deg

DESIGN OF MACHINERY SOLUTION MANUAL 7-15a-2 8.638 9.126 AB ABA t Acceleration Scale 0 100 IN/S/S Y ABA n X AA 5. From the graphical solution above, Acceleration scale factor ka 100 in sec 2   AB 9.126ka AB 912.6in sec 2   at an angle of 239 deg ABAt 8.638ka ABAt 863.8 in sec 2   6. Calculate 3 using equation 7.6.  ABAt b   479.9rad sec 2   CW 7. For point C, equation 7.4 becomes: AC = AA + (ACA t + ACA n ) , where ACAn p  2   ACAn 254.262in sec 2   APAn 180deg APAn 259.000deg ACAt p ACAt 691.040 in sec 2   APAt 90deg APAt 169.000 deg

DESIGN OF MACHINERY SOLUTION MANUAL 7-15a-3 8.Repeat procedure of step 4 for the equation in step 7. -7.215 0100IN/S/S B Ac Acceleration Scale 170g A 9.From the graphical solution above, Acceleration scale factor ka 100.in Sec? Ac=7.215-ka Ac=721.5 insec-2 at an angle of-170.61 deg

DESIGN OF MACHINERY SOLUTION MANUAL 7-15a-3 8. Repeat procedure of step 4 for the equation in step 7. CA 170.609° Y A 7.215 AC A t CA n Acceleration Scale 0 X AA 100 IN/S/S 9. From the graphical solution above, Acceleration scale factor ka 100 in sec 2   AC 7.215ka AC 721.5 in sec 2   at an angle of -170.61 deg