Velocity and Acceleration Analysis

Velocity and Acceleration Analysis
Unit-III Velocity and Acceleration Analysis Dr. M Krishna Professor and Dean R&D

Course Content Determination of velocity and acceleration of a point / link in simple mechanism by relative method (graphical) Coriolis component of acceleration Instantaneous Centre Centrodes Kennedy’s theorem To determine linear velocity and angular velocity of links of simple mechanism by instantaneous center method Klien’s Construction for velocity and acceleration of slider crank mechanism Analysis of velocity and acceleration of single slider crank chain and four bar chain by complex algebra method

Prerequisite Terminology Definition Motion
A change in position of link Absolute Motion : Motion described with respect to the frame Relative motion: Motion described with respect to another moving body Plane Motion A body is said to have plane motion if all its points move in planes which are parallel to some reference plane. Translation: A body has translation if it moves so that all straight lines in the body move parallel to each other Rectilinear translation is a motion where in all points of the body move in straight line paths Curvilinear translation in which points in the body move along curved path Rotation: all points in a body remain at fixed distance from a line which is perpendicular to the plane motion Translation and Rotation: Machine parts have motions which are a combination of rotation and translation. Spherical Motion A point has spherical motion if it moves in 3-D space and remains at a fixed distance from some fixed point. Helical Motion A point which rotates about an axis at a fixed distance and at the same time moves parallel to the axis describes a helix.

Prerequisite Terminology Definition Scalar
Scalar quantities are those which have magnitude only Eg: Distance, volume, area and time Vector Vector quantities have both magnitude and direction Eg: displacement, velocity, acceleration and force Addition and Subtraction of vector Addition vectors : A+ B Subtraction vector: A B (subtraction of B from A) Velocity Rate of change of Displacement Linear Velocity: Rate of change of linear displacement (= 𝑑𝑠 𝑑𝑡 ), m/s Angular Velocity: Rate of change of angular displacement ( 𝑑 𝑑𝑡 ). Rps Relation between linear velocity and Angular velocity 𝒗=𝒓  and = 𝒗 𝒓 Absolute velocity : the velocity of a point or body relative to frame (VA) Relative velocity: The velocity of a point or body relative another moving point or body (VAB)

Prerequisite Terminology Definition Acceleration
It is the time rate change of velocity Linear Acceleration : It is time rate of change of linear velocity (a= 𝑑𝑣 𝑑𝑡 = 𝑑 2 𝑠 𝑑𝑡 2 )m/s2 Angular acceleration: It is the time rate of change of angular velocity ( = 𝑑 𝑑𝑡 = 𝑑 2  𝑑𝑡 2 ), 𝑟𝑎𝑑/ 𝑠 2 Normal and tangential acceleration Relative Acceleration ABA = AnBA +At BA Normal acceleration: it is due to the change of magnitude. It the point has curvilinear motion, it will have normal acceleration resulting from a change in direction of its linear velocity ( 𝐴 𝑛 = 𝑣 2 𝑟 = 𝜔 2 𝑟) Tangential acceleration is due to the change in direction of velocity (if the direction of its linear velocity changes then the point will have a tangential acceleration. 𝐴 𝑇 =𝛼𝑟

Velocity Analysis Graphical method Relative velocity method
Instantaneous method Graphical Method The following points are to be considered while solving problems by this method. Draw the configuration design to a suitable scale. Locate all fixed point in a mechanism as a common point in velocity diagram. Choose a suitable scale for the vector diagram velocity. The velocity vector of each rotating link is r to the link. Velocity of each link in mechanism has both magnitude and direction. Start from a point whose magnitude and direction is known. The points of the velocity diagram are indicated by small letters.

Angular velocity of the link BC and CD
A four bar mechanism ABCD is made up of four links, pin jointed at the ends. AD is a fixed link which is 180 mm long. The links AB, BC and CD are 90 mm, 120 mm and 120 mm long respectively. At certain instant the link AB makes an angle of 60 with the link AD. If the link AB rotates at a uniform speed of 100 rpm clockwise determine. Angular velocity of the link BC and CD Angular acceleration of the links CD and CB Displacement Diagram (scale 2: 1) A D 60 B C Solution NB = 100 rpm and  𝐵 = 2𝑁 60 = 𝑟 𝑠 AD= 180 mm AB = 90 mm BC = 120 mm CD=120 mm

Angular velocity of the link BC and CD
A four bar mechanism ABCD is made up of four links, pin jointed at the ends. AD is a fixed link which is 180 mm long. The links AB, BC and CD are 90 mm, 120 mm and 120 mm long respectively. At certain instant the link AB makes an angle of 60 with the link AD. If the link AB rotates at a uniform speed of 100 rpm clockwise determine. Angular velocity of the link BC and CD Angular acceleration of the links CD and CB C 20 B Solution NB = 100 rpm and  𝐵 = 2𝑁 60 = 𝑟 𝑠 AD= 180 mm AB = 90 mm BC = 120 mm CD=120 mm 100 60 A D Displacement Diagram (scale 2: 1)

Angular velocity of the link BC = BC = 5.25 r/s
Velocity Analysis Sl Link length Angular velocity Linear Velocity Sense 01 AD= 180 mm Fixed link (AD = 0 and VAD = 0) 02 AB = 90 mm B = r/s VB = B x AB = m/s (ab)  to AB 03 BC = 120 mm CB = 𝑉 𝐶𝐵 𝐶𝐵 = =5.25 𝑟/𝑠 VCB = cb X scale= 3.15 x 0.2 = 0.63 m/s (cb)  to BC 04 CD = 120 mm C = 𝑉 𝑐 𝐶𝐷 = =5.25 𝑟/𝑠 VC = cd X scale= 3.15x0.2 = 0.63 m/s (cd) 110  To BC (VBC) A D 60 B C 20 100 dc= 3.15 10  To DC (VC) c VC a.d 30 b  To AB (VB ) bc= 3.15 VBC Scale 1cm = 0.2 m/s i) Results Angular velocity of the link BC = BC = 5.25 r/s Angular velocity of the link CD = C = 5.25 r/s

Acceleration Analysis
Sl. Link Magnitude Direction Sense 01 AB 𝐴 𝐵 𝑛 =  𝐵 2 AB = x 0.09 =9.87 m/s2 (a1b1) Parallel to AB A 𝐴 𝐵 𝑇 = AB x AB = 0 Constant angular velocity 02 BC 𝐴 𝐵𝐶 𝑛 =  𝐵𝐶 2 BC =5.252 x 0.12 = m/s2 (b1y’) Parallel to BC  𝐴 𝐵𝐶 𝑇 =c1y1 x scale = 2.45 x 2 = 4.9 m/s2 (y’c1)  To BC 03 CD 𝐴 𝐶 𝑛 =  𝐶 2 CD =5.252 x 0.12 = m/s2 (a1x’ ) Parallel to CD 𝐴 𝐶 𝑇 = c1x1 x scale = 5.15 x 2 = 10.3 m/s2 (x’c1)  To CD A D 60 B C 20 100 Scale 1cm = 2 m/s2 a1,d1 le to BC 𝐴 𝐵𝑐 𝑇 b1 ||le to AB 𝐴 𝐵 𝑛 ||le to CD 𝐴 𝐶 𝑛 x1 Ac le to CD 𝐴 𝐶 𝑇 c1 Abc ii) Results Angular acceleration of the link BC 𝛼 𝐵𝐶 = 𝐴 𝐵𝐶 𝑇 𝐵𝐶 == =40.83 𝑟 𝑠 2 Angular acceleration of the link CD r/s 𝛼 𝐶 = 𝐴 𝐶 𝑇 𝐷𝐶 == =85.83 𝑟 𝑠 2 ||le to BC 𝐴 𝐵𝑐 𝑛 y1 c1y1 = 2.45 cm and c1x1 =5.15cm

The Figure shows a four bar mechanism
The Figure shows a four bar mechanism. Crank O2A rotates at 200 rpm and an angular acceleration of 150 rad/sec2 at the instant when the crank makes an angle of 45 to the horizontal. Find the acceleration of points B and C and angular velocities and angular acceleration of the link 3 and 4. Solution NA = 200 rpm  𝐴 = 2𝑁 60 = 𝑟 𝑠 A = 150 r/s2 O2O4 = 120 mm O2A= 45 mm AB = 90 mm AC = 40 mm BO4 = 60 mm VA= A x O2A = m/s O2 O4 45 A B C Scale 1: 2 18 93

The Figure shows a four bar mechanism
The Figure shows a four bar mechanism. Crank O2A rotates at 200 rpm and an angular acceleration of 150 rad/sec2 at the instant when the crank makes an angle of 45 to the horizontal. Find the acceleration of points B and C and angular velocities and angular acceleration of the link 3 and 4. Solution NA = 200 rpm  𝐴 = 2𝑁 60 = 𝑟 𝑠 A = 150 r/s2 O2O4 = 120 mm O2A= 45 mm AB = 90 mm AC = 40 mm BO4 = 60 mm VA= A x O2A = m/s B C 45 A 18 93 O2 O4 Scale 1: 2

Locate c on ab (velocity diagram)
Velocity Analysis Sl Link length Angular velocity Linear Velocity Sense 01 O2O4 = 120 mm Fixed link (AD = 0 and VAD = 0) 02 O2A = 45 mm A = r/s (oa) VA = A x AB = x = m/s  to O2A 03 AB = 90 mm AB = 𝑉 𝐴𝐵 𝐴𝐵 = =8 𝑟/𝑠 VAB = ab X scale= 3.6 x 0.2 = 0.72 m/s (ab)  toAB 04 BO4 = 60 mm B = 𝑉 𝐵 𝐵 𝑂 4 = =8 𝑟/𝑠 VB = ob X scale 2.4 x 0.2 = 0.48 m/s (ob)  to O4B 05 Vc = oc x scale = 3.4 x 0.2= 0.68 m/s (oc) O2 O4 45 A B C Scale 1: 2 18 93  To O4B (VB) b o 3 45 VC to O2A VA  to AB (VAB) c 108 a Scale 1cm = 0.2 m/s Locate c on ab (velocity diagram) 𝑎𝑐 𝑎𝑏 = 𝐴𝐶 𝐴𝐵 then 𝑎𝑐 36 = then ac = 16 mm Angular velocity of link 3-AB = 𝑉 𝐴𝐵 𝐴𝐵 =8 𝑟 𝑠 Angular velocity of link 4- B = 𝑉 𝐵 𝐵 𝑂 4 =8 𝑟/𝑠

Sl. Link Magnitude Direction Sense 01 OB= 𝐴 𝐴 𝑛 =  𝐴 2 OA = x =19.74 m/s2 (ox’) le to O2A O2 𝐴 𝐴 𝑇 = A x O2A = 150x0.045 =6.5 m/s2 (x’a’)  to O2A a’ 02 AB 𝐴 𝐴𝐵 𝑛 =  𝐴𝐵 2 AB =82 x 0.09 = 5.76 m/s2 (a’y’) le to AB  𝐴 𝐴𝐵 𝑇 = y’b’ x scale = 8.7 x 2 =17.4 m/s2 (y’b’)  To BC 03 O4B 𝐴 B 𝑛 =  B 2 O4B =82 x = 3.84 m/s2 (o’z’) le to O4B 𝐴 B 𝑇 = z’b’ x scale =9.6 x 2 = 19.2 m/s2 (z’b’)  To O4B o’ AB AA || to O4B ( 𝐴 𝐵 𝑛 ) z’ O2 O4 45 A B C Scale 1: 2 18 93  To AB ( 𝐴 𝐴𝐵 𝑇 ) b’ to O4B ( 𝐴 𝐵 𝑇 ) || to O2A ( 𝐴 𝐴 𝑛 ) AC AAB c’ x’ Locate c’ on a’b’ 𝒂 ′ 𝒄′ 𝒂′𝒃′ = 𝑨𝑪 𝑨𝑩 𝒕𝒉𝒆𝒏 𝒂 ′ 𝒄′ 𝟖𝟕 = 𝟒𝟎 𝟗𝟎 a’c’ =38.6 mm to O2A ( 𝐴 𝐴 𝑇 ) Scale 1cm = 2m/s2 a’ II to AB ( 𝐴 𝐴𝐵 𝑛 ) y’

ABA = y’b’ x scale = 8.7 x2 =17.4 m/s2
Ac =o’c’ x scale = 9.15 x 2 = 18.3 m/s2 AB = o’b’ x scale = 9.7x2 = 19.2 m/s2 ABA = y’b’ x scale = 8.7 x2 =17.4 m/s2  𝑨𝑩 = 𝑨 𝑩𝑨 𝑻 𝑩𝑨 = 𝟏𝟕.𝟒 𝟎.𝟎𝟗 =𝟏𝟗𝟑.𝟑 𝒓 𝒔  𝑩 = 𝑨 𝑩 𝑻 𝑩 𝑶 𝟒 = 𝟏𝟗.𝟐 𝟎.𝟎𝟔 =𝟑𝟐𝟎 𝒓 𝒔 o’ || to O2A ( 𝐴 𝐴 𝑛 ) to O2A ( 𝐴 𝐴 𝑇 ) x’ a’ AA II to AB ( 𝐴 𝐴𝐵 𝑛 ) y’  To AB ( 𝐴 𝐴𝐵 𝑇 ) || to O4B ( 𝐴 𝐵 𝑛 ) z’ to O4B ( 𝐴 𝐵 𝑇 ) b’ AAB AB c’ AC

In a slider crank mechanism, the crank OB = 30 mm and the connecting rod BC= 120 mm. The crank rotates at a uniform speed of 300 rpm clockwise. For the crank position shown in Figure. Find a) velocity of piston C and angular velocity of connecting rod BC, b) Acceleration of piston C and angular acceleration of connecting rod BC. Solution NB = 300 rpm  𝐵 = 2𝑁 60 = 𝑟 𝑠 B = 0 r/s2 VB= B x OB = x .0.03= m/s OB = 30 mm BC = 120 mm Scale 1:1 C O 60 B

In a slider crank mechanism, the crank OB = 30 mm and the connecting rod BC= 120 mm. The crank rotates at a uniform speed of 300 rpm clockwise. For the crank position shown in Figure. Find a) velocity of piston C and angular velocity of connecting rod BC, b) Acceleration of piston C and angular acceleration of connecting rod BC. Solution NB = 300 rpm  𝐵 = 2𝑁 60 = 𝑟 𝑠 B = 0 r/s2 VB= B x OB = x .0.03= m/s OB = 30 mm BC = 120 mm Scale 1:1 60 B 13 C O

Velocity of piston = VC =0.92 m/s
Velocity Analysis Sl Link length Angular velocity Linear Velocity Sense 01 OB = 30 mm B = r/s VB = B x OB = x 0.03 = m/s (ob)  to OB 02 BC = 120 mm BC = 𝑉 𝐵𝐶 𝐵𝐶 = = 4 𝑟/𝑠 VBC = bc X scale= 2.4 x 0.2= 0.48 m/s (bc)  To AB 03 VC = oc X scale= 4.6x = 0.92 m/s (oc) le to OC  to BC (VBC) O 60 B C 13 || to OC (VC) o c Vc VBC b  to OB (VB) Scale; 1cm = 0.2 m/s Results Velocity of piston = VC =0.92 m/s Angular velocity of link BC-BC = 𝑉 𝐵𝐶 𝐵𝐶 = = 4 𝑟/𝑠

Sl. Link Magnitude Direction Sense 01 OB 𝐴 𝐵 𝑛 =  𝐵 2 OB = x 0.03 =29.6 m/s2 (o’b’) le to OB O2 𝐴 𝐵 𝑇 = B x OB = 0 m/s2 02 BC 𝐴 𝐵𝐶 𝑛 =  𝐵𝐶 2 BC =42 x 0.12 = m/s2 (b’x’) le to BC  𝐴 𝐵𝐶 𝑇 =x’c’ x scale = 6.5 x 4 =26 m/s2 (x’c’)  To BC 03 𝐴 C = o’c’ x scale = 2.7 x 4 =10.8 m/s2 (o’c’) le to OC  to BC ( 𝐴 𝐵𝐶 𝑇 ) O 60 B C 13 || to OC (Ac) o’ || to OB ( 𝐴 𝐵 𝑛 = 𝐴 𝐵 ) b’ c’ AC ACB  𝐵𝐶 = 𝐴 𝐶𝐵 𝑇 𝐶𝐵 = = r/s Results Acceleration of piston C = AC = 10.8 m/s2 Angular acceleration of connecting rod BC Scale 1cm = 4 m/s2 x’ || to BC ( 𝐴 𝐵𝐶 𝑛 )

Find the velocity and acceleration of the slider D and the angular velocity and angular acceleration of link CD for the engine mechanism shown in Fig. The crank O2A rotates at a uniform speed of 20 rad/sec in the clock wise direction. The various lengths of the given mechanism are O1A =0.5 m, AB = 1.0 m, O2B=0.75m, BC = 1m and CD = 1.75 m. Solution A = 20 r/s VA= A x O2A = 20 x 0.5 = 10 m/s O1A = 0.5m AB = 1.0 m O2B=0.75m BC=1m CD = 1.75 m 0.6 m 1.0 m O1 O2 50 A B C D

Find the velocity and acceleration of the slider D and the angular velocity and angular acceleration of link CD for the engine mechanism shown in Fig. The crank O2A rotates at a uniform speed of 20 rad/sec in the clock wise direction. The various lengths of the given mechanism are O1A =0.5 m, AB = 1.0 m, O2B=0.75m, BC = 1m and CD = 1.75 m. Solution A = 20 r/s VA= A x O2A = 20 x 0.5 = 10 m/s O1A = 0.5m AB = 1.0 m O2B=0.75m BC=1m CD = 1.75 m 0.6 m 0.6 m D O2 1.0 m 33 B C 99 O1 50 A 112

Velocity Diagram (scale 1cm = 2 m/s)
Velocity Analysis Sl Link length Angular velocity Linear Velocity Sense 01 O1A = 0.5 m A = 20 r/s VA = A x O1A = 10 m/s  to O1A 02 O2B = 0.75 m B = 𝑉 𝐵 𝑂 2 𝐵 = =10.93 𝑟/𝑠 VB = ob X scale= 4.1 x 2= 8.2 m/s  To O2B 03 O2C=1.75 m B = 𝑉 𝐶 𝑂 2 𝐶 = =10.93 𝑟/𝑠 VC= oc X scale=9.56 x 2=19.12 m/s  To O2C 04 AB=1.0 m AB= 𝑉 𝐴𝐵 𝐴𝐵 = =2.5 𝑟/𝑠 VAB= ab X scale=1.25 x 2=2.5 m/s  To AB 05 DC=1.75 m DC= 𝑉 𝐷𝐶 𝐷𝐶 = =5.37 𝑟/𝑠 VDC= dc X scale=4.7 x 2= 9.4 m/s  To DC 06 VD = odxscale =7.375x2 =14.75 m/s ||le to axis c o Velocity Diagram (scale 1cm = 2 m/s)  To AB (VAB) Vd || to piton axis (VD)  To O1A (VA) a  To AB (VAB) b Results Velocity of the slider D = VD = m/s Angular velocity of link CD =CD = 5.37 r/s Locate c on o2b 𝑂 2 𝐶 𝑂 2 𝐵 = 𝑜𝑐 𝑜𝑏 then 𝑜𝑐= 𝑂 2 𝐶 𝑂 2 𝐵 𝑋 𝑜𝑏= 𝑋 41=95.6 𝑚𝑚 Vc VCD d  to CD (VCD)

Sl. Link Magnitude Direction Sense 01 O1A 𝐴 𝐴 𝑛 =  𝐴 2 O1A =202 x 0.5 =200 m/s2 (o’a’) le to O1A O1 𝐴 𝐴 𝑇 = A x O1A = 0 m/s2 02 AB 𝐴 𝐴𝐵 𝑛 =  𝐴𝐵 2 AB =2.52 x 1 = m/s2 (a’x’’) le to AB  𝐴 𝐴𝐵 𝑇 =x’b’ x scale = 3.9x 40 = m/s2  To AB 03 O2B 𝐴 𝐵 𝑛 =  𝐵 2 𝑂 2 B = x 0.75 = m/s2 (o’y’) le to O2B 𝐴 𝐵 𝑇 = 𝑦 ′ 𝑏 ′ 𝑋 𝑠𝑐𝑎𝑙𝑒=3.4 𝑥 40= 𝑚/ 𝑠 2  To O2B 04 𝐴 𝐷𝐶 𝑛 =  𝐷𝐶 2 𝐷𝐶 =5.372 x 1.75 = m/s2 le to DC 𝐴 𝐷𝐶 𝑇 = 𝑧 ′ 𝑑 ′ 𝑋 𝑠𝑐𝑎𝑙𝑒=0.5𝑋40=20 𝑚/ 𝑠 2  To DC AD = o’d’xscale = 8.5 x40 = 340 m/s2 c’ ADC || to DC (𝐴 𝐷𝐶 𝑛 ) z’ 9 AD  to DC (𝐴 𝐷𝐶 𝑇 ) Scale 1 cm = 40 m/s2  to O2B (𝐴 𝐵 𝑇 ) b’  To AB ( 𝐴 𝐴𝐵 𝑇 ) x’ || to AB ( 𝐴 𝐴𝐵 𝑛 ) a’ || to O1A ( 𝐴 𝐴 𝑛 ) Locate c’ on o’b’ 𝑂 2 𝐶 𝑂 2 𝐵 = 𝑜′𝑐′ 𝑜′𝑏′ 𝑜 ′ 𝑐= 𝑂 2 𝐶 𝑂 2 𝐵 𝑋 𝑜 ′ 𝑏= 𝑋 0.041=95.6𝑚𝑚 || to O2B (𝐴 𝐵 𝑛 ) y’ o’

Results Acceleration of the slider D = AD = 340 m/s2 Angular Acceleration of link CD = 𝛼 𝐶𝐷 = 𝐴 𝐷𝐶 𝑇 𝐶𝐷 = = r/s2

In the toggle mechanism as shown in Fig
In the toggle mechanism as shown in Fig., the slider D is constrained to move on a horizontal path. The crank OA is rotating in the counter clockwise direction at a speed of 180 rpm. Increasing at the rate of 50 rad/sec2. The dimensions of the various links are as follows; OA=180 mm, CB = 240 mm, AB=360 mm and BD =540 mm. For the given configuration find 1) Velocity of slider D and angular velocity of BD and 2) Acceleration of slider D and angular acceleration of BD A B D C O 105 mm 360 mm 45 A A Solution NA = 180 rpm A = 𝟐 𝑵 𝑨 𝟔𝟎 =18.85 r/s VA= A x OA = x 0.18= 3.4 m/s OA 180 mm AB = 360 mm CB= 240 mm BD = 540 mm A = 50 r/s2 𝑨 𝑨 𝑻 =  𝑨 𝑿 𝑶𝑨 = 50 x 0.18 = 9 m/s2

Solution NA = 180 rpm A = 𝟐 𝑵 𝑨 𝟔𝟎 =18.85 r/s
VA= A x OA = x 0.18= 3.4 m/s OA 180 mm AB = 360 mm CB= 240 mm BD = 540 mm A = 50 r/s2 𝑨 𝑨 𝑻 =  𝑨 𝑿 𝑶𝑨 = 50 x 0.18 = 9 m/s2 A 45 O 360 mm B 105 mm D C

Velocity Diagram (scale 1cm =1 m/s)
Velocity Analysis Sl Link length Angular velocity Linear Velocity Sense 01 OA = 180 mm A = r/s VA = A x O1A = 3.4 m/s  to OA 02 BA = 360 mm BA = 𝑉 𝐴𝐵 𝐴𝐵 = =2.5 𝑟/𝑠 VB = ab X scale= 0.9 x 1= 0.9 m/s  To AB 03 CB= 240 mm B = 𝑉 𝐵 𝐶𝐵 = =11.45 𝑟/𝑠 VB= cb x scale= 2.75 x 1=2.75 m/s  To CB 04 BD = 540 mm BD= 𝑉 𝐵𝐷 𝐵𝐷 = =4.44 𝑟/𝑠 VBD= bd X scale=2.4 x 1 =2.4 m/s  To BD 05 VD = od x scale = 2.1 x 1 = 2.1 m/s ||le to CD Velocity Diagram (scale 1cm =1 m/s) C O 105 mm 360 mm A 45 B D  to CB (VB)) b VAB VA  to OA a  to AB (VAB) VB  to BD (VBC) VDB || to CD (VD) d VD o,c Results 1) Velocity of slider D = VD = 2.1 m/s Angular velocity of BD = BD = 4.44 r/s

Sl. Link Magnitude Direction Sense 01 OA 𝐴 𝐴 𝑛 =  𝐴 2 OA = x 0.18 = 64.2 m/s2 (o’x’) le to O1A O1 𝐴 𝐴 𝑇 = A x O1A = 50 x 0.18 = 9 m/s2 (x’a’) 02 AB 𝐴 𝐴𝐵 𝑛 =  𝐴𝐵 2 AB =2.52 x 0.36 = m/s2 (a’y’) le to AB  𝐴 𝐴𝐵 𝑇 = y’b’ x scale =3.8 x10 =38 m/s2 (y’b’)  To AB 03 CB 𝐴 𝐵 𝑛 =  𝐵 2 CB = x 0.54 = m/s2 (c’z’) le to CB 𝐴 𝐵 𝑇 = 𝑧 ′ 𝑏 ′ 𝑋 𝑠𝑐𝑎𝑙𝑒=2.3𝑥 10=23 𝑚/ 𝑠 2 (z’b’)  To CB 04 BD 𝐴 𝐵𝐷 𝑛 =  𝐵𝐷 2 𝐵𝐷 =4.442 x 0.54 = m/s2 le to BD 𝐴 𝐵𝐷 𝑇 = 𝑞 ′ 𝑑 ′ 𝑋 𝑠𝑐𝑎𝑙𝑒=3.85 𝑋 10=38.5 𝑚/ 𝑠 2  To BD AD = c’d’ x scale = 1.1 x10 = 11 m/s2 𝐴 𝐵𝐷 𝑇  BD Scale 1 cm = 10 m/s2 AD || to DC AD d’ o’ c’ C O 105 mm 360 mm A 45 B D AA 𝐴 𝐵 𝑛 || to CB z’ ADB AB 𝐴 𝐵 𝑇  CB y’ 𝐴 𝐵𝐴 𝑛 ||𝑡𝑜 𝐴𝐵 𝐴 𝐴 𝑛 || to OA x’ 𝐴 𝐵𝐷 𝑛 ||𝐵𝐷 q’ 𝐴 𝐴 𝑇  OA a’ 𝐴 𝐴𝐵 𝑇  AB AAB b’ 2) Acceleration of slider D = AD = 11 m/s Angular Acceleration of BD = BD = 𝐴 𝐷𝐵 𝑇 𝐷𝐵 = = rad/sec2

Coriolis Component of Acceleration
Definition Whenever a point in one body moves along a path on a second body and if the path rotating, then the acceleration of the point in the first body relative to a coincident point in the second body will have a Coriolis components. B point on link 3(slider) A Fixed point on link 2 Let B and A be coincident at the instant. Let link 2 have a constant angular velocity 2 in its motion from OP to OP1 in small interval of time ‘dt’ and the slider moves B to B2. VBA is constant velocity of the slider. Consider the motion of the slider from B to B2 in the following three stages B to A1 due to the rotation of the link 2 A1 to B1 due to the outward velocity VBA B1 to B2 due to acceleration perpendicular to the link 2 d B & A Q P P1 B2 A1 B1 2 1 2 3

B to A1 due to the rotation of the link 2 A1 to B1 due to the outward velocity VBA B1 to B2 due to acceleration perpendicular to the link 2 Third component is the Coriolis component of acceleration Arc B1B2 = Arc QB2 – Arc QB1 = Arc QB2 – Arc AA1 = OQ d - OA d = (OQ-OA) d = AQ d = A1B1 d (1) Where A1B1 = VBA dt (ds = v dt) d = 2 dt Substitute in 1 Arc B1B2 = VBA2 dt (2) and also Arc B1B2 = ½ a dt (3) From 2 and 3 ½ a dt2 =VBA2 dt2 a = 2 VBA 2 Which is called he Coriolis Component of Acceleration d B & A Q P P1 B2 A1 B1 2 1 2 3

The Fig. shows a crack and slotted lever type quick return mechanism. The crank rotates at a uniform speed of 60 rpm clockwise. The line of stroke of the ram is perpendicular to OA. Determine velocity and acceleration of D. 450 mm O A B C D E on link OC OA= 250 mm AB = 100 mm OC = 400 mm CD = 150 mm 120 Scale 1: 10 NB = 60 rpm B = 2 𝑁 𝐵 60 = 260 60 =6.28 𝑟 𝑠 VB = B x AB = 6.28 x 0.1 = m/s

The Fig. shows a crack and slotted lever type quick return mechanism. The crank rotates at a uniform speed of 60 rpm clockwise. The line of stroke of the ram is perpendicular to OA. Determine velocity and acceleration of D. NB = 60 rpm B = 2 𝑁 𝐵 60 = 260 60 =6.28 𝑟 𝑠 VB = B x AB = 6.28 x 0.1 = m/s D 450 mm E C 120 B A O Scale 1: 5

Locate convenient point ‘a’ and ‘o’ 2) Draw ab = VB=0.628m/s  to AB (suitable scale) 3) Draw line from o  to AB (VB) 4) From b draw line parallel to OB and meets at ‘e’ (VBE) 5) Locate ‘c’ on the velocity diagram (VC) 𝑂𝐶 𝑂𝐸 = 𝑜𝑐 𝑜𝑒 becomes = 𝑜𝑐 46 then oc = 57.5 mm VB  to AB b VBE e || to OB Vc 6) Draw the  to CD from point ‘c’ (VCD) VE  to OB c  To CD 7) Draw horizontal and parallel line from o (piston movement) until it meets pervious line (VD). d o,a VD VE = oe x scale = 4.6 x 10 = 46 cm/s= 0.46 m/s VEB = be x scale = 4.2 x 10 = 42 cm/s = 0.42 m/s VC = oc x scale = 5.75 x 10 = 57.5 cm/s = m/s VDC = cd x scale = 1.8 x 10 = 18 cm/s = 0.18 m/s VD = od x scale = 6.35 x 10 = 63.5 cm/s = m/s (stroke velocity ) Scale 1cm =0.1 m/s

The acceleration equation of point E on the link OC is = AE = AB+AEB + 2 EO VEB VBE BE Direction of Coriolis ace 2 BE VBE 𝐴 𝐵 𝑛 = 𝑉 𝐵 2 𝐴𝐵 = =3.94 𝑚 𝑠 ( 𝐴 𝐵 𝑇 =0) 𝐴 𝐸 𝑛 = 𝑉 𝐸 2 𝑂𝐸 = =0.66 𝑚 𝑠 2 Coriolis acceleration = 2 x BE x VBE = 2 𝑉 𝐸 𝑂𝐸 𝑉BE = m/s2 𝐴 𝐸𝐵 𝑛 =0 (𝑛𝑜 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑙𝑖𝑑𝑒𝑟 𝐴 𝐸𝐵 𝑇 =AEB a’ o’ 𝐴 𝐵 𝑛 || to BA b’ x’ 𝐴 𝐸 𝑛 || to BO 𝐴 𝐸 𝑇  BO e From any convenient point a’, draw a’b’= 𝐴 𝐵 𝑛 =3.94 𝑚 𝑠 2 parallel to BA 2. From b’ draw b’y’=2BEVBE = m/s2 in the direction of Coriolis acceleration (OC) 3. From o’ draw o’x’ = 𝐴 𝐸 𝑛 =0.66m/s2 parallel to BO 4. Draw perpendicular line form x’ (direction of 𝐴 𝐸 𝑇 ) y’ 2BEVBE 5. From y’ draw a line parallel to link OC (direction of slider), which intersects the  line drawn fro x’ at e’ Scale 1 cm = 0.5 m/s2

Locate c’ 𝑂𝐶 𝑂𝐸 = 𝑜 ′ 𝑐′ 𝑜 ′ 𝑒′ ’ = 𝑜 ′ 𝑐′ 2.2 o’c’ = 2.75cm 7. Acceleration equation for D as AD = AC + ADC i.e 𝐴 𝐷 𝑛 + 𝐴 𝐷 𝑇 = 𝐴 𝐶 𝑛 + 𝐴 𝐶 𝑇 + 𝐴 𝐷𝐶 𝑛 + 𝐴 𝐷𝐶 𝑇 8. 𝐴 𝐷𝐶 𝑛 = 𝑉 𝐷𝐶 2 𝐷𝐶 = m./s2 (From c’ draw c’z’ = 𝐴 𝐷𝐶 𝑛 =0.216 m/s2 parallel to DC) a’ o’ 𝐴 𝐵 𝑛 || to BA b’ y’ 2BEVBE x’ 𝐴 𝐸 𝑛 || to BO 𝐴 𝐸 𝑇  BO e 𝐴 𝐸𝐵 𝑇 =AEB 𝐴 𝐷𝐶 𝑇 9. From z’ draw a line to c’z’ (direction of 𝐴 𝐷𝐶 𝑇 ) d’ AD 10. From o’ draw a horizontal line indicating the path of D which intersect previous line (AD) 𝐴 𝐸𝐵 𝑇 =AEB c’ Z’ 𝐴 𝐷𝐶 𝑛 Acceleration of D = AD = o’d’x scale = 3.45 x 0.5 = /s2 Scale 1 cm = 0.5 m/s2

The Fig. shows a quick return mechanism. Link 2 rotates uniformly at 20 r/s in clockwise direction. Determine the angular acceleration of link 3 C O B on link 3 A on link 2 and link 4 3 2 4 1 Scale 1:5 2 OC = 350 OA = 150 mm CB= 250 mm 2=20 r/s VA = 2 x OA = 20 x 0.15 = 3 m/s

The Fig. shows a quick return mechanism. Link 2 rotates uniformly at 20 r/s in clockwise direction. Determine the angular acceleration of link 3 C O OC = 350 OA = 150 mm CB= 250 mm 2 A/B 2=20 r/s VA = 2 x OA = 20 x 0.15 = 3 m/s 1 3 Scale 1cm :5 cm

A/B 2 3 1 Velocity equation is VA = VB + VAB 2) From a convenient point ‘o’ draw oa = 3 m/s (VA) to link 2 is same direction of 2 3) From ‘a’ draw a lie parallel to the path of slider (link 3) (VAB) 4) Draw a line  to link 3 (VB), which cuts the previous line (VB) VAB = ab x scale = 2.6 x 1 = 2.6 m/s VB = cb x scale = 1.55 x 1 = 1.55 m/s || to slider Velocity diagram  Link 3 b vAB VB o, c a VAB 2 3 VAB 3 Scale 1 cm = 1 m/s

Acceleration Diagram The acceleration equation AA = AB + AAB + 23 VAB 𝐴 𝐴 𝑛 + 𝐴 𝐴 𝑇 = 𝐴 𝐵 𝑛 + 𝐴 𝐵 𝑇 + 𝐴 𝐴𝐵 𝑛 + 𝐴 𝐴𝐵 𝑇 + 23 VAB 𝐴 𝐴 𝑇 = 0 (link 2 rotates uniform velocity) 𝐴 𝐴𝐵 𝑛 = 0 (there is no change in direction for the slider 𝐴 𝐴 𝑛 = 𝑉 𝐴 2 𝑂𝐴 = =60 m/s2 𝐴 𝐵 𝑛 = 𝑉 𝐵 2 𝐶𝐵 = =9.61 m/s2 𝐴 𝐵 𝑇 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑖𝑠 𝑢𝑛𝑘𝑜𝑤𝑛 𝑏𝑢𝑡  𝑡𝑜 𝑙𝑖𝑛𝑘 3 𝐴 𝐴𝐵 𝑇 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑖𝑠 𝑢𝑛𝑘𝑜𝑤𝑛 𝑏𝑢𝑡 ||𝑙𝑒 𝑡𝑜 𝑙𝑖𝑛𝑘 3 2  3 𝑉 𝐴𝐵 =2 𝑉 𝐵 𝐶𝐵 𝑉 𝐴𝐵 = 𝑥 2.6=32.34 𝑚/𝑠 3) For a convenient point o’ draw o’a’ = 𝐴 𝐴 𝑛 =60 m/s2 parallel to link 2 4) From c’ (o’) draw c’x’ = 𝐴 𝐵 𝑛 =9.61 m/s2 parallel to link 3 From x’ draw a line  to c’x’ indicating the direction of 𝐴 𝐵 𝑇 From a’ draw the Coriolis component 2  3 𝑉 𝐴𝐵 =32.24 m/s2 = a’y’  to the link 3. 2  3 𝑉 𝐴𝐵 Y’ o’c’ a’ 𝐴 𝐴 𝑛 =AA y’ b’ 𝐴 𝐴𝐵 𝑇 = 𝐴 𝐴𝐵 C O A/B 2 3 1 𝐴 𝐵 𝑇 x’ Scale 1 cm = 10 m/s2

y’ b’ 𝐴 𝐴𝐵 𝑇 = 𝐴 𝐴𝐵 C O A/B 2 3 1 From y’ draw the line parallel to the link 3 indicating the direction 𝐴 𝐴𝐵 𝑇 which intersects the  drawn from x’ to b’. Scale 1 cm = 10 m/s2 o’c’ a’ 𝐴 𝐴 𝑛 =AA x’ 𝐴 𝐵 𝑇 2  3 𝑉 𝐴𝐵 Y’ 𝐴 𝐴𝐵 𝐴 𝐵 𝑇 = x’b’ x scale = 8.2 x 10 = 82 m/s2 𝐴 𝐵 𝑇 =  3 𝐶𝐵 3 = 𝐴 𝐵 𝑇 𝐶𝐵 = =328 𝑟 𝑠2 Angular acceleration of link 3 = 328 𝑟 𝑠2 b’ 𝐴 𝐵 𝐴 𝐵 𝑛

Instantaneous Centre Two bodies moves in a planar motion, there is two coincident points, the absolute velocities of which are the same. At the instant, the pair of coincident points with the same velocities is defined as the instantaneous center of relative motion Properties of Instantaneous Centre One rigid link rotates instantaneously relative to anther for configuration of the Mechanism considered. Two body with relative motion The instant center may be changing all the time The Velocity Vector is perpendicular to the rotating radius If one link is a frame, the instant center is the rotating center Number of Instantaneous centre in a mechanism The number of instantaneous Centre 𝑁= 𝑛(𝑛−1) 2 n is number of links

Instantaneous Centre Type of instantaneous Centres
The are three types of instantaneous centres Fixed Instantaneous centre – I12 and I14 b. Permanent Instantaneous centre – I23 and I34 Neither fixed nor permanent instantaneous centre- I24 and I13 I13 I34 I23 3 I24 4 2 I41 I12 1

Instantaneous Centre Arnold-Kennedy Theorem or three centres in line theorem If three bodies have motion relative to each other, their instantaneous centres should lie in a straight line. Proof: Consider a three link mechanism with link 1 being fixed link 2 rotating about I12 and link 3 rotating about I13 Hence, I12 and I13 are the instantaneous centres for link 2 and link 3. Let us assume that instantaneous centre of link 2 and 3 be at point A i.e. I23. Point A is a coincident point on link 2 and link 3. Considering A on link 2, velocity of A with respect to I12 will be a vector VA2  to link A I12. Similarly for point A on link 3, velocity of A with respect to I13 will be  to A I13 𝑁= 𝑛(𝑛−1) 2 = 3(3−1) 2 = 3 I12 = Instantaneous centre of 1 relative to 2 I13 = Instantaneous centre of 1 relative to 3 I23 = Instantaneous centre of 2 relative to B

Instantaneous Centre Arnold-Kennedy Theorem or three centres in line theorem If three bodies have motion relative to each other, their instantaneous centres should lie in a straight line. Proof: It is seen that velocity vector of VA2 and VA3 are in different directions which is impossible. Hence, the instantaneous center of the two links cannot be at the assumed position. It can be seen that when I23 lies on the line joining I12 and I13 the VA2 and VA3 will be same in magnitude and direction. Hence, for the three links to be in relative motion all the three centers should lie in a same straight line. Hence, the proof 𝑁= 𝑛(𝑛−1) 2 = 3(3−1) 2 = 3 I12 = Instantaneous centre of 1 relative to 2 I13 = Instantaneous centre of 1 relative to 3 I23 = Instantaneous centre of 2 relative to B

Instantaneous Centre- Problem
1. Locate all the instantaneous centres for the four bar mechanism shown in fig. A B C D I34 Number of Instantaneous centre 𝑁= 𝑛(𝑛−1) 𝑛 = 4(4−1) 2 = 6 3 I23 1 2 3 4 5 6 I13 2 4 I12 I14 I24 1 1 2) Instantaneous Table Links 1 2 3 4 IC 12 13 14 23 24 34

2. Locate all the instantaneous centres for the slider mechanism shown in fig. I14 at  I14 at  Number of Instantaneous centre 𝑁= 𝑛(𝑛−1) 𝑛 = 4(4−1) 2 = 6 I13 I24 1 2 3 4 5 6 I23 1 2 3 4 A B C I12 I34 2) Instantaneous Table Links 1 2 3 4 IC 12 13 14 23 24 34

A pin jointed four bar mechanism ABCD is shown in Fig. Link AB = 150 mm, BC = 180 mm, CD = 180 mm and the fixed link AD= 300 mm. Link AB makes 60 with the link AD, and rotates uniformly at 100 rpm. Locate all the instantaneous centres and find the angular velocity of link BC and the linear velocity of link CD. I13  𝐵 = 2𝑁 60 = 2 = r/s VB = B x AB = x 150 = mm/s or m/s From Fig. I13B = 49x scale = 49 x 5 = 245 mm BC = 𝑉 𝐵 𝐼 13 𝐵 = =6.41 r/s L Linear velocity of link CD I13C = 35 x scale = 35 x5= 175 mm Vc = BC I13C = 6.41 x = m/s Scale 1:5 A B C D 60 1 2 3 4 I34 I23 I12 I14 I24

A slider crank mechanism has lengths of crank and connecting rod equal to 200 mm and 800 mm respectively locate all the instantaneous centers of the mechanism for the position of the crank when it has turned through 30 from IOC. Also find velocity of slider and angular velocity of connecting rod if crank rotates at 40 rad/sec. A

I14 at  I13  𝐴 = 40 r/s VA = A AI12 = 40x0.2 =8 m/s Also VA = 3 AI13 3 = 𝑉 𝐴 𝐴𝐼 13 = = =8.7 𝑟.𝑠 VB = 3 x BI13 = 8.7 x 0.57 =5 m/s Velocity of slider I14 at  I24 I23 30 A 2 3 B o I12 I34 1 4 Scale 1 cm = 5 cm

In the toggle mechanism shown in figure the slider D is constrained to move in a horizontal path the crank OA is rotating in CCW direction at a speed of 180 rpm the dimensions of various links are as follows: OA = 180 mm, CB = 240 mm, 1= 360 mm and BD = 540 mm. Velocity of slider and Angular velocity of links AB, CB and BD. n = 6 links 𝑁= 𝑛(𝑛−1) 2 = 15 1 2 3 4 5 6 12 13 14 15 16 23 24 25 26 34 35 36 45 46 56

OA = 180 mm, CB = 240 mm, AB = 360 mm and BD = 540 mm. 1 I61 I12 I51 I14 I13 6 I13 I25 I36 2 I46 I24 I56 I23 I35 5 3 45 A 1 2 3 4 5 I45 I34 4 360 mm O B D C

OA = 180 mm, CB = 240 mm, AB = 360 mm and BD = 540 mm.     I24 I14 I34, I35 I45 I56 I23 I12 I36 2 I25 I51 1 3 I36 I46 5 4 1 I13

VA = A x AI12 = 3.4 m/s Link -3 VA = 3 x AI13  3 = 𝑉 𝐴 𝐴𝐼 13 =2.44 𝑟 𝑠 VB = 3 x BI13= 4 BI14 4 = r/s VB = 5 x BI18  5 = 𝑉 𝐵 𝐵𝐼 15 =4.37 𝑟 𝑠 VD = 5 DI15 = 2 m/s

Locate all the instantaneous centre of the mechanism shown in Fig. I14 at  I16 at  I15, I13 I56 I36 5 1 2 3 4 6 I43 I35 I46 I23 I35 I25 n = 6 links 𝑁= 𝑛(𝑛−1) 2 = 15 I26 I24 1 2 3 4 5 6 12 13 14 15 16 23 24 25 26 34 35 36 45 46 56 I12

Klein’s Construction for slider-crank Mechanism
This is a simple graphical method for drawing the acceleration diagram. The line Diagram for the mechanism is OAB with OA being crank and AB is connecting rod. Draw a circle centre O radius OA Draw a Circle with centre mid point on AB to pass through point A E C 11 O A B F C D Extend AB to pass through the centre line of the first circle to locate C Draw a circle centre point A to pass through C Locate points D and E and connect with a lin Triangle OAC- velocity diagram The scale may be determined OA=2 crank radius The red outline OAFG is the acceleration diagram OG-represents to scale the acceleration of piston OA represents Centripetal acceleration of A to O.

Klein’s Construction for slider-crank Mechanism
1. Determine the velocity and acceleration of the piston by Kelein’s construction to the following specification. Stroke = 300 mm, Ratio of length of connecting rod to crank length = 4, speed of engine = 300 rpm, Positon of crank = 45  with dead centre. Length of crank r = stoke /2 = 300/2 =150 mm and length of the connecting rod = l = 4r = 600 mm, = 45  and N = 300 rpm and A = 2N/60 = r/s Scale 1:5 M A B 45 P C O Triangle OAM-Klein’s velocity diagram Velocity of piston = VP=  x OM x scale = x 25 x 5 = 3927 mm/s = m/s D OABC quadrilateral – Klein’s acceleration diagram Acceleration of piston Ap = 2 OC x scale = x 21.5 x 5 = mm/s2 = m/s2

Complex Algebra Method
Vector Analysis: For the position vectors r shown below, the positive angle is measured counter-clock wise (ccw) from the horizontal line, which is drawn through the beginning of the vector. r = rx + I ry where I = −1 r = r (cosθ + isinθ) = reiθ, which is known as Euler formula. Vector Loop or Loop-Closure Equation Vector-loop equations are: For four-bar linkage: r1 + r2 + r3 + r4 = 0

Vector Loop or Loop-Closure Equation Vector-loop equations are: For slider-crank mechanism: r2 + r3 − r4 = 0

Raven’s Approach for the slider crank Mechanism- position 4= 270 r4 =  and 4=4=0 1=0 r1= r2+r3 r1 𝑒 𝑖  1 = 𝑟 2 𝑒 𝑖  2 + 𝑟 3 𝑒 𝑖  3 r1 = 𝑟 2 𝑒 𝑖  2 + 𝑟 3 𝑒 𝑖  3 −−−−−−(1) r1 = 𝑟 2 ( cos  2 + 𝑖 𝑠𝑖𝑛  2 )+ 𝑟 3 ( cos  2 + 𝑖 𝑠𝑖𝑛  2 ) r1 = 𝑟 2 cos  2 +𝑖 𝑟 2 𝑠𝑖𝑛  𝑟 3 cos  3 +𝑖 𝑟 3 𝑠𝑖𝑛  3 Real part r1 = 𝑟 2 cos  2 + 𝑟 3 cos  (2) Imaginary part 0 = 𝑟 2 𝑠𝑖𝑛  2 + 𝑟 3 𝑠𝑖𝑛  3 −−−−−− 3 𝑠𝑖𝑛  3 =− 𝑟 2 𝑟 3 sin  2 (3 can be determine) A B C r1 r2 r3 r4 at  2 3 4= 270

Raven’s Approach for the slider crank Mechanism- Velocity 4= 270 r4 =  and 4=4=0 1=0 A B C r1 r2 r3 r4 at  2 3 4= 270 r1 = 𝑟 2 𝑒 𝑖  2 + 𝑟 3 𝑒 𝑖  3 −−−−−−(1) Differentiate equation (1) with respect to time 𝑟 1 = 𝑟 2 𝑒 𝑖  2 + 𝑖𝑟 2 𝑒 𝑖  2  𝑟 3 𝑒 𝑖  3 + 𝑖𝑟 3 𝑒 𝑖  3  3 Where 𝑟 1  0, 𝑟 2 =0 and 𝑟 3 =0 ( r2 and r3 are constant lengths) 𝑟 1 = 𝑉 𝑃 = 𝑖𝑟 2 𝑒 𝑖  2  𝑖𝑟 3 𝑒 𝑖  3  (2) Rewrite in polar form 𝑉 𝑃 = 𝑖𝑟 2  2 (cos 2 + i sin 2) 𝑖𝑟 3  3 (cos 3 + i sin 3) Vp = ir22 cos 2 - r22 sin 2 + ir33 cos 3 – r33 sin 3 Real part: Vp = - r22 sin 2– r33 sin  (3) Imaginary part: 0 =r22 cos 2 + r33 cos  (4) From (3)  3 =− 𝑟 2  2 cos  2 𝑟 3 cos  (5) substitute 3 in (3) we get velocity of piston

Raven’s Approach for the slider crank Mechanism- Acceleration 4= 270 r4 =  and 4=4=0 1=0 A B C r1 r2 r3 r4 at  2 3 4= 270 𝑟 1 = 𝑉 𝑃 = 𝑖𝑟 2 𝑒 𝑖  2  𝑖𝑟 3 𝑒 𝑖  3  3 −−−−−(6) Differentiate equation with respect to time 𝑟 1 = 𝑖 𝑟 2 𝑒 𝑖  2  2 + 𝑖 2 𝑟 2 𝑒 𝑖  2  2  2 + 𝑖𝑟 2 𝑒 𝑖  2  2 + 𝑖 𝑟 3 𝑒 𝑖  3  3 + 𝑖 2 𝑟 3 𝑒 𝑖  3  3  3 + 𝑖𝑟 3 𝑒 𝑖  3  3 Where 𝑟 1  0, but , 𝑟 2 =0, 𝑟 3 =0, 𝑟 2 =0 and 𝑟 3 =0 ( r2 and r3 are constant lengths) 𝑟 1 = 𝐴 𝑃 =− 𝑟 2  2 2 𝑒 𝑖  2 + 𝑖𝑟 2  2 𝑒 𝑖  2 − 𝑟 3  3 2 𝑒 𝑖  3 + 𝑖𝑟 3  3 𝑒 𝑖  3 rewrite in polar form 𝐴 𝑃 = 𝑟 1 =− 𝑟 2  2 2 ( cos  2 +𝑖 sin  2 )+𝑖 𝑟 2  2 ( cos  2 +𝑖 sin  2 ) − 𝑟 3  3 2 ( cos  3 +𝑖 sin  3 )+𝑖 𝑟 3  3 ( cos  3 +𝑖 sin  3 ) 𝐴 𝑃 =− 𝑟 2  2 2 cos  2 − 𝑖 𝑟 2  2 2 sin  2 +𝑖 𝑟 2  2 cos  2 +𝑖 𝑟 2  2 (𝑖 sin  2 ) − 𝑟 3  3 2 cos  3 −𝑖 𝑟 3  3 2 sin  3 +𝑖 𝑟 3  3 cos  3 +𝑖 𝑟 3  3 (𝑖 sin  3 ) 𝐴 𝑃 =(− 𝑟 2  2 2 cos  2 − 𝑟 2  2 sin  2 − 𝑟 3  3 2 cos  3 − 𝑟 3  3 sin  3 ) +𝑖(− 𝑟 2  2 2 sin  2 + 𝑟 2  2 cos  2 − 𝑟 3  3 2 sin  3 + 𝑟 3  3 cos  3 ) Real part = 𝑨 𝑷 =− 𝒓 𝟐  𝟐 𝟐 𝐜𝐨𝐬  𝟐 − 𝒓 𝟐  𝟐 𝒔𝒊𝒏  𝟐 − 𝒓 𝟑  𝟑 𝟐 𝐜𝐨𝐬  𝟑 − 𝒓 𝟑  𝟑 𝒔𝒊𝒏  𝟑 𝑰𝒎𝒂𝒈𝒊𝒏𝒂𝒓𝒚 𝒑𝒂𝒓𝒕=− 𝒓 𝟐  𝟐 𝟐 𝒔𝒊𝒏  𝟐 + 𝒓 𝟐  𝟐 𝒄𝒐𝒔  𝟐 − 𝒓 𝟑  𝟑 𝟐 𝒔𝒊𝒏  𝟑 + 𝒓 𝟑  𝟑 𝒄𝒐𝒔  𝟑 =𝟎 r 3  3 cos  3 = r 2  2 2 sin  2 + r 2  2 cos  2 + r 3  3 2 sin  3  3 = r 2  2 2 sin  2 + r 2  2 cos  2 + r 3  3 2 sin  𝑟 3 cos  3 𝑨 𝑷 =− 𝒓 𝟐  𝟐 𝟐 𝐜𝐨𝐬  𝟐 − 𝒓 𝟐  𝟐 𝒔𝒊𝒏  𝟐 − 𝒓 𝟑  𝟑 𝟐 𝐜𝐨𝐬  𝟑 − 𝒓 𝟑  𝟑 𝒔𝒊𝒏  𝟑

1. Using complex algebra derive expression for velocity and acceleration of the piston, angular acceleration of the connecting rod of a reciprocating engine. With these expression determine the above quantities, if the crank length is 50 mm, connecting rod 200 mm, crank speed is constant at 3000 rpm and crank angle is 30. Problem r2 = 50 mm and r3 = 200 rpm N2= 3000 rpm (constant) 2 =30   2 = 2  𝑁 = 2  = rad/s 2 =0 A B C r1 r2 r3 r4 at  2 3 4= 270 Determine 3  3 = 𝑠𝑖𝑛 −1 (− 𝑟 2 𝑟 3 sin  2 ) = 𝑠𝑖𝑛 −1 − sin 30 =352.82 Determine 3(Angular velocity of CR)  3 =− 𝑟 2  2 cos  2 𝑟 3 cos  3 =− 50 𝑥 𝑥 cos cos = rad/s Derive the previous equations such as Determine Vp( velocity of piston) Vp = - r22 sin 2– r33 sin 3 = -50 x sin 30 – 200 x sin = m/s 𝑠𝑖𝑛  3 =− 𝑟 2 𝑟 3 sin  2 (position)  3 =− 𝑟 2  2 cos  2 𝑟 3 cos  3 (Angular velocity CR) Determine 3 ( Angular acceleration of CR)  3 = r 2  2 2 sin  2 + r 2  2 cos  2 + r 3  3 2 sin  𝑟 3 cos  3 Vp = - r22 sin 2– r33 sin 3 (Velocity of piston) Angular acceleration of CR 50 𝑥 sin 𝑥 0 cos (−68.56) 2 sin cos  3 = r 2  2 2 sin  2 + r 2  2 cos  2 + r 3  3 2 sin  𝑟 3 cos  3 = rad/s2 Acceleration of slider (piston) Determine Ap ( Acceleration of piston) 𝑨 𝑷 =− 𝒓 𝟐  𝟐 𝟐 𝐜𝐨𝐬  𝟐 − 𝒓 𝟐  𝟐 𝒔𝒊𝒏  𝟐 − 𝒓 𝟑  𝟑 𝟐 𝐜𝐨𝐬  𝟑 − 𝒓 𝟑  𝟑 𝒔𝒊𝒏  𝟑 𝑨 𝑷 =− 𝒓 𝟐  𝟐 𝟐 𝐜𝐨𝐬  𝟐 − 𝒓 𝟐  𝟐 𝒔𝒊𝒏  𝟐 − 𝒓 𝟑  𝟑 𝟐 𝐜𝐨𝐬  𝟑 − 𝒓 𝟑  𝟑 𝒔𝒊𝒏  𝟑 Ap = - 50 x cos 30 – 0 – 200 x (-68.56)2cos – 200 x x sin = m/s2

2 3 4 Position Vector Equation- Four Bar Linkage Position Analysis: The vector loop equation is written as 𝑟 1 𝑒 𝑖  1 + 𝑟 2 𝑒 𝑖  2 + 𝑟 3 𝑒 𝑖  3 + 𝑟 4 𝑒 𝑖  4 = (1) r1(cos 1+i sin 1) + r2(cos 2+i sin 2) + r3(cos 3+i sin 3) + r4(cos 4+i sin 4) = 0 Collecting real and imaginary parts yields r1cos 1 + r2cos 2 + r3cos 3 + r4cos 4 =0 r1 sin 1+ r2sin 2 + r3sin 3 + r4 sin 4 = 0 The problem of the position analysis is defined as Given: r1, r2, r3, r4, θ1 and θ2 Find: θ3 and θ4

Velocity Vector Equation- Four Bar Linkage Velocity Analysis: Taking time derivative of the position vector r = reiθ in general yields 𝑑𝑟 𝑑𝑡 = 𝑟= 𝑟 +𝑖𝑟 𝑒 𝑖 where ω is known as the angular velocity of the link. 𝑟 𝑖 + 𝑟 2 +𝑖 𝑟 2  2 𝑒 𝑖  𝑟 3 +𝑖 𝑟 3  3 𝑒 𝑖  𝑟 4 +𝑖 𝑟 4  4 𝑒 𝑖  4 =0 where 𝑟 𝑖 = 0, and also 𝑟 2 = 𝑟 3 = 𝑟 4 =0, since r1 is frame, and the lengths of links 2, 3 and 4 are fixed. 𝑖 𝑟 2  2 𝑒 𝑖  2 +𝑖 𝑟 3  3 𝑒 𝑖  3 +𝑖 𝑟 4  4 𝑒 𝑖  4 =0 i r2 2 (cos 2+i sin 2) + i r3 3(cos 3+i sin 3) + ir4 4(cos 4+i sin 4) = 0 Real: -r22 sin 2 - r33 sin 3 - r4 4 sin 4 = (1) Imaginary: r22 cos 2 + r33 cos 3 + r4 4 cos  (2) Given: All r’s and θ ’s, and ω2 and Find 3 and 4

Velocity Vector Equation- Four Bar Linkage (-r22 sin 2 - r33 sin 3 - r4 4 sin 4 = 0) x cos 4 (r22 cos 2 + r33 cos 3 + r4 4 cos 4 =0 ) sin 4 -r22 cos 4 sin 2 - r33 cos 4 sin 3 - r4 4 cos 4 sin 4 = 0 r22 sin 4 cos 2 + r33 sin 4 cos 3 + r4 4 sin 4 cos 4 =0 r22 (sin 4 cos 2 - cos 4 sin 2) + r33(sin 4 cos 3 - cos 4 sin 3 )=0 r22sin (4 - 2) +r33 2sin (4 - 3) = 0 3 =- 𝑟 2  2 sin⁡(  4 −  2 ) 𝑟 3 sin⁡(  4 −  3 ) (-r22 sin 2 - r33 sin 3 - r4 4 sin 4 = 0) x cos 3 (r22 cos 2 + r33 cos 3 + r4 4 cos 4 =0 ) sin 3 -r22 cos 3sin 2 - r33 cos 3 sin 3 - r4 4 cos 3sin 4 = 0 r22 sin 3 cos 2 + r33 sin 3 cos 3 + r4 4 sin 3 cos 4 =0 r22 (sin 3 cos 2 - cos 3sin 2) + r4 4 (sin 3 cos 4 - cos 3sin 4 )=0 r22sin (3 - 2) +r44 sin (4 - 3) = 0 4 =- 𝑟 2  2 sin⁡(  3 −  2 ) 𝑟 4 sin⁡(  4 −  3 )

Acceleration Vector Equation- Four Bar Linkage 𝑑 𝑟 𝑑𝑡 = 𝑟 =( 𝑟 +𝑖 2 𝑟 + ir - r2) ei Where 𝑟 1 = 0, and also 𝑟 2 = 𝑟 3 = 𝑟 4 =0 since r1 is frame, and the lengths of links 2, 3 and 4 are fixed ( 𝑟 𝑖 = 0, and also 𝑟 2 = 𝑟 3 = 𝑟 4 =0) 𝑑 𝑟 𝑑𝑡 = (ir - r2 ) ei (𝑖 𝑟 2 𝛼 2 − 𝑟 2  2 2 ) 𝑒 𝑖  2 +(𝑖 𝑟 3 𝛼 3 − 𝑟 3  3 2 ) 𝑒 𝑖  3 +(𝑖 𝑟 4 𝛼 4 − 𝑟 4  4 2 ) 𝑒 𝑖  4 =0 (𝑖 𝑟 2 𝛼 2 − 𝑟 2  2 2 ) ( cos  2 +𝑖 sin  2 ) +(𝑖 𝑟 3 𝛼 3 − 𝑟 3  3 2 )( cos  3 +𝑖 sin  3 ) +(𝑖 𝑟 4 𝛼 4 − 𝑟 4  4 2 )( cos  4 +𝑖 sin  4 ) =0 Real: − 𝑟 2  2 sin  2 − 𝑟 2  2 2 𝑐𝑜𝑠  2 − 𝑟 3  3 sin  3 − 𝑟 3  3 2 𝑐𝑜𝑠  3 − 𝑟 4  4 sin  4 − 𝑟 4  4 2 𝑐𝑜𝑠  4 =0 im: 𝑟 2  2 𝑐𝑜𝑠  2 − 𝑟 2  2 2 sin  2 + 𝑟 3  3 𝑐𝑜𝑠  3 − 𝑟 3  3 2 sin  3 + 𝑟 4  4 𝑐𝑜𝑠  4 − 𝑟 4  4 2 sin  4 =0 The problem of the acceleration analysis is defined as Given: All r’s, θ ’s, and ω ’s, and α2 Find: α3 and α4  3 =  3 ∝ 2  2 − 𝑟 2  2 2 cos  2 −  𝑟 3  3 2 cos  3 −  4 + 𝑟 4  𝑟 3 𝑠𝑖𝑛  3 −  4  4 =  4 ∝ 2  𝑟 2  2 2 cos  2 −  3 + 𝑟 3  𝑟 4  4 2 cos  3 −  𝑟 4 𝑠𝑖𝑛  3 −  4

1. In a four bar mechanism ABCD, link AB=300 mm, CD= 360 mm and the fixed that AD= 600 mm. The angle BAD = 60. The link AB has in angular velocity of 10 rad/sec and an angular acceleration of 30 rad/sec2, both clockwise. Determine the angular velocity and angular acceleration of link BC and CD by using Rave’s approach (complex Algebra Method) 2 = 60  A B C D 3 4  +  Given AD= r1 =600 mm, AB = r2= 300 mm, BC = r3 = 360 mm CD = r4 = 360 mm 2 = 60  2 = -10 rad/sec (CW) 2 = -30 rad/sec2 (CW)     From ABD BD2 =AB2 + AD2-2AB AD cos 60 = (300)(600) cos 60 BD = 519.6mm sin  2 𝐵𝐷 = sin  𝐴𝐵  sin = sin    = 30  From BDC BD2 =BC2 + DC2-2BC DC cos  = = (360)(360) cos  then  = 43.8  3 =  -  = 43.8 – 30 =13.8  sin  𝐶𝐷 = sin  𝐵𝐶  sin = sin    = 43.8  +  = = 73.8 and 4 =  +  = 360 – 73.8 =286.19

Given AD= r1 =600 mm, AB = r2= 300 mm, BC = r3 = 360 mm CD = r4 = 360 mm 2 = 60  2 = -10 rad/sec (CW) 2 = -30 rad/sec2 (CW) 2 = 60,3 =  and 4= 286.2 Angular velocity 3 =- 𝑟 2  2 sin⁡(  4 −  2 ) 𝑟 3 sin⁡(  4 −  3 ) = 300 −10 sin − sin⁡(286.2−13.806) =−6.02 𝑟𝑎𝑑/𝑠 4 =- 𝑟 2  2 sin⁡(  3 −  2 ) 𝑟 4 sin⁡(  4 −  3 ) = 300 −10 sin − sin⁡(286.2−13.806) = rad/s Angular Acceleration  3 =  3 ∝ 2  2 − 𝑟 2  2 2 cos  2 −  𝑟 3  3 2 cos  3 −  4 + 𝑟 4  𝑟 3 𝑠𝑖𝑛  3 −  4  3 = (−6.02)(−30) −10 − 300 − cos 60− − cos 13.8− (−6.02) 𝑠𝑖𝑛 13.8− = rad/sec2  4 =  4 ∝ 2  𝑟 2  2 2 cos  2 −  3 + 𝑟 3  𝑟 4  4 2 cos  3 −  𝑟 4 𝑠𝑖𝑛  3 −  4  4 = (−6.02)(−30) −10 − 300 − cos 60− − cos 13.8− 𝑠𝑖𝑛 13.8− = rad/sec2

Velocity and Acceleration Analysis

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Velocity and Acceleration Analysis

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