ACCELERATION ANALYSIS MMJ32103 MECHANISMS & MACHINES INTRODUCTION • Once the velocity is done, next is to find the accelerations • F=ma • The dynamic forces will contribute to the stresses in the links DEFINITION OF ACCELERATION • Acceleration is defined as the rate of change of velocity with respect to time. Angular Acceleration πΌ= ππ ππ‘ Linear Acceleration π΄= ππ ππ‘ DEFINITION OF ACCELERATION • Acceleration is defined as the rate of change of velocity with respect to time. Position π ππ΄ = ππ ππ Velocity ππ ππ πππ΄ = ππ΄ = ππ ππ =ππππ ππ ππ‘ ππ‘ Acceleration ππππ΄ π(ππππ ππ ) π΄ππ΄ = = ππ‘ ππ‘ ππ ππ π΄ππ΄ = ππ π ππ + πππ ππ ππ‘ ππ‘ π΄ππ΄ = ππΌππ ππ − ππ2π ππ π΄ππ΄ = π΄π‘ππ΄ + π΄πππ΄ Acceleration- tangential component and normal (centripetal component) Fig 7-1. pivot at A is fixed/stationary. NOTE: Tangential acceleration –perpendicular to the radius of motion, tangent to path of motion DEFINITION OF ACCELERATION π΄π = π΄π΄ + π΄ππ΄ Fig 7-2, pivot A is not stationary. GRAPHICAL ACCELERATION ANALYSIS Given θ2, θ3, θ4, ω2, ω3, ω4, α2. Find α3, α4, AA, AB, AP by graphical methods. 3 eqn’s: Eqn (7.4), (7.6) and (7.2) π΄π‘ = ππΌ π΄π = ππ2 π΄π΄π‘ = (π΄π2 )πΌ22 π΄π΄π = (π΄π2 )π22 GRAPHICAL ACCELERATION ANALYSIS Given θ2, θ3, θ4, ω2, ω3, ω4, α2. Find α3, α4, AA, AB, AP by graphical methods. π΄π‘ = ππΌ π΄π = ππ2 π΄ππ΅ = (π΅π4 )π42 π΄π΅ = π΄π΄ + π΄π΅π΄ [π΄π‘π΅ +π΄ππ΅ ] = [π΄π‘π΄ +π΄π΄π ] + [π΄π‘π΅π΄ +π΄ππ΅π΄ ] π΄ππ΅π΄ = (π΅π΄)π32 GRAPHICAL ACCELERATION ANALYSIS Given θ2, θ3, θ4, ω2, ω3, ω4, α2. Find α3, α4, AA, AB, AP by graphical methods. π‘ π π‘ π [π΄π΅ +π΄π΅ ] = [π΄π΄ +π΄π΄ ] + [π΄π‘π΅π΄ +π΄ππ΅π΄ ] GRAPHICAL ACCELERATION ANALYSIS [π΄π‘π΅ +π΄ππ΅ ] = [π΄π‘π΄ +π΄π΄π ] + [π΄π‘π΅π΄ +π΄ππ΅π΄ ] π΄π‘π΅ πΌ4 = π΅π4 π΄π‘π΅π΄ πΌ3 = π΅π΄ AnA AnB AtB AtBA AtA AnBA GRAPHICAL ACCELERATION ANALYSIS π΄πΆ = π΄π΄ + π΄πΆπ΄ π΄π‘πΆπ΄ = ππΌ3 π΄ππΆπ΄ = ππ32 ANALYTICAL SOLUTIONS FOR ACCELERATION ANALYSIS The vector loop equation: R2 + R3 – R4 – R1 = 0 Vector equation: ππ ππ2 + ππ ππ3 − ππ ππ4 − ππ ππ1 = 0 Differentiate (Velocity): πππ ππ2 π2 + πππ ππ3 π3 − πππ ππ4 π4 = 0 Second derivative: (οaο·2 e jο± 2 ο« aο‘ 2 je jο± 2 ) ο« (οbο·3 e jο±3 ο« bο‘ 3 je jο±3 ) ο (οcο·4 e jο± 4 ο« cο‘ 4 je jο± 4 ) ο½ 0 2 2 2 ANALYTICAL SOLUTIONS FOR ACCELERATION ANALYSIS Substitute Euler equation οaο‘ 2 ο(cο‘ ο ο ο j (cos ο± 2 ο« j sin ο± 2 ) ο aο· 22 (cos ο± 2 ο« j sin ο± 2 ) ο« bο‘ 3 j (cos ο± 3 ο« j sin ο± 3 ) ο bο·3 (cos ο± 3 ο« j sin ο± 3 ) ο 2 ο 4 j (cos ο± 4 ο« j sin ο± 4 ) ο cο· 4 (cos ο± 4 ο« j sin ο± 4 ) ο½ 0 2 Multiply by j οaο‘ (ο sin ο± 2 2 ο(cο‘ (ο sin ο± 4 ο ο ο ο« j cos ο± 2 ) ο aο· 22 (cos ο± 2 ο« j sin ο± 2 ) ο« bο‘ 3 (ο sin ο± 3 ο« j sin ο± 3 ) ο bο·3 (cos ο± 3 ο« j sin ο± 3 ) ο ο 4 ο« j cos ο± 4 ) ο cο· 4 (cos ο± 4 ο« j sin ο± 4 ) ο½ 0 2 2 ANALYTICAL SOLUTIONS FOR ACCELERATION ANALYSIS Separate into real and imaginary parts realpart ο aο‘ 2 sin ο± 2 ο aο· 22 cos ο± 2 ο« bο‘ 3 sin ο± 3 ο bο·3 cos ο± 3 ο cο‘ 4 sin ο± 4 ο« cο·4 cos ο± 4 ο½ 0 2 2 imaginarypart aο‘ 2 cos ο± 2 ο aο· 22 sin ο± 2 ο« bο‘ 3 cos ο± 3 ο bο·3 sin ο± 3 ο cο‘ 4 cos ο± 4 ο« cο·4 sin ο± 4 ο½ 0 2 Therefore πΆπ = πͺπ« − π¨π π¨π¬ − π©π« πΆπ = πͺπ¬ − π©π π¨π¬ − π©π« 2 π΄ = ππ πππ4 π΅ = ππ πππ3 2 πΆ = ππΌ2 π πππ2 + ππ2 πππ π2 + ππ32 πππ π3 − ππ42 πππ π4 π· = ππππ π4 πΈ = ππππ π3 πΉ = ππΌ2 πππ π2 − ππ22 π πππ2 − ππ32 π πππ3 + ππ42 π πππ4 ANALYTICAL SOLUTIONS FOR ACCELERATION ANALYSIS Linear Acceleration AA, AAB, AB Eqn.’s (7.13a), (7.13b) and (7.13c) ANALYTICAL SOLUTIONS FOR ACCELERATION ANALYSIS • The fourbar Crank-Slider • α3 = πα2 πππ π2 −a ω22 π πππ2 +πω23 π πππ2 π πππ π3 • π = −πα2 π ππ π2 −πω22 πππ π2 +πα3 π ππ π3 +πω23 πππ π2 ANALYTICAL SOLUTIONS FOR ACCELERATION ANALYSIS • The fourbar Slider-Crank • α3 = α2 = πα2 πππ π2 −a ω22 π πππ2 +πω23 π πππ3 π πππ π3 πω22 (πππ π2 πππ π3 +π πππ2 π πππ3 )−πω23 +π π πππ π3 π πππ π3 EXAMPLE • Given a fourbar linkage with the link lengths L1=d=100 mm, L2=a=40 mm, L3=b=120 mm, L4=c=80 mm. For π2 =40, π2 = 25 rad/s and πΌ2 =15rad/s2 . Given π3 = 20.298 deg and π4 =57.325 deg. π3 = -4.121 and π4 =6.998 rad/s. EXAMPLE • Given a fourbar linkage with the link lengths L1=d=100 mm, L2=a=40 mm, L3=b=120 mm, L4=c=80 mm. For π2 =40, π2 = 25 rad/s and πΌ2 =15rad/s2 . Given π3 = 20.298 deg and π4 =57.325 deg. π3 = -4.121 and π4 =6.998 rad/s. π¨ = ππππ½π − π²π − π²π ππππ½π + π²π π© = −πππππ½π π« = ππππ½π − π²π + π²π ππππ½π + π²π π¬ = −πππππ½π π = π²π + π²π − π ππππ½π + π²π πͺ = π²π − (π²π + π)ππππ½π + π²π π½ππ,π −π© ± π©π − ππ¨πͺ = πππππππ ππ¨ π½ππ,π −π¬ ± π¬π − ππ«π = πππππππ ππ« EXAMPLE • Given a fourbar linkage with the link lengths L1=d=100 mm, L2=a=40 mm, L3=b=120 mm, L4=c=80 mm. For π2 =40, π2 = 25 rad/s and πΌ2 =15rad/s2 . Given π3 = 20.298 deg and π4 =57.325 deg. π3 = -4.121 and π4 =6.998 rad/s. ππ2 sin(π4 − π2 ) π3 = π sin(π3 − π4 ) ππ2 sin(π2 − π3 ) π4 = π sin(π4 − π3 ) EXAMPLE • Given a fourbar linkage with the link lengths L1=d=100 mm, L2=a=40 mm, L3=b=120 mm, L4=c=80 mm. For π2 =40, π2 = 25 rad/s and πΌ2 =15rad/s2. • Find the values of A, B, C, D, E, F, πΌ3 and πΌ4 for the open circuit of the linkage. π΄ = ππ πππ4 π΅ = ππ πππ3 πΆ = ππΌ2 π πππ2 + ππ22 πππ π2 + ππ32 πππ π3 − ππ42 πππ π4 π· = ππππ π4 πΈ = ππππ π3 πΉ = ππΌ2 πππ π2 − ππ22 π πππ2 − ππ32 π πππ3 + ππ42 π πππ4 Therefore πΆπ = πͺπ« − π¨π π¨π¬ − π©π« πΆπ = πͺπ¬ − π©π π¨π¬ − π©π« • https://docs.google.com/forms/d/e/1FAIpQLSeyPd17C0EgJ8xOz7np QYt6l0OqSx2Ygb_Qva6-0SsvfoILjQ/viewform?usp=pp_url
