Two-mass Equivalent Link

Transcript

1 Notes_08_0 1 of 0 Two-ass Equivalent ink B G JG C B G C = total ass B centroid location CG B = = BC BG BC check approxiate ass oent J J = ( BG ) ( CG ) G G _ APP (for slender rod J = J ) G _ APP G _ ACTUA

2 Notes_08_0 of 0 Shaking Force for Slider Crank A=G T1 B θ G φ + C P 1 1 BA assue crank is statically balanced G = A constant crank speed θ = 0 split link into and assue φ is sall Case I - static force analysis for only d Alebert inertial force P with no friction ( cosθ + cos θ) µ = 0 sin φ = θ P = ( )s s = θ sin T 1 on P sin( θ + φ) X Y X Y = F1on = P F1on = P tan φ F1on = 0 F1on = P tan φ cosφ Case II - static force analysis for only inertial force ω T 1 on X Y = 0 F = ω cosθ F = ω sin θ 1on 1on Case III - place additional counterbalance BA on the crank at radius and 180º fro pin B Superposition = Case I + Case II + Case III 1 T1 on = ( ) θ ( sin θ sin θ sin θ) ( ) θ cosθ ( )s x F on1 = BA ( ) θ sin θ + ( ) s φ y F on1 BA = tan ( ) s φ y F on1 = tan

3 { F } = { F } + { F } on1 on1 = θ Notes_08_0 of 0 ( BA ) cosθ + ( )( cosθ + cos θ) ( ) sin θ BA

4 Notes_08_0 of 0 Model 1 no additional balancing ass BA = 0 Model unidirectional in-line shaking force { F } = θ ( BA ) cosθ + ( )( cosθ + cos θ) ( ) sin θ BA BA = { F } = θ ( )( cosθ + cos θ) 0 Model iniize the axiu agnitude of shaking force (approxiate) { F } = θ ( BA ) cosθ + ( )( cosθ + cos θ) ( ) sin θ BA BA = { F } = θ cosθ + ( ) cos θ ( ) sin θ Model iniize in-line shaking force (approxiate) { F } = θ ( BA ) cosθ + ( )( cosθ + cos θ) ( ) sin θ BA BA = = { F } = θ ( ) cos θ sin θ

5 Notes_08_0 5 of 0 0 BA = 0 0 BA = BA = 0 BA = Maxiu shaking force [lbf] Balancing ass [lb]

6 Notes_08_0 6 of 0 % shake. - single cylinder shaking force Notes_08_0 % HJSIII, clear % constants dr = pi / 180; % geoetry [inches] = 0.985; =.; % crank speed [rad/sec] w = ; % asses [lb] = 0.51; = 0.111; = 0.781; % crank angle th_deg = (0:60)'; th = th_deg * dr; % piston acceleration [inch/s/s] sdd = -*w*w*(cos(th) +*cos(*th)/); % plot four odels figure( 1 ) clf res = []; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Model 1 - BA = 0 subplot(,,1 ) BA = 0; % shaking force [lb.in/s/s] * [lbf*s*s /.17 lb.ft] * [ ft / 1 in ] Fx = ( (-BA)**w*w*cos(th) - (+)*sdd ) / 86; % [lbf] Fy = ( (-BA)**w*w*sin(th) ) / 86; Fs = sqrt( Fx.*Fx + Fy.*Fy ); Fs_ax = ax( Fs ); res = [ res ; BA Fs_ax ]; % plot shaking force curve plot( Fx, Fy ) axis square axis( [ ] ) title( 'BA = 0' ) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Model - BA = subplot(,, ) BA = ; % shaking force [lb.in/s/s] * [lbf*s*s /.17 lb.ft] * [ ft / 1 in ] Fx = ( (-BA)**w*w*cos(th) - (+)*sdd ) / 86; % [lbf] Fy = ( (-BA)**w*w*sin(th) ) / 86; Fs = sqrt( Fx.*Fx + Fy.*Fy ); Fs_ax = ax( Fs ); res = [ res ; BA Fs_ax ]; % plot shaking force curve plot( Fx, Fy ) axis square axis( [ ] ) title( 'BA = ' )

7 Notes_08_0 7 of 0 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Model - BA = subplot(,, ) BA = ; % shaking force [lb.in/s/s] * [lbf*s*s /.17 lb.ft] * [ ft / 1 in ] Fx = ( (-BA)**w*w*cos(th) - (+)*sdd ) / 86; % [lbf] Fy = ( (-BA)**w*w*sin(th) ) / 86; Fs = sqrt( Fx.*Fx + Fy.*Fy ); Fs_ax = ax( Fs ); res = [ res ; BA Fs_ax ]; % plot shaking force curve plot( Fx, Fy ) axis square axis( [ ] ) title( 'BA = ' ) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Model - BA = subplot(,, ) BA = ; % shaking force [lb.in/s/s] * [lbf*s*s /.17 lb.ft] * [ ft / 1 in ] Fx = ( (-BA)**w*w*cos(th) - (+)*sdd ) / 86; % [lbf] Fy = ( (-BA)**w*w*sin(th) ) / 86; Fs = sqrt( Fx.*Fx + Fy.*Fy ); Fs_ax = ax( Fs ); res = [ res ; BA Fs_ax ]; % plot shaking force curve plot( Fx, Fy ) axis square axis( [ ] ) title( 'BA = ' ) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % plot axiu shaking force versus balancing ass keep = []; for BA = 0 : 0.01 : (++), % shaking force [lb.in/s/s] * [lbf*s*s /.17 lb.ft] * [ ft / 1 in ] Fx = ( (-BA)**w*w*cos(th) - (+)*sdd ) / 86; % [lbf] Fy = ( (-BA)**w*w*sin(th) ) / 86; Fs = sqrt( Fx.*Fx + Fy.*Fy ); Fs_ax = ax( Fs ); keep = [ keep ; BA Fs_ax ]; end % plot results figure( ) clf plot( keep(:,1),keep(:,),'b', res(:,1),res(:,),'ro' ) axis( [ ] ) xlabel( 'Balancing ass [lb]' ) ylabel( 'Maxiu shaking force [lbf]' ) % botto of shake

8 Notes_08_0 8 of 0 Model 5 iniize the axiu agnitude of shaking force (exact) does not work????? F ( F ) = θ = θ BA BA BA BA ( ( BA ) + ( BA )( ) cosθ( cosθ + cos θ) + ( ) ( cosθ + cos θ) ) ( ( ) ( ) cosθ( cosθ + cos θ) ) = 0 BA ( )( cos θ + cosθcos θ) = + = + ( )( cos θ + cos θ cosθsin θ) ( )( cos θ + cos θ θ) = + cos ( ) cos θ + cos θ cosθ θ = 6 cos θsin θ cosθsin θ + sin θ = 0 ( )( ) sin θ = 0, θ = 0 + sin θ = 0, θ = 180 BA = + 1 ( )( ) BA = + 1 cosθ = 1± ( ) BA = + ( )( cos θ + cos θ cosθ) % odel5. - in/ax shaking force for slider crank % does not work??????????????? % HJSIII, clear % geoetry [inches] = 0.985; =.; % asses [lb] = 0.51; = 0.111; = 0.781; % cosine solution rho = / ; c1 = (-1 + sqrt( 1 + 6*rho*rho ) ) /6 /rho; c = (-1 - sqrt( 1 + 6*rho*rho ) ) /6 /rho; % values v1 = *rho*c1*c1*c1 + c1*c1 - rho*c1; v = *rho*c*c*c + c*c - rho*c;

9 % balancing asses BA1 = + (+)*v1 BA = + (+)*v Notes_08_0 9 of 0

10 Notes_08_0 10 of 0 Shaking Force for In-line Two Cylinder Air Copressor x x C sae crank, connecting rods and pistons C a E E y B z out z B y in θβ = θ θd = θ A = G only ABC { F } D = θ + ( BA ) + ( ) ( ) cosθ cosθ cos θ ( ) sin θ BA A = G D both { F } cosθ + cos cos θ + cos sin θ + sin both { F } = θ + ( BA )( cosθ + cos( θ )) + ( )( cosθ + cos( θ )) ( ) ( cos θ + cos( θ + 60 )) ( )( sin θ + sin( θ )) BA ( θ ) = cosθ + cosθcos180 sin θsin180 = ( θ + 60 ) = cos θ ( θ ) = sin θ + sin θcos180 + cosθsin180 = 0 = θ ( ) X 0 cos θ M = a F + F M = + a F shaking oents ( ) X Y on1 Y on1 0 Y on1

11 Notes_08_0 11 of 0 In-line Four Cylinder Engine

12 Notes_08_0 1 of 0 Shaking Force for In-line Four Cylinder Engine x x C E sae crank, connecting rods and pistons y in 1, y z out z θ θ 1 = θ = θ A = G = θ = θ + 180, 1 sae as two two-cylinder copressors irrored end to end { F } = θ ( ) 0 cos θ X Y shaking oents M = 0 M 0 =

13 Notes_08_0 1 of 0 In-line Six Cylinder Engine

14 Notes_08_0 1 of 0 Shaking Force for In-line Six Cylinder Engine x x y in sae crank, connecting rods and pistons 1,6 z out y A = G, z, θ θ θ 1 = θ = θ = θ 6 5 cosθ + cos cos θ + cos sin θ + sin = θ = θ + 10 = θ + 0 ( θ + 10 ) + cos( θ + 0 ) = 0 ( θ + 0 ) + cos( θ + 80 ) ( θ + 10 ) + sin( θ + 0 ) = = 0 X Y { F } = M = 0 M 0 =

15 Notes_08_0 15 of 0 V8 separate cranks V8 dual cranks

16 Notes_08_0 16 of 0 V8 split cranks V8 with pistons

17 Notes_08_0 17 of 0 Shaking Force for Four Bar assue crank is statically balanced G = A split link into and assue θ is sall link becoes two-force eber static force analysis with d Alebert inertial forces

18 Notes_08_0 18 of 0 ( ) ΣM on about D CCW + ( F sin γ) CD = θ J ( DG ) ( ) F = on G CD ( J ( DG ) ( CD) )/ CD ( θ θ ) on θ G sin ( ( DG ) ( CD) ) θ cosθ + ( ( DG ) ( CD) ) θ sin θ Fon θ x F on1 = cos ( ( DG ) ( CD) ) θ sin θ ( ( DG ) ( CD) ) θ cosθ Fon θ y F on1 = sin

19 Notes_08_0 19 of 0 ΣM on about A CCW + ( F sin( θ θ )) AB 0 T1 _ on _ on = T 1on = θ ABsin( θ θ ) ( J G ( DG ) ( CD) ) CDsin( θ θ ) T ( J ( DG ) ( ) ) 1on = θ G CD fro Notes_0_0 θ θ and θ fro Notes_0_0 F θ ( BA )( AB) θ cosθ + Fon θ = x on1 cos ( BA )( AB) θ sin θ + Fon θ y F on1 = sin

20 Notes_08_0 0 of 0 { F } F F + F x x = on1 on1 y + y on1 Fon1 { F } cosθ = sin θ sin θ cosθ cosθ sin θ ( )( ) BA AB θ ( ( DG ) ( CD) ) θ ( ( ) + ( )) θ DG CD