Link 3 (A to B) b := in Offset c := 1.00 in. m 4 := 0.20 blob. a G4 := in sec 2 θ AG4 := deg
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1 DESIGN O MACHINERY SOLUTION MANUAL -b- PROBLEM -b Stteent: Tble P- shows ketic nd geoetric dt for severl slider-crnk lkges of the type nd orienttion shown igure P-. The pot loctions re defed s described the text. or row the tble, solve for forces nd torques t the position shown. Also, copute the shkg force nd the shkg torque. Consider the coefficient of friction µ between slider nd ground to be zero. Units: blob := sec Given: Lk lengths: Lk (O to A) :=. Lk (A to B) b :=. Offset c :=. riction: µ := Crnk ngle nd otion: θ := deg ω := 5 rd sec θ := 77. deg Coupler pot: R P :=. δ RP :=. deg Mss: :=.5 blob :=. blob :=. blob Moent of erti: I G :=. blob I G :=. blob Mss center: R CG :=. δ := deg R CG :=. δ := deg orce nd torque: P := δ P := 5 deg T := 5 Accelertions: α := 5 rd sec G := 5.6 sec θ AG :=.7 deg α :=. rd sec G := 589. sec θ AG :=.5 deg G := 7.97 sec θ AG := 8. deg Solution: See igure P-, Tble P-, nd Mthcd file P.. Clculte the x nd y coponents of the position vectors. R x := R CG cos θ + δ + 8 deg R x =.6 R y := R CG s θ + δ + 8 deg R y =.766 ( ) ( R CG cos( δ )) R := R CG s δ + R =.89 ( ) δ := tn R CG cos δ, R CG s δ δ = 9.5 deg R x := R cos θ δ R x =.955 R y := R s θ δ R y =.7 R x := R CG cos θ + 8 deg + δ R x =.6 R y := R CG s θ + 8 deg + δ R y =.7
2 DESIGN O MACHINERY SOLUTION MANUAL -b- ( ) + ( b R CG cos( δ )) R := R CG s δ ( ) δ := tn b R CG cos δ, R CG s δ R x := R cos θ + 8 deg δ R y := R s θ + 8 deg δ R Px := R P cos θ + 8 deg + δ RP R Py := R P s θ + 8 deg + δ RP R = 6.85 δ = 7. deg R x = 6.68 R y =.67 R Px =.56 R Py =.8. Clculte the x nd y coponents of the ccelertion of the CGs of ll ovg lks the globl coordte syste (GCS). Gx := Gy := G cos( θ AG ) G s( θ AG ) Gx =.89 sec Gy = sec Gx := Gy := Gx := G cos( θ AG ) G s( θ AG ) G cos( θ AG ) Gx = sec Gy =.8 sec Gx = 7.97 sec. Clculte the x nd y coponents of the externl force t P on lk the CGS. Px := P cos δ P Py := P s( δ P ) Px = 7.7 Py = 7.7. Substitute these given nd clculted vlues to the trix eqution.g, odified for this proble. Note tht Mthcd requires tht ll eleents trix hve the se diension. Thus, the trix nd rry eqution.g will be de diensionless nd the diensions will be put bck fter solvg it. C := R y R x R y R y R x R x R y R x µ
3 DESIGN O MACHINERY SOLUTION MANUAL -b- := Gx Gy I G α Gx Px Gy Py I G α R Px Py + R Py Px T Gx R := C C = = R = x := R x =.9 y := R y = 9.8 x := R x =.8 y := R y =. x := R 5 x =. y := R 6 y = 6. y := R 7 y = 6. T := R 8 T =. 5. Clculte the shkg force nd shkg torque usg equtions.5.
4 DESIGN O MACHINERY SOLUTION MANUAL -b- := x j y := j y s := + s = j Mgnitude: s := s s =.9 Angle: θ s := rg( s ) θ s = deg T s T := T s =.
5 DESIGN O MACHINERY SOLUTION MANUAL -5b- PROBLEM -5b Stteent: Tble P- shows ketic nd geoetric dt for severl p-joted fourbr lkges of the type nd orienttion shown igure P-. All hve θ =. The pot loctions re defed s described the text. or row the tble, solve for forces nd torques t the position shown. Also, copute the shkg force nd the shkg torque. Work ny units syste you prefer. Units: blob := sec Given: Lk lengths: Lk (O to A) Lk (B to O ) := c :=.. Lk (A to B) Lk (O to O ) b := d := 6 Crnk ngle nd otion: θ := deg ω := rd sec Other lk ngles: θ := 9.5 deg ω :=. rd sec θ := 6.6 deg ω := 7.66 rd sec Coupler pot: Rocker pot: R P := R P :=.. δ RP := δ RP := 5. deg. deg Mss: :=.5 blob :=. blob :=. blob Moent of erti: I G :=. blob I G :=. blob I G :=. blob Mss center: R CG :=. δ := deg R CG :=. δ := deg R CG := 6. δ := deg orce nd torque: P := δ P := deg P := 5 δ P := 55 deg T := T := Accelertions: α := 5 rd sec G :=. sec θ AG :=.86 deg α :=.96 rd sec G := sec θ AG := 9.75 deg α := 6.75 rd sec G :=. sec θ AG := 56.5 deg Solution: See igure P-, Tble P- nd Mthcd file P5.. Clculte the x nd y coponents of the position vectors. R x := R CG cos θ + δ + 8 deg R y := R CG s θ + δ + 8 deg R x =.6 R y =.766 ( ) + ( R CG cos( δ )) R := R CG s δ ( ) δ := tn R CG cos δ, R CG s δ R x := R cos θ δ R y := R s θ δ R =.89 δ = 9.5 deg R x =.955 R y =.7
6 DESIGN O MACHINERY SOLUTION MANUAL -5b- R x := R CG cos θ + δ + 8 deg R x =.99 R y := R CG s θ + δ + 8 deg R y =.69 ( ) ( b R CG cos( δ )) R := R CG s δ + R = 6.85 ( ) δ := tn b R CG cos δ, R CG s δ δ = 7. deg R x := R cos θ δ R x =.7 R y := R s θ δ R y = 6.5 R x := R CG cos θ + δ + 8 deg R x = 5.9 R y := R CG s θ + δ + 8 deg R y =. ( ) ( c R CG cos( δ )) R := R CG s δ + R = 8.8 ( ) δ := tn c R CG cos δ, R CG s δ δ = 7.56 deg R x := R cos θ δ R x =.58 R y := R s θ δ R y = 8.97 R Px := R P cos θ + δ RP R Px = 7.9 R Py := R P s θ + δ RP R Py = 7.5 R Px := R P cos θ + δ RP R Px =.8 R Py := R P s θ + δ RP R Py =.5. Clculte the x nd y coponents of the ccelertion of the CGs of ll ovg lks the globl coordte syste (GCS). Gx := G cos( θ AG ) Gx = 6.9 sec Gy := G s θ AG Gy = 79.8 sec Gx := G cos θ AG Gx = 95.8 sec Gy := G s θ AG Gy = 5.85 sec Gx := G cos θ AG Gx =.6 sec Gy := G s θ AG Gy =. sec. Clculte the x nd y coponents of the externl force on lks nd the CGS. Px := P cos δ P Px =.6 Py := P s δ P Py =. Px := P cos δ P Px = 8.6 Py := P s δ P Py =.87
7 DESIGN O MACHINERY SOLUTION MANUAL -5b-. Substitute these given nd clculted vlues to the trix eqution.9, odified for dditionl ters. Note tht Mthcd requires tht ll eleents trix hve the se diension. Thus, the trix nd rry eqution.g will be de diensionless nd the diensions will be put bck fter solvg it. C := R y R x R y R y R x R x R y R y R x R x R y R x C = := Gx Gy I G α Gx Px Gy Py I G α R Px Py + R Py Px T Gx Px Gy Py I G α R Px Py + R Py Px T
8 DESIGN O MACHINERY SOLUTION MANUAL -5b- = R := C R = x := R x = 6. y := R y = 55.6 x := R x = 6. y := R y = 5.6 x := R 5 x = 8.5 y := R 6 y = 9.5 x := R 7 x = 95.9 y := R 8 y = 6.7 T := R T 9 = Clculte the shkg force nd shkg torque usg equtions.5. := x j y := x j y s := + s = j Mgnitude: s := s s = Angle: θ s := rg( s ) θ s = 5.75 deg T s := T T s =.
9 DESIGN O MACHINERY SOLUTION MANUAL -9- PROBLEM -9 Stteent: Given: igure P- shows fourbr lkge nd its diensions eters. The steel crnk nd rocker hve unifor cross sections 5 wide by 5 thick. The luu coupler is 5 thick. In the stntneous position shown, the crnk O A hs ω = rd/sec nd α = 5 rd/sec. There is verticl force t P of = N. d ll p forces nd the torque needed to drive the crnk t this stnt. Lk lengths: Lk (O to A) :=. Lk (A to B) b :=.6 Lk (B to O ) c :=. Lk (O to O ) d :=. Coupler pot: R p :=.6 δ := deg := N T := N Crnk ngle nd otion: θ := 6 deg ω := rd sec α := 5 rd sec Lk cross-section dis: w := 5 t := 5 t := 5 w := 5 t := 5 Mteril specific weight: steel D s := 78 kg luu D := 8 kg Solution: See igure P- nd Mthcd file P9.. Use progr OURBAR to detere the position, velocity, nd ccelertion of lks nd. θ :=.7 deg ω :=.669 rd sec α := rd sec θ := 96. deg ω :=. rd sec α := 67. rd sec. Detere the distnce to the CG the LRCS on ech of the three ovg lks. Lks nd : R CG :=.5 R CG =.5 R CG :=.5 c R CG =.65 R p cos δ + b Lk : R CGx' := R CGx' =.56 R p s δ R CGy' := R CGy' =.55 R CG := R CGx' + R CGy' R CG =.67 At n ngle with respect to the locl x' xis of. Detere the ss nd oent of erti of ech lk. δ := tn R CGx', R CGy' δ = 8.6 deg := w t D s := b R p s( δ ) t D := w t c D s = 9.75 kg =.6 kg =.78 kg nd Edition, 999
10 DESIGN O MACHINERY SOLUTION MANUAL -9- I G := w + I G =.85 kg ( ) ( ) I G := b + R p s δ R p cos δ b + R p cos δ 8 I G = 5.79 kg I G := w + c I G =.8 kg. Set up n LNCS xy coordte syste t the CG of ech lk, nd drw ll pplicble vectors ctg on the syste s shown igure -. Drw free-body digr of ech ovg lk s shown igure -. Y B y y R CG R P P y Gy A x R Gx x A R CG O R CG O () The coplete lkge with GCS X x R x (b) BD of Lk y y R y Gy B R y Gy x x Gx R P P R y Gx x x A R x y (d) BD of Lk (c) BD of Lk 5. Clculte the x nd y coponents of the position vectors. R x := R CG cos θ + 8 deg R y := R CG s θ + 8 deg R x := R CG cos θ R y := R CG s θ R x := R CG cos δ + θ + 8 deg R x =.5 R y =. R x =.5 R y =. R x =.79 nd Edition, 999
11 DESIGN O MACHINERY SOLUTION MANUAL -9- R CG cos( θ + δ ) ( ( + ) ) R y := R CG s δ + θ + 8 deg R y =.75 R x := b cos θ R x =.5 R y := R CG s θ δ b s θ R y =.7 R x := R CG cos θ R x =.8 R y := R CG s θ R y =.58 R x := R CG cos θ + 8 deg R x =.8 R y := R CG s θ + 8 deg R y =.58 R Px := R p cos θ + δ R x R Px =.9 R Py := R p s θ + δ R y R Py = Clculte the x nd y coponents of the ccelertion of the CGs of ll ovg lks the globl coordte syste (GCS). G := R CG α ( s( θ ) + j cos( θ )) R CG ω ( cos( θ ) + j s( θ )) Gx := Re( G ) Gx = 7.65 sec Gy := I( G ) Gy =.5 sec ( ) A := α s θ + j cos θ ω cos θ ( + j s( θ )) ( )... ( j s( θ δ )) CGA := R CG α s θ + δ + j cos θ + δ + R CG ω cos θ + δ + + G := A + CGA Gx := Re G Gx =.678 ( ) sec Gy := I( G ) Gy =.9 sec G := R CG α s θ + j cos θ R CG ω cos θ Gx := Re( G ) Gx = 77. sec Gy := I( G ) Gy =.6 sec 7. Clculte the x nd y coponents of the externl force t P the CGS. Px := N Py := ( + j s( θ )) nd Edition, 999
12 DESIGN O MACHINERY SOLUTION MANUAL Substitute these given nd clculted vlues to the trix eqution.9. Note tht Mthcd requires tht ll eleents trix hve the se diension. Thus, the trix nd rry eqution.9 will be de diensionless nd the diensions will be put bck fter solvg it. C := R y R x R y R y R x R x R y R y R x R x R y R x C = nd Edition, 999
13 DESIGN O MACHINERY SOLUTION MANUAL -9-5 := Gx N Gy N I G α N N Gx Px N Gy Py N I G α R Px Py + R Py Px Gx N Gy N N I G α T = R := C x := x = 68 N y := y = 97 N x := x = 7 N y := y = 886 N x := 5 x = N y := 6 y = 766 N x := 7 x = 7 N y := 8 y = 7 N T := 9 T = 767 N nd Edition, 999
14 DESIGN O MACHINERY SOLUTION MANUAL -- PROBLEM - Stteent: Given: igure P-5b shows fourbr lkge nd its diensions eters. The steel crnk, coupler, nd rocker hve unifor cross sections of 5 dieter. In the stntneous position shown, the crnk O A hs ω = - rd/sec nd α = rd/sec. There is horizontl force t P of = N. d ll p forces nd the torque needed to drive the crnk t this stnt. Lk lengths: Lk (O to A) :=.86 Lk (A to B) b :=.85 Lk (B to O ) c :=.86 Lk (O to O ) d :=. Coupler pot: R p :=. δ := deg := N T := N Crnk ngle nd otion: θ := 6 deg ω := rd sec α := rd sec Lk cross-section dis: d lk := 5 Mteril density: D := 78 kg Solution: See igure P-5b nd Mthcd file P.. Use progr OURBAR to detere the position, velocity, nd ccelertion of lks nd. θ := 6.8 deg ω :=.85 rd sec α := 9.87 rd sec θ := 6.89 deg ω :=.7 rd sec α :=.6 rd sec. Detere the distnce to the CG the LRCS on ech of the three ovg lks. Lks nd : R CG :=.5 R CG =. R CG :=.5 c R CG =. Lk : R CG :=.5 b R CG =.95. Detere the ss nd oent of erti of ech lk. π d lk π d lk π d lk := D := b D := c D =.7 kg = 8. kg =.7 kg I G := d lk + I G =.8 kg I G := d lk + b I G = 8.85 kg I G := d lk + c I G =.8 kg. Set up n LNCS xy coordte syste t the CG of ech lk, nd drw ll pplicble vectors ctg on the syste s shown igure -. Drw free-body digr of ech ovg lk s shown igure -. nd Edition, 999
15 DESIGN O MACHINERY SOLUTION MANUAL -- y y x B y Gy P R R y Gx x R P R CG x O R CG R CG O x (d) BD of Lk A y () The coplete lkge with GCS y x y R x y Gy Gx y x R Gy P R P Gx R x R x x (b) BD of Lk y (c) BD of Lk 5. Clculte the x nd y coponents of the position vectors. R x := R CG cos θ + 8 deg R y := R CG s θ + 8 deg R x := R CG cos θ R y := R CG s θ R x := R CG cos θ + 8 deg R y := R CG s θ + 8 deg cos( θ + 8 deg) R x := R CG b s( θ + 8 deg) R y := R CG b R x =.8 R y =.5 R x =.8 R y =.5 R x =.6 R y =.666 R x =.6 R y =.666 nd Edition, 999
16 DESIGN O MACHINERY SOLUTION MANUAL -- R x := R CG cos θ R x =. R y := R CG s θ R y =. R x := R CG cos θ + 8 deg R x =. ( ) cos( θ ) ( ) s( θ ) R y := R CG s θ + 8 deg R y =. R Px := R p R CG R Px =.8 R Py := R p R CG R Py =.9 6. Clculte the x nd y coponents of the ccelertion of the CGs of ll ovg lks the globl coordte syste (GCS). G := R CG α ( s( θ ) + j cos( θ )) R CG ω ( cos( θ ) + j s( θ )) Gx := Re( G ) Gx =.6 sec Gy := I( G ) ( ) Gy = 8.75 sec A := α s θ + j cos θ ω cos θ ( )... ( j s( θ )) CGA := R CG α s θ + j cos θ + R CG ω cos θ + ( + j s( θ )) G := A + CGA Gx := Re( G ) Gx =. sec ( ) Gy := I( G ) Gy = 9.86 sec G := R CG α s θ + j cos θ R CG ω cos θ Gx := Re( G ) Gx =.56 sec Gy := I( G ) Gy = 8.6 sec 7. Clculte the x nd y coponents of the externl force t P the CGS. Px := Py := N ( + j s( θ )) 8. Substitute these given nd clculted vlues to the trix eqution.9. Note tht Mthcd requires tht ll eleents trix hve the se diension. Thus, the trix nd rry eqution.9 will be de diensionless nd the diensions will be put bck fter solvg it. nd Edition, 999
17 DESIGN O MACHINERY SOLUTION MANUAL -- C := R y R x R y R y R x R x R y R y R x R x R y R x := Gx N Gy N I G α N N Gx Px N Gy Py N I G α R Px Py + R Py Px Gx N Gy N N I G α T R := C x := x = 77 N y := y = 5 N x := x = N y := y = 6 N x := 5 x = 9. N y := 6 y = 7 N x := 7 x = 8 N y := 8 y = 6 N T := 9 T = 8. N nd Edition, 999
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