Mathcad Machines 7.4b Reborn
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- Words: 1,200
- Pages: 6
Ryan Sullenberger Oct 17, 2008
MECH324
7.4b, open circuit only: The link lengths, coupler point location, and the values of θ2 , ω2 and α2 for the same fourbar linkages as used for position and velocity analysis in Chapters 4 and 6 are redefined in Table P7-1, which is the same as Table P6-1 (p. 371). The general linkage configuration and terminology are shown in Figure P7-1. For the row(s) assigned, draw the linkage to scale and graphically find the accelerations of points A and B. Then calculate α3 and α4 and the acceleration of point P. Repeat this problem solving by the analytical vector loop method of Section 7.3. Link1 := 7
Link2 := 9
α2 := 5
Link3 := 3
R PA := 9
R 1 := Link1
R 2 := Link2
Link4 := 8
θ2 := 85deg
ω2 := −12
δ3 := 25deg R 3 := Link3
R 4 := Link4
c := Link4
d := Link1
R 2 + R 3 − R4 − R1 = 0
a := Link2
b := Link3
INPUTS Intermediate Var K1 :=
K2 :=
K3 :=
( )
d a d c
Intermediate Var (cont) K1 = 0.778
K4 :=
K2 = 0.875
K5 :=
(a2 − b2 + c2 + d2) 2a⋅ c
( )
A := cos θ2 − K1 − K2⋅ cos θ2 + K3
d
K4 = 2.333
b
( c 2 − d 2 − a 2 − b 2) 2 ⋅ a⋅ b
K5 = −1.389
K3 = 1.285
A = 0.518
( )
( )
D := cos θ2 − K1 + K4⋅ cos θ2 + K5
( )
B := −2 ⋅ sin θ2
D = −1.876
B = −1.992
( )
C := K1 − ( K2 + 1 ) ⋅ cos θ2 + K3
( )
E := −2 ⋅ sin θ2
C = 1.899
( )
F := K1 + ( K4 − 1 ) ⋅ cos θ2 + K5 2 F = −0.495 E − 4 ⋅ D⋅ F = 0.256
2
B − 4 ⋅ A⋅ C = 0.036
OUTPUTS (
)
−E − E2 − 4 ⋅ D⋅ F 2⋅ D
θ31 := 2 ⋅ atan
θ31 = −0.755
180 θ31DEG := θ31⋅ π
E = −1.992
(
θ32 := 2 ⋅ atan22 ⋅ D , −E +
θ31DEG = −43.232
)
2
E − 4 ⋅ D⋅ F = 5.109
θ32 = 5.109
180 θ32DEG := θ32⋅ π
(
θ32DEG = 292.696 2
θ41 := 2 ⋅ atan2 2 ⋅ A , −B − B − 4 ⋅ A⋅ C
)
θ41 = 2.099
180 θ41DEG := θ41⋅ π
(
θ42 := 2 ⋅ atan2 2 ⋅ A , −B +
θ41DEG = 120.247
2
B − 4 ⋅ A⋅ C
)
θ42 = 2.255
180 θ42DEG := θ42⋅ π
θ42DEG = 129.217
O2X := 0 O2Y := 0
( )
AX := a⋅ cos θ2
( )
AY := a⋅ sin θ2
( )
B X1 := d + c⋅ cos θ41
( )
B X2 := d + c⋅ cos θ42
( )
( )
B Y1 := c⋅ sin θ41
B Y2 := c⋅ sin θ42
O4X := d O4Y := 0
O2X AX X1 := BX1 O4 X
0 0.784 X1 = 2.97 7
O2X AX X2 := BX2 O4 X
0 0.784 X2 = 1.942 7
O2Y AY Y1 := BY1 O4 Y
0 8.966 Y1 = 6.911 0
O2Y AY Y2 := BY2 O4 Y
0 8.966 Y1 = 6.911 0
Open
Crossed 10
8
8
6
6
Y1
Y2 4
4
2 0
2 0
2
4
6
X1
( (
0
2
4 X2
VELOCITY ANALYSIS
ω3 :=
0
) )
a⋅ ω2 sin θ41 − θ2 ⋅ = 73.06 b sin θ31 − θ41
6
( (
) )
a⋅ ω2 sin θ2 − θ31 ⋅ = −37.292 c sin θ41 − θ31
ω4 :=
( ( )
( ))
VA := a⋅ ω2 ⋅ cos θ2 i − sin θ2 VA = 107.589 − 9.413i VA = 108
(
( )
( ))
VB := c⋅ ω4 ⋅ −sin θ42 + cos θ42 i VB = 231.135 + 188.623i VB = 298.332
CONTINUED ANALYSIS θ1 := 0deg
a⋅ e
i⋅ θ2
+ b⋅ e
i ⋅ a⋅ ω2 ⋅ e
i⋅ θ2
i⋅ θ31
− c⋅ e
+ i ⋅ b ⋅ ω3 ⋅ e
i⋅ θ41
i⋅ θ31
− d⋅ e
i⋅ θ1
=0
− i ⋅ c⋅ ω4 ⋅ e
i⋅ θ41
=0
i⋅ θ2 i⋅ θ i⋅ θ 2 i⋅ θ2 + b⋅ α3 ⋅ i⋅ e 31 − b ⋅ ω3 2⋅ e 31 a⋅ α2 ⋅ i ⋅ e − a⋅ ω2 ⋅ e
− c⋅ α4 ⋅ i ⋅ e
i⋅ θ41
2 i⋅ θ41
− c⋅ ω4 ⋅ e
=0
Guesses α3 := 1
α4 := 1
Given
( )
2
( )
( )
( )
2
( )
( )
2
( )
( )
2
( )
( )
( )
2
( )
−a⋅ α2 ⋅ sin θ2 − a⋅ ω2 ⋅ cos θ2 − b ⋅ α3 ⋅ sin θ31 − b ⋅ ω3 ⋅ cos θ31 + c⋅ α4 ⋅ sin θ41 + c⋅ ω4 ⋅ cos θ41 = 0
2
a⋅ α2 ⋅ cos θ2 − a⋅ ω2 ⋅ sin θ2 + b ⋅ α3 ⋅ cos θ31 − b ⋅ ω3 ⋅ sin θ31 − c⋅ α4 ⋅ cos θ41 + c⋅ ω4 ⋅ sin θ41 = 0
(
ACCELS := Find α3 , α4
)
−2.983 × 104 ACCELS = 4 1.139 × 10
4
α3 := ACCELS = −2.983 × 10 0
α4 := ACCELS = 1.139 × 10 1
4
Find the acceleration of point P:
p := RPA Ap = AA + APA AA := −e
i⋅ θ2
( ) 2 + i⋅ e
⋅ a⋅ ω2
APA := p ⋅ ACCELS ⋅ i ⋅ e 0
(
i⋅ θ2
3
⋅ a⋅ α2
AA = −157.783 − 1.287i × 10
) − p⋅ ω
i⋅ θ31+ δ3
(
)
2 i⋅ θ31+δ3
3 ⋅e
5
5
APA = −1.296 × 10 − 2.399i × 10
Ap := AA + APA 5
Ap = −1.298 × 10 − 2.412i × 10 5
Ap = 2.739 × 10
5
(
5
5
atan2 −2.412⋅ 10 , −1.298⋅ 10
) = −151.713⋅deg
linkage configuration
MECH324
7.4b, open circuit only: The link lengths, coupler point location, and the values of θ2 , ω2 and α2 for the same fourbar linkages as used for position and velocity analysis in Chapters 4 and 6 are redefined in Table P7-1, which is the same as Table P6-1 (p. 371). The general linkage configuration and terminology are shown in Figure P7-1. For the row(s) assigned, draw the linkage to scale and graphically find the accelerations of points A and B. Then calculate α3 and α4 and the acceleration of point P. Repeat this problem solving by the analytical vector loop method of Section 7.3. Link1 := 7
Link2 := 9
α2 := 5
Link3 := 3
R PA := 9
R 1 := Link1
R 2 := Link2
Link4 := 8
θ2 := 85deg
ω2 := −12
δ3 := 25deg R 3 := Link3
R 4 := Link4
c := Link4
d := Link1
R 2 + R 3 − R4 − R1 = 0
a := Link2
b := Link3
INPUTS Intermediate Var K1 :=
K2 :=
K3 :=
( )
d a d c
Intermediate Var (cont) K1 = 0.778
K4 :=
K2 = 0.875
K5 :=
(a2 − b2 + c2 + d2) 2a⋅ c
( )
A := cos θ2 − K1 − K2⋅ cos θ2 + K3
d
K4 = 2.333
b
( c 2 − d 2 − a 2 − b 2) 2 ⋅ a⋅ b
K5 = −1.389
K3 = 1.285
A = 0.518
( )
( )
D := cos θ2 − K1 + K4⋅ cos θ2 + K5
( )
B := −2 ⋅ sin θ2
D = −1.876
B = −1.992
( )
C := K1 − ( K2 + 1 ) ⋅ cos θ2 + K3
( )
E := −2 ⋅ sin θ2
C = 1.899
( )
F := K1 + ( K4 − 1 ) ⋅ cos θ2 + K5 2 F = −0.495 E − 4 ⋅ D⋅ F = 0.256
2
B − 4 ⋅ A⋅ C = 0.036
OUTPUTS (
)
−E − E2 − 4 ⋅ D⋅ F 2⋅ D
θ31 := 2 ⋅ atan
θ31 = −0.755
180 θ31DEG := θ31⋅ π
E = −1.992
(
θ32 := 2 ⋅ atan22 ⋅ D , −E +
θ31DEG = −43.232
)
2
E − 4 ⋅ D⋅ F = 5.109
θ32 = 5.109
180 θ32DEG := θ32⋅ π
(
θ32DEG = 292.696 2
θ41 := 2 ⋅ atan2 2 ⋅ A , −B − B − 4 ⋅ A⋅ C
)
θ41 = 2.099
180 θ41DEG := θ41⋅ π
(
θ42 := 2 ⋅ atan2 2 ⋅ A , −B +
θ41DEG = 120.247
2
B − 4 ⋅ A⋅ C
)
θ42 = 2.255
180 θ42DEG := θ42⋅ π
θ42DEG = 129.217
O2X := 0 O2Y := 0
( )
AX := a⋅ cos θ2
( )
AY := a⋅ sin θ2
( )
B X1 := d + c⋅ cos θ41
( )
B X2 := d + c⋅ cos θ42
( )
( )
B Y1 := c⋅ sin θ41
B Y2 := c⋅ sin θ42
O4X := d O4Y := 0
O2X AX X1 := BX1 O4 X
0 0.784 X1 = 2.97 7
O2X AX X2 := BX2 O4 X
0 0.784 X2 = 1.942 7
O2Y AY Y1 := BY1 O4 Y
0 8.966 Y1 = 6.911 0
O2Y AY Y2 := BY2 O4 Y
0 8.966 Y1 = 6.911 0
Open
Crossed 10
8
8
6
6
Y1
Y2 4
4
2 0
2 0
2
4
6
X1
( (
0
2
4 X2
VELOCITY ANALYSIS
ω3 :=
0
) )
a⋅ ω2 sin θ41 − θ2 ⋅ = 73.06 b sin θ31 − θ41
6
( (
) )
a⋅ ω2 sin θ2 − θ31 ⋅ = −37.292 c sin θ41 − θ31
ω4 :=
( ( )
( ))
VA := a⋅ ω2 ⋅ cos θ2 i − sin θ2 VA = 107.589 − 9.413i VA = 108
(
( )
( ))
VB := c⋅ ω4 ⋅ −sin θ42 + cos θ42 i VB = 231.135 + 188.623i VB = 298.332
CONTINUED ANALYSIS θ1 := 0deg
a⋅ e
i⋅ θ2
+ b⋅ e
i ⋅ a⋅ ω2 ⋅ e
i⋅ θ2
i⋅ θ31
− c⋅ e
+ i ⋅ b ⋅ ω3 ⋅ e
i⋅ θ41
i⋅ θ31
− d⋅ e
i⋅ θ1
=0
− i ⋅ c⋅ ω4 ⋅ e
i⋅ θ41
=0
i⋅ θ2 i⋅ θ i⋅ θ 2 i⋅ θ2 + b⋅ α3 ⋅ i⋅ e 31 − b ⋅ ω3 2⋅ e 31 a⋅ α2 ⋅ i ⋅ e − a⋅ ω2 ⋅ e
− c⋅ α4 ⋅ i ⋅ e
i⋅ θ41
2 i⋅ θ41
− c⋅ ω4 ⋅ e
=0
Guesses α3 := 1
α4 := 1
Given
( )
2
( )
( )
( )
2
( )
( )
2
( )
( )
2
( )
( )
( )
2
( )
−a⋅ α2 ⋅ sin θ2 − a⋅ ω2 ⋅ cos θ2 − b ⋅ α3 ⋅ sin θ31 − b ⋅ ω3 ⋅ cos θ31 + c⋅ α4 ⋅ sin θ41 + c⋅ ω4 ⋅ cos θ41 = 0
2
a⋅ α2 ⋅ cos θ2 − a⋅ ω2 ⋅ sin θ2 + b ⋅ α3 ⋅ cos θ31 − b ⋅ ω3 ⋅ sin θ31 − c⋅ α4 ⋅ cos θ41 + c⋅ ω4 ⋅ sin θ41 = 0
(
ACCELS := Find α3 , α4
)
−2.983 × 104 ACCELS = 4 1.139 × 10
4
α3 := ACCELS = −2.983 × 10 0
α4 := ACCELS = 1.139 × 10 1
4
Find the acceleration of point P:
p := RPA Ap = AA + APA AA := −e
i⋅ θ2
( ) 2 + i⋅ e
⋅ a⋅ ω2
APA := p ⋅ ACCELS ⋅ i ⋅ e 0
(
i⋅ θ2
3
⋅ a⋅ α2
AA = −157.783 − 1.287i × 10
) − p⋅ ω
i⋅ θ31+ δ3
(
)
2 i⋅ θ31+δ3
3 ⋅e
5
5
APA = −1.296 × 10 − 2.399i × 10
Ap := AA + APA 5
Ap = −1.298 × 10 − 2.412i × 10 5
Ap = 2.739 × 10
5
(
5
5
atan2 −2.412⋅ 10 , −1.298⋅ 10
) = −151.713⋅deg
linkage configuration
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