Slider Crank Mechanism: Practice Problems and Solutions
Practice problem-1
based on today class.
Practice problem -1
: In a slider – crank mechanism, the crank is 480 mm long and rotates at
20 rad/s in the counter-clockwise direction. The length of the connecting rod is 1.6 m. When the
Crank turns 60
0
from the inner-dead centre, determine the
(i)
Velocity of the slider (answer : 9.7 m/s)
(ii)
Velocity of a point E located at a distance 450 mm on the connecting rod extended
(answer : 10.2 m/s)
(iii) Angular velocity of the connecting rod (answer: 3.28 rad/s clockwise)
OA = 480 mm
AB = 1.6 m
Velocity of point A
Magnitude
V
A
= OA x
ω
OA
= 0.48 x 20
= 9.60 m/s
Direction
Velocity of A is perpendicular to OA
In OAB,
Angle B = 15.06 degree
Angle A = 180 – 60 – 15.06 = 104.94 degree
o
A
B
60
0
1.6 m
15.06
0
104.94
0
V
A
75.06
0
14.94
0
V
B
Velocity component of A along AB = V
A
Cos ( 14.94
0
)
Velocity component of B along AB = V
B
Cos ( 15.06
0
)
Velocity component of A along AB = Velocity component of B along AB
V
A
Cos ( 14.94
0
) = V
B
Cos ( 15.06
0
)
V
B
= 9.6 x Cos ( 14.94
0
) / Cos ( 15.06
0
) = 9.605 m/s
o
A
B
60
0
1.6 m
15.06
0
104.94
0
V
A
75.06
0
14.94
0
V
B
DIRECTION PERPENDICULAR TO AB
DIRECTION
PERPENDICULAR TO AB
Velocity component of A perpendicular to AB
= V
A
Sin ( 14.94
0
) = 9.6 x Sin ( 14.94
0
)
Velocity component of B perpendicular to AB
= V
B
Sin ( 15.06
0
) = 9.605 x Sin ( 15.06
0
)
Angular velocity of Link AB = (V
A
Sin ( 14.94
0
) + V
B
Sin ( 15.06
0
) ) / AB =
3.11 rad/s
Velocity of E along AB , V
ex
= Velocity of A along AB = Velocity of B along AB.
Velocity of A along AB ,V
Ex
= = V
A
Cos ( 14.94
0
) = 9.28 m/s
V
A
sin ( 14.94
0
) =2.47 m/s
V
B
sin ( 15.06
0
) =2.50 m/s
V
by
= 2.5 m/s
V
Ay
= 2.47 m/s
1.6 m
V
Ey
A
B
E
0.45 m
m/s
m/s
m/s
Explore a practice problem based on a slider crank mechanism, involving calculations for velocity of the slider, velocity of a point on the connecting rod, and angular velocity. Detailed steps and solutions provided to understand the concepts clearly.
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Presentation Transcript
Practice problem-1 based on today class.
Practice problem -1: In a slider crank mechanism, the crank is 480 mm long and rotates at 20 rad/s in the counter-clockwise direction. The length of the connecting rod is 1.6 m. When the Crank turns 600 from the inner-dead centre, determine the (i) Velocity of the slider (answer : 9.7 m/s) (ii) Velocity of a point E located at a distance 450 mm on the connecting rod extended (answer : 10.2 m/s) (iii) Angular velocity of the connecting rod (answer: 3.28 rad/s clockwise)
A 1.6 m 104.940 20 rad/s 600 15.060 o B OA = 480 mm In OAB, AB = 1.6 m Velocity of point A Magnitude VA = OA x OA = 0.48 x 20 = 9.60 m/s Direction Velocity of A is perpendicular to OA Angle B = 15.06 degree Angle A = 180 60 15.06 = 104.94 degree
VA 14.940 A 1.6 m 75.060 104.940 15.060 600 o VB B Velocity component of A along AB = VA Cos ( 14.940 ) Velocity component of B along AB = VB Cos ( 15.060 ) Velocity component of A along AB = Velocity component of B along AB VA Cos ( 14.940 ) = VB Cos ( 15.060 ) VB = 9.6 x Cos ( 14.940 ) / Cos ( 15.060 ) = 9.605 m/s
DIRECTION PERPENDICULAR TO AB VA 14.940 A 1.6 m 75.060 104.940 15.060 600 o VB B Velocity component of A perpendicular to AB = VA Sin ( 14.940) = 9.6 x Sin ( 14.940) DIRECTION PERPENDICULAR TO AB Velocity component of B perpendicular to AB = VB Sin ( 15.060) = 9.605 x Sin ( 15.060) Angular velocity of Link AB = (VA Sin ( 14.940) + VB Sin ( 15.060) ) / AB = 3.11 rad/s
Velocity of E along AB , Vex = Velocity of A along AB = Velocity of B along AB. m/s Velocity of A along AB ,VEx = = VA Cos ( 14.940 ) = 9.28 m/s VA sin ( 14.940 ) =2.47 m/s VB sin ( 15.060 ) =2.50 m/s VEy VAy = 2.47 m/s 1.6 m 0.45 m B E A Vby = 2.5 m/s m/s m/s
