Solved: Balancing Rotating Masses & Friction Clutch Assignment

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Added on  2023/06/04

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Homework Assignment
AI Summary
This assignment solution covers two main topics in mechanical engineering: balancing of rotating masses and friction clutches. The first question involves determining the mass required to balance a rotating system, calculating the necessary balancing mass (MB) and its angular position. The second question deals with a numerical series related to rotational speeds, calculating unknown speeds. The third question calculates the length of an open belt drive and the power transmitted, considering the angle of contact and friction coefficient. The final question focuses on a single-plate friction clutch, calculating the required force under both uniform pressure and uniform wear conditions. Desklib offers a variety of solved assignments and past papers for students.
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Q1)
Radius into the x and y component
r1x =r1x = r1cos45 = 60cos45 = 42.42
r1y = 60sin45 = 42.42
r2x = 60*co115 = -25.35
r2y = 60*sin115 = 54.37
r3x = r2cos315 = 42.42
r3y = 60*sin315 = -42.42
find the mass MB to be added in the plane R at radius RBL distance LB
take summation of moments about plane L along x – axis
MBRBxLB = -m1r1x*L1 – m2*Rr2x – m3r3x*L3
= -15.1*42.42 *20 -39.35*25.35*23 -29.5*42.42*11
= -49519.15
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MBRBx = -49519.15/LB
Therefore, LB = R2 – L2
= 23- 14
= 9
MBRBx = -49519.15/9
= -5502.13
Similarly taking the summation of moments about L along y – axix
MBRBxLB = -m1r1y*L1 – m2*Rr2y – m3r3y*L3
= -15.1*42.42 *20 -39.35*54.37*23 -29.5*42.42*11
= -75783.70
MBRBx = -75783.70/LB
Therefore, LB = R2 – L2
= 23- 14
= 9
MBRBx = -75783.70/9
= -8420.41
The resultant of the above
MBRB = M B R Bx2+ M B R By2
= 5502.132 +8420.412
= 10,058.66
MB = 10,058.66/RB = 10,058.66/60 = 167.64 kg
ƟB = tan1
( MBRBy
MBRBx )=tan1
( 8420.41
5502.13 ) =
= 56.840
Since m2 is on right side its sign will change
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Now taking moment about R along y – axis
MARAxLA = -m1r1y*L1 +m2*Rr2y – m3r3y*L3
= -15.1*42.42 *20 + 39.35*54.37*23 +29.5*42.42*11
= 3093.91
Therefore, MaRAx = -109.962/9 = -12.218
= 3093.91/9 = 343.77
The resultant of the above
MARA = M A R Ax2 +M A R Ay2
= 343.772 +12.2182
= 343.99
MA = 343.99/RB = 343.99/100 = 3.4399 kg
ƟA = tan1
( MARAy
MARAx )=tan1
( 343.99
12.218 ) = -87.970
Hence the mass of 3.4399 kg at an angle of -87.970 or 87.970 is to be added on plane Lat 100 radius to
balance
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Q2)
N3 = 2N2 + N1
N3 = 2*70 + 50 = 190
N6 = 2N5 + N4
365 = 2*83 + N4
N4 = 199
w 1wH 1
w 2wH 1 =N 2
N 1
1 => fixed
W1 = 0, WH1 = -72
0 (72 )
w 2 (72 ) =70
50
-70(w2 + 72) = 72 *50
W2 = -123.43 rpm
w 2wH 1
w 3wH 1 =N 2
N 1
123.43 (72 )
w 2 (72 ) = 190
70
190(w2 + 72) = 70 *(-123.43 + 72)
W2 = -90.95 rpm
W3 = W6 = -90.95 rpm
4 => fixed
W4 = 0
w 6wH 2
w 5wH 2 = N 5
N 6 …1
w 5wH 2
w 4wH 2 =N 4
N 5 …2
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Multiply 1 and 2
w 6wH 2
w 5wH 2 =
N 5
N 6 N 4
N 5 =N 4
N 6
W4 = 0, W6 = -90.95 rpm
90.95wH 2
0wH 2 =199
365
199WH2 = 365(-90.95 –WH2)
WH2 = -58.86
Arm H2 will rotate with w = 58.86 rpm in counter clockwise direction WH2 = -58.66 rpm
Q3)
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Let the length of open belt be L0
β=angle submitted by each common tangent AB CD with OZ
O is center of big pulley and z is center of small pulley
Angle of contact of Big pulley = π +2 β
Angle of contact of small pulley = π2 β
sinβ= Rr
L =
301
2 100
2
2500
sinβ=0.0402
β=2.3040
Angle of contact of bigger pulley is = 180 + 2*2.304 = 184.610
Angle of contact of smaller pulley is = 180 - 2*2.304 = 175.3920
Length of Belt
L0 = π ( R+ r ) + ( Rr ) 2
L + 2 L
= π( 301
2 + 100
2 ) +
301
2 100
2
2500 +22500
L0 = 629.89 + 4.0401 + 5000
= 5633.9301 mm
= 5.633 m
Calculation for power transmission
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Tension in tight side
T1 = 3 kN
Tension slack side
T2 = ?
μ = 0.3
angle of contact Ɵ = 184.61 *π/180 = 3.222
now
T1/T2 = eμθ = 2/T2 = e0.33.222
T2 = 1.1411 kN
Power transmitted = (T1 – T2)V = (3 – 1.1411)*
2 π 81000
60 301
2
P = 29.296 KN
Q4)
Single plate friction clutch
Power transmission P = 20.03 *103 N
Initial radius = r1 = 0.0045 m
Final radius = r0 = 0.135 m
Speed N = 501 rev/min
N = 501/60 = 8.333 rev/sec
Coefficient of friction = 0.3
a) For uniform pressure
P = TW
Torque = T = P/W = 20.03 *103/2πN
= 20.03 *103/2π*8.333
T = 382.56 N-m
But Torque T = μFr
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F = thrust force
R = mean radius of of friction surface
R =
2
3r 03r 13
r 02r 12
=
2
30.13530.0453
0.13520.0452
= 0.0975
382.56 = 0.3 * F*0.0975
F = 13078.98 N
b) Pressure is inversely proportion to radius or uniform wear condition
T = μFR
R = r 0+r 1
2
R = 0.135+0.045
2
= 0.09
Since
T = μFR
382.56 = 0.3 * F * 0.09
F = 14168. 89 N
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