Given: m = 300 kg ;r = 0.3 m ; D = 1.8 m or R = 0.9 m ; a = 0.65 m ; Hammer blow = 46 kN = 46 x 103 N ; v = 96.5 km/h = 26.8 m/s
1. Fraction of the reciprocating masses to be balanced
Let c = Fraction of the reciprocating masses to be balanced, and B = Magnitude of balancing mass placed at each of the driving wheels at radius b. We know that the mass of the reciprocating parts to be balanced = c.m = 300 c kg
(b) Position of cranks
The position of planes of the wheels and cylinders is shown in Fig.(a), and the position of cranks is shown in Fig(b). Assuming the plane of wheel A as the reference plane, the data may be tabulated as below:
Plane (1) |
Mass (m) kg (2) |
Radius (r) m (3) |
Cent. force/ω2(m.r) kg-m (4) |
Distance from plane A (l) m (5) |
Couple/ω2(m.r.l.) kg-m2 (6 |
A (R.P.)
B
C
D |
B
300 c
300 c
B |
b
0.3
0.3
b |
B.b
90 c
90 c
B.b |
0
0.45
1.1
1.55 |
0
40.5 c
99 c
1.55 B.b |
Now the couple polygon, to some suitable scale, may be drawn with the data given in Table (column 6), as shown in Fig.. The closing side of the polygon (vector c0) represents the balancing couple and is proportional to 1.55 B.b. From the couple polygon,
1.55 B.b = v((40.5c)2 + (99c) 2) = 107c
We know that angular speed, 00=vIR = 26.8/0.9 = 29.8 rad/s
Hammer blow,
46 x 103 = B.m2 .b = 69 c (29.8)2 = 61275 c
c = 46 x 103/61275 = 0.751
2. Variation in tr active effort
We know that variation in tr active effort
= ± v2(l-c)m .. 002.r = ±v 2(1-0.751) 300 29.8f
0.3 = 28 140 N = 28.14 kN
Maximum swaying couple
We know the maximum swaying couple