The primary force polygon from the data given in Table 2.9 (column 6) is drawn as shown in Fig

2.26 (a). Since the primary force polygon is a closed figure, therefore there are no unbalanced primary forces,

∴ Unbalanced Primary Force, U.P.F. = 0.

(a) (b) Fig. 2.27 Primary couple polygon

The primary couple polygon from the data given in Table 2.9 (column 8) is drawn as shown in Fig. 2.27 (a). The closing side of the polygon, shown dotted in the figure, represents unbalanced primary couple. By measurement, the unbalanced primary couple is proportional to 2.1 kg-m

.

∴Unbalanced Primary Couple,

U.P.C = 2.1 ×ω

= 2.1 (65.97)

² U.P.C = 9139.3 N-m.

Unbalanced secondary forces and couples

(a) (b)

Fig. 2.28 Secondary force polygon

The secondary crank positions are shown in Fig. 2.28 (b). From the secondary force polygon as shown in Fig. 2.28 (a), it is a closed figure. Therefore there are no unbalanced secondary forces.

∴ Unbalanced Secondary Force, U.S.F. = 0.

(a) (b) Fig. 2.29 Secondary couple polygon

The secondary couple polygon is shown in Fig. 2.29 (a). The unbalanced secondary couple is shown by dotted line. By measurement, we find that unbalanced secondary couple is proportional to 3.4 kg-m

.

∴Unbalanced Secondary Couple,

2 2

(65.97) . . . 3.4 3.4

0.45 / 0.1

U S C

 



 

U.S.C. = 3288.2 N.m

(n = l / r)

Example 2.13: A four stroke five cylinder in-line engine has a firing order of 1-4-5-3-2-1. The

centers lines of cylinders are spaced at equal intervals of 15 cm, the reciprocating parts per cylinder have a mass of 15 kg, the piston stroke is 10 cm and the connecting rods are 17.5 cm long. The engine rotates at 600 rpm. Determine the values of maximum primary and secondary unbalanced forces and couples about the central plane.

l = 10cm = 0.1 m or r = 0.05 m

m = 15 kg N = 600 r.p.m.

n = l / r = 17.5/5 = 3.5 2 2 600

60 60





  ^{ }

= 62.83 rad/s

Fig. 2.30 Cylinder plane position

Table 2.10

Angle

2

Angle



Plane Mass (m) kg

Radius (r) m

Cent.force ÷ ω

² (mr) kg-m

Distance from Ref. Plane (l) m

Couple ÷ ω

² (mrl) kg-m²

0° 0° 1 15 0.05 0.75 -0.3 -0.225

216° 288° 2 15 0.05 0.75 -0.15 -0.1125

72° 216° 3 15 0.05 0.75 0 0

144° 72° 4 15 0.05 0.75 0.15 0.1125

288° 144° 5 15 0.05 0.75 0.3 0.225

Unbalanced primary forces and couples

(a) (b) Fig. 2.31 Primary force polygon

The position of the cylinder planes and cranks is shown in Fig. 2.30 and Fig. 2.31(b) respectively. With reference to central plane of the engine, the data may be tabulated as above:

The primary force polygon from the data given in Table 2.10 (column 6) is drawn as shown in Fig. 2.31 (a). Since the primary force polygon is a closed figure, therefore there are no unbalanced primary forces,

∴ Unbalanced Primary Force, U.P.F. = 0.

(a) (b)

Fig. 2.32 Primary couple polygon

The primary couple polygon from the data given in Table 2.10 (column 8) is drawn as shown in Fig. 2.32 (a). The closing side of the polygon, shown dotted in the figure, represents unbalanced primary couple. By measurement, the unbalanced primary couple is proportional to 0.53 kg-m

.

∴Unbalanced Primary Couple,

U.P.C. = 0.53 ×ω

= 0.53 (62.83)

² U.P.C. = 2092.23 N-m.

Unbalanced secondary forces and couples

(a) (b) Fig. 2.33 Secondary force polygon

The secondary crank positions are shown in Fig. 2.33 (b). From the secondary force polygon as shown in Fig. 2.33 (a), it is a closed figure. Therefore there are no unbalanced secondary forces.

∴ Unbalanced Secondary Force, U.S.F. = 0.

(a) (b) Fig. 2.34 Secondary couple polygon

The secondary couple polygon is shown in Fig. 2.34 (a). The unbalanced secondary couple is shown by dotted line. By measurement, we find that unbalanced secondary couple is proportional to 0.18 kg-m

.

∴Unbalanced Secondary Couple,

. . . 0.18

(62.83) . . . 0.18

3.5

 



 

U S C

n U S C

U.S.C. = 203 N.m

(n = l / r)

Example 2.14: In an in-line six cylinder engine working on two stroke cycle, the cylinder centre

lines are spaced at 600 mm. In the end view, the cranks are 60° apart and in the order 1-4-5-2-3-6. The stroke of each piston is 400 mm and the connecting rod length is 1 m. The mass of the reciprocating parts is 200 kg per cylinder and that of rotating parts 100 kg per crank. The engine rotates at 300 r.p.m. Examine the engine for the balance of primary and secondary forces and couples. Find the maximum unbalanced forces and couples.

L = 400 mm or r = L / 2 = 200 mm = 0.2 m

l = 1 m N = 300 r.p.m.

m

= 200 kg 2 2 300

60 60





  ^{ }

= 31.42 rad/s m

= 100 kg

Fig. 2.35 Positions of planes of cylinders

Table 2.11

Angle 2

Angle



Plane Mass (m) kg

Radius (r) m

Cent. force ÷ ω

² (mr) kg-m

Distance from Ref. Plane (l) m

Couple ÷ ω

² (mrl) kg-m²

0° 0° 1 300 0.2 60 -1.5 -90

360° 180° 2 300 0.2 60 -0.9 -54

120° 240° 3 300 0.2 60 -0.3 -18

120° 60° 4 300 0.2 60 0.3 18

240° 120° 5 300 0.2 60 0.9 54

240° 300° 6 300 0.2 60 1.5 90

Unbalanced primary forces and couples

(a) (b)

Fig. 2.36 Primary force polygon

The position of the cylinder planes and cranks is shown in Fig. 2.35 and Fig. 2.36(b) respectively. With reference to central plane of the engine, the data may be tabulated as above:

The primary force polygon from the data given in Table 2.11 (column 6) is drawn as shown in Fig. 2.36 (a). Since the primary force polygon is a closed figure, therefore there are no unbalanced primary forces,

∴ Unbalanced Primary Force, U.P.F. = 0.

(a) (b) Fig. 2.37 Primary couple polygon

The primary couple polygon from the data given in Table 2.11 (column 8) is drawn as shown in Fig. 2.37 (a). Since the primary couple polygon is a closed figure, therefore there are no unbalanced primary couples,

∴Unbalanced Primary Couple, U.P.C. = 0

Unbalanced secondary forces and couples

(a) (b) Fig. 2.38 Secondary force polygon

The secondary crank positions are shown in Fig. 2.38 (b). From the secondary force polygon as shown in Fig. 2.38 (a), it is a closed figure. Therefore there are no unbalanced secondary forces.

∴ Unbalanced Secondary Force, U.S.F. = 0.

(a) (b) Fig. 2.39 Secondary couple polygon

The secondary couple polygon is shown in Fig. 2.39 (a). The unbalanced secondary couple is shown by dotted line. By measurement, we find that unbalanced secondary couple is proportional to 249 kg-m

.

∴Unbalanced Secondary Couple,

. . . 249

(31.42) . . . 249

1 / 0.2

 



 

U S C

n U S C

U.S.C. = 49163.37 N.m

(n = l / r)

Example 2.15: The firing order in a 6 cylinder vertical four stroke in-line engine is 1-4-2-6-3-5.

The piston stroke is 100 mm and the length of each connecting rod is 200 mm. The pitch distances between the cylinder centre lines are 100 mm, 100 mm, 150 mm, 100 mm, and 100 mm respectively. The reciprocating mass per cylinder is 1 kg and the engine runs at 3000 r.p.m.

Determine the out-of-balance primary and secondary forces and couples on this engine, taking

In document Balancing of Rotating Masses (Page 60-66)