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MEC 3101 DYNAMICS OF MACHINERY
DATE
MODULE III
BALANCING / PART 1
Rotatting Masses
Dr. S. RASOOL MOHIDEEN, Dean, SMS
THANK YOU!
TOPICS COVERED IN MODULE III
Balancing of Rotating masses
- Balancing of single revolving mass
- Balancing of several masses revolving the same plane
- Balancing of several masses revolving in different planes
Balancing of reciprocating masses
- In the multi cylinder inline engine
- In the multi cylinder V engine
- In the locomotive engine
INTRO
31
INTRO
Forces Acting in Engine
ROTATING MASSES
Force acting along the connecting rod:
Thrust on the sides of the cylinder walls or normal reaction on the guide bars:
Thrust on crank shaft bearings:
Crank-pin effort:
Crank effort or turning moment or torque on the crank shaft:
PROBLEM 2
The crank-pin circle radius of a horizontal engine is 300 mm. The mass of the reciprocating parts is 250 kg. When the crank has travelled 60° from I.D.C., the difference between the driving and the back pressures is 0.35 N/sq.mm. The connecting rod length between centres is 1.2 m and the cylinder bore is 0.5 m. If the engine runs at 250 r.p.m. and if the effect of piston rod diameter is neglected, calculate : (i). Piston Effort , (ii). Thrust on slide bars, (iii). Thrust in the connecting rod, (iv). Tangential force on the crank-pin, and (v). Turning moment on the crank shaft.
Problem 2
Given: r = 300 mm = 0.3 m ; mR = 250 kg; θ = 60°; p1 – p2 = 0.35 N/mm.sq; l = 1.2 m ; D = 0.5 m = 500 mm ; N = 250 r.p.m. or ω = 2 π × 250/60 = 26.2 rad/s
(i) To find the piston effort (FP):
We know that net load on the piston,
Ratio of length of connecting rod and crank,
n = l / r = 1.2 / 0.3 = 4
and accelerating or inertia force on reciprocating parts,
Piston effort, FP = FL – FI = 68 730 – 19 306 = 49 424 N = 49.424 kN
Solution
(ii) Thrust on slide bars:
Let φ phi = Angle of inclination of the connecting rod to the line of stroke
φHence Phi = 12.5°
We know that thrust on the slide bars,
FN = FP tan (phi)φ = 49.424 × tan 12.5° = 10.96 kN
(iii) Thrust in the connecting rod:
thrust in the connecting rod,
Solution (Contd.)
(iv) Tangential force on the crank-pin:
We know that tangential force on the crank pin,
FT = FQ sin (θ theta + phi φ) = 50.62 sin (60° + 12.5°)
= 48.28 kN Ans.
(v) Turning moment on the crank shaft:
We know that turning moment on the crank shaft,
T = FT × r = 48.28 × 0.3 = 14.484 kN-m Ans.
Solution (Contd.)
Dynamically Equivalent System