The following data relate to the connecting rod of reciprocating engines Mass=50 kg. Distance between bearing centers = 900 mm, Diameter of the big end bearing = 100 mm, Diameter of small end bearing = 80 mm, Time of oscillation when a connecting rod is suspended from big end = 1.7 sec and small end = 1.85 sec,

Determine

1. The radius of gyration ‘k’ of the rod about an axis through center of mass perpendicular to plane of oscillation,
2. the moment of inertia of the rod about same axis, and
3. The dynamically equivalent system of the connecting rod comprising two masses, one at the small end bearing center.

Mumbai University > Mechanical Engineering > Sem 5 > Theory of Machines-II

Marks: 10 Marks

Year: May 20165

Let La=length of the equivalent simple pendulum when suspended from top of the big end bearing

Lb=length of the equivalent simple pendulum when suspended from top of the small end bearing

a=distance of centre of mass ‘G’ from top of the big end bearing

b= distance of centre of mass ‘G’ from top of the small end bearing

$ta=2Π \sqrt{\dfrac{La}{g}} \ \ \ \ \& \ \ \ \ tb=2Π \sqrt{\dfrac{Lb}{g}} \\ 1.7=2Π \sqrt{\dfrac{La}{g}} \ \ \ \ \& \ \ \ \ 1.85=2Π\sqrt{\dfrac{Lb}{g}} \\ \therefore La=0.7181m \& Lb=0.8505m \\ a+k^2/a=0.7181 \& b+k^2/b=0.8505 \\ k^2=0.7181a-a^2=0.8505b-b^2 \ \ \ \ ----------------------------- (I) \\ but, a+b=900+100/2+80/2 \\ =990mm \\ =0.99m \\ a=0.99-b \ \ \ \ --------------------------------(II) \\ \text{substituting the value of a in equation} (I) \\ 0.7181*(0.99-b)-(0.99-b)^2=.8505b-b^2 \\ \text{On solving}, b=0.654m \\ a=0.99-0.654 \\ a= 0.336m \\ k^2=0.8505*0.654-(0.654)^2 \\ =0.1286 \\ k=0.358m \\ MOI=I=mk^2 =50*(0.358)^2 \\ =6.4kg-m^2$

The distance of centre of mass of the rod from the centre of the small end bearing

b’=654-80/2

=614mm

Let the second mass be placed at ‘D’

$\therefore$ GD=d and md=mass at D

d=k2/b’=0.1285/0.614=.209m

md=mb’/b’+d = 50.614/.614+.209 = 37.3kg

Mb’= 50-37.3=12.7kg

Where Mb’ is the mass at the smaller end bearing.