0
1.4kviews
The following data related to a connecting rod of reciprocating engine mass =40kg. Distance between centre =800mm, big and bearing diameter =80mm, small and bearing diameter =60mm.
1 Answer
0
33views

Time of oscillation when suspended from small end =175sec and from big and =1.6sec. Determine m – I about the axis passing there mass centre (ii) The dynamically equivalent system of 2 mass one located at small and centre.

Solution: L=L1+L2=800mm

tp1=1.75 sec/cycle.

tp1=2πk2+(L1)2g.L1

1.75=2πk2(L1+0.03)2g(L1+0.03)

k2=0.701L1+0.0237L21

tp2=2πk2+(L2)2gL2

Where,

L1=L1+0.03

L2=L2+0.04

But L2=0.8L1

So L2=0.8L1+0.04

L2=0.84L1

{For big end}

1.6=2πk2+(0.84L1)29.81(0.84L1)

k2=1.044L10.1717L21

0.701L1+0.0237L21=1.044L10.1717L21

L1=0.569 m

L1=0.569+0.030

L1=0.599

L2=0.8000.569

L2=0.2303

Substituting value of L1

K=0.31m

I_G = mK^2

=40 (0.31)^2

= 43.9 \ kgm^2

2]

enter image description here

k^2 = l_1 . l_2

(0.314)^2 = 0.569 . l_2

l_2 = 0.173 \ m

m_2 = \frac{mL_1}{L_2} = \frac{40\times 0.5694}{0.569 + 0.173} = 30.69 \ kg

m_1 = \frac{mL_2}{L_1 + L_2} = 9.31kg

Please log in to add an answer.