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A Single slider crank mechanism:

Given: Crank AB = 20 mm = 0.02 m, C. R. BC = 80 mm = 0.08 m

N = 1000 rpm, ωBA= 2πN/60 = 2π x 1000/60 = 104.7 rad/sec

VBA = ωBA x AB = 104.7 x 0.02 = 2.09 m/s

From velocty diagram: Velocity of C w.r.t. B -

VCB = vector cb = 1.15 m/s

Angular velocity of Connecting rod ‘BC’ ωCB = VCB / CB = 1.15/0.08 = 14.375 rad /sec

Velocity of slider ‘C’

VC= vector ac = 2 m/sec

Numerical Problem-A casting weighing 9 kN hangs freely from a rope which makes 2.5 turns ............

Given: W= T1= 9 kN =9000N, d= 0.3 m, N = 20 rpm , µ= 0.25

(i) Force reqd. by a man 

T2- force reqd. by man

As rope makes 2.5 turns,

Therefore angle of contact ,

θ =2.5x2π = 5 π rad.

We know that,

2.3 log {T1/T2} = µ θ = 0.25 x 5 π = 3.9275

log {T1/T2} = 3.9275/2.3 = 1.71 or T1/T2 = 51

T2 = 9000/51 = 176.47 N

(ii) Power to raise casting : 

For a single slider crank mechanism , state the formulae to calculate by analytical method – Also state the meaning of each term.

i) Velocity of slider:

                                Vp = w.r [sinθ + sin2θ/2n ]

              where,

                  Vp - velocity of slider

w- angular velocity

θ – angle of crank to line of stroke ‘PO’

n- l/r = ratio of length of connecting rod to crank radius.

ii) Acceleration of slider:

               fp = w2 r(cos θ + cos2θ/n)

                 where, fp – acceleration of slider

iii) Angular velocity of connecting rod.:

Draw the neat sketch of epicyclic gear train and explain how it works.

In an epicyclic gear train, the axes of the shafts, over which the gears are mounted, may move relative to a fixed axis. A simple epicyclic gear train is shown in Fig. where a gear A and the arm C have a common axis at 1 about which they can rotate. The gear B meshes with gear A and has its axis on the arm at O2, about which the gear B can rotate. If the arm is fixed, the gear train is simple and gear A can drive gear B or viceversa, but if gear A is fixed and the arm is rotated about the axis of gear A (i.e.

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