Theory of Machines Test 1

Concept:

Center distance between two gear in mesh = Average of their diameters

Module = Diameter/Teeth

 Calculation:

Given that module = 8mm

m = D/T

Then,

D = mT

Now,

Diameter of pinion = \(8 \times 20 = 160mm\)

Diameter of gear = \(8 \times 36 = 288mm\)

\({\rm{Center\;distance\;}} = {\rm{\;}}\frac{{288\; + \;160}}{2} = 224mm = 0.224m\)

Concept:

Centre distance = \(\frac{m}{2}\left[ {{Z_g} + {Z_p}} \right]\)

Z_g = No of teeth on gear

Z_p = No of teeth on pinion

Calculation:

120 = \(\frac{4}{2}\left[ {40 + {z_p}} \right]\)

60 = 40 + Z_p

Concept:

Maximum arc of approach of pinion,

\({\rm{Max\;arc\;of\;approach}} = {\rm{}}\frac{{Max\;Path\;of\;contact}}{{Cos\;\emptyset }}\;\)

\({\rm{Max\;arc\;of\;approach}} = \frac{{{r_p}sin\emptyset }}{{cos\emptyset }}\)

 Max arc of approach = r_p tanθ

Where, r_p = radius of pinion

Given condition,

\({\rm{Arc\;of\;approach\;}} = {\rm{\;}}\frac{{\pi {d_p}}}{{{Z_p}}}\)

\({r_p}\tan \emptyset \ge \frac{{2\pi {r_p}}}{{{Z_p}}}\)

\({Z_p} \ge \frac{{2\pi }}{{tan\emptyset }}\)

For ∅ = 18°

\({Z_p} \ge \frac{{2\pi }}{{tan18^\circ }}\)

 Z_p ≥ 19.32

∴ Z_p = 20 

The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed.

ΔN = N_max - N_min

The difference between the maximum and the minimum energies is known as maximum fluctuation of energy

ΔE = E_max - E_min

Coefficient of fluctuation of speed:

\(C = \frac{{{\omega _{max}} - {\omega _{min}}}}{{{\omega _{mean}}}} = \frac{{2\left( {{\omega _{max}} - {\omega _{min}}} \right)}}{{\left( {{\omega _{max}} + {\omega _{min}}} \right)}}\)

Let I be the moment of inertia, ω_mean be the mean speed in rad/sec, C_s be the coefficient of fluctuation of speed, E be the fluctuation energy, N be rpm

Maximum fluctuation of energy: ΔE = Iω²_meanC_s

wmean = 2π × N_mean /60 = 2π × 180/60 = 6π = 18.84 rad/s

ΔE = Iω²_meanC_s

2094 = 11.8 × (18.84)² × C

C = 0.5

\(C = \frac{{{N_{max}} - {N_{min}}}}{{{N_{mean}}}} = \frac{{{\rm{\Delta }}N}}{{{N_{mean}}}}\)

Concept:

Unbalance is the unequal distribution of weight of a rotor about its rotating axis. When a rotor is unbalanced, it imparts vibratory forces of the rotor. Unbalance causes vibrations on the machine.

There are two types of balancing:

Static balancing: If the combined mass centre of the system lies on the axis of rotation.

\(\begin{array}{l} \sum F = 0 \Rightarrow {m_1}{r_1}{\omega ^2} + {m_2}{r_2}{\omega ^2} + {m_3}{r_3}{\omega ^2} = 0\\ \Rightarrow {m_1}{r_1} + {m_2}{r_2} + {m_3}{r_3} = 0 \end{array}\)

Dynamic Balancing: When serval masses rotate in different planes, the centrifugal forces, in addition to being out of balance, also from couples.

\(\begin{array}{l} \sum F = 0 \Rightarrow {m_1}{r_1} + {m_2}{r_2} + {m_3}{r_3} = 0\\ \sum C = 0 \Rightarrow {m_1}{r_1}{l_1} + {m_2}{r_2}{l_2} + {m_3}{r_3}{l_3} = 0 \end{array}\)

We can see that magnitude of balancing mass is independent of the speed of the shaft.

Concept: 

Gyroscopic couple C = Iωωp

Calculation:

Given:

R = 50 m; v = 50 m/s; m = 400 kg; k = 0.3 m; N = 2400 r.p.m. or 251.3 rad/s

Now,

Mass moment of inertia (I) of the engine and the propeller

I = mk² = (400) (0.3)² = 36 kg.m²

Now,

Angular velocity of precession (ωP): 

ω_P = v/R = 50/50

∴ ωP = 1 rad/s

Gyroscopic couple (C) = I.ω.ωP

C = 36 × 251.3 × 1

C = 9046.8 N.m

∴ C = 9.05 kN.m

Explanation:

Work done per revolution:

\(W = \mathop \smallint \limits_0^{\frac{2\pi }{3} } Td\theta = \mathop \smallint \limits_0^{\frac{2\pi }{3}} \left( {25000 - 7500\sin 3\theta } \right)d\theta \)

\(W = \left[ {25000\;\theta + 7500\;\frac{{\cos 3\theta }}{3}} \right]_0^{\frac{2\pi }{3} }\)

\(W = 25000\left( {\frac{2\pi }{3} } \right) + \frac{{7500}}{3}\left( {\cos 2\pi - \cos 0} \right)\)

W = 50000(2π/3) N - m

Mean Resisting torque:

\({T_{mean}} = \frac{W}{{\frac{2\pi }{3} }} = 25000\;N - m\)

\(P = \frac{{2\pi NT}}{{60}} = \frac{{2\pi \times 300 \times 25000}}{{60}}\)

Explanation:

Let the total time taken be T_o

Now,

T_o= (angle turned)/ (angular velocity) = 0.3/6 sec = 0.05 sec

Now using the 2^nd equation of motion for the half acceleration:

S= ut + ½ at²

S = 80/2 mm = 40 mm, u  =0 and t= T_o/2

0.04= ½ x a x 0.025²

a = 128 m/s²   

For maximum velocity, it will occur at the end of the half acceleration and after that the velocity will start to decrease as deceleration will start.

v² – u² = 2aS   [u=0, S = 0.04 m]

v² = 2 x 128 x .04 = 10.24

∴ v= 3.2 m/s 

Explanation:

Given:

Power (P) = 1.2 kW, I = 160 kg.m², N_m = 240 rpm

Now,

\({\omega _m} = \frac{{2\pi N}}{{60}} = 8\pi \;rad/s\)

Crank of shaper speed = 20 rpm

i.e. 20 rev = 60 seconds

∴ 1 rev = 3 seconds (cycle time)

\(QRR = 2 = \frac{{Cutting\;time}}{{Return\;time}}\)

∴ Cutting time = 2s

Return time = 1s

Fluctuation of energy.

ΔE = 1.2 × 1 = 1200 J

Also,

\({\rm{\Delta E\;}} = I\omega _m^2{C_s}\)

\(\therefore {C_s} = \frac{{1200}}{{160\; \times \;{{\left( {8\pi } \right)}^2}}} = 0.0118\)

∴C_s= 1.18 %