Theory of Machines Test 2

Concept:

Center distance is the distance between the centers of rotation of the two gears in mesh.

\(C = \frac{{{d_1} + {d_2}}}{2} = \frac{{m\left( {{T_1} + {T_2}} \right)}}{2}\)

Where d is pitch circle diameter.

Calculation:

Given: T₁ = 14, T₂ = 36, m = 3 mm

\(C = \frac{{m\left( {{T_1} + {T_2}} \right)}}{2} = \frac{{3\left( {14 + 36} \right)}}{2} = 75\;mm\)

Important Point:

Base circle diameter (d_b) = Pitch circle diameter (d) × cos ϕ

ϕ is pressure angle.

Concept:

Gyroscopic couple acting on the disc is given by,

T = I ω ω_p

I = mk², m = mass, k = radius of gyration, ω = spin speed about axis, ω_p = spin speed of axis

Calculation:

\({\rm{\omega \;}} = \frac{{2\pi N}}{{60}} = \;120\pi \;rad/s\)

\({\rm{\omega_p\;}} = {\rm{\;}}2\pi \times \frac{{240}}{{60}} = 8\pi \;rad/s\)

∴ Gyroscopic couple = 0.5 × 0.09 × 8π × 120π

Explanation:

Net Torque on the system in zero since shafts are rotating at constant speeds

We assume clockwise as - ve

Anti-clockwise as + ve

∴ T_A + T_B + T_C = 0

Assuming no loss in transmission.

T_A N_A + T_B N_B + T_C N_C = 0

N_C = 0, N_B = - 240 rpm, N_A = -120 rpm

T_A = - 40 kN.m

Now,

\({T_B} = \frac{{ - {T_A}{N_A}}}{{{N_B}}}\)

\({T_C} = - \left( {{T_A} + {T_B}} \right) = - \left( { - 40 + 40 \times \frac{{120}}{{240}}} \right)\)

T_c = - (- 40 + 20)

Concept: 

Gyroscopic couple C = Iωωp

Calculation:

Given:

mass (m) = 450 kg, Radius of gyration (k) = 0.32 m, v = 240 km/h

Now,

\(\begin{array}{l} \omega = \frac{{2\pi \times 2000}}{{60}} = 209.4\;rad/s\\ \nu = \frac{{240 \times {{10}^3}}}{{3600}} = 66.67\;m/s \end{array}\)

Now,

Moment of Inertia \(\left( {\rm{I}} \right) = m{k^2} = 450 \times {\left( {0.32} \right)^2} = 46.08\;kg.{m^2}\)

\({\omega _p} = \frac{\nu }{r} = \frac{{66.67}}{{60}} = 1.11\frac{{rad}}{s}\)

Gyroscopic couple C = Iωω_p

Gyroscopic couple C = 46.08 × 209.4 × 1.11  

Concept:

∴ V_sliding = path of approach × (ω₁ + ω₂)

Where, ω ₁ = angular velocity of pinion, ω₂ = angular velocity of gear

Calculation:

Let r → radius of pinion, R → radius of gear , t → teeth of pinion, T → teeth of gear

\(r = \frac{{mt}}{2} = \frac{{4\; \times \;20}}{2} = 40mm\)

\(R = \frac{{mT}}{2} = \frac{{4\; \times \;80}}{2} = 160\;mm\)

Maximum possible path of approach is possible when pinion is driver,

∴ Path of approach = r sin θ

∴ Path of approach = 40 sin 20°

∴ Path of approach = 13.68 mm

Now,

ω₁ = 5 rad/s

\({{\rm{\omega }}_2}{\rm{\;}} = {\rm{\;}}{{\rm{\omega }}_1}{\rm{\;}} \times {\rm{\;}}\frac{t}{T}\)

\({{\rm{\omega }}_2} = 5 \times \frac{{40}}{{160}}\)

∴ ω₂ = 1.25 rad/s

Now,

V_sliding = path of approach × (ω₁ + ω₂)

V_sliding = 13.68 × (5 + 1.25)

∴ V_sliding = 85.5 mm/sec

Concept:

For rack and pinion.

Addendum to avoid interference \( \le rsi{n^2}\phi \)

r = radius

ϕ = pressure angle

Using this find the minimum pressure angle

Path of contact

\(L = \;\sqrt {R_a^2 - {{\left( {rcos\phi } \right)}^2}} + \frac{{addendum}}{{sin\phi }} - rsin\phi \)

Calculation:

Given: 

d = 130 mm ⇒ r = 65 mm, addendum = 6.5 mm

Addendum to avoid interference ≤ rsin² ϕ

\({\sin ^2}\phi \ge 0.1\;\;\)

Φ ≥ 18.43°,

∴ Φ = 20°

R_a = 65 + 6.5 = 71.50 mm