Theory of Machines Test 1

• If there is all pairs are lower pair of a kinematic chain then 4 pairs required to make a chain.
• But, if there is atleast one higher pair involved then minimum 3 pairs required to make a kinematic chain. Cam and follower is example of a kinematic chain and it includes total 3pairs.

One higher pair = two lower pair.

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 = \(8 \times 36 = 288mm\)

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

Flexible Link: A flexible link is one that while transmitting motion is partly deformed in a manner not to affect the transmission of motion, for example, belts, ropes, chains, springs, etc.

Fluid Link: A fluid link is one that is deformed by having an incompressible fluid in a closed vessel and the motion is transmitted through the fluid by pressure, as in the case of a hydraulic press, hydraulic jack, and hydraulic brake.

Rigid Link: A rigid link is one that does not undergo any deformation while transmitting motion. Links in general are elastic in nature.
They are considered rigid if they do not undergo appreciable deformation while transmitting motion, for example connecting rod, crank, tappet rod, etc.

Resistant Link: A link that is rigid for the purposes it serves.
Apart from rigid links, there are some semi-rigid links that are normally flexible, but under certain loading conditions act as a rigid link for limited purposes and thus are resistant links.
These days resistant links are usually referred to as rigid links.

Floating Link: This is a link that is not connected to the frame.

Concept:

The schematic of mechanism,

At the instant shown in figure, the linear velocity of point B and C will be same,

∴ V_BA = V_CD

Calculation:

ω₁AB = ω₂CD

5AB = 2CD       ----(1)

Since the difference between AB and CD

CD – AB = 30

From equation (1)

\(CD = \frac{5}{2}AB\)

\(\therefore \frac{5}{2}AB - AB = 3\)

\(\frac{3}{2}AB = 30\)

\(AB = \frac{{2 \times 30}}{3} = 20\;cm\)

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

A kinematic chain is a group of links either joined together or arranged in a manner that permits them to move relative to one another. 

Now, let's take an example of such a mechanism

The above-shown figure is a cam and follower mechanism. 

In this mechanism, there are 2 lower pairs i.e. between links (1,2 and 3,1) and 1 higher pair which is between link (2,3). 

Hence, the minimum number of links that can make a mechanism is 3.

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

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

Whatever may be the number of links and joints Grubler's criterion applies to mechanism with only single degree freedom. Subject to the condition 3l-2j-4=0 and it satisfy this condition.
Degree of freedom is given by

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.

Mechanism: A mechanism is a set of machine elements or components or parts arranged in a specific order to produce a specified motion.

When one of the links of a kinematic chain is fixed, the chain is called a mechanism.

Mechanisms are of the following types:

Simple mechanism

• This is a mechanism which has four links.

Compound mechanism

• This is a mechanism which has more than four links.

Complex mechanism

• This is formed by the inclusion of ternary or higher-order floating link to a simple mechanism.

Planar mechanism

• This is formed when all the links of the mechanism lie in the same plane.

Spatial mechanism

• This is formed when all the links of the mechanism lie in the different plane.

Equivalent mechanism

• Turning pairs of plane mechanisms may be replaced by other types of pairs such as sliding pairs or cam pairs. The new mechanism thus obtained having the same number of degrees of freedom as the original mechanism is called the equivalent mechanism.

• If one of the links is fixed in a kinematic chain, it is called a mechanism.
• So, we can obtain as many mechanisms as the number of links (n inversions from a kinematic chain having ‘n’ number of links) in a kinematic chain.
• A slider-crank is a kinematic chain having four links so four inversions.
• It has one sliding pair and three turning pairs.
• Link 1 is a frame (fixed).
• Link 2 has rotary motion and is called a crank.
• Link 3 has got combined rotary and reciprocating motion and is called a connecting rod.
• Link 4 has reciprocating motion and is called a slider.
• This mechanism is used to convert rotary motion to reciprocating and vice versa.

Inversions of the slider-crank mechanism are obtained by fixing links 1, 2, 3, and 4.

• First inversion: This inversion is obtained when link 1 (ground body) is fixed.
  Application-  Reciprocating engine, reciprocating compressor, etc.
• Second inversion: This inversion is obtained when link 2 (crank) is fixed.
  Application- Whitworth quick returns mechanism, Rotary engine, etc.
• Third inversion: This inversion is obtained when link 3 (connecting rod) is fixed.
  Application - Slotted crank mechanism, Oscillatory engine, etc.
• Fourth inversion: This inversion is obtained when link 4 (slider) is fixed.
  Application- A hand pump, pendulum pump or Bull engine, etc.

Concept:

Velocity of slider,

V_s = ℓ_DE × ω_DE

Calculation:

As shown in configuration, link AB and CD form parallel linkages

∴ V_B = V_C

i.e. \(\frac{{{\omega _{BA}}}}{{{\omega _{CD}}}} = \frac{{{\ell _{CD}}}}{{{\ell _{AB}}}}\)

\(\therefore \frac{{{\omega _{AB}}}}{{{\omega _{CD}}}} = \frac{{30}}{{10}}\)

\(\therefore {\omega _{CD}} = \frac{4}{3}\;rad/s\)

As, link CD and DE are single link, ω_CD = ω_DE = 4/3 rad/s

Velocity of slider,

V_s = ℓ_DE × ω_DE

\(= 20\left( {cm} \right) \times \frac{4}{3}\;\left( {rad/s} \right)\)

∴ V_S = 26.67 cm/sec

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

Concept:

To determine angular velocities of links, we can use angular velocity theorem

\(\frac{{{\omega _x}}}{{{\omega _y}}} = \frac{{{I_{1y}}\;{I_{xy}}}}{{{I_{1x}}\;\;{I_{xy}}}}\)

Where, 1 = fixed link

\(\frac{{{\omega _2}}}{{{\omega _3}}} = \frac{{{I_{13}}\;{I_{23}}}}{{{I_{12}}\;{I_{23}}}}\)

Using geometry, I₁₃ I₂₃ = 100 mm

\(\frac{2}{{{\omega _3}}} = \frac{{100}}{{100}}\;\;\;\;\;\therefore {\omega _3} = 2\;rad/s\)

Now,

\(\frac{{{\omega _3}}}{{{\omega _4}}} = \frac{{{I_{14}}\;{I_{34}}}}{{{I_{13}}\;{I_{34}}}}\)

\({I_{34}}\;{I_{14}} = 100\sqrt 2 = 141.42\;mm\)

\({I_{13}}\;{I_{34}} = 100\sqrt 2 = 141.42\;mm\)

\(\frac{2}{{{\omega _4}}} = \frac{{100\sqrt 2 }}{{100\sqrt 2 }} \Rightarrow\)

∴ ω₄= 2 rad/s

According to Grashoff’s Law for a four bar mechanism, the sum of shortest and longest link lengths should not be greater than the sum of the remaining two link length.

i.e. S + L ≤ P + Q

For Double crank mechanism Shortest link is fixed.Here shortest link is ‘P’.

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 

• A machine is a tool containing one or more parts that use energy to perform an intended action.
• A simple machine is a device that simply transforms the direction or magnitude of a force, but a large number of more complex machines exist.
• Examples include Levers, Screw Jack, Wheel and axle, Pulleys, Wedge, Inclined plane, etc.
• Spring stores and releases energy. it neither’s changed direction or magnitude of the force. To qualify as a simple machine, a device must exchange the magnitude of a Force.

Crank and lever mechanism:

This is the first inversion of four bar chain.

It is used to convert rotary motion of input link into oscillatory motion of output link.

Beam engine mechanism is an example of this inversion and it consists of four links.

The crank rotates about a fixed centre and the lever oscillates at another fixed centre.

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 %