A V-engine is a two cylinder engine, which has a common crank and the axis of cylinder makes a "V" shape.
Since V-engines have a common crank and the crank revolves in one plane, there is no primary or secondary couple acting on the engine.
Consider a V-engine as shown in fig.1 having common crank OC and two connecting rods CP and CQ. The lines of stroke OP and OQ are inclined to vertical axis OY at an angle ‘α’.
Let,
$\begin{aligned} m &= \text{mass of reciprocating parts per cylinder, kg}\\ l &= \text{length of connecting rod, m}\\ r &= \text{radius of crank, m}\\ n &= \text{Obliquity ratio }= l/r\\ θ &= \text{Crank angle, measured from vertical axis OY, at any instant}\\ ω &= \text{Angular velocity of crank, rad/s}\\ 2 \alpha &= \text{V-angle i.e. angle between lines of strokes of two cylinders}\\ \end{aligned}$We know that,
$\begin{aligned} F_{PV} &= F_{P1} \cos\alpha + F_{P2} \cos\alpha\\ &= mr\omega^2 \cos (\alpha -\theta)\cos\alpha \ - mr\omega^2 \cos (\alpha +\theta)\cos\alpha\\ &= mr\omega^2 \cos\alpha [\cos (\alpha -\theta) + \cos (\alpha+\theta)]\\ &= mr\omega^2 \cos\alpha \sdot (2 \cos\theta \cos\alpha)\\ &= 2mr\omega^2 \cos^2\alpha \cos\theta \end{aligned}$As both components $(F_{P1}\sdot \cos\alpha)$ and $(F_{P2} \sdot \cos\alpha)$ are acting in same direction;
$\begin{aligned} F_{PH} &= F_{P1} \sin\alpha + F_{P2} \sin\alpha\\ &= mr\omega^2 \cos (\alpha -\theta)\sin\alpha \ - mr\omega^2 \cos (\alpha +\theta)\sin\alpha\\ &= mr\omega^2\sin\alpha [\cos (\alpha -\theta)-\cos (\alpha +\theta)]\\ &= mr\omega^2\sin\alpha \ \sdot (2\sin \theta \sin \alpha)\\ &= 2mr\omega^2\sin^2\alpha\sin \theta \end{aligned}$As both components $(F_{P1} \sdot \sin \alpha)$ and $(F_{P2}\sdot \sin \alpha)$ are acting opposite to each other;
The angle made by resultant primary force with vertical axis is;
$\begin{aligned} \beta_P &= \tan^{-1} \bigg ( {F_{PH} \over F_{PV}}\bigg)\\ &= \tan^{-1} \bigg ( {2mr\omega^2\sin^2\alpha\sin \theta \over 2mr\omega^2 \cos^2\alpha \cos\theta} \bigg)\\ &= \tan^{-1} (\tan^2\alpha \ \sdot \ \tan\theta) \end{aligned}$$\begin{aligned} F_{SV} &= F_{S1} \cos \alpha + F_{S2} \cos \alpha\\ &= mr \omega^2 {\cos2(\alpha-\theta) \over n}\cos\alpha + mr \omega^2 {\cos2(\alpha+\theta) \over n}\cos\alpha\\ &= mr \omega^2\cos\alpha \bigg[{\cos2(\alpha-\theta) \over n} + {\cos2(\alpha+\theta) \over n}\bigg]\\ &={1\over n} mr \omega^2\cos\alpha \sdot[2\sdot\cos2\theta\sdot\cos2\alpha]\\ &= {2 \over n}mr \omega^2\cos\alpha\sdot\cos2\alpha\sdot\cos2\theta \end{aligned}$As both components $(F_{S1}\sdot\cos \alpha)$ and $(F_{S2}\sdot \cos \alpha)$ are acting in same direction;
$\begin{aligned} F_{SH} &= F_{S1} \sin \alpha - F_{S2} \sin \alpha\\ &= mr \omega^2 {\cos2(\alpha-\theta) \over n}\sin\alpha - mr \omega^2 {\cos2(\alpha+\theta) \over n}\sin\alpha\\ &= mr \omega^2\sin\alpha \bigg[{\cos2(\alpha-\theta) \over n} - {\cos2(\alpha+\theta) \over n}\bigg]\\ &= {1\over n} mr \omega^2\sin\alpha \sdot[2\sdot\sin2\theta\sdot\sin2\alpha]\\ &= {2 \over n}mr \omega^2\sin\alpha\sdot\sin2\alpha\sdot\sin2\theta \end{aligned}$As both components $(F_{S1}\sdot\sin\alpha)$ and $(F_{S2}\sdot\sin\alpha)$ are acting opposite to each other;
The angle made by resultant secondary force with vertical axis is;
$\begin{aligned} \beta_S &= \tan^{-1} \bigg( {F_{SH} \over F_{SV}}\bigg)\\ &= \tan^{-1} \bigg( {{2 \over n}mr \omega^2\sin\alpha\sdot\sin2\alpha\sdot\sin2\theta \over {2 \over n}mr \omega^2\cos\alpha\sdot\cos2\alpha\sdot\cos2\theta} \bigg)\\ &= \tan^{-1}\ (\tan\alpha \sdot \tan2\alpha \sdot \tan2\theta) \end{aligned}$fig-1: displacement V/s time curve for under damped system
Logarithmic decrement is defined as the natural logarithm of the ratio of successive amplitude on the same side of mean position.
The rate of decay in the amplitudes of under-damped system is measured by the parameter known as logarithmic decrement.
Rate of decay in amplitudes depends on the amount of damping present in the system. So if the damping is more, then the rate of decay will also be more.
Let A and B are the two points on the successive cycles which shows maximum deflection as shown in figure.
The periodic time:
$\begin{aligned} t_p &= t_2 − t_1 \\ &= {2 \pi \over \omega_d} \\ &= {2 \pi \over \big(\sqrt {1-\xi^2}\big) \ \omega_n} \end{aligned}$The amplitude at time $t_1$ and $t_2$ are:
$x_1 = Xe^{-\xi \omega_n t_1} [sin(\omega_d t_1 + \varnothing)]$And,
$\begin{aligned} x_2 &= Xe^{-\xi \omega_n t_2} [sin(\omega_d t_2 + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin \{ \omega_d (t_1 + t_p) + \varnothing \}] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + \omega_d t_p + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + \omega_d \bigg({2 \pi \over \omega_d}\bigg) + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + 2 \pi + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin \{ 2 \pi + (\omega_d t_1 + \varnothing) \}] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin (\omega_d t_1 + \varnothing)] \\ \end{aligned}$Now,
Taking ratio, we get;
Now,
The logarithmic decrement is given by;
The logarithmic decrement can also be determined as follows;
$\begin{aligned} \delta &= log_e \bigg({x_0 \over x_1}\bigg)= log_e \bigg({x_1 \over x_2}\bigg)= log_e \bigg({x_2 \over x_3}\bigg) = \dotso = log_e \bigg({x_{n-1} \over x_n}\bigg) \\ \text {Adding upto 'n' terms}\\ n\delta &= log_e \bigg({x_0 \over x_1} \bigg) + log_e \bigg({x_1 \over x_2}\bigg) + log_e \bigg({x_2 \over x_3}\bigg) + \dotso + log_e \bigg({x_{n-1} \over x_n}\bigg) \\ \therefore \ \ n\delta &= log_e \bigg({x_0 \over x_1} \ . {x_1 \over x_2} \ . {x_2 \over x_3} \ . \dots \ . {x_{n-1} \over x_n}\bigg) \\ \text {Or} \qquad \\ n\delta &= log_e \bigg({x_0 \over x_n}\bigg) \\ \therefore \quad \delta &= {1 \over n} log_e \bigg({x_0 \over x_n}\bigg) \end{aligned}$where,
$x_0$ = amplitude at the starting position
$x_n$ = amplitude after ‘n’ cycles
Fig.1(Critically Damped System)
ξ
is equal to one, or the damping coefficient c
is equal to critical damping coefficient "c_{c}", then the system is said to be a critically damped system.Now, let at
$t = 0$ : $x = X_0$
$t = 0$ : $\mathring x =0$
Substituting these values in equation (1):
From above equation (5), it is seen that as time t
increases, the displacement x
decreases exponentially.
The motion of a critically damped system is aperiodic (aperiodic motion motions are those motions in which the motion does not repeat after a regular interval of time i.e non periodic motion) and so the system does not shows vibrations.
For critically damped systems, if a system is displaced from its initial position, it will try to reach its mean position in a very short time.
Critically damped systems are generally seen in hydraulic doors closer as it is necessary for the door to come to its initial position in a very short time.
We know that the characteristic equation of the damped free vibration system is,
$mS^2 + cS + K = 0$This is a quadratic equation having two roots $S_1$ and $S_2$;
$S_{1,2} = {-c \over 2m} \pm \sqrt{\bigg({-c \over 2m}\bigg)^2 - {K \over m}}$In order to convert the whole equation in the form of $\xi$ , we will use two parameters, critical damping coefficient '$c_c$' and damping factor '$\xi$'. So the roots $S_1$ and $S_2$ can be written as follows;
$\begin{aligned} \xi= {c \over c_c} \quad \text {OR} \quad \xi &= {c \over 2m\omega_0} \quad (\text {as} \quad c_c=2m\omega_n)\\ \therefore {c \over 2m} &= \xi \omega_n \end{aligned}$where, $\omega_n$ = natural frequency of undamped free vibration = $\sqrt{K \over m}$ rad/s
$\therefore \omega_n^2 = {K \over m}$So we can write roots $S_1$ and $S_2$ and as;
$\begin{aligned} S_{1,2} &= -\xi \omega_n \pm \sqrt{(\xi \omega_n)^2 - \omega_n^2}\\ \therefore S_{1,2} &= [-\xi \pm \sqrt{\xi^2 - 1}]\omega_n\\ \therefore \ \ S_{1} &= [-\xi + \sqrt{\xi^2 - 1}]\omega_n\\ \text{And, }\\ S_{2} &= [-\xi - \sqrt{\xi^2 - 1}]\omega_n \end{aligned}$If the damping factor ‘$\xi$’ is greater than one or the damping coefficient ‘$c$’ is greater than critical damping coefficient ‘$c_c$’, then the system is said to be over-damped.
$\xi > 1 \quad \text{Or} \quad {c \over c_c}\quad \text{Or} \quad c > c_c$In overdamped system, the roots are given by;
$\begin{aligned} S_{1} &= [-\xi + \sqrt{\xi^2 - 1}]\omega_n\\ \text{And, }\\ S_{2} &= [-\xi - \sqrt{\xi^2 - 1}]\omega_n \end{aligned}$For $\xi > 1$ , we get $S_1$ and $S_2$ as real and negative so we get,
$\begin{aligned} x &= Ae^{S_1t} + Be^{S_2t}\\ \therefore x &= Ae^{[-\xi + \sqrt{\xi^2-1}]\omega_nt} + Be^{[-\xi - \sqrt{\xi^2-1}]\omega_nt} \quad \dots \dots \text{(1)} \end{aligned}$Now differentiating equation (1) with respect to ‘t’, we get;
$\begin{aligned} \mathring x = &Ae^{[-\xi + \sqrt{\xi^2-1}]\omega_nt}[-\xi + \sqrt{\xi^2-1}]\omega_n \ + \\ &Be^{[-\xi - \sqrt{\xi^2-1}]\omega_nt} [-\xi - \sqrt{\xi^2-1}]\omega_n\quad \dots \dots \dots \text{(2)} \end{aligned}$Now, let at
$t = 0$ : $x = X_0$
$t = 0$ : $\mathring x =0$
Substituting this value in equation (1) we get;
$X_0 = A+B\quad \dots \dots \dots \text{(3)}$Substituting this value in equation (2) we get;
$0= A[-\xi + \sqrt{\xi^2-1}]\omega_n + B[-\xi - \sqrt{\xi^2-1}]\omega_n \quad \dots \dots \text{(4)}$From equation (3), $B= X_0 - A$ and putting value of B in (4);
$\begin{alignedat}{2} &\therefore \qquad 0 &=& \ A[-\xi + \sqrt{\xi^2-1}]\omega_n + (X_0 - A)[-\xi - \sqrt{\xi^2-1}]\omega_n\\ &\therefore \qquad 0 &=& \ -A\xi + A\sqrt{\xi^2-1} + X_0 [-\xi - \sqrt{\xi^2-1}]+A\xi+A\sqrt{\xi^2-1}\\ &\therefore \qquad 0 &=& \ 2A\sqrt{\xi^2-1} + X_0 [-\xi - \sqrt{\xi^2-1}]\\ &\therefore 2A\sqrt{\xi^2-1} \ &=& \ X_0 [\ \xi + \sqrt{\xi^2-1}]\\ &\therefore \qquad A &=& \ {X_0 [\xi + \sqrt{\xi^2-1}] \over 2\sqrt{\xi^2-1} }\quad \dots \dots \text{(5)} \end{alignedat}$Now,
Putting $A=X_0-B$, in equation (4);
Now putting equation (5) and (6) in equation (1), we get;
$\begin{aligned} x = &{X_0 [\xi + \sqrt{\xi^2-1}] \over 2\sqrt{\xi^2-1} } e^{[-\xi + \sqrt{\xi^2-1}]\omega_nt} +\\ &{X_0[-\xi + \sqrt{\xi^2-1}] \over 2\sqrt{\xi^2-1}} e^{[-\xi - \sqrt{\xi^2-1}]\omega_nt}\\ x = &{X_0 \over 2\sqrt{\xi^2-1}} \bigg [\bigg (\xi + \sqrt{\xi^2-1} \bigg) e^{[-\xi + \sqrt{\xi^2-1}]\omega_nt} + \\ &\bigg (-\xi + \sqrt{\xi^2-1} \bigg ) e^{[-\xi - \sqrt{\xi^2-1}]\omega_nt} \bigg ] \end{aligned}$Above equation represents the equation of motion for overdamped system.
The motion obtained by above equation is aperiodic (aperiodic motion motions are those motions in which the motion does not repeats after a regular interval of time i.e non periodic motion) and so the system does not shows vibrations.
This type of system does not show much damping, so this systems are used very rarely.
When machine is in working condition different forces are acting on it which may cause machine to vibrate and cause damage to machine parts. The different forces acting on machine parts are static forces and dynamic forces.
The force which depends on weight of a body, is known as static force. (Generally, a static force acts when vibrations occurs in same plane.)
The force which depends on acceleration of a body, is known as dynamic force. (Generally, dynamic force acts when vibrations occurs in different planes.)
Due to these forces, the efficiency of the system decreases and life span of the system also decreases.
Due to these forces, the machine starts vibrating and sometimes when the vibrations increases, the machine would lift from it’s position and cause damage to other machine or human. So to avoid this, foundation is made below the machine (as you can see in the Fig-1 below), which absorbs the vibration and protects the machine against causing damage. Thus, balancing of the machine is required.
Fig-1
Balancing is the process of eliminating, the effect of static forces and dynamic forces acting on machine components.
In any system with one or more rotating masses, if the centre of mass of the system does not lie on the axis of rotation, then the system is said to be unbalanced.
As you can see in the Fig-2 we have a rotor which is mounted on a shaft and the shaft has its own axis of rotation. Now you can see that the C.G(centre of gravity) of the rotor is at a distance r from the axis of rotation of the shaft, so when the rotor will start to rotate, a centrifugal force will act on it which is in the outward direction as you can see in Fig-3. Due to this force our system will become unbalanced and it will start to vibrate.
Fig-2
A system is said to be statically balanced, if the centre of masses(C.G) of the system lies on the axis of rotation.
The resultant of all the centrifugal forces (dynamic forces) acting on the system during rotation must be zero.
$\begin{aligned} \small ∑ \text {Centrifugal forces acting on the system} &= \text {zero}\\ i.e. \ +mrω^2 – mrω^2 &= 0 \end{aligned}$A system is said to be dynamically balanced, if it satisfies following two conditions:-
Fig-1
If the damping factor $\ \xi$ is less than one or the damping coefficient $c$ is less than critical damping coefficient $c_c$, then the system is said to be an under-damped system.
$\xi < 1 \quad \text OR \quad {c \over c_c} < 1\implies c < c_c$Where $i = \sqrt{-1}$ is the imaginary unit of complex root
Above equation shows the equation of motion for an underdamped system, and the amplitude reduces gradually and finally becomes zero after some time.
Amplitude decreases by $\ X.e^{-\xi \omega_nt}$
The natural angular frequency of damped free vibrations is given by:
→ Assignment model is a special application of Linear Programming Problem (LPP), in which the main objective is to assign the work or task to a group of individuals such that;
i) There is only one assignment.
ii) All the assignments should be done in such a way that the overall cost is minimized (or profit is maximized, incase of maximization).
→ In assignment problem, the cost of performing each task by each individual is known.
→ It is desired to find out the best assignments, such that overall cost of assigning the work is minimized.
Suppose there are 'n' tasks, which are required to be performed using 'n' resources.
The cost of performing each task by each resource is also known (shown in cells of matrix)
→ Assignment Model is a special application of Linear Programming (LP).
→ The mathematical formulation for Assignment Model is given below:
→ Let, $\text {C}_{ij}$ denotes the cost of resources 'i' to the task 'j'; such that
$\begin{aligned} x_{ij} &= 1 \ ; \text{if} \ \ i^{th} \text{resource is original to } j^{th} \text{task}\\ x_{ij} &= 0 \ ; \text{if} \ \ i^{th} \text{resource is not original to } j^{th} \text{task} \end{aligned}$→ Now assignment problems are of the Minimization type. So, our objective function is to minimize the overall cost.
$\therefore \text{Minimize Z}= \displaystyle\sum_{j=1}^n \ \sdot \ \displaystyle\sum_{i=1}^n \ C_{ij} \ \sdot \ x_{ij}$→ Subjected to constraint;
(i) For all $j^{th}$ task, only one $i^{th}$ resource is possible:
$\therefore \displaystyle\sum_{j=1}^n \ x_{ij} = 1 \quad \text{(i=1,2,3, ...,n)}$(ii) For all $i^{th}$ resource, there is only one $j^{th}$ task possible;
$\therefore \displaystyle\sum_{i=1}^n \ x_{ij} = 1 \quad \text{(j=1,2,3, ...,n)}$(iii) $x_{ij}$ is '0' or '1'.
In assignment problem, either allocation is done to the cell or not.
So this can be formulated using 0 or 1 integer.
While using this method, we will have n x n decision varables, and n+n equalities.
So even for 4 x 4 matrix problem, it will have 16 decision variables and 8 equalities.
So this method becomes very lengthy and difficult to solve.
As assignment problem is a special case of transportation problem, it can also be solved using transportation methods.
In transportation methods (NWCM, LCM & VAM), the total number of allocations will be (m+n-1) and the solution is known as non-degenerated. (For eg: for 3 x 3 matrix, there will be 3+3-1 = 5 allocations)
But, here in assignment problems, the matrix is a square matrix (m=n).
So total allocations should be (n+n-1), i.e. for 3 x 3 matrix, it should be (3+3-1) = 5
But, we know that in 3 x 3 assignment problem, maximum possible possible assignments are 3 only.
So, if are we will use transportation methoods, then the solution will be degenerated as it does not satisfy the condition of (m+n-1) allocations.
So, the method becomes lengthy and time consuming.
It is a simple trail and error type method.
Consider a 3 x 3 assignment problem. Here the assignments are done randomly and the total cost is found out.
For 3 x 3 matrix, the total possible trails are 3!
So total 3! = 3 x 2 x 1 = 6 trails are possible.
The assignments which gives minimum cost is selected as optimal solution.
But, such trail and error becomes very difficult and lengthy.
If there are more number of rows and columns,
( For eg: For 6 x 6 matrix, there will be 6! trails. So 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 trails possible)
then such methods can't be applied for solving assignments problems.
It was developed by two mathematicians of Hungary. So, it is known as Hungarian Method.
It is also know as Reduced matrix method or Flood's technique.
There are two main conditions for applying Hungarian Method:
(1) Square Matrix (n x n).
(2) Problem should be of minimization type.
Solution:
Given data:
mass $\ m = 85$ kg
Static deflection, $\ x = 18\ \text{mm} = 0.018\ \text{m}$
Let,
$x_0 =$ initial amplitude
$x_2=$ final amplitude after two complete cycle $= {1\over4}$
Taking ratio,
${x_0 \over x_2}={x_0 \over {1 \over 4} x_0}=4$Now,
$\begin{aligned} \xi= {c \over c_c} \quad \text {OR} \quad &\xi= {c \over 2m\omega_0} \quad (\text {as} \quad c_c=2m\omega_n)\\ \therefore \ &c=(2m\omega_n)\xi \ \dots\dots\dots\dots \ (1) \end{aligned}$The natural circular frequency of vibration is,
$\begin{aligned} \omega_n &= {\sqrt{k \over \ m}} \\ &= {\sqrt{m \ g \over \ x \ m}}\\ &= {\sqrt{g \over x}} \\ &= {\sqrt{9.81 \over 0.018}}\\ &= 23.34 \ \text {rad/s} \end{aligned}$From equation (1); we get,
Damping force at unit velocity;
The time period of damped vibration is,
$\begin{aligned} t_p &= {2\pi \over \omega_d} \\ &= {2\pi \over \omega_n \ \sqrt {1 -\xi^2}} \\ &= {2\pi \over (23.34) \ \sqrt {1 -(0.1091)^2}} \\ &=0.2708 \ \text {sec} \end{aligned}$PERT is the technique used to find project completion time of “variable activities”. In PERT, the time is combination of three different time estimations. Following are the three different time estimation:
The North West corner method is one of the methods to obtain a basic feasible solution of the transportation problems (special case of LPP).
We will now see how to apply this very simple method to a transportation problem. We will study steps of this method while applying it in the problem itself.
Note that all the explanation is provided in “CYAN” colour. You have to write in examination the only thing which are given in this regular colour under each steps(if any), else you can directly solve matrix of the problem as explained here
Solution:
Balance the problem meaning we need to check that if;
$\color{#32c5d4} \Sigma \text { Supply} = \Sigma \text { Demand}$If this holds true, then we will consider the given problem as a balanced problem.
Now, what if it’s not balanced?
If such a condition occurs, then we have to add a dummy source or market; whichever makes the problem balanced. You can watch a video on this type of numerical, which is known as Unbalanced Transportation Problems.
$\to$ The given transportation problem is balanced.
We will start the allocation from the left hand top most corner (north-west) cell in the matrix and make allocation based on availability and demand.
Now, verify the smallest among the availability (Supply) and requirement (Demand), corresponding to this cell. The smallest value will be allocated to this cell and check out the difference in supply and demand, representing that supply and demand are fulfilled, as shown below.
As we have fulfilled the availability or requirement for that row or column respectively, remove that row or column and prepare a new matrix, as shown below.
Repeat the same procedure of allocation of the new North-west corner so generated and check based on the smallest value as shown below, until all allocations are over.
Once all allocations are over, prepare the table with all allocations marked and calculate the transportation cost as follows.
$\begin{aligned} \to \ \text {Transportation cost} &= (40 \times 40) + (3 \times 30) + (4 \times 30) + (2 \times 10) + (8 \times 60) \\ &= \text {Rs } 870 \end{aligned}$
Find solution of same numerical by:
Activity | t _{ o } | t _{ m } | t _{ p } |
---|---|---|---|
1-2 | 6 | 9 | 12 |
1-3 | 3 | 4 | 11 |
2-4 | 2 | 5 | 14 |
3-4 | 4 | 6 | 8 |
3-5 | 1 | 1.5 | 5 |
2-6 | 5 | 6 | 7 |
4-6 | 7 | 8 | 15 |
5-6 | 1 | 2 | 3 |
Solution: First of all draw the network diagram for given data as shown below:
Here the time for completion of activities are probabilistic. So, using given values of time we will find the expected time to completion the activities and variance.
$\text {Expected time } \quad t_e = {t_o + 4t_m + t_p \over 6}$