THE IMPORTANCE OF MANUFACTURING ENGINEERING

If we carefully think about human civilization, one shall notice an wonderful fact about human beings. The thing that made us different from other hominids is the skill to manufacture tools. We just triumphed due to our ability to make primitive tools out of stone and metals during the dawn of the civilizations. Since then much time has passed and we have entered into a Machine Era and man has been still continuously engaged in converting the natural resources into useful products by adding value to them through machining and other engineering activities applying on the raw materials. Manufacturing is the sub branch of Engineering which involves the conversion of raw materials into finished products.

The conversion of natural resources into raw materials is normally taken care of by two sub branches of engineering viz. Mining and Metallurgy Engineering. The value addition to the raw materials by shaping and transforming it to final products generally involves several distinct processes like castings, forming, forging, machining, joining, assembling and finishing to obtain a completely finished product.

Understanding Manufacturing Engineering largely based upon three engineering activities and they are Designing,  Production and Development of new more efficient techniques.

At the Design stage, engineering design mainly concentrates on the optimization of engineering activities to achieve most economical way to manufacture a goods from raw materials. It also chooses the raw materials and impart the requisite engineering properties of materials like hardness, strength, elasticity, toughness by applying various heat treatment to them.

During the production stages, the selection of the important process parameters to minimize the idle time and cost, and maximizing the production and its quality is very important.

The New Technologies must be implemented to adapt to the changing scenarios of the markets and demands to make the sales competitive and sustainable.

PARALLEL AXIS THEOREM AND IT'S USES IN MOI

Moment Of Inertia of an Area.

download pdf file of solved numerical

MOI or MOMENTS OF INERTIA is a physical quantity which represents the inertia or resistances shown by the body against the tendency to rotate under the action external forces on the body. It is a rotational axis dependent function as its magnitude depends upon our selection of rotational axis. Although for any axis, we can derive the expression for MOI with the help of calculus, but still it is a cumbersome process.


Now suppose we take a different issue. We know MOI of an area about its centroidal axis is easily be obtained by integral calculus, but can we find a general formula by which we can calculate MOI of an area about any axis if we know its CENTROIDAL MOI.

We shall here find that we can indeed derive an expression by which MOI of any area (A) can be calculated about any Axis, if we know its centroidal MOI and the distance of the axis from it's Centroid G.


If I_GX be the centroidal moment of inertia of an area (A) about X axis, then we can calculate MOI of the Area about a parallel axis (here X axis passing through the point P) at a distance Ŷ-Y'=Y from the centroid if we know the value of I_GX and Y, then I_PX will be
I_PX = I_GX + A.Y² where Y=Ŷ-Y'


I_XX = I_OX = I_GX + A.Ŷ²

<sup>Where I_XX is the moment of inertia of the area about the co-ordinate axis parallel to X axis and passing through origin O, hence we can say,

I_XX = I_OX

 IMPORTANT: The notation of Moment of Inertia

MOI of an area about an axis passing through a point B will be written as I_BX</sup>

Q: Find the Centroidal Moment of Inertia of the figure given above. Each small division represents 50 mm.

To find out Centroidal MOI

QUESTION BANK : ENGINEERING MECHANICS PART-2

TOPICS: NUMERICALS ON FORCE SYSTEM- UNIT-1

5) A bar of AB 12 m long rests in horizontal position on two smooth planes as shown in the figure. Find the distance X at which 100 kN is to be placed to keep the bar in equilibrium.

 

6) A light string ABCDE whose extremity A is fixed, has weights W₁ & W₂ attached to it at B & C. It passes round a small smooth pulley at D carrying a weight of 300 N at the free end E as shown in figure. If in the equilibrium position, BC is horizontal and AB & CD make 150° and 120° with BC, find (i) Tensions in the strings and (ii) magnitudes of W₁ & W₂  

 

7) Find reactions at all the contact points if weight of P is 200 N & diameter is 100 mm, where as weight of Q is 500 N and diameter is 180 mm.

 

8) Determine the force P required to begin rolling the uniform cylinder of mass (m) over the obstacle of height (h) as shown in the figure.  

 

9) A roller of weight 500 N has a radius of 120 mm and is pulled over a step of height 60 mm by a horizontal force P. Find magnitudes of P to just start the roller over the step.

 

10) Two identical rollers each of weight 100 N are supported by an inclined plane of 30° with horizontal and a vertical wall as shown in the figure. Find all the reactions at each contact point.

 

11) A smooth cylinder of radius 500 mm rests on a horizontal plane and is kept from rolling by a rope OA of 1000 mm length. A bar AB of length 1500 mm and weight 1000 N is hinged at point A and placed against the cylinder of negligible weight. Determine the tension in the rope.

 

12)      A flat belt connects pulley B, which drives a pulley A; attached to an electric motor. μ_s =  0.25 and μ_k = 0.2 between both the pulleys and the belt. If maximum allowable tension in the belt is 600 N, determine the largest torque which can be exerted by belt on pulley B.

      

13)       Two blocks of mass MA & MB are kept at equilibrium as shown in the figure. The friction between the block B & the floor is 0.35 and between the blocks is 0.3, then find the minimum force P to just move the block B.

WHAT IS QUALITY OF A PRODUCT OR SERVICES?

EME-072: QUALITY MANAGEMENT
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|||---- WHAT IS QUALITY?
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Everyone has had experiences of poor quality when dealing with business organizations. These experiences might involve an airline that has lost a passenger’s luggage, a dry cleaner that has left clothes wrinkled or stained, poor course offerings and scheduling at your college, a purchased product that is damaged or broken, or a pizza delivery service that is often late or delivers the wrong order. So, what is the exact definition of Quality.

Although Quality is a vague concept up to some extent, but we can still define it. So, we define "Quality of a Product" as the degree of its excellence and fitness for the purpose.
Although, some of the quality characteristics can be specified in quantitative terms, but no single characteristics can be used to measure the quality of a product on an absolute scale. 

Quality of a product means all those activities which are directed to

  (i) Maintaining and improving such as setting of quality targets,

           (ii) Appraisal of conformance

          (iii) Taking corrective action where any deviation is noticed
          (iv) And planning for improvements in quality.

Quality is a measure of the user satisfaction provided by a product, it includes
            (i) Functional efficiency
           (ii) Appearance
          (iii) Ease of installation and operation
          (iv)  Safety reliability
           (v) Maintainability
          (vi) Running and maintenance cost
         (vii) Continued fault free service/ after-sales service.

There are two elements of quality, namely 

(i) Quality of Design
(ii) Quality of Conformance.

Quality is initially created by the designer in the form of product specifications and manufacturing instructions where as the design provides user satisfaction, the product must be conformed to the design.

Making quality a priority means putting customer needs first. It means meeting and exceeding customer expectations by involving everyone in the organization through an integrated effort. Total quality management (TQM) is an integrated organizational effort designed to improve quality at every level.

So, to be a successful brand a product must possess the best quality. But, how does one build quality into a product?

It is obvious that inspection alone can not build quality into a product unless quality has been designed and manufactured into it.

The quality of a product in a company is determined by the philosophy, commitment, and the quality policy of the top management and the extent to which these policies can be put into actual practice.

TQM is about meeting quality expectations as defined by the customer; this is called customer-defined quality. However, defining quality is not as easy as it may seem, because different people have different ideas of what constitutes high quality. Let’s begin by looking at different ways in which quality can be defined.

► Total quality management (TQM):

"An integrated effort designed to improve quality performance at every level

of the organization."

► Customer-defined Quality:

"The meaning of quality as defined by the customer."

► Conformance to Specifications:

"How well a product or service meets the targets and tolerances determined by its designers."

► Fitness for Use:

"A definition of quality that evaluates how well the product performs for its intended use."

► Value for Price Paid:
"Quality defined in terms of product or service usefulness for the price paid."


Quality Control and User-defined Characteristics of Quality:

The perception of quality is heavily dependent upon the types of processes adopted to maintain the quality of the product during manufacturing and distribution of the product. Those processes are called as Quality Control processes. In modern concept of quality control, mainly TQC or Total Quality Control, Quality Assurance and Quality Management have been termed as "QUALITY CONTROL".

Quality of a product is determined by the combined effects of various departments such as Design, Engineering, Purchase, Production and Inspection.

Quality is perceived differently by different people, but understood by almost everyone. The customer as a user takes the quality of fit, finish, appearance and performance in a manufactured product where as service quality may be evaluated on the basis of the "degree of satisfaction".

As the customer has the final saying about the quality of a product; therefore, the measurable characteristics in a product or service are basically translation of the customer needs.

Once the specifications are developed depending upon the customer satisfaction, next the ways to measure as well as monitor the characteristics should be devised.

This becomes the basis of further improvement or continuous improvement of the product or the service.

The ultimate objective of all the processes is to ensure the customer satisfaction so that they become ready to pay for the product or the service.
 

CENTROIDS OF LINES

CENTROID OF A STRAIGHT LINE

If we take a straight line of length (L), then its midpoint will be at a distance (L/2) from either end of the line. Let us denote the centroid as the point G(X_g,Y_g).

Suppose we have a straight line AB of length (L) that makes an angle θ with X axis. Let the coordinate of point A is (X_o,Y_o). Let the centroid be at G(X_g,Y_g), then

X_g = X_o + (Lcos θ)/2
Y_g = Y_o + (Lsin θ)/2

                                                                                                                                                             

     Again, suppose the coordinate of B is given instead of point A. Let it is given as B(Xi,Yi). Then, it will be                                                       

Xg = X₁ - (Lcos θ)/2
Yg = Y₁ - (Lsin θ)/2

For Horizontal lines θ = 0° and for Vertical lines θ = 90°

CENTROID OF A CURVED LINE

The steps to derive the centroid of a quarter circular arc of radius R. 

 

Centroid of a curved line can be derived with the help of calculus.

i) Draw the figure in a X-Y coordinate system. Let the curved line has been represented by a function θ

ii) Take an arbitrary point P(X,Y) on the curve. Join the line OP, where O is the origin of the coordinate system. Let OP has a length L and makes an angle (θ) with X axis. Therefore, we can write

                  

                                   X = Rcosθ ----- (a)
                             Y = Rsinθ ----- (b)

iii) Let another point Q, such that PQ = dL where dL is very very small. Let the angle subtended by (dL) be (dθ). So we can write

                                   dL = Rdθ ------ (c)

                             X_g = (1/L) ∫(XdL)
                                  = (1/L) ∫ Rcosθ.Rdθ
                                  = (1/L).R²  ∫ sinθ.dθ ------- (d)

                             Y_g = (1/L) ∫ YdL
                                  = (1/L) ∫ Rsinθ.Rdθ
                                  = (1/L).R²  ∫ sinθ.dθ -------- (e)

CENTROID OF A QUARTER CIRCULAR ARC OF RADIUS R

Suppose we have a quarter circular arc in a co-ordinate system as shown in the figure. Total length of the arc AB = (πR)/2 . We take an arbitrarily small length of the arc CD and denote it as dL.

So,

           dL = Rdθ  ------ (iii)    [ as s=Rθ ]

where R = Radius of the quarter circular arc.

Let the co-ordinate of the point D be D(x,y) where

   
               X = Rcosθ -----(iv) and

            Y = Rsinθ -----(v)

Hence   X_g = (1/L)∫x.dL  ;  here  L = (πR)/2  ;        

                                           X = Rcosθ      

                                          dL = Rdθ

    

             X_g = (2/πR)   ₀∫^π/2Rcosθ.Rdθ 

     =    (2/πR) R²  ₀∫^π/2cosθ.dθ

 =      2R/π
   

   Y_g = (2/πR)   ₀∫^π/2Rsinθ.Rdθ 

     

 =      (2/πR) R²  ₀∫^π/2sinθ.dθ

 =      2R/π

Hence, for a quarter circular arc of radius R will be G(2R/π,2R/π)                                                                

                                                                                      

CENTROID OF A COMPOSITE LINE

In the figure, a composite line A-B-C-D is made of three straight lines AB, BC, CD as shown in the figure.So, a composite line is consisted of several straight or curved lines.


Let a composite line is made of n number of lines, which may straight or curved lines.

STEP-ONE:

Draw the composite line and break it into n number of parts. Let the lengths of these lines are L_1, L_2,  L_{3 ........} L_n. Let the centroids of these lines are G₁(X₁,Y₁),G₂(X₂,Y₂), G₃(X₃,Y₃) ........ G_n(X_n,Y_n).

Calculate length (L_i), and coordinates (X_i,Y_i) for each and every parts.

 

Now, if the centroid of the composite line be G(X_g,Y_g)

X_g = (∑L_iX_i)/(∑L_i) 

    => (L₁X₁ + L₂X₂ + L₃X₃)/(L₁ + L₂ + L₃)
   

Y_g = (∑L_iY_i)/(∑L_i)

    => (L₁Y₁ + L₂Y₂ + L₃Y₃)/(L₁ + L₂ + L₃)
   

CENTROID OF A COMPOSITE LINE

In the figure, a composite line A-B-C-D is made of three straight lines AB, BC, CD as shown in the figure. If the coordinate of point A is (5,5), find the centroid of the composite line.


Solution: At first, the composite line is divided into three parts.








Part -1 : The line AB : Let the centroid of the line be G₁(X₁,Y₁)

length, L₁ = 40 mm;                  


X₁ = 4 + (40*cos 60⁰)/2 = 14  
Y₁ = 4 + (40*sin 60⁰)/2 = 21.32








Part -2 : The line BC : Let the centroid of the line be G₂(X₂,Y₂)


length, L₂ = 15 mm; 


X₂ = 4 + (40*cos 60⁰) + 15/2 = 31.5 
Y₂ = 4 + (40*sin 60⁰) = 38.64




Part -3 : The line CD : Let the centroid of the line be G₃(X₃,Y₃)


length, L₃ = 20 mm; 

X₃ = 4 + (40*cos 60⁰) + 15 = 39 
Y₃ = 4 + (40*sin 60⁰) - 20/2 = 28.64



If the centroid of the composite line be G  (X_g,Y_g)

X_g = (∑L_iX_i)/(∑L_i) 

    = (L₁X₁ + L₂X₂ + L₃X₃)/(L₁ + L₂ + L₃)
    = (40 x 14 + 15 x 31.5 + 20 x 39)/(40 + 15 + 20) 
    = 24.17
     

Y_g = (∑L_iY_i)/(∑L_i) 

    = (L₁Y₁ + L₂Y₂ + L₃Y₃)/(L₁ + L₂ + L₃)
    = (40 x 21.32 + 15 x 38.64 + 20 x 28.64)/(40 + 15 + 20) 
    = 26.74

CENTROID OF AN AREA

 

CENTROID OF AN AREA

Engineering Mechanics EME-102

Geometrical Center of an area (A) is often termed as Centroid or Center of an Area.

Suppose we have an area A in a certain X-Y coordinate system, we divide the area into n parts and named them as A₁, A₂, A₃, .... A_n,. Let the coordinates of those tiny elemental areas are as (X₁,Y₁), (X₂,Y₂), (X₃,Y₃) ..... (X_n,Y_n).

As area can be represented by a vector, hence, Area A can be treated as the resultant of the tiny elemental vectors A₁, A₂, A₃, .... A_n... Let the direction of the resultant vector passes through the point G(X_g,Y_g) on the plane of the area. The point G(X_g,Y_g) is called the CENTROID of the area A. (The direction of any area is along the perpendicular to the area drawn at the centroid of the area).

Like other vectors, an area has a moment about an axis and be represented by the product of the radial distance between the area and the axis and the area itself. So if an elementary area A₁ has a coordinate (X₁,Y₁) it means the area is at a distance X₁ from the Y axis and Y₁ from the X axis. Therefore the moment produced by A1 about Y axis is X₁A₁ and about X axis is Y₁A₁.

 Therefore the summation of all the moments produced by each and every elemental area about Y axis will be ∑A_iX_i and about X axis will be ∑A_iY_i.

Again, the resultant area A passes through the point G(X_g,Y_g). Therefore the moment produced by the area A about Y axis will be AX_g and about X axis will be AY_g.

Like other vectors, it will obey the Moment Theorem which states the total moment produced by individual vectors will be exactly equal to the moment produced by the resultant vector about a certain axis.

Therefore,
AX_g = A₁X₁ + A₂X₂ + A₃X₃ + ...... + A_nX_n

and
AY_g = A₁Y₁ + A₂Y₂ + A₃Y₃ + ...... + A_nY_n

           

                 For an area, a centroid G(X_g,Y_g) can be defined using calculus by the equations,

X_g = (1/A)∫x.dA   ------ (i)

Where dA = elemental area and A= total area.

Y_g = (1/A)∫y.dA   ------ (ii)

HOW TO DERIVE THE VALUES OF X_g and Y_g FOR BASIC GEOMETRIC FIGURE:

STEPS TO FIND X_g


i) Draw the figure in a Coordinate System.

ii) Draw a thin strip of area of thickness (dx) parallel to Y axis and at a distance (x) from Y axis.

iii) Find the height of the strip. Either the height will be constant or the height is a function of (x), that can be calculated from the equation of the figure.

iv) Calculate the elemental area of the strip, and named as dA. Hence, dA = hdx

v) integrate the expression ∫xdA, but dA = hdx. Therefore, we shall integrate  ∫hxdx over the total area.

vi) X_g = (∫ xdA)/A = (∫ hxdx)/A ; where A = total area = ∫dA = ∫hdx


STEPS TO FIND Yg

i) Draw the figure in a Coordinate System.

ii) Draw a thin strip of area of thickness (dy) parallel to X axis and at a distance (y) from X axis.

iii) Find the length (b) of the strip. Either the length will be constant or the length is a function of (y), that can be calculated from the equation of the figure.

iv) Calculate the elemental area of the strip, and named as dA. Hence, dA = bdy

v) integrate the expression ∫ydA, but dA = bdy. Therefore, we shall integrate ∫bydy over the total area.

vi) Y_g = (∫ ydA)/A = (∫bydy)/A





CENTROID OF A COMPOSITE AREA:




HOW TO FIND THE CENTROID OF A COMPOSITE AREA

(a composite area consists of several straight or curved lines.)

(i) Draw the figure in a coordinate system. Draw the dimensions too. Every dimensions will be measured with respect to origin of the coordinate system

(ii) Divide the composite area into several parts of basic geometric areas. Lebel them as part-1, part-2, part-3, .......part-n. Let the corresponding areas are A₁, A₂, A₃, .... A_n. Let the centroids are G₁(X₁,Y₁), G₂(X₂,Y₂), G₃(X₃,Y₃), ...... G_n(X_n,Y_n).

(iii) Let the centroid of the composite area be G(Xg,Yg). Hence,
Xg = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)

Yg = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)

(a) Suppose we have certain area of magnitude (A) in a coordinate system. The centroid of the area will be at its mid-point. A centroid is denoted by G.
 
                       In the figure we have a complex geometrical area composed of three basic geometrical areas. A rectangle, a semi circle and a isosceles triangle. Let us denote the centroids as G₁, G₂, G₃ for the given areas in the figure.

We shall have to find the Centroid of the entire area composed of  A₁, A₂, A₃. 

At first, the composite line is divided into three parts.

Part -1 : The semi-circle : Let the centroid of the area A₁ be G₁(X₁,Y₁)

Area, A₁ = (π/2)x(25)² mm² = 981.74 mm²                  
          X₁ = { 25 -  (4x25)/(3xπ)} mm = 14.39 mm
          Y₁ = 25 mm

Part -2 : The Rectangle : Let the centroid of the A₂ be G₂(X₂,Y₂)

Area, A₂ = 100 x 50  mm² = 5000 mm²                 
          X₂ = 25 + (100/2) = 75 mm
          Y₂ = 25 mm
Part -3 : The Triangle : Let the centroid of the area Area, A₃ be G₃(X₃,Y₃)

Area, A₃ = (1/2) x 50 x 50 mm² = 1250 mm²                 
          X₃ = 25 + 50 + 25 = 100 mm
          Y₃ = 50 + (50/3) =  66.67 mm



If the centroid of the composite line be G  (X_g,Y_g)

X_g = (∑A_iX_i)/(∑A_i) 

    = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)
    = (981.74 x 14.39 + 5000 x 75 + 1250 x 100)/( 981.74 + 5000 + 1250)
    = 71.09
     

Y_g = (∑A_iY_i)/(∑A_i) 

    = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)
    = (981.74 x 25 + 5000 x 25 + 1250 x 66.67)/ ( 981.74 + 5000 + 1250)
    = 32.20