Wednesday, September 19, 2018

AMIE Material Science Solved Question Papers Summer 2005: Q3

SUMMER-2005 MATERIALS SCIENCE ENGINEERING (AN 202JAD 302)
  (Answer FIVE questions, taking AA'Y TWO from Group A      ANY TWO from Group B and ALL from Group C        Irigures in the bracket indicate full marks)
                       
                            Group A  
Q.3. (a) Explain Lever Rule with a Tie Line.       
Find the weight percentage of pro-eutectoid ferrite just above the eutectoid temperature of a 0-3 C-steel.                              (2  2)       
(b) Derive the relationship between True Strain and Engineer- ing Strain. What is Resilience ? Why is it important for spring material ?                                                                        2  (1  1)1       
(c) Describe Yield Point Phenomenon. Draw the engineering stress-strain diagram of Glass. Why does necking occur during ten- sion test of a ductile material ?                                    (2  2  2) 
(d) Justify :
(i) Zinc is not as ductile as copper  (2 x 3)        
(ii) Cold working increases hardness of materials       
(iii) Steel is a brittle material at sub-zero atmosphere.    

   
Ans. (a) Lever rule : Lever rule is an adaptation of the mechani- cal lever force calculations, applied to thermal equilibrium diagrams to obtain the proportion of constituents present for a given composition.  
If two constituents A and B, are separated by a distance x in the diagram, then any composition between them, say distance y from A, will have x- y proportion of constituent A and y proportion of B. 


Fig. 7. Determination of amount of phases by lever rule 
Percentage of liquid phase= XZ/XY *100.
Percentage of solid phase= ZY/XY*100
The numerical values of  X Y, ZY and XZ are inserted  and the amount of phases is determined. The values of phases are as under.
        XZ=13,ZY÷15;XY=28. 
Hence,
the percentage of liquid phase= 13/28×100= 45.40
  The percentage of solid phase = 15/28×100 = 53.60
Hence the amount of solid is proportional to the distance from the fulcrum to be end of lever marking the liquid composition.

(B) Engineering Strain = dl/l
True Strain= l (Integration)lo dl/l
True Strain l (Integration)lo (Engineering Strain).
where l is the instantaneous length of the specimen and lo is the original length.   
    
Resilience :  Resilience is the capacity of a material to absorb or store energy, and to resist shock and impact.  It is measured by the amount of energy absorbed per unit volume. in stressing a material upto elastic limit.  This property is important in materials used for springs.       
           The maximum energy which can be stored in a body upto elastic limit is called proof resilience. Proof resilience per unit volume is called modulus of resilience.  Thus. the energy stored per unit volume at elastic limit is the modulus of resilence.        
          The materials having high resilience are used for springs.  The elastic limit of annealed copper is very low, thus it is not used for springs.  But cold- worked copper has much high elastic limit (and resilience) and thus it is used for springs.  Hence resilience is associated with high elastic limit.  Resilience is also of importance for materials required to bear shocks and vibrations.     
  
(c) yield point phenomenon : A specimen of mild steel during tensile deformation behaves elastically up to a certain high load and then it suddenly yields plastically to a lower value. The first higher point at which yield starts js called the upper yield point and the lower point at which considerable srrain occurs is termed as the lower yield point.  The important feature to notice from this curve is that the stress required to maintain plastic flow immediately after yielding has started is lower than that required to start it. After which the specimen work hardens and the curve rises steadily and smoothly.  This yield point phenomenons in mild steel is explained by the dislocation theory. 
The engineering stress-strain diagram of Glass.


Necking occurs during tension test of ductile muterial :       
The stages in the development of a ductile "cup-and-cone" fracture are illustrated in Fig. 9. Necking begins at the point of plastic instability where the increase in strength due to strain-hardening fails to compensate for the decrease in cross-sectional area Fig.  9 (a)).  This occvrs at maximum load.  The formation of a neck introduces a triaxial state of stress in the region.  A hydrostatic component of tension acts along the axis of the specimen at the centre of rhe necked region.  Many fine cavities form in this region Fig.  9 (b)). and under continued straining these grow the coalesce into a central crack Fig.  9 (c)J. This crack grows in a direction perpendicular to the axis of the specimen until it approaches the surface of the specimen.  It then propagates to the surface of the specimen in a direction'roughly 45o to the tensile axis to form the cone part of the fracture Fig.  9 (d)).



(d) (i) Zinc is not as ductile as copper : Zinc is bluish white metal. It is brittle at ordinary temperatures, but is malleable and ductile between lOOoC to 150oC. At 200oC it becomes brittle again and can be easily powdered. On the other hand copper is red in colour and has a crystalline structure and forms FCC structure. It is highly ductile and malleable and has a very high tenacity. It is very good conductor of heat and electricity, its conductivity being almost as much as that of silver. So. that Zinc is not as ductile as Copper.
(ii) The important change in properties is tl increase in strength of metals.  As deformation goes on increasing, the resistance of metal to further deformation increases constantly. The ductility of metal goes on decreasing simultaneously. The phenomenon is called strain hardening. Again. as the temperature of cold working increases the rate of strain hardening decreases. The hardness of all metals increases with cold work, Fig. 10.       
        Cold working produces elongation of grains in principal direction of working, which leads to changes in tensile properties.  Usually there is decrease in ductility, with cold working.


(iii) In slowly cooled carbon steels, the overall hardness and duc- tility of the steel are determined by the relative proportions of the soft, ductile ferrite and the hard, brittle cementite. The cementite content in- creases with increasing carbon content, resulting in an increase of hard- ness and a decrease of ductility, as we go from low carbon to high carbon steels. A rivet should have good deformability, and hence has a low carbon content. In contrast, a file should have high hardness and wear resistance and hence has a high carbon content. Even though we may increase the hardness by increasing the carbon further, the alloy becomes two brittle to be useful above 1-4 carbon. A rail has 0-6 carbon. It combines some toughness with some hardness and wear resistance.

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Tuesday, September 18, 2018

AMIE Material Science Solved Question Papers Q2.

SUMMER-2005 MATERIALS SCIENCE ENGINEERING (AN 202JAD 302)
  (Answer FIVE questions, taking AA'Y TWO from Group A      ANY TWO from Group B and Al,I_ from. Group C        Irigures in the bracket indicate full marks)
                          Group A      
Q  2. (a) State Fick's laws of Diffusion.  How can it help you in the prohlems of Case Carbu rising ?           
Given an activation energy, Q of 142 kJ/mol, for the diffusion of carbon in FCC iron and an initial temperature of 1000 K. find the tem- perature that will increase the diffusion coefficient by a factor 10.  {R =8.314 J/(mol. K)}.            
Will you use a very high temperature ?  (2+ 2+(3+1)         
(b) What is a Phase ?  What is the difference between Alpha-Iron and ferrite ? Define an invariant reaction with an example. 
(c) Differentiate between 
i) Phase Rule and Phase Diagram, (ii) Solvus Line and Solidus Line.
Ans. (a) Fick's First law of Diffusion : The rate at which particles diffuse in steady state conditions i. e., when there are no changes in the system  with time, the rate of diffusion is proportional to the concentration gradient =dC/dx
this is known as Fick's firs' law. See Fig. 4 
Fig. 4   


   Concentration gradient causes diffusion from left to right.
Rate of diffusion=  -D (dC/dx)
  where. D is difusion coefficient. 
The (-) ve sign shows that a positive flow of particle goes in the direction of falling concentration i.e. a negative  value of dC/dx.
   The rate of diffusion is in atoms moved per square metre per second a kilogram's per square metre per second the concentration in atoms per cubic metre, x in metres and hence D in Sqm/s. 
Fick's Second Law of Diffusion : Fick's second law is concerned with diffusion under unsteady condition; i.e. the flux is a function of both space and time. The flux J varies from section to secticn and varies with time at a given section. In this case the concentration profile will be a function of time. This kind of situation normally exists in practical systems and so Fick's second  law is applicable to real systems.        
Case Carbunising :  Diffusion of carbon into the surface of iron to harden its surface, the concentration of carbon atoms in the surface will change  with time as more and more carbon moves into the surface. If the rate of  diffusion into a small region is J1 and the rate, out of the region is beyond J2  then in the time dt number of particles in the region will increase by (J2 - J1) A delta t.
Thus, the change in concentration delta C is given by the formula
 
Delta C .A.delta x = ( J2 - J1) A. Delta.t
Hence  dC/dt=    (J2 - J1)/  delta x = -(Delta J/delta x)
 
where, - Delta J is the reduction in ratio of diffusion.
Thus it can be concluded that diffusion helps in case carbonization.


  (b)  Phase :  A phase is any physically distinct, chemically homogeneous  mechanically separable portion of a substance.                
In cornrnon terms. a phase has a well defined structure. uniform composition and distinct boundaries or interfaces. 
Alpha Iron :  An allotropic modification of Iron vvhich crystallises in the b_c_c. system and is stable below 912℃. 
Ferrite is the term applied to substantially pure Alpha -Iron occuring in iron-carbon alloys. It is precipitated durng the cooling of steels containing less than 0'85C, arid is so called to distinguish it from the iron of the eutectoid.
Irivariant Reaction:
Cooling---->
Lymda <----->A+B
<------Heating
                  The eutectic characteristics of the Pb-Sn system.               
We have.                
eutectic temperature = Te  180oC             Composition of liquid=Ce  62 Sn (38% Pb)             Composition of  or, Ca.e =1 8 Sn (82% Pb)             Composition of p, Cpe = 97 Sn (3% Pb) 
        The phase rule is readily applied in the single-phase and the two-phase  regions of the phase diagram.
At the eutectic temperature Te. three phases are  in equilibrium. The eutectic temperature Te and the compositions of the three phases, Ce, Cae Cbe are all fixed and none of them can be varied arbitrarily. On slightly increasing the temperature above T., either one or both of or. and 3 phases vvould disappear. On slight decrease of temperature below T:., the liquid phase vvould transform as per Eq. (i) above to a mixture of o- and L3- To denote the zero degree of freedom, the eutectic reaction is called an invarient reaction_ The eutectic temperature is known as aninvariun Temperature.

(c) (i) Phase Rule (Gibbs rule)       
The number of phases present in any alloy depends upon the numher of elements of which the alloy is composed. From thermodynamics considerations of equilibrium. Gibbs derived the following phase rule.         
F = C-P+2 
where, F= Degrees of freedom of system (temperature, pressure, concentration, composition of phases).
C=  Number of components forming the system                 (i.e., elements or compounds)        
P  = Number of phases in the alloys (in equilibrium sysrem)          
2  = Number of external factors
       Generaliy temperature and pressure are considered as external factors which determine the state of alloy. Metals are mostly used at atmospheric pressure. Thus, the pressure has not any appreciable effect on equilibrium of alloys in solid and liquid states. 

Phase Diagram or Equilibrirun Diagram or Constitution Diagram       
In chemistry and material science. a phase diagram is a type of graph used to show the equilibrium conditions between the thermodynarnically distinct phases.
        Phase diagrams show in graphical form, the constitution of alloys as a function of temperature under equilibrium conditions.  By equilibrium conditions we mean extremely slow heating or 'ooling conditions i.e., if any change js to occur, sufficient time must be allowed for it to take place. Phase change tends to occur at slightly higher or lower temperature.  Rapid variation in temperature, prevents the phase change that occur under equilibrium conditions, will distort these diagrams. 
b  METAL  Fig. 5. Phase Diragram or   Equilibrium diagram.
     
        
For pure metals. the diagram will be a verticaI Straight line.  The melting temperature, the boiling temperarure and allotropic transformation are shown as points on this line.  Fig. 5  shows that the metal melts at temperature 'a', and boils at temperature 'b'.  Any allotropic change will take place at temperature in between 'a' and 'b'.      
Generally phase diagrams are used for alloys.  Phase diagrams are called binary diagrams, temary diagrams or multi-phase diagrams when there are rwo. three or many ciemenrs present in the alloy.      
(ii) Solvus line is a line on an equilibrium diagram defining the limit of solid solubiliry.
On the other hand the solidus line is a line on an equilibrium diagram indicating the temperature at which a metal or alloy becomes completely solid on cooling, or at which melting begins on heating under equilibrium conditions. This line indicates the compositions of the solid that can co-exist in equilibrium,      
      When only solid solution of components can precipitate and no pure component can precipitate solid solution of B in A is a and A in B is b.


See Fig No.6. Equilibrium diagram for partial solid solubility.



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Monday, September 17, 2018

Material Science AMIE Solved Question Papers Summer 2005. Q1

SUMMER-2005 MATERIALS SCIENCE ENGINEERING (AN 202JAD 302)
  (Answer FIVE questions, taking AA'Y TWO from Group A      ANY TWO from Group B and Al,I_ from. Group C        Irigures in the bracket indicate full marks) 
                          Group A      
Q  I. (a) What is a Burger vector ? Show it by drawing a Burger circuit ? What is Frank Read source ? State its importance in plastic deformation.                                                               (2  2  2),
(b) Distinguish between :   (2 x 2)  (2 x 2)
(i) Slip and Cross slip (ii) Sessile dislocation and Glissile dislocation.
(c) What is Critical Resolved Shear Stress ?  Derive its formu            (2  2)          (d) Calculate the degree of freedom of ice and water kept in a beaker at 1 atmosphere pressure.                                         (2)       
Ans. (a) Burger Vector : The magnitude and direction of the slip resulting from the motion of a single dislocation is called the Burger Vector.


Fig. 1. Burger's vector's  (a) in Edge dislocation (b) in Screw dislocarions.
        
Frunk Read Source : Frank read source is best possible mechanism for multiplication of dislocations passing through a unit area of a polycrystalline material.     
The number of dislocation is of the order of 10 to 100 million per square centimetre under fabricarion process. 
       If the material is subjected to a plastic deformation this dislocation density increases to as much as  million million per square centimeter. A possible mechanism for this multiplication is called frank read source.

Fig. 2. Frank Read Source
          But this process does not continue indefinitely. The back stress produced by the dislocation piling-up along the slip plain opposes the applied stress.  When this back stress becomes equal to critical  stress, the process will stop.
    Thus process of Frank-Read is mostly observed in semiconductors and salts. In metals, it has been observed only rarely. In metals the majority of dislocation are generated at grain boundaries.
(b)
(i) Slip 
(i) In slip, movement occurs along slip plane i.e. slip plane is fixed 
(ii)    Slip direction is not changed
(iii)  The magnitude of slip vector remains same in slip 
(ii)  Sessile dislocation 
(i)   Burger vector and dislocation  
line both not lie in one active slip plane.
(ii) This is an immobile dislocation.
(iii) It is formed when two parallel dislocation on different plane meet.
Cross slip 
(i) Slip direction changes to avoid obstacle in path i.e. slip plane is not fixed. (ii) Slip direction changes in case of cross slip.
(iii) The magnitude of slip vector changes due to obstacle.
 
Glissile dislocation 
(i) In Glissile dislocation. Burger vector and dislocation line both lie in a slip plane.
(ii)This is an immobile dislocation.
(iii)  Its preferred directions are the glide planes, along which slipping may occur without fracture. The translatory motion may take place along a number of planes parallel to one another.
(C)
Critical Resolved Shear Stress:   At  room temperature occurs very easily by slip  mechanism. There arc two restrictions on this mechanism.
( 1) This slip occurs only on certain planes called slip planes and in cenain directions called slip directions,
(2) Slip begins, when the stress on the slip plane in the slip direction reaches a critical value called as critical resoIved  shear stress.

Critical stress for slip       
Consider the cylindrical single crystal having cross sectional area A (Fig. 3). An axial force P is acting on it. The angle between the slip direction and tensile axis is A(lymda). and the angle between the normal slip plane and tensile axis is q,. The component of axial force acting on the slip plane in slip direction is P cos A(lymda) Only this component of force is effective in moving the dislocation. The area of slip plane (which is inclined at a angle q ) is A/cos q .      
Mathematically critical resolved shear stress 
= Component of load along slip plane /        Area of slip plane         
Let P = Load applied along the axis of the cylindrical crystal      
  A  = Cross-sectional area of the crystal       
Let due to axial load, the stepping take place along the shaded plane     
   Now Let  A (Lymda)  Angle which the slip plane makes in the direction of the load.
      q=   Angle which the perpendicular line to the slip plane makes with the direction of load.       
Such that   A(Lymda) + q  = 90°                                   
Area of the slip plane= A/cos q
Component of the axial load along the slip plane=  P cos A(lymda).
Then shear stress or critical resolved shear stress.
  Te =Load/Area =   P cos A(lymda)/  (A/cos q) 
Te= P/A ( cos A(Lymda)cos q 
 
This cquation is also known as Schmid's Law.
(D)
Given :  lce and water kept in a breaker at 1  atmosphere pressure According to phase rule or Gibbs rule.       
  F= C - P+1                                                                                                      where,  F= Degree of freedom, 
C=  The number of components (elements or components).
P=  The number of phases.
Here, in this case,
     C=2, P=2     
So, F : 2-2+1=1.      
The degree of freedom of ice and water at pressure is 1.
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