Sep 17, 2018

Material Science AMIE Solved Question Papers Summer 2005. Q1

  (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.
(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.
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.
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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