A_GRAIN-GRAIN BOUNDARIES

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Unit-I :

1 Unit-I Topics covered present class Single crystal and Polycrystal Introduction to crystal formation. Grain and grain boundaries Effect of grain boundaries on the mechanical properties of metal/alloys Determination of grain size

Single crystal:

2 Single crystal

Polycrystalline :

3 Polycrystalline

SINGLE CRYSTAL:

4 SINGLE CRYSTAL Crystal axis are same in the direction, everywhere atoms are oriented in one direction only. There are no grain boundaries. Examples: Some naturally occurring galena, quartz, fluorite It is possible to grow single crystals in the laboratory (Bridgemen method). Extensively research purpose. Recent development in turbine blade and Si and Ge. Manufacturing is costlier affair.

Polycrystalline:

5 Polycrystalline Crystals are separated by a line, we call it as grain boundary. Everywhere crystal follow the structure except at grain boundary , at grain boundary there is misfit. It does not follow the crystal symmetry. Crystals are differently oriented. Most of our engineering applications belong to polycrystalline only .

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Nucleation and Growth of Crystals:

7 Nucleation and Growth of Crystals

Nucleation:

8 Nucleation Nucleation, the first step… First process is for microscopic clusters (nuclei) of atoms or ions to form Nuclei possess the beginnings of the structure of the crystal

Growth:

9 Growth It is the increase in size of these stable nuclei particles by atoms joining them from liquid metal finally formation of rigid solid.

Nucleation and Growth curve:

10 Nucleation and Growth curve Nucleation curve Growth curve T m T

GRAINS AND GRAIN STRUCTURE:

11 GRAINS AND GRAIN STRUCTURE Definition: Grain is the region of space occupied by a continuous crystal lattice . It is the arrangement of grains in a metal ,with a grain having particular crystal structure.

GRAIN BOUNDARY:

12 GRAIN BOUNDARY It is the outside area of grain that separates it from other grains . The dark areas are grain boundary and each light area is grain

END OF INSTRUCTION:

13 END OF INSTRUCTION

STRUCTURE OF SOLIDIFIED METALS POLYCRYSTALLINE:

14 STRUCTURE OF SOLIDIFIED METALS POLYCRYSTALLINE

EFFECT OF GRAIN BOUNDARIES ON PROPERTIES IN METALS AND ALLOYS:

15 EFFECT OF GRAIN BOUNDARIES ON PROPERTIES IN METALS AND ALLOYS Grain boundary effect in Polycrystals Grain boundaries act as barriers to the motion of dislocations in polycrystalline metal during plastic deformation .The metal undergoes strain hardening/work hardening the increase in strength of metal we observe in Cold worked metals.

GRAIN STRUCTURE AND BOUNDARY:

16 GRAIN STRUCTURE AND BOUNDARY Metals contain grains and crystal structures. The individual needs a microscope to  see  the  grains .The grain size is related to the mechanical properties of metals.

EFFECT OF GRAIN SIZE ON PROPERTIES OF METALS :

17 EFFECT OF GRAIN SIZE ON PROPERTIES OF METALS The smaller grain size, the better mechanical properties than bigger grain size metal/or alloys. The relationship between yield strength and grain size of metal/or alloy is furnished in the next slide.

THE RELATIONSHIP BETWEEN YIELD STRENGTH AND GRAIN SIZE:

18 THE RELATIONSHIP BETWEEN YIELD STRENGTH AND GRAIN SIZE The equation is popularly known as Hall-Petch.The relationship between Y.Stg and grain size of metal/or alloy is defined σ y = σ o +Kd -½ , the parameters . σ y = Yield strength of metal/or alloy σ o = Constant (Frictional stress for opposing the motion of dislocation) K= Constant d=Grain size

PROBLEM:

19 PROBLEM The yield strength of mild steel with an average grain size of 0.05 mm is 20,000 psi. The yield stress of the same steel with a grain size of 0.007 mm is 40,000 psi. What will be the average grain size of the same steel with a yield stress of 30,000 psi? Assume the Hall-Petch equation is valid and that changes in the observed yield stress are due to changes in dislocation density

SOLUTION TO THE PROBLEM:

20 SOLUTION TO THE PROBLEM Thus, for a grain size of 0.05 mm the yield stress is 20  6.895 MPa = 137.9 MPa. (Note:1,000 psi = 6.895 MPa). Using the Hall-Petch equation

SOLUTION (CONTINUED) :

21 SOLUTION (CONTINUED) For the grain size of 0.007 mm, the yield stress is 40  6.895 MPa = 275.8 MPa. Therefore, again using the Hall-Petch equation:

SOLUTION (CONTINUED) :

22 SOLUTION (CONTINUED) Solving these two equations K = 18.43 MPa-mm1/2, and σ 0 = 55.5 MPa. Now we have the Hall-Petch equation as σy = 55.5 + 18.43 d-1/2 If we want a yield stress of 30,000 psi or 30  6.895 = 206.9 MPa, the grain size will be 0.0148 mm.

SUMMERY OF THE TOPIC :

23 SUMMERY OF THE TOPIC

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Necleation:

26 Necleation

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