1. Engineering materials lecture #12
Professor Martinez
Engineering materials lecture #12

2. Quiz
A steel bar 100 mm long and having a square cross section 20 mm on an edge is pulled in tension with a load of 89,000 N and experiences an elongation of 0.10 mm. Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel

3. Design/Safety Factors
Safe stress or working stress
σw = σy / N
N = factor of safety
σy = yield strength
Design Example:
A tensile-testing apparatus is to be constructed that must withstand a max load of 220,000 N. The design calls for two cylindrical support posts, each of which is to support half of the maximum load. Furthermore plain-carbon (1045) steel ground and polished shafting rounds are to be used; the minimum yield and tensile strengths of this alloy are 310 MPa and 565 MPa, respectively. Specify a suitable diameter for these support posts.

4. Dislocations & Plastic Deformation
Edge Dislocation:
Lattice distortion along the end of an extra half-plane of atoms
Screw Dislocation:
Resulting from shear distortion, dislocation line passes through the center of a spiral

5. Plastic Deformation
Corresponds to the motion of large numbers of dislocations.
Edge: dislocation moves in response to a shear stress applied perpendicularly to dislocation line
Top halves of planes are forced to the right (Fig. 6.20, pg. 176)
Before and after the movement of dislocation through material, the atomic arrangement is ordered and perfect

6. Slip
Process by which plastic deformation is produced by dislocation motion
Fig. 6.20: plane along which the dislocation line traverses is the slip plane
Plastic deformation corresponds to permanent deformation that results from the movement of dislocations (slip as result of shear stress)

7. Dislocation Motion Similar to movement of a caterpillar
Caterpillar hump corresponds to extra half-plane
Screw Dislocation (Fig. 6.21, pg. 176):
Direction of movement is perpendicular to stress direction
Edge Dislocation (Fig. 6.20):
Direction of movement is parallel to shear stress

8. Dislocation Density The number of dislocations in a material:
The total dislocation length per unit volume or area
Carefully solidified metal crystals (103 mm-2)
Heavily deformed metals (106 mm-2)
Ceramic materials (104 mm-2)
Silicon single crystal (1 mm-2)

9. Characteristics of Dislocations
Strain fields: exist around dislocations; used to determine the mobility of the dislocations and ability to multiply
Plastic deformation: fraction (5%) of deformation energy is retained internally; the rest is dissipated as heat

10. Lattice Strain
The fraction of energy that remains is strain energy associated with dislocations
Edge: atoms immediately above and adjacent to the dislocation line are squeezed together (compressive strain)
Atoms below the half-plane, tensile strain is imposed
Screw: lattice strains are pure shear only

11. Slip Systems Combination of slip plane and the slip direction
For a particular crystal structure, the slip plane is the plane that has the most dense atomic packing (greatest planar density)
Slip direction has the highest linear density
Slip planes may contain more than one slip direction (Table 6.8, pg. 178)

12. Slip in Single Crystals
Dislocations move in response to shear stresses applied along a slip plane and in a slip direction
Shear components exist at all but parallel or perpendicular alignments to the stress direction
Resolved Shear Stresses: magnitude depends on applied stress and orientation of slip plane and direction

13. Resolved Shear Stress (Eq 6.14, pg 179)
τR = σ (cos φ) (cos λ)
σ = applied stress
φ = angle between normal to slip plane and applied stress direction
λ = angle between slip and stress directions
φ + λ ≠ 90°

14. Slip in Single Crystals
One slip system is generally oriented most favorably (largest resolved shear stress)
τR (max)= σ (cos φ cos λ)max

15. Critical Resolved Shear Stress
Slip in a system starts on the most favorably oriented slip system when shear stress is at a critical value (τcrss)
Minimum shear stress required to initiate slip
Single crystal plastically deforms when:
τR (max) = τcrss
σy = τcrss / (cos φ cos λ)max

16. Slip in Single Crystals
The minimum stress necessary to introduce yielding occurs when a single crystal is oriented (φ = λ = 45°):
σy = 2τcrss

17. Example
Consider a single crystal of BCC iron oriented such that a tensile stress is applied along a [010] direction
Compute the resolved shear stress along a (110) plane and in a [Ī I I] direction when a tensile stress of 52 MPa is applied.
If slip occurs on (110) plane and in [Ī I I] direction, and the critical resolved shear stress is 30 Mpa, calculate the magnitude of applied tensile stress necessary to initiate yielding.

18. Plastic Deformation of Polycrystalline Materials
Random crystallographic orientations or numerous grains causes the variation of slip from one grain to another
Dislocation motion occurs along the slip system that has the most favorable orientation

19. Plastic Deformation of Polycrystalline Materials
During deformation, mechanical integrity and coherency are maintained along the grain boundaries (grain boundaries do not come apart)
Individuals grains are constrained by neighboring grains
Grains have approximately the same dimension in all directions
Grains become elongated in direction of tensile force

20. Plastic Deformation of Polycrystalline Materials
Polycrystalline metals are stronger than their single-crystal equivalents
Even though a single grain may be favorably oriented with the applied stress for slip, cannot deform until adjacent and less favorably oriented grains slip

21. Mechanisms of Strengthening in Metals
Materials engineers have to design alloys with high strengths, ductility and toughness.
Several hardening techniques are at the disposal of an engineer
The ability of a metal to plastically deform depends on the ability of dislocations to move

22. Mechanisms of Strengthening in Metals
Reducing the mobility of dislocations (limiting plastic deformation) enhances mechanical strength
Restricting or hindering dislocation motion renders a material harder and stronger

23. Strengthening by Grain Size Reduction
Since two grains are of different orientations, a dislocation passing into grain B will have to change its direction of motion (more difficult as # of crystals increases)
The atomic disorder within a grain boundary region will result in a discontinuity of slip planes from one grain into another

24. Homework
Chapter 6
6.15, 6.16, 6.19, 6.27, 6.28, 6.43
Due this Friday to my mailbox
You can also turn it in this Thursday.