Engineering materials lecture #12 Professor Martinez Engineering materials lecture #12

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

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.

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

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

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)

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

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)

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

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

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)

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

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°

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

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

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

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.

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

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

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

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

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

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

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.