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Home , Cold working

9-17-2014 Wednesday, September 17, 2014 6:50 AM Strengthening mechanisms •stretching is due to plastic deformation

•plastic deformation is due to Motion of

• to strengthen Materials, make it harder for dislocations to move.

ENGR45-strengthening mech Page 1 Example: Calculate 0 and Ky and estimate YS of a polyxrystalline with ASTM number 8

From interactive graph we find:

Solve:

Use M=1 and n=8, you get N=1.28x106

•It means

•So

ENGR45-strengthening mech Page 2 ENGR45-strengthening mech Page 3 3)Work , strain hardening, Cold working, more later

Effect of cold work on tensile stress-strain curve for low- bars.

Pasted from

ENGR45-strengthening mech Page 4 4)ppt hardening (or Age hardening) more Later.

ENGR45-strengthening mech Page 5 Dislocations create Plastic deformation by "slip" "SLIP" occurs as shear in slip system consisting of SLIP PLANE and SLIP DIRECTION. Both SLIP PLANE and SLIP DIRECTION are closed pack. In BCC, SLIP PLANE is (110) and SLIP DIRECTION is [111] In FCC, SLIP PLANE is (111) and SLIP DIRECTION is [110]

(6planes)*(2 directions) =12 systems

ENGR45-strengthening mech Page 6 Slip system in FCC, (111) and [110)

(4 planes)*(3 directions) =12 systems

ENGR45-strengthening mech Page 7 Schmid's factor

ENGR45-strengthening mech Page 8 ENGR45-strengthening mech Page 9 More on Cold working, Strain Hardening, .

These are all examples of C.W: Bending Angle Tube Slitting  bending • Blanking  Roll •Spinning Piercing Sizing bending •Embossing Lancing Riveting Draw and •Stretch Perforatin Staking compressio forming g  n •Sheet Notching  Roll forming drawing Nibbling Burnishing Seaming •Ironing Shaving •Superplastic Heading Flanging Trimming forming  Straightenin  Thread g Cutoff rolling Dinking

ENGR45-strengthening mech Page 10

*100

ENGR45-strengthening mech Page 11 ENGR45-strengthening mech Page 12 ENGR 45, CW EXAMPLES A 3mm diameter brass wire of 85Cu-15Zn is cold worked 35%. What are the final diameter and the mechanical properties of the final product? Answer: 0.35 = (32-d2)/32 solving for d: d= 2.4 mm From graphs, the final product will have the following mechanical properties: Hardness = 73 RB, T.S. = 475 MPa, = 5%

You are given a 2.5 mm diameter 70Cu-30Zn brass. It must a have a final diameter of 1mm with the following mechanical properties: T.S>450 Mpa, Hardness>75RB, and a minimum ductility of 10%. Describe a processing history to provide this results.

From Graph: to have T.S.>450Mpa, %C.W should be more that 20%. to have Hardness of more than 75 RB, the % C.W. should be more than 27% to have ductility of at least 10%, the %C.W should be less than 35%. So we should pick a %C.W. between 27 and 35. Lets decide on 30%C.W. If Final diameter has to be 1mm with 30%C.W., we can calculate the starting diameter: 0.30 = (d2-12)/d2 solving for d: d=1.2 mm So the process is: to cold work in several steps from 2.5 mm to 1.2 mm, followed by hot work, then cold work to final diameter of 1mm. OR: Hot work from 2.5 mm to 1.2 mm. Then cold work form 1.2 mm to 1 mm.

A Cu-Ni is to be made with the following mechanical properties: T.S.>55 ksi, ductility of at least 42% and Y.S of approximately 28ksi. Suggest an alloy.

For T.S. to be more than 55ksi, the %Ni has to be between 45 to 85%. For ductility to be more than 42%, Ni has to be at least 50%. For the Y.S. to be 28 ksi, the Ni content has to be between 50 to 75 percentage.

ENGR45-strengthening mech Page 13 For ductility to be more than 42%, Ni has to be at least 50%. For the Y.S. to be 28 ksi, the Ni content has to be between 50 to 75 percentage.

So to have all the listed mechanical properties, the overlapped % C.W is between 50 to 75% Ni. We will choose 75% Ni in order to have all the required properties and room for batch variation. NOTE: you also have to be aware of price of Ni!

CW example.

1-An 85-15 brass sheet (initial thickness = .5 “) is cold work to a thickness of 0.43”. What are the mechanical properties of the final products? 2- A 3.7” diameter brass rod (made from 70Cu-30Zn) is desired to be drawn into a 1.8” diameter rod with the following properties: Hardness  70 RB, T.S  400MPa %EL  10% Is it possible to do this in one shot? If not, why and if yes, how? Show your reasoning.

This is a take-home quiz. It will be emailed to you

Examples listed in the manual,

ENGR45-strengthening mech Page 14 NOTE: Cold-working followed by annealing is called Hot-Working

ENGR45-strengthening mech Page 15 • Formation of polygonized subgrains • density is unchanged • Reduction in residual stress • Electrical resistivity decreases for AL and CU • Improvement in corrosion resistance • No change in grain size

• Initiated by Nucleation and Growth (N&G) • Results in smaller grains • Has important impact in welding • Rex temperature is between Tm/2 and Tm/3 depending on severity of CW.

Diffusion controlled (function of T and t)

Grain Growth:

Example: ENGR45-strengthening mech Page 16 Example:

Using n=2, a} calculate K and d0 at 600C b} estimate grain diameter if sample is heated for 24 hrs at 600C

Note: t=24x60=1440 min.

Crystal growth Recovery Recrystallization

Semester project from many moons ago

ENGR45-strengthening mech Page 17 Semester project from many moons ago

Extra resources: 1- Use of CW to improve fatigue properties: https://www.youtube.com/watch?v=_yOTga_3eTM 2- Cold drawing tubes: https://www.youtube.com/watch?v=J3aLT2B2m3Y 3- How it is made: nails and staples: https://www.youtube.com/watch?v=CUTTmtLKD2g 4- steel sheet: https://www.youtube.com/watch?v=CUTTmtLKD2g

ENGR45-strengthening mech Page 18