ch 11 powder metallurgy kc part 2(1)

8/10/2019 Ch 11 Powder Metallurgy KC Part 2(1)

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Project 1

Average 82%

Range 52-95

Top Teams

95% Team 1 The Industrious Mechys

95% Team 11 Tainted Monkey

93% Team 21 Future of Engineering 91% Team 15 Captain Crunch and the Cereal Killers

91% Team 4 Mountaineers

90% Team 10 The Mountaineers

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CHAPTER-11

POWDER PROCESSING

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POWDER PROCESSING


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Advantages:

Produces a high yield compared to other primaryprocessing methods - usually over 95 percent.

Complex and thin shapes can be producedrelatively easily.

Few machining operations are required, because

the part already has a good finish.

High production levels and good reproducibility.

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Process Limitations

Tooling costs are high - because high pressures are

involved and product shapes are intricate.

Material cost are high - because the powder form is

more costly than bars, sheets, etc. Mechanical properties (such as strength) are lower if

the part is not 100 percent dense.

Solid state shrinkage will occur during sintering andcooling.

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Design Limitations

Small, intricate shapes produced - maximum weight of 70 lbs.

Uniform sections preferred - grooves, tapers, and threads aregenerally not possible (since powder cannot be compacted to

uniform density).

Minimum section thickness is 0.06 in (1.5mm).

Length/diameter ratio should be under 5 for cylinders.

Thin shapes, two dimensional best (gears, knife blades, etc.)

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Design Considerations

The total volume of the partVT = Va +Vb + Vc+ void space.

WhereVT = Total volume of part.Va = Volume of part from material A.Vb = Volume of part from material B.

Vc = Volume of part from material C.void space = volume of part not filled.

The total weight of the partMT = Ma + Mb + Mc.

Where

MT=Total weight of part.Ma=Weight of part due to material A.Mb= Weight of part due to material B.Mc= Weight of part due to material C.

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Volume of the partVx = Mx / dx

Where:Vx = volume of material X in final component

Mx = Mass of material Xdx = density of material X

But Mass of Material X isMx = WFx*MWFx = weight fraction of material XM = Total mass of part M

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The Total Material cost would be the sum of

the individual mass components times their

respective unit costs.

CU=(Ca*Ma)+(Cb*Mb)+(Cc*Mc).

The strength of the component can be

estimated by means of the volume fraction.

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Example -1

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Example -2

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Summary

Virtually unlimited choice of alloys, composites, and associated

properties.

Refractory materials are popular by this process.

Controlled porosity for self lubrication or filtration uses.

Can be very economical at large run sizes (100,000 parts).

Long term reliability through close control of dimensions and

physical properties.

Very good material utilization.

Limited part size and complexity

High cost of powder material.

High cost of tooling.

Less strong parts than wrought ones.

Less well known process.

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Homework

HW 8 Due Tuesday, September 30

Problem 1. Work problem 11-7 with the finalvolume of the part = 8 in3and will be 85% dense. Allother data remains the same. Problem 11-7 issimilar to Example problem 11.1.

Problem 2. The density of a part is increased to 98%from 72% and the original part shape was 6.0 in x 3.0

in x 2.0 in. What are the final part dimensions? Thisproblem similar to Example problem 11.2.

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ch 11 powder metallurgy kc part 2(1)

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