Forging

1998 Bob Sturges -1-

100%xh

hh

0

10 −

Notes On Forging

Forging has been evidently used since ~500bc. It is the process of metaldeformation through compression.

3 kinds of forging: Cold < 0.3 T meltingWarmHot >0.6 T melting

Forging Processes: Open Die (Upsetting)Closed Die (Impression, Coining, Orbital, Cold

Heading, etc.)Simplest Case = A Compression Test

− Reduction in Height =

− Strain e1=

εε1=

Where the subscript means "a specific value."

Review and

for strain-rate sensitive materials

FORCES AND WORK OF DEFORMATION

since yielding is unconstrained,

F1=YA1 for and ideally plastic material.

From volume constancy: A1=A0h0/h1

Work in the process is =

How does this compare to toughness ?

=1e& =1εε&&

ε

∫ εσ1

0

d


Suppose σσ=Kεεn, then the F=YfA1

Where Yf = Flow Stress at the given strain εε1:

Recall that Area under the Yf, εε1 curve,

so

Now, the work done =Vol x Y x εε1 (for whatever material model we

choose.)

FORCES IN FORGING WHEN FRICTION IS SIGNIFICANT:

• Use an analytic method like

slab analysis and find:

σσy=p=Y'e2µµ(a-x)/h

Where Y'=1.15Y

µµ=friction co-efficient.

a=1/2 width of block

h =height of block

• Q: How do we account for strain hardening?

• Or use an approximate result:

pavg=Y'(1+µµa/h)

Q:what is pavg when µµ=0?

What is the upsetting force?

• For a cylinder, pavg=Y(1+2µµr/3h)

What is the upsetting force?

=1Yε

=Y


Q: Why don't we use Y' in this formula?

Important Point: The upsetting force is inversely proportional to the

height as the part compresses:

Example: A 4135 steel cylinder 6'' in diameter is forged from 4'' h0 to 2''

h1, while cold. What's the forging force if µµ =0.2?

Step1: Get material properties from Kalpakjian's table: K= 147 ksi;

n=0.17;

since it strain - hardens, we need to find Yf (at what strain?)

So εε1=ln(4/2)=0.693 and Yf = Kεεn=147ksi *(0.693)0.17=138 ksi

Step2: Use pavg=Y(1+2µµr/h)

What is r?

So, pavg=138k(1+(2)(0.2)(4.24)/(3*2))=177ksi

and F=pavgA1=177ksi ππ = 107#=5,000T

Also, the above steps are already calculated in a graph in Chapter 6:

Select 2r/h and just read off

pavg/Y !


Example: Forging a block with sticking friction:

Y=150Mpa

What value should we use for µµ in

pavg=Y'(1+µµa/h) ?

Recall that if F=µµN, ττ=µµp,

but ττ must be <=k, the shear yield , so ∞∞ won't work!

Ans: find the peak pressure from another slab analysis which gives:

p = Y'(1+(a-x)/h)

So Y'= =(1.15*150)=172MPa

a= h=

x= at peak pressure

so, p= 172(1+0.1/0.1)=345MPa

We could then find the force by integration over

the whole area.

APPROXIMATE RELATIONS FOR CLOSED-DIE FORGING:

− Large redundant work

− Too hard to calculate exactly ( see c-clamp)

• Estimate F= Kp Yf A where Kp ≈≈ 3~5 Simple, No Flash

Kp ≈≈ 5~8 Some Flash

Kp ≈≈ 8~12 complex with Flash

Y'3

2

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Forging