Forging
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Transcript of 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
1998 Bob Sturges -2-
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
1998 Bob Sturges -3-
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 !
1998 Bob Sturges -4-
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