Forging work ratio 3

As far as what can be certified I would base the reduction (work) ratio of the material as the cross-sectional area of the material as-received in your shop divided by the CSA of the as-cast strand or ingot.

In our shop we specify a minimum 5:1 reduction for components that will be significantly hot worked in their entirety; 10:1 if not, but there are always exceptions (ex. knuckles, yokes are similar to your material).

You did not mention what size stock you are working with, but consider that rolled billets are typically around 6", blooms range from 12-14", and ingots much larger.

If what you are receiving specifies the casting method, you may have to go to a larger bloom or ingot, or demonstrate through destructive sectioning that your existing material has the soundness for the application after the additional forging operations.

Thanks dbooker630,

Appreciate the help.

Our bloom for the part in question is a 24" dia. preforged round (with a work ratio of 2.2 already in it). on finish the size will be 11.5" that gives a work ratio of about 2.087(say 2.1).

With a 2.2 work already put on the billet, will the 2.1 work ratio add to 2.2 to give 4.3 or will the 2.1 work ratio multiply with 2.2 to give 4.62? Thats the debate here.

Thanks in advance.............cheers

Looks like the round is already made from about a 35.5" ingot so you cannot get any larger original material.

I would use the original ingot CSA divided by the final work CSA to calculate the reduction. Incidentally I think I reversed the terms in my first post.

Just the work that you are doing in your shop is a 4.4:1 reduction. This value is what multiplies with the original reduction to give you a new value of 9.5:1.

I believe you should be fine.

I think your house's division comes from the definition of terms used in plastic deformation. There is engineering strain and true strain (lab tests or large objects), reduction in area (tension testing), area reduction (during processing), and reduction ratio. Engineering strain is not additive, but true strain is additive. Area reduction (and reduction ratio) is multiplicative. For example, say you have a 100 mm diameter object that must be reduced to 10 mm diameter. If you do it in one step, the reduction ratio is A₀/A_f = 100²/10² = 100. If you do it in two steps (100 to 50 and 50 to 10), then you get reductions of 4 (100²/50²) and 25 (50²/10²), which must be multiplied to get the same value as you do during the one-stage deformation.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.