Recent content by 1350-F

My inclination was to find P/2k from a hodograph and then use that pressure to solve for strain using th the formula I posted above. However I wasn't sure whether there was a commonly used formula or something based on the individual strains.

Here you go

Well no matter how I calculate the stress, I still need some sort of stress-strain relationship. Any method I can think of to find strain otherwise has to include reduction of length, area, etc. I could indeed calculate the new area created by the cavity but if my object is semifinite I wouldn't...

I thought about doing σ = Ymε, but I don't have a value for Y! [Edit] I suppose I could look it up, however

I'm trying to figure out the effective strain of a frictionless punch on a deep plate. For simplicity's sake let's say it's in plane strain. Don't quite know where to start. Closest thing I can think of is the strain from a bulge test, but that involves a thin sheet. Looked at some indentor...

Rotated image!

Homework Statement Homework Equations dW/dt = k ∑ (S⋅V*) dW/dt = P (w/2) Vp The Attempt at a Solution Attached: img005.pdf I have some idea about the orientation of V* 1-4 based on the upper-bound field provided. Unsure if I should include a V5. Is there shear on CE? (Wrote DE in my work...

Taking another look at C1161, there's no provision for hot tests. I can see also how finding the yield wouldn't be the point. At the root of my question was: How can I evaluate a yield estimate for a brittle material? I am, like you say, interested in a yield at high temperature which will be...

Hmm. I suppose you could do an incremental bending test. ASTM C1161 calls for a constant loading rate. I'll go looking for a standard that includes it. Thanks.

It seems to me that the ultimate strength of a brittle material can be easily determined by a bending test, but what about the yield? In the brittle regime, I can see how you couldn't, since the sample would fail before it would flow significantly. However, brittle materials can be made to flow...

If we do it that way P = 25ksi + 1.154*25ksi(0.95/3*0.060) ~180 ksi F = 180ksi * pi * 0.95^2 = 500 000 lb = 250 tons Seems like a lot for a little coin Thanks for your help!

Good Point! The pressure to overcome friction would increase up until the end of the "strike," when R/h is greatest. I suppose in that case I can just use the final geometry. Seems intuitive I guess, but every problem I've encountered so far uses the initial geometry.

Homework Statement You are asked to figure out the force required to coin a 25-cent piece and are given the final dimensions and an average flow stress. Sticking friction is "reasonable" Hosford and Caddell 2nd Ed. Q 7-3 Homework Equations Pa = Y + 2kR0/3h0 The Attempt at a Solution I...

Reading this confirmed my suspicions that the axial yield would equal the differential yield and from there you would divide by √3 to get k. For tresca k would equal σd/2. In that case why report it as a differential stress? I think that's what was throwing me off. Assuming we would know to add...

If I have the differential yield stress σd=(σ1-σ3) for a material, what can I derive from that for use with the von mises criterion? Do I have k=0.5×(σ1-σ3) or σy in axial tension, where σ3 = 0? Essentially is my axial yield stress = σd or √3 × σd /2? (in plane stress)