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Message #12908
Re: [Question #295301]: calculate external work in Uniaxial Strainer
Question #295301 on Yade changed:
https://answers.launchpad.net/yade/+question/295301
Jan Stránský proposed the following answer:
Hi Jabrane,
you should put "this->dAX = dAX" to a place such that dAX is not changed
afterwards. e.g. it is changed at
https://github.com/yade/trunk/blob/master/pkg/dem/UniaxialStrainer.cpp#L106
So after line 121 (in the original file) it should be the correct position
Jan
2016-06-14 19:47 GMT+02:00 Jan Stránský <
question295301@xxxxxxxxxxxxxxxxxxxxx>:
> Question #295301 on Yade changed:
> https://answers.launchpad.net/yade/+question/295301
>
> Jan Stránský proposed the following answer:
> Hi Jabrane,
>
> But i have a question how do you know that dAX is a local variable ?
>
>
> it just comes from how C++ works / is defined, nothing else.
>
> "Real dAX" in cpp file inside a function simply declares a local variable.
> No matter if the class has a member with the same name.
>
> cheers
> Jan
>
>
> 2016-06-14 19:37 GMT+02:00 Yor1 <question295301@xxxxxxxxxxxxxxxxxxxxx>:
>
> > Question #295301 on Yade changed:
> > https://answers.launchpad.net/yade/+question/295301
> >
> > Yor1 posted a new comment:
> > Hello Jerome and Jan
> >
> > Thank you of your response.
> >
> > Jerome, I simulate the tensile test with TriaxialStressController using a
> > positive strainRate but i doesn't work.
> > I get sigma=sigma2=sigma3=0 and that is anormal.
> >
> > Jan, your solution work but i can't have a definitive opinion until i
> > calculate Wext.
> > But i have a question how do you know that dAX is a local variable ?
> >
> > Best regards.
> > Jabrane.
> >
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