Local response

[yfeng]

Hi Vesna,
I am using a beamcolumn element to model a RC cantilever. I tried beam with hinge element, nonlinear beam column element and displacement based element. The material constitutive model is calibrated also based on material test result.
Meanwhile, we are doing some experimental tests in the lab. After calibration of the model, the force displacement response match with test result really good. But the strain predicted by the model is always lower than the test result.
It looks like that the steel material constitutive model dominates the strain we got from the model. In material test, we see a lot of strain hardening for the rebar. We match the constitutive model with the material test result by involving strain hardening in Steel02.
As soon as we involve the strain hardening in Steel02, the strain at a specific displacement for a cyclic analysis is much lower than the one without strain hardening that it is also lower than the experimental result.
Is there any way to increase the strain in the model with keeping the strain hardening in the material model?

Thank you

Yuhao Feng

[vesna]

Hi Yuhao,

At what location along your cantilever beam-column element are you recording the strain? Is it in the plastic hinge region? How about your experimental element? Did you have any cracks in concrete at the location where you recorded strains in your bars?

[yfeng]

Hi Vesna,
Yes, I recorded the strain at plastic hinge region. In the experiment, cracks were all over the length of the whole column and concentrates in plastic hinge region. We block out the cover concrete of our column and put a series of LED sensors along the bar to cover the whole plastic hinge region. We use camera to track the movement of the LEDs and calculate the engineering strain based on the elongation of distance between two LED sensors. Basicly, spacing(gage length) of the LED sensors are around 2". So we got the strain distribution along the plastic hinge region.
Strain is usually higher at the gage which have a crack acrossing it.
The strain in fiber model is even lower than the strain from a uncrack gage of experiment.

Yuhao


vesna wrote:
> Hi Yuhao,
>
> At what location along your cantilever beam-column element are you
> recording the strain? Is it in the plastic hinge region? How about your
> experimental element? Did you have any cracks in concrete at the location
> where you recorded strains in your bars?

[vesna]

Where are your integration points with respect to the plastic hinges? The way I usually decide about the number of integration points and and number of elements needed to model the column using force-based beam column element is to by imposing constraint that the two integration points encompass the plastic hinge region. Most of the time this gives good results.

[yfeng]

Vesna,
With one beam with hinge element, I imposed the calculated plastic hinge length to the weight of the base integration point which is at the fix end.
I also tried one nonlinear beam column element to mode the column. For this case, I used 5 integration points along the element that the weight of end integration point will represent the plastic hinge length.
Do you mean that you use 2 integration points to represent the plastic hinge length? Do you set plastic hinge length equal to the sum of weights of this two integration points?

Thank you

Yuhao

vesna wrote:
> Where are your integration points with respect to the plastic hinges? The
> way I usually decide about the number of integration points and and number
> of elements needed to model the column using force-based beam column
> element is to by imposing constraint that the two integration points
> encompass the plastic hinge region. Most of the time this gives good
> results.

[vesna]

What I do is the following:
since I know position of the last two integration points (e.g., if #IP=5, one point is going to be at location 0, and another point is going to be at 0.173L, where L is the length of the element) I make sure to encompass the plastic hinge region with this two points (for this example the Lp~0.173L).

[yfeng]

Nice. Let me try your method and see what I get.

Thank you.

Yuhao

vesna wrote:
> What I do is the following:
> since I know position of the last two integration points (e.g., if #IP=5,
> one point is going to be at location 0, and another point is going to be at
> 0.173L, where L is the length of the element) I make sure to encompass the
> plastic hinge region with this two points (for this example the Lp~0.173L).