Experimental Force-Displacement curve: different measurement methods

Hi guys,

Talking about a compression test on a sample with an hydraulic press: what's the real difference between using 4 LVDTs to measure displacement (like the one of the photo below) compared to using only the actuator's internal displacement transducer? Would it cause a higher displacement measure to use the second method, and hence, a loss in stiffness?

I'm wondering about this, because I carried this same compression test, but measuring the displacement by the second method, and I'm thinking it might be the cause of the big nonlinear-elastic part of my experimental curves:

The black lines are FEM modelling. I would really appreciate if you know any literature/article discussion this.

Thanks in advance


  • akhemkaakhemka Forum Coordinator

    Hi @Rodrigo28 ,

    Please see if the discussion below helps:


    Ashish Khemka

  • Hi @akhemka ,

    Maybe I wasn't clear enough. My concern is purely experimental. I already know how to plot a numerical load-displacement curve.

    My question is: does using the press internal displacement transducer (relating to the actuator's movement) might affect the load-displacement curve? Instead of using LVDTs as the photo I posted above.

    This issue came to my mind as my linear elastic numerical modelling is much stiffer than the experimental curves.

  • akhemkaakhemka Forum Coordinator

    Hi @Rodrigo28 ,

    I have no comments on this query. But may be other people can comment further.


    Ashish Khemka

  • peteroznewmanpeteroznewman Member
    edited August 22

    Hi @Rodrigo28 ,

    Using the internal displacement measurement of a testing machine actuator to output displacement generates larger values than using instrumentation that is directly measuring the test sample displacement. The reason for this can be understood if you think of everything as a spring. The sample is a spring and you are trying to find the spring rate for the elastic modulus. But the testing machine frame is also a spring and these two springs are in series. When the actuator moves to compress the sample, it also applies the same force to the machine frame which stretches a little. In that way more displacement is needed by the actuator than goes into compressing the sample. If the sample has a low stiffness, then there is very little stretching of the machine frame and the actuator movement is almost all sample compression, but if the sample has high stiffness, then much more actuator movement goes into stretching the machine frame. That is why independent instrumentation on the sample is required.

    An experimental error such as having displacements that are larger than the true displacement of the sample will result in lower slopes on the elastic portion of the force-displacement graph.

  • Hi @peteroznewman,

    Thank you so much sir, for your valuable insight on this matter. Actually the sample is Cross-Laminated timber (wood). For this kind of test and sample, the expected stiffness is aproximately between 300 and 500 MPa.

    This was my test setup for this same mechanical experiment:

    As you can see, I didn't use LVDTs to measure the displacements because the laboratory didn't have them available. The only way was to use the internal displacement transducer.

    If i understood correctly your explanation, when using this method of measurement, the stiffness of the transducer (the spring analogy) ends up messing with the actual behaviour of the sample, right?

    I think that would be one of the main reasons I'm not being able to match the stiffness of the samples with the numerical simulation.

    Thanks again Peter


  • It is the Frame that stretches. The wheel on the piston is the transducer on the machine. If you squash the sample by 1 mm and stretch the frame by 1 mm, the transducer measures 2 mm. You must put an instrument on the sample to measure the 1 mm. You don't want to use the 2 mm value because it is wrong.

  • @peteroznewman The drawing was very helpful to better understand this phenomenon. Again, thank you very much.

    If there's a way to close this topic or to mark is as solved, please tell me.


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