Calculating stiffness of a beam of non-constant cross section

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This is a hard problem. Most likely you will not have uniform deflection, but a localized dimple at the point of contact. Theoretically the deflection at the point load is infinite but because in reality the load is spread over a small contact area the deformation becomes finite.

Combining contact mechanics with thin wall membrane deflection is a supremely complicated subject and I wish you luck in finding a relevant reference for a parabolic shape.

What you are attempting to do is consider only the radial stresses along the membrane, with no consideration of the hoop or shear components also. In doing so, the radial stress is perpendicular to the applied force at the center, causing an infinite deformation and hence zero stiffness. Mentally replace the surface of a chain mail material without resistance to bending, but with resistance to pulling. When a force is applied perpendicular to the surface there is nothing there to support the force. You need to do a far more in-depth analysis, and to get a finite answer, you need to move the point loading into a distributed loading over a small area.

You might want to look into this BOOK for more details.

Edit 1

Seems like I misread the question, and the body is a solid. To handle this situation imagine the paraboloid as a series of concentric thin cylinders, each with an infinitesimal thickness ${\rm d}r$, a radius $r=0\ldots R$ and a height $$z(r) = H \left(1-\frac{r^2}{R^2}\right)$$ where $H$ is the overall height of the paraboloid, and $R$ is the base radius.

If the paraboloid is pressed down by a plane, and deflects downwards a distance $\alpha$ from the tip (the approach distance as it is called), then each cylinder @r inside the load zone will see a deflection $$u(r)=\alpha - H \frac{r^2}{R^2}$$ and contribute to a fraction of the reaction force of

$${\rm d}F = \frac{E\,(\nu-1)}{(2 \nu-1)(\nu+1)} \frac{{\rm d} A }{z(r) } u(r)= \frac{2\pi\,E\,(\nu-1)}{(2 \nu-1)(\nu+1)} \frac{ r\, u(r) }{z(r) } \,{\rm d} r $$

Here $E$ is the modulus of elasticity and $\nu$ the Poisson's ratio. The formula comes from the axisymmetric hook's law and the fact that each cylinder has zero radial deflection (constrained by surrounding cylinders).

The load zone where ${\rm d}F>0$ is defined by $r=0\ldots R \sqrt{\frac{\alpha}{H}}$. So as the body is pressed down the load zone increases by $\sqrt{\alpha}$.

The total force applied is then

$$ F = \int_0^{R \sqrt{\frac{\alpha}{H}}} \frac{ 2\pi r E (H r^2 - R^2 \alpha) (1-\nu)} {H (R^2-r^2)(1+\nu) (2\nu-1)}\,{\rm d} r $$

The force displacement relationship is then

$$ \boxed{ F = \frac{\pi E R^2 (\nu-1)}{(\nu+1)(2\nu-1)} \left[ \left(1-\frac{\alpha}{H}\right) \ln \left(1-\frac{\alpha}{H}\right) + \frac{\alpha}{H} \right] } $$

For small deflections the above corresponds to a non-linear stiffness of

$$ k_{eff} = \frac{F}{\alpha} = \frac{\pi E R^2 (\nu-1)}{2 H^2 (\nu+1)(2\nu-1)} \alpha $$

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Updated on June 14, 2020

Comments

  • Hugh
    Hugh over 3 years

    I am considering a paraboloid shape which is fixed at its base and is being compressed downwards. I am trying to find its stiffness so I can calculate how much it deforms. When I attempt to do this I get a result that the stiffness is zero. Can anyone point out where I have made a mistake?

    First I'll define more clearly what paraboloid I am talking about.

    Paraboloids have a shape like this:

    enter image description here

    Which can be expressed by the equation $\frac{z}{c}=\frac{x^2}{a^2}+\frac{y^2}{b^2}$

    But that equation is only valid for paraboloids with the origin at their peak. The paraboloid I am considering has the origin at the base so it looks like this:

    enter image description here

    I hope that diagram is clear enough.

    The base is a circle so the scaling factors $a^2$ and $b^2$ are equal. The peak has been displaced by a distance $h$ so the equation becomes

    $\frac{h-z}{c}=\frac{x^2}{a^2}+\frac{y^2}{a^2}$

    The square of the radius of the cross-section is $x^2+y^2$. At $z=0$ we know that $x^2+y^2=a^2$

    Putting that information into the equation I get

    $\frac{h-0}{c}=\frac{a^2}{a^2}$

    $c=h$

    So the final equation of the paraboloid is

    $\frac{h-z}{h}=\frac{x^2+y^2}{a^2}$

    Now, the way I will calculate the stiffness is to cut the paraboloid into slices along the axis of symmetry and sum the stiffness of all the slices.

    For a slice of width $l$, cross-sectional area $A$, and Young's modulus $E$; the stiffness $k$ is given by:

    $k=\frac{EA}{l}$

    To sum $N$ slices in series I treat them like $N$ springs in series. The equivalent stiffness $k_e$ is given by the formula:

    $\frac{1}{k_e}= \sum_{i=1}^N \frac{1}{k_i}$

    In the limit as the thickness of each slice becomes very small and $N$ tends to infinity the sum becomes an integral.

    The slices I am using have a width $dz$ and a cross-sectional area which is a function of $z$ so the stiffness of a slice is

    $k=\frac{EA(z)}{dz}$

    The area is just a circle with $r=\sqrt{x^2+y^2}$

    $A(z)= \pi r^2$

    $A(z)= \pi (x^2+y^2)$

    From the equation of the paraboloid I know that $x^2+y^2= \frac{a^2}{h}(h-z)$

    $A(z)= \pi \frac{a^2}{h}(h-z)$

    So the stiffness of a slice is

    $k= E \pi \frac{a^2}{h} \frac{h-z}{dz}$

    For the integral the paraboloid extends from $z=0$ to $z=h$. Therefore,

    $\frac{1}{k_e}= \int_0^h \frac{h}{E \pi a^2} \frac{dz}{h-z}$

    $\frac{1}{k_e}= \frac{h}{E \pi a^2} (-\ln(h-h)- -\ln(h-0))$

    $\frac{1}{k_e}= \frac{h}{E \pi a^2} (-\ln(0)+ \ln(h))$

    Since $\ln(0)$ is undefined or $-\infty$ in the limit the stiffness is zero in the limit. I do not understand why the integral does not converge.

  • Hugh
    Hugh over 9 years
    Thanks a lot for checking my calculations. It does make sense that there would be zero resistance to compression at the tip and then as soon as it deforms the area increases and there is then some resistance. I appreciate the suggestions for other topics to look into
  • Hugh
    Hugh over 9 years
    Sorry I have a question about your answer. Did you interpret the paraboloid as being a hollow body? The first image makes it look like that and I should have made it more clear that the image is just the surface of the body but it is filled in so that the cross sectional area exists
  • John Alexiou
    John Alexiou over 9 years
    Oh! So it is like a column? I will take a crack at it again then.
  • John Alexiou
    John Alexiou over 9 years
    You still have a singularity at the tip, resulting in infinite deflection. You need to consider the contact properties, but approximating the tip as a sphere and do the sphere to plane contact found in emtoolbox.nist.gov/publications/…
  • John Alexiou
    John Alexiou over 9 years
    Since it is a non linear problem, there is no singular stiffness value, but a force displacement function.
  • Hugh
    Hugh over 9 years
    The solution you provided was very interesting for me, I am fascinated by contact mechanics. This will be so useful for my thesis :D
  • John Alexiou
    John Alexiou over 9 years
    My answer may be far off. It needs to be validated with an FEA program. Did you look at the NIST paper I linked a few comments up? There a lot of the basics of contact mechanics.
  • Hugh
    Hugh over 9 years
    I will take a look through that NIST document, it has a lot of cases detailed in it! I will also try to verify your formula in abaqus. I checked the integration you did and I think that the log term should be $ln((1+a/H)/(1-a/H))$ which is the log representation of the inverse hyperbolic tangent which appears in the integration. For alpha/H less than 1 the stiffness becomes negative. The way you defined u, alpha, and F were they all in the same direction?
  • John Alexiou
    John Alexiou over 9 years
    If $r' = R \sqrt{\frac{\alpha}{H}}$ then $$\int_0^{r'} \frac{r}{H \left( 1 - \frac{r^2}{R^2}\right)} \frac{H (r'^2-r^2)}{R^2}\,{\rm d}r = R^2 \int_0^Z \left(\zeta + \frac{\zeta}{\zeta^2+1} \left(1-Z^2\right) \right)\,{\rm d}\zeta =\frac{R^2}{2} \left( Z^2 + (1-Z^2) \ln \left( 1-Z^2\right) \right)$$ where $\zeta = \frac{r}{R}$ and $Z = \frac{r'}{R}$. All values are positive.
  • John Alexiou
    John Alexiou over 9 years
    In the end I got $$ F = \frac{\pi E (\nu-1)}{(\nu+1)(2\nu-1)} \frac{R^2}{H} \left(\alpha + (H-\alpha) \ln \left(1-\frac{\alpha}{H}\right) \right)$$ So since $\alpha<<H$ all parts are positive.