Mixing
Equation describing the deformation of a beam with large displacement
Short version of the question
Which is the equation describing the shape of a beam with large displacement knowing the starting and ending points and tangents?
My considerations:
The Eulero-Bernulli Beam theory says that the equation of a deformed beam must satisfy the differential equation:
$$ frac{d^2}{dx^2} left( EI frac{d^2w}{dx^2}right) = q(x)$$
where $EI$ is the bending stiffness of the beam, $w$ is the vertical displacement and $q(x)$ is the distributed load.
For the case $q(x) = 0$ and $frac{dEI}{dx}=0$ the solution is a third order polynomial.
If we consider the case of a beam with large displacements this is no longer true. Imagine a thin metal film, bent to form a almost closed loop. In this case the deformation is close to a circle. This, clearly, is not a polynomial. Same holds true if we consider a beam and we impose a $pi/2$ rotation at one end as no polynomial will ever have a vertical tangent point.
I think the solution should be the curve minimizing the elastic strain energy i.e. :
$$ mathrm{argmin}_{f(s)} frac{1}{2} int_0^L EI kappa(s) ds = mathrm{argmin}_{f(s)} int_0^L kappa(s)ds$$
where $kappa(s)$ is the curvature of the parametric curve $f(s)$ in $s$.
I suspect this to be somehow related to the splines (also considering that the splines where born to describe the shape of deformed beams link). Splines are however piece-wise polynomials, so I don't understand how they could describe something that is (see above) not a polynomial.
curves curvature
asked Nov 22 '18 at 0:55
Luca AmerioLuca Amerio
63
add a comment |
Short version of the question
Which is the equation describing the shape of a beam with large displacement knowing the starting and ending points and tangents?
My considerations:
The Eulero-Bernulli Beam theory says that the equation of a deformed beam must satisfy the differential equation:
$$ frac{d^2}{dx^2} left( EI frac{d^2w}{dx^2}right) = q(x)$$
where $EI$ is the bending stiffness of the beam, $w$ is the vertical displacement and $q(x)$ is the distributed load.
For the case $q(x) = 0$ and $frac{dEI}{dx}=0$ the solution is a third order polynomial.
If we consider the case of a beam with large displacements this is no longer true. Imagine a thin metal film, bent to form a almost closed loop. In this case the deformation is close to a circle. This, clearly, is not a polynomial. Same holds true if we consider a beam and we impose a $pi/2$ rotation at one end as no polynomial will ever have a vertical tangent point.
I think the solution should be the curve minimizing the elastic strain energy i.e. :
$$ mathrm{argmin}_{f(s)} frac{1}{2} int_0^L EI kappa(s) ds = mathrm{argmin}_{f(s)} int_0^L kappa(s)ds$$
where $kappa(s)$ is the curvature of the parametric curve $f(s)$ in $s$.
I suspect this to be somehow related to the splines (also considering that the splines where born to describe the shape of deformed beams link). Splines are however piece-wise polynomials, so I don't understand how they could describe something that is (see above) not a polynomial.
curves curvature
asked Nov 22 '18 at 0:55
Luca AmerioLuca Amerio
63
add a comment |
Short version of the question
Which is the equation describing the shape of a beam with large displacement knowing the starting and ending points and tangents?
My considerations:
The Eulero-Bernulli Beam theory says that the equation of a deformed beam must satisfy the differential equation:
$$ frac{d^2}{dx^2} left( EI frac{d^2w}{dx^2}right) = q(x)$$
where $EI$ is the bending stiffness of the beam, $w$ is the vertical displacement and $q(x)$ is the distributed load.
For the case $q(x) = 0$ and $frac{dEI}{dx}=0$ the solution is a third order polynomial.
If we consider the case of a beam with large displacements this is no longer true. Imagine a thin metal film, bent to form a almost closed loop. In this case the deformation is close to a circle. This, clearly, is not a polynomial. Same holds true if we consider a beam and we impose a $pi/2$ rotation at one end as no polynomial will ever have a vertical tangent point.
I think the solution should be the curve minimizing the elastic strain energy i.e. :
$$ mathrm{argmin}_{f(s)} frac{1}{2} int_0^L EI kappa(s) ds = mathrm{argmin}_{f(s)} int_0^L kappa(s)ds$$
where $kappa(s)$ is the curvature of the parametric curve $f(s)$ in $s$.
I suspect this to be somehow related to the splines (also considering that the splines where born to describe the shape of deformed beams link). Splines are however piece-wise polynomials, so I don't understand how they could describe something that is (see above) not a polynomial.
curves curvature
asked Nov 22 '18 at 0:55
Luca AmerioLuca Amerio
63
Short version of the question
Which is the equation describing the shape of a beam with large displacement knowing the starting and ending points and tangents?
My considerations:
The Eulero-Bernulli Beam theory says that the equation of a deformed beam must satisfy the differential equation:
$$ frac{d^2}{dx^2} left( EI frac{d^2w}{dx^2}right) = q(x)$$
where $EI$ is the bending stiffness of the beam, $w$ is the vertical displacement and $q(x)$ is the distributed load.
For the case $q(x) = 0$ and $frac{dEI}{dx}=0$ the solution is a third order polynomial.
If we consider the case of a beam with large displacements this is no longer true. Imagine a thin metal film, bent to form a almost closed loop. In this case the deformation is close to a circle. This, clearly, is not a polynomial. Same holds true if we consider a beam and we impose a $pi/2$ rotation at one end as no polynomial will ever have a vertical tangent point.
I think the solution should be the curve minimizing the elastic strain energy i.e. :
$$ mathrm{argmin}_{f(s)} frac{1}{2} int_0^L EI kappa(s) ds = mathrm{argmin}_{f(s)} int_0^L kappa(s)ds$$
where $kappa(s)$ is the curvature of the parametric curve $f(s)$ in $s$.
I suspect this to be somehow related to the splines (also considering that the splines where born to describe the shape of deformed beams link). Splines are however piece-wise polynomials, so I don't understand how they could describe something that is (see above) not a polynomial.
curves curvature
curves curvature
asked Nov 22 '18 at 0:55
Luca AmerioLuca Amerio
63
asked Nov 22 '18 at 0:55
Luca AmerioLuca Amerio
63
asked Nov 22 '18 at 0:55
Luca AmerioLuca Amerio
63
asked Nov 22 '18 at 0:55
Luca AmerioLuca Amerio
63
asked Nov 22 '18 at 0:55
Luca AmerioLuca Amerio
63
63
add a comment |
add a comment |
1 Answer
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The physical consideration at the base of Eulero-Bernoulli beam theory is that the cross-sections of the beam normal to the axis remain flat upon deformation, and that their distance on the neutral axis remains constant.
If the deformation is small, we can approximate $ds$ with $dx$ and get the equation that you report.
If the deformation is large, but the thickness of the beam is small enough and it has good elastic behaviour to allow the physical assumption that the cross-sections remains straight, then the distance along the axis cannot be any longer made $ds approx dx$ and you shall instead integrate in $ds$.
In the general case of clamped edges , even without distributed load, the solution is not simple, re. for instance to this article as a start.
answered Nov 22 '18 at 1:12
G CabG Cab
18k31237
That's more or less what I was thinking of, but is there a solution for this case? Do you know, for the case where the distributed load is null, which kind of equation will the result be? Is there a "fast" way to find the deformation of a beam knowing starting position, ending position and the two tangents?
– Luca Amerio
Nov 22 '18 at 10:24
@LucaAmerio, unfortunately, in general it is not a simple formula: added a note to my answer.
– G Cab
Nov 22 '18 at 15:54
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
The physical consideration at the base of Eulero-Bernoulli beam theory is that the cross-sections of the beam normal to the axis remain flat upon deformation, and that their distance on the neutral axis remains constant.
If the deformation is small, we can approximate $ds$ with $dx$ and get the equation that you report.
If the deformation is large, but the thickness of the beam is small enough and it has good elastic behaviour to allow the physical assumption that the cross-sections remains straight, then the distance along the axis cannot be any longer made $ds approx dx$ and you shall instead integrate in $ds$.
In the general case of clamped edges , even without distributed load, the solution is not simple, re. for instance to this article as a start.
answered Nov 22 '18 at 1:12
G CabG Cab
18k31237
That's more or less what I was thinking of, but is there a solution for this case? Do you know, for the case where the distributed load is null, which kind of equation will the result be? Is there a "fast" way to find the deformation of a beam knowing starting position, ending position and the two tangents?
– Luca Amerio
Nov 22 '18 at 10:24
@LucaAmerio, unfortunately, in general it is not a simple formula: added a note to my answer.
– G Cab
Nov 22 '18 at 15:54
add a comment |
The physical consideration at the base of Eulero-Bernoulli beam theory is that the cross-sections of the beam normal to the axis remain flat upon deformation, and that their distance on the neutral axis remains constant.
If the deformation is small, we can approximate $ds$ with $dx$ and get the equation that you report.
If the deformation is large, but the thickness of the beam is small enough and it has good elastic behaviour to allow the physical assumption that the cross-sections remains straight, then the distance along the axis cannot be any longer made $ds approx dx$ and you shall instead integrate in $ds$.
In the general case of clamped edges , even without distributed load, the solution is not simple, re. for instance to this article as a start.
answered Nov 22 '18 at 1:12
G CabG Cab
18k31237
That's more or less what I was thinking of, but is there a solution for this case? Do you know, for the case where the distributed load is null, which kind of equation will the result be? Is there a "fast" way to find the deformation of a beam knowing starting position, ending position and the two tangents?
– Luca Amerio
Nov 22 '18 at 10:24
@LucaAmerio, unfortunately, in general it is not a simple formula: added a note to my answer.
– G Cab
Nov 22 '18 at 15:54
add a comment |
The physical consideration at the base of Eulero-Bernoulli beam theory is that the cross-sections of the beam normal to the axis remain flat upon deformation, and that their distance on the neutral axis remains constant.
If the deformation is small, we can approximate $ds$ with $dx$ and get the equation that you report.
If the deformation is large, but the thickness of the beam is small enough and it has good elastic behaviour to allow the physical assumption that the cross-sections remains straight, then the distance along the axis cannot be any longer made $ds approx dx$ and you shall instead integrate in $ds$.
In the general case of clamped edges , even without distributed load, the solution is not simple, re. for instance to this article as a start.
answered Nov 22 '18 at 1:12
G CabG Cab
18k31237
The physical consideration at the base of Eulero-Bernoulli beam theory is that the cross-sections of the beam normal to the axis remain flat upon deformation, and that their distance on the neutral axis remains constant.
If the deformation is small, we can approximate $ds$ with $dx$ and get the equation that you report.
If the deformation is large, but the thickness of the beam is small enough and it has good elastic behaviour to allow the physical assumption that the cross-sections remains straight, then the distance along the axis cannot be any longer made $ds approx dx$ and you shall instead integrate in $ds$.
In the general case of clamped edges , even without distributed load, the solution is not simple, re. for instance to this article as a start.
answered Nov 22 '18 at 1:12
G CabG Cab
18k31237
edited Nov 22 '18 at 15:53
answered Nov 22 '18 at 1:12
G CabG Cab
18k31237
answered Nov 22 '18 at 1:12
G CabG Cab
18k31237
answered Nov 22 '18 at 1:12
G CabG Cab
18k31237
18k31237
That's more or less what I was thinking of, but is there a solution for this case? Do you know, for the case where the distributed load is null, which kind of equation will the result be? Is there a "fast" way to find the deformation of a beam knowing starting position, ending position and the two tangents?
– Luca Amerio
Nov 22 '18 at 10:24
@LucaAmerio, unfortunately, in general it is not a simple formula: added a note to my answer.
– G Cab
Nov 22 '18 at 15:54
add a comment |
That's more or less what I was thinking of, but is there a solution for this case? Do you know, for the case where the distributed load is null, which kind of equation will the result be? Is there a "fast" way to find the deformation of a beam knowing starting position, ending position and the two tangents?
– Luca Amerio
Nov 22 '18 at 10:24
@LucaAmerio, unfortunately, in general it is not a simple formula: added a note to my answer.
– G Cab
Nov 22 '18 at 15:54
That's more or less what I was thinking of, but is there a solution for this case? Do you know, for the case where the distributed load is null, which kind of equation will the result be? Is there a "fast" way to find the deformation of a beam knowing starting position, ending position and the two tangents?
– Luca Amerio
Nov 22 '18 at 10:24
That's more or less what I was thinking of, but is there a solution for this case? Do you know, for the case where the distributed load is null, which kind of equation will the result be? Is there a "fast" way to find the deformation of a beam knowing starting position, ending position and the two tangents?
– Luca Amerio
Nov 22 '18 at 10:24
@LucaAmerio, unfortunately, in general it is not a simple formula: added a note to my answer.
– G Cab
Nov 22 '18 at 15:54
@LucaAmerio, unfortunately, in general it is not a simple formula: added a note to my answer.
– G Cab
Nov 22 '18 at 15:54
add a comment |
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