Anyone have a good reference for a simple solution derivation for a beam on elastic foundation for a infinite beam with single point load. All I am interested in is peak soil pressure. I want to go in with an E & I and soil modulus apply single point load and come out with peak soil pressure.
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Beam on Elastic Foundation 1
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For hand calculations I use the method in Reynolds Concrete Designer Handbook (page 408 in my copy of the 10th edition) to calculate peak soil pressures.
Alternatively if you know the soil modulus or at least the range I would just use a simple 2D FE analysis with winkler springs. I think Bridge Deck Analysis - Hambly has the general equations/graphs but its much easier, quicker and less error-prone to use a simple FE program to do the numbers for you.
Alternatively if you know the soil modulus or at least the range I would just use a simple 2D FE analysis with winkler springs. I think Bridge Deck Analysis - Hambly has the general equations/graphs but its much easier, quicker and less error-prone to use a simple FE program to do the numbers for you.
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JSteven's suggestion is definitely what you need if you want the numerical answer.[ ] However your post asks for the "derivation", for which, if you ARE actually after the derivation, you will do no better than Hetenyi's original 1946 classic text (whose title happens to be the same as that of your post).
I missed the "derivation" part- I believe Timoshenko's book on Strength of Materials covers beams on elastic foundations as well- quite a few other Strength of Materials books may as well. The derivation of the governing differential equation is for sure given there, the solution for this case should be relatively straight-forward. I would consider it as 2 semi-infinite beams for the solution.
Roark (7th Ed, section 8.5) has details of the derivation (which oddly doesn't obviously ask for the support stiffness) and references. Table 8.6, case 10 seems to show your case; he has the equation for beam shear V(x) and from this you can quickly deduce the support reaction forces and so derive the maximum pressure.
another day in paradise, or is paradise one day closer ?
another day in paradise, or is paradise one day closer ?
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![[smile]](/data/assets/smilies/smile.gif "[smile] [smile]")
...Found it in Roark's as suggested exactly what I was looking for. Unfortunately went ahead and did an FEA model before read reply.
Settingsun
Structural
Did you get the same answer both ways?
I would like to second Steveh49's question.
I'd also be interested in which FEA system you used.[ ] Several FEAs claim to offer a "beam on elastic foundation" element, but in my (very limited) testing quite a long time ago I came to the conclusion that some of these were not a proper rigorous implementation of BEF theory.[ ] Some gave answers that were dependent upon the number of elements you used in the modelling.
And how you ensured you were adequately achieving infinite length if not by trial & error.
I'd also be interested in which FEA system you used.[ ] Several FEAs claim to offer a "beam on elastic foundation" element, but in my (very limited) testing quite a long time ago I came to the conclusion that some of these were not a proper rigorous implementation of BEF theory.[ ] Some gave answers that were dependent upon the number of elements you used in the modelling.
And how you ensured you were adequately achieving infinite length if not by trial & error.
Beams of infinite length do not occur in practice. For a beam of finite length, the point of application of load is important but was not specified.
An iterative method using fictitious springs spaced out along the beam will provide results sufficiently accurate. Newmark's Numerical Procedures would provide an easy solution.
BA
An iterative method using fictitious springs spaced out along the beam will provide results sufficiently accurate. Newmark's Numerical Procedures would provide an easy solution.
BA
normm,
Eight years ago, I posted a 25 page summary of Newmark's Procedures on thread507-267603 but it appears to have been deleted. Essentially, it is a numerical application of the Conjugate Beam Method for determining slopes and deflections of a beam at a number of selected points along its length.
Examples of engineering problems which have been solved using Newmark Procedures may be found on the internet using Google. One example is found in "Theory of Elastic Stability" by Timoshenko and Gere, Article 2.15 which uses the Newmark procedure to find the elastic buckling load of a bar of variable EI.
BA
Eight years ago, I posted a 25 page summary of Newmark's Procedures on thread507-267603 but it appears to have been deleted. Essentially, it is a numerical application of the Conjugate Beam Method for determining slopes and deflections of a beam at a number of selected points along its length.
Examples of engineering problems which have been solved using Newmark Procedures may be found on the internet using Google. One example is found in "Theory of Elastic Stability" by Timoshenko and Gere, Article 2.15 which uses the Newmark procedure to find the elastic buckling load of a bar of variable EI.
BA
Normm:
The text book I learned Newmark’s Methods from, many years ago, was “Numerical Analysis of Beam and Column Structures,” by William G. Godden, Pub. by Prentice-Hall. It has been a good ref. for years, and has a whole chapter on Columns and Beams with Elastic Support. BA and I learned these methods, Newmark’s Methods, at about the same time, back in the 1960’s, when we didn’t have computers to do the heavy lifting, on special analysis problems. It has been a useful engineering tool over the years.
The text book I learned Newmark’s Methods from, many years ago, was “Numerical Analysis of Beam and Column Structures,” by William G. Godden, Pub. by Prentice-Hall. It has been a good ref. for years, and has a whole chapter on Columns and Beams with Elastic Support. BA and I learned these methods, Newmark’s Methods, at about the same time, back in the 1960’s, when we didn’t have computers to do the heavy lifting, on special analysis problems. It has been a useful engineering tool over the years.
Many thanks BAretired and dhengr for the references.
I will also visit the library to look up another reference ' Numerical Procedure for computing deflections, moments and buckling loads. by Newmark Transactions ASCE Vol 108 1943.'
Looks it will be interesting trip.
I will also visit the library to look up another reference ' Numerical Procedure for computing deflections, moments and buckling loads. by Newmark Transactions ASCE Vol 108 1943.'
Looks it will be interesting trip.
dhengr
I got a copy of this book by Godden.
Although this book looks easy to follow, at present I am trying hard to fathom out what is ' Step 6 , Nodal Concentration. How exactly the values in the row ( -20,-80) ( -100,-85) ( -100, -72) ( -76.5) worked out from the M / EI values. It is supposed to be from trapezoidal formula (h/6) * ( 2wBa + wab).But I cannot see how that will give the values in the row. See the attachment marked red.
Any light you can throw will be much appreciated.
I got a copy of this book by Godden.
Although this book looks easy to follow, at present I am trying hard to fathom out what is ' Step 6 , Nodal Concentration. How exactly the values in the row ( -20,-80) ( -100,-85) ( -100, -72) ( -76.5) worked out from the M / EI values. It is supposed to be from trapezoidal formula (h/6) * ( 2wBa + wab).But I cannot see how that will give the values in the row. See the attachment marked red.
Any light you can throw will be much appreciated.
I do not have the Godden book but it refers to a factors column which does not show up on your copy of the page. It is probably to the right hand side of the page. The values you are questioning should be found by multiplying each value by a common factor shown in the factors column to the right.
BA
BA
Normm:
#1. w[sub]BA[/sub] = (10h/6)(-200 + 0)(w[sub]o[/sub]h/EI) = (10[sup]2[/sup]/6)(-20)(w[sub]o[/sub]h[sup]2[/sup]/EI)
#2. w[sub]BC[/sub] = (20h/6)(-200 + -200)(w[sub]o[/sub]h/EI) = (10[sup]2[/sup]/6)(-40 x 2)(w[sub]o[/sub]h[sup]2[/sup]/EI)
The first -200 above is (2)(w[sub]BC[/sub]), and the second is w[sub]CB[/sub] = +(-200). You do have to do some math gymnastics to work the common factor out of the formula.
The common factor that BA mentions is (10[sup]2[/sup]w[sub]o[/sub]h[sup]2[/sup]/6EI), and as he suggested it is shown in the righthand column on page 11.
And as an aside, our very own BA here on E-Tips, holds the patent on the first formula above, thus, the w[sub]BA[/sub] =…
.
#1. w[sub]BA[/sub] = (10h/6)(-200 + 0)(w[sub]o[/sub]h/EI) = (10[sup]2[/sup]/6)(-20)(w[sub]o[/sub]h[sup]2[/sup]/EI)
#2. w[sub]BC[/sub] = (20h/6)(-200 + -200)(w[sub]o[/sub]h/EI) = (10[sup]2[/sup]/6)(-40 x 2)(w[sub]o[/sub]h[sup]2[/sup]/EI)
The first -200 above is (2)(w[sub]BC[/sub]), and the second is w[sub]CB[/sub] = +(-200). You do have to do some math gymnastics to work the common factor out of the formula.
The common factor that BA mentions is (10[sup]2[/sup]w[sub]o[/sub]h[sup]2[/sup]/6EI), and as he suggested it is shown in the righthand column on page 11.
And as an aside, our very own BA here on E-Tips, holds the patent on the first formula above, thus, the w[sub]BA[/sub] =…
.![[blush]](/data/assets/smilies/blush.gif "[blush] [blush]")
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