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I am solving the free vibration of a beam with partially elastic support. I have to solve the following equation

2B^3*Cosh(BL)*Cos(x)-4(K^2/B^3)*Sinh(BL)*Sin(BL)-4K*Sinh(BL)*Cos(BL)+4K*Cosh(BL)*Sin(BL)-2B^3 = 0 

K and L are constant and B is the variable. I need roots of B.

I evaluated the following expression:

FindRoot[2*B^3*Cosh[B*L]*Cos[B*L] - 4*(K^2/B^3)*Sinh[B*L]*Sin[B*L] - 4*K*Sinh[B*L]*Cos[B*L] + 4*K*Cosh[B*L]*Sin[B*L] - 2*B^3, {B,4 \[Pi]}] 

It dose not give a solution — it gives the error:

FindRoot::nlnum: The function value {-3968.8+3968.8 Cos[12.5664 L] Cosh[12.5664 L]+4. K Cosh[12.5664 L] Sin[12.5664 L]-4. K Cos[12.5664 L] Sinh[12.5664 L]-0.00201572 K^2 Sin[12.5664 L] Sinh[12.5664 L]} is not a list of numbers with dimensions {1} at {B} = {12.5664}.

I am a Mathematica beginner and an answer will be very helpful to me.

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  • $\begingroup$ FindRoot is a numerical solver. You should give values to the constants K and L before you use it. Also avoid using uppercase single-letter variable names, as many of them are reserved and may lead to conflicts. $\endgroup$
    – MarcoB
    Jun 11, 2020 at 18:11
  • $\begingroup$ I need a general form of roots in terms of K and L. so is there is any other way or function to solve? $\endgroup$ Jun 11, 2020 at 18:18
  • $\begingroup$ You could try Solve or Reduce then, perhaps with some restrictions since your equations contain periodic functions, but beware: the fact that you want an analytical solution does not imply that one exists... $\endgroup$
    – MarcoB
    Jun 11, 2020 at 18:20
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    $\begingroup$ It is unlikely that analytic solution exists. In fact, it is quite typical for transcendental equations that numerical solution is the only option. $\endgroup$
    – yarchik
    Jun 11, 2020 at 20:03
  • $\begingroup$ I am familiar with solving beam equations. The peculiar term you have in here is the x in Cos[x]. What is this term? Is this a Bernoulli Euler beam with a spring at one and and another type of boundary condition at the other? It is usual to form non dimensional groups to simplify the problem. I can help if you give more information on what you are trying to do. $\endgroup$
    – Hugh
    Jun 13, 2020 at 17:16

1 Answer 1

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I've found a family of solutions by looking at the second derivatives of your equation and finding a fixed $k$ in terms of $l$. This does not cover all values of $b$ so it doesn't give you a general analytic solution. It might not be possible to represent in terms of nice elementary and transcendental functions. But this may be helpful in visualizing where some more non-trivial solutions lie:

eqn = 2*b^3*Cosh[b*l]*Cos[b*l] - 4*(k^2/b^3)*Sinh[b*l]*Sin[b*l] - 
    4*k*Sinh[b*l]*Cos[b*l] + 4*k*Cosh[b*l]*Sin[b*l] - 2*b^3 == 0;

D[eqn[[1]], k, k]
(* returns: -((8 Sin[b l] Sinh[b l])/b^3) *)

Suppose $\sin(b l)=0$ then $b=2\pi/l$. Plug this back into the original equation and simplify and solve for $k$:

tmp=FullSimplify[eqn[[1]] /. b -> 2 π/l]
(* returns: 8 Sinh[π] (-k Cosh[π] + (4 π^3 Sinh[π])/l^3) *)
Solve[tmp==0,k]
(* returns: (4 π^3 Tanh[π])/l^3 *)

(* verify it gives zero *)
FullSimplify[eqn[[1]] /. {b -> 2 π/l, k -> 4 π^3 Tanh[π]/l^3}] == 0
(* returns: True *)

So we have a family of solutions along a parametric curve in $b$ and $k$: $$ (l=0) \ \lor \ \left( b = \frac{2\pi}{l}\ \bigg| \ k=\frac{4 \pi ^3 \tanh (\pi )}{l^3},\ l\neq0\right) $$ If you choose $b=4\pi$, we have $l=1/2$ and $k=32 \pi^3 \tanh(\pi)$

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  • $\begingroup$ To me, it’s clearer to think of it this way: Instead of talking about partial derivatives, you set the leading coefficient equal to zero (of the quadratic polynomial in k). $\endgroup$
    – Michael E2
    Jun 14, 2020 at 15:04

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