Solving HW Geometrically Nonlinear Problem

Hi group, I need help on solving a geometrically nonlinear engineering problem. As the title states, it *is* a homework problem. I have to turn it in a week from tomorrow, so it's likely I will have figured it out by then, but I've had issues with these problems before.

Anyways, The problem involves a weight W suspended between two identical springs of unstressed length L and spring constant k. I have to solve the system for equilibrium. I've got it down to W = 2*k*x*sin(alpha). I tired to solve a triangle with sides L, sin(alpha)*(L+x), and hypotenuse (L+x). alpha is the angle between horizontal and the equilibrium position of either spring (With weight W). Any help would be appreciated. Thanks in advance.

- Doug -

Reply to
Doug Smith
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Apparently your springs started life lined up horizontally, attached to the roof at the outside ends, and now have W hanging down from their common intersection point, right? Why not use

sin(alpha) = sqrt((L+x)^2 - L^2)/L

for your second equation, with which you can eliminate alpha and, at least in principle, solve for x? Looks to be a 4th degree which you could solve numerically if you know L and k. Engineers like that stuff, right?

--Lynn

Reply to
Lynn Kurtz

Typo alert:

should be

sin(alpha) = sqrt((L+x)^2 - L^2)/(L+x)

--Lynn

Reply to
Lynn Kurtz

So the long leg of the triangle should be L = (L+x)*cos(alp). That gives a 2nd equation.

Solve one equation for x and plug into the other equation. Maybe there's some trig identities to use after that. I don't see anything at the moment.

You can't use small angles on problem, right?

Reply to
Jeff Finlayson

That seems fine, based on your definitions.

What do you mean you "tried to solve a triangle"? What did you do with this triangle.

Try writing down an expression for cos(alpha). That gives you a relationship between alpha and x. Then you can do a substitution of one for the other and solve.

What you said above also gives this relationship. Notice that by the Pythagorean Theorem:

L^2 + sin(alpha)^2 * (L+x)^2 = (L+x)^2 L^2 = (L+x)^2 * [1 + sin(alpha)^2]

But doing what I said will get you there quicker.

- Randy

Reply to
Randy Poe

Hint: If you do differential equations: Let 2a be distance between ponts of attachment, let L,L+x be y, y+dy and let F be force in spring. There are 3 situations dF = k dy ; y = a sec(alpha) , differentiate this ; and W = 2F sin(alpha).. Now tire again to solve it further eliminating dy,dF to get (W,alpha,F,y) relations with initial value of y as L ...

Reply to
Narasimham G.L.

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