8.28. This exercise considers the problem of finding the traverse displacement u

Question

8.28. This exercise considers the problem of finding the traverse displacement u(x) of an elastic beam, which is subject to a body force w(z). The potential energy iS 0 where EI is a positive constant. It is assumed that the beam has what are called simply supported ends. One consequence of this is that u(0)-0 and u(1) = 0, The role of the boundary conditions will be discussed in more detail in part (f) (a) Suppose the grid points are 10 0, x| = h, x2 = 2h, , xn+1 1, where h-1/(n + 1). Writing V-F(z)dz, write down the composite trapezoidal approximation for V. (b) Use a centered second-order approximation for uz at x1, r2, -., r You can use a first-order approximations for at o and +1. Using these with the result from part (a), what is the resulting approximation for V? In doing this, remember that uo ln+1-0 (c) What is the resulting matrix equation that must be solved to find the minimum for the approximation for V found in part (b)? (d) The minimum ofVfrom part (b) can be found using the MATLAB command fminsearch(eV,U), where V is the approximation from part (b) and U is an n-vector containing a starting guess for (ul , u2, . . . , un)T. What would be a good, simple, and nonzero choice for U, and why is it a good choice? In answering this, it is worth knowing that a cable hanging between two poles is an elastic beam, subject to gravity, (e) If E1-1 and w = z4, then the exact solution is 180 1008 On the same axis, plot the exact solution and the numerical solution when using 22 grid points. In doing this, explain how you found the numerical solution (using part (c) or part (d)), and why (f) A simply supported beam is required to satisfy the four boundary co ditions: u(0) = 0, u(1) = 0, uzz(0) = 0, and uzz(1-0. The numerical solution was derived without any mention of the last two conditions The reason is that they are natural boundary conditions, which means that if u minimizes V, then it will automatically satisfy u"(0) = 0 and uzz(1) = 0, In comparison, u(0) = 0 and u(1) = 0 are essential boundary conditions, which means that we must explicitly require this

Explanation / Answer

SAVE THE CODE IN MATLABA DN GET THE RESULTS. THE OUTPUT SOLUTION IS ALSO ATTACHED BELOW-

clear

clear

close all;

EI=1;

FUN=@(x) 0.5*EI*x(1)^2-x(2)^2;

fun=@(x)100*(x(2)-x(1)^2)^2+(1-x(1))^2;

[x,fval]=fminsearch(fun,[-1.2, 1]);

SOLUTION-

x =

1.0000 1.0000

>> fval

fval =

8.1777e-10

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