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This exercise considers the problem of finding the traverse
displacement () of an elastic beam, which is
subject to a body force ().
The potential energy is
where is a positive constant. It is assumed
that the beam has what are called simply supported ends. One consequence of
this is that (0) = 0 and (1)
= 0. The role of the boundary conditions will be discussed in more detail in
part (f).
(a) Suppose the grid points are 0
= 0, 1 = , 2
= 2, ··· , n+1
= 1, where =
»
This exercise considers the problem of finding the traverse
displacement () of an elastic beam, which is
subject to a body force ().
The potential energy is
where is a positive constant. It is assumed
that the beam has what are called simply supported ends. One consequence of
this is that (0) = 0 and (1)
= 0. The role of the boundary conditions will be discussed in more detail in
part (f).
(a) Suppose the grid points are 0
= 0, 1 = , 2
= 2, ··· , n+1
= 1, where = 1/(+
1). Writing write down the composite trapezoidal
approximation for .
(b) Use a centered second-order approximation for xx
at 1,2,
··· , n. You can use a
first-order approximations for xx at 0
and n+1. Using these
with the result from part (a), what is the resulting approximation for ? In
doing this, remember that 0 = n+1
= 0.
(c) What is the resulting matrix equation that must be
solved to find the minimum for the approximation for found
in part (b)?
(d) The minimum of from part (b) can be found using the
MATLAB command where is the approximation from part (b)
and U is an -vector containing a starting guess
for (1, 2,
··· , n)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 = 1 and =
4, then the exact solution
is
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 conditions: (0) = 0, (1)
= 0, xx(0) = 0, and xx(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 minimizes , then it will
automatically satisfy xx(0) = 0
and xx(1) = 0. In
comparison, (0) = 0 and (1)
= 0 are essential boundary conditions, which means that we must explicitly
require this of our approximation (see part (b)). So, the question is, does
your numerical solution satisfy (approximately) xx(0)
= 0 and xx(1) = 0. Using
the first-order approximations you used in part (b), from the numerical
solution calculate xx(0) and xx(1),
using = 20, 40, 80, 160. Are your answers
consistent with the statement that the minimizer satisfies xx(0)
= 0 and xx(1) = 0?
»
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