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One applied math question talks about an elastic bar with a displacement equation $du^2/dx^2 = 1$ for a uniform load and fixed on both ends. I solved this using the Toeplitz matrix K and u comes out to be a parabola. I am not however able to visualize this uniform load in my head: Teacher says imagine small round balls connected to eachother by threads, hanging vertically. But then if both ends are fixed, arent these balls piled up on eachother? The result being a parabola suggests that there is more displacement in the middle. I also cannot visualize this. A math book had this:

enter image description here

I guess this is not a fixed-fixed situation but it shows the displacement $u(x)$ at least. Should I think about this bar horizontally instead of vertically? Maybe it would make more sense that way.

Any ideas?

Addition: I simulated the system using KineticsKits and VPython

http://www.vpython.org

http://kineticskit.sourceforge.net

And using this code

from KineticsKit import *
from visual import vector

system = System(timestep=0.04, gravity=1)
## generate some masses
mass1 = Mass(m=0.1, pos=(0.0, 0.0, 0.0), fixed=1)
mass2 = Mass(m=0.1, pos=(0.0, 0.5, 0.0))
mass3 = Mass(m=0.1, pos=(0.0, 1.0, 0.0))
mass4 = Mass(m=0.1, pos=(0.0, 1.5, 0.0))
mass5 = Mass(m=0.1, pos=(0.0, 2.0, 0.0))
mass6 = Mass(m=0.1, pos=(0.0, 2.5, 0.0))
mass7 = Mass(m=0.1, pos=(0.0, 3.0, 0.0), fixed=1)
## insert them into the system
system.insertMass(mass1)
system.insertMass(mass2)
system.insertMass(mass3)
system.insertMass(mass4)
system.insertMass(mass5)
system.insertMass(mass6)
system.insertMass(mass7)

spring1 = SingleHelixSpring(m0=mass1, m1=mass2, k=1, damping=0.5)
system.insertSpring(spring1)
spring2 = SingleHelixSpring(m0=mass2, m1=mass3, k=1, damping=0.5)
system.insertSpring(spring2)
spring3 = SingleHelixSpring(m0=mass3, m1=mass4, k=1, damping=0.5)
system.insertSpring(spring3)
spring4 = SingleHelixSpring(m0=mass4, m1=mass5, k=1, damping=0.5)
system.insertSpring(spring4)
spring5 = SingleHelixSpring(m0=mass5, m1=mass6, k=1, damping=0.5)
system.insertSpring(spring5)
spring5 = SingleHelixSpring(m0=mass6, m1=mass7, k=1, damping=0.5)
system.insertSpring(spring5)

count = 0

loc_1 = [mass2.sphere.pos.y, mass3.sphere.pos.y, mass4.sphere.pos.y, mass5.sphere.pos.y, mass6.sphere.pos.y]

while 1:
    system.step()
    count += 1
    if count == 100: break

loc_2 = [mass2.sphere.pos.y, mass3.sphere.pos.y, mass4.sphere.pos.y, mass5.sphere.pos.y, mass6.sphere.pos.y]

from itertools import izip
for x,y in izip(loc_1, loc_2):
    print x-y

I do see the output as

0.237135416331
0.377720360592
0.42427376458
0.377720360592
0.237135416331

which shows more displacement in the middle. And starting from equal positions, after gravity effects are complete, the balls look like this (before left, after right)

enter image description here

How geeky is that? :)

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2 Answers 2

up vote 4 down vote accepted

At each end, the displacement is $0$ because it is attached at those points, but in the middle, it is sagging. So the material is compressed towards the bottom (sort of like the balls being piled up on each other). Note that the solution of $\frac{d^2u}{dx^2}=1$ (which I assume is the correct equation) is a parabola: $u=\frac{1}{2}x^2+b x+c$.

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So there is more displacement toward the middle because the middle part feels the effect of all the other balls on top of it, and it is furthest away from the fixed ends? –  BB_ML Aug 6 '11 at 12:58
    
@user6786: right you are. It is similar to a horizontal bar supported at both ends, which sags in the middle. We all have experience with this. –  Ross Millikan Aug 6 '11 at 14:39

The result depends on the boundary conditions. The equation $u''(x)=1$ is appropriate for a rod hanging vertically and subject to a fixed force (e.g. gravity) at every point. The solution is $u(x)=\frac{1}{2}x^2+bx+c$ with $b$ and $c$ determined by boundary conditions.

If the rod is of length $L$ and is fixed at both ends, then the boundary conditions are $u(0)=u(L)=0$, i.e. no displacement at either end. These are satisfied by $c=0$ and $b=-L/2$, so the solution is $u(x)=\frac{1}{2}x(x-L)$, which has its maximum when $x=L/2$, in the middle of the rod as you pointed out. The fact that the rod is fixed at both ends prevents there from being too much displacement at the lower end.

If the rod was free hanging then the appropriate boundary condition at the lower end would be $u'(L)=0$ (the rod is stationary) which is satisfied by $b=c=0$, giving a solution $u(x)=\frac{1}{2}x^2$, so the maximum displacement is at $x=L$.

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