New answers tagged continuum-mechanics
2
Continuum mechanics, but it has too many topics except for the rope. The principals are the same.
Besides, there does exist some books about rope only. Such as Theory of wire rope . Just try googling with book about rope/wire/string mechanics and you will find more.
I did some research about this before and if you are into codes, youcan find some programs in ...
1
This is not the Taylor series expansion for E(x). The derivation uses the Taylor series expansion for f(x) to determine the strain in the material at material location x.
f(x) is the location at time t of the material point that was at location x at time zero. The focus of this analysis is the small element of material that was situated between locations x ...
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Without making use of Taylor series what the textbook is computing is
\begin{align}
\require{cancel}
E(x)
&= \lim_{dx\to0} \frac{(f(x+dx)-f(x))^2-(dx)^2}{2(dx)^2} \\
&= \frac12\lim_{dx\to0} \left(\left(\frac{f(x+dx)-f(x)}{dx}\right)^2-\frac{(dx)^2}{(dx)^2}\right) \\
&= \frac12\left(\left(\lim_{dx\to0} \frac{f(x+dx)-f(x)}{dx}\right)^2-\lim_{dx\to0}...
5
The tension is due to the weight of portion of the rope below the point $x$ so as you go up there’s a greater portion of the rope and thus a greater portion of the mass of the rope to pull the little portion of the rope at $x$. Note that since the tension is “local” (it changes at every $x$ rather than being constant throughout), you also need the “local” ...
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Assuming that the sheet is homogenous & isotropic (i.e., it doesn't stretch more easily on one direction than another, and all points are equally stretchy), then the rubber sheet will form a minimal surface: it will minimize its surface area assuming the boundary is held fixed. If the surface can be expressed in the form $z = z(x,y)$, then an ...
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Trying to extend Chet Miller's answer considering an initial configuration where the surface lies in the x-y plane to an arbitrary plane z(x,y) I obtain following initial area:
$dA_i=|\mathbf{t_X}\times \mathbf{t_Y}|dXdY\tag{1}$
where
$\mathbf{t_X} = \frac{\partial X}{\partial X}\mathbf{i_x}+\frac{\partial Y}{\partial X}\mathbf{i_y}+\frac{\partial Z}{\...
0
Assuming that in the initial configuration, the entire surface lies in the x-y plane, the locations of the material points in the deformed configuration of the surface can be expressed parametrically in terms of their coordinates in the undeformed configuration as:
$$x=x(X,Y)\tag{1-a}$$$$y=y(X,Y)\tag{1-b}$$$$z=z(X,Y)\tag{1-c}$$An arbitrary differential ...
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The simplest answer is to consider the stress-strain relation as:
$σ = Eε + ηε_t$
This is the stress relation for a simple damped spring (Kelvin Voigt).
Then you can define your equation of motion as:
$ρAy_{tt} = M_{xx}$
Where Moment can be derived from the stress equation by substituting $I*y_{xx}$ (bend) for the strain.
Thus you have:
$M = (-EIy_{xx} - ...
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