Advanced elasticity/Curvature

It helps to a know a bit about curvature when you start learning how to do buckling analysis. The following discussion goes through the derivation of some useful elementary results relating to curvature. You have already learned these in your introductory calculus course. However, you may have forgotten the details. So this is a refresher lesson.

Tangent vector to a curve
Let $$\mathbf{x}(t)$$ be a vector valued function (curve) of the parameter $$t$$. The unit tangent vector to the curve traced by the function $$\mathbf{x}$$ is given by

\mathbf{t}(t) = \cfrac{\mathbf{x}'}{\|\mathbf{x}'\|} \qquad \text{where} \qquad \mathbf{x}' := \cfrac{d\mathbf{x}}{dt} ~. $$ Note that the "velocity" of a point on the curve is in the direction of the tangent. Therefore, the unit tangent vector and the unit velocity vector have the same value

\mathbf{t}(t) = \cfrac{\mathbf{v}}{\|\mathbf{v}\|} ~. $$

A straight line has the equation

\mathbf{x}(t) = \mathbf{x}_0 + t~\mathbf{x}_1 ~. $$ Taking the derivative with respect to $$t$$ we see that the tangent vector is constant, i.e., it does not change direction. Alternatively, we may say that the condition $$\mathbf{t}' = 0$$ implies that the unit tangent vector does not change direction.

If the curve is not a straight line, then the quantity $$\mathbf{t}'$$ measures the tendency of the curve to change direction.

Normal vector to a curve
The unit normal to the curve is defined as

\mathbf{n} = \cfrac{\mathbf{t}'}{\|\mathbf{t}'\|} ~; \qquad \|\mathbf{t}'\| \ne 0 $$

Curvature vector of a curve
The curvature vector is defined as the rate of change of the unit tangent vector with respect to the arc length. If $$s$$ measures the arc length, then the curvature vector is given by $$d\mathbf{t}/ds$$. Now, the "velocity" is given by

\cfrac{ds}{dt} = \|\mathbf{v}(t)\| ~. $$ Then

\cfrac{d\mathbf{t}}{ds} = \cfrac{d\mathbf{t}}{dt}~\cfrac{dt}{ds} = \cfrac{1}{\|\mathbf{v}\|}~\cfrac{d\mathbf{t}}{dt} = \cfrac{1}{\|\mathbf{v}\|}~\mathbf{t}' =\cfrac{\|\mathbf{t}'\|}{\|\mathbf{v}\|}~\mathbf{n} ~. $$ Therefore the curvature vector has the same direction at the unit normal vector.

Curvature
The curvature ($$\kappa$$) of the curve is the length of the curvature vector. That means,

\kappa(t) := \cfrac{\|\mathbf{t}'\|}{\|\mathbf{v}\|} ~. $$

Radius of curvature
To get a feel for the radius of curvature, consider the equation of a circle

\mathbf{x}(t) = r~\cos t~\mathbf{e}_x + r~\sin t~\mathbf{e}_y $$ where $$r$$ is the radius of the circle and $$\mathbf{e}_x, \mathbf{e}_e$$ are the unit basis vectors in the $$x, y$$ directions. Then the "velocity" is given by

\mathbf{v}(t) = -r~\sin t~\mathbf{e}_x + r~\cos t~\mathbf{e}_y \qquad \implies \|\mathbf{v}\| = r $$ and the unit tangent vector is

\mathbf{t}(t) = -\sin t~\mathbf{e}_x + \cos t~\mathbf{e}_y ~. $$ Differentiating with respect to $$t$$,

\mathbf{t}'(t) = -\cos t~\mathbf{e}_x - \sin t~\mathbf{e}_y \qquad \implies \|\mathbf{t}'\| = 1 ~. $$ Therefore, the curvature of the circle is

\kappa = \cfrac{\|\mathbf{t}'\|}{\|\mathbf{v}\|} = \cfrac{1}{r} ~. $$ This shows that the radius of the circle is the reciprocal of the curvature of the circle. The radius of curvature of any curve is defined in an analogous manner as the reciprocal of the curvature of the curve at a point.

Curvature of plane curves
Let us now consider a curve in a plane $$x-y$$. Let $$\theta(t)$$ be the angle that the tangent vector to the curve makes with the positive $$x$$-axis. Then we can write

\mathbf{t}(t) = \cos\theta(t)~\mathbf{e}_x + \sin\theta(t)~\mathbf{e}_y $$ where $$\mathbf{e}_x, \mathbf{e}_e$$ are the unit basis vectors in the $$x, y$$ directions.

Taking the derivative we have

\mathbf{t}' = -\sin\theta~\theta'~\mathbf{e}_x + \cos\theta~\theta'~\mathbf{e}_y = \theta'~[-\sin\theta~\mathbf{e}_x + \cos\theta~\mathbf{e}_y] ~; \theta' := \cfrac{d\theta}{dt} ~. $$ Therefore

\|\mathbf{t}'\| = |\theta'| ~. $$ Using the chain rule

\theta' = \cfrac{d\theta}{ds}~\cfrac{ds}{dt} = \|\mathbf{v}\|~\cfrac{d\theta}{ds} ~. $$ The curvature can then be expressed as

\kappa(t) := \left|\cfrac{d\theta}{ds}\right| ~. $$

Useful relation for the curvature of plane curves
If the plane curve is parameterized as

\mathbf{x}(t) = x(t)~\mathbf{e}_x + y(t)~\mathbf{e}_y $$ the curvature of curve can also be expressed as

\kappa(t) = \left|\cfrac{x'~y - x~y'}{(x'^2 + y'^2)^{3/2}}\right| ~; x' := \cfrac{dx}{dt},~ x := \cfrac{d^2x}{dt^2}, ~y' := \cfrac{dy}{dt}, ~y := \cfrac{d^2y}{dt^2} ~. $$ If, in addition, $$ y = g(x) $$, we have

 Proof: The tangent vector to the curve is given by

\mathbf{t} = x'~\mathbf{e}_x + y'~\mathbf{e}_y = \cos\theta~\mathbf{e}_x + \sin\theta~\mathbf{e}_y ~. $$ Therefore

\tan\theta = \cfrac{y'}{x'} ~. $$ Differentiating both sides with respect to $$t$$,

(1)\qquad\sec^2\theta~\theta' = \cfrac{x'~y - y'~x}{x'^2} ~. $$ Now,

(2)\qquad \sec^2\theta = 1+\tan^2\theta = 1 + \left(\cfrac{y'}{x'}\right)^2 = \cfrac{x'^2+y'^2}{x'^2} ~. $$ Plugging (2) back into (1) we get

(3) \qquad \theta' = \cfrac{x'~y - y'~x}{x'^2+y'^2} ~. $$ The curvature is given by

(4) \qquad \kappa = \left|\frac{d\theta}{ds}\right| = \left|\theta'~\cfrac{dt}{ds}\right| ~. $$ Also

(5)\qquad \cfrac{dt}{ds} = \cfrac{1}{\|\mathbf{v}\|} = \cfrac{1}{\sqrt{x'^2 + y'^2}} $$ since

\mathbf{t} = x'~\mathbf{e}_x + y'~\mathbf{e}_y = \cfrac{v_x}{\|\mathbf{v}\|}~\mathbf{e}_x + \cfrac{v_y}{\|\mathbf{v}\|}~\mathbf{e}_y \implies \|\mathbf{v}\| = \sqrt{x'^2 + y'^2} ~. $$ Plugging (3) and (5) into (4) gives

\kappa = \left|\cfrac{x'~y - x~y'}{(x'^2 + y'^2)^{3/2}}\right| ~. $$ For the situation where $$ y = g(x)\,$$ we can parameterize the curve using $$ x = t, y = y(t)\,$$ to get $$ x' = 1, x'' = 0 \,$$. Then,

\kappa = \left|\cfrac{y''}{(1 + y'^2)^{3/2}}\right| \qquad \square $$