Elegant solutions, Column buckling, and the hole through the middle of the Earth

The previous post presented some examples of solution of differential equations using an iterative process, including the simple harmonic motion problem of a ball falling through an evacuated hole through the middle of the Earth.  The final example was concerned with finding the deflection and buckling load of a column subject to an axial load with a small eccentricity.  This problem, which appears entirely unconnected at first sight, was solved with exactly the same differential equation as the oscillating ball problem, suggesting that the exact analytical solution of the two problems should also have the same form.

The equivalent terms in the applicable differential equations for the two problems are shown below:

Simple Harmonic Motion Column Buckling
Period, T Column Length
Half amplitude, R Transverse Deflection
Velocity Slope
Acceleration, g Curvature

For the case of the ball through the middle of the Earth it was shown that the period of oscillation was given by:

T = 2π(R/g)0.5

This can be rearranged as:

g = R(2π/T)2

To give the initial acceleration required for a given radius and period.

If we substitute in the equivalent terms for column buckling, noting that:

  • Curvature = M/EI = F.Delta/EI (where Delta is the eccentricity of the applied load, F, relative to the centroid of the base of the column, E is the elastic modulus of the column, and I is the second moment of area of the cross section)
  • In simple harmonic motion the period is the time for the ball to travel the distance from the surface of the Earth to the centre four times.  In the case of the column the column length is therefore equivalent to the Period/4.
  • We wish to find the axial force at the top of the column, F, at which the deflection due to the force (ignoring second order effects) is equal to the initial eccentricity at the base.

F.Delta/EI = Delta(π/2L)2

F = EI(π/2L)2   which is the Euler equation for the buckling load of a cantilever column.

But what is the physical significance of this equation, and how does it relate to the acceleration of a ball falling through the Earth, and the orbital velocity of a body with the same period?

This can be  seen if we consider a cylinder of radius delta, centred on the top of the column, with a vertical axis, as shown below:

This diagram shows the deflected column centroidal axis (red line), and the imaginary cylinder.

Consider a line coincident with the centroid of the column at the base (green line), and spiralling up the surface of the cylinder at a constant slope, so that at the top of the column it has passed through a right angle around the cylinder.  If this line is projected onto the XY plane (the plane of the deflected column) it will exactly follow the deflected shape of the column centroid.  It can be seen that:

  • The slope of the line from the vertical, around the surface of the cylinder = (Delta.π/2)/L =  (Delta.π)/2L
  • The curvature of the line in the XY plane at the base = Slope2/Delta =  ((Delta.π)/2L)2/Delta = Delta(π/2L)2
  • Therefore at the buckling load F.Delta/EI =  Delta(π/2L)2
  • Buckling load, F = EI(π/2L)2
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This entry was posted in Beam Bending, Differential Equations, Newton and tagged , , . Bookmark the permalink.

4 Responses to Elegant solutions, Column buckling, and the hole through the middle of the Earth

  1. Pingback: More on buckling columns « Newton Excel Bach, not (just) an Excel Blog

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