Euler Buckling of Columns

So, here we are, our last big `S' in Mechanics of Materials ... Stability. Or, I should say, IN-stability. It's not `stability' that keeps us (engineers) awake at night - it's the fright of something not being stable (in-stability).

We will look at 3 `kinds' of (in-) stability ...

1. Stability of Columns (Compression Members)

2. Stability of Beams (Bending Members) ... Lateral Torsion (In-stability) ...

3. (In-) Stability of Parts of Members

Let's think if a ½ in. diameter wood (or anything) dowel, say ½ in. long. (Yeah, it's real short.) We can push and push and push on that (loading it in compression) and maybe we'll get it to crush. Probable we'll hurt our hand first. Now, make that a ½ in. diameter dowel 10 feet long, and we'll pretty easily be able to `wobble' it, maybe even snap it. Why? Because of (in-) stability.

Let's look at it.

And let's start with the simple case of a `column' (compression member), that is axially loaded and supported by `pins' at ends ... allowing rotation but no translation. We then apply axial compression load `P', which we assume to be centric.

And now let's consider a column that is bent into a curve under the load P.

Note, now, that, at any location along the column there is a non-zero `v' (some deviation from the `straight' condition) ... the axial load P now generates a moment at the location of M = Pe = Pv.

Also, at any section of the beam a moment (M) is generated due to the bent shape (M = EI v'').

And, now, for the beam to be in equilibrium, these two moment effects at any and every section must be in equilibrium,

E I v'' + P v = 0.

You people in Diffy-Q (and/or Dynamics) may recognize this equation. It's like our spring equation,

m x'' + k x = 0.

Alternately we could say,

v + p² v'' = 0,

where p² = P/EI

The solution to this is ... (from Diffy-Q) ...

v(x) = C₁ sin (p x) + C₂ cos (p x) .

By dumping in some boundary conditions we can get C₂ ...

At x = 0, v = 0 ... so ...

0 = C₁ 0 + C₂ (1) ... C₂ = 0 (causing the second trig function to disappear).

At x = L, v = 0, so ...

0 = C₁ sin (p L) ...

This is possible if C₁ = 0 but that doesn't really get us where we want since it also gives us a non-bent shape.

But we can also satisfy the above if ...

... p L = 0, π, 2π, 3π, and so on.

Where, remember,

... p² = P/EI,

then,

... p = √(P/EI)

or,

[√(P/EI) ] L = 0, π, 2π, 3π, and so on.

OR,

P = (EI / L²⁾ (0² , π² , 2² π² , 3² π² ...) and so on.

For any real values of EI and L,

P = 0 isn't really of any value to us (zero compressive load).

But the next one ... P = EI π² / L² ... is of HUGE value to us.

We will call this P ... P_cr (P `critical).

P_cr = EI π² / L²

This P is `critical' in that ...

... for P < P_cr ... Pv < EI v'' ... and the stiffness (springy-ness) of the column causes it to get back in a straight line.

... for P = Pcr ... the curved shape is exactly in equilibrium ...

... for P > P_cr (even a tiny bit) ... Pv > EI v'' ... and it keeps getting more and more bent ... the stiffness of the column is unable to spring it back into straight, or keep it stably bent ...

SNAP!!!

So, at values of P greater than P_cr ... even the slightest bit of bent-ness, unstraightness, vibration ... GROWS!

SNAP!!!

But note a couple things ...

C₁ has not been evaluated yet, and could be ANYTHING! ...

... meaning, the column is in theoretical equilibrium with a HUGE curve (bent-ness) ...

so much so that it is serviceable ...

or probably even worse ...

so much so that the bending stresses are so high - as to cause the material(s) to fail.

AND !!!! ...

This is a `knife-edge' value.

Let's also look at the .... 2² π² , 3² π² ... and so on terms.

They really aren't of any practical value to us, since they elude to P values greater than the P_cr at π² ... and so `to get there' we already wreck the column.

First Example

Let's look at a 6 x 8 Timber column. Suppose it is `pinned' at both ends, and has an E of 1,600,000 psi. Let's find P_cr for a length (height) of 12 ft.

Using our equation ...

P_cr = EI π² / L² ...

P_cr = the thing we are after ...

E = 1,600,000 psi ...

π = 3.1416 blah, blah ...

L = 12 ft or 144 in.

Oh, what about I?

Everything else equal, the column will first `blow up' at the lowest possible value of I.

If this is a round section ... cool ... we calc our I and be done with it.

But if the section is other than round ... we have various possible I values.

The minimum I for any shape is a `principle I' ... in fact, the I _min.

For a rectangular shape it will be the I for `weak axis bending' ... typically ... hb³/12 ...

From the back of the book ... b = 5.5 in, h = 7.5 in.

So,

I = I_min = 7.5 (5.5)³/12 = 104 in.⁴ ...

So,

P_cr = EI π² / L² = 1,600,000 psi (104 in.⁴) π² / (144 in.)² = 79,200 lb.

This is a pretty big load.

NOTE THAT IT HAS NO F.O.S. ...

Okay,

Some things to add / think about ...

1. We have not looked at the strength of the wood itself ... in other words ... before we hit P_cr, we may start crushing wood (due to the brute compressive stress).

2. Often in columns we are able to provide some stability, or bracing. For example, if our column is at a wall, attachment to the wall can be used to brace it in the weak direction (using the in-plane rigidity of the wall). Thus, the appropriate I would be the `other' I ... in this example, 5.5(7.5)<sup>3</sup>/12 = 193 in.<sup>4</sup> ... nearly twice the value from before, giving nearly twice the P<sub>cr</sub> ... (and increasing the possibility that we will hit `wood crushing', first).

3. We'll deal with the `crushing' thing later ... for `stockier' (less slender) columns.

4. We'll also deal with other `end conditions' later.

5. We'll also look at eccentric axial loads, later.

6. We may or may not look at the stability of parts ...

7. And we will probably NOT look in depth into axial + bending, with regard to (in-) stability.

8. And we will also talk about `slenderness' (slenderness ratio).

Continued ... here.

References

Column Stability Part 2, Jeff Filler, Associated Content.