Stability of Beams

The Theory Part

Okay, here we go; the `second' kind of buckling. Consider a beam in `positive' bending. The bottom zone of the beam is in tension, and the top is in compression. Compression = possible instability. Especially since beams can be long and slender. In this case what we `worry' about (design against) is `Lateral Torsional Buckling' (LTB) ... the top of the beam `buckling' to either side, at the same time twisting, since the bottom is being `pulled' on.

Ref: AWC TR 14 ... here.

The moment that will `just' buckle the beam is ...

M _critical = (π/L_e)√(E I_y G J)

where

M _critical = M _cr as sometimes appears in the literature, but to be confused with the moment the cracks the beam (concrete) ...

L_e = unbraced length (of beam) ...

E = MOE with respect to bending to the side (and there is a difference, for example, in glu-lam, with regard to the E for strong axis bending) ...

I_y = MOI with respect to bending to the side

G = Shear Modulus ...

... often taken to be E/16 with wood (but ranges from ½ to 1/32 of E ...)

J = Torsional Constant for the Section ...

... for rect sections J = 4 I_y [1 - 0.63 (b/d) ]

where b is the width ...

and d is the depth ...

And presumably we are loading the beam in the strong direction and it is buckling in the weak (transverse) direction.

Example

Let's consider a beam that is 21 ft long, 5-1/8 x 12, wood, with a rupture stress (flexural tension) of 5000 psi, and E = 1,700,000 psi (weak axis bending).

Let's find the load P (conc load at midspan) with regard to a) LTB and also with regard to b) strong axis flexural tension rupture. The beam is not supported laterally between the ends, and, c) considering a factor of safety of 2.5 find the allowable P.

Strategy: a) solve for M _crit and back-calc P; b) solve for M = Fr S and back calc P. Use the assumptions above for G and J; for c) divide each P by the corresponding FOS and then choose the lesser.

Solution

Part a) ...

for G we get ... 1,700,000 psi /16 = 106,250 psi

for J we get ... 394 in.⁴ ...

Iy = 12 (5.125)³ /12 = 135 in.⁴ ...

J = 394 in.⁴ ...

Therefore ...

M _critical = π/L_e √(E I_y G J) = 1,220,000 lb-in. = 102,000 lb-ft.

For the corresponding P ...

M = PL/4

P = 4M/L = 4 (102,000 lb-ft) / 21 ft = 19,400 lb (LTB)

Part b)

M = Fb S

S = bh²/6 = 5.125 (12)²/6 = 123 in.³ ...

M = 5000 psi (123) in.³ = 615,000 lb-in. = 51,250 lb-ft

This corresponds to a P of ... 9762 lb (Flexural Tension)

Part c)

Dividing each by the corresponding FOS ... which are the same in this example ...

LTB 19,400 lb / 2.5 = 7760 lb

Flexural Tension ... 9752 / 2.5 = 3900 lb.

Answer: ... P _allow = 3900 lb based on flexural tension.

Design

Actual design of structural members is more complicated than what I have provided here. But now we have been exposed to the basics. For actual design we draw upon the `Design Provisions' in the building codes and standards. These `provisions' are wrought from experience, theory, testing, and include what can be learned from structural failures. Generally the design provisions are materials specific. In other words the steel people give us equations, etc. for the design of steel, the wood people for the design of wood members, and so on.

But, let me say this ... an easy way to deal with LTB is to `brace' the compression part of the beam against lateral movement, and this is actually very common. Such can be accomplished by attaching the compression zone to something `rigid' in the transverse (lateral) direction, such as a floor system, roof system, ceiling system, and so on. The system being attached to does not need to be particularly strong ... but it doesn't need to be pretty stiff (relatively rigid).

Update: ... a cool example of LTB using a steel beam follows, ... here.

References

Designing for Lateral Torsional Stability in Wood Members, Technical Report 14, American Forest & Paper Association / American Wood Council, 1111 19th St., NW, Suite 800, Washington, DC 20036.

Basic Lateral Torsional Buckling Example (Steel Beam), Jeff Filler, Associated Content.