Column Stability Part 3

Let's return to our Secant Formula ... for maximum stress and for deflection. Cast in terms of P/P_cr they are ...

σ _max = (P/A) [ 1 + (ec/r²) sec [(π/2)(√P/P_cr)]

δ = e { sec [ π/2(√P/P_cr) - 1] }

where

P_cr = π² EI / L²

And now let's crunch some number examples ...

Example ...

Consider a 3-1/8 x 4.5 glulam column with E = 1,700,000 psi and an axial compression load P of 5000 lb, and it's eccentric by 1.0 in. in the weak direction of the column.

For a length (L) of 10 ft let's determine the following:

1. Check the Slenderness Ratio of the column (with regard to both directions);
2. Check P with regard to Pcr in the strong direction;
3. Check P with regard to Pcr in the weak direction;
4. Axial Compression;
5. σ _max for the x direction buckling.(Secant Formula);
6. σ _max for the y direction bucking (Secant Formula);
7. δ for x direction buckling (Secant Formula); and
8. δ for y direction buckling (Secant Formula).

Let's use a maximum Slenderness Ratio of 50 for either (ANY) direction;

let's use Factors of Safety of 3 for `Buckling / Instability' checks (Items 2 and 3);

let's use FOS of 2 for axial compression / crushing (Item 4);

let's use FOS of 2.5 and the compressive strength for the `combined' buckling/crushing Secant Formula checks (Items 5 and 6);

and we generally use no FOS with deflections (Items 7 and 8), in this case we'll just calc the numbers and disregard some `limit'.

For axial compression let's use a compressive strength of 3000 psi.

NOTE: current design standards for wood use a different approach than this. If you are doing a design of an actual wood column - use the appropriate design standards. However, your answers should not differ hugely that from what we will get.

Let's consider pinned-pinned ... k = 1.0.

Let `x' denote strong direction stuff, and `y' weak direction ...

I_x = 3.125 (4.5)³ / 12 = 23.74 in.⁴ ...

I_y = 4.5 (3.125)³ / 12 = 11.44 in.⁴

A = (3.125)(4.5) = 14.06 in.²

For L = 10 ft ...

Item 1... First check slenderness ...

x-direction

kL/d = 10 x 12 in. / 4.5 in. = 26.67 ≤ 50 ... good.

y-direction

kL/d = 10 x 12 / 3.125 = 38.4 ... ≤ 50 ... good.

Item 2 ... P_cr in the strong direction ...

P_{cr, x} = π² EI / L² = π² (1,700,000 psi)(23.74 in.⁴) / (120 in.)² = 27,660 lb.

Divide by FOS ...

27,660 / 3 = 9220 lb.

P = 5000 lb ≤ 9220 lb ... good.

Item 3... P_cr in the weak direction ...

P_{cr, y} = π² EI / L² = π² (1,700,000 psi)(11.44 in.⁴) / (120 in.)² = 13,330 lb.

Divide by FOS ...

13,330 / 3 = 4443 lb.

P = 5000 lb NOT ≤ 4443 lb ... NOT good. (At least not good at this level of FOS.)

Item 4 ... Axial Compression (centric)

... f_c = P/A = 5000 lb/14.06 in.² = 356 psi.

Compression strength / FOS is 3000 psi / 2.0 = 1500 psi.

356 psi ≤ 1500 psi ... good.

Item 5 ... σ _max for the x direction buckling (Secant Formula) ...

σ _max = (P/A) [ 1 + (ec/r²) sec [(π/2)(√P/P_cr)] ... but in the x-direction the eccentricity, e, is zero ... so this becomes ... P/A ...

σ _allow = 3000 psi / 2.5 = 1200 psi

356 psi ≤ 1200 psi ... good.

Item 6... σ _max for the y direction buckling (Secant Formula) ...

e = 1.0 in. (given)

c = 3.125/2 = 1.5625 in. ... (`c' is the distance to the `extreme fiber')

r = √(I_y/A) = √(11.44/14.06) = 0.90 in.

ec/r² = 1.92 (dimensionless)

P/P_cr = 5000/ 13,330 = 0.375

√(P/P_cr) = 0.612

sec [(π/2)(0.612)] = sec (.962 rad) = sec (55 deg) = 1.75

So,

σ _max = (P/A) [ 1 + (ec/r²) sec [(π/2)(√P/P_cr)] =

σ _max = 356 psi [ 1 + (1.92)(1.75)] = 356 psi [ 1 + 3.361] = 1550 psi.

Is σ _max = 1550 psi ≤ 1200 psi ... NO, not good.

Item 7 ... δ for x direction buckling (Secant Formula)

δ = e { sec [ π/2(√P/P_cr) - 1] } = 0 since e = 0 ...

δ_x = 0.

Item 8 ... δ for y direction buckling (Secant Formula)

δ = e { sec [ π/2(√P/P_cr) ] - 1 } =

δ = 1.0 in. { 1.75 - 1 } = 0.75 in.

δ_y = 0.75 in.

Summary:

1. Slenderness ratios are 27 and 38 ... both less than 50 ... good.
2. Strong axis Euler Buckling P = 5000 lb ≤ P_cr / 3 = 9220 lb ... good.
3. Weak axis (centric) Euler Buckling P = 5000 lb NOT ≤ P_cr / 3 = 4443 lb ... NOT good.
4. Axial compression ... P/A = 356 psi ≤ 3000 / 2 = 1500 psi ... good!
5. Strong Axis Secant Formula ... 356 psi ≤ 3000 / 2.5 = 1200 psi, ... good!
6. Weak Axis Secant Formula ... 1550 psi NOT ≤ 1200 psi, ... NOT good!
7. δ for x direction ... 0.0 in.
8. δ for y direction ... 0.75 in. SCARY!

Discussion:

If we take away our factors of safety, the only "bad" thing is weak axis crushing as the moment generated by the eccentrici load bends the column too much ... axial compressive stress and flexural compressive stress add and exceed the compressive strength. Not good.

Also, a 0.75 in. deflection would probably `look' unsafe (as indeed it is in this case).

We either need to brace this column in the weak direction, or get a bigger section column.

Note that I used a bigger FOS for the Euler buckling. I did this to illustrate a couple things: 1) buckling is a failure we really want to avoid, and 2) in wood there is more uncertainty in E ... and since I didn't build it in anywhere else, I built it into the FOS here for buckling.

The FOS for the eccentric loading is in between the FOS for crushing and buckling ... I did that because the eccentric buckling is kind of a combined (`in-between') phenomenon.

Note that I didn't include FOS values for the deflection. Deflections are not safety issues but serviceability issues. We do in general have deflection limits, but I did not include them here. Normally we would look at the deflection and deflection limit on a column as being due to transverse loads, not (wildly) eccentric axial loads.

Note that we looked at kL/d for both directions with regard to slenderness. In this example, since kL was the same in both directions, we could have just looked at the smallest `d'. In other problems, however, the kL for the two directions may be different, so we look at both.

I did not use a FOS along with the slenderness ratio limits. Slenderness ratio limits already address safety.

AGAIN, THIS STUFF IS FOR A COLLEGE LEVEL MECH. OF MAT. CLASS, NOT ACTUAL DESIGN OF COLUMNS IN BUILDINGS. FOR ACTUAL COLULMN DESIGN, FOLLOW THE CODE ( ... THOUGH THE ANSWERS THE CODE GIVES YOU SHOULD NOT BE HUGELY DIFFERENT.)