## Saturday, January 12, 2013

### Alternate Calculation of the Suitability of a Drilled Vertical Hole in a Wood Beam

In previous articles we evaluated the suitability of drilled vertical holes in wood beams by comparing the beams’ loss in capacity due to the drilled hole and any excess capacity in the beam to begin with.  Excess capacity may arise from the beam having excess strength overall, or due to the hole being in regions of the beam where less capacity is demanded (regions of lower stress).  Obviously if a hole is drilled at a location where the beam has no excess capacity, the condition is `not good’.  Our investigations were cast in terms of ... “Does the loss in capacity caused by the hole exceed the excess capacity at the hole location?”  If the answer is `No’, then the hole is `allowed’.  If the answer is `Yes’, then a smaller hole must be used, or the hole must be drilled elsewhere, the proposal of drilling the hole must be abandoned, or, perhaps, a larger beam is required.
In those articles the American Institute of Timber Construction (AITC) Technical Note 19 was used as a basis for our `good’ versus `not good’ or `yes’ versus `no’ calculations, but the calculations we did were cast in terms of `excess capacity’ and `loss in capacity’, not as directly presented in the Technical Note.   In this article we will look at one of the same examples, but will use the AITC equations directly.  And, with a little math, we will show that we get the same end results.

Consider, then, the proposal to drill a ¾ in. diameter vertical hole at the midspan of a 6-3/4 inch (in.) x 24 in. glulam beam.  The beam spans 26 feet (ft) and carries a uniform load of 1485 pounds per linear foot (plf).  AITC Tech Note 19 requires `first of all’ that the hole not be closer than 3 hole diameters to the nearest side face of the beam (measured to center of hole).  Assuming the hole is centered, the distance to the side face, either direction, is 6.75 in. / 2 = 3.375 in.  In terms of `diameters’, this is ...

3.375 in. / (3/4 in. per diameter) = 4.5 diameters.

Since 4.5 is at least as great as the minimum, 3 ... good! (in terms of the hole `fitting’).

Now let’s do the `capacity’ calculation.

In the Tech Note this is cast in terms of ... (Item 2, Page 5) ...

“The maximum bending moment must not exceed the allowable fiber stress in bending multiplied by the reduced (net) section modulus.”

The maximum bending moment, in this context, is the bending moment at midspan, wL2/8 = 1485 plf (26 ft)2/8 = 125,483 lb-ft, as before, or 1,505,790 lb-in.

The Allowable fiber stress is Fb CD, CM, Ct, (CV or CL) and was simply given as Fb’ = 2454 psi, in the previous article.   The determination of the Allowable stress (Fb’ = Fb CD, CM, Ct, (CV or CL) is complicated and beyond this discussion.  We attack it in other articles.

The reduced (net) section modulus is (Equation 8, Page 5 of the Note) ...
Snet, v = (b – 1.5h) d2/6.
For our example this is,

Snet, v = [6.75 – 1.5(0.75) in.] (24 in.)2/6 = 540 in.3
The corresponding Allowable Moment is ... Fb’ Snet, v ... Fb CD, CM, Ct, (CV or CL) Snet, v =

... 2454 psi (540 in.3) = 1,325,160 lb-in.

Our `check’ then is,

[Is] ... M = 1,505,790 lb-in. ≤ Fb CD, CM, Ct, (CV or CL) = Fb’ Snet, v = 1,325,160 lb-in? ... NO!  NOT GOOD!
That’s the same `answer’ we got before.  (NO! Don’t drill a ¾ in. vertical hole at midspan!)

Let’s see how they are the same.

The ratio of Snet, v and the original S is the same as the ratio of the effective (reduced) capacity to the capacity of the section before drilling the hole (648 in.3), or
Snet, v / S = 540 / 648 = 0.83.  (This is that same 17% loss!)

The Allowable Bending Stress multiplied by the original (undrilled) section is the Allowable (undrilled) bending moment capacity of the beam,

Mallow = Fb’ S = 2454 psi (648 psi) = 1,590,192 lb-in.

The applied moment at midspan was, recall, 1,505,790 lb-in.
The ratio of the applied midspan moment to the original (undrilled) beam’s bending capacity is (was),
1,505,790 lb-in. / 1,590,192 lb-in. = 0.95 ... giving the 5% excess.

Since the loss due to the hole is 17%, and the excess only 5% (from which to draw), ... NOT GOOD!

Finally, in terms of a Unity Check,

M applied, midspan / M allowable, reduced = 1,505,790 lb-in. / 1,325,160 lb-in. = 1.14.

Since this number exceeds 1.00 ... NOT GOOD!

We could come up with the same ratio by taking ... 0.95 / 0.83 ... the ratio of demand to available for the un-drilled beam divided by the ratio of the available capacity left by the hole to the original capacity ...

0.95 / 0.88 = 1.14!

One more `finally’; we could have also done the design check in terms of `stress’, dividing the `M’s by `S’s’ ...

[Is] fb, net section = M applied / Snet, v ≤ Fb’ = Fb CD, CM, Ct, (CV or CL)?

Let’s see:

[Is] fb, net section = 1,505,790 lb-in. / 540 in.3 = 2789 psi ≤ Fb’ = 2454 psi?  NO!

In terms of a unity check:

[Is] fb, net section / Fb’ = 2789 / 2454 = 1.14 ≤ 1.00?  No!

And, again, notice the 1.14.

References

Evaluation of Vertical Holes in Glulam Beams, Jeff R. Filler, Yahoo! Contributor Network (Submitted).

Vertical Holes in Wood Beams, Jeff R. Filler, Yahoo! Contributor Network.

Beam Design Formulas with Shear and Moment Diagrams, Design Aid No. 6, American Wood Council, Washington,  DC, 2007.

Guidelines for Evaluation of Holes and Notches in Structural Glued Laminated Timber, Technical Note 19, American Institute of Timber Construction, Centennial, CO, July 2012.