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.
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.

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, wL

^{2}/8 = 1485 plf (26 ft)^{2}/8 = 125,483 lb-ft, as before, or__1,505,790 lb-in__.
The Allowable fiber stress is F

_{b}C_{D}, C_{M}, C_{t}, (C_{V}or C_{L}) and was simply given as F_{b}’ = 2454 psi, in the previous article. The determination of the Allowable stress (F_{b}’ = F_{b}C_{D}, C_{M}, C_{t}, (C_{V}or C_{L}) is complicated and beyond this discussion. We attack*in other articles.*__it__
The reduced (net) section modulus is (Equation 8, Page 5 of the
Note) ...

S

_{net, v}= (b – 1.5h) d^{2}/6.
For our example this is,

S

_{net, v}= [6.75 – 1.5(0.75) in.] (24 in.)^{2}/6 =__540 in.__^{3}
The corresponding Allowable Moment is ... F

_{b}’ S_{net, v}... F_{b}C_{D}, C_{M}, C_{t}, (C_{V}or C_{L}) S_{net, v}=
... 2454 psi (540 in.

^{3}) =__1,325,160 lb-in.__
Our `check’ then is,

[Is] ... M = 1,505,790 lb-in. ≤ F

_{b}C_{D}, C_{M}, C_{t}, (C_{V}or C_{L}) = F_{b}’ S_{net, 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 S

_{net, 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
S

_{net, 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

*of the beam,*__capacity__
M

_{allow}= F_{b}’ 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’ ...

_{b, net section}= M

_{applied}/ S

_{net, v}≤ F

_{b}’ = F

_{b}C

_{D}, C

_{M}, C

_{t}, (C

_{V}or C

_{L})?

Let’s see:

[Is] f

_{b, net section}= 1,505,790 lb-in. / 540 in.

^{3}= 2789 psi ≤ F

_{b}’ = 2454 psi? NO!

In terms of a unity check:

[Is] f

_{b, net section}/ F_{b}’ = 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.

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