Perforated Shear Wall Bracing for a Timber Frame

The knee bracing provided by the timber frame (on this particular bent) is insufficient to resist the design lateral load (on this bent), which, in this case, is that due to a big gust of wind. As with many modern timber frames (and unlike post and beam and `stick' framing) the walls are not in the same vertical planes as the frames. In this particular case the wall is `outside' the frame; the wall is an exterior wall; and, also, there is a lot of `opening' (doors and windows).

Solution

I will use a Perforated Shear Wall, the design of which is provided in Section 2305.3.7 of the International Building Code. This solution will utilize common framing materials (likely already considered for the wall construction) with a minimum of `fancy' detailing.

Calculations

Here we go ...

1. Determine Lateral Force

The design lateral force for this frame on this structure has already been determine to be 3600 lb (wind!), using ASCE-7, etc.. It is taken to be delivered via the roof diaphragm. The resistance of the timber frame itself (via knee braces), though real, will not be included (at least not at first).

2. Load Path: roof sheathing to frame

The lateral load from the roof plane shall be transmitted from the roof sheathing or panel via contact with the `rafters' of the frame.

Total approximate contact length is 20 ft.

So, 3600 lb / 20 ft = 180 plf.

The Z for 8d nail, 7/16 in. structural panel, ... is 56 lb (S.G. Wood 0.50) ... (from Table 11Q of the National Design Specification for Wood Construction.

Using CD = 1.6 ... Z' = 56 x 1.6 = 90 lb.

I need, then, 3600 lb / 90 lb per nail or 40 nails. But, actually, I need more than that, since they act along a sloped surface, and I only get a `component' of the 90 to add up to my 3600. Indeed, I will need 40 x 20/14 = 57 nails (same as 40 divided by the cosine of the angle that gives me a 20 ft slope on 14 ft horizontal distance). And, now, in terms of framing, this is 57 nails / 20 ft = about 3 nails per foot, or 4 in. o.c.

... 8d (box, dia. 0.113 in.), 4 in. o.c. ... or equivalent, roof sheathing or panel to frame (min. thickness 7/16 in.).

If the Owner wants to go with a structural insulated panel system (SIP) then a spiking or screwing system must be provided to deliver the equiv of the above (3600 lb horizontal).

3. Frame infill sheathing/framing/nailing

Use 2 x framing (infill studs and plates/sills) ... 8d @ 6 in. o.c. all panel edges, all edges backed and blocked, one side, will be sufficient.

All plates/sills shall be fastened per IBC Table 2304.9.1 ... namely, 3 - 16d @ 16 in. o.c. (or equiv.) ...

1.

Load Path: frame to top plate of perf shear wall

Here is where it gets fun! Now we take the lateral force from the plane of the frame to the plane of the wall. Here's how ...

Oversize top lam of double top plate so comes over and is directly adjacent to bottom of frame.

Transfer lateral load to double plate using A35 or equivalent.

A35 rated at ... 450 lb (Simpson Wood Construction Connectors ...)

So, 3600 lb / 450 lb = ... 8 connectors, total, evenly spaced, over 20 ft wall length.

4. Perf Shear Wall: sheathing/framing/nailing

Let's start with 7/16 in. Rated Sheathing, one side ...

Total wall length ... 14 ft.

Length of full height sheathing ... 10-1/2 + 27 + 10-1/2 = 48 in. ... (4 ft).

Percentage of full height sheathing ... 4 ft / 14 ft = 0.20 = 29 % ...

Maximum opening height ... Ughhhh ...

Counting only the doors ... 6-8 vs 10-2 is 66%

Adding windows ... 30 in. ... 110 / 122 is ... 90%

Let's use the 5H/6 column in the Table (Table 2305.3.7.2).

Gives ... Co = 0.49.

Now let's go to Table 2306.4.1 ...

Allowable shear ... 7/16 in. Rated Sheathing, 8d @ 4 in. o.c. ... 350 plf. But let's use studs @ 16 in. o.c., so the footnote d allows us to go to 380 plf. Also, since it's wind, 2306.4.1 allows us to increase by 40% ...

So, 380 plf x 1.4 = 532 plf.

Now we need to find the shear resistance of the wall ...

The total length of full-height wall is ... 4 ft.

The total resistance is 4 ft x 532 plf x 0.49 = 1043 lb. Ughhhh ... not enough ( ... need 3600 lb).

We need about 4 x as much resistance.

Try 19/32 Rated, 10d @ 3 in. o.c. ...

Allowable shear is 665 x 1.4 = 931 plf.

931 plf per side x 2 sides x 4 ft x 0.49 adjustment = 3650 lb.

Sweet!

Note: footnote h to Table ...

Where panels are applied to both faces ... and nail spacing is less than 6 in. o.c on either side, paneljoints shall be offset to fall on different framing members. Or framing shall be 3 inch nominal or thicker and nails on each side shall be staggered.

So, this does not require that all the framing be 3 in. nominal ... just studs that panel pieces join on.

Nails may be spaced up to 12 in. o.c. on intermediate framing members.

5. Anchorage: anchor bolts

Use a continuous rim to hold (tie) things together ...

Then we have a shear at the bottom of the wall of 3600 lb / 20 ft = 160 plf. We should be able to accommodate this with the fastening schedule of Table 2304.9.1.

Anchor bolts ... ½ in. diameter bolt in shear parallel to grain, 1-1/2 in. plate ... Table 11E of NDS ... 650 lb. So, 650 x, say, 1.33 = 865; 3600 lb / 865 lb per a.b. = 4.2 bolts ... round to 5 ... OR ... 4 - 5/8 in. dia. A.B., embedded 6 in. min. ...

6. Tie Down: holdowns or tie to frame

Neglecting the benefit of Dead load ...

The uplift at the end of the wall will be ... T x width = V x height ... or T = V x h / width ... 3600 lb x say 10 ft / 20 ft = 1800 lb.

Wall ends must be tied to foundation with `holdown' hardware rated at 1800 lb or greater uplift. Holdowns may be installed in end framing members of wall ... OR ... the end of the wall may be tied to the adjacent timber post provided the post is tied down (to resist 1800 lb uplift) in a manner similar to or equivalent to the transfer of load from frame to top of wall (4 - A35 each timber post).

YEAH!!!

7 Additional Comments

Note: we could argue that there is some rotational stuff going on at the transfer of shear from the frame to the (out-of-plane) shear wall. We could! But we could also argue that there is enough overall rotational resistance in the rest of the structure to `take care of it'.

1) use full-height studs (engineered or dimension lumber)

2) where necessary, pre- drill fastener holes to avoid splitting of wood members

Well, we didn't quite stay within the realm of using what the Contractor would have used otherwise ... (3 in. nominal framing vs. 2 in.; 5/8 in. sheathing vs ½ in.) ... but it's still very do-able. And, though the design is `robust' ... the robust-ness is on a relatively small and affordable scale.

Wood Stove pipe may exit through hole in in-fill of timber frame or roof, but NOT through the shear wall.

References

International Building Code, International Code Council, 4051 West Flossmoor Road, Country Club Hills, IL 60478.

Minimum Design Loads for Buildings and Other Structures, ASCE Standard ASCE/SEI 7, American Society of Civil Engineers, www.asce.org.

National Design Specification for Wood Construction, American Forest & Paper Association / American Wood Council, 1111 Nineteenth St., NW, Suite 800, Washington, D.C., 20036, www.awc.org.

Wood Construction Connectors, C-2008, Simpson Strong-Tie, 5956 W. Las Positas Blvd., Pleasanton, CA 94588, www.strongtie.com.