Required Knowledge in a Component Designer’s Tool Bag – Part 1

After several years of designing floor, wall and roof components followed by extensive job site consultations, I realized there a few items that should be committed to memory to provide optimal value to contractors and homeowners. The purpose of this article is to share those with you and to help continue to grow your skill set.

In this day and age it’s easy to grab our phones and Google an answer to any question. It’s even easier to rely on the design software to solve your problem without understanding why it failed to begin with. However, I’ve found that clients become confident in your knowledge and skill set if you can easily answer what they believe to be seemingly difficult questions.

Here are three areas of consideration that are vital to becoming an expert in the eyes of your current and potential wood component customers:

Floors

A question you will often here from architects, contractors and homeowners is “What’s the farthest I can span a floor joist?.” Here’s your first opportunity to show off your “mad” skills! Without reaching for that EWP catalog or truss manual you say it boils mainly down to two things and two easy formulas.

L/D<20

  • This is the span to depth ratio allowed for a floor joist. It is the “starting” point when deciding which depth of joist you want to use. Here’s an example:
    • Your client wants to free span over a basement 25′.
      • L= Length (Span) in inches (25×12=300)
      • D= Depth of the joist in inches. In this case, we are solving for this variable
      • 20= the maximum allowed result in inches
      • Therefore, 300/20=D
      • D=15”

Remember, this is a starting point. Loads applied and conditions will dictate the spacing of the joists along with requirements for chord width for EWP or top chord plies for floor trusses. Other considerations like “bounciness”, vibration and deflection will also come into play at the time of final analysis. Which brings us to the second item for floors, deflection.

L/360 Deflection

  • This is the maximum allowed deflection allowed for a floor joist
    • For the same situation you can also tell your customer that under those conditions we could expect the following results:
      • L= Span in inches (25×12=300)
      • 360 is the maximum allowed result of the ratio. This can be adjusted higher for even less deflection but cannot be less than 360.
      • 300/360= .83 or 13/16”

Solutions to these two formulas will give you a basis to continue the conversation to either deeper trusses or adding a bearing wall to break up the span. Converting the “engineer’s jargon” into quantifiable tangible terms will be valuable to your customer.

Walls

I spent 11 years in the heavy snow country of Idaho where the snow load ranged from 150-185 pounds per square foot (PSF). In this type of environment, the interaction between the roof and wall systems was a critical one. For starters, the width of the wall will determine how much insulation can be installed. A pretty important consideration when the temperatures often spend weeks or months at a time well below freezing.

The other is the structural performance of the walls themselves. Here we are going to discuss the depth only and not the height restrictions of framed walls. Because of lateral and vertical loading factors, that would be a whole other article in and of itself!

How the top plate of a wall performs depends on the species of the plate and the load being applied to it from other structural members such as:

  • Beams
  • Joists
  • Trusses

The main area of focus here is the compression capacities or “crushing capacities” of the top plates of the wall. Here is a chart that lists the different criteria for various species of wood along with EWP:

lumber-crushing-factors

In the Northwest, Douglas-Fir (DF) was the primary material used. Therefore, 625 was the number I committed to memory. Now that I am designing all over the country with GDI, I’ve had to memorize a few more on the list. Here is an example of how to apply this information:

Your customer wants to know if a 2×6 DF wall will satisfy the conditions when carrying a 30′ free-span truss with 2′ over hangs. The snow load is 150psf with an additional 20psf of dead load on the top and bottom chords

  • 34(span+overhangs)x170= 5,780#. The same can be achieved by converting the PSF to a PLF and dividing by 2 for the tributary load (if the spacing is 24” o.c.) This is the reaction at each end of the truss
  • 1.5”x5.5”= 8.25”. This is the square inches of the top plate that will have the load applied to it from a single ply truss.
  • 8.25×625=5,156. This is the maximum allowable load.

The potential for crushing of the top plate is likely which could result in structural failure. Most notably will be the site of drywall cracking under the truss where the wall board meets the ceiling board like such:

bearing-crushing-cracked-drywallHere are some solutions to suggest:

  • Truss Bearing Enhancers
    • This is a mechanical connection that will transfer some of the load to more of the plate. A commonly used application is the Simpson TBE. In this case, with a TBE6 installed on both sides of the truss, the capacity of the top plate is increased to 6,975#

simpson-tbe6

  • Bearing Blocks
    • Blocks specified by the truss manufacturer would be installed on one or both sides of the truss as specified. This will provide more surface area for the load to be transferred to by increasing the width

truss-bearing-blocks

  • Wall Thickness Increase
    • It is becoming fairly common in heavy snow load areas to frame the exterior walls with 2×8 walls or with double 2×4 walls with a thermal break and 2×8 top plate. In the example above, we would satisfy the compression capacity of the wall. 1.5×7.25=10.875. 10.875×625=6,796#

double-2x4-wall

Roofs

The criteria for roofs are similar to those mentioned above with floors. Let’s say a contractor is doing an addition with a carport and wants to free span 32′ with a flat truss or rafter. He naturally wants to know what the smallest depth he can use. Here we go, “mad” skill #3 is about to make you a valuable asset to your customer for his meeting later with his architect!

L/D<24

  • Like in the floor scenario, this will give you a good “starting” point for your rafter options.
    • L= Span in inches (32×12=384)
    • D= the depth of the rafter or truss we are solving for
    • 24= the maximum allowed result in inches
    • 384/24=D
    • D=16”

Once again, additional factors like snow and construction loading will need to be accounted for final evaluation

L/240 Deflection

  • Same as the floor situation, this is the minimum allowed deflection in inches for a roof application
    • With a 16” flat truss or 16” EWP, we could expect the following results:
      • 384/240= 1.6 or 1 5/8”
      • More than likely your customer will not want to see that much deflection and have you increase the depth of your component to get it within a more acceptable performance

There you go! Remember these small but yet vital formulas for floor, wall and roof criteria and you will prove yourself to be a valuable member of your company’s team. This will build confidence in the relationship and allow your customer to be better informed to make a decision based off of your expertise and knowledge.

Stay tuned for Part 2 in this series.

What other tools should a designer be required to have? I look forward to your comments.

Jake Caufield – Design Professional

Gould Design, Inc.