wheel loads on heavy timber decking

[BecomingCurmudgeon]

I am trying to analyze a existing 3x8 heavy timber deck floor being used as a passenger garage. I found resources for checking the bending in the deck but nothing for shear.

The decking is rough 3x8 Doug Fir planks laid next to each other with a small 1/4" gap. Not tongue and groove. The deck supports are spaced at 2 ft on center. There does not appear to be a mechanism for load transfer between planks.

The Owner has been parking cars on it for decades. I would like to do a shear check to answer the Owner's question about a Tesla's wheel load. When the wheel is just off of the support. there should be punching shear and horizontal shear in the deck. The same loads can occur if the Owner is using a jack to change the tire.

I can demonstrate that bending is OK by calculation using normal procedures. I am puzzled by the shear/punching checks. I can't find any shear values for decking in the AITC manual or a sample calculation for decking.

Anyone done something similar?

 

[OldDawgNewTricks]

I can't find any shear values for decking in the AITC manual

The AITC manual says "Heavy timber decking essentially forms a beam and must be analyzed as such. The basic criteria for bending, deflection, and shear must be satisfied. However, due to the typical flat-wise orientation of the decking pieces, shear rarely controls the design. The equations in Table 10.3-1 may be used to compute the allowable uniform load for the standard patterns. Additional checks might be necessary for point loads and short spans with heavy loads where shear becomes significant."

Refer to NDS for the required additional shear checks.

 

[Ron247]

I do not have an AITC manual, but do they have a formula for shear parallel to grain similar to (3*fv)/(2*b*d) from NDS?

It sounds like you are looking for shear perpendicular to the grain. I do not know of published values on that.

 

[Tomfh]

Just check the shear capacity. A 3x8 Douglas fir should be fine for a car wheel

 

[BecomingCurmudgeon]

I am having trouble finding an allowable shear value. I am using Doug Fir Larch Commercial Dex values for the bending check. Bending, bearing, and deflection are fine. Table 4E in the NDS provides nothing for shear. A Doug-Fir-Larch value from Table 4D gives you an Fv = 170 psi. The shear stress with an IBC wheel load just off of the supporting beam would be 3 x 3,000 lbs / 2 / 7.25" / 2.5" = 248 psi. The IBC wheel load is applied to a 4.5" x 4.5" area or just one plank.

Something isn't right with the analysis because the car wheels do not punch thru.

What am I missing?

 

[XR250]

I am having trouble finding an allowable shear value. I am using Doug Fir Larch Commercial Dex values for the bending check. Bending, bearing, and deflection are fine. Table 4E in the NDS provides nothing for shear. A Doug-Fir-Larch value from Table 4D gives you an Fv = 170 psi. The shear stress with an IBC wheel load just off of the supporting beam would be 3 x 3,000 lbs / 2 / 7.25" / 2.5" = 248 psi. The IBC wheel load is applied to a 4.5" x 4.5" area or just one plank.

Something isn't right with the analysis because the car wheels do not punch thru.

What am I missing?

Because the wheel load is not actually 3,000 lbs and the design values are conservative. Also, draw a free body diagram of the clear span of the plank as the length and your point load 1/2 a tire's width from the end. The shear won't be 3,000 lbs there. It will be slightly less. You also can ignore any load within 2 1/2" of the support.

 

[BecomingCurmudgeon]

I was able to draw the FBD and show that when the point load is applied at 2.5", the shear drops to 2,688 lbs resulting in fv = 222 psi. Then if I use a Table 4A Doug-Fir shear value of 180 psi. And add in a 1.15 repetitive use factor, a 1.15 flat use factor, a 0.97 wet use factor to get Fv' = 231 psi and a DCR of 0.96.

Not sure if it is kosher to use both Tables 4A and 4E at the same time. One for shear and the other for bending. Seems like picking design strengths just to make the check work.

 

[Tomfh]

Use the actual wheel loads.

 

[OldDawgNewTricks]

Flat Use Factor and Repetitive Member Factor are not applicable to shear.

Something isn't right with the analysis because the car wheels do not punch thru.
What am I missing?

The Reference Design Value of 180 psi is not the failure stress. It has a factor of safety built in. So it looks like you would be using up some of your factor of safety if the design load of 3,000 lb was actually applied. Note that this is not likely the actual wheel load, but also accounts for the point load from a jack.