Composite Wood Beam 1

[TimmyH76]

Hello all,

I have a timber frame job here that the designer wants to use two smaller beams to carry a pretty large load. How do I calculate the shear stress for the connection between the two beams so that they act as one beam. The maximum moment is ~20 K*Ft. Take a look at the attached drawing. Thanks,

 

[Lion06]

If they are side-by-side like that, you really don't need much, just enough to make sure they deflect together. They won't be acting truly composite. As long as they deflect together, and the stresses work out, then you're fine. I would stagger lag bolts at about 2'-0" o.c. and call it a day.
This is assuming that they are oriented the way they are shown and not so that one is on thop of the other. In that case, you'll need to do a shear flow calc.

 

[TimmyH76]

Sorry..I meant for them to be shown one on top of the other...Sorry that was my mistake with ADOBE

 

[JKStruct]

VQ/I

 

[youngstructural]

As brief as the message may be, JKStruct is right... You need to check (and fasten composite against) the longitudinal shear in the beam...

SO, you need the V to be the maximum shear felt by your composite section, the Q is the first moment of area (SUM A Ybar) about either section and I is the second moment of area (aka moment of inertia) of your resultant composite section. Note that I like to take the Q twice, once about each component... They must match, otherwise you have miss calculated your Neutral Axis (NA).

Example (from an actual design I did):

_________
| |
| |
| | <-- Existing Conc. Beam
| | (508mm deep x 406mm)
y | |
| |_________|
_|__x | | <-- Toes down Parallel Flange Channel (PFC)

Note: Using a 200 PFC, so Area = 2920mm^2
Flange = 75mm
Height = 200mm
__
|
|· (Dot represents the Centroid)
|__

Centroid is 24.2mm from the back (Left hand side) face of the web. Given that the section will be toes down and it has 75mm flanges, this means the centroid of the 200PFC will be at 75mm-24.4mm = 50.6mm from our reference.

Step 1) Solve for your neutral axis of the composite section, in my case it's at ybar = 325.1mm from the tips of the PFC.

Q (for 200PFC) = Area of 200PFC (ybar - 50.6)
= 801.6 x 10^3mm^3
Q (ex conc beam) = Area of existing conc beam (ybar - 329)
= 802.3 x 10^3 mm^3

Step 2) Solve for your composite section's second moment of area (moment of inertia) using the parallel axis theorem:

Ix' = SUM(Ix + Ad^2) of all component parts

Step 3) Plug your numbers into the VQ/I formula along with your maximum shear in the beam.

Also I prefer to use:

fv = (VQd)/I where d is the distance between fastenings, which yields a directly useful result to check against the shear strength of the fastener you want to use.

Note that this is not often done in timber the way you are going because to get enough shear strength you normally need bolting, which reduces your section (ie: you're drilling out material and need to substract it away for your effective section). See what your numbers tell you, but you might find it's not all that great. I would next try adding a steel plate top and bottom and using the bolts to make the whole thing composite. You would also only need this through some portion of your beam (often the middle 1/3 to 1/2).

Good luck,

YS

B.Eng (Carleton)
Working in New Zealand, thinking of my snow covered home...

 

[Lion06]

Yes, do the shear flow calc, VQ/I where:
V = the max vertical shear force
I = moment of intertia for the composite section
Q = (A')*(ybar'), where
A' = the area of the section above (or below) the section where the shear flow is being calculated
ybar' = the distance from the neutral axis of the composite section to the centroid of the area above (or below) the section where the shear flow is being calculated.

 

[TimmyH76]

Thanks everyone..this place is great!!