Composite Wood Section (Shear Flow)

[STR04]

I have (2)6x6 timber posts that are stacked on top of one another (acting in bending for a simple span beam as a 6x12), stitched together with lag screws from the top down (connection not shown for architectural purposes). What is that appropriate shear flow to design the lag screw for if the plane stress where to two members come together is zero because it is at member’s centroid for the composite section?

TIA

 

[csd72]

v = VQ/It

 

[STR04]

Agree, but Q = 0 because y-bar = 0. Shear flow changes direction at the centroid thus zero.

 

[DRC1]

at the center Ybar = d/4 and shear flow is a maximum. Ybar is the distance from the nutetral axis to the centroid of th area defined by the outer edge to the point of application of the shear flow. i.e. for the neutral axis the area would be the entire shape above the axis and ybar is the distance from the NA to the center of that shape.

 

[STR04]

DRC1,

As you define the "point of application of shear the flow" is at the centroid where the splice occurs. I agree that there is shear flow within the member at d/4 (point of application) but not at the centroid just as you look at the section as a simple beam where V=0 at the midspan. Again, I'm not saying your wrong but I'm looking at a the FBD and just cannot convince myself that shear flow exists at the centroid (point of application).

 

[swearingen]

To align your intuition with what DRC1 is saying, think of a softcover book being bent into a U shape. Notice that the pages of the book don't line up any more at the edges. The shear plane between them is sliding. Glue the pages together and it doesn't slide - now the book can't be bent into a U any more. Your book has two pages and the shear plane between them is where the action takes place. It is indeed the place of maximum internal shear which is why it needs the fasteners. If there were no shear there, why would they slide relative to each other? There must be a force that provides that motion...

As DRC1 stated, the ybar is the distance from the shear plane (the centroid of the ENTIRE section in this case) to the outer fibers. Pull your mechanics of materials book off of the shelf - it'll have a pictorial description of how to apply the VQ/it formula.



If you "heard" it on the internet, it's guilty until proven innocent. - DCS

 

[Lion06]

V=0 at midspan of beam, not centroid of cross section. Shear flow is the MAXIMUM at the centroid of the cross section.
Use VQ/(It)
Q= the area above the centroidal axis of the composite section (5.5" * 5.5")) times the distance from the cetroidal axis of the composite section to the centroid of the area above (or below) the centroid of the composite section (5.5" - 5.5"/2).
t= 5.5"
V= max shear force
I = moment of inertia of the composite section.

Just think about what you are saying. You are saying that if you lay (2) 6x6 on top of one another they will be as strong and stiff whether you join them or not. That is just not true. There is a difference between (2) 6x6 and (1)12x6.
Also, you know that the shear flow (horizontal shear) at any point must equal the vertical shear at the same point, correct? You also know that that the vertical shear stress is maximum at the centroid of the section. By inspection you can say that the horizontal shear (shear flow) is maximum at the centroid of the section.

 

[Lion06]

swearineng-
ybar is the distance from the shear plane under consideration (centroidal axis in this case) to the centroid of the area above (or below) the shear plane (again, centroidal axis in this case). It is not from the shear plane to the outermost fiber.

 

[zaes73]

ST04 --- i uploaded a greating training module to clarify everything related to shear, i hope this is helpfull.

http://rapidshare.com/files/74271319/shear.exe.html

 

[AlmostPE]

I would just look at it if they were side by side and call it good. If the value is not strong enough then I would have done GL, LVL, Wrapped steel or deeper timber.

 

[STR04]

Thanks all for the help. I knew something didn't feel right.

 

[ctcray]

Understanding shear flow slips out of my mind sometimes. I go back to this website for an excellent explanation of mechanics with illustration to refresh my memory. A great resource:

http://web.umr.edu/~mecmovie/index.html

 

[msquared48]

The only time I like shear flow is when I slice the lime for the margaruita.

Mike McCann
McCann Engineering

 

[swivel63]

Q doesn't equal 0. y equals the 5.5/2 (quarter the section depth), and A equals 30.25 (5.5^2) thus making Q = 83.2 in^3. shear flow and stress in a rectangular member is ALWAYS highest at the center.