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Asymmetric bending in wood roof purlins

Asymmetric bending in wood roof purlins

Asymmetric bending in wood roof purlins

(OP)
What is the best way to approach asymmetric bending forces in wood?  For instance, if you have a pitched-roof purlin with the narrow face against the sheathing, such that dimension "d" is not vertical, how do you calculate Fbmax for snow load?

Would you break the vertical load into components perpendicular to the sawn member dimensions, calculate two Fb's, then add them together for a combined Fbmax at the most extreme fiber, which is at the vertex of the two member faces?

RE: Asymmetric bending in wood roof purlins

Consider the snow load to be the (level) ground snow load, but applied to (level) members that have an assumed length equal to the roof's horizontal projection.

Take a look at these two free .pdf downloads:

"Structural Design Loads for One- and Two-Family Dwellings", Paragraph 7.
http://www.huduser.org/publications/destech/strdes...

"Residential Structural Design Guide, 2000 Edition", Paragraph 3.7
http://www.huduser.org/publications/destech/reside...

You may find both of these documents (and others at this website) useful for many situations.

RE: Asymmetric bending in wood roof purlins

(OP)
I'm afraid I wasn't clear enough.  I mean that the members are "purlins," not rafters.  The members are level, but canted at an angle, so that they look more like a diamond than a rectangle in section.  Hmmmm, try connecting these dots:
                          .
                          .
             LOAD:        .
                        . . .
                          .
                       .
                      .    .
 MEMBER:             .    .
                    .    .
                        .
                    




RE: Asymmetric bending in wood roof purlins

(OP)
Well, my dots didn't work too well.  They looked good in composition; guess I should have previewed.  Oh well.  Maybe you can still get the idea.  The members have been twisted such that a load is very likely to induce lateral-torsional buckling without bracing (the sheathing, in the case of roofs).

RE: Asymmetric bending in wood roof purlins

I think I get it--the axis of the member along its length is horizontal, the member is "tilted" about this axis such that plane of member edge coincides with roof plane, plane of member face is perpendiular. Question: is sheathing supported by purlins also fastened to whatever is supporting the purlins (e.g. purlins flush-framed with top chord of trusses)? If so, member is subject only to strong-axis bending from component of vertical load acting perpendicular to roof plane: resistance to the parallel component provided by weak-axis bending of the purlin is insignificant compared to diaphragm action of sheathing, because the in-plane stiffness of latter is so much greater. Load path for parallel component goes straight from sheathing to the purlins' supports. If, instead, sheathing is not fastened to purlin supports (e.g. purlins on top of trusses without blocking between trusses), then a bigger problem is preventing "roll-over" of the purlins.

RE: Asymmetric bending in wood roof purlins

(OP)
Yep, that's what I was thinking.  The sheathing will definitely have to be attached often enough to turn each purlin into a tee along the weak axis.  Fair enough, strong axis component and bending it is!

RE: Asymmetric bending in wood roof purlins

I agree with doejohn up to a point...

The shathing will only act as a diaphragm IF it is capable of doing so. If it cannot act as a diaphragm them one of two things can be considered.

Case one - sheathing is in compression due to parallel load component. A support is needed at eaves and is usually provided by the wall for vertical component and ceiling ties for horizontal. In this system the purlins are designed for major axis bending (as described by doejohn).

Case two - If there is no support at eaves AND no diaphragm action in the sheathing, the purlins are subject to biaxial bending. The exact approach to this will be specific to your design code. My normal method is to split the load into major and minor axis components. The stress at the outermost fibre is the sum of the two axis stresses. Don't forget to check deflection (using pythagoras theorem) as this is frequently the most onerous condition.

RE: Asymmetric bending in wood roof purlins

What's the roof pitch? What is the purlin span?
If pitch is small then you can probably ignore minor axis bending. Conservatively ignore the diaphragm action and design for biaxial bending. The Australian code has provision for this so I would suppose your local code does too.
If spans are long you may need bridging between the purlins.

RE: Asymmetric bending in wood roof purlins

(OP)
The pitch in a case I'm casually looking at right now is 10:12.  I believe that the force can broken into two components, one with the major axis, and one with the minor.  The force along the minor axis will act along the face of the sheathing, being resisted by the deep beam action of the diaphragm.  The other component will be resisted by the major axis of the purlins.  

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