NDS wood diaphragm Table 4.2--what if inclined roof?
NDS wood diaphragm Table 4.2--what if inclined roof?
(OP)
In the NDS 2005 wood diaphragm Tables 4.2A, B, and C, values are given for different nailing patterns, sheathing thickness, grade, etc...but what if my roof is a 30 degree gable? Does anyone know where I can read about how inclines affect unit shear resistance values, etc., for a sloped roof diaphragm?
Thanks.
Thanks.






RE: NDS wood diaphragm Table 4.2--what if inclined roof?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
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RE: NDS wood diaphragm Table 4.2--what if inclined roof?
It just seems that a wood paneled diaphragm wouldn't work that well in diaphragm action if it's too steep.
thanks
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
I would agree with CJSchwartz about resolving it into components when calcing the shearwall loads. I would add this, however: Add in the downward vertical force (from the diaphragm) for designing the compression chord, Add in the upward vertical force (from the diaphragm) for designing the tension chord and the hold downs.
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
....but the diaphragm also should be checked using its sloped length (which is also longer) to get the shear per foot of diaphragm.
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
It seems to me that if the horizontal shear is the resultant shear, resolving into normal and parallel to the plane shears would result in respectively smaller values of shear. What am I missing here?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
If you look at the forces as a right triangle (angle between a and b is 90 degress), with the wind blowing left-to-right, the wind is the hypotenuse (c), while the components are the in-plane (a) and normal-to-plane forces (b).
/`
a / ` b
/ _____`
c
No?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
HORIZ component of the diaphragm shear = HORIZ lateral wind force.
The DIAGONAL shear along the plane of the diaphragm is the RESULTANT force (R) made up of a horizontal component (A) and a vertical component (B) such that R2 = A2 + B2.
A = horizontal component of the resisting diaphragm
B = vertical component of the resisting diaphragm
R = Resulting in-plane diaphragm Shear
H = lateral wind load
Where H = A
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
Do you then design a chord member at the ridge or is the chord force negated by combining chord forces from each side of the sloped roof? Do the length-to-width calcs use the full roof width or just the (projected) width from eave to ridge?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
This has been discussed in the past here:
thread726-131641: Roof Diaphragm
thread507-178730: Ridge Vents in a Steel Diaphragm Roof
And a whole bunch of links to threads here that might help: thread1066-95784: * Welcome *
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
We have 6 forces/force components here, say H = 20 kips
1. The applied wind, which is horizontal, H = 20 kips
2. The parallel-to-diaphragm component of H...equal to H*sin 60=17.32
3. The perpendicular-to-diaphragm component of H...equal to H*cos 60=10
4. The parallel-to-diaphragm resisting force R
5. The horiz. resisting component of R = R cos 30
6. The vert. resisting component of R = R sin 30
CJ says I need to make sure 4 is greater than 2, and I need to make sure that the sheathing in bending and shear works for 3
JAE says I need to make sure 5 is greater than 1
I think it depends how we assume we resist "H". JAE is right if I depend on membrane action (in-plane) only from the sheathing, and depend on no bending resistance from the sheathing. CJ is right if we assume the sheathing acts both in bending and membrane action.
Sort of like JAE is making a truss analogy and CJ is making a frame analogy.
Technically either way is right.
No?
No?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
I am curious now as to which way is the correct way. This, I think is similar to resolving a vertical force acting on a sloped rafter into parallel and normal to the slope components. The vertical force is the resultant, just as the wind load (horizontal) is the resultant in this case.
If someone has any more difinitive reasoning on this, please advise. Thanks.
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
Doesn't your method require a vertical wind force? If the resulting in-plane diaphragm shear is the resultant of the vectors then there is a net vertical reaction from the wind. That is not the case. The Horizontal wind force is the resultant, with no vertical reaction.
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
Draw a free body diagram of the node at the wall - sloped diaphragm interface. You have a horizontal wind force (say 10k) coming from left to right into the node. That must be balanced by a net force from the diaphragm coming from right to left into the node.
If we say the sloped diaphragm force is the resultant (and not a component), then there is clearly a net downward force from the diaphragm that is not balanced by the horizontal-only wind load.
Statics says this can't work.
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
I think we are all kind of talking about the same thing in different ways.
I think as long as you either do either of the following you're okay:
1. Assume your wind is the resultant, and properly design for the resultant only (don't need to resist the wind components)
or
2. Resolve your wind into components, and properly design for ALL wind compenents (don't need to resist the resultant)
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
Using your example, I have a 10 kip horizontal force coming in at the node.
The node has only two elements attached to it: the sloping diaphragm and the vertical wall.
Free body diagram
The 10 kip force can only be resisted by the horizontal COMPONENT of the diaphragm. There is nothing else to resist it.
Thus, there is a sloping RESULTANT which is comprised of a horizontal COMPONENT (equal to the 10 kips and is LESS than the sloping Resultant) and a vertical COMPONENT.
So (Horiz 10kips)2 + Vertical2 = Sloping Resultant2
The vertical component is a downward force resisted by axial force in the vertical wall. Both the vertical wall and the sloping diaphragm have axial compression.
ΣF = 0 in both directions.
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
OK. I see what you are saying now. I think both ways are correct as hippo11 notes.
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
You can legitimately resolve any resultant into components on of any set of axes you want, called "P" and "Q", oriented at 22.32345 degrees and 112.32345 degrees (as long as they're perpendicular to each other).
I think as long as you fully design for either
1. your resultant OR
2. ALL your components
that you are being geometrically correct.
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
How does the shear from the windload on the wall get into the roof diaphragm if the wall can only take axial load?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
Here's a layout of the FBD:
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
1. Calculate the horizontal unit shear (H divided by the projected horizontal length) and compare it to the tabulated allowable.
2. Calculated the unit shear along the lenth of the diaphragm (R divided by actual length of panel) and compare it to the tabulated allowable.
Correct?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
(my sketch above shows it in the form of a concentrated load in kips, but in reality, the lateral wind comes into the diaphragm along the wall in lbs./ft.)
This is like a horizontal uniform load on a "horizontal" beam...the beam being the diaphragm spanning from end-wall to end-wall.
Then you calculate the "R" value in lbs/ft (= H/cosΘ). This is your uniform load on the diaphragm.
Calculate your diaphragm end shear = 0.5 x (R x Length of wall)/(depth of diaphragm along the slope). This is your diaphragm shear to be compared with the IBC tables or equivalent.
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
key on the "Process TGML" button below the message window and it will show you how to insert an image (Under the "Image" section).
You have to have a website where you can upload pictures and link to them. I use VillagePhotos.com but there are a number of sites out there. Then you just insert the URL of the photo on the site into a message box on Eng-Tips.
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
Does this normally happen - adding in that compression?
RE: NDS wood diaphragm Table 4.2--what if inclined roof?
Wouldn't the FBD shown assume the diaphragm is a support as drawn? I thought the idea of a diaphragm is the support is the shearwall which typically is on a wall perpendicular to the wall drawn. I would think the diaphragm would need to be thought of more as a 3d plate element with horizontal supports along the shearwalls.
Of course this discussion is neglecting any actual uplift or downward forces from the wind directly applied to the diaphragm