More thinking on trusses

[Don P]

Not saying its necessarily good thinking 

http://www.forestryforum.com/members/donp/44axbend.htm

I've worked up a calculator to show the interrelationship between allowable axial (lengthwise) load when combined with a bending load in a truss top chord. This is not for figuring your trusses, its old... 1944 math. Buddy Showalter put up a downloadable copy of the 44 NDS on their website. It's been helpful for me to see the math evolve. I think I've got a handle on it, the newer versions of these formulas are much more involved and I'm not comfortable there yet. I'm not sure I'm comfortable here, but I guess if I was comfortable I wouldn't be learning.

I've lightly reworked a couple of drawings to use as an example here, don't worry about scale or particulars. The concept of how the top chord is loaded is what I'm showing. Just the most basic truss, a triangle being loaded first at a node, or panel point, and second, uniformly along the top chord.

First shot is of a pair of trusses supporting a ridge, that ridge in turn supports a set of common rafters. The ridge is supporting half the roof weight, the plates are supporting 1/4 of the roof weight each. So each truss is carrying 1/4 of the total roof load at the peak. The top of each truss is in the same plane as the common rafters and the sheathing for the roof is attached to it. So it has a small, uniform bending load of the sheathing that is the area from halfway to each common rafter on either side of it.


The second picture shows a pair of trusses supporting purlins that pretty uniformly distribute all of the roof load to the trusses as a bending load. We'll work it so the total load on the trusses is the same, to generate the same axial loads. The second truss is carrying a large relatively uniform bending load along the top chords. Using the same axial load but different side bending loads should show how much larger a timber would be required if significant bending loads are applied to a truss.


Call the top chord length 144" and assume that the axial load in each truss top chord is 8000 lbs.

The first truss supports a small span of sheathing, assume it can deliver a bending load to the top edge of the truss under full snow load totalling 750 lbs. I can make that work with a white pine 6x8

Truss set 2 has the same 8000 lb axial load but now the purlins are delivering the entire roof weight to the truss along its length. I'm figuring each top chord has 7500 lbs loading it in bending to produce that. Find the smallest timber.

[Don P]

Never end a post with a quiz I was coming up with a 14x10 if the truss is loaded all along the top chord and a 6x8 if it was just loaded on the panel points.

I was reading in the Timber Construction Manual last night.

"The design of truss chord members will be governed primarily by axial tension or compression if chords are not continous and if all secondary framing transmits loads into panel points only. However, if distributed loads are applied to chord members or if chord members are continous over  panel points or panel points are fixed, then considerable bending stresses along with axial stresses could result and a combined stress unity check should be made. If chords are curved, moments resulting from eccentricity and axial forces must be included in the design."

" If chords are continuous or any connections are not pinned, the distribution of forces will be affected and a more complex analysis should be considered. In reality most truss top and bottom chords have continuity and may be pinned only at the middle of the truss span and at the ends, although webs are usually pinned to the chords.
  If the span is relatively short, the span-depth ratio is in accordance with table 5.37 (1/6 or deeper for triangular trusses) and the loads are fairly low, the designer may decide that a manual analytical or graphical analysis will be satisfactory, although not exact. In such cases, the chords should be analyzed as multispan beams in order to determine support reactions at panel points. Apply these reactions to the truss as panel point loads and analyze for for axial member forces."

[Thehardway]

Nice "points" Don.

Here is a little bit of a tricky one.  In the following drawing, are the SIPS loading the truss uniformly across the top chord? Or are the purlins (location indicated by the square pockets in top chord. loading the truss at panel points?  The span between trusses is 8' and the distance between purlins is 4'.  The SIPS are structurally rated to withstand 8' span and with roof load.  Purlins are rated to withstand SIP and roof load but actually do nothing other than serve as a fastening point for SIPS. They are dovetailed and wedged in.

Is this a uniform load or a panel point load?

Seems like a situation similar to wedging.  Mating surfaces must match up or a load point is created. If a purlin is slightly high it assumes all load at that point and transmits it to the point on truss.  If purlin is slightly low, panel assumes all load and distributes uniformly. If mating surface is perfect, load is shared and it becomes a ridged truss itself and the load is distributed evenly on the truss.

Am I thinking right?








 

[Don P]

I'm playing around my edges of understanding so take it from that perspective.

Load goes to stiffness, if that purlin is a slight bit high it will deflect until it can bear the load, the resulting higher load goes to that section of truss which will bear the load or deflect until it shares the load with the next high spot. Within reason I think it tends to become pretty uniform. These formulas allow for some of that.

I see a uniformly distributed load on the top chords of your truss. There are 4 top chord members, heel to web, web to ridge, ridge to web, web to heel. In other words there are 4 shorter spans being checked for combined loading rather than two longer ones, but in total 2 situations, upper and lower parts of a top chord.  There is also an eccentric tension load caused by the raised tie beam. In the bottom member of the top chord there is a compressive axial load running down the grain of the member. Then there are two bending loads on the side of the chord, the uniform load of the Sip and the tension pull from the raised tie. That one gets really interesting. Quantifying all that is beyond me.

At its simplest it is this. If a member is loaded axially as a column, and in bending as a beam, the combined effects of both loads cannot equal more than one member.

(P/A)/c +(M/S)/fb <1
P= the axial load
A= breadth x depth of member, area
c= allowable compressive load.
Looking at that side of the equation, we've taken the load, P, and divided it by the area of the column, that gives us the load in pounds per square inch. Dividing it by the allowable load, in psi, gives the percentage of the allowable column load is being used.

M= bending moment in inch pounds
S= section modulus of the beam, bd²/6
fb= allowable bending stress
This is a beam bending check, find how hard the side bending load is and check it against the allowed bending load for that sized beam. Dividing gives the percentage of allowed bending.

Putting the two halves of the formula together, if the top chord is using half its available strength taking care of the compressive load then it can only use half of its strength or less resisting a bending load. Adding the percentage of compressive strength used and the percentage of bending strength used has to equal less than the 1 member you propose or you need a bigger timber. (P/A)/fc +(M/S)/fb <1




The formula gets more complicated as the chords get longer and more slender. You'll see Euler formulas mentioned often. This is the best mathmatical model for long slender columns buckling stress, it begins modifying Fc in the simple formula basically coming into play when length/depth ratio gets above 1/11. It is this output on the calc "Max allowable buckling stress (psi)
P/A=(.329*E))/(Span/D1)2". I have you visually double checking Fc and FcE, the Euler buckling stress.

Notice it uses the stiffness of the column (E) in considering its buckling strength. Actually it uses about 1/3 of the stiffness (.329). Thats a safety. That's the 1944 formula, in '05 they switched to using Emin, a new design value for use in columns... it equals about 1/3  E .

To look at it another way, short fat columns crush, Fc is the crushing strength of the wood. Longer or more slender columns buckle, FcE. Check both. Euler may say the column can take 6000 psi and not buckle. Fc might say yes, but it will have crushed. or vice versa.

A post with a brace coming into it is another situation of a member with combined axial and bending loads.

[Don P]

I called your truss 24' wide and the tie raised 2'. at a 2/12 pitch and 30 pounds per square foot total load on 8' spacing I came up with 5760 lbs on the truss.
 Roughly figuring the particulars of your truss I came up with an axial compression load of about 3100 lbs in the lower half of the top chord. I called the length 135"
 
If the tension in the bottom chord is about 2200 lbs and if it is raised 2' it produces a 4400 ft lb bending moment x 12 = 52,800 in lb bending moment.
The uniform load on the lower half of the bottom chord produces about 32,062 in lb bending moment.
combining those, 52800 +32062 =84,862 in lbs total bending moment in the lower half of the top chord. (enter a 5000 lb uniform side load in the calc and look at the bending moment, there's my best guess)

That's just from thinking about it today  . Plugging that in when I got home it came out at about a 7x10 red oak not accounting for joinery. I'd sure want someone to confirm it though.

[Thehardway]

Don,

Did you mean 2/12 or 12/12?  When you say 135" what points is that taken from?

The actual pitch is 12/12

Your calculation of 7X10 is exactly what I have from another.   The plans call for rafters and kingpost 7X10, and horizontal tie 7X8 in White Oak. Struts 4X6 black cherry and purlins 4X6 yellow pine.  The spline is KD 2"x10" X 5'.

Actual unsupported span (inside dimension at rafter feet) is 23' 1"

Was not attempting to get you to do my math for me here, just provoking thought about how looks can be deceiving on how a particular member is functioning in a given structure.  Purlin supporting panel or panel supporting purlin may determine whether loading is panel point or uniform distribution.  Get rid of the SIP and use a standard t&g planked roof and there are no questions about what is doing what.  Engineered products can make things appear to function different than they are such as use of LVL ridgebeams in conjunction with other timberframe elements.

It is nice to know you come up with about the same figures though.

[Don P]

Yup, I meant 12/12.
This should show where I guessed at the 135" span length. I've also drawn in where the purlins would be if it was loaded at panel points. So the left is drawn uniformly loaded and the right side is loaded on the panel points. The panel points, or nodes, are the heel joints, the web joints, the kingpost joint and the ridge. Load over them and the stress in the truss is axial. Loading elsewhere combines a bending stress with the axial stress. I was happy to run a scenario just for fun, as long as you don't build off of it. If there is a purpose for this type of calc amongst us it is for rough design to help speed things through at the engineer's review.

[Thehardway]

Got it Don.  Good info and my thanks.