Rafter Loading

[Guest090822]

Help! A young coworker is studying for the PE exam and asked me for help with this problem. I did what I thought was correct in a few minutes only to find out my answer was not correct. Just curious what everyone else gets for an answer. See attachment.

 

[WARose]

I ran it in a program I have with pinned-pinned supports and got 420 lbs vert. (like I mentioned above) and 67 lbs horiz. (both reactions the same at each support).

 

[gte447f]

WARose, I agree with the 67 lb horiz, but I think the vert reaction would be 400 lb. You just have to project the perpendicular load onto the vert and horiz projected lengths of the inclined surface, right?

 

[WARose]

IIRC, when I did it by hand, I did it kind of like SRE did except that the 20 #/ft was the hypotenuse. So it was: [(20lbs*17.33')/2]*(12/13) added to the load from the 30 lbs projected.

 

[gte447f]

Yep, that's right, but it's 400, not 420.

 

[WARose]

I don't know why we are getting different numbers. Are you projecting the 30#/ft against 17.33' or 16'? I was projecting it against 17.33. (which I feel is right but could be wrong.)

 

[JStephen]

I don't think there is a solution. Specifically, they show a load component in the horizontal direction and don't show any horizontal restraint. If that restraint is at one end or the other, that gives you different answers. If there's no restraint, it's accelerating to one side and possibly rotating as well. I'd say the question says more about the test writer than the test taker.

 

[gte447f]

OK. I see where your extra 20 lbs is coming from.
You are doing (30*17.33/2) + (20*17.33/2)*(12/13) = 420
I am doing (30*16/2) + (20*17.33/2)*(12/13) = 400
I think for a live load the 30 is typically projected on the horizontal plane, so 16 instead of 17.33, right?, but your number is more conservative, so no argument here. 30*16/2 is also consistent with the original problem statement if the answer for the original problem is 413.

 

[WARose]

Yeah, it is kind of a screwy problem. The "projected" vertical load is kind of a issue: do you put it against the hypotenuse of the horizontal plane? Breyer's wood book has a similar type problem. (Where he gets a "conservative" answer for one case.)

 

[gte447f]

JStephen, it's a dead load; it acts in the vertical direction. The way the arrows are drawn perpendicular to the member is misleading.

 

[Agent666]

Jstephen, yes there is a missing horizontal restraint (but it would be assumed by most engineers for global stability), however the vertical reaction will still be the same each end irrespective of what end the horizontal restraint is applied (it has to be for moment equilibrium, keeping in mind the horizontal reaction is actually zero as no horizontal loads are applied (unless I'm missing something obvious!)).

 

[dhengr]

They are apparently trying to teach on a subject which is a fairly common problem/error for young engineers, drafters/designers and builders. And, that is, that LL’s are expressed as loads in lbs./sq.ft. (whatever units) projected on a horiz. plane. However, we typically calc. (sum up) DL’s w.r.t. their own primary plane, such as the fl. sheathing for a fl.system, or that roof system w.r.t. its sheathing plane. Then, the DL must be converted into a slightly greater load (13/12) in lbs./sq.ft. projected on the horiz. plane, so the LL and DL can be combined. It is fairly common practice then, to use the horiz. length as the beam length to do normal beam calcs., with due consideration for the potential of horiz. thrust or axial loads on the members and supports. This is basically what BA and SRE and some others have done, without the above discussion.

 

[hokie66]

I think we all understand now. But it is interesting that the first couple of people who reckoned 428 is the answer are in Australia. We concern ourselves mostly with wind load on roofs here, and that is applied normal to the plane. But I suppose the dimensions should have given it away. Feet and pounds are something else we don't deal in.

 

[civeng80]

So let me see if I understand it. The Dead load shown perpendicuklar to the rafter should really be pointing vertically downward. Is that the trick here ?

 

[hokie66]

civeng80,
They are saying the DL is not a load, but the mass per lineal foot on the rafter. Why the academic showed the arrows that way is a mystery, or as you say, trickery.

 

[civeng80]

OK, I see, thanks Hokie66. I think it would be clearer just to state that the rafter self weight is 20 pounds per foot and erase the arrows.

 

[JStephen]

My guess as to what actually happened is that they got the load labels interchanged.
Generally, a test question won't be a "trick" question, and this is one that should have been removed.
Once you assume there's an error in the presentation, you really have no basis for answering the question- the error could be in the arrow orientation or an omitted horizontal reaction or in the labels, or the whole diagram could be skewed off vertical, etc.

 

[Agent666]

Maybe I'm the only one who thinks it's a very valid question, but if it mimics a real world error that people seem to do in real life by not accounting for the projected loads (I've certainly seen people do it in the peer review work), then what's the problem with the question?

It teaches a valuable lesson to understand and interpret the problem in front of you and question the information as it presented to you, aren't these valuable skills for any engineer to possess (those who have answered and followed along and came up with an incorrect answer are unlikely to make the same error again are they not irrespective if there are arrows or not in future?).