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Been a long time since college, but I always thought gravity was vertical direction.

(OP)
I don’t like the problem at all. If it’s a dead load then the arrows on the sloped load should be pointed down. They also don’t give the support type (pinned or roller).

I came out with none of the above. (I.e. 420 lbs.)

I get 432

Agree with OG and Rick. Also, why have 5 significant figures on a problem that should only have 3? maximum (arguably 2 or even 1 in the real world). They taught us about appropriate use of sig figs first semester, first year of engineering school.

all logic aside I get D. 400 lb. (on my second try)

It's not 413 pounds?

I get 428 pounds.

LL = 16*30/2 = 240
DL = 20*16*13/12*1/2 = 173 (assuming 20 plf per foot of sloping roof)

DL + LL = 413

BA

(OP)
The answer is 413 lbs according to the book the problem came from.

The solution exactly as BAretired shows above.

I arrived at 400, 413, and 428 taking different approaches.

I got 409, 427 and 454

Mike McCann, PE, SE (WA)

voting for 413

S&T

I thought you'd have to ignore the designations WLL and WD and just go with the free body diagram, exactly as BARetired said, and I computed. I think that's the lesson they are trying to teach. Screw gravity, or the orientation the drawing happens to be in.

The horizontal component of the sloping load has to go somewhere. Taking moments about the support, I get 428, so agree with civeng80.

This OG got side tracked with those arrows. 413 is mine now. Darn stuff like this kept me awake after bed time and I had to getup and fix it. Why is it your best thinking comes after bed time? Anyhow isn't student help frowned upon here and for good reason as proved here? Now back to bed. Another lesson learned. No Eng Tips after dinner, like coffee.

So maybe this is a trick question. The arrows perpendicular to the rafter would lead one to think the force is applied in that direction, but dead load is gravitational. If that load is a wind load as indicated by the arrows, I stick with 428.

I thought the loading perpendicular to the rafter was a Wind Load, but after reviewing the diagram it is Dead Load. It is a confusing question because Dead load acts vertically down all the time so to show it perpendicular to the rafter is plain wrong.
I worked out the reaction by taking moments about the right support.

So, the question is: did they intend you to rotate the FBD so the DL component is vertical or simply compute the support vector component as shown. Is this a FBD exercise or a trick question? Are we reading DL to be Dead Load or could it be Design Load, Dumb Load, etc? My vote originally was to ignore the LL and DL designations since their definitions were not included with the problem. If LL and DL were not written there would be no question - the answer is 413. I think the real-life answer is: what is meant by LL and DL since it makes a difference to the answer. And petition the test giver to fix this idiot problem so there is only one correct answer.

Agree with BAretired, 413 lb:

#### Quote (TheRick109)

I don’t like the problem at all. If it’s a dead load then the arrows on the sloped load should be pointed down. They also don’t give the support type (pinned or roller).

IMHO, this problem is presented in an unusual way but is a good one for the PE exam:

A student is given initial conditions need to solve a problem (so they can learn).
An engineer does not necessarily have the initial conditions, or the info may not be obvious or straightforward.

In "Design of Wood Structures" 3rd Ed., Donald Breyers covers two methods for solving rafter loads:
"Sloping Beam Method" and "Horizontal Plane Method". See the attachment (Note: I consider this use of the book's example to be "fair use" for copyright purposes).

Live Load is presented in "Horizontal Plane Method" format.
Type of support - An engineer qualified to take the PE exam should be able to figure that out.

Bonus question. What would the vertical reaction be if it really was acting perpendicular to the member (like a wind load)?

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).

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?

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.

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

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.)

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.

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.

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.)

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

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!)).

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.

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.

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 ?

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.

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.

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.

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?).

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