Roof Live Load 1

[hubba]

I don't work in this stuff everyday, so please forgive me for the elementary nature of this question. ARE ROOF LIVE LOADS APPLIED VERTICALLY (GLOBAL -Y) TO SLOPED ROOF MEMBERS? (i.e. not perpendicular to sloped member?) Seems like they should bc big part of roof live is weight (i.e. vertical load) of various construction incidentals.

However, I get confused bc codes/stds/packages make point of distinguishing horizontal projection vs real length. If it's vertically applied, isn't it always the horz projection? Thanks.

 

[ToadJones]

Might not be a single answer to your question.
If you look at it simply as the direction that loads are acting, then gravity loads are globally vertical and downward.
Pressures act normal to the surface.
so, say for a purlin design on a sloped roof, gravity loads would cause bi-axial bending in the purlin, but pressure loads would cause only uni-axial bending.

agreed?

 

[hubba]

Agreed. This case is a truss w/ 20psf specified LL on top chord. Only looking at truss, no purlin w/ biaxial. Just trying to tune a RAM Elements model to match reactions, etc provided on output from a MiTek model. Thanks for chiming in.

 

[ToadJones]

If someone chimes in about shear centers and purlins that are channel shapes, I will go home immediately and start the Holiday weekend early!!!!

 

[csd72]

Dont forget the torsion induced by...darn ToadJones has beeten me to it

 

[ToadJones]

csd,
don't do it man, I'm serious!
If BARetired starts making diagrams about torsion, I'm outta here!!!!

 

[WillisV]

The answer to the actual question is per IBC 1607.11 Live and Snow loads both act vertically on the horizontal projection of the beam.

You would multiply the live load by cosine of the rise to run angle to get the vertical load acting along the slope.

You multiply that value by cosine of the rise to run angle to get the component of load acting perpendicular to the member, or by sine of the angle to get the component acting parallel (axial force) to the member.

So...in summary:

Component perpendicular to member: LL(cos theta)^2
Component parallel to member: LL(cos theta)(sine theta).

where theta = rise to run angle in degrees.

 

[ToadJones]

So....
Say you have:

A truss with a 4:12 pitch, a 10 ft tributary width, and live load of 20 psf.

Say you want to model this truss with a uniform distributed load acting in the Global Y (down) direction, then you are saying that load would be:

rise = 4'
run = 12'
hypot.= 12.65'

vertical load (12/12.65)* 20psf * 10' = 189.7 plf?

And, from here you'd break that into forces perpendicular and parallel to the top chord?

 

[WillisV]

ToadJones - correct - you would take the 189.7 and multiply it by cos or sin for perpendicular and parallel components.

 

[ToadJones]

hmmmm...
I honestly never interpreted in that way.
I guess it is a good thing most of the roofs I have done are 4:12 and under.
Even at 4:12 it seems to only represent about a 6% change.

 

[OldPaperMaker]

hubba,
For a detailed examples of the Sloping Beam and Horizontal Plane Methods of analysis see pages 2.19 to 2.21 of the 6th edition of Design of Wood Structures (ASD/LFRD) by Donald E. Breyer, et al.
Basically the two approaches yield the same results (monents & shear).

 

[JLNJ]

This is an east mistake to make in specifying the load in an analysis program. You need to pay attention to the defaults and make sure you apply it locally or globally on purpose. RISA3D for instance can do it either way, you just need to specify it properly.

 

[msquared48]

In higher pitched roofs, it does make a difference in the rafter size.

It is particularly applicable in Hip and Valley member design.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[Lion06]

If you have a truss
with a 4:12 pitch and 10' trib width with 20 psf LL why wouldn't the line load simply be 20 psf * 10' = 200plf ; then use that to determine the load acting perpendicular to the beam (causing bending) and in line with the beam (causing axial load)? The 10' trib isn't sloped so the horizontal projection is 10'.

 

[WillisV]

SEIT - LL is specifically given by code to act on the horizontal projection of the beam, so the 20psf has to be placed along the slope of the beam before you can calculate the parallel and perpendicular components. What you stated is correct for dead load etc. (loads that are already along the length of the beam don't need that initial multiplying by cosine step from my above post).

 

[Lion06]

What am I missing? Isn't the LL already acting vertically? Why do you need to do anything to it to get it to act vertically? I see the part about taking the components perpendicular to and parallel to the beam, but I'm missing something when it comes to the acting vertical part.

If you have a beam that spans horizontally 10' with a trib width of 10' and a LL of 20psf and is flat, the vertical reaction at each end is 1000#. Of you keep the horizontal span of 10', but offset the ends vertically by 3', the vertical reaction at each end is still 1000#, no? In both cases the horizontal projection is 10'.

 

[Lion06]

The internal forces will be different for each, because they have different lengths and different components acting perpendicular to and parallel to the axis of the beam, but in both cases the vertical reaction is the same.

 

[WillisV]

SEIT yes, you are missing something - it's not the vertical term that is key, it's that it needs to be vertical along the length of the beam because the code gives you the live load values in terms of horizontal projections - I'll post a diagram next week when I can - basically for a 45deg roof the code roof horizontal projection LL value of 20 psf results in an actual load of 14.14plf acting vertically on a slope along the beam.
You can get correct moments without dealing with this, but this is the only way to get correct axial forces.

 

[BAretired]

I'm with StructuralEIT. Snow is a gravity load and is specified by code on a horizontally projected area. If it is 20 psf, then the total snow load acting on plan area A is 20*A, irrespective of roof slope (unless it is steep enough for the snow to slide off).

Dead load is also a gravity load, but if the dead weight of deck plus roofing materials weighs 'w' psf and the slope is [θ] then the dead weight on plan area A is w*A/cos[θ].

BA

 

[BAretired]

Toad,

Sorry to disappoint you. No diagrams about torsion.

BA