Live load on a slope 2

[glass99]

If you have a steep ramp carrying people with a live load of 100psf, does the 100psf apply to the projected length or on the length along the slope? For example, if I have a plank 10ft long and I tilt it at 45 degrees, is the total vertical live load 1000lb or 707lb? Wind loads are clear because they are always normal to the surface.

This may sound like splitting hairs, but it makes a difference for my design. Assume US codes (IBC etc).

 

[jayrod12]

All gravity loads are on projected areas in my understanding but I may get corrected in this thread. I know that snow load is considered only over projected area.

 

[msquared48]

With a 45 degree tilt, I would be more worried about impact, as the people are more likely to be rolling than standing.

Definitely projected length. I always reduce load to sloping rafter members, but use the longer slope too in doing so. Gives a much more reliable deflection.

Mike McCann, PE, SE (WA)

 

[glass99]

msquared: OK thanks. Just to double check: For my 10ft long 45 degree plank example, the deflection would be calculated based on 50psf acting normal to the plank (100psf times cos45 ^2), and L in the deflection formula is 10ft? There is a double benefit with the slope.

I am designing a slippery slide.

 

[jayrod12]

Your deflection would be based on 70.7 psf over the slope distance. If my math is correct.

 

[steellion]

Live load : projected length
Dead load : actual length

 

[glass99]

If you have 1 ft of rafter measured along the slope your projected length is 0.707ft, so the load per ft is 70.7lb vertically down. The component perpendicular to the rafter is 70.7 / sqrt2 = 50psf.

Am I screwing that up?

 

[jayrod12]

The way I look at it, you have 100 psf vertically down. Therefore the load perpendicular to the slope is 70.7 psf. Regardless of length or spacing of rafter.

But I've been wrong before.

 

[Bagman2524]

The 100 psf load is vertical. For stairs I separate the loads into one perpendicular to the stringer (produces bending) and one parallel to the stringer (axial load). DL can be done similarly andnd steellion is correct the dead load has to use the entire/actual length of the stringer.

Faith is taking the first step even when you can't see the whole staircase. -MLK

 

[glass99]

OK, thanks guys. Seems like its unanimous that live load is on the projected length.

 

[Cyrus46]

Actually there are two ways to design the sloped elements.
1- apply the load on the projected length or surface and continue like regular.
2- you can project the applied load to two loads one is perpendicular to the sloped
length or surface and the second will be parallel to the sloped length and then design
like regular situation. In this case one of the load will act like a friction that we
always have in design of piles. keep in mind in this situation this load creates axial
forces in the member too so the supporting reactions have to be calculated too.

 

[azcats]

But what if there is a treadmill on the sloped rafter?

 

[msquared48]

No problem, as long as it is mounted underneath.

Mike McCann, PE, SE (WA)

 

[JLNJ]

A vertical load will produce axial load in a sloping member, but the reactions will still be just vertical.

This, of course, is true whether you apply it as a verticla force or break it into perpendicular- and along-the-member components.

 

[Cyrus46]

azcats,
If we ignore dynamic load on the rafter for now there will be no difference when a person standing on
a treadmill or a roof top unit on the rafters.
the treadmill is just a transformer of load to the rafter, like a roof curb for roof top units.
If you solve the problem with the second approach then the friction between the treadmill and the
rafter has to be greater than the sloped force projection along the rafter in order to keep the treadmill in place.