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vertical loads

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发表于 2009-9-16 18:20:26 | 显示全部楼层 |阅读模式
vertical loads
on page 59 of the april edition of modern steel construction magazine, does anyone know why the flat roof snow load is multiplied by cos^2 instead of just cosine to get the sloped roof equivalent, like the dead load?
check out our whitepaper library.
fatalxception,
the span of a parallel chod sloped joist is the length along the slope.    the gravity loads have two components, one perpendicular to the joist that produces the moment and shears, and the other parallel to the slope that produces axial force in the top chord.
the dead load is given over the slope length of the joist, and is multiplied by the cosine of the slope to obtain the load perpendicular to the joist.
the live and snow loads are given over the horizontal projection of the slope, and are multiplied by cos^2; the first cosine to spread the load over the sloped span, and the second cosine to obtain the load perpendicular to the joist.
aef
i guess i'm missing something, but if i have a snow load i have calculated with a dead load, i would multiply each one by the cosine one time to find the perpendicular component on a sloped joist.
wait, i think i see what you mean now.it's like the snow height is getting shorter as the joist span increases with increasing slope. it's more like they give you a total snow load and it gets spread over a longer length like you said. thanks.......
fatal, i think it has nothing to do with the snow load being less for a given higher slope.
you start with q psf of snow adequate for the situation on one notional horizontal plane.
the first multiplication by cos slope gets the horizontal load spreaded along the length.
but then, if you are to calculate the forces with the span along the slopes, the vertical forces are not orthogonal to your inclined beam. by vectorial decomposition of the force, the component parallel to the beam does not cause any bending, and only the perpendicular to the beam, got by multiplying again per cos slope does.
so it is that you multipliy by cos^2 slope.
the effect you refer to, the snow loads being lesser for roofs of higher slope is cared of at the establishment of the initial load as if projected on one horizontal surface.
snow is applied in force per unit length of horizontal projection... plf.  but you may want pounds per length "along the span".  two different "feets".
i sent this email before i posted on this forum, here is the reply:

we apologize for the delay in responding to you, as we had to contact the author.  the author of that article left me a voice message regarding your inquiry.  he mentioned that there are errors in both figure 19 and the sl calculation.

in figure 19, it appears that "ll x cos^3" should really be "ll x cos^2".  in the text, the sl calculation uses a cos^2 term, which according to the author, should be just cos.

i have forwarded this message to modern steel construction, as they may want to confirm these typos with "(i deleted name here)" and possibly print an editorial correction sometime in the future.

regards,
"(i deleted name for anonymity here also)", ph.d.
aisc solutions center
american institute of steel construction, inc.
one east wacker drive, suite 3100
chicago, il  60601-2001
this document has been prepared in accordance with recognized engineering principles and information made available to aisc at the time of its preparation. while it is believed to be accurate, it has not been prepared for conventional use as an engineering or construction document and should not be used or relied upon for any specific application without competent professional examination and verification of its accuracy, suitability and applicability by a licensed engineer, architect or other professional. aisc disclaims any liability arising from the unauthorized use of the information contained in this document.


-----original message-----
on page 59 of the april edition of modern steel construction, why is the snow load multiplied by cos^2 instead of just cosine like the dead load?
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