It was quite interesting to learn what hourly revenue was at stake with that particular structure, as to whether the store was open, and bringing in revenue, or not. And this particular store was one in which shoppers turned away at the door simply went down the street to the competitor store. (Turned out to be the best day of the year for one competitor.) But my fees were not based on what revenue was saved from being lost, which is good, in the matter of public safety. I gave the nod for building re-occupancy when it was safe to do so (after particularly distressed portions of the roof were relieved of excess snow).
In that article I describe how to calculate the snow load (vertical weight). (And parts of the structure involved in the paragraphs above were experiencing loads greater than what they were designed for.) More recently I have been intrigued by the thing called `snow density'. The context is this: if snow has a constant density, then we could simply measure the depth of snow at any time and from that come up with the snow load. Growing up I was told (I think by my mother, from whom we learn most of what we learn growing up) that `ten inches of snow equals one inch of water'. And so for decades I took that `rule of thumb' (?) for granted. And maybe the rule holds true if you live in a place that at most gets an inch or two of snow and then it melts before noon (like where I grew up), but if you mess around in snow much that sticks around for a while you may begin to suspect the notion of a `one size fits all' approach to how dense the snow it.
So, this morning I couldn't resist. We got another three-fourths of an inch of snow on top of the twelve already there, but this last inch is so light that it is basically CLEAR. You can see right through it. So, I got out shovel, measuring tape, and garbage bag, to measure how dense the snow is.
I picked an undisturbed area on the ground, and measured. The snow shovel I used has a blade that is 17 inches wide, perfect! ... as a 17 in. by 17 in. square is 2.0 square feet, making number crunching easier. Actually, when done the hole was 18 in. by 20 in. And the weight of the snow is 19 lb. So, following the procedure from before, for kicks and giggles, the ground snow load is ...
... σ = W / A = 19 lb / (18/12 ft x 20/12 ft) = 19 lb / (2.5 sq ft) = 7.6 psf.
The value is way less than the design snow loads for around here, and much less than last year (but this winter isn't over yet).
AND, I took one more measurement; I measured the depth of snow also. Currently our snow depth is at 13 inches.
So, now I have the means to calculate the current snow density, since now I can also calculate the volume the snow occupies.
(Actually, the more precise term is specific weight.)
In equation form ... γ (specific weight) = W (Weight) / Vol (Volume).
In our case, then, γ = 19 lb / (2.5 sq ft x 13/12 ft) = 19 lb / 2.7 cu ft = 7.0 pcf (lb per cu. ft.).
What did mommy say? Mommy said, essentially, that snow weighs one-tenth that of water. (Snow for the same weight is ten times as deep). Well, water weighs 62 lb per cubic feet. One tenth of that is 6 lb per cubic foot. And I just calculated 7. SHE WAS RIGHT! Or really close anyway! Precisely the snow out there right now is one-ninth the density of water. (Or, if it melts, and just stands there, the 13 inches of snow will give us about an inch and a half of water.)
But this snow is all relatively fresh! I am suspicious that this ten-to-one (or nine-to-one) thing may break down on me as the winter progresses ... as the snow depth increases, and / or as warmer weather melts or ripens it. So, stay tuned!
Oh, yeah, that top three-quarters of an inch. I cannot think of a practical way to weigh just it ... it's simply too light.
References
Measuring Snow Loads on Roofs, Associated Content
Published by Jeff Filler
Consulting Engineer, Educator, Aspiring Writer and Photographer, Husband, Father, and Serious Hunter. View profile
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1 Comments
Post a CommentExcellent article. I liked it!