Yesterday (2/16) still using copies of DS15 and SD18 without issue.
On my 15 version, I get the reaactivation message saying it is going to expire. I figure that is how they will get people off it. I will try registry edit to see how long it lasts.
Today is 1/3/20.
I just opened my free version of DS18 without a hit. did a print, a save, and a save as without issue.
Will it last? I don't know. just wanted to pass that along.
If it is a means of egress from the building, I'd have to go with chapter 12.
If it's just a way to access a roof for periodic maintenance, chapter 15.
I argued with an architect one time about the underside of an awning at a grocery store. What he wanted cost 3x what was needed. He won the argument. I'd have to say the owner lost.
My degree is architectural engineering. I can appreciate what an architect is trying to achieve. He(the architect) just usually doesn't see the innate difficulties in achieving his dream.
add a couple of outcroppings and returns to a box. you still have a box but now it has texture. It's amazing...
I hadn't tried adding nodes into the model. to me, it's their(RISA's) biggest shortcoming. p-little delta can not be accurately modeled with less than about 20 sections on a column.
thanks for checking it out. In empirical units w/g = slugs. That's pretty much what was stumping me in the first...
It was easier(or quicker, at least) to post the hand written solution.
If anything needs clarification, let me know.http://files.engineering.com/getfile.aspx?folder=89fe5670-e143-4659-b672-3f3d40ad3797&file=frequency__resolution.pdf
I always boggles my mind when engineers leave one of the most critical links of the structural system up to someone else to design. Often that other person is not an engineer or is an engineer with very limited knowledge of all the forces going into the connection.
I'm not saying you shouldn't...
While his diagram shows fixity at the bottom, both his equations imply pinned at the base. I don't believe his intent was to fix the base. How much good will a <=3' wide footing do in that situation.
Yes that is exactly what I did. It is tedious to convert everything over to metrics however. After I found my answer (8.7 Hz), I backed that out in english units. mass is given in slugs. Of course it didn't match the RISA results (6.3Hz), but both at way over 1 so I'll just go with Gf=0.85...
I understand where the 386 comes from.(gravity in in/sec^2) What I am asking is for clarification as to why you're ignoring the empirical conversion factor.
In English units:
lbf=lbm*a/gc
or
lbf=slug*a
a slug = 32.174 lbm.
to further clarify:
11.96lbf*0.225=53.2N /9.807(acceleration due to...
I follow your equation.
could you explain why you're only multiplying by 32.174 and not dividing by 32.174 also? one is acceleration due to gravity, the other is G sub c, an empirical conversion factor to convert pound force to pound mass.
Thanks for your efforts.
TA=24^2/2/20
Unless you're doing some kind of unique design, the wall reinforcing will make the wall act as a cantilevered beam. pinned at bottom and at roof.
Now if you're tying bars into the roof or in some other way making the roof a moment connection, then all bets are off.
F=m*a/gc therefore m=F*gc/a
11.96(lbm/ft)=11.96(lbf/ft)*32.174(lbm-ft/lbf-sec^2)/32.174(ft/sec^2)
11.96*32.174=384.8 (close to what you said) but why don't you divide by acceleration due to gravity to get pure lbm?
You've got to love the english units. . .
That is my question. everywhere else in the book, h is defined as feet. E and I as near always expressed in inches.
No matter how I solve the equation, the result is 1/length.
So there has to be a conversion factor in the 0.56. Thus what are the proper units to use?
from ASCE 7-05, pg 294
eqn (c6-22a) reads : n1=(0.56/h^2)*sqrt(EI/m)
can anyone please tell me the units for
h, E, I, m?
I had h=ft, E=psi, I=in^4, m=lb/ft(mass per unit height according to the book)
however the answer yields in/ft^2 or 1/ft (w/ conversion)
my problem:
hss5x5x3/16
I=12.6, h=13'...
the rods are tension only.
stiffeners won't work. the rod penetrates the web and sets on a hillside washer. I could use a large plate washer i suppose, but I'm still pondering how to calculate if its needed.
If it is, you still have to consider the effects on the web at the edge of the washer.
if 0.6D-W controls your design and you want to use D-W in that combination, couldn't you multiply D X 1.667 and use that in all you other combinations. if the 0.6D-W still controlled with you amplified dead loads, you'll be good to go, still within the code requirements, and be able to sleep...
