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Natural frequency in ASCE question

Natural frequency in ASCE question

Natural frequency in ASCE question

(OP)
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', wt=11.96lb/ft
my answer: n1=0.56/13^2*sqrt(29e6*12.6/11.96)=18.32in/ft^2 =1.54/ft

thanks for you input.
 

RE: Natural frequency in ASCE question

Why are you mixing feet & inches?  

RE: Natural frequency in ASCE question

(OP)
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?

RE: Natural frequency in ASCE question

no,
the m is mass, which is different
386 something
so the units of n1 is 1/s

RE: Natural frequency in ASCE question

(OP)
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. . .

RE: Natural frequency in ASCE question

sqrt(EI/M)=

sqrt[lb/in^2*in^4/

(lb*s^2/in^2)]= sqart(in^4/s^2)=in^2/s

 

RE: Natural frequency in ASCE question

(OP)
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.

RE: Natural frequency in ASCE question

Force= mass*g
so
EI/M= mass*g*in^2/(mass/in)=g*in^3

g has units of in/s^2, it is the standard gravity

So for EI/M, the units should be in^4/s^2

the same thing.

I may confuse you by saying 386

RE: Natural frequency in ASCE question

(OP)
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 gravity)=5.4kg
5.4kg*2.205=11.952lbm.
Therefore, the seconds are still eliminated.

Can anybody else shed some light on the subject?

RE: Natural frequency in ASCE question

what? I see SI and imperial all over.
You seem to be mixing soup from Barbados with soup from Ireland.
Try SI, has a more intuitive feel to it wink

peace
Fe

RE: Natural frequency in ASCE question

(OP)
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.
Thanks for everyone's help.

RE: Natural frequency in ASCE question

Grizz-
I still don't see how you came to your answer?
Can you explain?  

RE: Natural frequency in ASCE question

(OP)
Bah! I wrote the unit resolution wrong. right answer, I just didn't drop the square from the second after taking the square root.
Sorry about that.

RE: Natural frequency in ASCE question

You're answer (8.7 Hz) is correct. To work this in imperial units, you merely need to make m=w/g, where w is in lb/in and g=386 in/sec².  Curious that RISA gives 6.3 Hz for a single member.  I notice that the RISA solution converges to 8.7 Hz is you split the member into a number of pieces.

RE: Natural frequency in ASCE question

(OP)
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 place.

RE: Natural frequency in ASCE question

Looks like RISA takes the mass and concentrates it at the nodes.  In this case, if you  put half the mass at the tip, you get the 6.3 Hz.  Adding nodes distributes the mass.  I don't see that as a shortcoming.

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