Frequency of Vibration of a water tank

AI Thread Summary
The discussion revolves around estimating the natural frequency of a water tank supported on a 200 ft pedestal, with weights ranging from 50,000 lbs to 20,000 lbs. Participants clarify the calculation of the spring constant using the tension in the cable and the deflection observed, leading to a value of approximately 231.965 lbs/in. There is confusion regarding the distinction between angular frequency and natural frequency, with participants confirming that natural frequency can be derived by dividing angular frequency by 2π. Additionally, the discussion touches on how to approach part b of the problem, suggesting the use of the magnification factor equation to determine the frequency range for horizontal Earth motion. Overall, the thread emphasizes the importance of understanding the relationship between forces, deflection, and frequency in vibration problems.
CGI
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Homework Statement


Assume that there is a tank on a 200 ft pedestal type support. When full the tank and contents have a weight of 50,000 lbs and it is never drained to a point where the tank and contents have a weight less than 20,000 lbs. Assume the pedestal weight is negligible compared to the tank's weight and that the dimensions of the tank are such that it might be assumed a lumped mass( a particle). A force transducer is attached to a truck, is connected by a 600ft long cable to the tank, such that there is an 18° angle relative to the horizontal, and that the cable's line of action passes roughly through the centroid of the tank and its contents. Careful measurements indicate that the deflection of the tank is proportional to the force and that the tank has moved horizontally 20.5 plus or minus 0.04 inches when the force in the cable is 5000 lbs.

a) Estimate the natural frequency for vibration of the tank at both its maximum and minimum weights. In other words at 50,000 lbs and 20,000 lbs.
b) Estimate the frequency range of horizontal Earth motion which "would" likely cause the tank to displace 2 times or more than the magnitude of the "earth's horizontal vibrational motion". Neglect damping.

Homework Equations



ω = 2π/P
ƒ = 1/P
ωn = √(k/n)
Magnification Factor = 1/(1 - (ωƒn)^2)

The Attempt at a Solution


I have some trouble with these kind of questions where I feel like there is a lot going on. I sometimes don't know where to start, so I apologize in advance if I ask dumb questions!

Like I said, I'm not sure where to start, but I drew a free body diagram of the tank where there are only two forces acting on it. The forces on the water tank are its weight and T, the tension in the cable. I have that the force of T is 5000cos18 since the cable is at an 18 degree angle relative to the horizontal. From there I'm not quite sure where to go. If there's anyone out there that could help me, I would sure appreciate it. Again, these "vibration" problems always seem to be so difficult for me so a thousand "thank yous" in advance!
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The original wording of the question is poor but I think you should be trying to work out the natural frequency of the pylon and tank only . The part about the truck and cable is only there to give you a way of getting the load - deflection characteristics of the pylon .

Can you work out the effective spring rate of the pylon from the information given ?
 
When you say Spring rate, are you talking about the spring constant?
 
Yes .
 
Oh okay. So if

F = kx and F = 5000cos18 and x = 20.5

Could we say that k = F/ x? So k = 231.965 lbs/in ?
 
That's it .

Can you do the rest of the problem now ?
 
I think so. I just have one question. I've been asking different people and they give me different answers. The omega that I find is the angular frequency, right? So if I want natural frequency I just divide that by 2pi?
I guess the real question is, what is the difference between angular and natural frequency?
 
CGI said:
I think so. I just have one question. I've been asking different people and they give me different answers. The omega that I find is the angular frequency, right? So if I want natural frequency I just divide that by 2pi?

Short Video to watch .
 
Okay awesome! One last thing, I'm having a hard time understanding what to do for part b. When they ask for a range, there's not an equation that comes to mind
 
  • #10
Oh wait, would I use the magnification factor equation, set it equal to -2, solve for omega sub f, and then set it equal to 2, then solve for omega sub f again?
 
  • #11
Look up Forced Vibrations and Base Excitation .
 
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