I am currently trying to optimize the location of a hydraulic cylinder on a wind tower which is being erected using that cylinder. Just wanted to know how high can the hydraulic cylinder(2) be connected so that the wind effect is ignored or if there is some guidelines which I should follow.
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Wind Load calculations
- Thread starter technever
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Hi
What’s i and I assume F is the cylinder force ?. Why is W a+b+c acting vertically?
Please put some labels on the diagram so people can understand your problem better.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
What’s i and I assume F is the cylinder force ?. Why is W a+b+c acting vertically?
Please put some labels on the diagram so people can understand your problem better.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
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- #4
hello, desertfox, Im sorry I was not able to put that up earlier,
Wabc is the weight of the turbine
Wt is the weight of the tower
i is the centroid for the turbine weight
e is the centroid for tower weight
h is the height of the tower
a is what I want to find, but right now I just need a constraint as to how high i can go without any significant impact from wind
y is the distance of connection
Wabc is the weight of the turbine
Wt is the weight of the tower
i is the centroid for the turbine weight
e is the centroid for tower weight
h is the height of the tower
a is what I want to find, but right now I just need a constraint as to how high i can go without any significant impact from wind
y is the distance of connection
apologies, but your sketch is awful, close to meaningless.
you're trying to erect the tower on a hill (slope B) ?
position of the lift would be limited by the bending moments in the tower.
lift on a windless day if you're that concerned about wind.
add a wind load (a distributed load ?) to the tower.
Weight acts down, but the hydraulic actuator would act up ...
"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.
you're trying to erect the tower on a hill (slope B) ?
position of the lift would be limited by the bending moments in the tower.
lift on a windless day if you're that concerned about wind.
add a wind load (a distributed load ?) to the tower.
Weight acts down, but the hydraulic actuator would act up ...
"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.
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- #6
Apologies for the sketch, Perhaps this picture would be of help.
Im trying to optimize the position of the hydraulic cylinder on the tower. with adding the hydraulic cylinder the area for wind will increase ( I believe, Im not sure) so im trying to attach the cylinder as low as possible but I want to know how high it could be attached.
Im trying to optimize the position of the hydraulic cylinder on the tower. with adding the hydraulic cylinder the area for wind will increase ( I believe, Im not sure) so im trying to attach the cylinder as low as possible but I want to know how high it could be attached.
Hi technever
Well without any details of any lengths or details of wind loads we can’t tell you anything about an height restriction.
You need to calculate the moments exerted by the equipment at various points in the lift which is fairly straightforward and bring them into equilibrium with the position of the cylinder, if you have a wind load then that needs to be resolved into a force and again balanced along with the other forces you have mentioned with the cylinder position.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
Well without any details of any lengths or details of wind loads we can’t tell you anything about an height restriction.
You need to calculate the moments exerted by the equipment at various points in the lift which is fairly straightforward and bring them into equilibrium with the position of the cylinder, if you have a wind load then that needs to be resolved into a force and again balanced along with the other forces you have mentioned with the cylinder position.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
The lower down you position the cylinder, the bigger the cylinder you will need, without the wind speed for the location you are in and not to mention the height and the projected area of the tower that the wind will act on how can we give you an answer?
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
I think he means the wing loading (laterally) on the tower. Like I said, I think tower bending moments are the key thing to look at.
"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.
"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.
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- #11
Hi guys sorry English is not my first language so im not sure ill be able to explain this properly but please refer to this figure may be this can help.
This is the tower in upright position, I know that the lower i go the size of the cylinder will be larger but the area created on Z direction to be influenced by wind will be less and conversely i can connect it higher and the cylinder size decreases but the area on Z direction increases. i want to know up to what height can I ignore the wind effects on the area. Once again the wind is coming from Z direction. Im sorry if my question is not clear
(English isnt my first language)
This is the tower in upright position, I know that the lower i go the size of the cylinder will be larger but the area created on Z direction to be influenced by wind will be less and conversely i can connect it higher and the cylinder size decreases but the area on Z direction increases. i want to know up to what height can I ignore the wind effects on the area. Once again the wind is coming from Z direction. Im sorry if my question is not clear
(English isnt my first language)
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- #13
but what about the cylinder, there would be suction and all sorts of things I assume, so I just want it to be below the height where I can ignore all the wind forces (If that is a thing)
Also thank you all for the help. Im really at my wits end with this so all the help is appreciated
Also thank you all for the help. Im really at my wits end with this so all the help is appreciated
So you are talking about the wind blowing in the Z-direction which is perpendicular to the plane of the cylinder and structure. I don't see what issue wind in this direction would have on the placement of the cylinder. Even a 100 MPH wind would only give a maximum pressure of 25 PSF which will be a negligible load on the cylinder itself due to the small projected area of the cylinder. There will no difference in the wind load on the structure no matter where you connect the cylinder. In any case the load on the structure in the off plane direction would need to be taken by the bottom connection pins. Also note that you are not going to be lifting the structure in 100 MPH winds or even 50 MPH wind. Are you?
SWComposites
Aerospace
Agreed. Wind loading on the cylinder is probably negligible.
I guess we have to recognise that this is the student forum, and maybe he's been told "consider wind effects".
If so, windage is a distributed load, the suggestion above is something like 25 pounds/sq ft (and a sq ft is 1 ft length * x ft diameter). Now you can use that number or you can calculate it (or maybe there's a code reference you can use ?) or maybe your can say "assume windage loads negligible" ? (what year are you in ?)
But the key thing to determine your actuator point is the bending moment of the tower. Also the length of your actuator (how much it extends)
If we're talking practical aspects, there'd be two guy ropes to control the out-of-plane (and so small loads (like windage can be managed this way.
"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.
If so, windage is a distributed load, the suggestion above is something like 25 pounds/sq ft (and a sq ft is 1 ft length * x ft diameter). Now you can use that number or you can calculate it (or maybe there's a code reference you can use ?) or maybe your can say "assume windage loads negligible" ? (what year are you in ?)
But the key thing to determine your actuator point is the bending moment of the tower. Also the length of your actuator (how much it extends)
If we're talking practical aspects, there'd be two guy ropes to control the out-of-plane (and so small loads (like windage can be managed this way.
"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.
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- #17
ASCE 7 covers wind loads on structures.
Basically the wind load is based on the velocity head pressure = V2/2g. So if you have a 100 mile per hour wind which is 146.7 feet per second the theoretical wind load on the projected area of any object when the wind acts on the object at a right angle is:
Velocity Head = (146.7)2/2(32.2) = 334 Feet of head
Converted to pressure when velocity come to complete stop on the object in accordance with the Benouli Equation:
Pressure(PSF) = (Density of Fluid #/ft3) x (Head - FT) = .076 x 334 = 25.4 PSF assuming air density at standard conditions.
So this is the maximum pressure on the projected surface. Now for a cylindrical shape the shape factor for wind loading is 0.6 meaning that the real effective load on a cyliner is 0.6 x 25.4 = 15.2 PSF.
However different codes such as ASCE 7 apply more safety and other factors etc. So if you are designing to Code you must follow ASCE but the real loads are indicated above.
Basically the wind load is based on the velocity head pressure = V2/2g. So if you have a 100 mile per hour wind which is 146.7 feet per second the theoretical wind load on the projected area of any object when the wind acts on the object at a right angle is:
Velocity Head = (146.7)2/2(32.2) = 334 Feet of head
Converted to pressure when velocity come to complete stop on the object in accordance with the Benouli Equation:
Pressure(PSF) = (Density of Fluid #/ft3) x (Head - FT) = .076 x 334 = 25.4 PSF assuming air density at standard conditions.
So this is the maximum pressure on the projected surface. Now for a cylindrical shape the shape factor for wind loading is 0.6 meaning that the real effective load on a cyliner is 0.6 x 25.4 = 15.2 PSF.
However different codes such as ASCE 7 apply more safety and other factors etc. So if you are designing to Code you must follow ASCE but the real loads are indicated above.
So the wind loading on the cylinder itself would not be that great I dont think and it would basically cause out of plane bending of the cylinder. However if you are designing for a significant wind load during lifting then the wind load on the structure will cause a deflection of the structure which is greater as the height above the base increases. So if you connect the cylinder to a point that is high and has a lot of defection the cylinder is going to deflect with it assuming the cylinder is not strong or rigid when bending sideway. This could damage the cylinder. The base of the structure pinned connection point will need to be designed for the full design wind load.
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