Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

Register Log in
  • Congratulations KootK on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Rotation from Transient Load 2

Status
Not open for further replies.
MegaStructures

MegaStructures

Structural
Sep 26, 2019
376
What would be the best way to calculate how far this object would rotate if a force acted on it for 3-seconds only?

image_tw7hcy.jpg


“The most successful people in life are the ones who ask questions. They’re always learning. They’re always growing. They’re always pushing.” Robert Kiyosaki
 
Replies continue below

Recommended for you

Sort by date Sort by votes
My first thought after looking at this for 5 minutes is that this is a Single DOF dynamic equilibrium system.
Capture_estt86.png

Instead of springs and dampers, you would have (mg)*y(t) as a restoring force, with y(t) as a function of the rotation.

Edit: After looking at this a bit more, if the force is assumed to parallel to the x-axis, below is what I think the equilibrium equation will be. If the theta is small, you may be able to simplify and get rid of the sin and cos terms. The trick is to integrate the equation to get theta(t), but this can probably be done numerically (i.e. Newmark-beta method).
Capture_kvmptn.png
 
Upvote 0 Downvote
Can you just use the displacement formula (Delta = 1/2*a*t^2 = 1/2*(F/M)*t^2) and then treat the distance as an arc length to get the rotation (Theta = Delta/radius)?

(I'm sure there's more to this that is not being accounted for like the polar moment of inertia of the plate and friction in the hinge. Also, does gravity act against the applied force once it rotates 90 degrees?)

Structural Engineering Software: Structural Engineering Videos:
 
Upvote 0 Downvote

I assume the force is applied tangential to the rotation .
You are expected to use the rules of rotational dynamics.

You may follow the following steps for calculation,

- Calculate the Moment of Inertia of the rotating body (I)
- Torque T = F Xr

- T=Iα ( α is angular acceleration )

- ω = ω0 + αt ( ω is angular velocity )

- θ = ( 1/2)*α* t**2
 
Upvote 0 Downvote
What would be the best way to calculate how far this object would rotate if a force acted on it for 3-seconds only?
Click to expand...

My approach would be a angular impulse momentum along with conservation of energy solution. It's a little long to lay out here. But any decent dynamics text can help. (I could run it myself to demonstrate....but your OP gives no numbers.)

 
Upvote 0 Downvote
I'd follow klbridge's post, eqn 2.32 (as modified by his last comment). and note y(t) = R*theta(t), where R is the CG position.



another day in paradise, or is paradise one day closer ?
 
Upvote 0 Downvote
Would need to know if the force remains in the same direction or rotates with the mass.
 
Upvote 0 Downvote
Three seconds is plenty of time to justify using statics. Then, the problem is easy.

BA
 
Upvote 0 Downvote
See attached for a problem I worked out from my old dynamics text. (It's a hanging, 5' long, 100 lbs rod (at rest) being hit on by a short transient pulse.) In the first part, I figure the rotational velocity generated by the impact. In the second, I figure the rise height that it gets to. (The rise height from Part #1 is, as you can see, virtually zero.....so I didn't add it to the rise height from part #2.) Since Part #2 happens after the application of the pulse.....there was no need to include the pulse's energy component (i.e. U1-2) in Part #2.

I know you have a longer pulse.....but hopefully this will (still) be useful.

Sorry for the sloppy handwriting by the way.
 
 https://files.engineering.com/getfile.aspx?folder=d3274863-0556-4d72-b3ff-205caba117ef&file=conservation.momentum_001.jpg
Upvote 0 Downvote
I assumed this would be as easy as comparing the energy of the wind to the rotational energy of the body, but I'm realizing that time of the forcing function isn't included in the kinematic energy equation

“The most successful people in life are the ones who ask questions. They’re always learning. They’re always growing. They’re always pushing.” Robert Kiyosaki
 
Upvote 0 Downvote
I assumed this would be as easy as comparing the energy of the wind to the rotational energy of the body, but I'm realizing that time of the forcing function isn't included in the kinematic energy equation
Click to expand...

Time is part of the solution I posted.
 
Upvote 0 Downvote
Thanks WARose that is perfect. I'm trying to verify/understand the equations used.

Part 1 to find the rotational velocity: I recognize that the right side of the equation is the rotational momentum of the bar. What is on the left side? The derivative of angular momentum is torque?

Where does the derivation of the equation come from used to find theta?



“The most successful people in life are the ones who ask questions. They’re always learning. They’re always growing. They’re always pushing.” Robert Kiyosaki
 
Upvote 0 Downvote
kwlbridge: I really like that method, but it’s been too long since I’ve taken a reasonable math course. I don’t even know where to start solving that equation at this point

“The most successful people in life are the ones who ask questions. They’re always learning. They’re always growing. They’re always pushing.” Robert Kiyosaki
 
Upvote 0 Downvote
Part 1 to find the rotational velocity: I recognize that the right side of the equation is the rotational momentum of the bar. What is on the left side? The derivative of angular momentum is torque?
Click to expand...

The whole equation is the equation of angular impulse and momentum. (It's equation 8.25 in my old dynamics text: 'Engineering Mechanics: Dynamics', by: Das, Kassimali, and Sami, 1994.)

Where does the derivation of the equation come from used to find theta?
Click to expand...

That's the equation of angular velocity: θ(t)=1/2(greek letter alpha)t2+ω0t+θ0

In what I posted, the initial angular velocity & [angular] position was zero.....ergo it just cut down to the first term.

See attached for another approach. (This is straight out of my old dynamics text. So you don't have to read my sloppy handwriting.) In this case, the force is constant.....and is always perpendicular to the bar. For your case, you may want to use this approach by having the angle as a unknown and selecting a minimal angular velocity (i.e. say 0.001 rad/s; what it will be when it approaches its peak rise height).

This may be more useful to you since your force may be [close to] constant (since you mentioned wind in one of your latest posts).

Hope this helps.
 
 https://files.engineering.com/getfile.aspx?folder=69bd7162-9a1a-406d-a25d-176e26b21114&file=rt_002.jpg
Upvote 0 Downvote
Would anybody care to explain the numerical integration required for kwlbridge's method? I learned Newmarks years ago, but can't quite remember how to get started for a specific equation above. I do have a nice step by step procedure to solve the general equation of motion, but unclear how the rotational nature of the problem changes the various k hat, p hat, a, and b formulas

“The most successful people in life are the ones who ask questions. They’re always learning. They’re always growing. They’re always pushing.” Robert Kiyosaki
 
Upvote 0 Downvote
What are the values of F, M, and the dimensions?
 
Upvote 0 Downvote
Assuming F does not follow a curved path (i.e. h is constant)
For equilibrium, F*h = W*e, so F = W*e/h

F = force at distance h below the pin.
W = weight of block = Mg
e = eccentricity after a long time (3 seconds is a long time)

BA
 
Upvote 0 Downvote
BA, maybe you're right. A big point of doing this calculation is to prove you're wrong though.. I don't think 3-sec's is enough to reach a significant portion of the max displacement.

Tomfh, see below:

[ul]
[li]Force=1500 lbf (always perpendicular to axis)[/li][/ul][ul]
[li]Weight:100,000 lbf[/li][/ul][ul]
[li]The weight and lateral force act at the same point, 60 ft from the rotation[/li][/ul][ul]
[li]length of rod is 120 ft[/li][/ul]


“The most successful people in life are the ones who ask questions. They’re always learning. They’re always growing. They’re always pushing.” Robert Kiyosaki
 
Upvote 0 Downvote
Are you trying to find the exact solution? Or are you happy with a conservative approach? ProgrammingPE's method will achieve that, and gives a displacement of 0.66m at the centre of the rod, i.e. about 2 feet.
 
Upvote 0 Downvote
Tomfh, conservative is ok, but that seems overly conservative. The inertia of the rod is quite large

“The most successful people in life are the ones who ask questions. They’re always learning. They’re always growing. They’re always pushing.” Robert Kiyosaki
 
Upvote 0 Downvote
Status
Not open for further replies.

Similar threads

EngDM
  • Question
Instantaneous center of rotation - CISC
Replies
3
Views
3K
Celt83
Celt83
A
  • Locked
  • Question
Load Distribution from CMU Wall on Footing
Replies
6
Views
3K
jerseyshore
jerseyshore
Aytacoglu
  • Locked
  • Question
Impact Load from Falling Concrete 1
Replies
3
Views
2K
human909
human909
Deener
  • Question
Points of beam support restrained against rotation
Replies
1
Views
2K
Lomarandil
Lomarandil
T
  • Question
Differential settlement, thermal load in load combination
Replies
1
Views
2K
canwesteng
canwesteng

Part and Inventory Search

Sponsor

Back
Top Bottom