What Factors Determine the Breaking Point of a Paddle Under Dynamic Forces?

  • Thread starter Thread starter meanswing
  • Start date Start date
  • Tags Tags
    Break
AI Thread Summary
The discussion focuses on determining the breaking point of a paddle under dynamic forces, particularly the drag force from water and the force from a pneumatic cylinder. Key calculations involve the drag force equation and the force exerted by the piston, with the paddle expected to break when these forces exceed its yield strength. The conversation emphasizes treating the paddle as a beam fixed at one end and considering the bending moment caused by the forces applied. Participants highlight the importance of calculating the bending stress and ensuring it remains below the paddle's yield strength, factoring in safety margins. Overall, understanding the interaction of these forces is crucial for predicting paddle failure.
meanswing
Messages
10
Reaction score
0
Hi Everyone. I am trying to find out when this paddle (in red) would break due to the force of the drag of the water and from the force of the pneumatic cylinder that is pushing it. I took intro to statics but as you can see with the figure the machine is not static. Can someone point me to the right direction.
______________________________________________________________________________
piston.JPG


Here is my attempt:

let the force of the water be the drag force of the water, F_w = C_d*rho*v^2*A_p , where C_d is the coefficient of drag of the paddle, rho is the density of water, v is the velocity of the paddel, and A_p is the cross sectional area of the paddle exposed to the water. The force of the pneumatic piston is, F_p = P*A_b = P*pi*d^2/4 , such that P is the pressure supplied to the cylinder and d is the diameter of the piston bore.

The paddle will break when the applied forces exceeds the paddles yield strength so. Let sigma be the yield strength resulting from the drag force and the force of the piston.

sigma = (F_w + F_p)/A_p

piston.jpg
 
Engineering news on Phys.org
Is the paddle fixed to the piston? There does not appear to be a pivot point. Does one consider the force of gravity, i.e., weight of paddle.

It is static in the sense that one is looking for the maximum force which would cause the maximum stress to exceed yield or critical shear stress if that is the criterion for failure. The force in the water occurs in conjunction with the force in the piston - but the two forces are not colinear, so there is a moment induced during motion of the paddle.
 
You can treat the problem as a beam. Consider the fixed end to be mounted and not moving. Then use the force of the water as you would a variably loaded cantilever beam. That should help you.
 
@ Astronuc , yes the paddle is fixed to the piston with a bracket (yellow) and the weight of the paddle is negligible compared to the forces applied to it. What do you mean there doesn't seem to be a pivot point?

@cstoos Shouldnt i consider the Force from the piston pushing the bracket. Wouldnt that increase maximum force being applied to the paddle?
 
meanswing: The pivot point (or rather, the point about which the paddle would try to rotate, if it could) is the centerpoint of the yellow bracket. The maximum possible force on the paddle, Fw, is Fw = Fp. Therefore, you can use Fp to compute Fw, unless you already have velocity v. You don't include Fp hereafter in the moment summation, because Fp causes no moment about the pivot point. Don't worry too much about shear force. Instead, you need to compute bending moment, M, on the paddle, which is Fw in your second diagram multiplied by the distance from Fw to the yellow bracket. After you obtain M, compute bending stress, sigma = M*c/I. Ensure sigma does not exceed Sty/FSy, where Sty = tensile yield strength, and FSy = yield factor of safety, such as 1.50 or 2.0.
 
Bending Moment is Fw*(2L-11/12L-L/2)=Fw*7L/12, and in your case depend on the velocity.
 
I need some assistance with calculating hp requirements for moving a load. - The 4000lb load is resting on ball bearing rails so friction is effectively zero and will be covered by my added power contingencies. Load: 4000lbs Distance to travel: 10 meters. Time to Travel: 7.5 seconds Need to accelerate the load from a stop to a nominal speed then decelerate coming to a stop. My power delivery method will be a gearmotor driving a gear rack. - I suspect the pinion gear to be about 3-4in in...
Thread 'Calculate minimum RPM to self-balance a CMG on two legs'
Here is a photo of a rough drawing of my apparatus that I have built many times and works. I would like to have a formula to give me the RPM necessary for the gyroscope to balance itself on the two legs (screws). I asked Claude to give me a formula and it gave me the following: Let me calculate the required RPM foreffective stabilization. I'll use the principles of gyroscopicprecession and the moment of inertia. First, let's calculate the keyparameters: 1. Moment of inertia of...
Thread 'Turbocharging carbureted petrol 2 stroke engines'
Hi everyone, online I ve seen some images about 2 stroke carbureted turbo (motorcycle derivation engine). Now.. In the past in this forum some members spoke about turbocharging 2 stroke but not in sufficient detail. The intake and the exhaust are open at the same time and there are no valves like a 4 stroke. But if you search online you can find carbureted 2stroke turbo sled or the Am6 turbo. The question is: Is really possible turbocharge a 2 stroke carburated(NOT EFI)petrol engine and...
Back
Top