Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Shear Force

  1. May 19, 2010 #1
    Hi Everyone. Im 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
     
  2. jcsd
  3. May 19, 2010 #2

    Astronuc

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  4. May 19, 2010 #3
    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.
     
  5. May 19, 2010 #4
    @ 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 doesnt 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?
     
  6. May 24, 2010 #5

    nvn

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  7. Sep 2, 2010 #6
    Bending Moment is Fw*(2L-11/12L-L/2)=Fw*7L/12, and in your case depend on the velocity.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook