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

Shear valve Design. Wall thickness at break point.

  1. Jan 9, 2012 #1
    Hi.

    Situation: A pipe with a cut out so it will break when a shear force is exerted at the top of the pipe. The break point is at the cut out and the the force is impact in lbs.

    What is the wall thickness(or cross sectional area) at the cut out point?

    I know i may be missing variables..

    Thank you for reading
     
  2. jcsd
  3. Jan 9, 2012 #2
    I've decided to try it without knowing if it'll work. I decided to use the modulus of rigidity which is in force per area. S=F/A I find the area and say its my cross sectional area. I have a given force, pipe inner radius, and want to find the cross sectional area of the break point. I used this equation A=pi(r+w.t.)^2-pi(r)^2 which can be arrange for wall thickness w.t.=sqroot(A*pi). I keep getting a very low wall thickness(.013 in). Is this calculation right?
     
  4. Jan 30, 2012 #3
    ** re written **

    FBD (probably useless)
    Pipe lengthwise cross section
    | | | |
    | | | |
    > | | < <-shear point, distance tip of cuts is the outer diameter (D_o)
    | | | |
    | | | |

    Problem

    A shear pipe is is intended to break while going under 400-600 lb-feet bending moment.
    What is the outside diameter of the shear point to break under this load?

    Variables
    Stainless steel 304
    Modulus of Rigidity(G)= 11e3 ksi
    Inner Diameter of pipe (D_i) = 1 in
    yield stress of 304 SS (σ_y) = 30,000 psi
    let Moment (M) = 500 lb-feet = 6000 lb-in
    Inertia(I) = ?
    Outside diameter at shear section(D_o)

    Formula
    Bending normal stress = σ_y= (M*(D_o/2))/I
    Inertia of pipe = I=(pi/64)(D_o^4 - D_i^4)

    Solve Attempt
    plugged in for I
    I = (pi/64)(D_o^4) - (pi/64)

    rearrange σ_y
    D_o = (2*σ_y*I)/M
    D_o = (2*(30000psi)*((pi/64)(D_o^4)-(pi/64)))/(6000 lb-in)
    rearranged and solved
    0= (10*pi/64)(D_o^4) - D_o - (10*pi/64)

    solve for D_o
    equation solver
    D_o= 2.44601217,-0.4088289

    didn't show but used ultimate strength instead of yield and got 25 and 0. way worse

    I'm looking for a number in between 1 and 1.5 in.
    I thought this calculation was the real deal.

    A mistake or bad equations?
    Is there an equation to figure out the outside diameter?

    Thank you for reading
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Shear valve Design. Wall thickness at break point.
  1. How to design a valve? (Replies: 2)

Loading...