1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spindle design for knife gate valve

  1. Nov 28, 2012 #1
    I have to assess the sizing of a spindle of a DN600 Knife gate valve.
    The spindle is subjected to two sources of loading acting in both shear and tensile.
    •Torque applied to the spindle.
    •Resulting thrust from line pressure.

    Known data:
    Required thrust at 2 bar line pressure: F = 53.2 kN
    Corresponding torque: T = 266 Nm
    Material: S30400 (BS 970 PART4 GR 304 S15) with the following assumes properties:
    Minimum ultimate tensile strength: Sut = 505 MPa
    Tensile yield strength: Syt = 205 MPa
    Shear ultimate strength: Ssu = 0.6x505 MPa = 303 MPa
    Shear yield strength: Ssy = 0.577x205 MPa = 118 MPa
    Modulus of elasticity: E =193 GPa

    Spindle thread: M40x7 Trapezoidal thread 2 starts; d_m = 36.5 mm
    Valve stroke: La = 650 mm

    Note: The clearance, fillet radii, surface finish and accuracy of machining have a significant effect on the actual stresses and has not been taken into account in the design calculations.

    I have done strength calculations on the thread and the root of the spindle, I also did some calcs to check the spindle for buckling. So here goes, haha...

    Stresses acting on threads
    Bending stress
    σ_bending=My⁄I
    with M=Ft⁄2
    y=h⁄2
    I=(bh^3)⁄12
    t=0.5p

    This gives:
    σ_bending=3Ft⁄(bh^2 )

    Where thread length is
    b=π*d_m*n

    Assuming only 2 threads take the load and there are 2 starts, n will be equal to 4

    b=π(36.5)4
    b=458.7 mm
    For TR thread
    h=0.634p
    h=0.634(7)
    h=4.438
    Hence
    σ_bending=(3(53200)(3.5))⁄((458.7)(4.438))
    σ_bending=61.83 MPa
    The safety factor against bending is:

    SF_yielding=S_ty⁄σ_bending
    SF_yielding=205⁄61.83
    SF_yielding=3.3

    Maximum shear stress
    Because bending is involved, the shear stress distribution is not uniform; it occurs at the centre of the thread and is 50% higher than the average stress. At this point the bending stress is zero because maximum bending stress occurs at the outer fibres.
    Tau_max=1.5F⁄bh

    With
    Tau_max=(1.5(53200))⁄((458.7)*4.438)
    Tau_max=39.2 MPa

    The safety factor against shear is:
    SF_yielding=S_sy⁄Tau_max
    SF_yielding=118.3⁄39.2
    SF_yielding=3.02

    Stresses acting on spindle root
    Axial compression stress in root

    σ_x=F⁄A_t

    Tensile stress area
    A_t=π⁄4 (D-0.938194p)^2
    A_t=π⁄4 (40-0.938194(7))^2
    A_t=878 mm^2

    Hence compression stress
    σ_x=((-53200))⁄878
    σ_x=-60.6 MPa

    Shear stress:
    Polar moment of inertia for spindle root
    J_spindle=(π*D^4)⁄32
    J_spindle=π(33)^4⁄32
    J_spindle=116427 mm^4

    Resultant torsional shear stress on shaft:
    Tau_torsional=T(D⁄2)⁄J_total
    Tau_torsional=((266×1000)(33))⁄(2(116427))
    Tau_torsional=37.7 MPa

    Combined stresses on spindle:
    σ_x = -60.6 MPa and Tau_xy=37.7 MPa
    From the maximum distortion energy theory and Mohr’s Circle we have

    Tau_maxinplane=sqrt(((σ_x-σ_y)⁄2)^2+Tau_xy^2 )
    σ_1=((σ_x-σ_y)⁄2)±sqrt(((σ_x-σ_y)⁄2)^2+Tau_xy^2 )

    Hence
    Tau_maxinplane)=sqrt((((-60.6))⁄2)^2+(37.7)^2 )
    Tau_maxinplane)=48.36 MPa

    And principal stresses
    σ_max=(((-60.6))⁄2)+sqrt((((-60.6)⁄2)^2+〖(37.7)〗^2 )
    σ_max=18.06 MPa

    σ_min=(((-60.6))⁄2)-sqrt((((-60.6)⁄2)^2+〖(37.7)〗^2 )
    σ_min=-78.66 MPa


    Shear safety factor:
    SF_ys=S_ys⁄Tau_maxinplane)
    SF_ys=118.3⁄48.36
    SF_ys=2.45

    Tension safety factor:
    SF_yt=S_yt⁄σ_max
    SF_yt=205⁄18.06
    SF_yt=11.35

    Compression safety factor:
    SF_yt=S_yt⁄σ_min
    SF_yt=205⁄78.66
    SF_yt=2.6  

    Buckling calculations
    Because the load in the spindle is compressive when the blade of the valve bottoms out, the spindle is effectively a column and the possibility of buckling should be checked.

    We assume that the spindle is flexible (pinned to blade) at one point and rigid at the other point. As per Mechanical Design Manual February 1999, the relationship between effective length Le and actual length La should be Le: 0.85La.

    Slenderness ratio
    S=L_e⁄k
    Where
    L_e=0.85L_a
    L_e=0.85(650)
    L_e=552 mm
    Radius of gyration for circular shafts
    k=d_m⁄4
    k=36.5⁄4
    k=9.125
    Hence
    S=((552))⁄((9.125) )
    S=60.5

    This value is less than the limiting slenderness ratio for steel, the Johnson formula applies for the critical buckling load.

    F_cr=A_t *S_ty [1-(S_ty*S^2)⁄(4*π^2*E)]
    F_cr=(878)(205)[1-((205)(60.5)^2)⁄(4π^2 (193×10^3))]
    F_cr=162.2 kN

    Buckling safety factor:
    SF_buckling=F_cr⁄F_thrust
    SF_buckling=162.2⁄53.2
    SF_buckling=3.05

    In this knife gate valve the blade bottoms out on the body casting, so I recommended that if the company are to fit an electric actuator that they make sure the torque settings and sizing of the actuator is set to the correct values to make sure that if the stroke has not been set right that the actuator does not exceed the maximum torque allowed without buckling...

    From the calculations the spindle should be adequate to transfer the required torque and resulting thrust with a minimum safety factor of 2.45.

    I know it is a lot to check but is someone would please help me check this it would be much apreciated, I am only in my 2nd year of working so would like some experienced input.

    Thanks guys/girls
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Spindle design for knife gate valve
  1. Magnet valve (Replies: 3)

  2. How to design a valve? (Replies: 2)

  3. Controllable Valve? (Replies: 10)

Loading...