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Strength calculations of threaded spindle.

by WillemBouwer
Tags: calculations, spindle, strength, threaded
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WillemBouwer
#1
Nov27-12, 05:21 AM
P: 82
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
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WillemBouwer
#2
Nov28-12, 02:47 AM
P: 82
Is there any person who can check this please, quite urgent. Thanks


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