- #1

WillemBouwer

- 81

- 1

In a 400mm pinch valve I have the driving spindle fastened to the top pinch bar through a M36x4 thread. To secure the thread a drilled a hole and fitted a smaller pin through the spindle to secure it in place on the pinch bar. I want to calculate whether the spindle will still be adequate to transfer the neccesary torque and force requirements with the hole weakening it and whether the spindle will hold without the help of the threads.

Known data:

The valve is rated for 10 bar working pressure; it will be operating at 3.45 bar line pressure.

At 3.45 bar the required thrust to close the valve is specified as 111.2 kN.

Pin material: BS 970 PART 4 GR 420 S37. It is assumed that the following properties apply to the material:

Minimum ultimate tensile strength: Sut = 695 MPa

Tensile yield strength: Syt = 435 MPa

Shear ultimate strength: Sus = 0.7x695 MPa = 486 MPa

Shear yield strength: Sys = 0.577x435 MPa = 251 MPa

Thread driven spindle: M44x7 Trapezoidal thread (Handwheel thread)

Pin diameter: 10 mm

Spindle to pinch bar thread diameter: M36x4 mm

First we need to calculate the amount of torque required to close the valve at 3.45 bar.

Trapezoidal thread: M44x7

Nominal diameter: d_m = 40.5 mm

Thread friction coefficient: u_t = 0.15

Lead: L = 7mm

Thread angle: ϕ = 15°

Torque formula:

T=F d_m/2000×(((L⁄(πd_m )+μ_t⁄cosϕ ))⁄((1-(μ_t.(L⁄(πd_m )))⁄cosϕ ) ))

With

tanα=L⁄(πd_m )

tanα=7⁄(π(40.5))

tanα=0.055017

And

ϕ=(15°×π)⁄180

ϕ=0.2618

Then

T=(111200)40.5/2000×(0.055017+0.15⁄cos(0.2618))⁄(1-(0.15×0.055017⁄cos(0.2618))

T=478 N.m

Add 20% safety factor

T=478 ×1.2

T=583 N.m

Pinned male thread (Spindle to pinch bar connection)

Thread: M36x4

Pin diameter: d = 10 mm

Pitch diameter: Dp = 33.4 mm

We will use the maximum distortion energy theory:

Stresses in pin (double shear):

Shear stress in pin:

τ_pin=F⁄A

With the force being

V=T⁄((Dp⁄1000) )

V=583⁄((33.4⁄1000) )

V=17454 N

hence

τ_pin=17454⁄((π〖(10)〗^2)⁄4)

τ_pin=222.23 MPa

This gives a Safety factor for the pin of:

SF_pin=S_ys⁄τ_pin

SF=251⁄222.23

SF=1.13 without shear yielding taking place assuming that all the forces will be working on the pin and not the threads of the spindle ( Hence this is a worst case scenario SF)

Stresses in spindle: Shear and tensile:

As a result of the pinhole, the spindle is substantially weakened.

Shear stresses:

Polar moment of inertia for shaft

J_spindle=(πD^4)⁄32

With

J_spindle=π(33.4)^4⁄32

J_spindle=122204 mm^4

Polar moment of inertia of square

J_square=(Dd(D^2+d^2))⁄12

J_square=((33.4)(10)(33.4^2+10^2))⁄12

J_square=33838.6 mm^4

Final polar moment of inertia

J_total=88365.4 mm^4

Resultant torsional shear stress on shaft:

τ_torsional=T(D⁄2)⁄J_total

τ_torsional=((583×1000)(33.4))⁄(2(88365.4))

τ_torsional=110.2 MPa

Tensile stresses:

Tensile stress area

A_t=π⁄4 (D-0.938194p)^2

A_t=π⁄4 (36-0.938194(4))^2

A_t=816.72 mm^2

Stress area of thread as reduced by the pin:

A_thread=A_t-A_pin

A_thread=816.72-((π×d^2)⁄4)

A_thread=738.46 mm^2

Tensile stress

σ_tensile=F_thrust⁄A_thread

σ_tensile=(-111200)⁄738.46

σ_tensile=-150.59 MPa

Combined stresses on thread:

σx = -150.59 MPa and τxy=110.2 MPa

From the maximum distortion energy theory and Mohr’s Circle we have

τ_(max_in_plane)=SQRT(((σ_x-σ_y)⁄2)^2+(τ_xy^2))

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

Hence

τ_(max_in_plane)=√((((-150.59))⁄2)^2+〖(110.2)〗^2 )

τ_(max_in_plane)=153.23 MPa

And principal stresses

σ_max=(((-150.59))⁄2)+√((((-150.59)⁄2)^2+(110.2)^2 )

σ_1=77.94 MPa

σ_min=(((-150.59))⁄2)-√((((-150.59)⁄2)^2+(110.2)^2 )

σ_min=-228.53 MPa

Hence with the material properties in mind we get the following safety factors

Shear safety factor:

〖SF〗_ys=S_ys⁄τ_(max_in_plane)

〖SF〗_ys=251⁄153

〖SF〗_ys=1.6 against yield

〖SF〗_us=S_us⁄τ_(max_in_plane)

〖SF〗_ys=486⁄153

〖SF〗_ys=3.2 against rupture

Tension safety factor:

〖SF〗_yt=S_yt⁄σ_max

〖SF〗_ys=435⁄77.94

〖SF〗_ys=5.6 against yield

〖SF〗_us=S_us⁄σ_max

〖SF〗_ys=486⁄77.94

〖SF〗_ys=6.2 against rupture

Compression safety factor:

〖SF〗_yt=S_yt⁄σ_min

〖SF〗_ys=435⁄228.53

〖SF〗_ys=1.9 against yield

〖SF〗_us=S_us⁄σ_min

〖SF〗_ys=486⁄228.53

〖SF〗_ys=2.1 against rupture

From these calcs I think the connection is adequate to transfer the torque with a safety factor in access of 1.6...

This is a lot of equations but thanks for your time to look them over for me, very important job, and seeing that I am only in my second year after graduating I think it is essential to get some help on this, haha, thanks