Bursting pressure test setup

I sized the power screw, frame and handle for a hydrostatic pipe test setup and built the assembly in SOLIDWORKS.

May 2024

Calculations

Design requirements

I selected a nominal diameter of 64 mm and a 6 mm pitch for the single-start screw. The axial load was based on the maximum pipe diameter.

\[F=p\frac{\pi D^2}{4}=10^6\frac{\pi(0.225)^2}{4}=39.761\ \mathrm{kN}\]

\[\begin{aligned}d_m&=d-\frac{p_s}{2}=61\ \mathrm{mm}\\d_r&=d-p_s=58\ \mathrm{mm}\\l&=n_s p_s=6\ \mathrm{mm}\end{aligned}\]

I assumed a thread friction coefficient of 0.125. The torque calculation considered thread friction.

\[T_R=\frac{Fd_m}{2}\left(\frac{l+\pi f d_m}{\pi d_m-fl}\right)=190.30\ \mathrm{N\,m}\]

\[\eta=\frac{Fl}{2\pi T_R}=0.1995=19.95\%\]

\(p\): fluid pressure; \(p_s\): thread pitch; \(d_m\): mean diameter; \(d_r\): root diameter; \(l\): lead.

I included torsion, axial compression and thread-root bending, assuming that one thread carried 38% of the axial load.

\[\begin{aligned}\tau&=\frac{16T_R}{\pi d_r^3}=4.967\ \mathrm{MPa}\\\sigma_a&=-\frac{4F}{\pi d_r^2}=-15.05\ \mathrm{MPa}\end{aligned}\]

\[\sigma_B=-\frac{2(0.38F)}{\pi d_rp_s}=-27.64\ \mathrm{MPa},\qquad \sigma_b=\frac{6(0.38F)}{\pi d_rp_s}=82.92\ \mathrm{MPa}\]

Using the assumed stress components \(\sigma_x=\sigma_b\), \(\sigma_y=\sigma_a\), \(\sigma_z=0\) and \(\tau_{yz}=\tau\):

\[\sigma_{\mathrm{VM}}=\sqrt{\sigma_b^2+\sigma_a^2-\sigma_b\sigma_a+3\tau^2}=91.78\ \mathrm{MPa}\]

\[n=\frac{S_y}{\sigma_{\mathrm{VM}}}=\frac{390}{91.78}=4.25\]

I treated the base as two load-sharing rectangular members, using \(H=300\) mm and \(b=70\) mm.

\[\begin{aligned}M&=\frac{F}{2}\left(\frac{H}{2}+\frac{h}{2}\right)\\\sigma&=\frac{6M}{bh^2},\qquad n=\frac{S_y}{\sigma}\end{aligned}\]

I selected a 70 × 52 mm section for the base-member calculation.

\[|\sigma_{\max}|=\frac{3pR_0^2}{4t^2},\qquad t=\sqrt{\frac{3npR_0^2}{4S_y}}\]

For the plate-sizing approximation, \(R_0\) represents the loaded radius. I used an 18.58 mm plate thickness in the CAD drawings.

I compared lever lengths from 200 to 400 mm and selected a 350 mm handle.

\[\begin{aligned}F_h&=\frac{T_R}{L}\\M_h&=F_h(L-r_s)\\\sigma_h&=\frac{32M_h}{\pi d_h^3}\end{aligned}\]

\[d_h=\left[\frac{32nT_R(L-r_s)}{\pi L S_y}\right]^{1/3}\]