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Screw Jack Design
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Self-locking & overhauling of screw

$T =W\tan(\phi-\alpha).\frac{dm}{2}$ --------------(1)

Where $T= \text{torque required to lower the screw}$

$\phi$ = angle of friction

$\alpha$ = helix angle

$W = \text{load to be lifted}$

$dm = \text{mean diameter}$

The torque required to lower the screw is given in equation (1). In equation (1) if $\phi \lt \alpha$ then the torque required to lower the load will be negative (-ve). In other words, load will start moving downward without application of any torque such condition is known as overhauling of the screw

If $\phi \gt \alpha$, then torque required to lower the load will be positive (+ve) indicating that an effort is applied to lower the load. Such a screw is called self-locking screw.

Q. Design a screw jack to lift load of $250\ KN$ and having a maximum lift of $270\ mm$ select proper material and draw a proportional sketch

Selection of material for screw

The screw is under compression due to steady load but undergoes tension when rotated. It is subjected to buckling also when the load is raised through the max lift. Therefore from strength consideration, let us select steel as the material of the screw.

Let us take C40

Selection of FOS

As screw is subjected to both compressive as well as torsional shear stress and to account for high-stress concentration in threads.

Hence consider higher FOS

Take $n = 4$ based on $\sigma_y$ value

Permissible stresses for screw From PSG 1.9

Take $\sigma_y = 330 \ N/mm^2$

$\sigma_t = \sigma_c=\frac{\sigma_y}{FOS}= \frac{330}{4} = 82.5\ N/mm^2$

Take $\sigma_c =82\ N/mm^2$

By maximum shear stress theory

$e =\frac{0.5 \times \sigma_y}{FOS}$

$e = 41.25\ N/mm^2$

Take $e=41\ N/mm^2$

Selection of material for nut

Considering nut material like cast iron which is easy to manufacture and cheap C.I also wear rest.

Taking Grey C.I. i.e. GCI 20

Form PSG 1.4

$\sigma _4 =200\ N/mm^2$

Selection of FOS

As the material is brittle and to account for high-stress concentration threads, consider high FOS=6 based on $\sigma_u$

Permissible stress for nut

$\sigma _t=\frac{\sigma _u}{FOS} = \frac{200}{6} = 33.33\ N/mm^2$

Considering $e = \sigma _t = 33\ N/mm^2$

As compressive strength of C.I. is more than tension strength consider $\sigma _c= 2\sigma _t=66\ N/mm^2$

Bearing stress selection

from PSG 7.8.7

From screw and C.I. nut combination

$\sigma _{\text{br}}=80\ Kgf/cm^2$

$= 80 \times 10\ N/cm^2$

$= 800 \times 10^{-2} \ N/mm^2$

$= 8\ N/mm^2$

Design of screw

Consider the compressive failure of the screw

$\sigma_c = \frac{W}{A_c}$

$A_c = \frac{2.50\times10^3}{82}$

$A_c =3.0487\times10^3\ mm^2$

$A_c =\frac{\pi}{4}d_c^2$ where $d_c = 62.3\ mm$

Above diameter is based on only direct compressive stress but the actual screw is also subjected to torsional shear stress. Hence to take into account the effect of torsional shear stress take a diameter 30% greater than the above value

$\therefore d_c = 1.3\times 62.3 = 80.9 \ mm$

$\therefore A_c = \frac{\pi}{4}d_c^2= 5151.72 \ mm^2$

from PSG 5.7 and 5.71 selecting normal series we can provide a standard screw of the cover area $A_c$

Take $A_c =5411 \ mm^2$

nominal diameter

$d_o = 95 \ mm$

cover dia $d_c = 23 \ mm$

pitch dia $p =12 \ mm$

Mean dia $d_m$ = $\frac{d_0+d_c}{2} =39 \ mm$

helix angle is selected by a single shot

$\tan(\alpha)$ = $\frac{pitch}{\pi d_m}$

$\therefore \alpha = \tan^{-1} (\frac{12}{\pi\times89})$

$\alpha = 2.45\degree$

Selecting the angle of friction

$\phi$ =$6 \ to \ 8\degree$ from PSG 7.87 as $\phi$ >$\alpha$, self-locking will be provided

$\therefore$ frictional torque acting on the screw, $T_1 = \frac {P\times d_m}{2}=w tan(\phi+\alpha)\frac{d_m}{2}$

$= 250 \times 10^3 \tan (6+2.45)\frac{89}{2}$

$T_1=1.65\times 10^6 \ N-mm$

$\therefore$ shear stress developed in screw '

$e = \frac{16T_1}{\pi d_c^3}$

as $\frac{T}{I_p}=\frac{e}{r}$

$I_p=\frac{\pi}{64}{d^4}+\frac{\pi}{64}{d^4}=\frac{\pi}{32}{d^4}$

$e = \frac{16 \times 1.65 \times 10^6}{\pi \times 83^2}$

$e= 14.89\ N/mm^2$

$\therefore$ the maximum principal stress in screw

$\sigma_{\text{cmax}}$ = $\frac{1}{2} [\sigma_c +\sqrt{(\sigma_c^2+4e^2)}]$

where $\sigma_c$ is the direct compressive stress developed in the screw

$\therefore$ $\sigma_c = \frac{w}{A_c}$ = $\frac{250 \times 10^3}{5411}$

$\sigma_c=46.20\ N/mm^2$

$\therefore \sigma_{cmax} = \frac{1}{2}[46.20+\sqrt{(46.20)^2+4(14.69)^2}]$

$\sigma_{\text{cmax}} = 50.47\ N/mm^2$

$\sigma_{\text{cmax}}=50.47 \lt \sigma_c=82\ N/mm^2$

$\therefore$ screw is safe in compressive stress Maximum shear stress developed in the screw,

$\tau_{\text{max}}$ = $\frac{1}{2}[\sqrt{(\sigma_c^2+4\tau^2)}]$

$\tau_{\text{max}}$ = $\frac{1}{2}[\sqrt{((46.2)^2+4(14.69)^2)}]$

$\tau_{\text{max}} = 27.37\ N/mm^2 \lt \tau = 41\ N/mm^2$

$\therefore$ screw is safe in shear stress

Let us check screw for buckling

Assume the nut is a solid

from PSG 7.87, for solid nut

Ratio $\frac{H}{d_p}$= 2 where H = 2$\times d_m=2\times 89=178\ mm$

Length of the screw for buckling purpose i.e. unsupport length of screw

$\therefore L =$ lifting height $+ H/2$

$L=270+178/2 = 359 \ mm$

Considering screw as a short column whose one end is fixed to nut and other end is free

The end fixing coefficient is $n= 0.25$ from PSG 6.8 using Johnson's parabolic formula,

$P_c= a_c \sigma_y[1-\frac{ \sigma_y}{4n\pi ^2E}(\frac{t}{k})^2]$

where $K =$ radius of gyration

$K = 0.25 \times d_c = 0.25 \times 83 = 20.75\ mm$

$E= 2.1 \times 10^5 N/mm^2$

$\therefore P_c = 5411 \times 330[1-\frac{330}{4(0.25)\pi^2\times 2.1\times 10^5}(\frac{359}{20.75})^2]$

$P_c = 1.7\times 10^6\ N \gt 250 \times 10^3 \ N$

As value of buckling load is greater than the load at which screw is designed so there is no chance of screw to buckle hence it is safe.

Design of nut

We know nominal dia of screw is $d_0= 95 \ mm$ Taking $d = d_0 + 10 \ mm = 105\ mm$

Considering crushing failure of nut column $W = \frac{\pi}{4}(D^2-d^2)\times \sigma_E$

$D =125.88\ mm$

Let n be the no of threads in engagement

$\therefore$ height of the nut ,

$H = pitch \times n$

$178 = p \times n = 12 \times n$

$n =14.83$ take $n= 15$

Let us check bearing stress developed in nut threads

W = $\frac{\pi}{4}(d_0^2-d_c^2 ) \times n \times \sigma _{\text{br}}$

$250 \times 10^3 = \frac{\pi}{4}(95^2-83^2 ) \times 15 \times \sigma _{\text{br}}$

$\therefore$ Nut threads fails in bearing stress.

$\therefore$ Increase no of threads in engagement

Taking $n=20$

W = $\frac{\pi}{4}(d_0^2-d_c^2 ) \times 20 \times \sigma _{\text{br}}$

$\therefore \sigma_{br}=7.45\ N/mm^2 \lt \sigma_{br}=8\ N/mm^2$

$\therefore$ nut threads are safe

Let us check shear stress developed in nut threads $\therefore$ W = $π× d_0×t_1×n×T$ induced

$t_1 = \frac{pitch}{2}$

$\therefore$ 250×$10^3$ = π×95×6×20×$T_{\text{induced}}$

$T_{\text{induced}}$=6.98 < $T_{\text{nut}}$ =33 $\therefore$ nuts and threads are safe

Let us check shear stress developed in the nut threads

$\therefore W = π \times d_0\times t_1\times n \times T_{\text{induced}}$

$t_1 =\frac{pitch}{2}= 12/2 =6mm$

$\therefore 250×10^3 = π×6×20×T_{\text{induced}}$

$T_{\text{induced}}=6.98\ N/mm^2\ltT_{\text{nut}}=33$

$\therefore$ nuts and threads are safe in shear.

Now calculate 't’ by considering the shear failure of nut column

$W= π×d×t×T$

$250\times 10^3=\pi \times 105 \times \ t \times \ 33$

$t=22.93 \ mm$

Take $t=23\ mm$

Height of nut $H = pitch×n$ $H = 240\ mm$

Design of handle

The diameter of the head $(D_1)$ on top of the screw is taken 1.5 to 1.7 of $d_0$

$\therefore D_1$ = 1.6× $d_0$ $=1.6×95$ $= 152\ mm$ $D_2 = 0.5 D_1$ $= 76mm$

Then find the torque required to overcome friction at the top of the screw assuming uniform pressing

$T_2 = \frac{2}{3} \mu N\frac{R_1^3-R_3^3}{R_1^2-R_2^2}$

$R_1 =D_1/2 =152/2=76$

$R_2 =D_2/2=76/2=38$

Assuming coefficient of friction $\mu = tan\phi$

$\mu = 0.1$

$T_2 = \frac{2}{3} \mu N\frac{R_1^3-R_3^3}{R_1^3-R_2^3}$

$T_2=\frac{2}{3} \times 0.1 \times 250 \times 10^3 [\frac{76^3 - 38^3}{76^2 - 38^2}]$

$= 1.47×10^6 \ N-mm$

$\therefore$ Total torque to lift load

$T=T_1+T_2 = 1.65 \times 10^6 +1.47 \times 10^6$ = $3.12×10^6\ N-mm$

Let $L_n$ be the length of the handle $T = F \times L_n$ $L_n = T / F = 7.8m$

Generally length of handle is 1m

Maximum bending moment will be $M = \text{force} \times {Length} =400×1000$

Consider material of handle C40 and FOS 2 $\sigma_c = \sigma_{\text{bending}}= \frac{\sigma_y}{FOS}=330/2=165\ N/mm^2$

$165 = \frac{400×10^3}{\frac{π}{32}×d_h^3}$

$d_h = 29.12\ mm$