The problem provides information about a power screw with a diameter of 25 mm and a pitch of 5 mm. It is subjected to a vertical load of 5 kN. The coefficients of friction for the collar and threads are 0.06 and 0.09, respectively. The frictional diameter of the collar is 45 mm. The tasks are to determine: a) thread depth, thread width, pitch diameter, minor diameter, and lead. b) the torque required to raise and lower the load. c) the efficiency during lifting the load.

1. Problem Description

The problem provides information about a power screw with a diameter of 25 mm and a pitch of 5 mm. It is subjected to a vertical load of 5 kN. The coefficients of friction for the collar and threads are 0.06 and 0.09, respectively. The frictional diameter of the collar is 45 mm. The tasks are to determine:

a) thread depth, thread width, pitch diameter, minor diameter, and lead.

b) the torque required to raise and lower the load.

c) the efficiency during lifting the load.

2. Solution Steps

a) Determining thread depth, thread width, pitch diameter, minor diameter, and lead.

Since square threads are used,

Thread depth = Pitch / 2

Thread width = Pitch / 2

$Thread \ Depth = p/2 = 5mm / 2 = 2.5mm$

$Thread \ Width = p/2 = 5mm / 2 = 2.5mm$

Pitch diameter = Major diameter - Thread depth

Pitch diameter = $d - p/2$

$d = 25 \ mm$ (Given)

$Pitch \ Diameter = 25 \ mm - 2.5 \ mm = 22.5 \ mm$

Minor diameter = Major diameter - 2 * Thread depth

Minor diameter = $d - 2*(p/2)$

Minor diameter = $d - p$

$Minor \ Diameter = 25 \ mm - 5 \ mm = 20 \ mm$

Lead = Pitch (for a single-threaded screw)

$Lead = 5 \ mm$

b) Determining the torque required to raise and lower the load.

Torque to raise the load, $T_{raise} = \frac{F d_m}{2} (\frac{l + \pi \mu d_m}{\pi d_m - \mu l}) + F \mu_c \frac{d_c}{2}$

Torque to lower the load, $T_{lower} = \frac{F d_m}{2} (\frac{\pi \mu d_m - l}{\pi d_m + \mu l}) + F \mu_c \frac{d_c}{2}$

Where:

$F = 5 \ kN = 5000 \ N$ (Load)

$d_m = 22.5 \ mm = 0.0225 \ m$ (Pitch diameter)

$l = 5 \ mm = 0.005 \ m$ (Lead)

$\mu = 0.09$ (Coefficient of friction for threads)

$\mu_c = 0.06$ (Coefficient of friction for collar)

$d_c = 45 \ mm = 0.045 \ m$ (Frictional diameter of collar)

$T_{raise} = \frac{5000 * 0.0225}{2} (\frac{0.005 + \pi * 0.09 * 0.0225}{\pi * 0.0225 - 0.09 * 0.005}) + 5000 * 0.06 * \frac{0.045}{2}$

$T_{raise} = 56.25 * (\frac{0.005 + 0.00636}{0.07069 - 0.00045}) + 5000 * 0.06 * 0.0225$

$T_{raise} = 56.25 * (\frac{0.01136}{0.07024}) + 6.75$

$T_{raise} = 56.25 * 0.1617 + 6.75$

$T_{raise} = 9.095 + 6.75 = 15.845 \ Nm$

$T_{lower} = \frac{5000 * 0.0225}{2} (\frac{\pi * 0.09 * 0.0225 - 0.005}{\pi * 0.0225 + 0.09 * 0.005}) + 5000 * 0.06 * \frac{0.045}{2}$

$T_{lower} = 56.25 * (\frac{0.00636 - 0.005}{0.07069 + 0.00045}) + 6.75$

$T_{lower} = 56.25 * (\frac{0.00136}{0.07114}) + 6.75$

$T_{lower} = 56.25 * 0.01912 + 6.75 = 1.075 + 6.75 = 7.825 \ Nm$

c) Determining the efficiency during lifting the load.

Efficiency, $\eta = \frac{F l}{2 \pi T_{raise}} = \frac{Wl}{2 \pi T_{raise}}$

$\eta = \frac{5000 \ N * 0.005 \ m}{2 \pi * 15.845 \ Nm}$

$\eta = \frac{25}{99.566} = 0.251$

$\eta = 25.1 \%$

Alternatively:

$\eta = \frac{l}{\pi d_m tan(\alpha + \phi)} = \frac{lead}{\pi d_m } / (\frac{\pi d_m tan(\alpha + \phi)}{\pi d_m}) = \frac{l}{\pi d_m} / (\frac{l + \pi \mu d_m}{\pi d_m})$

$\eta = \frac{F l}{2 \pi T_{raise}}$ where the torque is to raise the load.

Ideal torque to lift the load = $F l / 2 \pi$

$\eta = \frac{F l / 2 \pi}{T_{raise}} = \frac{F l / 2 \pi}{\frac{F d_m}{2} (\frac{l + \pi \mu d_m}{\pi d_m - \mu l}) + F \mu_c \frac{d_c}{2}}$

$\eta = \frac{F l }{ \pi (\frac{F d_m}{2} (\frac{l + \pi \mu d_m}{\pi d_m - \mu l}) + F \mu_c \frac{d_c}{2})}$

$\eta = \frac{l }{ \pi ( \frac{d_m}{2} (\frac{l + \pi \mu d_m}{\pi d_m - \mu l}) + \mu_c \frac{d_c}{2})}$

If there is no collar friction, then the previous formula would become

$\eta = \frac{l }{ \pi ( \frac{d_m}{2} (\frac{l + \pi \mu d_m}{\pi d_m - \mu l}))}$

$\eta = \frac{2 l }{ \pi d_m (\frac{l + \pi \mu d_m}{\pi d_m - \mu l})}$

$\eta = \frac{2 l (\pi d_m - \mu l)}{ \pi d_m (l + \pi \mu d_m)}$

Without collar friction

$\eta = \frac{2(0.005)(\pi * 0.0225 - 0.09*0.005)}{\pi * 0.0225 (0.005 + \pi * 0.09* 0.0225)} = \frac{0.01(0.07068583 - 0.00045)}{0.07068583 * (0.005 + 0.00636173)} = \frac{0.01(0.07023583)}{0.07068583 * 0.01136173} = \frac{0.0007023583}{0.000803122868} = 0.8745$

With collar friction and using the first efficiency formula

$\eta = \frac{5000 * 0.005}{2 \pi * 15.845} = 0.2512$ or $25.12\%$

3. Final Answer

a) Thread depth: 2.5 mm, Thread width: 2.5 mm, Pitch diameter: 22.5 mm, Minor diameter: 20 mm, Lead: 5 mm

b) Torque to raise the load: 15.845 Nm, Torque to lower the load: 7.825 Nm

c) Efficiency during lifting the load: 25.1%