SCREW J3010/5/1 UNIT 5 SCREW OBJECTIVESGeneral Objective : To understand the concept of the working of a screw.Specific Objectives : At the end of this unit you should be able to : state the terms that are important for the study of screw. match the principle of screw working to that of friction on an inclined plane. use these suitable concepts to solve problem involve. calculate the answer using these concepts correctly.
SCREW J3010/5/2 INPUT 5.0 INTRODUCTION The screws, bolt, studs, nuts etc. are widely used in various machines and structures for fastenings. These fastenings have screw threads, which are made by cutting a continuous helical groove on cylindrical surface. If the threads are cut on the outer surface of a solid rod, these are known as external threads. But if the threads are cut on the internal surface of a hollow rod, these are known as internal threads. We use The principle of the inclined plane in the screw thread concept/applications. 5.1 THE SQUARE-THREADED SCREW (a) A Single-Start Thread Let W be the axial force against which the screw is turned, Fig. 5.1, and P the tangential force at the mean thread radius necessary to turn the nut. The development of a thread ia an inclined plane, Fig. 5.2, and turning the nut against the load is equivalent to moving this load up the plane by the horizontol force P applied at the mean radius of the thread.
SCREW J3010/5/3 Fig. 5.1 Fig. 5.2 The angle α, p Tan α = πD (b) A Two-Start Thread For a two-start thread the distance moves axially by the screw in its nut in one revolution is the lead , which is twice the pitch. Tan α = πD 2p Tan α = πD
SCREW J3010/5/4 5.2 THE VEE-THREADED SCREW For a V-thread, the normal force between the nut and the screw is increased since the axial component of this force must equal W. Thus, if the semi-angle of the thread is β, Fig. 5.3., then normal force = W sec β . The friction force is therefore increased in the ratio sec β : 1, so that the V-thread is equivalent to a square thread having a coefficient of friction µv sec β Fig. 5.3
SCREW J3010/5/5 INPUT Turning the screw is equivalent to moving a mass of weight W along the inclined plane by a horizontal force P. 5.3 RAISING LOAD When the load is raised by the force P the motion is up the plane.(Fig.5.4). Motion R R α +φ P W α α φ P W Fig. 5.4 From the triangle of forces: P Tan (α + φ) = W P = W tan (α + φ)
SCREW J3010/5/6 Torque The torque T required to rotate the screw against the load is: 1 T = P( D) 2 +O 1 T = PD 2 1 T = WD tan (α + φ) 2 1 P D 2 Fig. 5.5: Cross-section of Screw Efficiency The efficiency of the screw is equal to: Forward efficiency : ηforward = force P required without friction (φ = 0) force P required with friction w tan α η = w tan(α + φ ) tan α η= tan(α + φ ) or, η = work done on load W in 1 revolution work done by P in 1 revolution w η= P (πD) w η= x P πD
SCREW J3010/5/7 But, w 1 = and tan α = P tan(α + φ ) πD tan α η= tan(α + φ ) Maximum Efficiency 1 − sin φ ηmak = 1 + sin φ Example 5.1 The helix angle of a screw tread is 10˚. If the coefficient of friction is 0.3 and the mean diameter of the square thread is 72.5 mm, calculate: (a) the pitch of the tread, (b) the efficiency when raising a load of 1 kN, (c) the torque required. Solution 5.1 Given: α = 10˚ μ = 0.3 Dmean = 72.5 mm p (a) tan α = πD p = tan 10˚ (3.14)(72.5) = 40.1 mm
SCREW J3010/5/8 (b) μ = tan α α = tan-1 0.3 α = 16.7˚ tan α Efficiency for raise load, η= tan(α + φ ) tan 10 0 η= tan(10 0 + 16.7 0 ) 0.1763 η= 0.5027 η = 0.35 x 100% η = 35% 1 (c) Torque, T = WD tan (α + φ) 2 1 T = (1000)(72.5) tan (10˚ + 16.7˚) 2 1 T = (1000)(72.5)(0.5027) 2 T = 18.22 x 103 kN mm T = 18.22 Nm
SCREW J3010/5/9 5.4 LOAD BEING LOWERED (i) When α > φ P is applied to resist the downward movement of the load. Under this condition the load would just about to move downwards. If P were not applied in this direction the load would overhaul, that is move down on its own weight. R P P Motion α α R α-φ W W Fig. 5.6 From the triangle of forces: P Tan (α - φ) = W P = W tan (α - φ) Efficiency Downward efficiency : ηdownward = work done by P in 1 revolution work done on load W in 1 revolution P(πD) η= w tan(α − φ ) η= tan α
SCREW J3010/5/10 Example 5.2 Calculate the pitch of a single-start square threaded screw of a jack which will just allow the load to fall uniformly under its own weight. The mean diameter of the threads is 8 cm and μ= 0.08. If the pitch is 15 mm, what is the torque required to lower a load of 3 kN? Solution 5.2 If the load is to fall uniformly under its own weight, α = φ Given : μ = 0.08 μ = tan φ φ = 4.57˚ p Tan α = then, α = φ πD p Tan 4.57˚ = (3.14)(8) p = 20 mm p If the pitch is 15 mm, Tan α = πD 15 Tan α = (3.14)(8) Tan α = 5.97 α = 80.49˚ 1 Torque, T = WD tan (α - φ) 2 1 T = (3000)(8)tan (80.49˚- 4.57˚) 2 1 T = (3000)(8)(tan 75˚ 92’) 2 T = 2.4 Nm
SCREW J3010/5/11 (ii) When φ > α If the angle of friction is greater than the angle of the plane, then it can be seen from the triangle of force, that force P must be applied to help lower the load. R Motion α φ-α R φ-α W P α α P W Fig. 5.7 From the triangle of forces: P Tan (φ - α ) = W P = W tan ( φ - α ) Efficiency tan(φ − α ) ηterbalik = tan α
SCREW J3010/5/12 Example 5.3 The mean diameter of a square-threaded screw jack is 5 cm. The pitch of the thread is 1 cm. The coefficient of friction is 0.15. What force must be applied at the end of a 70 cm long lever, which is perpendicular to the longitudinal axis of the screw to lower a load of 2 tonnes? Solution 5.3 Given: Mean diameter, d = 5 cm Pitch, p = 1 cm Coefficient of friction, μ = 0.15 = tan φ Then, φ = 8.32˚ Length of lever, L = 70 cm Load W = 2 t = 2000 kg Force required to lower the load, P = W tan (φ – α) = 2000 tan (8. 32˚-3.39˚) = 2000 tan 4. 53˚ = 2000 x 0.0854 = 170.8 kg
SCREW J3010/5/13 Activity 5A TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 5.1 A nut on a single-start square thread bolt is locked tight by a torque of 6 Nm. The thread pitch is 5 mm and the mean diameter 6 cm. Calculate : (a) the axial load on the screw in kilogrammes (b) the torque required to loosen the nut. μ = 0.1. 5.2 The total mass of the table of a planning machine and its attached work piece is 350 kg. The table is traversed by a single-start square thread of external diameter 45 mm and pitch 10 mm. The pressure of the cutting is 600 N and the speed of cutting is 6 meters per minute. The coefficient of friction for the table is 0.1 and for the screw thread is 0.08. Find the power required. 5.3 The mean radius of a screw of a square thread screw jack is 2.5 cm. The pitch of the thread is 7.5 mm. If the coefficient of friction is 0.12, what effort applied at the end of a lever 60 cm length is needed to raise a weight of 2000 kg?. 5.4 A screw press is used to compress books. The thread is double thread (square head) with a pitch of 4 mm and a mean radius of 25 mm. The coefficient of friction for the contact surfaces of the threads is 0.3. Determine the torque required for a pressure of 500 N.
SCREW J3010/5/14 Feedback to Activity 5A Have you tried the questions????? If “YES”, check your answers now 5.1 161 kg, 3.47 Nm 5.2 191 W 5.3 14.07 kg 5.4 4450 Nmm
SCREW J3010/5/15 INPUT 5.5 OVERHAULING OF THE SCREW When the load moves down and overcomes the thread friction by its own weight, it is said to overhaul. When it moves down against a resisting force P we found that P = W tan ( α - φ ) When the load overhauls, P = 0 Therefore, tan ( α - φ ) = 0 α-φ =0 α=φ Hence, when the angle of the inclined plane is equal to the angle of the friction the load will overhaul. The efficiency of such a screw when raising a load is given by: tan α η= tan(α + φ ) but α = φ tan α η= tan 2α
SCREW J3010/5/16 Example 5.4 A load of 6 kN is lifted by a jack having a single-start square thread screw of 45 mm core diameter and pitch 10 mm. The load revolves with the screw and the coefficient of friction at the screw thread is 0.05. Find the torque required to lift the load. Show that the load will overhaul. Solution 5.4 Given: 1 W = 6000 N D mean = 45 + x 10 = 50 mm μ = 0.05 p = 10 2 mm Then, p tan φ = μ tan α = πD 10 tan φ = 0.05 tan α = (3.14)(50) φ = 2.86˚ tan α = 0.0637 α = 3.64˚ 1 Torque required, T = WD tan (α + φ) 2 1 = (6000)(50) tan (3.64˚ + 2.86˚) 2 = 3000(50) tan (6.50˚) = 3000(50) (0.114) = 17.1 Nm tan α Efficiency, η= tan(α + φ ) tan 3.64 0 η= tan(3.64 0 + 2.86 0 ) 0.0636 η= 0.114 η = 0.558
SCREW J3010/5/17 η = 56% > 50% (overhauling) Activity 5B TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 5.5 The drum of a windlass is 10 cm in diameter and the effort is applied to the handle 60 cm from the axis. Find the effort necessary to support a weight of mass 120 kg. 5.6 A wheel and axle is used to raise a load of mass 50 kg. The radius of the wheel is 50 cm and while it makes seven revolutions the load rises 3.3 m. What is the smallest force that will support the load?. 5.7 A table carrying a machine tool is traversed by a three-start screw of 6 mm pitch and external diameter 23 mm. The mass of the table is 200 kg and the coefficient of friction between the table and its guides is 0.1. The screw is driven by a motor at 12 rev/s. Find the speed of operation of the tool, the power required and the coefficient of friction for the thread if the efficiency is 70 percent. 5.8 A differential pulley, the two parts of which have respectively twenty four and twenty five teeth, is used to raise a weight of mass 500 kg. Show by sketch how the apparatus is used, and determine its velocity ratio. Find also what effort must be exerted if the efficiency is 60 percent.
SCREW J3010/5/18 Feedback To Activity 5B Have you tried the questions????? If “YES”, check your answers now 5.5 10 gN 1 5.6 7 gN 2 5.7 12.96 m/min, 60.5 W, 0.11 2 5.8 50, 16 gN. 3
SCREW J3010/5/19 SELF-ASSESSMENT 5 You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback on Self-Assessment 5 given on the next page. If you face any problems, discuss it with your lecturer. Good luck. 1. Find the torque to raise a load of 6 kN using a screw jack with a double-start square thread, containing two threads per centimeter and a mean diameter of 60 mm. μ = 0.12. What is the torque required to lower the load? 2. A lathe saddle of mass 30 kg is traversed by a single-start square-thread screw of 10 mm pitch and mean diameter 40 mm. If the vertical force on the cutting tool is 250 N, find the torque at the screw required to traverse the saddle. The coefficient of friction between saddle and lathe bed and for the screw thread is 0.15. 3. A double-start square-thread screw has a pitch of 20 mm and a mean diameter of 100 mm, µ = 0.03. Calculate its efficiency when raising a load. 4. A load of 2500 N is to be raised by a screw jack with mean diameter of 75 mm and pitch of 12 mm. Find the efficiency of the screw jack, if the coefficient of friction between the screw and nut is 0.075.
SCREW J3010/5/20 Feedback to Self-Assessment 5 Have you tried the questions????? If “YES”, check your answers now 1. 31.4 Nm , 12 Nm 2. 0.376 Nm 3. 80.5 % CONGRATULATIONS!!!! …..May success be with 4. 41.5 % you always….