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J3010 Unit 5

on Feb 22, 2010

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J3010 - Mechanics of Machines 1

J3010 - Mechanics of Machines 1

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J3010 Unit 5Document Transcript

• SCREW J3010/5/1 UNIT 5 SCREW OBJECTIVES General 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 J3010/5/3 Fig. 5.1 Fig. 5.2 The angle , p Tan  = D (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  = D 2p Tan  = D
• 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 , Fig. 5.3., then normal force = W sec  . The friction force is therefore increased in the ratio sec  : 1, so that the V-thread is equivalent to a square thread having a coefficient of friction v sec  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 P  + W    P W Fig. 5.4 From the triangle of forces: P Tan ( + ) = W P = W tan ( + )
• 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 ( + ) P 2 1 D 2 Fig. 5.5: Cross-section of Screw Efficiency The efficiency of the screw is equal to: Forward efficiency : forward = force P required without friction ( = 0) force P required with friction w tan  = w tan(   ) tan = tan(   ) or,  = work done on load W in 1 revolution work done by P in 1 revolution w = P(D) w  = x P D
• SCREW J3010/5/7 But, w 1   and tan  P tan(   ) D tan = tan(   ) Maximum Efficiency 1  sin  mak = 1  sin  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 = D 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 Efficiency for raise load, = tan(   ) tan10 0 = tan(10 0  16.7 0 ) 0.1763 = 0.5027  = 0.35 x 100%  = 35% 1 (c) Torque, T = WD tan ( + ) 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  >  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   R - W W Fig. 5.6 From the triangle of forces: P Tan ( - ) = W P = W tan ( - ) Efficiency Downward efficiency : downward = work done by P in 1 revolution work done on load W in 1 revolution P(D) = w tan(   ) = tan
• 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,  =  Given : μ = 0.08 μ = tan   = 4.57˚ p Tan  = then,  =  D p Tan 4.57˚ = (3.14)(8) p = 20 mm p If the pitch is 15 mm, Tan  = D 15 Tan  = (3.14)(8) Tan  = 5.97  = 80.49˚ 1 Torque, T = WD tan ( - ) 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  >  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  - R - W P   P W Fig. 5.7 From the triangle of forces: P Tan ( -  ) = W P = W tan (  -  ) Efficiency tan(   ) terbalik = tan
• 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  Then,  = 8.32˚ Length of lever, L = 70 cm Load W = 2 t = 2000 kg Force required to lower the load, P = W tan ( – α) = 2000 tan (8. 32˚-3.39˚) = 2000 tan 4. 53˚ = 2000 x 0.0854 = 170.8 kg