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Unit1 designof levers

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DME-1_uNIT-1 Mechanical Engineering, SPPU, Pune

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Unit1 designof levers

  1. 1. UNIT 01 LEVERS AND ITS DESIGN •Introduction •Types of levers •Design of Hand lever •Design of Foot Lever •Design of Bell Crank lever 1Dr Somnath Kolgiri SBPCOE, Indapur,Pune.
  2. 2. Introduction • A lever is a rigid rod or bar pivoted at a point and capable for turning about the pivot point called fulcrum. • The levers are used to lift a load with the small effort. • The ratio of lifted load to an effort called mechanical advantages. • The ratio of length of effort arm to the length of load arm is called as leverage. 2 Dr Somnath Kolgiri SBPCOE, Indapur,Pune.
  3. 3. 3 Dr Somnath Kolgiri SBPCOE, Indapur,Pune.
  4. 4. Types of Levers • According to the application of load and effort, the levers are classified as 1) One Arm Lever 2) TwoArm Lever 3) Angular Lever 4) Bell Crank Lever 4 Dr Somnath Kolgiri SBPCOE, Indapur,Pune.
  5. 5. 1. One Arm Lever • The one arm lever is an example of hand lever, foot lever and cranking lever. • It has only one arm and that is effort arm. • This type of lever is used to apply external torque. 5 Dr Somnath Kolgiri SBPCOE, Indapur,Pune.
  6. 6. 2. Two Arm Lever • Depending upon the position of fulcrum pin, load and effort the two arm lever are of three types as explain below. 1) Fig. Shows two arm lever, in which the load arm and effort arm are of equal length. The fulcrum pin is pivoted in between the load and effort arm. The examples of such lever are rocket arm of I.C. engine, beam of abalance, handle of hand pump. These lever have mechanical advantage is equal to one. 6 Dr Somnath Kolgiri SBPCOE, Indapur,Pune.
  7. 7. 2) Fig. shows the two arm lever in which the effort arm is longer than the load arm and mechanical advantages than one. Such types of lever used in boiler safety valve. 7 Dr Somnath Kolgiri SBPCOE, Indapur,Pune.
  8. 8. 3) Fig. shows the two arm lever in which the effort arm is smaller than the load arm and mechanical advantage is less than one. Such types of levers are used in stapler and forceps. 8 Dr Somnath Kolgiri SBPCOE, Indapur,Pune.
  9. 9. Why levers are tapered at the end? • The thickness of lever is kept uniform throughout. • The width of the lever is tapered from boss to the handle because the arm is subjected to varying bending moment which is maximum near the boss and decreases to the end and also for easy gripping of the hand on lever. 9 Dr Somnath Kolgiri SBPCOE, Indapur,Pune.
  10. 10. Cross Section of Levers / Handles Section mod ulus for differentsections 6 th2 For rec tan gular section Z  2 32 For Elliptical sec tion Z  ba  10 Dr Somnath Kolgiri SBPCOE, Indapur,Pune.
  11. 11. Design of Hand Lever 11Dr Somnath Kolgiri
  12. 12. Let, P  Forceof effort applied at the handlein N Le  Effective length of thelever armin mm. l  Overheadlength of the shaft in mm. t  Permissibletensile stressof thelever in N / mm2 s  Permissible shear stress of thelever in N / mm2 1) The maximum effort or force applied by a man may be assume as 300 N to 400 N 12Dr Somnath Kolgiri
  13. 13. 2) Due to the force applied at handle at a length ‘Le’ the shaft is subjected to twisting moment (torque). T  P Le    (1) The diameter of shaft (d) is obtained by considering the shaft under pure torsion. 16 d3    (2)sT    Equating equation (1) & (2), the diameter of shaft (d) is obtained 13Dr Somnath Kolgiri
  14. 14. 3) The diameter of shaft at the center of bearing (d1) is obtained by considering the shaft in combined twisting and bending. Also equivalent twisting momentis e s e 1 l2 M 2 Fromequation (3)and (4),d1 is obtained 16   (3)T  (Pl )2 (PL )2  P  L2 e e e T2 Equivalent twisting moment  T  B.M.  M  Pl and Twisting moment  T  P Le T    d3     (4) 14Dr Somnath Kolgiri
  15. 15. 4) Diameter of the boss of the lever = db=1.6d 4 4 3 kk kk k  d b) for squarekey, w  t a) For rec tan gular key,w  d and t  2 w 5) Length of boss = lb= d or 1.5d 6) Dimensions of key – Let, wk Width of keyin mm tk  Thicknessof keyin mm lk  lb  Length of keyin mmlength of boss 15Dr Somnath Kolgiri
  16. 16. Fromaboveequation wk and tk isobtained crk k 2 2 ii)Considering keyunder crushing failure 2 i)Considering keyunder shear failure duetotorsional moment  d T  l  tk   d T  l  wk k 16Dr Somnath Kolgiri
  17. 17. 7) Dimensions of lever cross section. 2 6 2 2 bt2 bt2 Z ee b e 6P(L  db )  P(L  db ) M P(L  db ) Consider the rec tan gular crosssection of lever. Let,b  width of lever in mm. t  depth orthicknessof thelever in mm. Takingb  2t or3t Thelever is subjected to bendingmoment. Themax imum B.M.ontheleveris takennear theboss.   M  17Dr Somnath Kolgiri
  18. 18. Design of Foot Lever • The foot lever is design and designated in the same way as the hand lever. • Only the difference is that hand is replace by foot plate. • The force exerted by a single person by foot is taken as 600 N to 800 N. 18Dr Somnath Kolgiri
  19. 19. 19Dr Somnath Kolgiri
  20. 20. Problems 1) A foot lever is 1 m long from centre of shaftto point of application of 800 n load. Find i) Diameter of shaft ii) Dimensions of key iii) Dimensions of rectangular arm of foot lever at 60 mm from the centre of shaft assuming width of arm as 3 times thickness. Allowable tensile stress may be taken as 73 N/mm2 and shear stress as 70N/mm2. 2) Draw a neat labelled diagram of a hand lever and state how diameter of shaft and boss is calculated. 20Dr Somnath Kolgiri
  21. 21. 3) A foot lever is 1m from the centre of the shaftto the point of application of 800 N load. Find diameter of shaft if shear stress is 70 MPa,. (4M) Weknowthat,torqueonthelever is T  P L  8001000 T  800103 Nmm Also weknowthat, 16 800103   70d3 16 d  40mm T    d3 21Dr Somnath Kolgiri
  22. 22. 4) Design a foot brake lever from the following data (8M) i) Length of lever from the C.G. of spindle to the point of application of load = 1m ii) Maximum load on the foot plate = 800 N iii) Overhang from nearest bearing = 100 mm iv) Permissible tensile & shear stress = 70 MPa 22Dr Somnath Kolgiri
  23. 23. Bell Crank Lever In a bell crank lever, the two arms of the lever are at right angles. Such type of levers are used in railway signalling, governors of Hartnell type, the drive for the air pump of condensors etc. The bell crank lever is designed in a similar way as discussed earlier. The arms of the bell crank lever may be assumed of rectangular, elliptical or I-section. 23Dr Somnath Kolgiri
  24. 24. Design of Bell Crank Lever 24Dr Somnath Kolgiri
  25. 25. 1) Determination of effort/load as per problem. Let W be the load and P is the effort at the load arm of length lw and effort arm le respectively. Final load/effort can be calculated by taking moment about the fulcrum. W lw  Ple 2) Determination of resultant fulcrum reaction ( RF) RF  W  P 25Dr Somnath Kolgiri 2 2
  26. 26. 3) Design of fulcrum pin – The fulcrum pin is supports the lever and allows to oscillate. Due to relative motion of lever on pin, fulcrum pin is subjected to bearing pressure and direct shear stress. a) Fulcrum pin is designed by considering under bearing pressure l Assume p 1.25 b Find the diameter of pin(dp ) RF lp dp P  dp lp length of pin d p  diameter of fulcrum pin 26Dr Somnath Kolgiri
  27. 27. b) Direct shear stress – Fulcrum pin is subjected to double shear. 2 pd RF 2A  4 2   RF  This equation is used for checking the shear stress. If calculated shear stress is less than given shear stress then the design of fulcrum pin is safe. 27Dr Somnath Kolgiri
  28. 28. 4) Diameter of Boss of Lever – The boss of lever is subjected to bending stress due to bending moment of lever. do= outer diameter of the boss of lever di= inner diameter of the boss of lever lb= length of boss There is relative motion between the fulcrum pin and the boss of a lever, a brass bush of 2 mm to 3 mm thickness should be insert in the boss of fulcrum lever as a bearing so that renewal became simple when wear occurs. 28Dr Somnath Kolgiri
  29. 29. di = dp + (2 x 3) if bush of 3 mm is used. di= dp if bush is not used. do= 2dp 29Dr Somnath Kolgiri
  30. 30. • The maximum bending stress induced in the boss of the lever is b b b I y IZ M  do  M  y  M  M where B.M.  M W lw  Ple 6Mdo 12 2 ( ) l [d3  d 3 ] b o i l [d3  d3 ] b o i    b  1   From this equation outer diameter of boss can be obtained or bending stress should be checked 30Dr Somnath Kolgiri
  31. 31. 5) Design of Lever to find dimensions – The lever is subjected to bending moment The maximum bending moment acts near the boss . Z b e WhereZ isthesection modulusof crosssection of lever that mayberec tan gular or elliptical Themax imumbending stressinduced is d  do  W  lo    e 2   2      M  Pl    M 31Dr Somnath Kolgiri
  32. 32. 6 2 hy  Z  bh2 Z  I  12 h  depth of lever b  thickness of lever Assumeh  2bto 4b 1 bh3 a) Consider the rectangular C/s of the lever - 32Dr Somnath Kolgiri
  33. 33. b) For Elliptical Section – 32 ( ) ( ) h 2 bh2 bh3 Z  Z 64 where,b  thickness of thelever i.e.min or axis h  Depth or height of thelever i.e.majoraxis h 2bto 2.5b Z  I y 33Dr Somnath Kolgiri
  34. 34. Problems 1) A right angled bell crank lever having one 500 mm long and another arm is 150 mm is used to lift a load of 5 KN. The permissible stresses for pin and lever is 80 MPa in tension and compression and 60 MPa in shear. The bearing pressure on pin is not to exceed 10 MPa. Determine the dimensions of rectangular cross section of the lever and pin diameter. 34Dr Somnath Kolgiri
  35. 35. 2) A right angle bell crank lever having one arm 700 mm and other 400 mm long. The load of 1.75 KN is to be raised acting on a pin at the end of 700 mm arm and effort is applied at the end of 400 mm arm. The lever consists of a steel forgings, turning on a point at the fulcrum. The permissible stresses are in tension and compression are 80 N/mm2 and 60 N/mm2 in shear. The bearing pressure on the pin is not to exceed 6 N/mm2. Find i) Diameter and length of fulcrum pin. ii) Thickness (t) and depth (b) of rectangular C/s of the lever (Assume b = 3t) 35Dr Somnath Kolgiri SBPCOE,
  36. 36. Design of C clamp & offset link • Some machine component are subjected to two or three types of stresses such a combination of stress known as combine stresses. • When the line of action of an external load is parallel but non co-axial with the centroidal axis of the component, this type of load called as eccentric load and the distance between the two axis is called as eccentricity (e). 36Dr Somnath Kolgiri
  37. 37. Design procedure of such types 1) Direct stress – The magnitude of the direct stress induced in the machine component. A d where A  crosssectional area  P  37Dr Somnath Kolgiri
  38. 38. 2) Bending Stress – The magnitude of bending stress induced in machine component is given by ZIb y M  Pe bending moment  M  M y whereZ  I  Section modulus  38Dr Somnath Kolgiri
  39. 39. 3) Resultant stress – The resultant stress are obtained by the principle of super position. R d b R d b  M      ( P ) A Z      ( P ) A Z  ( M ) The positive sign indicate the tensile stress while negative sign indicate the compressive stress. 39Dr Somnath Kolgiri
  40. 40. Dr Somnath Kolgiri 40 Example1.The frame of a hacksaw is shown in Fig. 4.26(a). The initial tension P in the blade should be 300 N. The frame is made of plain carbon st eel 30C8 with a tensile yield strength of 400 N/mm2 and the factor of safety is 2.5. The cross section of the frame is rectangular with a ratio of depth to width as 3, as shown in Fig. 4.26(b).Determine the dimensions of the cross-section.
  41. 41. Dr Somnath Kolgiri 41
  42. 42. Dr Somnath Kolgiri 42
  43. 43. Dr Somnath Kolgiri 43 Example2.An offset link subjected to a force of 25 kN is shown in Fig. 4.27. It is made of grey cast iron FG300 and the factor of safety is 3. Determine the dimensions of the cross-section of the link.
  44. 44. Dr Somnath Kolgiri 44 Step II Calculation of direct tensile and bending stresses The cross-section is subjected to direct tensile stress and bending stresses. The stresses are maximum at the top fibre. At the top fibre,
  45. 45. 45Dr Somnath Kolgiri

DME-1_uNIT-1 Mechanical Engineering, SPPU, Pune

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