04. Design of Power

Marks – 12 Hours - 10 Introduction • The power screws (also known as translation screws) are used to convert rotary motion into translatory motion. • For example, in the case of the lead of lathe, the rotary motion is available but the tool has to be advanced in the direction of the cut against the cutting resistance of the material. • In case of screw jack, a small force applied in the horizontal plane is used to raise or lower a large load. • Power screws are also used in vices, testing machines, presses, etc. • In most of the power screws, the has axial motion against the resisting axial force while the screw rotates in its bearings. • In some screws, the screw rotates and moves axially against the resisting force while the nut is stationary and in others the nut rotates while the screw moves axially with no rotation.

Types of Thread Profile Used in Power Screws 1) Square Thread – This threads is adopted for the transmission of power in either direction.  It is difficult to cut with taps and dies. It is generally cut on lathe machine Application – Screw jack, with single point Mechanical Press, Clamping devices. cutting tool. • Advantages – 1. It has maximum efficiency. 2. Minimum radial or bursting pressure on nut. 3. These are of self locking type. • Limitations – 1. Strength of square thread is less as compared with other forms of threads. 2. These threads can not be used conveniently with split nut because of wear compensation is not possible and it is difficult in engagement & disengagement. 2) Acme Threads – The acme thread are not efficient as square thread. The acme threads are easier to cut and are stronger than square threads. Application – Lead screw of a lathe machine. • Advantages – 1. These permits the use of split nut which can be easy for engagement & disengagement and compensate wear. 2. These are stronger than the square threads in shear because of the larger C/s at the root. • Limitations – 1. The efficiency is lower than that of square thread due to slop given to the sides. 2. The slop on the sides introduces some brusting pressure on the nut. 3) Trapezoidal Threads –

The trapezoidal threads are similar to Acme threads except the thread angle is 300 in trapezoidal threads. 4) Buttress Threads – A buttress thread is used when large forces act along the screw axis in one direction only. This thread combines the higher efficiency of square thread and the ease of cutting and the adaptability to a split nut of acme thread.  It is stronger than other threads because of greater thickness at the base of the thread. The buttress thread has limited use for power transmission. It is employed as the thread for light jack screws and vices. Torque Required to Raise Load for Square Threaded Screws W  Load to be raised in N. P  Effort applied at the circumference of screw p  Pitch of the screwin mm d  meandiameter of the screwin mm   Helix angle   coefficien t of friction between screwand nut   tan   friction angle L  Lead of screw

Fromthe geometry of fig. pitch p tan   d  d

The force of friction F  RN will acts downwards Re solving the forces along the path

 FH  0 P cos W sin  F  0

P cos W sin   RN  0

P cos  W sin   RN    (1)  FV  0

RN  Psin W cos  0

RN  Psin W cos    (2)

Put the valueof RN in eq.1 P cos  W sin   (Psin W cos) P cos  W sin   Psin  W cos P cos  Psin  W sin  W cos P(cos   sin)  W (sin   cos) (sin   cos) P  W  (cos   sin) But   tan (sin  tan.cos) P  W  (cos  tan.sin) sin But tan  cos sin (sin  cos) cos P  W  sin (cos  sin) cos (sin.cos  sin.cos) P  W  (cos.cos  sin.sin) But sin(A  B)  sin A.cos B  cos A.sin B &cos(A  B)  cos A.cos B  sin A.sin B sin( ) P  W  cos( ) P  W  tan( ) Torquerequired to overcomethread friction d T  P 2 d T  W  tan( ) 2 If effort is applied by lever of length 'l' then,T  Pl T  W  tan( )l Torque Required to Lower the Load • Resolving the forces in horizontal & vertical direction.

 FH  0 P cos  F W sin

P cos  RN W sin    (1)

 FV  0

RN  W cos  Psin    (2) Put eq.2in eq.1 P cos  (W cos  Psin) W sin P cos  W cos  Psin W sin P cos  Psin  W cos W sin P(cos   sin)  W ( cos  sin) ( cos  sin) P  W  (cos   sin) (tan cos  sin) P  W  .....(as  tan) (cos  tan sin) Multiply and divideby cos (sin cos  sin cos) P  W  (cos cos  sin sin) sin( ) P  W  cos( ) P  W  tan( ) Torquerequired to lower the load in overcomin g the friction (T ). d T  P 2 d T  W tan( ) 2 Self Locking & Overhauling of Screw • Self Locking Screw – The torque required at the circumference of the screw to lower the load is given by d T W tan( ) 2 In the above equation, if friction angle is greater than helix angle (Ø > α) then the torque required to lower the screw is positive i.e. some effort is required to lower the load, such a screw is known as self locking screw. Self Locking & Overhauling of Screw • Overhauling Screw – The torque required at the circumference of the screw to lower the load is given by d T W tan( ) 2 In the above equation, if friction angle is less than helix angle (Ø < α) then the torque required to lower the screw is negative i.e. the load will start moving downward without application of any effort, such a screw is known as overhauling screw. Design of Screw Jack • The various parts of screw jack are 1. Screw spindle having square thread screw. 2. Nut and collar for nut. 3. Head at the top of the screw spindle for handle. 4. Cup at the top of head for the load. 5. Body of the screw jack. 6. A handle or Tommy bar. Animation of Screw Jack

1) Design of screw – i) Find the core diameter of screw by considering the screw under pure compression. W W    c  A  d 2 4 c

Fromthis equation dc canbeobtained ii) Select the standard square threads for screw

Take(dc  6)  dc for design in exam d Then d  c &d  d  p o 0.84 c o p d  d meandia.  d  d   o c o 2 2 iii) In addition to direct compressive load, the screw

is subjected to twisting moment (T1), so for that the core diameter is increased and appropriately find the torque required to rotate the screw. d T W tan( ) 1 2 A) Checking of Screw – i) Shear stress due to torque  T   d 3 1 16 c fromthis eq. find shear stress( ) ii) Direct compressive stress dueto axial load  W   c   d 2 4 c Then,use Maximum principle stresstheory 1   [   2  4 2 c(max) 2 c c Maximum shear stresstheory 1     2  4 2 max 2 c For safety

 c(max)   (Given) &

 max  (Given) 2) Design of nut – i)Theheight of nut 'h'canbe found by considering the bearing pressureon nut W P  b  [d 2  d 2 ] n 4 o c where,n  no.of threads in contact with screwspindle Height of nut  h  n p where, p  pitch of threads ii)Check for shear stressesin screwand nut threads W Shear stressin screw  s   dc t  n W Shear stressin nut  n   do t  n p where,t  2 For safety of the screw&nut threads

 s and n  (Given) 3) Design of Nut Collar –

Let, D1 = Inner Diameter of nut collar or outer diameter of nut.

D2 = Outer diameter of nut collar t = thickness of nut collar 1 a) a)Considering tearing of nut dueto tensile strength W   t(nut)  (D2  d 2 ) 4 1 o

Find out D1 b)Considering the crushing of nut and collar W   cr (nut)  (D2  D2 ) 4 2 1

Find D2 c)Considering the shearing of nut collar W  (nut)   D1 t1

Find t1 4) Design of (Spindle) Screw Head –

To find the diameter of spindle head D3 and diameter of cup Useempirical relation as givenbelow 

Diameter of head  D3 1.75do D Diameter of cup  D  3 4 4 5) To find the torque to overcome due to friction at the top of the screw (Collar friction) – a) Assumeuniform pressurecondition 3 3 2 R3  R4 Tc   c W [ 2 2 ] 3 R3  R4 b) Assumeuniform wear condition

Tc  c W  R R  R Where R  3 4 2 6) Design of Handle – a) To design the handle, consider the total torque to which the handle will be subjected is given by –

Ttotal = T1 + Tc Assume that one person can apply a force of 300 N to 400 N Effective length of handle required T L  total E 300to 400 N

Length of handle  L  LE  Gripping length b) To find the diameter of handle (dh) – Consider the handle under bending

Ttotal  Total torque req.to raisethe load M  MaximumB.M.acting on handle

 b  Bending stressinduced in handle M   b Z M   b   d 3 32 h

Find dh

Height of nut  H  2dh 7) Buckling load of the screw – Check the safety of the screw against buckling failure – According to J.B.Johnson formula  L Buckling load  W  A  [1 yc ( )2 ] cr c yc 4C 2 E k Where,C  end fixity coefficien t  0.25

k  radius of gyaration  0.25dc

 yc  yield point stress L  unsup ported length of the screw  A  d 2  c / s area of screw c 4 c E  mod ulus of elasticity Factor of safety against buckling failure is W N  cr cr W For safe design of screwunder buckling

Ncr  N f 8) Find the efficiency of the screw jack  T   o 0 T 9) Find dimensions of screw jack body by empirical relation –

a) diameter of body at top  D5 1.5 D2 b)Thicknessof body  t3  0.25do c) Insidediameter at bottom  D6  2.25 D2 d)Outside diameter at bottom  D7 1.75 D6 e)Thicknessat thebase  t2  2t1 f ) Height of body  Hb  Max.lift  Ht.of nut 100mmextra for clearance

Hb  (Hl  h 100) Design of Toggle Jack

• The various parts of the toggle jack are i) The compound screw ii) Two nuts iii)Eight links iv)The pins v) The head  Animation of Toggle Jack

1) Find the maximum tensile force in screw – The maximum load on the occurs when the jack in bottom position. From figure W tan  2 F W F  2 tan Where  angle of link with horizontal  Maximum valueof  is taken as350 Dueto force ' F'links aresubjected to tension and squarethread screwunder pull Total tensile force onthread screw  2F  W Considering screwunder tensile failure W   t  d 2 4 c

dc canbeobtained .

Then d0  dc  p d  d &d  o c 2 2) Find the torque required to overcome the thread

friction (T1) – d T  W tan ) 1 2  &T    d 3 1 16 c Fromthis eq. find ' ' As the screwis subjected to both direct tensile stressand torsional shear stress i) max imum principle stress 1   [   2  4 2 ] t(max) 2 t t ii) max imum shear stress 1   [  2  4 2 ] max 2 t For safety

 t(max)   t & max  3) Design of nut - a) Find the Height of nut by considering bearing pressure – W P  b  (d 2  d 2 ) n 4 o c Find 'n' For providing good stability &avoid tearing Taken' n  2 height of nut  n'p

Width of nut  b1 1.5do b) Check the stresses induced in nut and screw thread W Shear stressin screw  s   dc t  n W Shear stressin nut  n   do t  n 4) Calculate length of screw portion – Length of screw = dist. between the centered line of nut when toggle jack at the top position + thickness of nut + (2 x thickness of ring) 5) Design of Pins – The pins are design under double shear. F   p 2A F   p  2  d 2 4 p

Where,d p  diameter of pin

 p  Shear stress for pin

THE END