Formulae for calculations A) Nomenclature

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Formulae for calculations A) Nomenclature do= Nominal diameter {Outside diameter of screw} dc= Core diameter of screw d= Mean Diameter of the screw p= Pitch of the screw.

1 Formulae for calculations A) Nomenclature do= Nominal diameter {Outside diameter of screw} dc= Core diameter of screw d= Mean Diameter of the screw p= Pitch of the screw. ) Thread angle ( α) α = tan 1 p or p or 3p ( ) Π d p for single start 3) Angle of Friction ( φ) 1) Mean diameter of Screw (d) Mean diameter p d = d o, core diameter d c = d 0 p p for double start 3p for triple start For Square Threads... ϕ = tan 1 (μ)

2 For Trapezoidal threads.. ϕ = tan 1 ( cos15 ) For ACME threads.. ϕ = tan 1 ( μ ) 4) Torque Required Torque required to raise load d T s = W [tan(ϕ + α)]. μ cos 14.5 Torque required to Lower load T d s = W [tan(ϕ α)]. 5) Collar Friction r 1+r T c = μ c.w.( ) N m m r 1+r ) is mean radius of collar ( μ c 3 1 T c =.W. [ ( )] N m m 3 r r 1 r r 3...Uniform wear...uniform Pressure {If anything not mentioned assume uniform wear} 6) Total Torque T raise = T s + T c T Lower = T s + T c 7) Force/Power required to operate T raise = F R adius (length of handle) Find force F

3 Find power using formula T raise = 60.P ΠN 10 6 Find Power P Cutting speed =Pitch RPM mm/min 8) Efficiency O verall Efficiency Output = workdone F orce displacemen in one revt at Input workdone = output W P F orce Displacement in one rev at input = itch π r F W P itch = Π T raise T orque required without considering Screw Efficiency = friction Actual torque required for screw and collar = W tanα d/ T s+t c Problems on Torque and force calculation of SCREW 1] A vertical two start square threaded screw of a 100 mm

4 mean diameter and 0 mm pitch supports a vertical load of 18 kn. The axial thrust on the screw is taken by a collar bearing of 50 mm outside diameter and 100 mm inside diameter. Find the force required at the end of a lever which is 400 mm long in order to lift and lower the load. The coefficient of friction for the vertical screw and nut is 0.15 and that for collar bearing is 0.0. {Ans : T raise = N-mm, Force=143 N, T lower = 3x35315 N-mm, Force=838.8 N} ) A vertical screw with single start square threads of 50 mm mean diameter and 1.5 mm pitch is raised against a load of 10 kn by means of a hand wheel, the boss of which is threaded to act as a nut. The axial load is taken up by a thrust collar which supports the wheel boss and has a mean diameter of 60 mm. The coefficient of friction is 0.15 for the screw and 0.18 for the collar. If the tangential force applied by each hand to the wheel is 100 N, find suitable diameter of the hand wheel. {T raise =1100 N-mm, Dia of wheel =11 mm} )The nominal diameter of a Triple threaded screw is 50mm & pitch 8mm.It is used with collar 100mm outer dia &65mm inner dia coefficient of friction for threads as well as collars is Screw is used to raise of load of 15 KN calculate

5 Using uniform wear theory i)torque required to lift the load. ii)torque required to lower the load. iii)force required at radius 500mm. {Ans : T raise= N-mm, T lower = N-mm, Force to raise = N } An electric motor driven power screw moves a nut in a horizontal plane against a force of 75 kn at a speed of 300 mm / min. The screw has a single square thread of 6 mm pitch on a major diameter of 40 mm. The coefficient of friction at screw threads is 0.1. Estimate power of the motor. {T raise = N m m,power=1.108 kw } ] The cutter of a broaching machine is pulled by square threaded screw of 55 mm external diameter and 10 mm pitch. The operating nut takes the axial load of 400 N on a flat surface of 60 mm and 90 mm internal and external diameters respectively. If the coefficient of friction is 0.15 for all contact surfaces on the nut, determine the power required to rotate the operating nut when the cutting speed is 6 m/min. Also find the efficiency of the screw. {T raise =4410 N-mm, Power=0.77 kw, efficiency 14.4%} ] A Machine vice has single start square threads with nominal diameter mm and pitch 5 mm.callar dia are 55mm & 45mm. Coefficient of friction for threads 0.15 & u for collar 0.17 The machinist can comfortably apply a force of

6 15 n on handle of mean radius 150 mm. Assuming uniform wear theory calculate, 1) The clamping force developed between the jaws, ) The overall efficiency of screw { W= N, eff=1.18%} ] The lead screw of a lathe has Acme threads of 50 mm outside diameter and 8 mm pitch. The screw must exert an axial pressure of 500 N in order to drive the tool carriage. The thrust is carried on a collar 110 mm outside diameter and 55 mm inside diameter and the lead screw rotates at 30 r.p.m. Determine (a) the power required to drive the screw; and (b) the efficiency of the lead screw. Assume a coefficient of friction of 0.15 for the screw and 0.1 for the collar. {Ans : Total torque 4565 N-m, Power=0.077 Kw, Efficiency =13%} ] The lead screw of a lathe has single start I.S.O. metric trapezoidal threads of 5 mm nominal diameter and 8 mm pitch. The screw is required to exert an axial force of kn in order to drive the tool carriage during the turning operation. the thrust is carried on a collar 100 mm outer diameter and 60 mm inner diameter. the values of coefficient of friction at the screw threads and the collar are 0.15 and 0.1 respectively. The screw rotates at 30 RPM. Calculate;1) The

power required to drive the lead screw and ) The overall efficiency of the screw. { P=0.

7 power required to drive the lead screw and ) The overall efficiency of the screw. { P= kw, overall efficiency =1.94%} B) Design of Screw jack { Screw and nut } Step I] Design of screw on basis of compressive failure

8 Compressive Failure of screw W = π 4 (d c) f c Find d c from this equation Nominal diameter d 0 = 1.19 d c = {Take round even number} Pitch of Screw P = d 0 - d c {Take Round figure } Assume single start square thread Step II] Check Stresses induced in screw due to combined comp. and torsion Torque required to raise the load α = tan 1 ( Π d ) ϕ = tan 1 (μ) d T s = W [tan(ϕ + α)]. p Torsional shear stress induced in screw T = π 16 f s dc 3...F ind f s Direct compressive stress induced in screw f c = Load Area = W π d 4 c Combined compressive and shear stress results in maximum principal stress and maximum shear stress in screw which should be less than allowable f (c)max = 1 [ than allowable f c f c + ]...This should be less f.f c + 4 s

9 f (s)max = 1 [ ]...This should be f.f c + 4 s less than allowable f s Step III] Design of Nut on basis of Allowable bearing pressure n= Number of threads in contact with the screw h= Height of nut = n p t = Thickness of screw = p Allowable bearing pressure in nut, W p b = π 4 o c [d d ] n equation find n and h. Step III] Check Stressed induced in Nut. Shear stress induced in Nut than allowable f s W = π.n.d o.t f s... From this...this should be less ============================================== ================ Problem 1: A screw jack carries a load of 5 kn. If the coefficient of friction between screw and nut is 0.15, Design the screw and nut. Neglect collar friction and column action take σ c = 4 N/mm and τ = 30 N/mm for screw and τ nut = 3 0 N/mm for nut Allowable bearing pressure = 14 start thread). N/mm (Use single

10 Problem : A screw jack carries a load of kn. If the coefficient of friction between screw and nut is 0.15, Design the screw and nut. Neglect collar friction and column action take σ c = 4 N/mm and τ = 8 N/mm for screw and τ nut = 1 N/mm for nut Allowable bearing pressure = 14 start thread). N/mm (Use single C) Calculation of Stresses induced in screw and nut for power screw Formulae: Stresses in screw Torsional shear stress T = π 16 τ dc3...f ind τ Compressive Stress σ c = Load Area =...Find σ c π 4 W d c f (c)max = 1 [ f c + ]...This should be less f.f c + 4 s

11 than allowable f c f (s)max = 1 [ f.f c + 4 s less than allowable f s ]...This should be 1) A power screw having double start square threads of 5 mm nominal diameter and 5 mm pitch is acted upon by an axial load of 10 kn. The outer and inner diameters of screw collar are 50 mm and 0mm respectively. The coefficient of thread friction and collar friction may be assumed as 0. and 0.15 respectively. The screw rotates at 1 r.p.m. Assuming uniform wear condition at the collar and allowable thread bearing pressure of 5.8 N/mm, find: 1. the torque required to rotate the screw;. the stress in the screw; and 3. the number of threads of nut in engagement with screw. {Ans T raise = n-mm, Stress=44.8 Mpa,number of th=9.76}

12 {Ans : T raise =10033 N-mm, Length of handle =15.4 mm, Max shear stress =39.83 Mpa, bearing pressure=9.6 Mpa}

Bhagwan mahavir college of Engineering & Technology, Surat.

Bhagwan mahavir college of Engineering & Technology, Surat. Department of automobile Engineering Assignment Subject: Machine Design & Industrial Drafting B.E. Second year Instructions: 1. This set of tutorial