Power screw application
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The document outlines the design of a power screw clamping mechanism. It includes: 1. An introduction to the project and screw thread design. 2. Details on the power screw design including material selection, size calculations, and stress analysis. 3. Details on the frame and arm design including force and stress analysis. 4. An overview of modeling the individual parts and full assembly in SolidWorks.
Power screw application
- 1. Members in the group: -MOHAMED OSAMA ALSAMSAM - YASSER MONJED - MOHAMED SALAH RAGAB - ALAA IBRAHIM -SHADY HESHMET - BADR SAYYID - SAYYID KHAIRY - MOHAMED KAMEL - ASSAYYID FATHY - ASSEM HUSSIEN CLAMPING MECHANISM&CRANE Power screw project
- 2. TABLE OF CONTENTS Contents: 1- INTRODUCTION. 1 2- PROJECT PROCEDURE 3 3- POWER SCREW DESIGN 4 4- FRAME DESIGN. 17 5- PARTS AND ASSEMBELY IN SOLIDWORKS. 22 6- STRESS ANALYSIS IN SOLIDWORKS. 25 7- PROJECT APPLICATION. 26 8- REFRENCES 27
- 3. 1-INTRODUCTION Page 1 1-INTRODUCTION If the point of contact between the product and people become a point of friction, then the industrial designer has failed. On the other hand, if people are made safer, more efficient, more comfortable, by contact with the product, then the designer has succeeded. Henry Dreyfuss, 1967. A screw thread is formed by cutting continuous helical groove on cylindrical face. A screw made by cutting a single helical groove in cylinder is known as (single threaded) . And similarly (double threaded and triple quadruple) May be cut either right hand or lift hand. A screwed joint is mainly compost of two elements i.e. a bolt and nut (you can see them in project paper sheet)
- 4. 1-INTRODUCTION Page 2 We will use metric standard to produce our thread: Metric Threads standard The steps which the team go on it: 1- The primary drawing (Sayyid khairy + M.kamel) 2- The power screw design (M.salah +Alaa Ibrahim) 3- The frame design (M.Osama + Yasser Monjed) 4- The solid works drawing (assayyid fathy + Assem Hussin) 5- The Report writing (Shady heshmet + Badr sayyid) So, The design of Machine clamp is done. Finally Working and revision is done by (assayyid fathy + M.Salah + Sayyid khairy) In 1/2/2018. The project was done. In 10/2/2018.The revision was done
- 5. 2- PROJECT PROCDURE Page 3 2- PROJECT PROCDURE A- Purpose of project (clamping or lifting) Its limitation (W=18KN; L=500mm) B- Product concept (3F)(Form+feature+function) C- Nature of Load (static load) D- Appropriate material (Next pages) E- Force analysis, stress analysis and mode of failure (Next pages) F- DFM and DFA Page22 G- Detail drawing (Paper sheet A1)
- 6. 3- POWER SCREW DESIGN Page 4 3- POWER SCREW DESIGN The thread is trapezoidal because: 1. The efficiency of trapezoidal is high 2. If there are any corrosion in thread. it’s no problem Given W = 18 KN Max. L = 500 mm Full design (screw) Full design (nut) Handle A- Screw (material, size, modes of failure) 1-material (P.175) AISI 1015 Condition: cold drawn
- 7. 3- POWER SCREW DESIGN Page 5 u = 420 Mpa f.S = 3 y = 314 Mpa f.S = 3 × 0.6 = 1.8 E = 205 Gpa 2-size dc Ac 3- Modes of failure: a) Compression failure c = 𝑤 𝐴𝑐 ≤ σy 𝑓. 𝑠 AT standard P.184 c = 𝑤 𝐴𝑐 ≤ σy 𝑓. 𝑠 18 x103 Ac ≤ 314 1.8 103.18 mm2 ≤ Ac AT table P.185
- 8. 3- POWER SCREW DESIGN Page 6 Trapezoidal Then: Ac = 105mm2 do = 16mm dc = 11.5 mm p = 4mm b) Buckling failure W cr ≥ wgiven wcr = π 2 EI 𝑙𝑒2 ≥ wgiven I = π 64 ( dcore )4 Le = K × L AT p. 159 K = 2 (fixed free)
- 9. 3- POWER SCREW DESIGN Page 7 Le = 2 × 500 = 1000mm Wcr = π 2 EI 𝑙𝑒2 = π 2 205∗103 10002 × π ×11.44 64 Wcr = 1737.06 N Wcr = .1.737 KN But wgiven = 18 KN Wcr ≥ wgiven Then Wcr = wgiven π 2 205×103 1000^2 × π(dc)4 64 = 18000 Dc = 20.633 mm AT P.185 do = 28mm dc = 22.5mm p = 5 mm A=389mm2
- 10. 3- POWER SCREW DESIGN Page 8 Safe design buckling failure C) combined stress Se = √σc2 + 3 τxy2 ≤ σall c = 𝑤 𝐴𝑐 xy = 𝑇.𝑟 𝑗 r=dc/2 Ttot= Tr +Tc Ttot= F.L = Tr +Tc Tr = 𝑊𝑑𝑚 2 tan (β+α) dm = dc+do 2 α = tan−1 𝑛 𝑝 𝜋dm n: no. of starts
- 11. 3- POWER SCREW DESIGN Page 9 β = tan−1 𝜇 cos 𝜃 θ=14.5 degree AT ( P. 186 ) 𝜇 = .18 β > α Rc = 𝑅1+𝑅2 2 R1=.5dc R2=dc Rc =16.875 𝜇c = 0.17 Cast iron Tc = w.Rc . 𝜇c =18*103 *16.875*.17=51.6375KN dm = dc +do 2 = 22.5+28 2 =25.25 mm α = tan−1 𝑛 𝑝 𝜋 dm = 3.606 degree
- 12. 3- POWER SCREW DESIGN Page 10 β = tan−1 𝜇 𝑐𝑜𝑠𝜃 = 10.54 degree β≥α safe Tr = 𝑊𝑑 2 tan ( β + α ) = 57.27 KN.mm Then : Ttot = Tr + Tc = 57.27+ 51.677=108.9075KN.mm Ttot = F.L=195.12 × 103 = 3000 L L = 650.4 xy = 𝑇.𝑟 𝑗 = 48.69 Mpa
- 13. 3- POWER SCREW DESIGN Page 11 where J= 𝜋 64 ∗ 22.54 =25161.11987mm4 all = 0.5×y 1.8 = .5×314 1.8 =87.22 Mpa xy < all c = 𝑤 𝐴𝑐 = 18∗103 389 =46.272 Mpa Check : Se = √σc2 + 3 τxy2 ≤ σall Se = 96.193 Mpa all = 314 1.8 = 174.44 Mpa
- 14. 3- POWER SCREW DESIGN Page 12 Se ≤ all Safe design B- NUT Material Select malleable cast iron p.178 ASTM class 47 y = 224Mpa 1-Bearing /crushing mode b = 𝑤 𝐴𝑝 ≤ all ≤ σy 𝐹∗𝑠 Ap = π/4 (do2 – dc2 ) * Z1 H=Z * p 4∗𝑤 𝜋(282−22.52)∗𝑍1 ≤ 224 6 F.S=10*.6=6 Z1 = 2.43 = 4 even number H1= Z1*P = 4*5=20mm 2 - shearing mode (strip off) transverse shear Τmax = 3 w π d h ≤ τall = 0.5 σy f.s
- 15. 3- POWER SCREW DESIGN Page 13 3 × 18× 103 π × 22.5 × 5× z ≤ 0.5 ×224 6 z2 = 8.18 ≈ 10 H2 = 10 × 5 = 50 mm b = 1.6 do b = 1.6 × 28 = 44.8 mm Through (Crashing, strip off) see its (safe or not) 3 – combined stress: Se = √σb + 3 τ ≤ σall = σy f.s σb = 4 w π ×(do2−dc2 ) z = 4 × 18 × 103 π ×( 282−22.52) 10 σb = 8.251 Mpa τmax = 3 w π d h = 3 × 18× 103 π ×22.5 ×50 τmax = 15.279 Mpa σall = σy f.s = 224 6 =37.33 Mpa check : se =√(8.251)2 + 3(15.279)2 = 27.72Mpa
- 16. 3- POWER SCREW DESIGN Page 14 ∴ se < σall ∴ its safe design c – handle: need to get 1- length of the arm 2- diameter of the arm 3- handle material 1 - human force: (200: 400) N Assume 300 N Τtot = f . LH LH = Τtot f = 108.9075 × 103 300 = 363.03mm arm length 2 - dh = 0.5 do dh = 0.5 × 28 = 14 mm 3 - material According to stress we select the material Se = √σb + 3 τ ≤ σall = σy f.s
- 17. 3- POWER SCREW DESIGN Page 15 σy = ? ? ? σb = 𝐌∗ 𝐲 𝑰 y=dh/2 M = F × LH = 300 × 363.025 = 108907.5 N.mm I = π 64 ( dh )4 = π 64 ( 14)4 = 1885.74 mm4 σb = 108907.5 × 7 1885.75 = 404.27 Mpa τ = 𝐹 𝐴 = 4 f π ×d (h)2 = 4 ×300 π × (14)2 = 1.949 Mpa (direct shear) se = √(404.27)2 + 3(1.949)2 = 404.284 Mpa then get σy then from table page 177 choose the material se ≤ σy 1.8
- 18. 3- POWER SCREW DESIGN Page 16 404.284 ≤ σy 1.8 σy ≥ 727.711 Mpa from table page 177 σy = 834 Mpa material ( AISI 1340 ) steel condition oil quenched
- 19. 4- FRAME DESIGN Page 17 4- FRAME DESIGN 1-ARM At section A-A: θ=zero Force analysis: ∑ fy = 0 pv = 9000 N ∑ mz = 0 mz = 9000 × 120 = 108000 N.mm Stress analysis: σb = m y 𝐼 , I = 1 12 × 10 × 1003 , y = 100 2 =50 mm
- 20. 4- FRAME DESIGN Page 18 σb = ± 108000×50 1 12 × 10 × 1003 = ±64.8 Mpa τ = 3V 2𝐴 = 3 × 9000 2 × 10 ×100 = 13.5 Mpa by mohr: σ1 =67.5 MPa , σ2 = - 2.7 MPa at section A-A: θ=45 ∑ fy = 0 pv= 9000cos(45) =6363.96N ∑ fx = 0 ph = 9000sin(45) =6363.96N ∑ mz = 0 mz = 6363.96 ×120= 7.6×105 N.mm
- 21. 4- FRAME DESIGN Page 19 Stress analysis: σb= ± 𝑀𝑦 𝐼 = ±45.6 Mpa σt = 𝑃ℎ 𝐴 = 6.364 Mpa τ = 3𝑉 2𝐴 = 3(6363.96) 2∗(10∗100) = 9.54 Mpa by mohr’s : σ1 =53.87 MPa , σ2 = 1.69 MPa
- 22. 4- FRAME DESIGN Page 20 AT sec A-A: θ = 90 ∑ fx = 0 ph= 9000 N Stress analysis: σt = 9000 10∗100 = 9 Mpa at section B-B: ∑ fy = 0 pv = 9000 N ∑ mz = 0 mz = 9000 × 350 = 3.1*106 N.mm Stress analysis : σb = m y 𝐼 , I = 1 12 × 10 × 2003 , y = 200 2 = 100 mm σb = ± 3.1∗106×100 1 12 × 10 × 2003 = ± 46.5 Mpa
- 23. 4- FRAME DESIGN Page 21 τ = 3V 2𝐴 = 3 × 9000 2 × 10 × 200 = 6.75 Mpa by mohr : σ1 =47 MPa , σ2 = -1 MPa Bearing failure: “bolts) σt = 𝐹 ( 𝐵−𝑛∗𝑑) 𝑡𝑚𝑖𝑛 ≤ σall = 9000 (100−1∗40)∗10 = 15 Mpa *steel Material (AISI1006) cold drawn σy =285 MPa F.S= 3*.6=1.8 σall= 285/1.8 = 158.3 Mpa Max stress at θ=zero 76.5 ≤ σall safe design F: applied force B: width of plate n: no. of bolts d: dia of bolt tmin: min. thickness of plate .
- 24. 5- PARTS AND ASSEMBLY IN SOLIDWORKS. Page 22 5- PARTS AND ASSEMBLY IN SOLIDWORKS. We used solid works 2016 to design parts and make assembly for this parts. The mission includes: 1- The sketching and feature: Many tool is used
- 25. 5- PARTS AND ASSEMBLY IN SOLIDWORKS. Page 23 2- The main parts are designed: - Support - Arm Screw 3- Selecting the material
- 26. 5- PARTS AND ASSEMBLY IN SOLIDWORKS. Page 24 5- Assembly working
- 27. 6- STRESS ANALYSIS IN SOLIDWORKS. Page 25 6- STRESS ANALYSIS IN SOLIDWORKS. Using a study advisor (static – Buckling. etc.)
- 28. 7- PROJECT APPLICATION. Page 26 7- PROJECT APPLICATION. - In the work shop machines. Is used to lifting working piece. - with some editing. In the production lines to transport the product from line to another (put on rotating disk) - with some editing. As machine clamp. - Fully controlled of large work piece.
- 29. 9- REFRENCES Page 27 9- REFRENCES Machine Design R.S Khurmi 2010 Design of machine elements for prof DR. Abdelhay M. Abdehay. Part and Assembly Modeling with SOLIDWORKS 2015 Huei-Huang Lee YouTube videos Shigley's Mechanical Engineering Design 9th Edition + Solutions MACHINE DRAWING BOOK Dr. K.L.Narayana Dr. P. kannaiash Dr. Venkata Reddy