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# DME Project

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### DME Project

1. 1. 1 A PROJECT REPORT ON “DESIGN OF LINE SHAFT FOR RICE MILL” SLOT – D2 SUBMITTED TO - PROF. MANOHARAN R SUBMITTED BY- ASUTOSH PADHY 12BME0478 ROHAN SUKLECHA 12BME0523 AKHAND PRATAP SINGH 12BME0042 VIPUL BANJARE 12BME0174
2. 2. 2 ACKNOWLEDGEMENTS We would first like to thank VIT University’s Management for providing us an opportunity to get a hands-on experience and gain practical knowledge while completing this PBL project under the course. We then thank our DME Faculty, Prof. Manoharan R, who allowed us to work over the design project and owe our gratitude towards him for his supportive hand in the successful completion of the project, for taking our review on the given date and suggesting necessary changes to be done in the project. We would also like to thank our Program Manager Prof. Antony Xavier and our Dean Prof. S.K. Sekar for helping us in obtaining some practical application of theoretical knowledge via this project. Thank you all.
3. 3. 3 TABLE OF CONTENTS SL. NO. TITLE PAGE NO. 1. Objective 4 2. Introduction/Literature 5 3. Methodology 6 4. Calculation 7-9 5. Analysis 10-15 6. Results 16-18 7. Summary and Conclusion 19 8. Bibliography 20
4. 4. 4 OBJECTIVE The objectives of this project were as follows:  Study the technical aspects related to a line shaft and investigate methods of designing shafts.  Apply the concepts of Strength of Materials and Design of Machine Elements.  Design a line shaft of a particular length and diameter and –  To select an appropriate material that can be used under the above conditions  To analyze the stresses built up in the shaft and to set a shaft diameter that can withstand these stresses with a minimum of factor of safety 2 and maximum deflection 1mm.  To analyze the bending stress, support reactions, moments and displacements occurring in the shaft under dynamic loading.  To allow a minimum of 5 belt pulley attachments on the shaft without bypassing any of the aforementioned objectives.
5. 5. 5 INTRODUCTION A line shaft is a power driven rotating shaft for power transmission that was used extensively from the Industrial Revolution until the early 20th century. Prior to the widespread use of electric motors small enough to be connected directly to each piece of machinery, line shafting was used to distribute power from a large central power source to machinery throughout a workshop or an industrial complex. A typical line shaft would be suspended from the ceiling of one area and would run the length of that area. One pulley on the shaft would receive the power from the parent line shaft elsewhere in the building. The other pulleys would supply power to pulleys on each individual machine or to subsequent line shafts. Power loss with line shafts vary widely and are typically 25% and often much higher; however, using roller bearings and good quality lubrication can minimize losses but the efficiency would still be lower compared to their single motor driven shafts. Line shafts had multiple operations in the past; ranging from lathes to compressors. It was the sole medium for transferring power from a Combustion/Steam engine to a driven element because electric motors weren’t prevalent back then. After the 1980s the usage of line shafts dwindled to select areas like Mills and Paper factories where large rolls were a necessity. Our project topic is on one of these applications of line shafts viz. the Rice mill. We intend to distribute the power from a single motor among 5 different rice mill apparatus and check the validity and viability of our design. In such cases, the design of line shaft is a secondary aspect of design, the primary being the designing of the crusher and milling apparatus. For a crusher/ mill to be able to separate husk from rice, a particular amount of force needs to be applied at the crusher apparatus, mostly decided by the size of grains, volume of mill, speed of rotation etc. This force is called the bio-yield force. Sagar H. Bagade et al(2014) published a paper on a similar ground; the design procedure for a mini dal machine recently. Taking that into consideration and formulae by Mutalubi Aremu Akintunde(2007) in their publication on rice polishing, a critical analysis of the line shaft has been performed and validated. The data for modeling the problem was taken from experimental results of I.K. Adegun (2012) and the power for the shaft has been taken in proportion to the number of machines connected to it. The problem has been analyzed in both 2D and 3D geometries with valid calculations and formulae to back them up.
6. 6. 6 METHODLOGY: The problem was considered as a typical design problem and a similar procedure was followed. Design Process Flow:  Recognition of Need  Problem Formulation  Approach strategy/plan  Model formation  Gathering data  Analysis  Verification  Validation  Optimization  Presentation of results Recognition of need: Our current topic has no pressing needs to attend to but it is a novel way of getting acquainted with the design process of a line shaft. The typical application of the problem has been rendered obsolete with the invention of portable motors. The problem still poses a uphill task with multiple loading conditions and end conditions. Problem Formulation: The main objective of the problem is to obtain a shaft diameter at which it can sustain whirling for long periods of time while carrying 5 belt drive loads. So the parameters that need to be checked while designing are –  Shaft length - Standardized to 5 m  Distance between loads – Taken as 1m for simplicity  End support condition – Cylindrical rotation support  Loads and their direction – Belt loads horizontal, weights vertical  Whether to apply supports at the middle – Depends on deflection Approach Strategy: There are 2 approaches to such a design problem: 1) Safe stress: To take the maximum allowable values of stresses and deflections and get the diameter. Typical of theoretical problems. 2) Diameter iteration: To take a diameter value and evaluate the stresses and deflection from the shaft to get the FOS and minimum deflection. Then subsequent adding of supports or increasing the diameter till the required value is reached. We used the 2nd approach while designing our shaft because it also allows us to check for deflections along with stresses.
7. 7. 7 MODEL FORMULATION: Assumptions and Constraints for theoretical calculations (2D):  Point loads and simple supports  Horizontal belt drive  No chuck length  No support length  Rankine and Guest Theories for stress  Macaulay’s theorem for deflection 20 N 20N 20N 20N 20 N 1m 1m 1m 1m ` 441.6N 441.6 N 441.6N 441.6N 441.6N 5m Input Data: The input parameters were initialized from data research and Journals in the field of Agricultural Science.  Pulley weight: 20N  Pulley Diameter: 0.12m  Pulley belt tension ratio : 9:1  Belt Thickness: 30 mm  Keyway slots: 2 at ends 100 mm each for motor chuck  Belt Speed requirement: 2.83 m/s  Motor Power: 1000W (200 W for each motor)  Material Choice: C-40 or AISI 1040 steel  Yield strength = 380 MPa UTS = 580 MPa  Shear Yield strength = 190MPa Elastic Modulus = 2.05x 10^11 Pa
8. 8. 8 2D CALCULATIONS: After iterations at 20mm and 30 mm diameter, a viable solution was found for 40mm diameter shaft. Taking diameter as 40mm and assuming point loads with point supports: Pulley diameter = 0.12m, radius = 0.06m ( rp) ω = v/r = 2.83/0.06 = 47.1 rad/s Input data: ω = 47.1 rad/s P = 1000 W Torque = P / ω = 1000 / 47.1 = 21.2 Nm Belt tension calculation: (T1 – T2) rp = Torque  (T1 – T2) = 21.2 / 0.06 = 353.3 N ---- (1) T1 = 9T2 ----- (2) T1 = 397.46 N; T2 = 44.162 N Assuming belts to take loads along the horizontal direction total tension T, T = 441.6 N Mass of shaft = 50.96 kg Forces acting on shaft: 1) Horizontal belt force – 441.6 N per pulley 2) Vertical pulley weight – 20 N per pulley 3) Self-weight of shaft – 500N There are two cases present before us, one case with supports and other without supports. Vertical Forces – Self-weight – Density x Volume x g = 7800 x 0.78539 x 0.042 x 9.81x 5.2 N W = 500 N distributed uniformly over the length 5 m or 100N/m UDL. Without support: Vertical case: Ra + Rb = 600 N at x=2.5m 5 x Ra = 20 x (12.5) + 500x 2.5
9. 9. 9 Ra = 300 N; Rb = 300 N Max bending moment M = 300x2.5 -20x 2-20x 1 –100 x 6.25/2 M = 377.5 N Horizontal case: Ra + Rb = 2208 N Ra =1104 N Rb = 1104 N (symmetric loading) Max bending moment M = 1104x 2.5 -441.6 x 3 = 1435 Nm Total bending moment = (14352 + 377.52 ) 0.5 = 1483.8 Nm Torque = 22.1 Nm From Rankine and Guests combined Theory; Fs max = 16 (M2 + T2 )0.5 /(π d3 ) Ft max = 32 [M + (M2 + T2 )0.5 ]/ ( πd3 ) Fs = 118.09 MPa (Shear stress) Ft = 472.334 MPa (tensile stress) The above values exceed the required limits. Iterating again with supports at 0.5 m offset from center load. With support: Vertical case: RA + RB + RC + RD = 600 2.5RA+0.5RB - 0.5RC - 2.5RD=0 5RA +3RB + 2 RC = 1500 0.5 RA – 1.5 RB - 2.5 RC – 4.5 RD = -1200 Solving we get, RA=100.21 N, RB= 199.79N, RC=199.79 N, RD =100.21 N Max bending moment M = 100.21 x 0.5 – 100x 0.52 /2 Nm M = 37.605 Nm Horizontal case: RA + RB + RC + RD = 2208 5RA+ 3RB+2 RC = 5520N 2.5RA+0.5RB - 0.5RC - 2.5RD=0
10. 10. 10 0.5 RA – 1.5 RB - 2.5 RC – 4.5 RD = 4416 Solving we get, RA = 358.81 N, RB = 745.19 N,RC= 745.19N RD = 358.81 N Max bending moment = 358.81x 0.5 = 179.405 Nm Total bending moment = 183.303 Nm Torque = 22.1 Nm Using Ranking and Guest combined theory for shafts, Fs max = 16 (M2 + T2 )0.5 /(π d3 ) Ft max = 32 [M + (M2 + T2 )0.5 ]/ ( πd3 ) F s max = 14.926 MPa(Shear stress) and Ft max = 58.768 MPa (tensile stress) Taking C- 40 as the material with hardened and tempered properties and shaft diameter 40 mm the shaft is designed with two supports. Deflection of the shaft : The shaft’s deflection is not possible to find analytically because of indeterminate problem. Otherwise Macaulay’s theorem is used to find the deflection for a simply supported beam. Since it is not possible to calculate deflections for a statically indeterminate beam we take the values from ANSYS to be correct anyways. Macaulay’s theorem is used to find deflection. EI y’’ = Bending Moment EI = 2 x 10 11 x π d4 /64 = 2x10 11 x π (0.04)4 / 64 = 25761.059Nm2 Vertical loads without support case: EI y’’= 300x – 50x2 - | 20 ( x-0.5) - |20(x-1.5) - |20(x-2.5) –| 20(x-3.5) – 20 (x-4.5) Double integrating, Y = (1/EI) (50 x3 + C1x – (50/12)x4 - (20/6) [ (x-0.5)3 + (x-1.5)3 +(x-2.5)3 +(x-3.5)3 + (x- 4.5)3 ] y=0 at x=0; y=0 at x=5m. C1 = -627 Y at x= 2.5 m Y = (1/EI) (50 x3 – 627x - 50x4 /12 - (20/6) [ (x-0.5)3 + (x-1.5)3 +(x-2.5)3 + (x-3.5)3 + (x- 4.5)3 ] Y = (1/25761.059) ( 50 x 15.625 – 627 x 2.5 – 50x 2.54 /12 – 3.33 x 9) Y= -978/ 25761.05 m
11. 11. 11 Y= - 0.03796m or 37.96 mm in simply support condition with point loads Horizontal Loads without support case: EI y’’= 1104x - | 441.6 (x-0.5) - |441.6(x-1.5) - |441.6(x-2.5) –| 441.6(x-3.5) – 441.6 (x- 4.5) EI y = (1104/6)x3 + C1x –(441.6/6) [ (x-0.5)3 + (x-1.5)3 +(x-2.5)3 +(x-3.5)3 + (x-4.5)3 ] C1 = -2346 Y = (1/EI) ((1104/6) x3 – 2346 x - (441.6/6) [ (x-0.5)3 + (x-1.5)3 +(x-2.5)3 + (x-3.5)3 + (x- 4.5)3 ] At x=2.5m Y = (1/25761.059) (1104/6 x 15.625 – 2346 x 2.5 – 73.6 x 9) Y= (1/25761.059)(-3652.4) Y = -0.1417m or 141.7mm in simply support condition with point loads Polar Moment of Inertia of Shaft : 2I = 2.513 x 10-7 m4 . Supported condition: Vertical case: EI y’’= 100.21x – 50x2 - | 20 ( x-0.5) - |20(x-1.5) + |199.79(x-2)- |20(x-2.5) +|199.79(x- 3)-| 20(x-3.5) – 20 (x-4.5) Y = (1/25761.059) [(100.21/6)x3 +C1x – (50/12) x4 –(1/6){ | 20 ( x-0.5)3 - |20(x-1.5)3 + |199.79(x-2)3 - |20(x-2.5)3 +|199.79(x-3)3 -| 20(x-3.5)3 – 20 (x-4.5)3 }] Y = 0 at x= 0, 5m 2087.77 +5c1 – 2604.166 – 510.4 + 1165.44 C1 = -27.728 Y = (1/25761.059) [(100.21/6)x3 -27.728x – (50/12) x4 –(1/6){ | 20 ( x-0.5)3 - |20(x-1.5)3 + |199.79(x-2)3 - |20(x-2.5)3 +|199.79(x-3)3 -| 20(x-3.5)3 – 20 (x-4.5)3 }] At x=0.9286m Y is maximum. Y = (-15.472)/(25761.059) Y = -0.6mm Horizontal case: EI y’’= 358.81x - | 441.6 (x-0.5) - |441.6(x-1.5) +745.19(x-2)- |441.6(x-2.5) +|745.19(x- 3)–| 441.6(x-3.5) – 441.6 (x-4.5) EI y = (358.81/6)x3 +C1x -(1/6) [| 441.6 (x-0.5)3 - |441.6(x-1.5)3 +745.19(x-2)3 - |441.6(x- 2.5)3 +|745.19(x-3)3 –| 441.6(x-3.5)3 – 441.6 (x-4.5)3 ] 7475.208 +5C – 11270 + 4346.94
12. 12. 12 Y=0 at x=0,5 m C1 = -110.4296 y = (1/25761.079){(358.81/6)x3 -110.4296 x -(1/6) [| 441.6 (x-0.5)3 - |441.6(x-1.5)3 +745.19(x-2)3 - |441.6(x-2.5)3 +|745.19(x-3)3 –| 441.6(x-3.5)3 – 441.6 (x-4.5)3 ]} At x=0.9286 m y is maximum. Y = (1/25761.079) (-60.45) Y = -2.346 mm 2D ANALYSIS IN ANSYS Parameter No extra supports Extra supports Horizontal deflection 142.09mm 2.3534mm Vertical deflection 38.08mm 0.6mm Bending moment 1484Nm 183.3Nm
13. 13. 13 3D ANALYSIS The software used for analysis of the shaft was ANSYS Workbench 15. The part model was designed in Solid Works for simplicity. Bearing loads of 30mm belt thickness; Diameter of shaft 40mm. End cylindrical supports – 100mm long each (chuck constraint) Middle steady rest supports – 20mm long (roller bearing thickness) The analysis of the shaft was iterated from a minimum diameter of 20mm to 40 mm where it satisfied the required limits set for usage. Above is a picture of the shaft’s loading conditions. Since the shaft is a cylindrical entity, it was required that sinusoidal bearing loads be applied on the belt drive attachment points to simulate the actual forces acting on it. The Moment applied is equivalent to the torque and the rotational velocity is set as calculated above.
14. 14. 14 The shaft failed in iterations at 20 and 30 mm diameters owing to large deformations and stresses due to heavy loads to Cross section area. The loads caused excessive bending and large deflections which would render the shaft useless in a finite no. of cycles. After iterating to 40mm diameter we got the results as below: The deflection order was significantly reduced from order of 1 to 10-2 but there was still a large deflection of 28.9mm. To reduce this deflection our faculty suggested using center steady rests to reduce the Bending moment and hence deflection. Again we used the steady rest exactly between the two applied belt pulleys to facilitate ease of calculation for theoretical results. The two rests were provided a thickness of 20mm each to support the shaft and prevent bending moment build up at the center of the shaft and reduce its deflection. The results after using steady rests were as follows:
15. 15. 15 Notice the drastic reduction in deflection to 0.7mm after addition of cylindrical supports from the previous diagram. The addition of supports has proven to be clinical in improving the working life of the shaft and hence has helped us arrive at a perfect design for the shaft i.e. at 40mm diameter with 2 cylindrical supports.
16. 16. 16 RESULTS 2D ANALYSIS RESULTS The 2D analysis results tally perfectly with each other with one or two stray errors due to calculation. The 2D model has been validated and verified with the tallying of calculations and ANSYS 2D results. Condition Parameter Calculations ANSYS % error 2 end supports Deflection vertical 37.96mm 38.08mm 0.31% 2 end supports Deflection Horizontal 141.7mm 142.09mm 2.74% 2 end supports Bending moment total 1483.8Nm 1484 Nm 0.013% Condition Parameter Calculations ANSYS % error 2 end + 2 mid support Deflection vertical 0.6mm 0.6mm 0% 2 end + 2 mid support Deflection Horizontal 2.346mm 2.3534mm 0.31% 2 end + 2 mid support Bending moment total 183.303Nm 183.3Nm 0%
17. 17. 17 3 D Analysis: REAL CASE SCENARIO In supported condition, designed shaft: Ansys: Rankine theory stresses: No mid support : Fs=118N/m2 Ft= 472N/m-2 Supported:F s max = 14.926 MPa(Shear stress) and Ft max = 58.768 MPa Real Case scenario: Diameter of shaft = 40mm Length of shaft = 5.2m (0.2m inside chuck) Total vertical force acting on shaft = 500+ 100 = 600N Total horizontal force acting on shaft = 441.6 x 5 = 2208 N Torque of shaft = 21.2Nm Maximum Bending Moment M = 118.0 Nm Maximum Tensile stress developed = 27.2 MPa Maximum Shear stress developed = 14.9 MPa Condition Parameter ANSYS 2 end supports Equivalent stress 159.11MPa 2 end supports Total deflection 29.073mm 2 end supports Shear stress 82.598 MPa Condition Parameter ANSYS 2 end + 2 mid support Equivalent stress 14.922MPa 2 end + 2 mid support Total deflection 0.7mm 2 end + 2 mid support Shear stress 27.2MPa
18. 18. 18 Maximum and Minimum Principal stress developed = 16. 38MPa tensile and 1.55 Mpa compressive Yield strength = 380 MPa Shear Yield strength = 190MPa Factor of Safety =10.36 Maximum deflection = 29.072 mm unsupported condition Maximum deflection = 0.7 mm in supported condition It isn’t possible to provide a perfect validation for 3D results or verify them through manual calculations because of the following reasons:  The forces on the belt drive are not point loads but they have been approximated as such for manual calculations. The actual impact of this is not known.  The supports cover a large portion of the shaft while in the question, a point support has been assumed.  The shaft has been given a cylindrical fixed rotation support but in our calculations the support is a simple support.  The shaft’s deflection is a statically indeterminate problem with a simply supported assumption taken to get a result. However we would like to point out the following facts –  The calculated tensile and shear forces 87.35 MPa and 21 Mpa lie in close proximity to the Ansys result and are way below the critical values of these parameters hence making the solution a dual verified one.  The deformation obtained from ANSYS is lower than the calculated one which signifies a very good design and superior supports.
19. 19. 19 SUMMARY AND CONCLUSIONS Our project was started with one aim in mind; to apply the concepts of Strength of Materials and Design of Machine Elements in a practical scenario and at the conclusion of it, we’re glad that we could do so with some help. The project is summarized as follows:  The shaft was taken up as a design problem and calculations were performed on it with assumptions.  The diameter of the shaft was varied from 20mm to 40 mm and subsequently the stresses and deflection were found out using Ansys and calculations.  There were two cases after that: Case 1 without supports and Case 2 with supports. In both these cases the parameters were calculated separately for both vertical and horizontal conditions.  The stresses calculated in ANSYS analysis were – Equivalent Von Mises stress, Max.Principal Stress, Max. Shear stress. Under deformation, Total deformation and directional deformation were calculated.  Under calculations, the reactions obtained at the supports from ANSYS were used to calculate bending moment.  The bending moment combined with torque was used to calculate the Bending stress and shear stress for the shaft.  Macaulay’s Theorem with an assumption for supports and loads was used to calculate deflection for the shaft.  The results from ANSYS were tabulated and the Safety factor was calculated. CONCLUSIONS: The calculated FOS value and Deflection for the 3D are a real case scenario representation of the shaft. The validation of the 2D model only goes to show the correlation between design and calculations but doesn’t address the original issue i.e.“Designing a practical shaft”. The 2D calculations are in no way related to the 3D models because of vast number of assumptions and theoretical constraints from various theories. In the end the purpose is anyways served and a shaft of required FOS and minimum deflection has been designed. SOFTWARES USED: ANSYS 15 Workbench, SolidWorks 13
20. 20. 20 BIBLIOGRAPHY  Journal of Agricultural Technology 2012 vol 8(14)  A mini rice processing machine for Nigerian farmers  Design, fabrication and Performance Evaluation of Polisher Machine of Mini  Dal Mill  Development of a Rice Polishing Machine  ―Design, Construction and Performance Evaluation of a Combined Coffee De hulling and  Polishing Machine‖ References http://en.wikipedia.org/wiki/Line_shaft Rice Mill – Wikipedia Finite Element Simulations ANSYS Workbench Huang Lee ANSYS MECHANICAL APDL VERIFICATION MANUAL PSG College of Engineering Data Book V.B Bhandari- Machine Design R.S Khurmi- Strength of Materials