Lecture 1.pdf
- 1. MACHINE DESIGN II MDP 2140 Instructor: Dr. Mohamed El-Shazly Associate Professor of Mechanical Design 1 24 March 2021 MACHINE DESIGN 2
- 2. INTRODUCTION 2 24 March 2021 MACHINE DESIGN 2
- 3. COURSE CONTENTS ،للنماذج والتصور ،االبتكار،االحتياج ،التصميم وطرق نظرية القدرة نقل لعناصر المبدئي التصميم : االدارة أعمدة – المسننة التروس – القارنات – القوابض – الفرامل – السيور – القدرة جنازير – بالحبال االدارة ت المتدحرجة التحميل وكراسي االدارة ألعمدة التفصيلي التصميم وطرق نظرية حت المختلفة االحمال المتدحرجة التحميل لكراسي المناسب التركيب والهيدروستاتيكية الهيدروديناميكية الكراسي تصميم و نظرية – االدارة كأعمدة الماكينات عناصر لتجميع االنشائية التفاصيل الماكينات عناصر اختيار في الكمبيوتر تطبيقات . . 24 March 2021 MACHINE DESIGN 2 3
- 4. Complete Machine Design-1 Threaded joints Design of bolted joints Mechanical springs; compression coil springs Machine Design 2 SHAFTS AND COUPLINGS SPUR GEARS CLUTCHES AND BRAKES Bearings Belts and Chains 24 March 2021 MACHINE DESIGN 2 4
- 5. Assessment 24 March 2021 MACHINE DESIGN 2 5 Schedule of Assessment Tasks for Students During the Semester Assessment Assessment task Week due Marks 1 In Class Assignment T.B.D 20 2 Mid Term Exam 6-8 20 3 Term Project 12 20 4 Final Exam 15 40
- 6. 6 Introduction A shaft is a rotating or stationary member usually having a circular section much smaller in diameter than the shaft length and having mounted on it such power transmitting elements as gears, pulleys, belts, chains, cams, flywheels, cranks, sprockets, and rolling bearing elements. In practical applications, the shaft may often be stepped instead of having a constant diameter with under cuts or grooves at the step. Some of the main consideration in designing a shaft are strength, using yield or fatigue (or both) as criterion; deflection; or the dynamics established by the critical speeds.
- 7. Shaft Design Procedure 1. Develop a free-body diagram by replacing the various machine elements mounted on the shaft by their statically equivalent load or torque components. 2. Draw a bending moment diagram in the x-y and x-z planes. The resultant internal moment at any section along the shaft may be expressed as 2 2 x xy xz M M M = +
- 8. Shaft Design Procedure 3. Develop a torque diagram. Torque developed from one power- transmitting element must balance the torque from other power- transmitting elements. 6. Establish the location of the critical cross-section, or the x location where the torque and moment are the largest. 7. For ductile materials use the maximum-shear-stress theory (MSST) or the distortion-energy theory (DET). 8. For brittle materials use the maximum-normal-stress theory (MNST), the internal friction theory (IFT), or the modified Mohr theory (MMT.
- 9. Example (a) Chain drive assembly The power from the motor of a front-wheel car is transmitted to the gearbox by a chain drive. The two chain wheels are the same size. The chain is not prestressed, so the loose chain exerts no force. The safety factor is 4. The shaft is to be made of AISI 1080 steel. The chain transmits 100 kW of power at the chain speed of 50 m/s when the motor speed is 6000 rpm. Find the appropriate shaft diameter by using the DET.
- 10. Example (a) Chain drive assembly; (b) free-body diagram; (c) bending moment diagram. ( ) The force transmitted through the chain is 100,000 2000 50 The chain radius is 50 0.07958 79.58 6000 2 / 60 From the free body diagram, the torque being applied by the chain is (2000) P h P N u u r m mm T P r = = = = = = = = = (0.07958) 159.15 . Force and monment equilibrium give 0 0 0.2 0 2500 , 500 A B A B B A N m R R P M P R R N R N = − + − = = → − = = =
- 11. Example 1 (a) Chain drive assembly; (b) free-body diagram; (c) bending moment diagram. ( ) ( ) 3 3 4 3 From the bending moment diagram, the critical section occurs at x=0.16 m where the moment is 80 N.m with steady torque of 159.15 N.m / 2 16 2546.4 2560 , / 64 The equation for principal xy x M d T d d d d = = = = ( ) ( ) ( ) ( ) ( )( ) y 2 2 2 2 1 2 3 3 1 2 3 3 1/2 2 2 3 stresses when 0 are 2560 1280 1 , 2546.4 2 4 4 4130 1570 , The von Mises stress for biaxial stress state is 1 4130 1570 4130 1570 0.02576 The shaft dia x x xy y e s d d d d S d n d m = = + = + = = − = + + mter should be 26 mm
- 12. 12 Design Against Static Loading I- Bending Moment and Torsion The transverse forces and torques applied to the shaft will produce both bending and shear stresses given by For a circular cross section For the plane stress state, the principal normal stresses when y is zero are given by
- 13. 13 Design Against Static Loading I- Bending Moment and Torsion Using the Distortion Energy Theory (DET)
- 14. 14 Design Against Static Loading I- Bending Moment and Torsion Using the Maximum Shear Stress Theory (MSST)
- 15. (a) Assembly drawing; (b) free-body diagram; (c) moment diagram in xz plane; (d) moment diagram in xy plane; (e) torque diagram. Example 2 An assembly of belts has a tensile forces applied as shown in the figure and frictionless journal bearings at locations A and B. The yield strength of the shaft material is 500 MPa and the safety factor is 2.
- 16. Example 2
- 17. 17 Design Against Static Loading II- Bending Moment, Torsion, and Axial Load The axial stress in this case will be given by For the plane stress state, the principal normal stresses when y is zero are given by
- 18. 18 Design Against Static Loading II- Bending Moment, Torsion, and Axial Load The Distortion Energy Theory The Maximum Shear Stress Theory ≈ ≈
- 19. 19 Design Against Cyclic Loading I- Ductile Materials
- 20. 20 Design Against Cyclic Loading I- Ductile Materials Soderberg line for shear stresses. Recall Soderberg line, for shear loading the end points are Sse=Se/(2ns) and Ssy=Sy/(2ns) Substitute the expressions for a and m Where
- 21. 21 Design Against Cyclic Loading I- Ductile Materials We are interested in finding the combination of stresses that produce the smallest safety factor , differentiate the equation below to find the minimum This yields
- 22. 22 Design Against Cyclic Loading I- Ductile Materials
- 23. 23 Design Against Cyclic Loading I- Ductile Materials
- 24. 24 Design Against Cyclic Loading II- Brittle Materials For brittle material, the forces in this case are assumed to be normal rather than tangent to the diagonal along the cut element. The maximum normal stress theory (MNST) extends from Se/ns to Su/ns. Following similar procedure, the safety factor and the shaft diameter will be given by