Design of machine elements- unit 1.pptx

The machine design may be classified as follows :
Adaptive design. In most cases, the designer’s work is concerned with adaptation of existing designs. This type of design needs no special knowledge or skill and can be attempted by designers of ordinary technical training. The designer only makes minor alternation or modification in the existing designs of the product.
Development design. This type of design needs considerable scientific training and design ability in order to modify the
existing designs into a new idea by adopting a new material or different method of manufacture. In this case, though the
designer starts from the existing design, but the final product may differ quite markedly from the original product.
New design. This type of design needs lot of research, technical ability and creative thinking. Only those designers who have personal qualities of a sufficiently high order can take up the work of a new design.
The designs, depending upon the methods used, may be classified as follows :
Rational design. This type of design depends upon mathematical formulae of principle of mechanics.
Empirical design. This type of design depends upon empirical formulae based on the practice and past experience.
Industrial design. This type of design depends upon the production aspects to manufacture any machine component in the industry.
Optimum design. It is the best design for the given objective function under the specified constraints. It may be achieved by minimising the undesirable effects.
System design. It is the design of any complex mechanical system like a motor car.
Element design. It is the design of any element of the mechanical system like piston, crankshaft, connecting rod, etc.
Computer aided design. This type of design depends upon the use of computer systems to assist in the creation, modification, analysis and optimisation of a design.
Classification of Engineering Materials
The engineering materials are mainly classified as :
Metals and their alloys, such as iron, steel, copper, Aluminium, etc.

Non-metals, such as glass, rubber, plastic, etc.

The metals may be further classified as :
(a) Ferrous metals, and (b) Non-ferrous metals.

1. DESIGN OF MACHINE ELEMENTS Dr. C.GOPINATH Assistant Professor St. Joseph college of Engineering

2. INTRODUCTION TO MACHINE DESIGN The subject Machine Design is the creation of new and better machines and improving the existing ones. A new or better machine is one which is more economical in the overall cost of production and operation. The process of design is a long and time consuming one. From the study of existing ideas, a new idea has to be conceived. The idea is then studied keeping in mind its commercial success and given shape and form in the form of drawings. In the preparation of these drawings, care must be taken of the availability of resources in money, in men and in materials required for the successful completion of the new idea into an actual reality. In designing a machine component, it is necessary to have a good knowledge of many subjects such as Mathematics, Engineering Mechanics, Strength of Materials, Theory of Machines, Workshop Processes and Engineering Drawing.

3. Classifications of Machine Design The machine design may be classified as follows : Adaptive design. In most cases, the designer’s work is concerned with adaptation of existing designs. This type of design needs no special knowledge or skill and can be attempted by designers of ordinary technical training. The designer only makes minor alternation or modification in the existing designs of the product. Development design. This type of design needs considerable scientific training and design ability in order to modify the existing designs into a new idea by adopting a new material or different method of manufacture. In this case, though the designer starts from the existing design, but the final product may differ quite markedly from the original product. New design. This type of design needs lot of research, technical ability and creative thinking. Only those designers who have personal qualities of a sufficiently high order can take up the work of a new design. The designs, depending upon the methods used, may be classified as follows : Rational design. This type of design depends upon mathematical formulae of principle of mechanics. Empirical design. This type of design depends upon empirical formulae based on the practice and past experience. Industrial design. This type of design depends upon the production aspects to manufacture any machine component in the industry. Optimum design. It is the best design for the given objective function under the specified constraints. It may be achieved by minimising the undesirable effects. System design. It is the design of any complex mechanical system like a motor car. Element design. It is the design of any element of the mechanical system like piston, crankshaft, connecting rod, etc. Computer aided design. This type of design depends upon the use of computer systems to assist in the creation, modification, analysis and optimisation of a design.

4. G e n e r a l C o n s i d e r a t i o n s i n M a c h i n e D e si g n T y p e o f lo ad a n d s t r e ss e s c a u s e d b y t h e lo ad Motion of the parts or kinematics of the machine. Selection of materials. Form and size of the parts. Frictional resistance and lubrication. Safety of operation. W o r k s h o p f a c ili t i e s . Number of machines to be manufactured. Cost of construction. Assembling.

5. Introduction The knowledge of materials and their properties is of great significance for a design engineer. The machine elements should be made of such a material which has properties suitable for the conditions of operation. In addition to this, a design engineer must be familiar with the effects which the manufacturing processes and heat treatment have on the properties of the materials. In this chapter, we shall discuss the commonly used engineering materials and their properties in Machine Design. Classification of Engineering Materials T h e e n g i n e e r i n g m a te r ial s a r e m ai n l y c la ss i f i ed a s : M e t al s a n d t h e i r a ll o y s , s u c h a s i r o n , s tee l , c o pp e r , A l u m i n i u m , et c . Non-metals , such as glass, rubber, plastic, etc. T h e m et a l s m a y b e f u r t h er c la ss i f i ed a s : ( a ) Ferrous metals, and ( b ) Non-ferrous metals.

6. Mechanical Properties of Materials The following are the mechanical properties of materials.

7. General Procedure in Machine Design

8. Simple stresses and strain Load It is defined as any external force acting upon a machine part. Types of Loads Dead or steady load Live or variable load Suddenly applied or shock loads Impact load Stress When some external system of forces or loads act on a body, the internal forces (equal and opposite) are set up at various sections of the body, which resist the external forces. This internal force per unit area at any section of the body is known as unit stress or simply a stress . mathematically σ = P / A P = Force or load acting on a body, and A = Cross-sectional area of the body.

9. S t r a i n When a system of forces or loads act on a body, it undergoes some deformation. This deformation per unit length is known as unit strain or simply a strain. Strain,  = δ l / l δ l = Change in length of the body, and l = Original length of the body Tensile Stress Compressive Stress Shear Stress Types of stresses

10. PROBLEM 1: A coil chain of a crane required to carry a maximum load of 50 kN , is shown in Fig. Find the diameter of the link stock, if the permissible tensile stress in the link material is not to exceed 75 MPa . σ = P/A

11. PROBLEM 2: A cast iron link, as shown in Fig . is required to transmit a steady tensile load of 45 kN. Find the tensile stress induced in the link material at sections A-A and B-B. All dimensions in mm.

12. FACTOR OF SAFETY

13. PROBLEM 3: A pull of 80 kN is transmitted from a bar X to the bar Y through a pin as shown in Fig. If the maximum p e r m i ss i b l e te n s i l e s t r e s s i n t he ba r s i s 10 N / mm 2 an d t he p e r m i ss i b l e s h e a r s t r e s s i n t he p i n i s 8 N / m m 2 , f i nd t he diameter of bars and of the pin. G i v e n : P = 80 kN = 80 × 10 3 N;  t = 100 N / mm 2 ;  = 80 N / m m 2

14. Torsional and Bending Stresses in Machine Parts Torsional equation When a machine member is subjected to the action of two equal and opposite couples acting in parallel planes (or torque or twisting moment), then the machine member is said to be subjected to torsion . The stress set up by torsion is known as torsional shear stress .  = Torsional shear stress induced at the outer surface of the shaft or maximum shear stress, r = Radius of the shaft, T = Torque or twisting moment, J = Second moment of area of the section about its polar axis or polar moment of inertia, C = Modulus of rigidity for the shaft material, l = Length of the shaft, and  = Angle of twist in radians on a length l . T/J = τ /r = C ϴ / l

15. Formulas for Torsional moment T o r s i o n a l m o m e n t e qu a t i o n ( 1 ) P=2  NT/60 J=  /32 x d4 T= POWER (2) polar moment of inertia for solid shaft (3) polar moment of inertia for hollow shaft (4) Torque for solid shaft (5) Torque for hollow shaft (6) T=

16. Problem 1: A shaft is transmitting 100 kW at 160 r.p.m. Find a suitable diameter for the shaft, if the maximum torque transmitted exceeds the mean by 25%. Take maximum allowable shear stress as 70 MPa. solution P=2  NT/60 T=

17. A shaft is transmitting 97.5 kW at 180 r.p.m. If the allowable shear stress in the material is 60 MPa, find the suitable diameter for the shaft. The shaft is not to twist more that 1° in a length of 3 meters. Take C = 80 GPa . P o w er T r a n s m i t t ed P = 2  N T / 6

18. Problem 3: hollow shaft is required to transmit 600 kW at 110 r.p.m., the maximum torque being 20% greater than the mean. The shear stress is not to exceed 63 MPa and twist in a length of 3 meters not to exceed 1.4 degrees. Find the external diameter of the shaft, if the internal diameter to the external diameter is 3/8. Take modulus of rigidity as 84 Gpa . P = 2  N T / 6 P O W E R

19. T=

20. Bending moment equation /Bending Stresses Bending Stress in Straight Beams  = M/Z - B e nd i n g s t r e s s

21. An axle 1 meter long supported in bearings at its ends carries a fly wheel weighing 30 kN. at the centre. If the stress (bending) is not to exceed 60 MPa, find the diameter of the axle. SOLUTION Be n d i n g s t r e s s (  ) = M / Z 1 m

22. A beam of uniform rectangular cross-section is fixed at one end and carries an electric motor weighing 400 N at a distance of 300 mm from the fixed end. The maximum bending stress in the beam is 40 MPa. Find the width and depth of the beam, if depth is twice that of width. G i v e n: W = 400 N ; L = 300 m m ;  b = 40 M P a = 40 N / m m 2 ; h = 2 b B e nd i n g s t r e s s (  ) = M / Z

23. A pump lever rocking shaft is shown in Fig. The pump lever exerts forces of 25 kN. and 35 kN. concentrated at 150 mm and 200 mm from the left and right hand bearing respectively. Find the diameter of the central portion of the shaft, if the stress is not to exceed 100 MPa . Reaction forces R A + R B = 2 5 + 3 5 = 6 k N. Taking moments about A , we have R B × 950 = 35 × 750 + 25 × 150 = 30 000 R B = 30 000 / 950 = 31.58 kN = 31.58 × 10 3 N R A = (25 + 35) – 31.58 = 28.42 kN. = 28.42 × 103 N Bending moment at C = R A × 150 = 28.42 × 10 3 × 150 = 4.263 × 10 6 N-mm Bending moment at D = R B × 200 = 31.58 × 10 3 × 200 = 6.316 × 10 6 N-mm 150 mm 200 mm 600 mm

24. maximum bending moment is at D , therefore maximum bending m o m e n t , M = 6.316 × 1 6 N - mm. bending stress ( σ b ), B e nd i n g s t r e s s (  ) = M / Z

25. Bending Stress in Curved Beams

26. Resultant stress on the inner surface = σ t + σ bi Resultant stress on the outer surface, = σ t – σ bo maximum bending stress at the inner surface maximum bending stress at the outer surface Distance from the neutral axis to the inner surface Distance from the neutral axis to the outer surface D i s t a n c e b e t w ee n t he ce n t ro i d a l a x i s a nd n e u t r a l a x i s R a d i us of c urv a t ure of t he ce n t ro i d a l a x i s , = R R a d i us of c urv a t ure of t he n e u t r a l = R n

27. The frame of a punch press is shown in Fig. Find the stresses at the inner and outer surface at section X-X of the frame, if W = 5000 N . S o l u t i o n

28. The crane hook carries a load of 20 kN. as shown in Fig. The section at X-X is rectangular whose horizontal side is 100 mm. Find the stresses in the inner and outer fibers at the given section. S o l u t i on

29. T e n s il e s t r e s s

30. A C-clamp is subjected to a maximum load of W, as shown in Fig. If the maximum tensile stress in the clamp is limited to 140 MPa, find the value of load W . S o l u t i on

31. V a r i a b le S t r e s s e s in M a c h i n e P a r t s

32. Goodman Method A straight line connecting the endurance limit (  e ) and the ultimate strength (  u ), as shown by line AB in Fig. follows the suggestion of Goodman. A Goodman line is used when the design is based on ultimate strength and may be used for ductile or brittle materials. G oo d m a n E qu a t i o n or

33. Soderberg Method Proceeding in the same way as discussed in Art 6.20, the line AB connecting σ e and σ y , as shown in Fig.is called Soderberg's failure stress line . If a suitable factor of safety ( F.S. ) is applied to the endurance limit and yield strength, a safe stress line CD may be drawn parallel to the line AB. Let us consider a design point P on the line CD . Now from similar triangles COD and PQD , Soderberg Equation or

34. Determine the diameter of a circular rod made of ductile material with a fatigue strength (complete stress reversal),  e = 265 MPa and a tensile yield strength of 350 MPa. The member is subjected to a varying axial load from Wmin = – 300 × 1 3 N t o W m a x = 700 × 1 3 N and h as a s t r e s s c on c e n t r a ti on f a ct or = 1.8. U s e f a ct or of s a f et y as 2.0. S O L U T I O N M e a n l o a d V a r ia b l e l o a d Soderberg Equation

35. Determine the thickness of a 120 mm wide uniform plate for safe continuous operation if the plate is to be subjected to a tensile load that has a maximum value of 250 kN and a minimum value of 100 kN. The properties of the plate material are as follows: Endurance limit stress = 225 MPa, and Yield point stress = 300 MPa. The factor of safety based on yield point may be taken as 1.5. Solution Area, A = b × t = 120 t mm2 M e a n l o a d V a r ia b l e l o a d S o d e r b e r g E qu a t i o n

36. A steel rod is subjected to a reversed axial load of 180 kN. Find the diameter of the rod for a factor of safety of 2. Neglect column action. The material has an ultimate tensile strength of 1070 MPa and yield strength of 910 MPa. The endurance limit in reversed bending may be assumed to be one-half of the ultimate tensile strength. Other correction factors may be taken as follows: For axial loading = 0.7; For machined surface = 0.8 ; For size = 0.85 ; For stress concentration = 1.0. S ol u t ion M e a n l o a d V a r i a b l e l o a d S o d e r b e r g E qu a t i o n

37. A circular bar of 500 mm length is supported freely at its two ends. It is acted upon by a central concentrated cyclic load having a minimum value of 20 kN and a maximum value of 50 kN. Determine the diameter of bar by taking a factor of safety of 1.5, size effect of 0.85, surface finish factor of 0.9. The material properties of bar are given by : ultimate strength of 650 MPa, yield strength of 500 MPa and endurance strength of 350 MPa. M i n i m u m b e nd i n g m o m e n t Av e r a g e b e nd i n g m o m e n t v a r i a b l e b e nd i n g m o m e n t v a r ia b l e b e n d i n g s t r e s s

38. G oo d m a n E qu a t i o n Soderberg Equation

39. A 50 mm diameter shaft is made from carbon steel having ultimate tensile strength of 630 MPa. It is subjected to a torque which fluctuates between 2000 N-m to – 800 N-m. Using Soderberg method, calculate the factor of safety. Assume suitable values for any other data needed. Variable torque Variable shear stress

40. A simply supported beam has a concentrated load at the centre which fluctuates from a value of P to 4 P. The span of the beam is 500 mm and its cross-section is circular with a diameter of 60 mm. Taking for the beam material an ultimate stress of 700 MPa, a yield stress of 500 MPa, endurance limit of 330 MPa for reversed bending, and a factor of safety of 1.3, calculate the maximum value of P. Take a size factor of 0.85 and a surface finish factor of 0.9.

41. Combined Variable Normal Stress and Variable Shear Stress When a machine part is subjected to both variable normal stress and a variable shear stress; then it is designed by using the following two theories of combined stresses : 1. Maximum shear stress theory, and 2. Maximum normal stress theory. A hot rolled steel shaft is subjected to a torsional moment that varies from 330 N-m clockwise to 110 N-m counterclockwise and an applied bending moment at a critical section varies from 440 N-m to – 220 N-m. The shaft is of uniform cross- section and no keyway is present at the critical section. Determine the required shaft diameter. The material has an ultimate s t r e ng t h o f 55 0 MN / m 2 an d a yiel d s t r e ng t h o f 41 0 MN / m 2 . T a k e t he e ndu r an c e li m i t a s ha l f t he u lti m a t e s t r e ng t h, factor of safety of 2, size factor of 0.85 and a surface finish factor of 0.62.

42. Mean shear stress variable shear stress, equivalent shear stress, average bending moment, variable bending moment Mean bending stress variable bending stress

43. maximum equivalent shear stress,