Strength of machine element of stress strain relation

This power point describe about stress and strain relation

1. DEBRE TABOR UNIVERSITY GAFAT INSTITUTE OF TECHNOLOGY Department of Electromechanical Engineering Course Name: Strength of Materials (5/6) Course Code: EMEg2081 By: Daniel Ketemaw 2016 E.C./2024 G.C.

2. Chapter-1 : Introduction – Concept of stress Contents of the Chapter Introduction Stresses in the members of a structure Axial loading; normal stress shearing stress, bearing stress in connections Stress on an oblique plane under axial loading Stress under general loading conditions; components of stress

3. Major Objectives of the Chapter To know the definitions of Stresses in the members of a structure To compute Axial loading; normal stress shearing stress, bearing stress in connections To compute on an oblique plane under axial loading To compute stress under general loading conditions; components of stress

4. 1.1. Introduction to Machine Tools Design The main objective of the study of the mechanics of materials is to provide the future engineer with the means of analyzing and designing various machines and load-bearing structures. Both the analysis and the design of a given structure involve the determination of stresses and deformations . This first chapter is devoted to the concept of stress .

5. The force per unit area, or intensity of the forces distributed over a given section, is called the stress and is denoted by the Greek letter (sigma).   1.2. Stresses in the members of a structure Fig.1.1. Member with an axial load

6. Cont. The stress in a member of cross-sectional area A subjected to an axial load P (Fig. 1.2) is therefore obtained by dividing the magnitude P of the load by the area A : … (1) Fig.1.2. Member with an axial load

7. Cont. A positive sign will be used to indicate a tensile stress (member in tension) and a negative sign to indicate a compressive stress (member in compression) Since P expressed in newtons (N) and A in square meters ( ), the stress will be expressed in => called a pascal (Pa).  

8. 1.3. Axial loading; normal stress, shearing stress, bearing stress in connections Axial Loading & normal stress Fig.1.3. (a) Beam used to support a 30-kN load & (b) load for specific point Q

9. Cont. Because of the forces and acting on its ends B and C (Fig. 1.3a) are directed along the axis of the rod => rod is under axial loading . The internal force in the rod and the corresponding stress was perpendicular to the axis of the rod. The internal force was therefore normal to the plane of the section (sectioned area) as shown in Fig. 1.1 and the corresponding stress is described as a normal stress .  

10. Cont. Therefore, gives us the normal stress in a member under axial loading . To define the stress at a given point Q of the cross section, we should consider a small area as shown in Fig. 1.3b. Dividing the magnitude of by , we obtain the average value of the stress over . Letting approach zero , we obtain the stress at point Q: … (2)  

11. Cont. In general, the value obtained for the stress at a given point Q of the section is different from the value of the average stress given by the above formula and is found to vary across the section. In a slender rod subjected to equal and opposite concentrated loads P and as shown in Fig. 1.4a.   Fig.1.4. Stress distributions at different sections along axially loaded member.

12. Cont. This variation is small in a section away from the points of application of the concentrated loads (Fig. 1.4c), but it is quite noticeable in the neighborhood of these points (Fig.1.4b and d). The magnitude of the resultant of the distributed internal forces is: expressed using P : The actual distribution of stresses in any given section is statically indeterminate .

13. Cont. In practice , it will be assumed that the distribution of normal stresses in an axially loaded member is uniform , except in the immediate vicinity of the points of application of the loads . When we assume that the internal forces are uniformly distributed across the section, it follows from elementary statics that the resultant P of the internal forces must be applied at the centroid C of the section as shown in Fig. 1.5a.

14. Cont. The uniform distribution of stress is possible only if the line of action of the concentrated loads P and passes through the centroid of the section considered => called centric loading .   Fig.1.5. (a) Stress distributions at different sections along axially loaded member & (b) load applied at the centroid C

15. Cont. If a two-force member is loaded axially , but eccentrically as shown in Fig. 1.6a, we find from the conditions of equilibrium of the portion of member shown in Fig.1.6b that the internal forces in a given section must be equivalent to a force P applied at the centroid of the section and a couple M of moment M = Pd . The distribution of stresses— cannot be uniform . Fig.1.6. Eccentric axial loading.

16. Cont. Shearing stress Shearing stress is obtained when transverse forces P and are applied to a member AB as shown in Fig. 1.7.   Fig.1.7. Member with transverse loads.

17. Cont. Let the two forces passing a section at C between the points of application as shown in Fig. 1.8a, we obtain the diagram of portion AC shown in Fig. 1.8b. The elementary internal forces are called shearing forces , and the magnitude P of their resultant is the shear in the section. Fig.1.8. Member with transverse loads.

18. Cont. Dividing the shear force P by the area A of the cross section, we obtain the average shearing stress in the section. Denoted by Greek letter (tau): … (3) Noted that, unlike normal stress the distribution of shearing stresses across the section cannot be assumed uniform .  

19. Cont. Thus the actual value of the shearing stress varies from zero at the surface of the member to a maximum value that may be much larger than the average value . Shearing stresses are commonly found in bolts, pins, and rivets used to connect various structural members and machine components (Photo 1.1).   Photo 1.1. Member with transverse loads.

20. Cont. Consider the two plates A and B, which are connected by a bolt CD as shown in Fig. 1.9. Let plates are subjected to tension forces of magnitude F , stresses will develop in the section of bolt corresponding to the plane .   Fig.1.9. Bolt subject to single shear.

21. Cont. The average shearing stress in the section is: … (4) Not only for single bolt, also for double bolts shear will be exhibit. Fig.1.10. Bolts subject to double shear.

22. Cont. Observing that the shear P in each of the sections is P = F/2 , we conclude that the average shearing stress is: … (5) Bearing Stress In Connections Bolts, pins, and rivets create stresses in the members they connect, along the bearing surface, or surface of contact.

23. Cont. As show in Fig. 1.11, the bolt exerts on plate A subjected by a force P equal and opposite to the force F exerted by the plate on the bolt. The force P represents the resultant of elementary forces distributed on the inside surface of a half-cylinder of diameter d and of length t equal to the thickness of the plate. Fig.1.11. Bearing stress on single bolt.

24. Cont. The stress , called the bearing stress , obtained by dividing the load P by the area of the rectangle representing the projection of the bolt on the plate section shown in Fig. 1.11b. Noted that: the area subjected by the load P is equal to td , where t is the plate thickness and d the diameter of the bolt. Thus, we have: … (6)  

25. Noted that, axial forces cause both normal and shearing stresses on planes which are not perpendicular to the axis of the member. Similarly, transverse forces exerted on a bolt or a pin cause both normal and shearing stresses on planes which are not perpendicular to the axis of the bolt or pin. 1.4. Stress on an oblique plane under axial loading Fig.1.12. (a) Axial forces & (b) Transverse forces

26. Cont. What if the section forming an angle with a normal plane? Consider the plate as shown in Fig. 1.13a, which is subjected to axial forces P and . Thus, the free-body diagram of the portion of member located to the left of that section become as shown in Fig. 1.13b.   Fig.1.13. free body diagram.

27. Cont. Resolving P into components F and V , respectively normal and tangential to the section (Fig. 1.13c), we have: … (7) The force F represents the resultant of normal forces distributed over the section, and the force V the resultant of shearing forces as shown in Fig. 1.13d.

28. Cont. From the normal force we will get the average normal stress and from the shearing force; the shearing stress. i.e. … (8) Notice that; … (9) denotes the area of a section perpendicular to the axis of the member.   =  

29. Cont. Substituting for F and V from (7) and from(9) into (8), to get: … (10) The above equation also rewrite as: … (11) From eq-11 the normal stress is maximum when , i.e., when the plane of the section is perpendicular to the axis of the member, and that it approaches zero as approaches .  

30. Cont. Thus, the value of when reaches maximum: … (12) And for shearing stress, is zero for and , and that for it reaches its maximum value: … (13)      

31. Cont. At the same is true for using eq(11); … (14) Thus, note that the same loading may produce either a normal stress, with no shearing stress (Fig. 1.14b). Or normal and a shearing stress of the same magnitude (Fig. 1.14c & d).   Fig.1.14. free body diagram.

32. Noted that the examples of the previous sections were limited to members under axial loading and connections under transverse loading as shown in Fig. 15. However, most structural members and machine components are under more involved loading conditions. 1.5. Stress Under General Loading Conditions; Components of Stress Fig.1.15. free body diagram.

33. Cont. Consider a body subjected to several loads , , and as shown in Fig. 1.15. Let the loads pass through the point Q within the body. We shall first pass a section through Q , using a plane parallel to the yz -plane as shown in Fig. 1.16.   Fig.1.16. free body diagram.

34. Cont. The sectioned part is subjected to some of the original loads ( ), and to normal ( ) and shearing forces ( ) distributed over the section and these forces acting on a small area surrounding point Q shown in Fig. 16. Note that the superscript x is used to indicate that the forces act on a surface perpendicular to the x axis .   Fig.1.17. free body diagram.

35. Cont. Notice the normal force has a well-defined direction, i.e. normal to the area at point-Q along x-axis as shown in Fig 1.17. However, the shearing force may have any direction in the plane ( yz -axis ) of the section as shown in Fig 1.17. Thus, resolve into two component forces, and , in directions parallel to the y and z axes as shown in Fig. 1.17. To get the stresses, by dividing each force by the limited area .  

36. Cont. Thus the normal stress by will be in the x-axis direction; … (15) The shearing stress by is: … (16) The shearing stress by is: … (17)  

37. Cont. The normal stress is positive if the corresponding arrow points in the positive x-direction , and negative to the opposite.   Positive: the body is in tension Negative: the body is in compression

38. End of the chapter Thank you