1. BFC 20903 (Mechanics of Materials) Chapter 1: Stress & Strain Shahrul Niza Mokhatar shahruln@uthm.edu.my

2. Chapter Learning Outcome 1.Defined the relationship between stress and strain 2.Analyse the stress and strain using related equations 3.Determine and analyse the deformation of a rod of uniform or variable cross section under one or several load 4.Determine the principal stress using equation and Mohr’s circle method BFC 20903 (Mechanics of Materials) Shahrul Niza Mokhatar (shahrul@uthm.edu.my

3. Introduction Mechanics of materials is a branch of mechanics that studies the internal effects of stress and strain in a solid body that is subjected to an external loading. Stress is associated with the strength of the material from which the body is made. Strain is a measure of the deformation of the body.

4. Type of forces Normal force, N – Acts perpendicular to the area. – Developed when the external loads to push or pull on the two segments of the body Shear force, V – Shear force lies in the plane of the area. – Developed when the external loads tend to cause the two segments of the body to slide.

5. Type of forces Torque or torsional moment, T -Developed when the external loads tend to twist one segments of the body with respect to the other. Bending moment, M -Cause by the external loads that tend to bend the body about an axis within the plane of the area.

6. Stresses Stress is the internal force exerted by one part of an elastic body upon the adjoining part. Stress has a dimension of Newton/m 2 Generally, they are 6 independent components of stress at each point. or

7. Type of stress – Normal stress or uniaxial stress, σ - stress is evenly distributed over the entire cross-section. Normal stress includes tensile and compressive stress – Shear stress, τ – results when a force tends to make part of the body or one side of a plane slide past the other.

8. Type of stress: Shear Stress

9. Bearing stress, τ – Bearing stress is a type of normal stress but it involves the interaction of two surfaces. The bearing stress is the pressure experience by the second surface due to the action from the first surface. Example: the pressure between bolt and plate at a joint. Type of stress: Bearing Stress

10. Example 1

11. Example 2

12. Example 2: Solution

13. Example 3

14. Example 3: Solution

15. Quiz 1 : Shear Stress

16. Quiz 1: Solution

17. Strain Measure of deformation representing the displacement between particles in the body relative to a reference length. Ratio of change in length due to deformation to the original length. It is dimensionless quantity. Numerical values of strain are usually very small, especially for structural materials, which ordinarily undergo only small changes in dimensions.

18. Normal strain - Normal strain (ε) is the deformation of a body which involved elongation or contraction. When a bar of length L and cross-sectional area A is subjected to axial tensile force P through the cross- section's centroid, the bar elongates. The change in length divided by the initial length is the bar's engineering strain. The symbol for strain is ε (epsilon). The strain in an axially loaded bar is: Types of Strain

19. Shear strain - Shear strain is a strain which involved a shear deformation i.e. body twist due to torsion and a distorted cuboid as shown in Figure 1.8. Strain changes the angles of an object and shear causes lines to rotate.

20. Stress & Strain relationships Stress & strain relationships Elastic Proportional Limit (Hooke's Law) From the origin O to the point called proportional limit, the stress-strain curve is a straight line.

21. Stress & Strain relationships Elastic Limit The elastic limit is the limit beyond which the material will no longer go back to its original shape when the load is removed, or it is the maximum stress that may be developed such that there is no permanent or residual deformation when the load is entirely removed. Yield Point/ Yield Strength Yield point is the point at which the material will have an appreciable elongation or yielding without any increase in load. The material is said to undergo plastic deformation. Strain hardening Point C to D is called as strain hardening region whereas the curve rises gradually until it flatten at D. The stress which correspond to point D is called ultimate strength/stress Ultimate Strength/Stress The maximum ordinate in the stress-strain diagram is the ultimate strength or tensile strength. Rapture Strength (Fracture) Rapture strength is the strength of the material at rupture. This is also known as the breaking strength (final point).

22. Offset method Beside steel, other materials such as aluminium, glass, brass and zinc, constant yielding will not occur beyond the elastic range. This metal often does not have a well defined yield point. Therefore, the standard practice to define yield strength for this metal is graphical procedure called the offset method. Normally a 0.2% (0.002 mm/mm) is chosen, and from this point on the strain (ε) axis, a line parallel to the initial straight-line portion of the stress-strain diagram is drawn. The point where this line intersects the curves defines the yield strength.

23. Example 5

24. Example 5: Solution From the graph: (a)E = 112.07/0.00125 (b)σ y = 230 N/mm 2 (c)σ max = 270 N/mm 2

25. Hooke’s Law Stiffness; Modulus Young Stiffness is a material's ability to resist deformation. The stiffness of a material is defined through Hooke's Law. Young's Modulus is the slope of the linear-elastic region of the stress-strain curve.

26. Hooke’s Law where E is Young's Modulus (the modulus of elasticity), a material property. Values of E for different materials are obtained experimentally from stress-strain curves. From Hooke’s Law, the displacement can be derived: Consider a homogenous rod BC of length L and uniform cross section of area A subjected to a centric axial load P (Figure). If the resulting axial stress σ = P/A does not exceed the proportional limit of the material, the Hooke’s law can be apply and write as follow:

27. Poisson ratio Poisson's ratio is the ratio of lateral contraction strain to longitudinal extension strain in the direction of stretching force. Tensile deformation is considered positive and compressive deformation is considered negative. The definition of Poisson's ratio contains a minus sign so that normal materials have a positive ratio. Poisson's ratio, also called Poisson ratio or the Poisson coefficient. Poisson's ratio is a materials property.

28. Example 6

29. Working stress, permissible stress and temperature stress Temperature stress  An object will expand when heated and contract when the temperature drops.  Since this is the effect of temperature on the member then the corresponding stress and strain are called temperature stress and temperature strain.  The variation of the length due to temperature change depends upon its coefficient of linear expansion or contraction α where α is the change in length for a unit change of temperature per unit original length.

30. Example 7: Tutorial

31. Example 7 : Solution

32. Assignment 1: Individu

33. Plane Stress and Mohr Circle

34. Stress Analysis Using Equation and Mohr Diagram Method Rotation angle, θ Clockwise –ve Counterclockwise +ve

37. Principal stresses occur on the principal planes of stress with zero shear stress Maximum shearing stress Normal stress on the y’ and x’axis Shear stress corresponds to x’ and y’

38. Example 8 For the state of plane stress shown, determine (a) the principal stress (b) principal planes (c)maximum shear stress

39. Example 8: Solution

41. Example 9: Tutorial The state of plane stress at a failure point on shaft as shown in Fig. Represent this stress state in terms of its principal stresses

42. Example 9: Solution From the established sign convention, we have σ x = -20 MPa, σ y = 90 MPa, τ xy = 60 MPa Orientation of element Recall that θ must be measured positive counterclockwise from the x axis to the outward normal (x’ axis) on the face of the element.

43. Principal stress Example 9: Solution

44. The principal plane on which each normal stress acts can be determined with θ p2 = -23.7 0 Example 9: Solution

45. The state of plane stress at a point on a body is represented on the element as shown in Fig. Represent this stress state in terms of maximum in-plane shear stress and associated average normal stress From the established sign convention, we have σ x = -20 MPa, σ y = 90 MPa, τ xy = 60 MPa Example 10

46. Orientation of element Note how these angles are formed between x and x’ axes. Maximum in-plane shear stress Example 10: Solution

47. The shear stress corresponding to θ s2 = 21.3 0 can be determined by This positive result indicates that τ max = τ xy’ acts in the positive y’ direction on this face (θ = 21.3 0 ). Average normal stress

48. Mohr Circle Mohr’s circle can be used to determine the principal stresses, the maximum in-plane shear stress and average normal stress or the stress on any arbitrary plane. By adding and squaring each equation, the value of θ can be eliminated τ τ

49. τ τ Mohr Circle: Two manners

51. Example 11 Using Mohr’s cirlce method. Determine: (a) normal and shearing stresses after rotated 40 0 (b)principal stress (c)maximum shear stress

52. Example 11: Solution Determine the centre line, C and radius of Mohr’s circle R Determine the coordinate A and A’ as a stress on x and y C = A (σ = 15 Mpa and τ = 4 Mpa) A’ (σ = 5 Mpa and τ = - 4 Mpa)

53. C Determine the angle of ACP, location of plane for the maximum stress Determine B as a plane of x’ corresponds to the angle of 40 0 (a) So, stress on x’is point B and stress on y’is point B’ (b) Principal stress is point P and Q (c) Max shear stress is point S and R A’ (5,- 4) τ (+ve) A (15, 4) tan 2 θ = 4/5 = 38.66 0 θ p = 19.33 0 τ xy = -4 P 2 θ = 80 0 from A Angle BCP = 80 0 - θ p = 60.67 0 60.67 0 B B’ σ x’ = 10 + 6.4cos 41.34 0 = 14.81 MPa τ x’y’ = -6.4sin 41.34 0 = -4.23 MPa σ y’ = 10 - 6.4cos 41.34 0 = 5.19 MPa τ x’y’ = 6.4sin 41.34 0 = 4.23 MPa σ x‘ = 14.81 σ min = 3.6 σ max = 16.4 R S Q τ (-ve) σ y’ = 5.19 σ x = 15 38.66 0

54. Stress on plane after rotated 40 0 θ = 40 0 5.19 MPa 4.23 MPa 14.81 MPa θ p = 19.3 0 3.6 MPa 16.4 MPa Stress on principal plane θ smax = 90 0 - 38.66 0 = 25.67 0 6.4 MPa Shear stress on the shear plane

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