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- 1. د / احمد مصطفى هالة الطبيعي العالج كلية األولى الفرقة الطبية الحيوية الفيزياء Medical Biophysics الفيزياء
- 3. CHAPTER (4) Lesson Objectives: By the end of this chapter, a student will be able to : 1-Explain the Stress &Strain 2-Define the Mechanical Tests 3-Use Hooks Low 4-Classify Elastic coefficients .
- 4. Biomechanics Elasticity. It is the ability of a substance to restore its original shape and size after deformation. Elastic materials. Materials that restore its original shape and size after deformation .
- 5. Inelastic materials. Materials that cannot restore its original shape and size after deformation λ = Stress/Strain Where: λ (lambda) is the elastic modulus
- 6. Stress It is the force per unit area Strees σ s =Force/ Area=F/Ao N/m2
- 7. The Types of stress. 1-Tensile (Longitudinal) Stress. Tension results in a body when the body is acted by two forces equal in a magnitude directed away from each other and acting along the same line. The tensile stress increases the body length Tensile Stress σ L = Fn A Where: Fn is the force acting perpendicular A is the cross sectional area
- 8. 2:Compression Stress. Compression Stress in a body when the body is acted by two forces equal in a magnitude directed toward from each other and acting along the same line .The compression stress decreases the body length. Compression Stress σ V = Fn A
- 9. 3:Shearing Stress. Shearing Stress is result applied to the body which are directed toward from each other but not acted along the same straight line . Shear Stress (σ s ) = Fn A h A B C D F F X h A B C D
- 10. 4:Complex Stress. Complex Stress exist at the same time acted on the body.
- 11. 4:Strain. It is the deformation produced in a body as a result of stress.
- 12. The Types of strain. 1:Longitudinal strain. It is produced due to the tensile stress ϵ L = ∆L/L0 2:Volume Strain. It is produced due to the compressive stress which causes a change in the volume without shape. ϵ V = ∆V/V0
- 13. 3:Shearing Strain. It is produced due to the shear stress which causes a change in the shape of the body without change in volume. ϵS = tanφ
- 14. 5: Hook's law Hook's law states that :in elastic region the force applied to the spring is directly proportional to the extension. F = KX Where: X is the distance that the spring has been stretched or compressed away from the equilibrium position, which is the position where the spring would naturally come to rest (usually in meters) F is the restoring force exerted by the material (usually in newton). K is the force constant (or spring constant).
- 16. Elastic coefficients. According to the types of stress applied to a body and the accompanied type of strain produced in the body, there are different types of elastic module The Types elastic coefficients. 1:Youngs modulus. It is the ratio between the longitudinal stress to the longitudinal strain E= Longitudinal stress σ L = F/A Longitudinal strain ϵ L ∆L/L0
- 17. 2:Shear modulus It is the ratio between the shear stress to the shear strain. G= Shear stress σ s = F/A Shear strain ϵs tanΦ tanΦ = x/h
- 18. 3:Bulk modulus. Bulk modulus is the ratio between the compressive stress to the volume strain B= Compressive stress σ V = F/A Volume strain ϵV ∆V/V0
- 19. 7: Mechanical Tests In order to study mechanical properties of materials, there are different mechanical methods or test should be followed .These are : 1-The stress – strain test 2-The fatigue 3-The hardness 4-The creep test 1-The stress –strain test Stress strain test involves relation between the stress and strain from which a stress strain curve is obtained
- 20. A:The proportional limit A The proportional limit A is the stress at which the material still obey Hooks law B:The elastic limit B Elastic limit is the maximum stress C:The Yield strengthC The Yield strengthC is the stress at which the material behaves plastically. D:The ultimate strength D The ultimate strength D is the maximum stress that a material can with stand just before fracture or rupture.
- 21. 2:The Fatigue It is deformation produced in amaterial under the application of cyclic stress. 3:The Hardness It is the ability of the surface of a material to resist penetration by a point under a specific load. 4:The Creep test Creep test involves the relation between stain and time at constant stress. It is usually occurs at temperature near to the softening point of the material.
- 22. Poissons Ratio (µ ) µ = Lateral strain Longitudinal strain µ = ∆r/r ∆L/L µ = ∆b/b ∆L/L µ = ∆t/t ∆L/L
- 24. 1-Wire 12.5 m long is stretched to a length 12.58 m A) What is the strain of the stretched wire? B) If the wire is copper whose Young's modulus (E) =120x109 N/m2 ,what is the stress required to produce this strain? C) If the cross sectional area of the wire is 4x10-5 m2 ,what is the tension stretched wire?
- 25. A) Strain(ϵL) = elongation/originallength=ΔL/ Lo =12.58-12.5 = 6.3 x10-3 12.5 B) E= Longitudinal stress σ L = F/A Longitudinal strain ϵL ∆L/L0 Stress= E X Strain =(120x109 N/m2 )x (6.4x10-3 )= 7.68x109 N/m2
- 26. C)Strees (σ L) =Force/ Area=F/Ao Force = AreaX Strees (σ L) = (4x10-5m2 )x (7.68x109 N/m2)= 30.72x104N
- 27. 2-How much does a steel wire of length 1.5 m and radius (r) is 3x10-3 m is stretched when a tension of 600 N is applied to it? Young's modulus (E) for steel =200x109 N/m2
- 28. A=π r2 =3.14x (3x10-3)2 = 28.26x10-6 m2 Strees (σ) =Force/ Area=F/Ao = 600 N 28.26x10-6 m2 = 2.12x106 N/m2 E= Longitudinal stress σ L = F/A Longitudinal strain ϵL ∆L/L0
- 29. Strain = Stress E = 2.12x106 (N/m2) =1.06x10-4 200x109
- 30. Strain (ϵL) =elongation/original length=ΔL/Lo original length ΔL= Strain (ε) X elongation Lo =(1.06x10-4)(1.5m)=1.59x10-4m