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# File Lecture2 - Lecture 2 1 Lecture 2 1 Outline-Poisson’s...

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Unformatted text preview: Lecture 2 1 Lecture 2 1 Outline: -Poisson’s ratio-Principle of superposition -Generalized Hooke’s Law -Stress and deformation in axially loaded members -Statically indeterminate axially loaded members -Composite axially loaded members 1. Poisson’s ratio At the end of the previous lecture we mentioned how the deformation is related with the stress through a certain type of material property. Due to the elasticity modulus when a body is subjected to axial tensile force, it elongates for a certain amount. The body may also laterally contract for a certain amount depending on a certain type of material property which is called Poisson ratio. Fig. 1 Change of the shape of a rod under tensile force The longitudinal strain of the bar is long L δ ε = The lateral strain of the bar due to lateral contraction is lat r δ ε ′ = − 1 Mechanics of Solids Dr Emre Erkmen Lecture 2 2 Poisson realized that within elastic range of the material, ratio of the two strains is a constant value which is called the Poisson’s ratio after his name. lat long ε ν ε = − Note that similarly, when it is subjected to an axial compressive force its sides expand laterally • Why negative sign? Longitudinal elongation cause lateral contraction and vice versa • Lateral strain is the same in all lateral (radial) directions • Poisson’s ratio is dimensionless, 0 ≤ ν ≤ 0.5 • Poisson’s ratio ν is unique for homogenous and isotropic material Homogenous material : The material has the same composition everywhere Isotropic material : The material has the same behaviour in all directions For example steel is a homogenous isotropic material, wood is a homogenous anisotropic material (because of the layers and thus wood is a laminated material). In the elastic range we will be using the three material constants that we have learnt so far, i.e. E modulus of elasticity, G shear modulus and ν Poisson’s ratio throughout this course. These three material constants are actually related by the equation ( ) 2 1 E G ν = + Example A bar is made of steel (whose modulus of elasticity is E = 200 GPa) and behaves elastically. Determine the change in its length and change in dimensions of its cross section after the load is applied. Fig. 2 A rod under tensile force Lecture 2 3 Normal stress in the bar is P A σ = 3 2 80 10 16 50 100 N MPa mm × = = × For, E = 200 GPa, strain in z-direction is E σ ε = 6 16 80 10 200 MPa GPa − = = × Axial elongation of the bar is 6 80 10 1.5 L m δ ε − = = × × 120 m μ = Using ν = 0.32, contraction strains in both x and y directions are 6 0.32 80 10 x y z ε ε νε − = = − = − × × 25.6 m m μ = − Thus, changes in the dimensions of the cross-section (along x and y direction) are 6 25.6 10 0.1 2.56 x x x z L m m δ ε νε μ − = = − = − × × = − 6 25.6 10 0.05 1.28 y y y z L m m δ ε νε μ − = = − = − × × = − 2. Principle of Superposition The effect of a given combined loading on a structure or machine part can be obtained by determining the...
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File Lecture2 - Lecture 2 1 Lecture 2 1 Outline-Poisson’s...

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