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Unformatted text preview: Lecture 2 1 Lecture 2 1 Outline: Poisson’s ratioPrinciple 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 zdirection 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 crosssection (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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 Three '11
 BROWN
 Force, Superposition, Deformation, Strain, Stress

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