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Unformatted text preview: Mechanics of Solids Lecture 2 by Dr Emre Erkmen Lecturer, School of Civil and Environmental Engineering, The University of Technology Sydney Office: 2.520 Phone:9514 9769 Email: [email protected] Lecture hours: Tuesdays 11:0014:00, Wednesdays 15:0018:00 Office hours: Tuesdays 15:0017:00 1 Mechanics of Solids  Lecture 2 Outline Outline Poisson’s ratio Stress and deformation in axially loaded members  Principle of superposition  Statically indeterminate axially loaded members  Composite axially loaded members 2 Mechanics of Solids  Lecture 2 • When a body is subjected to axial tensile force, it elongates and laterally contracts • Similarly, when it is subjected to an axial compressive force its sides expand laterally Poisson’s ratio 3 Mechanics of Solids  Lecture 2 • Strains of the bar are: ε long = δ L ε lat = δ’ r Poisson’s ratio Poisson’s ratio Poisson’s ratio, ν = − ε lat ε long • Early 1800s, Poisson realized that within elastic range, ratio of the two strains is a constant value, since both are proportional. 4 Mechanics of Solids  Lecture 2 • Why negative sign? Longitudinal elongation cause lateral contraction and vice versa • Lateral strain is the same in all lateral (radial) directions Poisson’s ratio Poisson’s ratio • Poisson’s ratio is dimensionless, 0 ≤ ν ≤ 0.5 • Poisson’s ratio is unique for homogenous and isotropic material 5 Homogenous material : The material has the same composition everywhere Isotropic material : The material has the same behaviour in all directions 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 Mechanics of Solids  Lecture 2 Bar is made of A36 steel and behaves elastically. Determine change in its length and change in dimensions of its cross section after load is applied. Example 6 Mechanics of Solids  Lecture 2 Normal stress in the bar is rom tables, 200 GPa, strain in irection is Example P A σ = 3 2 80 10 16 50 100 N MPa mm × = = × From tables, E = 200 GPa, strain in zdirection is Axial elongation of the bar is, 7 E σ ε = 6 16 80 10 200 MPa GPa − = = × 6 80 10 1.5 L m δ ε − = = × × 120 m μ = Mechanics of Solids  Lecture 2 Using ν st = 0.32, contraction strains in both x and y directions are Example 6 0.32 80 10 x y z ε ε ν ε − = = − = − × × 25.6 m m μ = − Thus changes in dimensions of crosssection are 8 6 25.6 10 0.1 2.56 x x x z L m m δ ε ν ε μ − = = − = − × × = − 6 25.6 10 0.05 1.28 y y y z L m m δ ε ν ε μ − = = − = − × × = − Mechanics of Solids  Lecture 2 Principle of superposition P=P 1 +P 2 L P 1 P 2 L L The effect of a combined loading can be obtained by determining the effects of each load separately and combining the results 9 After applying the first load P 1 , we did not consider the length of the bar as 1 L δ + Why?...
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This note was uploaded on 08/22/2011 for the course ENG 48331 taught by Professor Brown during the Three '11 term at University of Technology, Sydney.
 Three '11
 BROWN
 Environmental Engineering

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