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Unformatted text preview: Mechanics of Solids Lecture 10 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 10 Outline Outline Inelastic bending c Partially Plastic Section c Fully Plastic Section c Shape Factor 2 Biaxial bending and axial load Mechanics of Solids  Lecture 10 Inelastic bending The elastic flexure formula derived earlier: My I σ = − is valid only while stress is proportional to strain No matter what the material properties are, the deformation of the crosssection is such that “plane sections remain plane” and therefore the strain distribution is “linear” . 3 y ε x N.A y σ x Stress distribution N.A Strain distribution x x E σ ε = The stress distribution is “not linear” . x d y dx θ ε = − 1 "curvature" d dx θ κ ρ = = = κ Mechanics of Solids  Lecture 10 Stress distribution for elastoplastic material P σ 1 2 3 4 σ y Crosssection 4 3 1&2 1 Consider the response of a simple rectangular beam loaded incrementally in four stages Zones of yielded material BMD t each stage 4 Strains Stresses Section ε ε y fully plastic ε >>> ε y σ = σ y M = M p 4 4 M p ε > ε y σ = σ y part plastic M > M y 3 3 ε = ε y σ = σ y all elastic M = M y 2 2 M y ε < ε y σ < σ y all elastic M < M y 1 at each stage of loading Mechanics of Solids  Lecture 10 Example 1 A beam of rectangular crosssection, 50 mm wide and 160 mm deep is made of mild steel with a yield stress of 350 MPa. Calculate the resultant moment when 20 mm of the section has yielded both at the top and the bottom. 5 Yielding StressStrain Diagram both in Tension & Compression σ y =350 MPa Yield Stress E ε σ ε y = = = = . 0 1 6 7 1 6 7 1 6 7 1 6 7 Mechanics of Solids  Lecture 10 Example 1 Elastic – Plastic Stress versus Strain Relationship for Steel 20 80 50 160 350 C 1 C 2 0.00167 6 Section 20 Stress T 2 T 1 Strain Mechanics of Solids  Lecture 10 Example 1 ( ) 1 1 C kN 350 20 50 350 = = × × = T ( ) 2 2 C kN 525 2 / 50 60 350 = = × × = T kN 875 2 1 = + = T T T 350 T C 1 C 2 50 60 20 80 7 1 2 20 2 60 2 3 60 2 91kNmm M T T = × + + × × = Stress 2 T 1 Mechanics of Solids  Lecture 10 Example 1 What is the “curvature” at the crosssection with this partial plastic moment. dx d θ κ = Curvature “rate of change of slope” 8 0.00167 20 60 Curvature = 0.00167/60 = 2.7833 x 105 mm1 Strain Distribution Mechanics of Solids  Lecture 10 Inelastic bending ε x The material is “nonlinear” and/or the stresses are outside the proportional range (i.e. yielded or plastic) 9 y N.A Strain distribution TENSION COMPRESSION Stress – Strain Diagram ε σ C T NA d n Mechanics of Solids  Lecture 10 Neutral axis for nonlinear material behaviour When the material is linearly elastic...
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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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