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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 11:00-14:00, Wednesdays 15:00-18:00 Office hours: Tuesdays 15:00-17:00 1
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Mechanics of Solids - Lecture 10 Outline - Inelastic bending checkbld Partially Plastic Section checkbld Fully Plastic Section checkbld Shape Factor 2 - Biaxial bending and axial load
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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 ε σ No matter what the material properties are, the deformation of the cross-section is such that “plane sections remain plane” and therefore the strain distribution is “linear” . 3 y x N.A y 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
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Mechanics of Solids - Lecture 10 Stress distribution for elasto-plastic material P σ ε 1 2 3 4 σ y Cross-section 4 3 2 M y 1&2 1 Consider the response of a simple rectangular beam loaded incrementally in four stages Zones of yielded material BMD at each stage 4 Strains Stresses Section 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 ε < ε y σ < σ y all elastic M < M y 1 of loading
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Mechanics of Solids - Lecture 10 Example 1 A beam of rectangular cross-section, 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. σ 5 Yielding Stress-Strain Diagram both in Tension & Compression σ y =350 MPa Yield Stress E ε ε y = 0 .00167 00167
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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
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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 2 C 1 C 2 50 160 20 80 7 1 2 20 2 60 2 3 60 2 91kNmm M T T = × + + × × = Stress T 1
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Mechanics of Solids - Lecture 10 Example 1 What is the “curvature” at the cross-section 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 10 -5 mm -1 Strain Distribution
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Mechanics of Solids - Lecture 10 Inelastic bending ε x The material is “non-linear” and/or the stresses are outside the proportional range (i.e. yielded or plastic) C d n 9 y N.A Strain distribution TENSION COMPRESSION Stress – Strain Diagram ε σ T NA
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