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1.033/1.57 Mechanics of Material Systems (Mechanics and Durability of Solids I) Franz-Josef Ulm 1.033/1.57

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If Mechanics was the answer, what was the question ? Traditional: – Structural Engineering – Geotechnics Structural Design -Service State (Elasticity) -Failure (Plasticity or Fracture) -Mechanism 1.033/1.57

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If Mechanics was the answer, what was the question ? Material Sciences and Engineering – New materials for the Construction Industry Micromechanical Design of a new generation of Engineered materials Concrete with Strength of Steel 1.033/1.57

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If Mechanics was the answer, what was the question ? Diagnosis and Prognosis – Anticipating the Future 1.033/1.57

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If Mechanics was the answer, what was the question ? Diagnosis and Prognosis – Anticipating the Future 1.033/1.57

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If Mechanics was the answer, what was the question ? Traditional: – Structural Engineering – Geotechnics –… Material Sciences and Engineering – New materials for the Construction Industry – Engineered Biomaterials,… 1.033/1.57 Diagnosis and Prognosis – Anticipating the Future – Pathology of Materials and Structures (Infrastructure Durability, Bone Diseases, etc.) – Give numbers to decision makers…

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If Mechanics was the answer, what was the question ? 1.033/1.57 – Fall 01 Mechanics and Durability of Solids I: –Deformation and Strain –Stress and Stress States –Elasticity and Elasticity Bounds –Plasticity and Yield Design 1.033/1.57 1.570 – Spring 01 Mechanics and Durability of Solids II: – Damage and Fracture – Chemo-Mechanics – Poro-Mechanics – Diffusion and Dissolution

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Content 1.033/1.57 Part I. Deformation and Strain 1 Description of Finite Deformation 2 Infinitesimal Deformation Part II. Momentum Balance and Stresses 3 Momentum Balance 4 Stress States / Failure Criterion Part III. Elasticity and Elasticity Bounds 5 Thermoelasticity, 6 Variational Methods Part IV. Plasticity and Yield Design 7 1D-Plasticity – An Energy Approac 8 Plasticity Models 9 Limit Analysis and Yield Design 1.033/1.57

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Assignments 1.033/1.57 Part I. Deformation and Strain HW #1 Part II. Momentum Balance and Stresses HW #2 Quiz #1 Part III. Elasticity and Elasticity Bounds HW #3 Quiz #2 Part IV. Plasticity and Yield Design HW #4 Quiz #3 FINAL 1.033/1.57

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Part I: Deformation and Strain 1. Finite Deformation 1.033/1.57

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Modeling Scales Λ d H B d l

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1.033/1.57 Modeling Scale (contd) d << l << H Material Science Scale of Continuum Mechanics

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1.033/1.57 LEVEL III Mortar, Concrete > 10 -3 m Cement paste plus sand and Aggregates, eventually Interfacial Transition Zone LEVEL II Cement Paste < 10 -4 m C-S-H matrix plus clinker phases, CH crystals, and macroporosity LEVEL I C-S-H matrix < 10 -6 m Low Density and High Density C-S-H phases (incl. gel porosity) LEVEL 0 C-S-H solid 10 -9 –10 -10 m C-S-H solid phase (globules incl. intra-globules nanoporosity) plus inter-globules gel porosity

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1.033/1.57 LEVEL III Deposition scale > 10-3 m LEVEL II (Micro) Flake aggregation and inclusions 10 -5 10 -4 m LEVEL I (Nano) Mineral aggregation 10 -7 10 -6 m LEVEL 0 Clay Minerals 10 -9 –10 -8 m Scale of deposition layers Visible texture. Flakes aggregate into layers, Intermixed with silt size (quartz) grains. Different minerals aggregate to form solid particles (flakes which include nanoporosity). Elementary particles (Kaolinite, Smectite, Illite, etc.), and Nanoporosity (10 – 30 nm).

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1.033/1.57 Modeling Scales Λ d H B d l

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1.033/1.57 Transport of a Material Vector ξ=x X dx=F·dX Deformation Gradient e1e1 e2e2 e3e3 X x

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1.033/1.57 Exercise: Pure Extension Test e1e1 e2e2 e3e3

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1.033/1.57 Exercise: Position Vector e 2 (e 3 ) [1β(t)]H L [1+α]L e1e1 x 1 =X 1 (1+α); x 2 =X 2 (1β); x 3 =X 3 (1β);

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1.033/1.57 Exercise: Material Vector / Deformation Gradient e 2 (e 3 ) e1e1 [1-β(t)]H L [1+α(t)]L F 11 = (1+α); F 22 = F 33 = (1-β)

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1.033/1.57 e1e1 e2e2 e3e3 X x Volume Transport d dtdt dX dX 1 dX 2 dx 2 dx 1 d t = det(F)d Jacobian of Deformation

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1.033/1.57 Transport of an oriented material NdA U u=F.U nda surface (a) (b) nda=J t F -1 NdA Chapter 1

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1.033/1.57 Transport of scalar product of two Material Vectors dY dX e2e2 e3e3 e1e1 dy π/2θ dx E = Green-Lagrange Strain Tensor dx·dy= dX·(2E+1)· dY

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1.033/1.57 Linear Dilatation and Distortion Length Variation of a Material Vector: Linear Dilatation λ(e α )=(1+2Ε αα ) 1/2 1 Angle Variation of two Material Vectors: Distortion

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1.033/1.57 (b)(a) e1e1 e1e1 e2e2 e2e2 dXdx Training Set: Simple Shear

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1.033/1.57 Problem Set #1 R R-Y α=α(X) eyey x exex Initial Fiber Deformed Fiber

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double shear X2X2 αX1X1

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