1.033/1.57 Mechanics of Material Systems (Mechanics and Durability of Solids I) Franz-Josef Ulm 1.033/1.57If Mechanics was the answer, what was the question ? • Traditional: – Structural Engineering – Geotechnics Structural Design - Service State (Elasticity) - Failure (Plasticity or Fracture) - Mechanism 1.033/1.57If 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 1.033/1.57If Mechanics was the answer, what was the question ? • Diagnosis and Prognosis – Anticipating the Future 1.033/1.57If Mechanics was the answer, what was the question ? • Diagnosis and Prognosis – Anticipating the Future 1.033/1.57If Mechanics was the answer, what was the question ? • Traditional: • Diagnosis and – Structural Engineering – Geotechnics –… • Material Sciences and Engineering – New materials for the Construction Industry – Engineered Biomaterials,… Prognosis – Anticipating the Future – Pathology of Materials and Structures (Infrastructure Durability, Bone Diseases, etc.) – Give numbers to decision makers… 1.033/1.57If Mechanics was the answer, what was the question ? • 1.033/1.57 – Fall 01 • 1.570 – Spring 01 Mechanics and Mechanics and Durability of Solids I: Durability of Solids II: – Deformation and Strain – Damage and Fracture – Stress and Stress States – Chemo-Mechanics – Elasticity and – Poro-Mechanics Elasticity Bounds – Diffusion and – Plasticity and Yield Dissolution Design 1.033/1.57Content 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 1.033/1.57Assignments 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 1.033/1.57Part I: Deformation and Strain 1. Finite Deformation1.033/1.57Λ R dΩ l B d H Modeling Scales 1.033/1.57Modeling Scale (cont’d) d << l << H 10−6 m Material Science d 10−3 m 10−2 m Scale of Continuum Mechanics l 1.033/1.57Cement paste plus sand and Aggregates, eventually Interfacial Transition Zone C-S-H matrix plus clinker phases, CH crystals, and macroporosity Low Density and High Density C-S-H phases (incl. gel porosity) C-S-H solid phase (globules incl. intra-globules nanoporosity) plus inter-globules gel porosity LEVEL III Mortar, Concrete > 10-3 m LEVEL II Cement Paste < 10-4 m LEVEL I C-S-H matrix < 10-6 m LEVEL ‘0’ C-S-H solid 10-9–10-10 m 1.033/1.57Scale 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 LEVEL III Deposition scale > 10-3 m LEVEL II (‘Micro’) Flake aggregation and inclusions 10-5 -4 m LEVEL I (‘Nano’) Mineral aggregation 10-7 -6 m LEVEL ‘0’ Clay Minerals 10-9–10-8 m x x x x x 1.033/1.57 Nanoporosity (10 – 30 nm). –10–10Λ R dΩ l B d H Modeling Scales 1.033/1.57Transport of a Material Vector e1 e2 X dX x dx ξ=x − X dx=F·dX Deformatione3 Gradient 1.033/1.57Exercise: Pure Extension Test e1 e2 e3 1.033/1.57Exercise: Position Vector e2 (e3) e1 X x L [1+α]L [1−β(t)]H x1=X1(1+α); x2=X2(1−β); x3=X3(1−β); 1.033/1.571.033/1.57e1e2 (e3)dXdxL[1+α(t)]L[1-β(t)]HXxF11= (1+α); F22= F33= (1-β) Exercise: Material Vector / Deformation GradientVolume Transport dΩ e1 e2 X x dX1 dX2 dX3 dx1 dx2 dx3 dΩt dX dx dΩt= det(F)dΩ Jacobian of Deformation e3 1.033/1.57Transport of an oriented material NdA U u=F.U nda surface (a) (b) nda=JtF-1 NdA Chapter 11.033/1.57Transport of scalar product of two Material Vectors e1 e3 e2 dX dY π/2 dy dx π/2−θ E = Green-Lagrange Strain Tensor dx·dy= dX·(2E+1)· dY 1.033/1.57Linear Dilatation and Distortion Length Variation of a Material Vector: Linear Dilatation λ(eα)=(1+2Εαα)1/2−1 Angle Variation of two Material Vectors: Distortion 2Εαβsinθ(eα,eβ)= [(1+2Εαα) (1+2Εββ)]1/2 1.033/1.57Training Set: Simple Shear e2 X x ξ e2 dX dx dx1 dx2 =dX2 e1 e1 (a) (b) 1.033/1.57ex ey R-Y α=α(X) Initial Fiber Deformed Fiber y x R Problem Set #1 1.033/1.57X2 α X1 double shear