Elasticity and Plasticity. 1.Basic principles of Elasticity and plasticity. 2.Stress and Deformation of Bars in Axial load 1 / 59
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1 Elasticity and Plasticity 1.Basic principles of Elasticity and plasticity 2.Stress and Deformation of Bars in Axial load 1 / 59
2 Basic principles of Elasticity and plasticity Elasticity and plasticity in building engineering theoretical basement for the theory of structures (important for steel, concret, timber structures design) - to be able design safe structures (to resist mechanical load, temperature load ) Statics: external forces, internal forces Elasticity and plasticity new terms: 1) stress 2) strain 3) stability 2 / 59
3 Principal terms Stress Strain Stability Material Elastic behavior of material Hooke s law - Elasticity Plastic behavior of material - Plasticity Load External force load (F, M, q) Temperature load 3 / 59
4 Stress and Strain Elasticity and plasticity - new terms stress and strain External forces, internal forces, stress V F V A... ormal component of the vector F... Tangential component of F... Element of cross section area A M A normal lim A 0 A stress (intensity of the internal forces distributed over a given section) shear lim A 0 V A 4 / 59
5 Basic principles of Elasticity and plasticity The initial presumptions of the clasic linear elasticity: 1) continuity of material, 2) homogenity(just one material) and isotropy (properties are the same in all directions), 3) linear elasticity (valid Hook s law), 4) the small deformation theory, 5) static loading, 6) no initial state of stress 5 / 59
6 Basic principles of Elasticity and plasticity 1. Continuity of material: A solid is a continuum, it has got its volume without any holes, gaps or any interruptions. Stress and strain is a continuous function. 2. Homogenity and isotropy 3. Linear elasticity 4. Small deformation 5. Static loading 6. o initial state of stress 6 / 59
7 Basic principles of Elasticity and plasticity 1. Continuity of material 2. Homogenity a isotropy: Homogeneous material has got physical characteristics identical in all places (concret, steel, timber). Combination of two or more materials ( concret + steel) is not homogeneous material. Isotropy means that material has got characteristics undependent on the direction (concret, steel yes, timber not). 3. Linear elasticity 4. Small deformation 5. Static loading 6. o initial state of stress 7 / 59
8 Basic principles of Elasticity and plasticity 1. Continuity of material 2. Homogenity and isotropy 3. Linear elasticity: Elasticity is an ability of material to get back after removing the couses of changes (for example load) into the original state. If there is a direct relation between stress and strain than we talk about Hooke s law = this is called physical linearity. 4. Small deformation 5. Static loading 6. o initial state of stress 8 / 59
9 Basic principles of Elasticity and plasticity Plasticity (on the contrary from linearity): This is an ability of material to deform without any rupture by non-returnable way. After removing the load there are staying permanent deformations. Plastic range Plastic range Ideally elastic-plastic material e 9 / 59
10 Basic principles of Elasticity and plasticity For axial load F Stress-strain diagram of an ideal elastic-plastic material: l Δl x + ε elast. ε plast εplast. dx dx e x dx e 1 dx Small-deformation theory Compression Plastic range Yield limit Y f y B arctg E = e a fy x x. Y ε plast Linear elastic range E ε elast. A=C B Plastic range Tension +e l/l - normal stress e strain 10 / 59
11 Basic principles of Elasticity and plasticity Hooke s law - physical relations between stress and strain it is valid only in the linear elastic range of the stress-stran diagram x e x.e Hooke s law + /A f y ε elast. Linear elastic range Y Tension Yield limit tana E e By substituting: x e x... Axial strain [-] a = arctg E x... ormal stress [Pa] A e x l l l l EA Other version of Hooke s law Hooke s law, another expression ε e = l/l E... Young s modulus of elasticity in tension and compression [Pa] 11 / 59
12 Basic principles of Elasticity and plasticity xz Hooke s law in shear xz xz G xz... Angle deformation xz... Shear stress [Pa] tana G G... modulus of elasticity in shear [Pa] 3 a = arctg G xz dz xz zx 3 dz Small-deformation theory: 1 Simplyfying: tan 12 / 59
13 Physical constants E, G, n In case of isotropic materila E, G are not mutual indipendent. E G Poisson s ratio 0 0,5 E 3 G E 2 Benchmark values of physical constants of some matherials: E G Steel MPa MPa 0,3 Glass MPa MPa 0,25 Granite to MPa - 0,2 Softwood E = MPa E^ = 300 MPa 600 MPa - Physical Relations between Strains and Deformations 13 / 59
14 Basic principles of Elasticity and plasticity Stress-strain diagrams steel concrete f e Elasticity limit f y... Yield limit f u... Ultimate limit Plasticity: the ability of material to get permanent deformations without fracture Ductility: plastic elongation of a broken bar (range /OT/), steel 15%). 14 / 59
15 Basic principles of Elasticity and plasticity 1. Continuity of material 2. Homogenity and isotropy 3. Linear elasticity 4. Small deformations theory: Changes of a shape of a (solid) structure are small with aspect to its size (dimensions). Than we can use a lot of mathematical simplifications, which usually lead to linear dependency. 5. Static loading 6. o initial state of stress 15 / 59
16 Basic principles of Elasticity and plasticity Theory of small deformations F d << l H b Theory of large deformation (finite deformation) H F b d l Theory of the I-st order l Theory of the II-nd order - Geometric nonlinearity d l a a M ay =H.l M ay =H.l+F.d The equilibrium conditions we set on the deformated construction (buckling of columns). 16 / 59
17 Basic principles of Elasticity and plasticity 1. Continuity of material 2. Homogenity and isotropy 3. Linear elasticity 4. Small deformations theory 5. Static loading: It means gradually growing of load (not dynamic effects) 6. o initial state of stress 17 / 59
18 Basic principles of Elasticity and plasticity 1. Continuity of material 2. Homogenity and isotropy 3. Linear elasticity 4. Small deformations theory 5. Static loading 6. o initial state of stress: In the initial state there are all stresses equal zerro. (Inner tension e.g. from the production). 18 / 59
19 Basic principles of Elasticity and plasticity 1. Continuity of material 2. Homogenity and isotropy 3. Linear elasticity 4. Small deformations theory 5. Static loading 6. o initial state of stress All these assumptions enable to use principal of superposition which is based on linearity of all mathematic relationship. 19 / 59
20 Saint - Venant principle of local effect F part without affect Jean Claude Saint-Venant ( ) q F F area of failure Makes possible to replace a given load by a simpler one for the purpose of an easier calculation of stresses in a member. the state of stress is influenced just in near surroundings of the load farther from this load we have nearly uniform distribution of stress ear surroundings Used for: replacement the surface load by the load statically equivalent but simpler for solution 20 / 59
21 Saint - Venant principle of local is not valid in these cases: a) concentrated loads on the end of bar: F part without affect area of failure part without affect q F area of failure const. const. x x b) bars with variable cross-section area: deduced conditions are valid for bars with gradual changes of cross-section area. Abrupt changes (announce by holes, nicks or narrowing) lead to no validity of condition. d q q b q q 21 / 59
22 Theorems of superposition and proportionality These theorems are valid when the basic principles are kept. Issac ewton ( ) Basic Theorems of Statics 1) Principle of action and reaction 2) Principle of superposition 3) Principle of proportionality 22 / 59
23 Stress and Strain Elasticity and plasticity - new terms stress and strain External forces, internal forces, stress V F V A... ormal component of the vector F... Tangential component of F... Element of cross section area A M A normal lim A 0 A stress (intensity of the internal forces distributed over a given section) shear lim A 0 V A 23 / 59
24 Stress Stress: vector, characterised by its components Unit: Pascal... [Pa] Format of the unit: Pa 2 m The Pascal is a small quantity, in practise we use multiplies of this unit MPa 10 6 Pa M 2 m mm 2 kpa 10 3 Pa k m 2 24 / 59
25 Basic (simple) types of mechanical stress 1. Axial loading 2. Bending 3. Torsion 4. Shear ormal force 0 a b + Rax tension F a b - Rax compression F 25 / 59
26 Basic (simple) types of mechanical stress 1. Axial loading 2. Bending 3. Torsion 4. Shear Bending moment My, Mz 0 a Raz M F compression tension tension M b Rbz + a Raz M compression F M b Rbz - 26 / 59
27 Basic (simple) types of mechanical stress 1. Axial loading 2. Bending 3. Torsion 4. Shear F3 F n v = 6 F1 Torsion moment Mx 0 +y +z +x 27 / 59
28 Basic (simple) types of mechanical stress 1. Axial loading 2. Bending 3. Torsion 4. Shear Shear force Vy, Vz 0 F V + V V - V a b Raz Rbz 28 / 59
29 Basic (simple) types of mechanical stress Type of the loading Axial Loading (tension, compression) Internal force Stress x - normal Bending M y, M z x - normal (bending stress) Shear V y, V z xy, xz - shear Torsion M x xy, xz - shear 29 / 59
30 Types of loading a) simple (axial loading, bending, torsion, shear) b) combined Combined loading: general bending (unsymmetric bending) eccentric axial loading torsion combined with tension or compression and with bending Due to the Principle of superposition, which is valid in a linear elastic range, we can solve the combined stresses. First by spliting up to basic stresses and then we can add these results together. 30 / 59
31 Axial loading tension, compression The only one inner force in each cross-section is an axial force. M x 0 V V 0 M y M 0 y z z 0 tension a b + Rax Tension F 0 compression a b - Rax compression F Basic principles and condition of solution 31 / 59
32 Conditions of solution a) deformated cross-sections stay on plane figure and it is vertically to the axis (Bernoulli hypothesis) Character of condition is deformationgeometrical. Cross sections stay mutually paralel without tapering. Outcome: Daniel Bernoulli ( ) dx dx before and after deformation xy x xz const. 0 0 xy xz for x = const. b) longitudinal fibres are not mutually compressed together y z 0 32 / 59
33 Stress and strain in external load F Axis x = axis of a member R l + x σ x Constant a) Stress under axial loading axial force F normal forces normal stress σ x [MPa] - Intensity of internal forces distributed over a given section -Tensile stress - positive sign compressive stress negative sign 33 / 59
34 Stress and strain in external load b) Strain under axial loading - deformation Deformations : elongation or contraction Axial strain Lateral strain e e l l Dimension s changes: l ν x y e 0,5 z e x Poisson s ratio l = l + l F e x Δl - elongation (dimensionless quantity [-]) y b b b = b+ b h = h+ h y e z z b b h h h h Circle - diameter d? 34 / 59
35 Stress and strain in temperature changes +ΔT a b a) stress If there is not defended the deformation of a member doesn t come up normal force and stress, later on indeterminate members b) Strain (thermal strain) e xt e yt e zt a T at - Coefficient of thermal expansion [ C -1 ] l T.T. l a ε xt = Δl/l = Δb/b = Δh/h = Δd/d T l = l + l b = b+ b h = h+ h 35 / 59
36 l = l0 m Example 1 The steel rod (see the picture) has a circle cross-sectional area of a diameter d = 0,025 m. E=2, MPa. ν =0,3 Determine σ x, elongation of the rod, the lateral changes ( in dimensions) and determine new dimensions of the rod). (Ignore the dead weight). Results: A = 490, m 2 σ x = 203,718MPa R Δl = 0,0097m =9, m = 9,7mm ε x = 9, l = 10,0097m + ε y = ε z = -2, Δd = -7, m = -7, mm d = 0,02499m P = 100 k 36 / 59
37 Example 2 R Ocelová tyč kruhového průřezu d = 0,01 m a délky l = 2 m je namáhána tahovou silou = 15 k. Determine the total deformation of the rod in its length (see the picture). Určete normálové napětí x, celkové S Fix prodloužení = 0: R - F1 l + F2 a příčné = 0 zkácení prutu. E = MPa, n = 0,3 R = F1 - F2 = = 30 k 1 = -R = -30k 2 = -R + F1 = = 10 k Result: l = Results: A1 = 314, m 2 A2 = 78, m 2 Δl1 = -1, m = -1,364mm Δl2 = 1, m = 1,212mm σ1 = -95,49MPa σ2 = 127,33MPa 37 / 59
38 Example 3 The concret column of a square cross-section of a side a =0,6m and the hight h = 3,6 m is uniformly warmed by T = 75 C. Determine the changes in dimensions of the column - cube. α T = C -1 x h = 3,6m z a = 0,6 m y a = 0,6 m Results: h = 3,6027m, a = 0,6005m 38 / 59
39 Example 4 Determine the stress in the circle rod and determine the change of the lenght of the rod. F =20k, t =15 C, d=0.02m, l =1,5m, l 1 = 1 m, l 2 = 2 m, E = 210GPa, α T = C -1 l =1,5 t Solution: 1- force from F 2- normal stress from F l 1 l 2 3- elongation of the rod (from + from temperature) Results: =30k, σ x = 95,5MPa (tension), l =0,682mm, l T =0,27mm 39 / 59
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