MECH2430 – Mechanics of Solids 1 Course Coordinator: A/Prof Chris Wensrich Lecturer/Tutor : Mr Lee Kam Choong MECH2430 Mechanics of Solids 1 Introduction MECH2430 – Teaching Staff Course Coordinator: - A/Prof Chris Wensrich Christopher.Wensrich@newcastle.edu.au ES319 #16203 (it will divert to mobile if you wait) Lecturer/Tutor: - Lee Kam Choong kamchoong.lee@newcastle.edu.au 3 MECH2430 – Mechanics of Solids 1 – What is in the course:- Definition of stress and strain - Axial loadings, torsion and bending - Introductory elasticity theory and elasto-plasticity - Deflection of beams - Thin-walled pressure vessels - Transformation of stress and strain - Superposition of stress and stress analysis - Failure Theories 4 MECH2430 – Mechanics of Solids 1 5 MECH2430 – Mechanics of Solids 1 6 MECH2430 – Mechanics of Solids 1 Assessment will be as follows; • 3x 2hr in-class quizzes (Week 4, 9 and 12) (75%) • 1 engineering report (Week 13) (25%) – The quizzes will be closed book (formula sheet provided) – A more detailed breakdown of the topics covered in each of the quizzes can be found in the subject handout. – The report is due at the end of semester (details to follow) – Optional Reference : “Mechanics of Materials”, Beer, Johnston, DeWolf & Mazurek 7 MECH2430 Mechanics of Solids 1 Concept of Stress and Strain Solid Mechanics – Key Concerns • Solid mechanics is the study of solid materials behavior under the action of forces, temperature changes, and other influences. • There are three key concerns when solving problems in solid mechanics: • Loading (forces, moments, temperature change etc.) • Stress and strain (structure integrity/failure/safety margin) • Deformation (deflection/alignment/ serviceability) 9 Free Body Diagram (FBD) • The first step when solving a solid mechanics problem is to establish a free body diagram (FBD) • FBDs are simplified diagrams to show the relative magnitude and direction of all loadings acting upon an isolated solid body in any given scenario • FBDs should be: • Free of all mechanical interfaces (supports, connections etc.) • Include all external loading and internal reactions • Key dimensions and reference axis 10 Free Body Diagram ๐ ๐ถ๐ฆ ๐ ๐ถ๐ฅ C C ๐ ๐ D A ๐ ๐ด๐ฅ E B ๐ ๐ ๐ ๐ ๐ A truss system ABC supporting weight ๐ D A ๐ ๐ด๐ฆ E ๐/2 ๐ B ๐/2 2๐ ๐ FBD for truss system 11 Free Body Diagram ๐ ๐ถ๐ฆ ๐ ๐ถ๐ฅ ๐ ๐ถ๐ฆ ๐ ๐ถ๐ฆ ๐ ๐ถ๐ฅ C C ๐ ๐ถ๐ฅ C FBD for partial truss BC ๐น๐ต๐ถ FBD for ๐ truss system ๐ ๐ด๐ฅ D A ๐ ๐ด๐ฆ ๐/2 ๐ ๐ ๐ด๐ฅ ๐ ๐ด๐ฆ E B ๐/2 2๐ ๐ ๐น๐ต๐ถ A D E ๐/2 ๐น๐ต๐ถ FBD for truss BC ๐/2 FBD for truss AB ๐ ๐ด๐ฅ A D B ๐ ๐ด๐ฆ ๐น๐ด๐ต ๐/2 FBD for partial truss AB 12 Concepts of Body in Static Equilibrium Static – not in motion (relative to any fixed frame) Only address static problems whereby there is no inertia forces or these forces are not significant Equilibrium – a state of force balance in the context of mechanics 6 DOF (degree of freedom) Conditions for static equilibrium ๏ฅF x =0 ๏ฅMx = 0 ๏ฅF y =0 ๏ฅMy = 0 Fz ๏ฅF z =0 ๏ฅMz = 0 z Mx x These are the Static Equilibrium Equations Fx Mz My y Fy 13 Concepts of Body in Static Equilibrium Statically Determinate System – static equilibrium equations are sufficient to determine the internal forces and reactions of a system Free Body Diagram (free of mechanical interface) L known force P y L MA RAx x wall A A P B B RAy For this 2D system, the 6 equilibrium equations are reduced to only 3 equations ๏ฅF ๏ฅF ๏ฅM x y Z ,A = 0 ๏ RAx + 0 = 0 ๏ RAx = 0 = 0 ๏ RAy − P = 0 ๏ RAy = P = 0 ๏ M A − PL = 0 ๏ M A = PL All unknown reactions can be solved 14 Concepts of Body in Static Equilibrium Statically Indeterminate System – static equilibrium equations are insufficient to determine the internal forces and reactions of a system known force P y L MA RAx x wall A Free Body Diagram (free of mechanical interface) L A P B B Hinged Roller Support ๏ฅF x ๏ฅF y ๏ฅM Z ,A RAy RBy = 0 ๏ RAx + 0 = 0 ๏ RAx = 0 = 0 ๏ RAy + RBy − P = 0 ๏ RAy + RBy = P = 0 ๏ M A + RBy L − PL = 0 ๏ M A + RBy L = PL 3 unknowns in 2 equations, cannot be solved 15 Concepts of Body in Static Equilibrium Identifying reaction forces at common mechanical interfaces A Built-in End or Fixed End A B Hinged Roller Support B Hinged Slider B’ B B Hinged Support B 16 Bending Moment & Shear Force Diagram Revisit • Determination of maximum normal and shearing stresses requires identification of maximum internal shear force and bending couple. • Shear force and bending couple at a point are determined by considering the free body diagram of a section of the beam and applying an equilibrium analysis on the beam portions on either side of the section. 17 Bending Moment & Shear Force Diagram Revisit Sign Convection Shear Force + sense reaction Positive Shear Negative Shear Bending Moment Sagging : +BM Hogging: -BM 18 Bending Moment & Shear Force Diagram Revisit SOLUTION: • Draw the Free Body Diagram (FBD) and identify all applied and reaction forces • Section the beam at points near supports and load application points. Apply equilibrium analyses on resulting free bodies to determine internal shear forces and bending couples. • Identify the maximum shear and bending moment from plots of their distributions. FBD is free of mechanical interfaces (e.g. supports) 19 Bending Moment & Shear Force Diagram Revisit SOLUTION: • Draw the Free Body Diagram (FBD) and identify all applied and reaction forces • Section the beam at points near supports and load application points. Apply equilibrium analyses on resulting free bodies to determine internal shear forces and bending couples. • Identify the maximum shear and bending moment from plots of their distributions. Section AB 0 ≤ ๐ฅ < 2.5๐ : 20 ๐๐ ๐ ๐ด ๐ต ๐ฅ ๐ σ ๐น๐ฆ = 0 โน ๐ + 20 = 0 โน ๐ = −20 ๐๐ σ ๐๐ง = 0 โน ๐ + 20 โ ๐ฅ = 0 โน ๐ = −20๐ฅ ๐ด๐ก ๐ฅ = 0, ๐ = −20 ๐๐, ๐ = 0 ๐ด๐ก ๐ฅ = 2.5๐, ๐ = −20 ๐๐, ๐ = −50 ๐๐๐ ๐ด Section BC 2.5 ≤ ๐ฅ < 5.5๐ : 20 ๐๐ ๐ ๐ต ๐ฅ 46 ๐๐ ๐ถ ๐ σ ๐น๐ฆ = 0 โน ๐ + 20 − 46 = 0 โน ๐ = 26 ๐๐ σ ๐๐ง = 0 โน ๐ + 20 โ ๐ฅ − 46 ๐ฅ − 2.5 = 0 โน ๐ = 26๐ฅ − 115 ๐ด๐ก ๐ฅ = 2.5๐, ๐ = 26 ๐๐, ๐ = −50 ๐๐๐ ๐ด๐ก ๐ฅ = 5.5๐, ๐ = 26 ๐๐, ๐ = 28 ๐๐๐ 20 Bending Moment & Shear Force Diagram Revisit SOLUTION: • Draw the Free Body Diagram (FBD) and identify all applied and reaction forces • Section the beam at points near supports and load application points. Apply equilibrium analyses on resulting free bodies to determine internal shear forces and bending couples. • Identify the maximum shear and bending moment from plots of their distributions. Section AB 0 ≤ ๐ฅ < 2.5๐ : ๐ด๐ก ๐ฅ = 0, ๐ = −20 ๐๐, ๐ = 0 ๐ด๐ก ๐ฅ = 2.5๐, ๐ = −20 ๐๐, ๐ = −50 ๐๐๐ Section BC 2.5 ≤ ๐ฅ < 5.5๐ : ๐ด๐ก ๐ฅ = 2.5๐, ๐ = 26 ๐๐, ๐ = −50 ๐๐ ๐ด๐ก ๐ฅ = 5.5๐, ๐ = 26 ๐๐, ๐ = 28 ๐๐๐ Section CD 0 ≤ ๐ข < 2.0๐ : ๐ด๐ก ๐ข = 0, ๐ = −14 ๐๐, ๐ = 0 ๐๐ ๐ด๐ก ๐ข = 2 ๐, ๐ = −14 ๐๐, ๐ = 28 ๐๐๐ ๐ ๐ ๐ข ๐ท 14 ๐๐ 21 Basic Concepts – Force and stress Truss - a structural member that can only take axial load 3 FBC 5 FBC ?? Eg:- a pin-jointed frame; B A From statics we can calculate forces, 4 FBC 5 30 kN FAB = −40kN FBC = 50kN (Note that –ve implies compression) This simple system has lots of different types of stress… Pin Double Clevis Single Clevis 22 Basic Concepts – Force and stress SI SI (mm) Eg:- Link BC… Force N N Length m mm – Mass kg ton Stress/ Modulus N/m2 or Pa N/mm2 or MPa Stress represents the concentration of force – Force in N, Area in – Force in N, Area in mm2 gives Stress in MPa – Note that this is AVERAGE stress and the actual stress is usually not uniform (varies across the section). – The meaning of stress is that each element of area within the material (dA) carries a force of dF = σdA m2 gives Stress in Pa F = ๏ฒ ๏ณdA or ๏ณ = A dF dA โ๐น โ๐ด→0 โ๐ด ๐ = lim โ๐ด Average Stress is the force intensity. Stress is a point parameter. So we speak of “Stress at a Point” 23 Basic Concepts – Force and stress Uniform stress can only be achieved with “Centric Loading” ๐≠ ๐ ๐ด ๐ ๐= ๐ด ๐≠ Stress contours ๐ ๐ด Saint-Venant’s Principle : As long as the point of interest is remote from supported boundaries and application of force, the stress in a body may be determined from equilibrium consideration. 24 Basic Concepts – Shear Stress – Shear stress from transverse loading. Eg, the pin joint at C ๐น ๐น – Note that (once again) the shear stress is an average value 25 Basic Concepts – Shear Stress – ๐น ๐น Shear stress from transverse loading. Eg, the pin joint at C ๐น ๐น ๐น ๐น ๐น ๐น ๐น – ๐น Note that (once again) the shear stress is an average value ๐น ๐ด ๐น Force tangential to surface gives shear ๐น stress ๐ = ๐ด Force normal to surface gives normal ๐น stress ๐ = ๐ด 26 Basic Concepts – Shear Stress – Consider the shear stress in the pin joint at A; ๐น/2 ๐น/2 Shear plane ๐ด ๐น/2 ๐น/2 ๐น/2 ๐น ๐น ๐น ๐น/2 ๐น/2 ๐น/2 – ๐น/2 ๐น/2 Double Shear 27 Basic Concepts – Bearing Stress – The pins also apply normal stresses to the inside of the holes (and vice versa) projected area ๐ด Bearing – NetTension tension ShearOut Cleavage Is this an actual stress?? Bearing stress is not the actual stress on the bearing surface but a nominal value to simplify design calculations. compression 28 Stress on an Oblique Plane – So far we have only considered simple cases of normal and shear stress. – However, normal and shear are closely related; – Consider the stress on an oblique plane in an axially loaded member… ๐ด = ๐ด๐ cos ๐ โน ๐ด๐ > ๐ด In general on every cut face, there is normal and tangential forces – P cos(๏ฑ) P sin(๏ฑ) On ๐ด๐ , the normal force is ๐๐๐๐ ๐ and the tangential force is ๐๐ ๐๐ ๐ On this plane, the axial force (P) can be broken up into normal (F) and shear (V) components… F = P cos(๏ฑ ), V = P sin (๏ฑ ) 29 Stress on an Oblique Plane – The normal and shear stress on this plane is then; ๏ณ๏ฑ = F V , ๏ด๏ฑ = A๏ฑ A๏ฑ ๐น = ๐๐๐๐ ๐ ๐ = ๐๐ ๐๐ ๐ A , or, cos(๏ฑ ) P P ๏ณ ๏ฑ = cos2 (๏ฑ ), ๏ด ๏ฑ = sin (๏ฑ ) cos(๏ฑ ) A A Where A๏ฑ = Stress is • point parameter • dependent on orientation of plane of consideration Normal and shear stresses at a point, change with orientation 30 General Stress in 3-Dimensions – Consider a body under some general 3-dimensional load and cut a section plane parallel to the y-z axes. – Any element of area (dA) on the exposed surface will have a resultant force that can be decomposed into a normal and two shear components; ๐๐น๐ฅ , ๐๐๐ฅ๐ฆ and ๐๐๐ฅ๐ง are perpendicular (orthogonal) 31 General Stress in 3-Dimensions – As dA becomes infinitesimally small we calculate the normal and shear stresses on this plane at that point; โ๐น ๐ = lim โ๐ด→0 โ๐ด ๏ณx = – dVxy dFx dV ,๏ด xy = ,๏ด xz = xz dA dA dA Similarly, we can determine stresses on all the faces of an infinitesimally small cube of material at any point… 32 Stress in 3-Dimensions – What sort of thing is stress? Vector or Scalar?? State of Stress at a Point 33 Why ๐๐๐ = ๐๐๐ ? • Consider an elemental volume of โ๐ฅ × โ๐ฆ × โ๐ง • Stresses on opposite faces are equal and opposite • Consider moment about z axis • Forces in the z direction do not generate moment about z • Forces without moment arm • Moments from forces in opposite face cancel out each other ๐๐ฅ๐ฆ โ๐ฆโ๐ง × โ๐ฅ − ๐๐ฆ๐ฅ โ๐ฅโ๐ง × โ๐ฆ = 0 ⇒ ๐๐ฅ๐ฆ = ๐๐ฆ๐ฅ State of Stress at a Point 34 Stress in 3-Dimensions – What sort of thing is stress? Vector or Scalar?? Neither!! – Stress is something you haven’t met before… Stress is a Tensor!! – Very briefly, a tensor is a kind of map. The stress tensor tells us the concentration of force (as a vector) on any plane we choose (from the normal vector). – The stress tensor can be written as a matrix; Second Order Tensor (vector is a First Order Tensor) ๏ฉ๏ณ x ๏ด xy ๏ด xz ๏น ๏ช ๏บ ๐ = ๐๐๐ = ๏ช๏ด yx ๏ณ y ๏ด yz ๏บ ๏ช๏ด zx ๏ด zy ๏ณ z ๏บ ๏ซ ๏ป Stress Tensor is Symmetrical (off diagonal terms are the same) ๐๐ฅ๐ฆ = ๐๐ฆ๐ฅ ๐๐ฅ๐ง = ๐๐ง๐ฅ ๐๐ฆ๐ง = ๐ง๐ฆ 35 Basic Concepts- Strain – When stresses are applied to solid objects they deform (change shape) – Strain is a measurement of deformation – In axial loading, normal strain is given by; ๏ฅ= โ๐ฅ ๏ค L ๏ฅ = deformation/original length – For a variable cross section or distributed load (or more precisely)… ๏๏ค d๏ค = ๏x →0 ๏x dx ๏ฅ = lim ≠ ๐๐ฟ ÷ ๐๐ฅ ๐๐ฟ and ๐๐ฅ are called differential โ๐ฟ is the elongation of an elemental length โ๐ฅ ๐๐ฟ ๐๐ฅ is called a derivative 36 Stress and strain – In all materials, stress and strain are related to each other (sometimes in a complicated way) – For solid materials this relationship is measured using a “tensile test” (or “compression test”); In the tensile test, the load ๐ required to cause an extension ๐ฟ on the specimen are measured. The results are typically plotted ๐ vs ๐ฟ and then converted to ๐ vs ๐ 37 ๐ = ๐/๐ด ๐ Stress and strain ๐ฟ ๐ = ๐ฟ/๐ฟ Typical behaviour… Ductile Materials “Necking” linear ๏ณy - yield strength ๏ณu - ultimate strength ๏ณb - breaking strength linear nonlinear 0.2% Proof Strength ≡ ๐๐ฆ ๐ = 0.002 38 Stress and strain Typical behaviour… No “necking” Very little deviation from straight line 39 Stress and strain Stress-strain curve from Engineering Report data 40 Constitutive Relationships (Stress – Strain Relationship) – In all materials, stress and strain are related to each other (sometimes in a complicated way) – A “constitutive law” is a model for the way that stress and strain are related to each other. – Below the yield stress, engineering materials obey Hooke’s Law (for elastic materials). – For one dimension only:- ๏ณ = E๏ฅ Linear elastic materials E – Modulus of Elasticity or Young’s Modulus Steel ~ 210GPa, Aluminum ~ 70GPa, Brass ~ 120GPa, Diamond ~ 1200GPa 41 Deformation Under Axial Loading – Prismatic ๐ = lim For a uniform section P = E๏ฅ A P ๏ค So, ๏ฅ = = or... EA L โ๐ฅ→0 โ๐ฅ = ๐๐ฟ ๐๐ฅ ๐๐ฟ ๏ณ= – โ๐ฟ ๐๐ฟ = ๐๐ฅ ๐๐ฅ ๐๐ฟ ๐ → ๐ฟ = เถฑ ๐๐ฅ = เถฑ ๐๐๐ฅ = เถฑ ๐๐ฅ ๐๐ฅ ๐ธ๐ด PL ๏ค= EA →๐ฟ= ๐ ๐ธ๐ด ๐๐ฟ โซ ืฌโฌ1 ๐๐ฅ = ๐ธ๐ด (for prismatic) For varying section/properties or distributed load; Non-Prismatic ๏ฅ= ๐ฟ ๐ ๐ฟ=เถฑ ๐๐ฅ 0 ๐ธ๐ด ๐ฅ P d๏ค = or... EA dx L P ๏ค =๏ฒ dx AE 0 ๐ ๐๐ ๐ฟ๐ ๐=1 ๐ธ๐ด๐ ๐ฟ=เท 42 Statically Indeterminate Problems 2D Problem: σ ๐น๐ฅ = 0 , σ ๐น๐ฆ = 0, σ ๐๐ง = 0 Eg, Calculate the reactions เท ๐น๐ฆ = 0 → ๐ ๐ด + ๐ ๐ต − ๐ = 0 → ๐ ๐ด + ๐ ๐ต = ๐ 2 unknown 1 equation Statically Determinate System – static equilibrium equations are sufficient to determine the internal forces and reactions of a system Statically Indeterminate System – static equilibrium equations are insufficient to determine the internal forces and reactions of a system 43 Superposition Method Valid for Small Displacement Theory An alternative systematic approach… – Statically indeterminate systems have redundant reactions – Eliminate redundant reactions and treat them as “unknown loads” – Consider separately the deformations caused by the given loads and the redundant reactions – Superimpose the results.. Conceptually ๐ฟ๐ฟ = ๐ฟ๐ ⇒ ๐ฟ๐ฟ = −๐ฟ๐ ⇒ ๐ฟ๐ฟ + ๐ฟ๐ = 0 ๐ฟ๐ฟ : deformation due to applied loads ๐ฟ๐ : deformation due to R B 44 Superposition Method ๐ ๐ด,1 ๐ฟ ๐ ๐ด,2 Consider deflection at B ๐ฟ๐ต,1 + ๐ฟ๐ต,2 = 0 ๐๐ฟ ๐ ๐ฟ ⇒ 1− ๐ต =0 ๐ธ๐ด + = ๐ฟ๐ต,1 ⇒ ๐ ๐ต = From vertical equilibrium: ๐ ๐ด = ๐ − ๐ ๐ต = ๐ ๐ด,1 − ๐ ๐ด,2 ๐ฟ๐ต,2 ๐๐ด๐ถ ๐๐ด๐ต ๐๐ต๐ถ ๐ถ ๐ฟ๐ต,1 ๐ ๐ต ๐๐ด๐ถ = ๐ ๐ธ๐ด ๐ฟ1 ๐ ๐ฟ ๐ต ๐๐ต๐ถ = 0 ๐๐ด๐ถ ๐ฟ1 ๐๐ต๐ถ ๐ฟ2 = + ๐ธ๐ด ๐ธ๐ด ๐๐ฟ1 ๐ฟ๐ต,1 = ๐ธ๐ด ๐๐ด๐ต ๐ฟ ๐ธ๐ด ๐ ๐ต ๐ฟ =− ๐ธ๐ด ๐ฟ๐ต,2 = ๐ต ๐๐ด๐ถ ๐ฟ๐ต,2 ๐ ๐ต = −๐ ๐ต 45 Consider deflections with redundant reactions ๐๐ด๐ถ ๐๐ต๐ถ ๐ฟ ๐ถ ๐ = ๐ต ๐ต ๐ ๐ต ๐ ๐ต ๐๐ด๐ถ = ๐ − ๐ ๐ต Consider deflection at B ๐ฟ๐ต = ๐ฟ๐ถ๐ด + ๐ฟ๐ต๐ถ + ๐ฟ๐ด = 0 ๐ฟ๐ต = 0, ๐ฟ๐ด = 0 ⇒ ๐ฟ๐ถ๐ด + ๐ฟ๐ต๐ถ = 0 ๐ ๐ฟ ๐ ๐ฟ ⇒ ๐ด๐ถ 1 + ๐ต๐ถ 2 = 0 ๐ธ๐ด ๐ธ๐ด ⇒ ๐ − ๐ ๐ต ๐ฟ1 − ๐ ๐ต ๐ฟ2 = 0 ⇒ ๐ ๐ต ๐ฟ1 + ๐ฟ2 ๐๐ฟ1 = 0 ๐ฟ ⇒ ๐ ๐ต = 1 ๐ (same as before) ๐ฟ ๐๐ต๐ถ = −๐ ๐ต From Vertical equilibrium ๐ ๐ด = ๐ − ๐ ๐ต 46 Thermal Expansion (Superposition) In general, materials expand as the get hotter; – Thermal expansion is proportional to temperature change โ๐ ๏คT = ๏ก (๏T )L or ๏ฅ t = ๏ก (๏T ) ๐ผ โถ thermal expansion coefficient – Eg, The bar has no stress or strain before raising the temp by ๏T • Remove the reaction at B and apply as an unknown force Stress free thermal expansion ๐ฟ๐ ๐ต ๐ฟ๐ = ๐ฟ๐ ๐ต consider magnitude change or ๐ฟ๐ + ๐ฟ๐ ๐ต = 0, ๐ฟ๐ is a negative strain 47 Thermal Expansion (Superposition) ๏ค T = ๏ก (๏T )L RB L EA ๏คR = B The fixed supports imply zero deformation; ๏ค T + ๏ค R = ๏ก (๏T )L + Stress free thermal expansion B ๐ฟ๐ ๐ต RB L =0 EA So, RB = − AE๏ก (๏T ) or ๏ณ = − E๏ก (๏T ) ๐= Negative sign Indicates compression ๐ ๐ต ๐ด 48 Poisson’s Ratio Consider the following axial loading; – No stress is applied in the y and z directions; ๏ณy =๏ณz = 0 – ๐๐ฅ ๐๐ง This does NOT mean that there is no strain in the lateral (y and z) directions; ๏ฅy = ๏ฅz ๏น 0 – Poisson’s ratio is the ratio of lateral to axial strain; ๏ฎ =− ๏ฅy ๏ฎ๏ณ ๏ฅ = − z or ε y = ๏ฅ z = − x ๏ฅx ๏ฅx E ๐ (nu) is a positive value ๐๐ฆ ๐๐ฆ = −๐๐๐ฅ = − ๐๐๐ฅ ๐ธ ๐๐ง = −๐๐๐ฅ = − ๐๐๐ฅ ๐ธ 49 Generalised Hooke’s Law (multi-axial Loading) Consider a cube subject to normal stresses only; (for isotropic materials) isotropic materials have infinite planes of material symmetry - ๐ and ๐ธ are the same in any direction ๐๐ฅ ๐๐ง ๏ฅx = ๏ณx E ๏ฅ y = −๏ฎ ๏ฅ z = −๏ฎ ๐๐ฆ −๏ฎ ๏ณx E ๏ณx E ๏ณy E + −๏ฎ ๏ณy −๏ฎ E ๏ณz E −๏ฎ ๏ณy E + ๏ณz E ๏ณz E For anisotropic materials we have ๐๐ฅ๐ฆ , ๐๐ฆ๐ฅ , ๐๐ฅ๐ง โฏ and ๐ธ๐ฅ , ๐ธ๐ฆ , ๐ธ๐ง and the equations for strain are more complicated 50 Generalised Hooke’s Law (multi-axial Loading) = −๐๐๐ฅ = −๐๐๐ฆ = −๐๐๐ง = −๐๐๐ฅ = −๐๐๐ฆ = −๐๐๐ง 51 Dilation Consider a unit cube subject to normal stresses; ๐ฟ′ = ๐ฟ + ๐ฟ = ๐ฟ + ๐๐ฅ ๐ฟ ๐ฟ′ = ๐ฟ 1 + ๐๐ฅ = 1 + ๐๐ฅ Original Vol ๐0 = 1 × 1 × 1 = 1 New Volume:- V = (1 + ๏ฅ x )(1 + ๏ฅ y )(1 + ๏ฅ z ) ๏ป 1 + ๏ฅ x + ๏ฅ y + ๏ฅ z Dilation (ratio):- e = ๏ฅ + ๏ฅ + ๏ฅ Volumetric Strain x y z Substitute strains from Hooke’s Law; e= 1 − 2๏ฎ (๏ณ x + ๏ณ y + ๏ณ z ) E ๐ − ๐0 ๐0 1 + ๐๐ฅ + ๐๐ฆ + ๐๐ง − 1 ๐= 1 ๐ = ๐๐ฅ + ๐๐ฆ + ๐๐ง ๐= 52 Dilation 53 Bulk Modulus Special Case:- Hydrostatic pressure; ๏ณ x = ๏ณ y = ๏ณ z = − p e=− 3(1 − 2๏ฎ ) p E p=− where k = E k 3(1 − 2๏ฎ ) – ๐ ๐ ≡ ๐ = k is known as the “bulk modulus” or modulus of compression ๐ ๐ Obviously if the pressure is positive then the volume must decrease (e is –ve) – It follows that k must be positive – Therefore… 1 − 2๏ฎ ๏พ 0 or 0 ๏ผ ๏ฎ ๏ผ – If – ๐ธ= ๏ฎ = , k =๏ฅ 1 2 1 2 Κโถ∞ and the material is incompressible (eg rubber) For most metallic materials ๐ ≈ 0.3 For rubber or incompressible material ๐ = 0.5 54 Shear Strain ๏ณ vs ๏ฅ ๏ด vs ๏ง – Plotting – Strength in shear is about half of the strength in tension for most materials – In the elastic region; – G is the “Shear modulus” or modulus or rigidity (Steel ~75GPa, Aluminum ~ 25GPa) is similar to ๏ด = G๏ง 55 General Hooke’s Law (3D) (for isotropic materials only) ๏ฅx = ๏ณx E ๏ฅ y = −๏ฎ ๏ฅ z = −๏ฎ ๏ง xy = ๏ด xy G −๏ฎ ๏ณx E ๏ณx E ๏ณy E + ๏ณy E −๏ฎ , ๏ง xz = −๏ฎ −๏ฎ ๏ณy ๏ด xz G ๏ณz E + E ๏ณz E ๏ณz E , ๏ง yz = ๏ด yz G 56
