THE GENERAL ELASTICITY PROBLEM IN SOLIDS

Transcription

1 Chapter 10 TH GNRAL LASTICITY PROBLM IN SOLIDS In Chapters 3-5 and 8-9, we have developed equilibrium, kinematic and constitutive equations for a general three-dimensional elastic deformable solid bod. In this chapter, we will summarize these equations for the one-, two- and three-dimensional states of stress. We will find that in each case, a sstem of equations will be obtained that must be solved with appropriate boundar conditions for the particular problem being addressed. The solution procedure for the general case will also be addressed in this chapter Necessar quations for a 3-D Stress State For a linear elastic solid under static equilibrium, we can now summarize the following three sets of equations for an 3-D bod: 1. Static quilibrium quations Conservation of Linear Momentum -component: -component: z-component: σ σ σ z + σ + σ + σ z + σ z + ρg =0 + σ z + ρg =0 3equations σ zz + ρg z =0 2. Constitutive quations for a linear elastic isotropic material are given b no thermal effects included: ε = 1 [σ ν σ + σ zz ] ε = 1 [σ ν σ + σ zz ] ε zz = 1 [σ zz ν σ + σ ] ε = 1+ν σ 6equations 10.2 ε z = 1+ν σ z ε z = 1+ν σ z 229

2 230 CHAPTR 10. TH GNRAL LASTICITY PROBLM IN SOLIDS Alternatel, 10.2 ma be inverted to ield: σ = σ = σ zz = σ = σ = σ z = σ z = σ z = σ z = 1 + ν1 2ν [1 ν ε + νε + νε zz ] 1 + ν1 2ν [νε +1 ν ε + νε zz ] 1 + ν1 2ν [νε + νε +1 ν ε zz ] 1+ν ε ν ε z 1+ν ε z 3. Kinematics Strain-Displacement equations for small strain are given b ε ε ε z [ε] = ε ε ε z ε z ε z ε zz = u 1 u u + u + uz 6equations 1 u u u + u + uz 1 uz u z + u u z + u 10.4 Consequentl, for the general 3-D linear elasticit problem, we have a sstem of 15 governing partial differential equations: 3 equilibrium linear momentum equations for σ 6 constitutive equations [σ] =[C] {ε} 6 kinematic strain equations {ε} = function of displacement gradients that relate the 15 unknown variables stresses 6, strains 6 and displacements 3. The 15 governing equations are coupled partial differential equations that must be solved simultaneousl. As in the solution of an differential equation, boundar conditions must be specified to solve this boundar value problem. These boundar conditions must be either displacements or tractions on ever point of the boundar Necessar quations for a 2-D Stress State There are man special cases of the general elasticit problem that are of practical interest and can be solved!. These include Plane Stress and Plane Strain problems.

3 10.2. NCSSARY QUATIONS FOR A 2-D STRSS STAT 231 z σ σ σ σ Figure 10.1: Plate in State of Plane Stress Plane stress occurs in thin bodies which have non-zero stresses in one plane onl and all out-ofplane stresses are zero. For eample, in the - plane, we have: Forageometr that is plane stress in the - plane, we assume σ zz = σ z = σ z =0. The stress σ σ 0 tensor reduces to [σ] = σ σ 0 and 1. Static quilibrium becomes -component: -component: σ σ + σ + ρg =0 + σ + ρg = Constitutive quations for a linear elastic isotropic material in plane stress condition become have to substitute plane stress requirement of σ zz = σ z = σ z = 0 into general equations for strain: Alternatel, 10.6 ma be inverted to ield: ε = 1 [σ νσ ] ε = 1 [σ νσ ] ε = 1+ν σ 10.6 ε zz = ν σ + σ σ = 1 + ν1 2ν [1 νε + νε + νε zz ] σ = 1 + ν1 2ν [νε +1 νε + νε zz ] σ zz = 0 σ = 1 + ν ε 10.7 σ z = 0 σ z = 0

4 232 CHAPTR 10. TH GNRAL LASTICITY PROBLM IN SOLIDS 3. Kinematics Strain-Displacement equations become [ε] = ε u 1 u ε u ε ε 0 = 1 u ε + u u zz u z 10.8 Note that for 2-D plane stress, we have 8 unknowns 3 stresses, 3 strains, and 2 displacements Necessar quations for a 1-D Stress State Consider the case where a uniaial load is applied to a uniaial bar oriented in the direction. For this one-dimensional state of stress, onl σ eists and σ = σ zz = σ = σ z = σ z =0. Wehave the following set of necessar equations: 1. Static quilibrium becomes dσ -component: d + ρg = Constitutive quation for a linear elastic isotropic material becomes: σ = 1 + ν1 2ν [1 νε + νε + νε zz ] ε = σ, ε = νσ, ε zz = νσ 3. Kinematics Strain-Displacement equation becomes u ε [ε] = 0 ε 0 = u ε zz 0 0 u z For the 1-D case, we have 3 unknowns: 1 stress σ, 1 strain ε and 1 displacement u. The other non-zero strain and displacement components appearing in equations and can be written in terms of the three unknown quantities listed previousl. It must be noted that while there is onl one non-zero stress σ, this stress produces non-zero strains normal to the direction of loading ε and ε zz due to the Poisson s ratio effect D Special Cases All of the general 3-D equations relating stress, strain and deformation ma be specialized to some important 1-D cases which involve long, slender geometries like beams, rods, bars, tubes, etc. in which the stress state is 1-D. In the following chapters, we will consider in detail three special cases: bars in tension, bars in torsion and beam bending. Tpical eamples of these are shown below. Bar with aial force onl Bar or pipe in torsion Beam in bending In each case, we will develop epressions for the appropriate displacement or twist and stress within the bod as function of position within the bod.

5 10.4. GNRAL SOLUTION PROCDUR 233 A 1, 1,L 1 A 2, 2,L P Arigidhorizontal bar is attached to the bottom of each vertical bar. The horizontal bar remains horizontal when pulled downward Figure 10.2: 25, 000 in-lbs 4 20 G = psi Figure 10.3: P a L Figure 10.4: 10.4 General Solution Procedure The governing equations are for an elastic solid bod are in general a sstem of coupled partial differential equations that must be solved simultaneousl. As in the solution of an differential equation, boundar conditions must be specified to solve this boundar value problem. These boundar conditions must be either displacements or tractions on ever point of the boundar. Consequentl, for ever problem we must satisf the following four 4 sets of equations: Four Sets of quations to be Satisfied for An Solid Deformable Bod

6 234 CHAPTR 10. TH GNRAL LASTICITY PROBLM IN SOLIDS 3-D 2-D 1-D Static quilibrium quation from COLM: Constitutive quation Stress-Strain: Kinematics strain-displacement: Boundar Conditions: Depends on problem geometr and loading for plane stress condition ample 10-1 Consider a uniaial bar that eperiences a simple etension in the direction under a uniform loading σ = σ 0 on the end surfaces as shown below: n σ 0 σ 0 L Figure 10.5: Tr σ = σ 0, σ = σ zz = σ = σ z = σ z =0 quilibrium equations: identicall satisfied Constitutive quations ε = 1 [σ νσ + σ zz ] = σ 0 ε = 1 [σ νσ σ zz ] = ν σ 0 ε zz = 1 [σ zz νσ + σ ] = ν σ 0 ε = ε z = 0 ε z = 0 Kinematics u = σ0, u = νσ0, u z = νσ0 z Substitute σ 0 = P A into the equation for u: u = P A 10.14

7 10.4. GNRAL SOLUTION PROCDUR 235 resultant force = P = A σ 0 da = σ 0 A σ 0 σ 0 Figure 10.6: Boundar Conditions: [ tn ] =[n] [σ] lateral surface : [ ] [ t n = 0 n2 n 3 ] end surfaces : [ ] [ t n = 1 0 ] 0 σ σ = [ ] = [ σ ] = L =0 [ ] [ t n = 1 0 ] 0 σ = [ σ ]

8 236 CHAPTR 10. TH GNRAL LASTICITY PROBLM IN SOLIDS Deep Thought Collaborative learning brings ou one step closer to the solution steps of BVP in continuum mechanics.