东南大学：《固体力学基础》课程教学课件（英文讲稿）11 Simple Elastoplastic BVPs

Simple Elastoplastic BVPsmi@se.edu.cn

Simple Elastoplastic BVPs

Outline·Assumptions（假设）·Introduction（引言）·Summary of governing equations（弹塑性控制方程）· Cylindrically symmetric elastoplastic solids （轴对称）· Hollow cylinder under monotonic internal pressure (空心圆筒受单调增载）· Spherically symmetric elastoplastic solids （中心对称）· Hollow sphere under monotonic internal pressure （球壳受单调增载)·Hollow sphere under cyclic internal pressure（球壳循环受载)2

Outline • Assumptions（假设） • Introduction（引言） • Summary of governing equations（弹塑性控制方程） • Cylindrically symmetric elastoplastic solids（轴对称） • Hollow cylinder under monotonic internal pressure（空心 圆筒受单调增载） • Spherically symmetric elastoplastic solids（中心对称） • Hollow sphere under monotonic internal pressure（球壳受 单调增载） • Hollow sphere under cyclic internal pressure（球壳循环受 载） 2

Assumptions?Body force density is given? Prescribed boundary tractions and/or displacements All displacements are small. This means that we can usethe infinitesimal strain tensor to characterizedeformation; we do not need to distinguish betweenstress measures, and we do not need to distinguishbetween deformed and undeformed configurations of thesolid when writing equilibrium equations and boundaryconditions? The material is isotropic, elastic-perfectly plastic solid.? Neglect temperature changes3

Assumptions • Body force density is given. • Prescribed boundary tractions and/or displacements • All displacements are small. This means that we can use the infinitesimal strain tensor to characterize deformation; we do not need to distinguish between stress measures, and we do not need to distinguish between deformed and undeformed configurations of the solid when writing equilibrium equations and boundary conditions. • The material is isotropic, elastic-perfectly plastic solid. • Neglect temperature changes. 3

Lntroduction. The elastic limit: This is the load required to initiate plastic flow in the solid.Theplasticcollapseload:Atthisload,thedisplacements inthe solidbecomeinfinite..Residual stress:If a solid isloadedbeyond theelasticlimitandthenunloaded,asystem of self-equilibrated stress is established in the material.:Shakedown: If an elastic-plastic solid is subjected to cyclic loading and themaximum load during the cycle exceeds yield, then some plastic deformationmust occur in the material during the first load cycle. However, residual stressesare introduced in the solid, which may prevent plastic flow during subsequentcycles of load. This process is known as shakedown," and the maximum load forwhich it can occur isknown as the shakedownlimit. The shakedown limit is oftersubstantially higher than the elastic limit, so the concept of shakedown can oftenbe used to reduce the weight of a design.Cyclic plasticity: For cyclic loads exceeding the shakedown limit, aregion in thesolid will be repeatedly plastically deformed.4

Introduction • The elastic limit: This is the load required to initiate plastic flow in the solid. • The plastic collapse load: At this load, the displacements in the solid become infinite. • Residual stress: If a solid is loaded beyond the elastic limit and then unloaded, a system of self-equilibrated stress is established in the material. • Shakedown: If an elastic-plastic solid is subjected to cyclic loading and the maximum load during the cycle exceeds yield, then some plastic deformation must occur in the material during the first load cycle. However, residual stresses are introduced in the solid, which may prevent plastic flow during subsequent cycles of load. This process is known as “shakedown,” and the maximum load for which it can occur is known as the shakedown limit. The shakedown limit is often substantially higher than the elastic limit, so the concept of shakedown can often be used to reduce the weight of a design. • Cyclic plasticity: For cyclic loads exceeding the shakedown limit, a region in the solid will be repeatedly plastically deformed. 4

Summary of Governing EguationsDisplacement-strain relation:L? Strain partition: de, = de, +dep. Incremental stress-stain relation:0,S1 +vVEE3 02dep302 0yV2? Equations of static equilibrium: ji,j + F, = 0.. Traction BCs on S: O,n, =tj? Displacement BCs on Su: u, = ü5

Summary of Governing Equations • Displacement-strain relation: 5  , ,  1 2 ij i j j i    u u • Strain partition: e p ij ij ij d d d      • Incremental stress-stain relation: 3 0, 1 2 ; 3 3 , 2 2 ij ij Y e p ij ij kk ij ij p ij ij ij Y Y d d d d E E d                                   , 0. • Equations of static equilibrium:  ji j i   F • Traction BCs on St : • Displacement BCs on Su : ij i j  n t  i i u u 

Cylindrically Symmetric Elastoplastic Solids? Cylindrically symmetric geometryand loading (i.e. internal bodyAerforces, tractions or displacements7BCs, nonuniform temperaturedistribution)022. Cylindrical-polar bases: (e, e.,e.). Cylindrical-polar coordinates: (r, 0, z).Position vector: x=re,? Displacement vector: u = u, [r]e,. Body force vector: F = F, [r]e,? Acceleration vector: a =-o’re,6

Cylindrically Symmetric Elastoplastic Solids 6 • Cylindrically symmetric geometry and loading (i.e. internal body forces, tractions or displacements BCs, nonuniform temperature distribution). • Cylindrical-polar bases: • Cylindrical-polar coordinates: • Position vector: • Displacement vector: • Body force vector: • Acceleration vector: e e e r z , ,   r z , ,   r x e  r u e  u r r r   F e  F r r r   2 r a e   r

Cylindrically Symmetric Elastoplastic Solids. Cauchy stress: o =o, [rle,e, +oe[rlege,+o, [rle.e. Infinitesimal strain: =e, [rle,e, +ee[rlege。+8. [rle.eduru,? Strain-displacement relation: ,80dr: Stress-strain relation in elastic region (plane strain orgeneralized plane strain):E(1-v)e, +V8 +Ve:)a(1+v)(1-2v)Eve, +(1-v)8p +ve:)6(1+v)(1-2v)Eve, + Ve。 +(1-v)8.)a(1+v)(1-2v) von Mises yield criterion:0. = /(0, -0.) +(a。-0.) +(a -0,) ) =0y7

Cylindrically Symmetric Elastoplastic Solids • Cauchy stress: 7 • Strain-displacement relation: • Stress-strain relation in elastic region (plane strain or generalized plane strain): • von Mises yield criterion: • Infinitesimal strain: σ     r r r z z z r r r e e e e e e + +        ε     r r r z z z r r r e e e e e e + +        d , d r r r u u r r                          1 , 1 1 2 1 , 1 1 2 1 . 1 1 2 r r z r z z r z E E E                                                   1 2 2 2 2         e r z z r Y         

Cylindrically Symmetric Elastoplastic Solids: Stress-strain relation in plastic region? Strain partition: de, =de +dep,de = dee +ds,de, =dee +ds?do,.v(do。+do.)?Elastic strain: deeEE: Flow rule:de = d31S=2gv2 0ydo, +,-Oe+F =-po"r· Equations of motion:drr. Traction BCs: Cr[a]=a,Or[b]= Ob.BCs: u,[a]=ua, u,[b]=ub;or ,[a]=oa, o,[b]=oi. There is no clean, direct, and general method for integrating theseequations. Instead, solutions must be found using a combination ofphysical intuition and some algebraic tricks8

Cylindrically Symmetric Elastoplastic Solids • Stress-strain relation in plastic region 8 • Traction BCs: • Equations of motion:     R a R b a b    , .   • Strain partition: • Elastic strain: • Flow rule: • BCs: • There is no clean, direct, and general method for integrating these equations. Instead, solutions must be found using a combination of physical intuition and some algebraic tricks. , , e p e p e p r r r z z z d d d d d d d d d                     , , e r z r d d d d E E             3 3 1 1 1 + + + , , 2 2 3 2 p p p p r r r r z r z Y Y Y d d d d                                    r r 2 r d F r dr r          u a u u b u a b r a r b r a r b       , ; or ,          

Hollow Cylinder under Monotonic Pressure: We consider a long hollow cylinder. The sphere is stress free before it isverloaded.7? No body forces act on the cylinder.The cylinder has zero angular velocity02C. The cylinder has uniform temperature.: The cylinder does not stretch parallel toits axis.. The inner surface r = a is subjected to monotonicallyincreasing pressure pa: The outer surface r = b is traction free. Strains are infinitesimal.. We aim to find9

Hollow Cylinder under Monotonic Pressure • We consider a long hollow cylinder. • The sphere is stress free before it is loaded. • No body forces act on the cylinder. • The cylinder has zero angular velocity. • The cylinder has uniform temperature. • The cylinder does not stretch parallel to its axis. 9 • The inner surface r = a is subjected to monotonically increasing pressure pa . • The outer surface r = b is traction free. • Strains are infinitesimal. • We aim to find

Hollow Cylinder under Monotonic Pressure(1+v) pα26·Elastic solutionE(b?-α?2vp.ap.ap.a2/0。0b?-α?62-: von Mises effective stress:3p.a?b?3b4Paa?+1+4v2-4v, V~0.5=6deb? - α?(b2 -α°)r2. We see that the hollow cylinder first reaches yield at r = a,With the elastic limit: pa/oy ~(1-a2 /b2)/ /3 If the pressure is increased beyond yield, we anticipate thata region a < r < c will deform plastically, whereas a regionc < r < b remains elastic10

Hollow Cylinder under Monotonic Pressure • Elastic solution 10 • von Mises effective stress: • We see that the hollow cylinder first reaches yield at r = a, with the elastic limit: • If the pressure is increased beyond yield, we anticipate that a region a < r < c will deform plastically, whereas a region c < r < b remains elastic.       2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 ; 2 1 , 1 , . a r a a a r z p a r b u E b a r p a p a p a b b b a r b a r b a                                         2 2 2 4 2 2 2 4 2 2 2 3 3 1 4 4 , 0.5 a a e e p a p a b b b a r b a r                 2 2 1 3 a Y p a b   