东南大学：《固体力学基础》课程教学课件（英文讲稿）12 Simple Dynamic Solutions for Linear Elastic Solids

Simple Dynamic Solutions forLinear Elastic Solidsmi@se.edl.cn

Simple Dynamic Solutions for Linear Elastic Solids

Outline· Surface under time-varying normal pressure（自由表面时变压力载荷）·Surfaceunderconstantnormal pressure（自由表面常压）·Surface under time-varying shear traction（自由表面时变剪应力载荷）·One-dimensional bar under end loading（端部受载杆件）·Planewavesininfinitesolids（无限固体中的平面波）·Wave speeds inisotropic solids（各向同性固体中的波速）·Reflection of longitudinal waves at a free surface （反射波) Reflection and transmission of waves at a bimaterialinterface（反射波与透射波）2

Outline • Surface under time-varying normal pressure（自由表面时 变压力载荷） • Surface under constant normal pressure（自由表面常压） • Surface under time-varying shear traction（自由表面时变 剪应力载荷） • One-dimensional bar under end loading（端部受载杆件） • Plane waves in infinite solids（无限固体中的平面波） • Wave speeds in isotropic solids(各向同性固体中的波速) • Reflection of longitudinal waves at a free surface(反射波) • Reflection and transmission of waves at a bimaterial interface（反射波与透射波） 2

Surface under Time-varying Normal Pressure. The solid is at rest and stressp(t)free at time t = O.Strain-displacement relation:=二i+uiienIsotropic Hooke's Law:1+vEVVO1EE(1+I1auagi: Linear momentum balance equations:OOt?Ox;? Symmetry condition indicates =g =O, z =z[x2,t]ou The only nontrivial strain component: ,Ox23

• The solid is at rest and stress free at time t = 0. • Strain-displacement relation: Surface under Time-varying Normal Pressure 3  , ,  1 2  ij i j j i   u u 2 2 ij j i u x t          1 ; . 1 1 2 ij ij kk ij ij ij kk ij E E E                           • Linear momentum balance equations: • Isotropic Hooke’s Law: • Symmetry condition indicates u u u u x t 1 3 2 2 2    0, ,   • The only nontrivial strain component: 2 2 2 u x    

Surface under Time-varying Normal Pressure? Nontrivial stress components:E(1-v)EvauEOu.o,[x2,t]=0,[x2,1] -,/x,t2-+1-2v2](1+)(1+v)(1-2v) ax2 (1+v)(1-2v) ax2 : The only nonzero linear momentum balance equation:a"uE(1-v)u'uz - 0002at?at?0x2(1+v)(1-2v) x2?u2E(1-v)a"uz1ax?at?p(1+v)(1-2v): The 1-D wave equation has the general solution:u2 [x2,t] = f [t - x2 /c, ]+g[t + x2 /c,]. Initial conditions:[u,[x2,0] = 0[f[-x2 /c, ]+g[x2 /cL]= 0>f[-x2 /c]=-g[x2 /c,]= AOu2 [x2, 0]['"[-x2 /c,]+g'[x2 /c,] = 00at4

• Nontrivial stress components: Surface under Time-varying Normal Pressure 4                 2 2 2 2 2 2 1 2 3 2 2 2 1 , , , , . 1 1 2 1 1 2 1 1 2 E E E u u x t x t x t x x                                               2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 2 1 1 , 1 1 2 L L u u u E x t x t u u E c x c t                                   • The only nonzero linear momentum balance equation: • The 1-D wave equation has the general solution: u x t f t x c g t x c 2 2 2 2  ,       L L    • Initial conditions:              2    2 2 2 2 2 2 2 2 2 ,0 0 0 , 0 0 0 L L L L L L u x f x c g x c u x f x c g x c t f x c g x c A                           

Surface under Time-varying Normal Pressure: The vertical displacement:g[x2 /c,]=-A =g[t+x2/c,]=-Au2 [x2,t]= f [t -x2 /c, ]+g[t +x2 /c, ] = f [t - x2 /c, ] - AE(1-v) _uz =__E(1-v)→0 [3, ]= (1+(-2) a -- (1+(-21)f'[t-x2 /c.] Boundary conditions at the plane surface:E(1-v)[(1+(-2)[]=-[] = [[]=(+(-2)1,[0,1] =-p[] =--p[1]E(1-v)= T[]=% (1+v)(1-2)T [] t+ = [-x / ]-(+ []dt+ EE(1-v)(1-v)· Determine B: t=0 = A= f[-x2 /c-]= B? Final solutions:[,-~(+2 []dt, ,[ --p[1- /]t-x2/c≥0E(1-v)5

• The vertical displacement: Surface under Time-varying Normal Pressure 5                           2 2 2 2 2 2 2 2 2 2 2 2 1 1 , 1 1 2 , 1 1 1 2 L L L L L L L u x t f t x g x c A g t x c A f t x c g t x c E E x t u f t x c x c A c                                   • Boundary conditions at the plane surface:                                         2 2 2 0 0 1 1 1 1 2 0, 1 1 2 1 1 1 2 1 1 2 1 1 L L L t t x c L L L E c t p t f t p t f t p t c E c c f t p d B f t x c p d B E E                                                  • Determine B: t A f x c B      0  2 L  • Final solutions:              2 2 2 2 0 2 2 2 1 1 2 , , , , 1 0 L t L x L L c c u x t p d x t p t x c t x c E                 