366 Mechanics of Materials s14.8 14.8.Effect of lateral restraint (a)Restraint in one direction only Consider a body subjected to a two-dimensional stress system with a rigid lateral restraint provided in the y direction as shown in Fig.14.4.Whilst the material is free to contract laterally in the x direction the "Poisson's ratio"extension along the y axis is totally prevented. 一可x m Rigid restraint Fig.14.4.Material subjected to lateral restraint in the y direction. Therefore strain in the y direction with a,and a,both compressive,i.e.negative, 1 =6,=-Ea,-G)=0 Gy=VGx Thus strain in the x direction =ex=- (ox-voy) E 】 一 E(0x-v2a,) =-1-2) (14.9) E Thus the introduction of a lateral restraint affects the stiffness and hence the load-carrying capacity of the material by producing an effective change of Young's modulus from E to E/(1-v2) (b)Restraint in two directions Consider now a material subjected to a three-dimensional stress system o,and o:with restraint provided in both the y and z directions.In this case, 1 8,=-Eo,-0x-o）=0 (1) and e:=-E(o.-ox-o)-0 (2)
366 Mechanics of Materials $14.8 14.8. Effect of lateral restraint (a) Restraint in one direction only Consider a body subjected to a two-dimensional stress system with a rigid lateral restraint provided in the y direction as shown in Fig. 14.4. Whilst the material is free to contract laterally in the x direction the “Poisson’s ratio” extension along they axis is totally prevented. UY I I Rigid restraint Fig. 14.4. Material subjected to lateral restraint in the y direction. Therefore strain in the y direction with a, and oy both compressive, i.e. negative, 1 ’E =E = -- (ay-va,) = 0 .. a, = va, Thus strain in the x direction 1 E = E, = --(a, - va,) 1 E = - - (a, - v%,) (14.9) Thus the introduction of a lateral restraint affects the stiffness and hence the load-carrying capacity of the material by producing an effective change of Young’s modulus from E to E/(1 -v*) (b) Restraint in two directions Consider now a material subjected to a three-dimensional stress system a,, a, and a, with restraint provided in both the y and z directions. In this case, (1) (2) 1 ’E 1 ‘E E = --(ay-va,-va,) =o and E = --(a,-vo,-vay)=0
$14.9 Complex Strain and the Elastic Constants 367 From (1), Gy=vOx+VO: :=(ay-va (3) Substituting in(2), 1 (y-vax)-vax-vay=0 0,-0x-v2ox-v2o,=0 6,(1-v)=0x(w+v2) v(1+ 0,=0x(1-2) Oxv =(1-0 and from (3), strain in x direction=- E v2 (14.10) Again Young's modulus E is effectively changed,this time to 14.9.Relationship between the elastic constants E,G,K and v (a）E,G and v Consider a cube of material subjected to the action of the shear and complementary shear forces shown in Fig.14.5 producing the strained shape indicated. Assuming that the strains are small the angle ACB may be taken as 45. Therefore strain on diagonal OA BC.AC cos45°AC1AC a/2a√W2√2"2a
$14.9 Complex Strain and the Elastic Constants From (l), .. 6, = vox + voz 6, = (0, - vox) - 1 V Substituting in (2), 1 - (oy -vox) - vox - Yoy = 0 V .. o, - vox - v20x - v20y = O o,(l-v2)= o,(v+v2) v(l +v) 0, = 6,- (1 - v2) fJXV =- (1 - v) and from (3), .. cz=- '[ -- vox vox] =ox[ (1-v) ] (1-v) v (1 -v) 1 - (1 - v) VOX =- ox o o strain in x direction = - - + v--li + v? EEE E (1-V) (1-V) 367 (3) (14.10) Again Young's modulus E is effectively changed, this time to 14.9. Relationship between the elastic constants E, G, K and v (a) E, G and v Consider a cube of material subjected to the action of the shear and complementary shear Assuming that the strains are small the angle ACB may be taken as 45". Therefore strain on diagonal OA forces shown in Fig. 14.5 producing the strained shape indicated. BC ACcos45" AC 1 AC OA- ad2 a~2~~2=2a =-A =-
368 Mechanics of Materials §14.9 Fig.14.5.Element subjected to shear and associated complementary shear. But AC ay,where y=angle of distortion or shear strain. strain on diagonal =} 2a2 Now shear stress=G shear strain y y= G strain on diagonalt 2G (10 From $13.2 the shear stress system can be replaced by a system of direct stresses at 45,as shown in Fig.14.6.One set will be compressive,the other tensile,and both will be equal in value to the applied shear stresses. 45° Fig.14.6.Direct stresses due to shear. Thus,from the direct stress system which applies along the diagonals: strain on diagonal= -V- =E1+ T (2) Combining (1)and (2), 2G-E1+) tt E=2G(1+V) (14.11)
368 Mechanics of Materials $14.9 Fig. 14.5. Element subjected to shear and associated complementary shear. But AC = ay, where y = angle of distortion or shear strain. .. Now .. .. From Q aY Y strain on diagonal = - = - 2a 2 shear stress z shear strain y =G z y=- G t strain on diagonal = - 2G 3.2 the shear stress system can be replaced by a system of direct stresses at 5", as shown in Fig. 14.6. One set will be compressive, the other tensile, and both will be equal in value to the applied shear stresses. u2=-r /_I ~ - 4x=T _- r- Fig. 14.6. Direct stresses due to shear Thus, from the direct stress system which applies along the diagonals: 01 02 strain on diagonal = -- v- EE (-4 =-- V- EE T = -(1 +v) E Combining (1) and (2), tz -=-(1+v) 2G E E = 2G(l+v) (14.11)
§14.9 Complex Strain and the Elastic Constants 369 (b)E,K and v Consider a cube subjected to three equal stresses a as in Fig.14.7 (i.e.volumetric stress =a). Fig.14.7.Cubical element subjected to uniform stress a on all faces ("volumetric"or "hydrostatic" stress). Total strain along one edge E E-E =E1-2y But volumetric strain=3 x linear strain (see eqn.14.5) 30 °E1-2) (3) By definition: bulk modulus K volumetric stress volumetric strain volumetric strain=K (4) Equating (3)and (4), 03 K=E(1-2v) E=3K(1-2v) (14.12) (c)G,K and v Equations(14.11)and(14.12)can now be combined to give the final relationship as follows: From eqn.(14.11), E v= 2G-1
$14.9 (b) E, K and v Complex Strain and the Elastic Constants 369 Consider a cube subjected to three equal stresses 6 as in Fig. 14.7 (Le. volumetric stress = 6). Fig. 14.7. Cubical element subjected to uniform stress u on all faces (“volumetric” or “hydrostatic” stress). 606 Total strain along one edge = - - v- - vEEE 6 = -(1 -2v) E But volumetric strain = 3 x linear strain (see eqn. 14.5) 30 E = -(I -2v) By definition: volumetric stress volumetric strain bulk modulus K = Equating (3) and (4), .. 0 volumetric strain = - K 0 30 KE - = -(1-2v) (3) (4) (14.1 2) (c) G, K and v Equations (14.1 1) and (14.12) can now be combined to give the final relationship as follows: From eqn. (14.1 l), E
370 Mechanics of Materials §14.10 and from egn.(14.12), 1E v=26K Therefore,equating, E 1 E 20-1=26 E+113 2G+6永2 6K+2G13 E 12KG 9KG i.e. E= (3K+G) (14.13) 14.10.Strains on an oblique plane (a)Linear strain Consider a rectangular block of material OLMN as shown in the xy plane(Fig.14.8).The strains along Ox and Oy are &x and y,and ysy is the shearing strain. Xyasing acos 8e.Ty a sin 8 e,acos e+y a sin 8 2L a sndey M 8才M a sin e、 a51r8 0 to OM M Movement of parallel N -acos 8 acos 8Ex Fig.14.8.Strains on an inclined plane. Let the diagonal OM be of length a;then ON a cos 0 and OL a sin 0,and the increases in length of these sides under strain are ea cos 6 and e,a sin 6(i.e.strain x original length). If M moves to M',the movement of M parallel to the x axis is Exacos0+Ysasine and the movement parallel to the y axis is Eya sin
370 and from eqn. (14.12), Mechanics of Materials Therefore, equating, .. i.e. 9KG (3K + G) E= 414.10 (14.13) 14.10. Strains on an oblique plane (a) Linear strain Consider a rectangular block of material OLMN as shown in the xy plane (Fig. 14.8). The strains along Ox and Oy are E, and cy, and yxy is the shearing strain. C, a cos 8+ xv a sin 8 / Q +.-acosR-+ + a cos 8 E, Fig. 14.8. Strains on an inclined plane. Let the diagonal OM be of length a; then ON = a cos 8 and OL = a sin 8, and the increases in length of these sides under strain are &,a cos 8 and E~U sin 8 (i.e. strain x original length). If M moves to M', the movement of M parallel to the x axis is &,a cos 8 + ?,,a sin 8 Eya sin 8 and the movement parallel to the y axis is