Inverting gives the relations among stresses and strains for plane stress 1b0 E D10 00 Plane strain In this case we consider situations in which 0 Then ∈33=0 O11+ 11 22 1=E{n-{p+(n+o2) E v(1+)a2 (1-2)2-D(1+ In matrix form: 12-u(1+u) (1+D) 0 Inverting gives the relations among stresses and strains for plane strain 1-D 0 E 1 (1+u)(1-2 0 Practice problem: Verify equations(32) and(39) using Mathematica 6Inverting gives the relations among stresses and strains for plane stress:      σ11 1 ν 0 �11 σ22 = E ν 1 0  �22  (32) σ12 1 − ν2 0 0 (1−ν) 2�12 2 Plane strain In this case we consider situations in which: �i3 = 0 (33) Then: 1 � � �� �33 = 0 = σ33 − ν σ11 + σ22 , or: E � � (34) σ33 = ν σ11 + σ22 (35) 1 � � � ��� �11 = σ11 − ν σ22 + ν σ11 + σ22 E 1 �� � � � � (36) = 1 − ν2 σ11 − ν 1 + ν σ22 E 1 �� � � � � �22 = 1 − ν2 σ22 − ν 1 + ν σ11 (37) E In matrix form:      �11 1 1 − ν2 −ν(1 + ν) 0 σ11  �22  = −ν(1 + ν) 1 − ν2 0 σ22 (38) E 2�12 0 0 2(1 + ν) σ12 Inverting gives the relations among stresses and strains for plane strain:      σ11 E 1 − ν ν 0 �11 σ22 =  ν 1 − ν 0  �22  (39) σ12 (1 + ν)(1 − 2ν) 0 0 (1−2ν) 2�12 2 Practice problem: Verify equations (32) and (39) using Mathematica. 6