residual stress -neutral axis G1(x) pplied stre applled stress Ior homogeneous specimen with elastic modulus e,ef surface Fig. 3. Schematic of residual and applied stress distribution in layered specimen moduli of the layers result in specific distribution of the applied stress along x-direction. Elastic material demonstrates continuous linear distribution of the applied strain under bending. This promotes piecewise-linear distribution of the applied stress, shown schematically in Fig. 3. To derive the applied stress distribution under bending we can use expressions (3),(4),(7a), (7b), and(12), taking into account that in this case Fa =0. If only the applied stress considered, we can take E(r)=0. Then it follows that the applied stress acting in the layer with number i is E [Lor-lu, ≤x≤ 6 Here xi is the coordinate of upper boundary of ith layer, E=E/(-vi), and E and vi are the elastic modulus and Poisson ratio of layer number i, respectively. Values of IL can be obtained from expression(8)accounting for re(Fig. 1) Residual stress distribution can be found from Eqs. (3),(4),(7a), and(7b) taking into account that Fa =0, E 0,(x)= [nJn1-112Jo+(uJLo-loJn)x],x-1≤x≤x 111-110/12 (15) where Jy can be obtained from the expressions(9)accounting for layered structure ∑Exr1-x) Here Ei is the strain of ith layer non-associated with stress. The thermal expansion or/and a volume change due to a crystallographic phase transformation can be the source of this strain. However, the case of phase transformation is out of the scope of this paper. In case of thermal expansion E,=B, (T)dT, where, (T)is the thermal expansionmoduli of the layers result in specific distribution of the applied stress along x-direction. Elastic material demonstrates continuous linear distribution of the applied strain under bending. This promotes piecewise-linear distribution of the applied stress, shown schematically in Fig. 3. To derive the applied stress distribution under bending we can use expressions (3), (4), (7a), (7b), and (12), taking into account that in this case Fa = 0. If only the applied stress is considered, we can take ~ε ( ) x = 0. Then it follows that the applied stress acting in the layer with number i is: σ σ a i L L L mL L i i x E w I II ( ) IxI x xx ( ) = [ ], . ′ − − ≤≤ − 2 1 2 0 2 01 1 6 (13) Here xi is the coordinate of upper boundary of ith layer, E E ii i ′ = − ( ), 1 ν and Ei and νi are the elastic modulus and Poisson ratio of layer number i, respectively. Values of I Lj can be obtained from expression (8) accounting for layered structure (Fig. 1): I j Ex x j Lj i i j i j i N = + ′ − = + − + = ∑ 1 1 012 1 1 1 1 ( ) ( , , ). (14) Residual stress distribution can be found from Eqs. (3), (4), (7a), and (7b) taking into account that Fa = 0, Ma = 0 (Fig. 3): σr i L L L LL L L LL L L x E I II () [ ( ) = IJ I J IJ I J ′ − −+ − 1 2 0 2 11 2 0 10 01 x], x xx i i −1 ≤ ≤ , (15) where JLj can be obtained from the expressions (9) accounting for layered structure: J j Ex x j Lj i i i j i j i N = + ′ − = + − + = ∑ 1 1 0 1 1 1 1 1 ~ε ( ) ( , ). (16) Here ~εi is the strain of ith layer non-associated with stress. The thermal expansion or/and a volume change due to a crystallographic phase transformation can be the source of this strain. However, the case of phase transformation is out of the scope of this paper. In case of thermal expansion ~ε β () , i i T T T dT j = ∫ 0 where βi ( ) T is the thermal expansion 295 Fig. 3. Schematic of residual and applied stress distribution in layered specimen