Don't use exact number since it turns symbolic calculation and probably force `LinearSolve` to get exact solution. Just put for instance `alpha = 45.001 (\[Pi]/180);` and then you get desirable result (I deleted line of code you repeat twice with `omega = Rectangle[{0, 0}, {a, b}]; mesh = ToElementMesh[omega, MaxCellMeasure -> 0.001];`) Needs["NDSolve`FEM`"]; G = 6.894745 10^9; E1 = 26.25 G; E2 = 1.49 G; G12 = 1.04 G; nu12 = 0.28; nu21 = (E2*nu12)/E1; t = 0.0050 .0254; Son = {{1/E1, -nu12/E1, 0}, {-nu21/E2, 1/E2, 0}, {0, 0, 1/G12}}; Qon = Inverse[Son]; Q11 = Qon[[1, 1]]; Q12 = Qon[[1, 2]]; Q22 = Qon[[2, 2]]; Q66 = Qon[[3, 3]]; U1 = (3 Q11 + 3 Q22 + 2 Q12 + 4 Q66)/8; U2 = (Q11 - Q22)/ 2; U3 = (Q11 + Q22 - 2 Q12 - 4 Q66)/ 8; U4 = (Q11 + Q22 + 6 Q12 - 4 Q66)/ 8; U5 = (Q11 + Q22 - 2 Q12 + 4 Q66)/8; alpha = 45.001 (\[Pi]/180); a = 1; b = 1; d = a Cos[alpha] + b Sin[alpha]; omega = Rectangle[{0, 0}, {a, b}]; mesh = ToElementMesh[omega, MaxCellMeasure -> 0.001, PrecisionGoal -> 5, AccuracyGoal -> 5]; u0 = 0.01; angle1 = 10; angle0 = 0; angles = {{angle0, angle1}, {-angle0, -angle1}, {angle0, angle1}, {-angle0, -angle1}, {angle0, angle1}, {-angle0, -angle1}, {-angle0, -angle1}, {angle0, angle1}, {-angle0, -angle1}, {angle0, angle1}, {-angle0, -angle1}, {angle0, angle1}}; num = Dimensions[angles][[1]]; h = num*t; pos = Table[0, num + 1]; pos[[1]] = -h/2; For[i = 2, i <= num + 1, i++, pos[[i]] = pos[[i - 1]] + t]; \[Xi]A = {0, 0, 0, 0, 0}; \[Xi]B = {0, 0, 0, 0, 0}; \[Xi]D = {0, 0, 0, 0, 0}; For[i = 1, i <= num, i++, T0 = angles[[i, 1]]; T1 = angles[[i, 2]]; func[s_] := Simplify@((2.0/d) (T1 - T0) Sqrt[(s - d/2)^2] + T0) (\[Pi]/180); theta[x_, y_] := alpha + func[x Cos[alpha] + y Sin[alpha]]; zA = pos[[i + 1]] - pos[[i]]; zB = pos[[i + 1]]^2 - pos[[i]]^2; zD = pos[[i + 1]]^3 - pos[[i]]^3; V1 = Cos[2 theta[x, y]]; V2 = Sin[2 theta[x, y]]; V3 = Cos[4 theta[x, y]]; V4 = Sin[4 theta[x, y]]; \[Xi]a = {1, V1, V2, V3, V4} zA; \[Xi]b = {1, V1, V2, V3, V4} zB; \[Xi]d = {1, V1, V2, V3, V4} zD; \[Xi]A = \[Xi]A + \[Xi]a; \[Xi]B = \[Xi]B + \[Xi]b; \[Xi]D = \[Xi]D + \[Xi]d;]; mU = {{U1, U2, 0, U3, 0}, {U4, 0, 0, -U3, 0}, {U1, -U2, 0, U3, 0}, {0, 0, U2/2, 0, U3}, {0, 0, U2/2, 0, -U3}, {U5, 0, 0, -U3, 0}}; mA = mU.\[Xi]A; mB = (mU.\[Xi]B)/2; mD = (mU.\[Xi]D)/3; A11[x_, y_] = mA[[1]]; A12[x_, y_] = mA[[2]]; A16[x_, y_] = mA[[4]]; A22[x_, y_] = mA[[3]]; A26[x_, y_] = mA[[5]]; A66[x_, y_] = mA[[6]]; D11[x_, y_] = mD[[1]]; D12[x_, y_] = mD[[2]]; D16[x_, y_] = mD[[4]]; D22[x_, y_] = mD[[3]]; D26[x_, y_] = mD[[5]]; D66[x_, y_] = mD[[6]]; Nx[x_, y_] = A11[x, y] D[u[x, y], {x, 1}] + A12[x, y] D[v[x, y], {y, 1}] + A16[x, y] (D[u[x, y], {y, 1}] + D[v[x, y], {x, 1}]); Ny[x_, y_] = A12[x, y] D[u[x, y], {x, 1}] + A22[x, y] D[v[x, y], {y, 1}] + A26[x, y] (D[u[x, y], {y, 1}] + D[v[x, y], {x, 1}]); Nxy[x_, y_] = A16[x, y] D[u[x, y], {x, 1}] + A26[x, y] D[v[x, y], {y, 1}] + A66[x, y] (D[u[x, y], {y, 1}] + D[v[x, y], {x, 1}]); PDEs = {D[Nx[x, y], {x, 1}] + D[Nxy[x, y], {y, 1}], D[Ny[x, y], {y, 1}] + D[Nxy[x, y], {x, 1}]}; gammaD = {DirichletCondition[{v[x, y] == u0, u[x, y] == 0}, y == 0],DirichletCondition[{v[x, y] == -u0, u[x, y] == 0}, y == b]}; {U, V} = NDSolveValue[{PDEs == {0, 0}, gammaD, Dirichlet Condition[u[x, y] == 0, x == a/2]}, {u, v}, {x, y} \[Element] mesh]; Visualisation {DensityPlot[U[x, y], {x, y} \[Element] mesh, ColorFunction -> "Rainbow", PlotLegends -> Automatic], DensityPlot[V[x, y], {x, y} \[Element] mesh, ColorFunction -> "Rainbow", PlotLegends -> Automatic]} [![Figure 1][1]][1] If we change boundary condition as `gammaD = {DirichletCondition[{v[x, y] == u0}, y == 0], DirichletCondition[{v[x, y] == -u0}, y == b]};`, then the solution looks as follows [![Figure 2][2]][2] [1]: https://i.sstatic.net/HmJPe.png [2]: https://i.sstatic.net/GMbm2.png