\text { Consider the } \mathrm{X} \text {-force equilibriu } \mathrm{m}, \sum \mathrm{F}_{\mathrm{x}}=0 \text {; } Cancelling out equal terms in opposite faces \frac{\partial \sigma_{x}}{\partial x} d x d y

d z+\frac{\partial \tau_{y x}}{\partial y} d x d y d z+\frac{\partial \tau_{z x}}{\partial z} d x d y d z+X d x d y d z=0 So we get the x- equilibrium equation \frac{\partial \sigma_{x}}{\partial x}+\frac{\partial \tau_{y x}}{\partial y}+\frac{\partial \tau_{z x}}{\partial z}+X=0 \frac{\partial \tau_{y x}}{\partial x}+\frac{\partial \sigma_{y}}{\partial y}+\frac{\partial \tau_{y z}}{\partial z}+Y=0 \frac{\partial \tau_{z x}}{\partial x}+\frac{\partial \tau_{z y}}{\partial y}+\frac{\partial \sigma_{z}}{\partial z}+Z=0 in short, the equilibrium equations in tensor notation Ou, +X, = 0 (i, j = x, y, z) Take moment quilibrium about an axis through the center and parallel to z - axis \tau_{x y} d y d z \frac{d x}{2}+\left(\tau_{x y}+\frac{\partial \tau_{x y}}{\partial x} d x\right) d y d z \frac{d x}{2}-\tau_{y x} d x d z \frac{d y}{2}-\left(\tau_{y x}+\frac{\partial \tau_{y x}}{\partial y} d y\right) d x d z \frac{d y}{2}=0 \tau_{x y} d x d y d z-\tau_{y x} d x d y d z=0 \therefore \tau_{x y}=\tau_{y x} \tau_{y z}=\tau_{z y} \tau_{z x}=\tau_{x z}