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The strain rate tensor actually comes from decomposing the velocity gradient tensor i.e. $$\frac{\partial u_i}{\partial x_j}=\frac{1}{2}(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i})+\frac{1}{2}(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i})$$

where on the LHS, former is the symmetric part and the latter is the antisymmetric part. It is similar to writing a matrix as the sum of a symmetric matrix and skew symmetric matrix.

If you don't understand decomposition of tensors then simply think of the above equation as writing A= 1/2(A+B) + 1/2(A- B).

Now if would have gone through the derivation of the rate of angular deformation of a 2-D fluid element then you will notice that,$$\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}$$is the rate of angular deformation of the fluid element and,$$\frac{1}{2}(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i})$$is the angular velocity of the fluid element. In solid objects, stress is the function of Modulus of elasticity and the strain in the element.

Now, stress is the function of velocity gradient tensor as a whole and not only the symmetric part as you've written above. What you've written is correct only if we impose a constraint on the fluid element that the Moment of the shearing forces about the centre of the fluid element is zero. This condition causes the rotational velocity of the fluid element to become zero and therefore we can write the shear stress as the function of symmetric part only.

Assuming your case in which there is x-direction translation of the fluid element, let's say there is a no slip condition at the lower surface such that x dir. velocity is zero. Moment about the centre(M) $$M= \sigma_{yx}.(lb)h-\sigma_{xy}.(bh)l$$ where lb is area of the face on which sigma yx is acting and h is distance between the the opp. faces and respectively for sigma xy. So if M is zero we get $$ \sigma_{yx}.(lb)h=\sigma_{xy}.(bh)l$$

And therefore the $$\sigma_{xy}$$ is a result of maintaining rotational equilibrium.

For more details watch this lecture.

Note:- In continuum mechanics, the stress in the fluid element is the function of viscosity coefficient and the rate of deformation (angular + linear).

The strain rate tensor actually comes from decomposing the velocity gradient tensor i.e. $$\frac{\partial u_i}{\partial x_j}=\frac{1}{2}(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i})+\frac{1}{2}(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i})$$

where on the Left Hand Side, former is the symmetric part and the latter is the antisymmetric part. It is similar to writing a matrix as the sum of a symmetric matrix and skew symmetric matrix.

If you don't understand decomposition of tensors then simply think of the above equation as writing A= 1/2(A+B) + 1/2(A- B).

Now if would have gone through the derivation of the rate of angular deformation of a 2-D fluid element then you will notice that,$$\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}$$is the rate of angular deformation of the fluid element and,$$\frac{1}{2}(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i})$$is the angular velocity of the fluid element. In solid objects, stress is the function of Modulus of elasticity and the strain in the element.

Now, stress is the function of velocity gradient tensor as a whole and not only the symmetric part as you've written above. What you've written is correct only if we impose a constraint on the fluid element that the Moment of the shearing forces about the centre of the fluid element is zero. This condition causes the rotational velocity of the fluid element to become zero and therefore we can write the shear stress as the function of symmetric part only.

Assuming your case in which there is x-direction translation of the fluid element, let's say there is a no slip condition at the lower surface such that x dir. velocity is zero. Moment about the centre(M) $$M= \sigma_{yx}.(lb)h-\sigma_{xy}.(bh)l$$ where lb is area of the face on which sigma yx is acting and h is distance between the the opp. faces and respectively for sigma xy. So if M is zero we get $$ \sigma_{yx}.(lb)h=\sigma_{xy}.(bh)l$$

And therefore the $$\sigma_{xy}$$ is a result of maintaining rotational equilibrium.

For more details watch this lecture.

Note:- In continuum mechanics, the stress in the fluid element is the function of viscosity coefficient and the rate of deformation (angular + linear).

The strain rate tensor actually comes from decomposing the velocity gradient tensor i.e. $$\frac{\partial u_i}{\partial x_j}=\frac{1}{2}(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i})+\frac{1}{2}(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i})$$

where on the LHS, former is the symmetric part and the latter is the antisymmetric part. It is similar to writing a matrix as the sum of a symmetric matrix and skew symmetric matrix.

If you don't understand decomposition of tensors then simply think of the above equation as writing A= 1/2(A+B) + 1/2(A- B).

Now if would have gone through the derivation of the rate of angular deformation of a 2-D fluid element then you will notice that,$$\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}$$is the rate of angular deformation of the fluid element and,$$\frac{1}{2}(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i})$$is the angular velocity of the fluid element. In solid objects, stress is the function of Modulus of elasticity and the strain in the element.

Now, stress is the function of velocity gradient tensor as a whole and not only the symmetric part as you've written above. What you've written is correct only if we impose a constraint on the fluid element that the Moment of the shearing forces about the centre of the fluid element is zero. This condition causes the rotational velocity of the fluid element to become zero and therefore we can write the shear stress as the function of symmetric part only.

Assuming your case in which there is x-direction translation of the fluid element, let's say there is a no slip condition at the lower surface such that x dir. velocity is zero. Moment about the centre(M) $$M= \sigma_{yx}.(lb)h-\sigma_{xy}.(bh)l$$ where lb is area of the face on which sigma yx is acting and h is distance between the the opp. faces and respectively for sigma xy. So if M is zero we get $$ \sigma_{yx}.(lb)h=\sigma_{xy}.(bh)l$$

And therefore the $$\sigma_{xy}$$ is a result of maintaining rotational equilibrium.

For more details watch this.

Note:- In continuum mechanics, the stress in the fluid element is the function of viscosity coefficient and the rate of deformation (angular + linear).

The strain rate tensor actually comes from decomposing the velocity gradient tensor i.e. $$\frac{\partial u_i}{\partial x_j}=\frac{1}{2}(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i})+\frac{1}{2}(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i})$$

where on the LHS, former is the symmetric part and the latter is the antisymmetric part. It is similar to writing a matrix as the sum of a symmetric matrix and skew symmetric matrix.

If you don't understand decomposition of tensors then simply think of the above equation as writing A= 1/2(A+B) + 1/2(A- B).

Now if would have gone through the derivation of the rate of angular deformation of a 2-D fluid element then you will notice that,$$\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}$$is the rate of angular deformation of the fluid element and,$$\frac{1}{2}(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i})$$is the angular velocity of the fluid element. In solid objects, stress is the function of Modulus of elasticity and the strain in the element.

Now, stress is the function of velocity gradient tensor as a whole and not only the symmetric part as you've written above. What you've written is correct only if we impose a constraint on the fluid element that the Moment of the shearing forces about the centre of the fluid element is zero. This condition causes the rotational velocity of the fluid element to become zero and therefore we can write the shear stress as the function of symmetric part only.

Assuming your case in which there is x-direction translation of the fluid element, let's say there is a no slip condition at the lower surface such that x dir. velocity is zero. Moment about the centre(M) $$M= \sigma_{yx}.(lb)h-\sigma_{xy}.(bh)l$$ where lb is area of the face on which sigma yx is acting and h is distance between the the opp. faces and respectively for sigma xy. So if M is zero we get $$ \sigma_{yx}.(lb)h=\sigma_{xy}.(bh)l$$

And therefore the $$\sigma_{xy}$$ is a result of maintaining rotational equilibrium.

For more details watch this lecture.

Note:- In continuum mechanics, the stress in the fluid element is the function of viscosity coefficient and the rate of deformation (angular + linear).

The strain rate tensor actually comes from decomposing the velocity gradient tensor i.e. $$\frac{\partial u_i}{\partial x_j}=\frac{1}{2}(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i})+\frac{1}{2}(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i})$$ where on the LHS, former is the symmetric part and the latter is the antisymmetric part. It is similar to writing a matrix as the sum of a symmetric matrix and skew symmetric matrix.

Now if would have gone through the derivation of the rate of angular deformation of a 2-D fluid element then you will notice that,$$\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}$$is the rate of angular deformation of the fluid element and,$$\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i}$$is the angular velocity of the fluid element. In solid objects, stress is the function of Modulus of elasticity and the strain in the element.

Now, stress is the function of velocity gradient tensor as a whole and not only the symmetric part as you've written above. What you've written is correct only if we impose a constraint on the fluid element that the Moment of the shearing forces about the centre of the fluid element is zero. This condition causes the rotational velocity of the fluid element to become zero and therefore we can write the shear stress as the function of symmetric part only.

Assuming your case in which there is x-direction translation of the fluid element, let's say there is a no slip condition at the lower surface such that x dir. velocity is zero. Moment about the centre(M) $$M= \sigma_{yx}.(lb)h-\sigma_{xy}.(bh)l$$ where lb is area of the face on which sigma yx is acting and h is distance between the the opp. faces and respectively for sigma xy. So if M is zero we get $$ \sigma_{yx}.(lb)h=\sigma_{xy}.(bh)l$$

And therefore the $$\sigma_{xy}$$ is a result of maintaining rotational equilibrium.

For more details watch this.

Note:- In continuum mechanics, the stress in the fluid element is the function of viscosity coefficient and the rate of deformation (angular + linear).

The strain rate tensor actually comes from decomposing the velocity gradient tensor i.e. $$\frac{\partial u_i}{\partial x_j}=\frac{1}{2}(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i})+\frac{1}{2}(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i})$$

where on the LHS, former is the symmetric part and the latter is the antisymmetric part. It is similar to writing a matrix as the sum of a symmetric matrix and skew symmetric matrix.

If you don't understand decomposition of tensors then simply think of the above equation as writing A= 1/2(A+B) + 1/2(A- B).

Now if would have gone through the derivation of the rate of angular deformation of a 2-D fluid element then you will notice that,$$\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}$$is the rate of angular deformation of the fluid element and,$$\frac{1}{2}(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i})$$is the angular velocity of the fluid element. In solid objects, stress is the function of Modulus of elasticity and the strain in the element.

Now, stress is the function of velocity gradient tensor as a whole and not only the symmetric part as you've written above. What you've written is correct only if we impose a constraint on the fluid element that the Moment of the shearing forces about the centre of the fluid element is zero. This condition causes the rotational velocity of the fluid element to become zero and therefore we can write the shear stress as the function of symmetric part only.

Assuming your case in which there is x-direction translation of the fluid element, let's say there is a no slip condition at the lower surface such that x dir. velocity is zero. Moment about the centre(M) $$M= \sigma_{yx}.(lb)h-\sigma_{xy}.(bh)l$$ where lb is area of the face on which sigma yx is acting and h is distance between the the opp. faces and respectively for sigma xy. So if M is zero we get $$ \sigma_{yx}.(lb)h=\sigma_{xy}.(bh)l$$

And therefore the $$\sigma_{xy}$$ is a result of maintaining rotational equilibrium.

For more details watch this.

Note:- In continuum mechanics, the stress in the fluid element is the function of viscosity coefficient and the rate of deformation (angular + linear).