## Example Stress Vector normal to the Tangent of a Cylinder

The stress tensor throughout a continuum is given with respect to Cartesian axes as

3xix2 5x2 0 5x2 0 2x3 0 2x3 0

Determine the stress vector (or traction) at the point P(2, ^VS) of the plane that is tangent to the cylindrical surface x2 + x2 = 4 at P, Fig. 3.9.

3.4 Gradi

Solution:

At point P, the stress tensor is given by

The unit normal to the surface at P is given from

At point P, and thus the unit normal at P is

Thus the traction vector will be determined from

3.4.2 Vector

20 We can also define the gradient of a vector field. If we consider a solid domain b with boundary Q, Fig. 3.5, then the gradient of the vector field v(x) is a second order tensor defined by

and with a construction similar to the one used for the divergence, it can be shown that dvi(x)

where summation is implied for both i and j.

21 The components of Vxv are simply the various partial derivatives of the component functions with respect to the coordinates:

dvœ |
dvy |
dvz | ||

dx |
dx |
dx | ||

[Vxv] = |
dvœ dy |
dvy dy |
dvz dy |
(3.33) |

dvœ |
dvy |
dvz | ||

_ dz |
dz |
dz _ | ||

' dvx |
dvx |
dvœ | ||

dx |
dy |
dz | ||

[Wx] = |
dvy dx |
dvy dy |
dvy dz |
(3.34) |

dvs |
dvz |
dvz | ||

_ dx |
dy |
dz _ |

that is [Vvjjj gives the rate of change of the ith component of v with respect to the jth coordinate axis.

22 Note the diference between vVx and Vxv. In matrix representation, one is the transpose of the other.

### 23 The gradient of a vector is a tensor of order 2.

24 We can interpret the gradient of a vector geometrically, Fig. 3.10. If we consider two points a and b that are near to each other (i.e As is very small), and let the unit vector m points in the direction from a to b. The value of the vector field at a is v(x) and the value of the vector field at b is v(x + Asm). Since the vector field changes with position in the domain, those two vectors are different both in length and orientation. If we now transport a copy of v(x) and place it at b, then we compare the differences between those two vectors. The vector connecting the heads of v(x) and v(x + Asm) is v(x + Asm) — v(x), the change in vector. Thus, if we divide this change by As, then we get the rate of change as we move in the specified direction. Finally, taking the limit as As goes to zero, we obtain v(x + Asm) — v(x) lim ---= Dv(x)m

As^Q As

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