Determine the strain rate for a material fiber

AI Thread Summary
To determine the strain rate for a material fiber in the direction of the surface normal, the velocity field is given as V = ((4y-3x)i + (5x+3y)j) ft/s. The discussion emphasizes the need to find the unit normal vector, which can be derived from the cross product of two vectors defined from the origin to specific coordinates. The correct unit normal vector is identified as sqrt(2)/2(-i+j), highlighting the importance of normalizing vectors to obtain unit vectors. Participants suggest that drawing a diagram can aid in visualizing the intersection of the plane and identifying appropriate vectors for the cross product. Understanding these concepts is crucial for calculating the rate of deformation tensor accurately.
LeFerret
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Homework Statement


Determine the Strain Rate for a Material Fiber in the direction of the surface normal.

The Velocity Field is
V=((4y-3x)i+(5x+3y)j) ft/s

http://puu.sh/9hQ7Q/2bda80620f.jpg is the picture

which describes a steady, planar flow

where i and j are unit vectors.

Homework Equations


n * (n * ∇) V

where n is the unit normal, and * are dot products.

The Attempt at a Solution


I know that the solution to this itself is very simple, it is just math, my biggest issue though is how do I find the unit normal vector? I have no idea where to begin, any hints would be greatly appreciated!
 
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In your figure, the z axis lies within the plane, and the point 1,1,1 lies within the plane. Do you know how to determine a unit normal to this plane?

Also, the rate of deformation tensor is 1/2 the velocity gradient tensor and its transpose, not just the velocity gradient tensor.

Chet
 
Chestermiller said:
In your figure, the z axis lies within the plane, and the point 1,1,1 lies within the plane. Do you know how to determine a unit normal to this plane?

Also, the rate of deformation tensor is 1/2 the velocity gradient tensor and its transpose, not just the velocity gradient tensor.

Chet

The only thing I can think of is a cross product of two vectors.
If I defined two vectors from the origin to the coordinates (0,0,1) and (1,1,0) and crossed them, this would give me -i+j
however the solution is sqrt(2)/2(-i+j) and I'm not sure where that common factor is coming from
 
LeFerret said:
The only thing I can think of is a cross product of two vectors.
If I defined two vectors from the origin to the coordinates (0,0,1) and (1,1,0) and crossed them, this would give me -i+j
however the solution is sqrt(2)/2(-i+j) and I'm not sure where that common factor is coming from
(1,1,0) is not a unit vector. Divide it by its magnitude, and you will see where the sqrt(2)/2 came from. Another way to get the unit vector normal to the plane is just to draw a diagram of the intersection of the plane with the x-y plane.

Chet
 
Chestermiller said:
(1,1,0) is not a unit vector. Divide it by its magnitude, and you will see where the sqrt(2)/2 came from. Another way to get the unit vector normal to the plane is just to draw a diagram of the intersection of the plane with the x-y plane.

Chet

In general, when given something like, this how would I know which vectors to use for my cross product?
 
LeFerret said:
In general, when given something like, this how would I know which vectors to use for my cross product?
Any two convenient in-plane unit vectors will do the trick. But often, it's easier to draw a diagram with a unit normal to the plane, and resolve it into components in the coordinate directions.

Chet
 
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Chestermiller said:
Any two convenient in-plane unit vectors will do the trick. But often, it's easier to draw a diagram with a unit normal to the plane, and resolve it into components in the coordinate directions.

Chet

Ah I see, thank you.
 
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