DISCUSSION THREAD
Hints: Review the section in Chapter 5 on Hooke's Law in 3D before starting this problem.
Any questions? Ask (and answer) questions here.
8 thoughts on “H05 Discussion - Su24”
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DISCUSSION THREAD
Hints: Review the section in Chapter 5 on Hooke's Law in 3D before starting this problem.
Any questions? Ask (and answer) questions here.
You must be logged in to post a comment.
For part a, is there also a rigid floor on the bottom of the block? The picture makes it look like there is one.
Yes, that is a rigid floor on the bottom. Note that there is no wall on either the front or the back of the block for Part a.
Because there is a rigid plate on top of the block, does that mean that the block cannot move in the positive y-direction as well?
The weight of the plate is supported by the block, creating a compressive normal stress on the top surface of the block. Since the block is elastic, the top surface of the block will move downward from the weight of the plate.
When writing my strain equations for a I have three equations and 4 unknowns. I know that the stress of z will be zero because it has no restraints and the strain of x will be 0, but I am confused what else is 0 because y is not completely restrained.
Due to the rigid walls, epsilon_x = 0.
Due to no constraint in the z-direction, you know that sigma_z = 0.
Due to the known weight of the plate and the known surface area of the plate, you know sigma_y (don't forget the sign).
That leaves you with three unknowns: two strains and one stress.
I noticed for part b that the container has a height of 3b but the block has a height of 2b. It says that it is snuggly fitted into the container and I assumed all 6 faces of the block were covered with restraints, so this is making me confused.
The height of the box is not relevant except that it is at least as much as the height of the block; that is, 3b > 2b. The top of the container is open, leaving top surface (the positive y-face) of the block to be stress-free.