New answers tagged stress-strain
1
The premise of your question is wrong: The trace of the stress tensor is not (in general) one of the eigenvalues.
The stress tensor $\sigma$ is symmetric, so it can be diagonalised with real eigenvalues, which are called principal stresses and usually denoted (in three dimensions) $\sigma_1$ to $\sigma_3$. What you have written ist the hydrostatic (or ...
2
Bulk modulus is defined as $$B=-V\frac{P}{\Delta V}$$
It is obvious that we can't change the volume of an ideal rigid body so $\Delta V$ tends to zero hence magnitude of bulk modulus tends to $\infty$
3
I think the question could be clearer (that wording leaves a lot of wiggle room); but I believe I understand what it is trying to say.
I find it's easy to picture the coil when we think of a helical spring. As you stretch the spring, the windings of the coil get further and further apart.
If you look at it from the perspective of the wire though, the ...
1
The shear modulus $G$ and Young's modulus $E$ for a homogeneous and isotropic material are related by the equation:
$$G=\frac{E}{2(1+ν)}$$
where $ν$ is Poisson's ratio, or -(lateral strain)/(longitudinal strain).
Poisson's ratio takes into account that when a material stretches longitudinally it also contracts laterally and when it is compressed ...
-1
I think you are looking for an estimate of the circumferential strain on the inside of the tread portion of the tire as the car is driving. Of course there will be an initial positive strain due to tire inflation. Below, I will estimate the additional positive strain generated when your piezo element rotates from the top position to the bottom (adjacent to ...
0
It is possible to impose a known stress and the strain be the result. For example: the strain at the rope of a plumb line with a known weight.
Or the opposite, as when a guitar wire is strained by a known displacement, and the stress can be calculated from the pitch.
But for 3D situations some displacements and some stresses, (or only some displacements) ...
0
So, note that here I'll use Einstein's convention on repeated indices with the usual tensor notation.
So, at the beginning, we all know that the forces $f_i$ are obtained from the stress tensor as follows
\begin{equation}
f_i=\frac{\partial\sigma_{ik}}{\partial x_k}
\end{equation}
Therefore, by the nature of the strain tensor $u_{ik}$ we have that the ...
1
I think there is no correct answer to this question. There are cases where strain occurs without stress (i.e. heating of an unrestrained bar) and others where stress occurs without strain (i.e. heating of a fully restrained bar).
If you are asking about uniaxial "engineering" stress-strain experimental curves, then the stress axis is obtained from reading ...
1
The confusion is in part the difference between making a theoretical force balance and determining in practice force from stress. A theoretical free-body force balance is made at a point. Stress $\sigma$ is converted in practice to force $F$ using area $F = \sigma A$. To convert the latter (area) reference frame to the former (force balance reference frame), ...
2
What is going on there is that there is an equal and opposite force acting on the interface
This force results in stress of $\sigma_1 = \frac{F}{A_1}$ on one side and $\sigma_2 = \frac{F}{A_2}$ on the other side far away from the interface.
The engineering assumption being that the contact pressure $P=\frac{F}{A_2}$ spreads to the full cross-section $A_1$ ...
-1
The reason you take the complete area on $A_{1}$ is because its an approximation: no matter how locally the stress on its surface is being applied, the whole object feels the stress as if it were applied evenly across its cross-section. Otherwise there would be local deformation.
This isn't true in reality though.
2
The carbon fiber composite should indeed have a higher stiffness (modulus of elasticity) and lower strain (because it is brittle). The first curve is also a typical curve for a composite: the stress increases until a critical point. At this point, there arises a crack that will propagate and cut the sample very fast. This causes an abrupt drop in the curve.
...
1
Dished heads on round pressure tanks provide excellent strength and durability. If pressure tanks had flat ends or sides, the material would have to be much thicker to withstand the stress. With flat ends constant pressure changes would bow thinner material back and forth creating fatigue, or stress cracks eventually.
0
The math behind bending even simple cases with rectangular or circular plates is very complex. You can look through the classic book of Roark Formulas for Stress and Strain Chapter 11, page 451 to see for your self.
In page 502, the simplest case of a rectangular plate is shown which is simply supported on all edges and a uniform load is applied ...
1
Apart from geometric measures and material characteristics, then together with the forces applied, you need to specify how and where the plate is fixed.
Then, mainly according to the ratio of thickness wrt width / length, various approximated approaches can be taken, and we are speaking of plates and shells.
0
You need to know the shape and exact thickness of the plate, the material it is made from, its processing history (heat treatment, cold work, etc.), the points of application of the forces, and their magnitudes.
In the mechanical engineering field, calculations like this are routine and are typically handled by modeling programs that run inside any of the ...
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