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Which of the following materials will be most suitable to manufacture the disc for the stiffest/strongest possible part.
It seems that you need a material that will transform a non-uniform load into a more-or-less uniform load by providing consistent bearing. As such, you're looking at something like a plate on an elastic foundation, albeit on a much ...
7
A tensor is a mathematical object which has to obey certain rules about how to transform it from one coordinate system to another.
Engineers started using and measuring strains a century or more before tensors were invented (by Ricci, in around 1900, and not in the context of continuum mechanics). They just happened to make a bad choice of how to measure ...
6
All these terms refer to the effect of loading on the deformation of materials. Let us assume that we start with zero load and zero deformation.
Elastic deformation
If you increase the load you get an increase in deformation. During the process of elastic deformation, if you decrease the load to zero you will not have any residual non-zero deformation.
...
5
The wall is experiencing rising damp.
Basically, moisture in the soil is being absorbed by the wall and traveling up the wall. This can become more prevalent during wet periods and subside during dry periods.
The photograph of the outside of the wall appears to show a of line salt crystallizing towards the top of the affected zone. This is not unusual as ...
5
When the load is removed, the elastic strain is recovered completely (by definition); the plastic strain is not recovered at all (by definition):
If the load is reapplied, the curve picks up where it left off (i.e., the system moves back up the unloading line and resumes plastic deformation). This point is explained here.
5
deflection of a constrained rectangular plate
By this I assume you mean that all edges are fixed.
My general go-to for these types of formulations is Roark's Formulas for Stress and Strain, 7th Edition. These formulations assume a flat plate with straight boundary conditions and constant thickness. Also, Poisson's ratio for the material is assumed to be $...
3
This answer expands upon the approach suggested by @setun-90 and shows that the algebra gets quite complicated. I'd suggest a Green's function approach instead.
Assumptions
isotropic
deflection of membrane is small
applied tension large so that its variation due to deflection is small
Governing equation
Let the tension in the $x, y$-directions be $T_x, ...
3
Yes.
In a perfectly elastoplastic material (where the stress-strain relationship is perfectly linear until the yield point and is then constant and equal to the yield point), the material will elastically recover all of its elastic deformation, but the plastic deformation will remain. This is the blue line in the figure below.
In (not perfectly) ...
3
Before we dig into practical equations, I'll just say that tires are surprisingly complex in their behavior. There are a variety of equations that try to fit the experimental data reasonably well. Hard to say that any one is right or wrong.
That having been said, the first equation is pretty inconsistent with the everything that I've ever seen. Weight on ...
2
Displacement field - a vector field, that assigns a displacement vector to each infinitesimal particle of the body over time.
The object, in this case will be usually defined by a density field (scalar field) with zero density being "outside the object."
This description works on all kinds of matter - gas, fluid mechanics, and non-rigid solids, their ...
2
It is not necessarily a typo, but you wouldn't normally call a thin walled cylinder 8mm high and 100mm diameter a "pipe".
However, the formulas for Euler buckling are most definitely not valid for a structure with those dimensions. The appendix of this rather old report http://www.dtic.mil/dtic/tr/fulltext/u2/a801283.pdf gives some basic theory and graphs ...
2
I will try to explain the solution:
QUESTIONS: 1. Why is the elongation distribution in the length of the beam a triangle?
The "real-life" visualization of the displacement of the system is supposed to be circular. In the picture, the red lines indicates the exaggerated actual movement of points in the system. As depicted, the end of beam, C', displaces ...
2
(edit 3/31/2018. Completed all equations)
I would split the trapezoidal load
into the sum of a uniform load
and triangular load
.
The reactions $R_1$, $R_5$, $R_4$, and $R_8$ can be calculated from statics.
$$R_4=Q_1L_{23}\frac{L_{12}+\frac 12L_{23}}{L_{14}}$$
$$R_1=Q_1L_{23}-R_4$$
$$R_5=\frac 12Q_2L_{67}(\frac 13L_{67}+L_{78})\frac{1}{L_{58}}$$
$$...
2
Between disc 1 & 2, there is a reaction force on disc 1 which is not going to act on the edge rather at the distance of 2nd disc edges.
This means you need to have the model of 1st disc bending developed for 2 cases,
Pressure drop
load acting at the radius of 2nd disc
For the 2nd disc, you have 2 cases
reaction load from the 1st disc at the edge of ...
2
Let's imagine a solid cylinder attached to a fix support at one end and free at other. We twist it at the other end by a torque.
It will rotate about it's axis without warping, ( small angles condition).
Now we cut the cylinder with laser longitudinally in 4 cuts into 8 triangle sections, radiating from the center of the circle, so if you look at the cross ...
2
The reason is symmetry.
For a circular rod under torsion, if you rotate the rod through any angle the warped shape has to look the same, because there is no "special" point around the circumference where you can start measuring angles from.
So, if the rod warps, the warped shape can only vary with the radius across the section.
Similarly, the warping must ...
2
For members of same material and same length, the one with larger moment of inertia is stiffer. If material is same, but their lengths differ, compare (I1/L1) and (I2/L2), the larger one is more stiff. Otherwise, the larger one is stiffer by comparing (E1I1) and (E2I2), if L is constant; or (E1I1/L1) and (E2I2/L2), if lengths differ.
2
I think its the other way around. I.e. the tensile testing takes place in an quasi static (relatively slow rate). This means that you shouldn't expect (usually) larger strains to failure.
However, when you are starting to compress time (i.e. do things faster), then you have adiabatic processes, inertial effects and/or stress waves that take place and lead to ...
2
This is not the answer but to remind you that you need to identify the material of concern (as not all materials have the same failure mode), and the UTS on the curve (whether it is the engineering stress-strain curve or the true stress-strain curve). The best bet is predicting its behavior from its own stress-strain curve.
1
There is an easy way you can calculate a god approximation to the bending strength of the guitar handle.
You can assume you guitar handle cross-section as a trapezoid and measure the moment of inertia, I of that as per this formula. $I_{Cx} \ $ is the moment of inertia about the trapezoid's y centroid parallel to the x axis, where you are interested in. All ...
1
deflection of a beam
Depending if the beam is cantilever, simply supporter or else you have different formulas.
For the simple case of a cantilever beam, with a concentrated load at the end the maximum deflection at the end is :
$$\delta_max = \frac{P \cdot l^3}{3EI}$$
for other cases you can find under deflection of beams tables with equations like the ...
1
It's straightforward to model very slight deformation of the sphere (i.e., slightly flattening its surface). It's also pretty straightforward to model the case when the sides are already fully flattened and meeting at the corners. It's trickier to bridge the gap.
The small-deformation case has an exact solution for a single side (Case 2 here): the ...
1
In a mechanical engineering curriculum at university, you would probably take a series of 3 classes. First would be called "Strengths of Materials" (or maybe "mechanics of materials"). This would be a sophomore level class. No tensors, just linear algebra (matrices). The second class would be "theory of elasticity". This is going to involve tensors. ...
1
The Radius of curvature on a beam bent under moment is $1/r = m/EI $
If we use this just as a rough model the m affecting each disk gradually becomes smaller as we go up and radius of curvature bigger. So it is safe to assume only the outer edge of each top disk touches the lower disk.
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In ductile materials there is a well defined straight area at the beginning of the stress/strain curve which is purely elastic, the slope of this part of the curve is Young's modulus. If you remove the load the sample will completely relax to its original length and can give back all the work put into it.
If you keep loading the specimen beyond this ...
1
If you want to bend a piece of metal to a particular angle then you have to bend it slightly more than the angle as when it is released it will spring back a bit.
Try it in a vise and see how it works.
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Very difficult to think that anodizing could cause distortion. I think there is a possibility that bead blasting could cause distortion with high pressure and large beads. Bead blasting is similar to shot peening which is used to put residual stresses and distortion into metals. Lower strength materials would be more likely to distort under extreme bead ...
1
Wooden tables have been doing that for a long time using an apron. In fact a steel clad wooden surface may be enough. But if you seal the top with a slab of steel you will also want to seal the underside to avoid moisture related warping.
Add a T-profile beam on the underside connecting the supports and your table will be much stiffer.
If necessary you can ...
1
1: The problem describes the beam as rigid, this means it will not deform and will always remain straight. Therefore when loaded in point C it only rotates about A. This is why you get a triangle when drawing the deformed shape together with the original position and the deformation at the points of interest in the same figure.
2: There is no reason they ...
1
If we can assume that the material behaves in an elastic linear way and that the joint is only loaded by bending, then as you commented the neutral line lies in the middle and that deformation on the two extreme fibres is equal but in opposite directions. Then, original length is just the average of both deformed lengths.
Of course, to get this result we ...
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