7

You could try a bath of concentrated acid. As long as you could maintain circulation so the concentration was fairly constant, a spring presents a very uniform cross-section and should be dissolved at a consistent rate. I would recommend keeping one of the springs as a reference so that you can measure the result after a short period and adjust your time ...


6

If we're staying within the realms of beam theory, we can go with this approach which is valid for any material that exhibits linear-elastic behaviour before yield: The curvature of a beam is related to the applied moment and it's flexural stiffness: $$ \begin{align} \kappa = \frac{M}{EI} \end{align} $$ where, for a cicular rod, $I = \frac{\pi d^4}{64}$. ...


5

You are asking for an unrealistic material. If you peruse the engineering toolbox you'll find that the Young's modulus of typical materials is measured in 109 Pa so the value you are asking for is 7 orders of magnitude smaller than typical materials. Even the Young's modulus of the rubber in rubber bands is ~107 Pa. Another simple thought experiment will ...


4

This has been asked and answered numerous times before: https://www.quora.com/Why-is-strain-energy-equal-to-1-2*force*displacement-What-about-the-remaining-half basically, it's because you're calculating the area under a triangle, because the strain energy increases linearly as the displacement increases. The Work term here refers to the final position, but ...


4

By "strength" do you mean stiffness/modulus? The modulus does not change with tensile or yield strengths. Make it with wire of 1/2 the diameter for 1/2 the modulus. Acid will hydrogen stress crack hardened steel ( This condition has many other names like "embrittlement".). Make it out of aluminum and get 1/3 the modulus or titanium and get 2/3 the modulus ( ...


3

Definitions Before we can answer your question, let us look at two standard definitions of terms (from ASME Guide for Verification and Validation in Computational Solid Mechanics) : 1) Verification: The process of determining that a computational model accurately represents the underlying mathematical model and its solution. 2) Validation: The process ...


3

The elastic modulus would give you how much it would twist in use, you would need the tensile strength: taken from engineering toolbox http://www.engineeringtoolbox.com/torsion-shafts-d_947.html : τ = T r / J (1) where τ = shear stress (Pa, psi) T = twisting moment (Nm, in lb) r = distance from center to stressed surface in ...


3

A simple first order approach would be to treat the plate like a composite material, with the holes acting as a medium with no modulus. The rule of mixtures , treating the "holes" as fibers with 0 modulus, would yield a modulus of 0. So, the Semi-Emperical Halpin Tsai would be better: $$ \eta = \frac{\frac{E_f}{E_m} -1}{\frac{E_f}{E_m}+1} $$ $$ E_c = \...


3

The answer to this lies in the defining equation for the strains that you have supplied. The full displacement gradient $u_{i,j}$ (where the $\bullet_{,j}$ represents the $j^{\rm th}$ derivative) can be linearly decomposed into a symmetric and an anti-symmetric component: $u_{i,j} = \epsilon_{i,j} + \omega_{i,j}$, in the usual way as for all rank-2 tensors. ...


3

Probably just another hypotetical idea: If you squeeze the wire from circular into a square shape, it will have about 54% of the original rate. If you squeeze it slightly more into a rectangle, it will at some point become 50%. And 1/2 the diameter of the wire does not gie you 1/2 the rate. It would give you 1/16 of the rate. 85% of the diameter gives you ...


3

Another hypothetical way to reduce the amount of material in the spring would be to fix it in a jig stretched to double its normal length. Then, heat it sufficiently to anneal it and remove the tension in it. Then, re-temper it and cut it in half, back to its original length. But just buying a new spring would be a much simpler method.


2

The fundamental beam equation is $$\dfrac{\text{d}^2}{\text{d}x^2}\left(EI\dfrac{\text{d}^2w}{\text{d}x^2}\right) = q$$ Which basically translates to "the fourth derivative of the deflection function is equal to the applied load". In fact the first derivative is the tangent of the deflection, which for small angles is approximately equal to the angle of ...


2

all tensile testers exhibit this accomodation range to a slight degree, but in your case the large accomodation range is most likely caused by worn parts in the drive mechanism that pulls on the bar- most likely the mechanical "grabbers" that dig into and engage the ends of the bar under test. this might be the reason that this machine was discarded in the ...


2

This is quite a poorly-worded question. After all, how do the layers interact with one another? Can they be considered to be fully bonded to one another? I assume not, because then they've behave as one element, which kind of beats the purpose. Should we assume that the load is evenly distributed between each of the layers (so, for a given load $q$, if you ...


2

Rather than thinking about a Young's modulus directly, the problem becomes easier to grasp if you think in terms of a relationship between the load and displacement, and subsequently between stress and the strain. If you consider the case where deformations are small, and just consider the stiff direction (tension/compression vertically and not horizontal ...


2

This is a link to a solved problem similar to your question. inflation of an annular cylinder In their example the two ends are fixed, no traction of cylinder, meaning the length of cylinder doesn't change. It is a partial deferential boundary value problem solved as part of their online course. NPTEL is one of India's online university institutes.


2

A good place to start is Ogden's book Nonlinear elastic deformations, section 5.2.3: "Extension and inflation of a circular cylindrical tube". For a material with a strain energy function: $$ W = F(I_1) - A I_2 + B I_3 $$ where $I_1$, $I_2$, $I_3$ are the three principal invariants of the stress tensor, Ogden shows that for the situation where a stretch $\...


2

The hyperelastic and viscoelastic material models are both constitutive relations that relate: Stress and strain, in the case of hyperelasticity. Stress, strain and strain rate, in the case of viscoelasticity. They are both empirical models, which means that you typically need to run experiments to find the necessary parameters to fit each model (there are ...


2

Let's look at some (very rough) definitions: 1) Viscoelasticity = elastic behavior that changes if the rate of application of the load is changed or, if the load is kept fixed, the change in the elastic behavior of a material over time. 2) Hyperelasticity = rate/time-independent elastic behavior beyond linear elasticity Of course, elastic behavior means ...


2

Material science can almost always be broken down into "generally useful" and "specifically useful". The link in the comments demonstrates a commonly used statistic that describes behavior of foams at higher strain values - the three dimensional strain energy density function. But this statistic is "generally useful". It generalizes to all foams. In ...


2

I would consider perhaps the easiest way would be to change the leverage... Failing that, I would get a new spring made with the details necessary and spare one 10% above and below your "new" value.


2

While I think the concentrated acid solution is brilliant, I'd like to suggest a refinement. Rather than use just any old acid, you should specifically use a chemical etchant. Typically, a chemical etchant's nucleophile strength does not come from Hydrogen, so it should have the benefit of much lower -if not entirely eliminated- hydrogen enbrittlement. (...


2

Stiffness: If you cut the spring by 1/2, it will have double stiffness (spring rate). This means your 2/3 springs rate will be 3/2 of the original rate. This is independent of the length of the spring, it is onyl the number of coils, that cause the difference. If you deform a material into plastic deformation, depending on the material anything can happen. ...


1

I would use a Shear Box test for this. Stress testing is for maximum load-bearing capacity... Shearing is what gives you the modulated graphic results you seek. Have a look at Smith's Soil Mechanics 9th ed for more details.


1

Unconfined compressive strength test is generally a undrained soils test. The issue with a constant load machine is that you will likely load it too quickly as well close to the failure point, which will produce inaccurate readings and measurements at failure (i.e. you will end up missing the point of failure, and needing a new sample again). Otherwise, the ...


1

According to J. E. Gordon's book "Structures or why things don't fall down" he explains The concept of Work of Fracture. During his exposition he explains that cylindrical shapes have a specific relationship of tensile strength to the thickness of the cylinder weigh is what a tube is. The general rule is that the tensile strength of any material is ...


1

It would be indetrminant structure if we release the axial constraints. In a very crude first estimate one could compare it to a syspended bridge with flexible end supports, or a balloon with built-in geometric constraints. I guess if you don't want to use FEM then, you would want to guess a curve for the expanded shape and solve by energy methods. It would ...


1

At the crushed configuration we can assume we are dealing with a flat beam. Width= L of pipe, span = 15pi. Height = 5mm. for simplicity we assume three plastic hinges ( we will refine this by assuming more hinges, such as one at either edge of pressure block if need be), one at either crushed edge of the pipe, one at the center of horizontal flatly ...


1

Edit after AgentP's comment: As AgentP pointed out, you have swapped your stress-strain axes in the zoomed image. This fact alone refutes your conjecture... Your graph should look like this: You see vertical increases in stress, and sudden jumps in strain. A reason the steps might occur: The steps in your data occur at very specific displacement/strain ...


1

There are two major effects that make the elasticity modulus of concrete vary with time: hardening and creep. In a Eurocode context $E_{cm}(t)$ refers to the hardening of concrete until it reaches its 28-day strength. This is section 3.1.3(3). If you need to calculate a crack width at an early stage, you should probably use $E_{cm}(t)$ instead of $E_{cm}$, ...


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