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Grain grinding windmills of the “classic” design ie grinding stones need lots of torque and low speeds. Classic as in those in Holland and England - windmills in old Amsterdam... The classic windmill has 4 blades which are not designed for high rotational speeds while the current wind turbines have blades designed such that the tip speed is just below ...

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CTE characterizes dimensional change, with no load. It corresponds to a relatively fundamental physical principle, making it easy to use in a design calculation Heat Deflection test characterizes the deformation under heat AND load, which includes a variety of phenomena happening at the same time. The result is dependent on the particular geometry and ...

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The coefficient of thermal expansion is the change in size of the material as the temperature changes, which is obvious when looking at the units: a length change per metre per degree. Heat deflection is not about a change in length.

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TLDR: look at Viton / FKM PTFE has poor mechanical stability, would be careful whether to use it in most sealing apps (unless it has mechanical backup). Look at other fluoroelastomers, in particular Viton (FKM) or Kalrez/Chemrez (FFKM). The other common elastomer substitutes for NBR are EPDM and silicone (PDMS). Those are the most common ones you can usually ...

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Another way to do it is Monel , 70 Ni : 30 Cu. It should be as available as cupronickel with tensile of 80,000 psi ( depending on thickness ). Essentially the same or stronger than a hull steel so the same thickness as steel. Also available is 70 Cu : 30 Ni cupronickel , tensile strength about 65,000 psi ( depending on thickness ). Again a direct substitute ...

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$I.$ - Assume a cantilever beam and deflection control $\Delta = P*L^3/3E*I, I = b*t^3/12$, where $\Delta$ is the deflection of a cantilever beam loaded on its free end, P is a unit force, L is the beam length, E is the elastic modulus of the plate, I is the moment of inertia, b is the beam width, and t is the beam depth (plate thickness). Let's plug I into ...

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For a rectangular beam under a load, P, at the end the deflection is $$\delta= \frac{PL^3}{3EI}$$ therefore if you have the E as half you need to double the I of the part. In rectangular beams and roughly rectangular tabs: $$I= \frac{BH^3}{12}$$ So you ned a tab thicker by the ratio of $T_{new}= T_{old}*\sqrt[3]{2}$ edit if we are concerned with a yield ...

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Strain is determined as $\epsilon = (l - l_o)/l_o$. The 0.2% offset strain is the point where $\epsilon = 0.002$. The 0.2% offset stress is determined by using a line that has the same slope as the initial stress strain ... the initial Young's modulus. In materials that are fully elastic up to and beyond this point, the initial slope remains constant. In ...

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if the material is mild steel then the easiest way to find the yield stress is to plot the stress strain and find the first knee in the curve. if the material does not have a clear yield point, what you do, is you start from 0.2% strain and draw a straight line parallel to the Elastic part of the line (see below). The proof stress 0.2% (equivalent to yield ...

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