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It is an intriguing question, the manufacturer of the resin won't disclose the details of the resin content, but could be interested in giving some advice. The only idea that comes to mind is that the steel powder in the mix could be from a hard magnetic steel that acquires some magnetic...

The answer to a) should simply be FV/c=r/2 The answer to c) should come from b) by calculating x from v(x)=0 Your solution to b) is not far from being correct. The energy stored in the spring is c x2/8, as the spring elongates by x/2. By replacing J1 and ω1 with their expressions you'll be...

The original formula can't be correct in general. Letting a=Dpipe/2, b=dn/2 and ρ=b/a, it may be written as [ΔPa4/(64D]plate)(1-ρ2)2. Calling P the total load acting on the plate, P=ΔPπ(a2-b2), that formula may be rearranged as [Pa2/(64D)](0.318-0.318ρ2). The right parenthesis may be considered...

You can find it here. prex http://www.xcalcs.com : Online engineering calculations https://www.megamag.it : Magnetic brakes and launchers for fun rides https://www.levitans.com : Air bearing pads

Yes, it is indeterminate, according to the accepted definition. About reinventing the wheel: when I was a student, long long ago, there was a story that was told to engineering students who took the monster exam on Theory of elasticity. The story read: Do you know who was the student who...

Owing to symmetry, if you cut the beam at B, you end up with a beam on two supports, one fixed and one simple (pinned). It's a statically determinate problem that's solved in many engineering books (remember: an engineer should never reinvent the wheel [wink]). With P midway of A and B the...

1) Think of a common belt drive: you have two pulleys with a tight side under tension and a slack side with nearly zero tension, excluding pretension. This explains how is it that the different sections of your cable have different tension forces: the moment applied to each pulley determines the...

3D printed air bearing casters. The key feature of the printing process is to obtain a separation surface between the upper and lower faces of the caster, in order to get the flexible inflatable torus that makes up the caster. prex http://www.xcalcs.com : Online engineering calculations...

The formulae are in the Roark in the Properties of Plane Area chapter. Those formulae are quite lengthy, though. You can more easily derive your result from: - inertia Jx of a sector of a hollow circle wrt the diameter: (R4-r4)*(α+sinαcosα)/4 - neutral axis distance to the diameter of the same...

If you start with a specific problem or example, it will be easier to help you. Often you don't separate the primaries from the total stress, you calculate the primaries by formula instead. prex http://www.xcalcs.com : Online engineering calculations https://www.megamag.it : Magnetic brakes and...

I usually take, for vertical walls, 10 W/m2K for calm air and 20 W/m2K for a fairly developed windy condition. Consider also that these figures include the effect of radiation (some 50% in the lower figure). So you can't state a single value for the exchange in ambient air, and an evaluation for...

Two sides: one is on ground, the other one is supported by the upper triangle: this is of course to check the bending behavior of the plates, before that you have to calculate how the loads distribute among the panels and the center pole, not difficult, only a bit complex because of the uncommon...

The structure is stable with all connections pinned, so you have a lower triangular plate with two supported sides and one free, and an upper triangular plate, having one supported side and one supported corner for transverse loads, and being essentially supported at the lower corner and pinned...

Please clarify, it is unclear what exactly you are after. And what's z ? prex http://www.xcalcs.com : Online engineering calculations https://www.megamag.it : Magnetic brakes and launchers for fun rides https://www.levitans.com : Air bearing pads

Take the integral in ∂z of both sides, then multiply both integrands by ∂T/∂z , so that the right integral is in ∂T, and you can calculate it as above, and on the left you have (∂T/∂z)2 I cannot explain the factor 2 in the result, though. prex http://www.xcalcs.com : Online engineering...

Hint: the indefinite integral of (1+kx)/(1+hx) is (kx+(1-k/h)log(1+kx))/h+C prex http://www.xcalcs.com : Online engineering calculations https://www.megamag.it : Magnetic brakes and launchers for fun rides https://www.levitans.com : Air bearing pads

I don't want to dispute the engineering toolbox site, but it doesn't seem to state correct values (and doesn't mention the radiative contribution). From long ago, I know that a ballpark figure for heat exchange coef in calm ambient air is 10 w/sqm K (including roughly 50% contribution from...

If I understand you correctly, that's correct, the vacuum is from outside the vessel wall, so it adds up to the internal pressure. But you should also have a calculation for the outer wall of the interspace. prex http://www.xcalcs.com : Online engineering calculations https://www.megamag.it ...

See faq391-224. However your vessel must be a very small one, to see a significnt change in temperature in an hour or two. Moreover the rain could cause first the cooling of the wall, thus causing initially a pressure rise. Are you trying to explain something you don't want to admit?[wink] prex...

The design pressure is defined in 3-2 as the "pressure used in the design of a vessel component", so it is conceptually local and must include the static head, if applicable. However the 'design pressure' as intended by UG-21 and customarily used for pressure vessels and marked on the nameplate...