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There's a quite simple proof of the original problem statement. As this proof is based on simmetry, it proves also that the same statement won't be generally true in the lack of simmetry. 1) Assume the angle alpha is zero: because of simmetry, the statement is true 2) Assume the angle alpha is...

The second moment of area, as the word iteself says, is area x distance squared. Pls post this kind of questions in the students forum

sigma_theta+sigma_r is not almost constant, it is exactly constant. To me the elongation due to end pressure only is 30 in, the contraction due to Poisson effect id 27 in, for a total elongation of 3 in

If you look at your equations, you'll find that sigmatheta+sigmar=const, so the longitudinal elongation (negative, it is a contraction) is, of course, independent of r and is easily found from the strain eq. for epsz (sigmaz being, of course, null everywhere for a free ends pipe). Pls post this...

Sorry guys, I need to make amends. I now see where is the point: Newton's law only holds, for an isolated system, if the mass in F=ma is constant. This problem has something in common with the well known phenomenon of ice skaters, when they spin on the spot at varying speeds by narrowing or...

btrue, I fully respect civil engineers, but I am not one of them. ;) And an inextensible chain is not that unreal, we engineers should be accustomed to the concepts of nearly..., substantially... etc. In our case an actual metallic chain would behave as inextensible, because the strain that it...

Look at it from another perspective. If you want the jumper to experience an acceleration greather than g, then you need to exert a force on the jumper, additional with respect to its weight. This force can't be the weight of the cord, as this one is used to accelerate the cord itself. This...

Greg, read their conclusions, and you'll see that they are (were) not positive on the demonstration of the calculated accelerations by the tests.

The energy argument is used incorrectly IMO. Take an inextensible chain suspended at one end, as in the bungee, falling under its own weight. When the chain is fully down there is no kinetic energy in the system (except for some pendulum oscillations): this means that all the initial potential...

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...