# Tag Info

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Bearing stress is essentially the contact pressure between two surfaces (the bolt and the joint in this example). I find illuminating the following picture that shows the force flow. figure (source chegg) how it is transferred though one part of the joint to the bolt and then to the other part of the joint. The bearing stress in this bolt correspond to ...

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Yes, I think you can say so if you reverse its concept exactly the way it is concerned. The explanation (of the concept) below is quite straightforward and precise: Saint-Venant's Principle simply states that the stress measured at any point on an axially loaded cross section is uniform given that the measured location is far enough away from the point of ...

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The meral bar is not likely to be "fixed" on a base that has less thermal expansion coefficient than it, when the heat reaches the support, it expands too. The source was mainly focused on the longitudinal displacement (1D), but you have a valid point for 3D volume expansion.

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I guess nobody knows the analytical answer but I hazard to say the stresses if the support is really kept cool or has a largely smaller thermal index, will be very significant. A large bar with a sudden increase in temperture can even crack at the support or damage the support Roark's formulas for stress and strain had some empirical formulas. I am flying ...

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I would like to add something to the already existing answer here. Yes, it would be considered as an application of Saint Venant's principle. Since the law just mentions about replacing a point force with an equivalent distributed force over the area and vice versa, it can be an externally applied force or it can be a reaction force, so it doesn't matter. ...

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Perhaps the support bracket accepts radial expansion. Or you could consider that the stress is only being evaluated over the free section and measurement starts a value of x distance from the support. At least that was how we set up our examination of the change in length of a copper bar. I think we had point zero about 2cm from the support and not only was ...

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You made a mistake in solving the reaction at joint "A". See calc below: $\sum M_G = 0$ $R_A = \dfrac{22.31*8}{12} = 14.873$ kN Solve internal member force using the method of section: Since there is only one unknown in the vertical direction, so we can solve the member force $F_{BC}$ directly by $\sum F_X = 0$ $\sum F_X = 0$ $-F_{BC}cos 30^{o}$ +...

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{I am going to assume you are applying a axial loading condition on the right end of the beam). That seems a singularity and formally, it should happen if you are constraining all the DOF's of all the nodes on the left boundary. But since as it appears that it is only occurring on the top left node and not on bottom left node, I am guessing there exists some ...

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The invariants affect the shape of the yield surface. The von Mises condition assumes that the yield surface remains cylindrical in principal stress space. If you want pressure-dependence (the circular cylinder becomes a circular cone), then you add the first invariant into the mix. If the yield surface varies depending on whether you are in pure triaxial ...

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