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The Fourier transform of $y\left(x,t\right)$ from the time to the frequency domain is given by $Y\left(x,\omega\right)=\int_{-\infty}^{\infty}y\left(x,t\right)e^{i\omega t}dt$ and satisfies the differential equation: $$ EI\frac{\partial^{4}Y\left(x,\omega\right)}{\partial ...


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To state the physics of this question: fire raises the temperature of load bearing steel. The latter's mechanical properties radically change - yield strength in particular drops radically - leading to structural failure under normally well bearable loads. The engineering stuff: Steel is a known, severe fire hazard in this way, but its strength and low ...


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Strength of materials is affected by defects. A perfect crystal of iron would be extremely strong. Once a crack starts, it is not so hard to make it advance one atom deeper. Think of tearing open a plastic bag. Much easier once the tear starts. Brittle materials can be easier to break because they stretch less. It is easier to tear a sheet of paper than a ...


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So, buckling is the bifurcation of static equilibrium. And thus: More technically, consider the continuous dynamical system described by the ODE $\dot x=f(x,\lambda)\quad > f:\mathbb{R}^n\times\mathbb{R}\rightarrow\mathbb{R}^n.$ A local bifurcation occurs at $(x0,λ0)$ if the Jacobian matrix $\textrm{d}f_{x_0,\lambda_0}$ has an ...


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Generalities The problem has spherical symmetry, so it makes sense to use spherical coordinates ($r$, $\theta$, $\phi$). We can divide the vessel into differential elements like the one shown in this post. Deformation and strain Only radial deformations are allowed by the spherical symmetry, so let's parametrize the deformation by $$r \rightarrow r + ...


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No, it's not possible. The general rule of strength in foams is $\rho^3\propto\sigma^2$ so if you cut the density by a factor of 4, then your strength would be cut by a factor of 8. From a conceptual standpoint slicing a organized shape under a tensile load by a plane perpendicular to the load will show that there are $n$ discrete elements, all under a ...


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It depends on the stress you need to support because of buckling as is visible in your first diagram. This also means it depends on how the force is applied - at a point, at the top, across the end? For small forces (no worries about buckling) you want the tallest thinnest beam since you are maximizing the second moment of area for constant area. However in ...


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The dominant load that a large structure has to face it's its own weight. The weight load scales as volume, which means that it goes as $\lambda^3$ The canonical strain $\epsilon$ upon a given load $F$ is $$ \epsilon = \frac{F}{k} = \frac{ F L }{ A E} $$ so the strain will scale as $\lambda^3 \lambda^{-1} = \lambda^2$, which means that scaling up the ...


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"There is a height I have to consider" Yes, that is where the factor $I$ in the equation comes in. It is known as the second moment of area, and for a rectangular beam of width $w$ and height $h$ it is $$I = \frac{h^3w}{12}$$ The rest then follows from the equation you found...


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Disproof by counterexample Let's see if we can construct a counterexample to the fully-symmetric solution, where the forces at $A$ and $C$ are equal to each other and the forces at $B$ and $D$ are equal to each other but there is a different force on each diagonal, $A \neq B$. Does such an arrangement exist where the rectangle is stable? It does. We can ...


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If I got it right, you probably mean the stress on a column - not a beam - caused by an axial load $F$. In that case your guess $F/A$ about the stress is right, otherwise the column should be moving!


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You cannot add the deflections between the two cases because the overall shape is different. See this answer to a similar question. Maybe you can see it like this. Each segment of beam between a max. curvature point and zero curvature zone will have the same deflection of the same length. As you can see you have zero curvature (max slope) at the load ...


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The go to formula in pressure vessel design is the semi-empirical NASA SP-8007. Page 7, & 10 tell you most of the annotation used in the rest of the document, page 14 has the equation for isotropic cylinders. Personally, I've mostly used the composites formulas in the back. The long explanation is that due to poor experimental results, a cylinder will ...


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Not sure you care anymore, but - for large deflections, this does not hold. As the pretty pictures at wikipedia show, an elemental does not remain perpendicular to the neutral axis when bending deep beams, so that goes out the window. The second is the neutral axis in general shifts towards the compression side, because there is less material on that side ...


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The bending stress is greatest in the middle; and when the ends are clamped, the stress is less than if they are free to bend. All this is captured in the equations for bending moment of a beam - a (pretty) complete set of these can be found at http://www.awc.org/pdf/DA6-BeamFormulas.pdf For part 1, you can use figure 7 (load at center): and when the ...


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You want the 2 flanges at either side to be bigger. The reason the I-Beam cross-section is shaped that way is because it is the material furthest from the centroid (i.e. the 2 flanges) that is providing the most stiffness against bending. The material closer to the centroid is not providing as much stiffness, so it makes sense to take most of that away to ...


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Actually the data presented by You show that iron/steel is more brittle than diamond. Precise tensile strength of diamond is unknown, however values of up to 60000 MPa have been observed. Typical values of tensile strength of iron/steel varies from 100 to 11000 MPa. Therefore diamond can withstand more than iron/steel.


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"different values of n represent different modes of buckling" (http://emweb.unl.edu/NEGAHBAN/Em325/21-Buckling%20of%20columns/Buckling%20of%20columns.htm )


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Buckling is not limited to thin columns, it is also important, e.g., for thin shells under compression; for example, if pressure in a poorly designed tank is below atmospheric pressure, the tank can buckle under atmospheric pressure (it happens to large oil tanks, railway car tanks - you name it; you can easily find a lot of impressive images on the net).


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The bucking formula comes from a stability analysis of the restoring moment inside the beam. Actually as compressive forces are applied to a beam, its natural frequency drops, and when you reach the Euler's limit the natural frequency of the beam becomes zero. By definition this is the point the beam will not behave at all in a static fashion and will move ...


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I don't think that the deflection of the beam in the centre should be the same. You use two different relations to compute curvature from the deflection. Since, you use the same load, the two different relation have to result in different solutions. In general it's hard if not impossible to say in advance what the differences in the solution will be. This ...


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I'm not sure if you are still interested, but I believe the equation you are looking for is: $$F = \frac{2\sin(θ)EI}{L^2}$$ where θ is the angle at the end of your cantilever. I base this equation #16 from the paper, "An integral approach for large deflection cantilever beams"


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Q1: How does one interpret tensile strength, yield strength, etc.? The answer is to interpret them as the result of a test that tells you what the material can withstand in an engineering application. The type of machine used to measure tensile strength is popularly called an Instron machine (the most famous manufacturer is Instron; kind of like how tissue ...


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The diagram in your edit would be true if it were pins at the ends, not walls. If we assume that the walls and beam are fused together at the ends, then it means that the walls now resist bending moment. Imagine for the cantilever case (one end to the wall, the other free), if you pull down on the free end, and the fixed end have none zero slope, the ...


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Yes you are right. To understand exactly why you are right, keep firmly in mind exactly what the "bending moment" and "shear" mean. When we say that the moment and shear are $M(x_0)$ and $\tau(x_0)$ at position $x_0$ we mean that: We imagine the beam cut at the position $x_0$ and we draw free body diagrams for the two sundered sections; In particular, we ...



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