# Tag Info

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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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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 ... 3 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 ... 2 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 ... 2 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 ... 1 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. 1 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 ... 1 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). 1 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 ... 1 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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"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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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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