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You should first identify the purpose of using fiber reinforced concrete (FRC), what it is for, and what it is to improve, then how to achieve the goal. In the attached article, it states: " Fiber-reinforced concrete is used to overcome the difficulty of plain cement concrete which gives very low tensile strength, low ductility strength, and a little ...

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This is very much platform specific. In general there are many ways of approaching this. A common approach among solvers for unidirectional fiber composite materials (until up to a last decade, because I am a bit rusty on recent developments), is that use 2d elements (see shell elements), and through the thickness they define layers with the orientation of ...

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IMHO this is a matter of convention more than anything else. However, I acknowledge that there are other more experienced and knowledgeable people in this forum (e.g. alephzero), that can offer a more scientific (and less based on experience) view on the subject. My view is that Even if the solid part's constitutive equations are elastic, the interaction ...

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Continuum mechanics has two branches - solid mechanics, and fluid mechanics. The subbranches of solid mechanics are Elasticity (linear) and Plasticity (nonlinear); the subbranches of fluid mechanics are Newtonian Fluids (linear) and Non-Newtonian Fluids (nonlinear). Depending on the branch of study, the nonlinear continuum mechanics involves matters with ...

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Although this is probably a gross oversimplification, I regard nonlinear continuum mechanics as a superset of linear continuum mechanics. To my experience, the basic difference is in the form of constitutive equations. I.e in the case of linear continuum mechanics you have the stress strain relationship $$\sigma = E \epsilon$$ where E is a constant and ...

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differential derivation The displacement vector is defined as : $$\vec{u} = u_{r}\mathbf{\vec{e}_r} + u_{\theta}\mathbf{\vec{e}_\theta}+u_{z}\mathbf{\vec{e}_z}$$ by derivation you can obtain all the relevant strains. geometrical derivation One way to derive the strain tensor is from geometry. The diagonal (normal) components $\epsilon_{rr}$ , \$\epsilon_{...

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Taylor is straightforward: $$\sqrt{1+2x} =\left.\sqrt{1+2x}\right|_0 +\left.{d \over dx}\sqrt{1+2x}\right|_0x +O(x^2) \\ =1 +\left.{d \over dx}{1 \over \sqrt{1+2x}}\right|_0x +O(x^2) \\ =1 +x +O(x^2) \\$$ Note that Taylor holds for matrices variables under some conditions. ps.If you can handle it, you also have the Generalized Binomial Expansion for ...

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Put a modest floor layer on the bottom (another piece or two of cardboard), and the failure location (with careful handling, no bouncing resulting in concentrated load) becomes the bends at the edges, so I would guess the max load should be proportional to the perimeter.

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Your question lacks many essential specifics required for a meaningful comparison. However, in the general sense, while the larger box have more volume/room, thus may hold more weight, but a smaller box will be more effective, and stronger, compared to the former. The reason is the smaller box posses the shorter span length, for everything else holding ...

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The answer is "Yes", and the direct source to find such relationship is from the technical publishing, that offer steel design tables/charts showing the most optimum beam/column sections for certain load with respect to the length of the beam column. The image below is an example of a chart comparing moment capacity (indicated on the left axis) of ...

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