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The stiffness of a rectangular cross section, be it steel, concrete, wood, or any other material, is related almost entirely to it's modulus of elasticity, $E$, and it's moment of inertia about the axis of bending, $I$. Since you already have your material set, steel, you cannot change $E$. What you can change is your $I$. The moment of inertia for a ...

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Stiffness is a murky term frequently used ambiguously in engineering. However, the most common definition of stiffness is the product of a beam's Young's Modulus $E$ (which is a function of its material) and its moment of inertia $I$ (which is a function of its cross-section). So $\text{Stiffness} = EI$. Loading has nothing to do with stiffness according ...

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Play around with a simple version of this structure, made from a sheet of paper fixed in a slight curve, and see what happens when you apply a load to the mid point. If the first example, any deflection of the beam will increase its length, which creates more stiffness that is included in the nonlinear analysis but not in the linear one. In the second ...

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This is feasible and can be used to modify a theoretical stiffness matrix calculated by the Finite Element method to match experimental results more accurately. The FE model can then be used to calculate things which would be impractical to measure directly. The simple approach you suggest is possible but not necessarily the best practical method. It may ...

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The symmetry itself is not a boundary condition. It is a property of your system which means that both the geometry and the load are symmetric with respect to an axis or a plane. It allows to reduce the computation to a downsized domain, which leads to considerable computational time saving. I guess you are using a FEA software and manually reduced the mesh. ...

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Let's start talking not about hinges, but supports. Specifically, why do supports generate the forces (including bending moments, if applicable) they do (or don't)? Think of a simply supported beam under a downward force. Its supports generate upward reaction forces. Why? Because if they didn't, the system would be unbalanced, and we wouldn't have a ...

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There are two basic types of structure. Statically determinate structures are those where you can calculate the forces at the restraints without knowing anything about the flexibility of the structure itself (you assume it is strong enough to carry the loads without breaking, of course). Statically indeterminate or redundant structures are those where the ...

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The translational stiffness is written as $$k_l = \frac{A\,E}{\ell},$$ where the stiffness $k_l$ is in $\left[\frac{N}{m}\right]$, the area $A$ is in $\left[m^2\right]$, the young's modulus is in $\left[\frac{N}{m^2}\right]$ and the length $\ell$ is in $\left[m\right]$. $$\,$$ The rotational stiffness is written as $$k_r = \frac{G\,J}{\ell},$$ where the ...

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A simple first order approach would be to treat the plate like a composite material, with the holes acting as a medium with no modulus. The rule of mixtures , treating the "holes" as fibers with 0 modulus, would yield a modulus of 0. So, the Semi-Emperical Halpin Tsai would be better: $$\eta = \frac{\frac{E_f}{E_m} -1}{\frac{E_f}{E_m}+1}$$ $$E_c = \... 2 k=F/d is a linear relationship. If you're doing a non-linear analysis, you shouldn't expect a linear response. For this beam, a geometrically non-linear analysis is appropriate if your deflections are any way significant. This is because a downwards load (if it is big enough) will cause snap-through buckling behaviour, whereas an upwards load will not. ... 2 Material science can almost always be broken down into "generally useful" and "specifically useful". The link in the comments demonstrates a commonly used statistic that describes behavior of foams at higher strain values - the three dimensional strain energy density function. But this statistic is "generally useful". It generalizes to all foams. In ... 2 Assuming that nothing else will give out first, what metric would I use to estimate how much weight it can support before flexing the bolts? Is it "bending stiffness" rather than "shear strength"? The metric that you should be interested in is "bending stiffness" instead of "shear strength". Shear strength has to do with opposing forces that are very close ... 2 The answer of alephzero is spot on. I just want to mention that the arch structure is particularly sensitive to 2nd order effects when it is shallow (say depth to span ratio ~0.1) . It's not acceptable to model it using 1st order theory. The figure below demostrates the type of behaviour expected (load vs vertical displacement curve) for an arch with pinned ... 2 My advice would be to forget about the mathematical theory and start thinking about what the engineering means. It's hard to think of any situation where the stiffness of a sensibly designed structure might vary "randomly" by a factor of a million, unless all the stiffness values are so high that the response is very insensitive to the stiffness value - and ... 2 You seem to be mixing up the conditions on the forces and moments acting across the hinge, and the displacements and slopes there. There is no bending moment transmitted across the hinge, but there can be equal and opposite shear forces on each side, and the two slopes can be different non-zero values. The fact that the slopes can be non-zero should be ... 2 Symmetry is used to reduce the size of the object and therefore the mesh or allow more mesh to represent that reduced object. So the “cut plane” of symmetry is not a boundary where the material changes in the real object. 1 For members of same material and same length, the one with larger moment of inertia is stiffer. If material is same, but their lengths differ, compare (I1/L1) and (I2/L2), the larger one is more stiff. Otherwise, the larger one is stiffer by comparing (E1I1) and (E2I2), if L is constant; or (E1I1/L1) and (E2I2/L2), if lengths differ. 1 The bending stiffness will be determined by the second moment of area (I). The formula you provide \int\int r^2 da is for the Polar Moment of area (J_p), and is valid for torsional problems. Apart from little issue you are on the right track. Assuming that: x is the horizontal axis y is the vertical axis then you are after I_{xx}. Additionall, I'm ... 1 The material has to be tested for each application as the fibre density, weave and number of layers all have effects on the stiffness. This means even each batch may have a variation so exact values may not be easily available. Some companies are very "expert" in this as they are using carbon fibre sheets in extreme situations and have found what ... 1 Here is an idea: try using a bundle of very thin wires and fuse the ends together at the tip. You could prototype this with small gauge, multi-strand, electrical wire. Solder the ends together so you have a solid tip for pressing. You can buy stainless steel wire down to 50 microns very inexpensively and fuse the ends together if corrosion is a concern. ... 1 One way to get this right is to start by ignoring the gearbox and modelling the system with three DOFs, i.e. the angular position of Shaft 1, and the positions of the two ends of Shaft 2. That gives the equations of motion as$$\begin{bmatrix}J_0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & J_1 \end{bmatrix} \begin{bmatrix} \ddot\theta_1 \\ \ddot\...

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The stiffness of a beam does not change with the loading if the equivalent loads and their points of action on the beam are equal. First lets do the stiffness of the beam under q uniform load. $$\delta = \frac{qL^4}{8EI}$$ Now let's load a cantilever beam with a point load equivalent to uniform load. in the distribuited load we have total load $\ P=qL \ ... 1 About the 2nd question after you read the answers: Actually it's the other way around. Imagine a load on your beam. The integration of the loading is the shear force. The integration of the shear force is the moment. The angle of deflection is the integration of the moment divided by E*I (this is where the material kicks in, E is Young's modulus and I is ... 1 As mentioned by other answers, when dealing with a statically determinate structure, the stiffness of each element is irrelevant when calculating the bending moment, but a key variable when calculating the deflection. Meanwhile, for statically indeterminate structures, even the calculation of bending moment requires the stiffness. In simplified terms, this ... 1 In your example the change in the cross section of the beam doesn't have any effect on the end moment even if the beam is a hollow section such as a pipe as long as the ratio of length to depth is greater than 10. When the beam is very deep and this ratio is less than 10, shear deformation and web warping effect could change the picture. We follow Euler- ... 1 Wooden tables have been doing that for a long time using an apron. In fact a steel clad wooden surface may be enough. But if you seal the top with a slab of steel you will also want to seal the underside to avoid moisture related warping. Add a T-profile beam on the underside connecting the supports and your table will be much stiffer. If necessary you can ... 1 Equations of Motion From what I understand from the question and your comments, you can write the equations of motion of the 3-dof oscillator in the form: $$M \frac{d^{2}x}{dt^{2}} + C \frac{dx}{dt} + K x = f$$ where the vector$x$represents the displacements,$M$,$C$and$K$are the mass, damping and stiffness matrix respectively, and$f$is some ... 1 Generalized Linear Elasticity Tensor In general, stress and strain can each be modeled by a 2nd-rank tensor with$3\times 3=9$elements. A full linear relationship between the stress and strain tensors requires each element to be paired with every other element, giving a 4th-rank tensor with$9\times 9=81$elements. By symmetry of the stress tensor, i.e.$\...

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