15

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 ...


11

IMHO there is an "misuse" of the words brittle/strong/stiff that the OP is using. My interpretation of brittle has to do with sudden failure during testing with little to none plastic deformation. (This usually --but not necessarily-- follows a linear region). In this sense the opposite of brittle is ductile. Additionally in the context of ...


8

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 ...


7

Let me see if I understood it correctly: You have a rubber block under a uniaxial load (compression or tension). That block may or may not be constrained on one pair of sides (the load is applied along the block's x-axis and the block can't deform in the y-axis but can in the z-axis, for example). If this is the case, then the 1.33 is easy enough to explain. ...


6

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 ...


6

These are some great questions to get started thinking about the field of material science! Most of these properties are imao explained very well by looking at a Stress vs strain graph of a material test. First however a quick primer on stress and strain: In material science we are trying to determine how materials behave regardless of their shape, but ...


5

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 ...


5

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. ...


5

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 ...


4

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 = \...


4

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 ...


3

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 ...


3

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 ...


3

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- ...


3

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 ...


3

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 ...


3

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.


3

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 ...


3

For concrete and asphalt, you would be more interested in the properties of creep, which is always a function of temperature. For example, there is an entire wikipedia page about this at this time, which mainly references ACI Committee 209 For polymers, there are many models for creep. Due to the extensive range of polymer chains, it would be difficult to ...


3

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 ...


3

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. ...


3

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 ...


2

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 ...


2

The modulus of elasticity is not going to significantly change for concrete in ambient temperatures but there are provisos. The assumption is that you are talking about set concrete that has achieved 28 day strength. If you are talking about fluid concrete then that is down to the slump test and results are highly variable between strength types. High ...


2

As mentioned in other answers, what controls the deflection is the second moment of area, and the easiest way to increase it is increase the thickness. Doubling the thickness would increase the 16 fold the second moment of area and it would increase the weight by 100% However below I am outlining, another way to improve the second moment of area is to ...


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 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 ...


2

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 \ ...


2

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 ...


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