17

TL;DR: The bending and compression (buckling) stiffness is so small because the second moment of area of the fibres is small. Bending stiffness It does have a bending stiffness however it is really really small. More precisely it has a really really small second moment of area per fibre. (see the numerical example below to understand why many more ...


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 ...


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

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 ...


3

Having low stiffness is part of the specification of a rope. It would be far cheaper to obtain the same tensile strength with a solid rod. Creating and manipulating multiple strands is expensive. It is designed to use multiple thin strands to achieve this lack of stiffness. The question of how thin strands, with their small moment of area, achieve this lack ...


3

The standard is that The cylinder part should be hatched but not the rib. So something like the following.


2

I'll try to be a bit more verbose than the other answers. Essentially, I believe we should arrive at the same results (although I haven't checked). Like you have surmised, the temperature difference of $\Delta T=50ºC$ will initially cause the member to expand and bridge the gap -between the wall and the copper member- up to $0,08$ mm . Then $\Delta T $ ...


2

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.


2

The stiffness is the same for both, as the properties of each are identical along its longitudinal (x) axis, and $(\dfrac{EI}{L})_A = (\dfrac{EI}{L})_B$. However, for the physical length is the same, given the same loading, beam A will experience a higher moment and deflect more, as its vertical projected length (the loading span) is longer ($L_B = L_A cos\...


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

It won't just keep itself straight You know a steel beam held at one end won't just keep itself straight either right? It also bends under it's own weight when supported by one end. Just because you can't perceive something with eyes doesn't mean it's not there. It just means your eyes aren't precise enough. The fact that the rope has a radius of curvature ...


2

(Spring) Stiffness $K$ is a property of a structure which includes geometric and material effects. On the other hand, Young's Modulus $E$ is a property of the material. Bottom line is that given the same material (i.e. same Young's modulus), changing the cross-section A or the length L could result in different deformation. So: for a given structure K is ...


2

"$E$", "$K$", and "hardness", all indicate stiffness but are measured and used in different manners. $E$ - Elastic modulus is defined as the slope of the tangent line to the stress-strain (elastic) curve. It is a material-specific quantity that measures the resistance to being deformed elastically when stress is applied to it, ...


1

Check out this link, it might help. Should we use Hooke's Law (that linearly relates stresses to strains) if the stiffness of body is changing during deformation? Now, the stiffness equation i.e. K = EA/L is only used for axial loading conditions. It is derived by dividing the load applied by max deflection. However, for bending cases, the bending stiffness ...


1

Engineering point of view: There are ropes that do have significant bending stiffness (and, as a consequence, compression hardness). E.g. steel wire ropes. This is almost never an advantage - such ropes tend to self-tangle easier (but making intentional knots is harder), tend to break at small bending radius (making most knots unusable) and tend to retain ...


1

They design the jumping rope this way. if it would have any moment strength or compression strength it woul act as a soft "column" at the strat where half of the cord is bent up to reach the Jumper's harness. Thus it could send the jumper into an unpredictable dangerous arc, as opposed to let them fall straight down into hopefully a clear vertical ...


1

The best way is to add another HSS c channel back to back because this way you are quadrupling the I in the weak axis. Just make sure the bolts are tight. If the bolts are not tight fusing the two channels as one, you get at least something more than 2*I.


1

There is an easy way you can calculate a god approximation to the bending strength of the guitar handle. You can assume you guitar handle cross-section as a trapezoid and measure the moment of inertia, I of that as per this formula. $I_{Cx} \ $ is the moment of inertia about the trapezoid's y centroid parallel to the x axis, where you are interested in. All ...


1

deflection of a beam Depending if the beam is cantilever, simply supporter or else you have different formulas. For the simple case of a cantilever beam, with a concentrated load at the end the maximum deflection at the end is : $$\delta_max = \frac{P \cdot l^3}{3EI}$$ for other cases you can find under deflection of beams tables with equations like the ...


1

I made mistake on my earlier response (deleted). Allow me to try it again. Please let me know if there is mistake.


1

We know that the spring and the copper bar reach an equilibrium at Fs =Fc. $ Fc = KcXc$ where: $Kc =\frac{(\text{Young modulus of copper})A}{L}$ $Fs =KsXs$ $KcXc=KsXs$ $ Xc/Xs= Ks/Kc$ and we already know $Xc-0.08-Xs= 1.325 -166.5*1.325/(Ks+Kc)$ I let you handle the rest.


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