5

So, we apply a point load to the end of the cantilever beam and can see that for external equilibrium, there must be shear reaction at the support. (Leaving aside for now the moment reaction that also exists.) We have an intuitive understanding of this Newtonian action-reaction from our daily lives. So, the next question is, how is the load 'getting' from ...


5

When trying to figure out whether or not a given reaction will exist at a given support, it's worth remembering what a reaction actually is. A reaction is the means by which the support resists the movement of the beam at that point. Force reactions resist the beam's attempts to deflect up-down or left-right at the support. Moment reactions resist the beam'...


5

A hinge is a point where there is no restriction on rotation. For other points on a beam, the rotation to the left of a point must be equal to the rotation to the right of that same point; that is, there can't be a discontinuity in the rotations along a beam. Hinges, however, don't have this restriction and therefore allow for discontinuities of rotation. ...


5

The derivation for bending stress is depended on the assumption that the strain distribution across the thickness is linear. i.e. $$\epsilon(z) = a_1 z + a_0$$ where: $a_0$ $a_1$ are coefficients of the slope. Because in the linear region the constitutive equation is $\sigma(z) = E\cdot \epsilon(z)$, the development of stress at distance z is proportional ...


5

$P$ is power. $P = \tau \omega$ $\tau = Torque$ $\omega = Angular Velocity (rad/s)$ https://byjus.com/physics/relation-between-torque-and-power/


4

Since the bending moment is constant across any section, by the bending equation, the beam will deform as a pure circular arc, i.e. with constant radius of curvature across any small element length, in other words, uniform deformation. This compares directly to the case when "M" varies along the length, then the radius of curvature, in turn the deformation ...


4

This is a statically indeterminate problem. I don't know what exactly you mean by moments develop at the fixed supports But I guess you mean if there exist any non-zero components of moment at the fixed points. The answer is yes. A simple and lengthy calculations yields: $$M_{\text{Fixed points}} = \left|\frac{FL}{8}\right|$$ The sign of the moments ...


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 ...


4

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 ...


4

TL;DR: Yes, any structure deforms if you put a load on it. Even adding an ant on top of a granite mountain will change (lower) the height of the mountain - imperceptibly so but it will still change it. The problem is that its not possible to measure it. That is the whole idea behind Young's Modulus (modulus of elasticity). Essentially, all materials behave ...


4

Because we're not interested in the moment of inertia or the "c" ordinate of a particular area element of the beam for their own sake - we're interested in them as intermediate steps on the way to computing the bending moment and/or the axial load. That involves multiplying by axial longitudinal stress and integrating. If you take the origin ...


3

Essentially, the problem is poorly conditioned. As drawn, it cannot be in static equilibrium, which is why the equations of equilibrium are returning inconsistent solutions. We can confirm this by summing the moments about the point where force "X" is applied. Let's take counter-clockwise moments as positive. $$\Sigma M = (-200*300) + (100*800) = 20,000 \...


3

You seem to be mixing up strain and deformation. The strain, $\epsilon$ or change in length per length, is clearly uniform throughout the beam, and therefore the deformation is not. You get the deformation by integrating the strain, so the deformation varies linearly with zero at one end and maximum at the other.


3

If we cut a section at any point of the beam, an equal and opposite force will be there (shear reaction) to maintain the equilibrium. So if we think it as a differential section, it will be easy to understand. Now imagine this differential section extended to length 'l', each fiber of the beam will have a shear resistance until it becomes equal to the ...


3

Actually, the pressure inside the vessel is uniform and constant everywhere. What happens is that geometry results in reduced stresses in one direction and not in the other. There may be some confusion here with respect to stress and pressure being the same thing. Although they share a few similarities ($\frac{F}{A}$ and units $[Pa] or [psi]$, they have a ...


3

First of all the bending moment, according to the usual convention (and the one you are presenting here) is positive. However, the problem originates when you add the area of the shear force. (Probably without realising it), what you are doing is you integrate the shear force over dx $\int_0^L V(x)dx$. I am using the mathematical notation because it is ...


3

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 ...


3

It hasn't disappeared. You have made your hand a little warmer because your own tissue has deformed. Your hand will return to its original shape and the generated heat will be moved to the environment as radiation and by convection. If you tried too hard you could see and remember the consequences at least a while. The concrete generally happens to be much ...


3

None of the three would occur if the material of the element holding the rod is "rigid", or is much stiffer than the rod. For such cases, my assessment is as shown below. For the case that the rod is much stiffer than the element holding it, the stress distribution is based on the flexibility of the rod. You can verify the result using a model ...


3

Stress stiffening is the work done to displace the structure that is caused by its internal stress. The simplest formulation of FEA (and classical theories like Euler-Timoshenko beam theory) assume that displacements and strains can be approximated to first order (i.e. "engineering strains") which means that the stress stiffening terms are second ...


3

$$M_t = \frac{P}{2\pi\cdot n}$$ where: $M_t$ is the torque transmitted through the shaft (unit in SI: Nm). P is the power transmitted through the shaft (unit in SI: W). $n$ are the rotations of the shaft in revolutions per minute (rps). (this is important for the units to be correct) If you want to use $n$ with revolutions per minute (rpm), you should use ...


3

The superposition principle is valid if the assumptions it makes are valid. A commonly used set of valid assumptions are small displacements and strains and linear elastic material behavior. For large displacements, large strains, nonlinear elasticity, plastic deformation, etc it is usually not valid.


3

In setting small-angle deformations as a limit we set the limit for strains in all configurations to be elastic and totally reversible. Meaning if the beam deflects under the load or twists under the torque, etc, it will recover its original shape fully after removing the load. As long as the strains and stresses are within this envelope, you can add or ...


2

A tensile test of a steel rod is typically used to demonstrate the difference. Necking occurs due to the lateral contraction according to poissons ratio. Because of this, the cross section changes in area. Because the load is constant, the true stress is increases in the area where necking occurs. Engineers base calculations off of the original cross ...


2

I presume you got these equations from a derivation similar to https://en.wikipedia.org/wiki/Deformation_(mechanics)#Normal_strain For small (infinitesimal) strains, the second order (squared) terms are negligible compared with $\partial u/\partial x$, and are simply ignored. This gives what is commonly called "engineering strain". For large strains, you ...


2

As member AB is orthogonal to Point C it cannot impart any vertical reaction, hence Point B provides the only vertical support, and hence has to have a vertical reaction equal to P. to find point B horizontal reaction, we set the sum of moments about point C equal to zero. $$\begin{align} \sum M_C &=0 \\ -PL +F_{h_{B}}H &=0 \\ F_{h_{B}} &= PL/H\...


2

Strain, $\epsilon = \Delta L/L$, is the ratio of deformation, not the deformation. Deformation along the beam even with the same $\epsilon$ is proportional to the length of the section we consider, like deformation for one foot of the beam is half of the deformation of two feet. It means deformation is not constant. The stress $\ \sigma $ is constant no ...


2

The general idea in both cases is similar. In the first case, we are basically summing the differential areas multiplied by the square of their distance from the axis of rotation, neutral axis. This is tightly related to the concept of the linear relationship of stress and strain in the flexural properties of beams and Young's modulus. This is usually ...


2

The calculation you are showing concerns the shear force in the top flange. The shear stress is calculated at a horizontal distance of $l$ from B. To use Zhuravskii's shear stress formula, you need the first moment of area of that part of the top flange with length $l$. It has an area of $tl$ and its centroid has a distance to the $X$-axis of the distance ...


2

Assuming your sketch has been drawn to scale, it won't be easy to make a hand calculation of the location of the shear center. The challenge is not the use of two different materials but that the usual assumption of a thin-walled cross section won't be very accurate. If you need the accurate location of the shear center, you will pretty much have to use a ...


Only top voted, non community-wiki answers of a minimum length are eligible