# Tag Info

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'...

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 ...

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

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 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 ... 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 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 ... 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 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 ... 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

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 ...

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

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. ...

2

In general what you are doing is ok. Assuming you have small enough deflections (either through bending or torsional), you can independently solve the problems. I.e.: Calculate the Force required to obtain the bending exactly as you have done. $$F = \frac{3 \delta \pi d^4 E}{32l^3}$$ Calculate the magnitude of the shear stress. Caveats However, from that ...

2

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 ...

2

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 ...

2

There are two aspects to this problem that are each significantly more complicated than what you seem to have tried so far: The angle of 45 degrees by which the cross section is changing dimension is just too much for the usual approximations to make sense. At such a large angle, plane section will not remain plane and the standard beam equations no longer ...

2

Maximum shear in a solid cylinder is $$\tau_{max}= 4/3 V_{average}$$ Maximum shear in a hollow cylinder is intuitively zero at the top and bottom of the cylinder and a maximum of 2V/A on the vertical sides. here is a link to analytical calculations: shear in a hollow cylinder. $$\tau_{max}=2V/A$$

2

Yes, your second paragraph is a pretty good natural-language explanation. The way I'd have put it is as follows: A partial derivative is a derivative with respect to (something), calculated on the assumption that (something else) is held constant. Specifically, the "local derivative" is defined as the derivative with respect to time, calculated on ...

2

There a few steps in this problem. The first one is to figure out what is the torque that each shaft is subjected. The second one is to determine the twisting angle. In this you are making the assumption that the disks do not deform. So the regarding the torque (if you know the basics about gear assemblies this is obvious): the short rod (L/2) is ...

2

How is this possible? Even if the velocity didn't depend on time, and was only changing with the change of position, it doesn't make sense for the streamline and trajectory to coincide because they hold different physical meaning. The only thing that represents a single physical thing is a trajectory. Pathlines, streaklines, and streamlines can all be ...

2

$$v_{max} = \frac{1}{EI}\left[ \frac{1}{2} M_0 \langle 0.5 a\rangle^2 - \frac{1}{2} M_0 \langle -0.5 a\rangle^2 - \frac{1}{2} M_0 a\langle 1.5 a\rangle \right]$$ $$v_{max} = \frac{1}{EI}\left[ \frac{1}{2} M_0 \langle \frac{1}{2} a\rangle^2 - 0 - \frac{1}{2} M_0 \langle a\frac{3}{2} a\rangle \right]$$  v_{max} = \frac{M_0}{2EI}\left[ \langle \frac{1}{...

2

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 ...

2

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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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 ...

1

If yes then the does this mean bending moment about point B (the hinge) is zero. You should re-read your notes, or look up "hinge" in your book's index, and start reading up, starting with the first mention. How can the bending moment around a hinge be anything but zero? What happens if you have a hinge and it has bending moment? What does it do?...

1

The equilibrium equations mean every small element of the material is in equilibrium , i.e. the forces balance at every point. The compatibility equations mean the deformed material is "continuous" everywhere, i.e. it doesn't have any internal holes, cracks, or overlapping regions.

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