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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 ...


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

In the old days we used the formula- Stress = PD/2t , and the hoop stress was twice the axial /longitudinal stress . Pressure vessel heads were usually half the thickness of the walls because of the lower stress relative to the walls.


2

I think the real point of the idea is that "if your model of the physics is not coordinate invariant, it is wrong, because the real world doesn't know anything about coordinate systems." It is certainly a very useful principle for checking computer software. Set up the same model in two different coordinate systems, and if the results are ...


2

Here is why coordinate invariance is mandatory: Imagine a lab containing a physical system upon which we make measurements in the interests of deriving a physical law. There is a special class of transforms we can apply to the system which leave its dynamics invariant. For each such special transform, called a symmetry transformation, there is associated ...


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 ...


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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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 ...


1

Spiral torsion springs and constant-force springs behave in the same manner, although that's by design. A material's stiffness is what makes it want to resume its least stressed state. See Hooke's law.


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The force of resistance of the latch varies. at the time the door touches the latch it is zero and when it is pushed fully back it is k*d. Edit why F*tan j is the horizontal component? after OP's comment. If we draw the perpendicular, $F_d*cos j,$ and $F_d* Sin j \ , $ components of Fd on the surface of latch, the hypotenuse of the triangle is the tan of ...


1

Let's accept the 100N as the top force or reaction and go from there. in order to maintain equilibrium in this beam there are set of prescribed positions and forces, or pairs of (Y, F)that will work. Any other position with wrong conjugate force will break the balance. The curve containing all the conjugate pairs of (Y,F) has this equation which is just a ...


1

You will find a large number of standard tests for the elastic properties of anisotropic engineering materials (ususally composites) in the ASTM standards. For rubber and other materials that can support large elastic stretches there are a few standard tests that can be extracted from Rubber standards(Mechanical Test Methods). Other standards are ...


1

@AndyT and @kamran explained how to technically solve this question. I would like to add a short comment about the thing seemed to confuse you the most. Indeed, there are 4 reactions in this problem. If the triangle was made of a solid body, you would deal with a statically indeterminate problem and find it quite hard to resolve those reactions. However, ...


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