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9

Consider an infinitesimal element of area $r d\theta dr$ which is at a distance $r \sin (\theta)$ from the $x$ axis. Its moment of inertia is $r d\theta dr (r \sin (\theta ))^2$. The moment of inertia about the $x$ axis of the complete sector: $$ I_x = \int _0^{r_0}\int _{-\frac{\alpha }{2}}^{\frac{\alpha }{2}}r^3\sin ^2(\theta) d \theta dr = \frac{...


8

Well, such an equilibrium (be it of moment, or shear or axial force) is necessary for any static system, and can be trivially demonstrated with Newton's second law. For forces, that is the classic $F=ma$ (force equals mass times acceleration), while for moments it is equal to $M = I\alpha$, where $M$ is moment (or torque), $I$ is the rotational inertia (also ...


7

no. The change in moment is zero, as you can see on your plot. I think you can see if you imagine sectioning the beam slightly to the right of the support and constructing a free body diagram, the moment there is not zero. the support does not provide any concentrated moment, so the moment does not change there and so the moment at the support can not be ...


6

Edit: There seems to be a much easier way I overlooked, which I'll explain. My first answer is kept below for reference. Your assembly consists of a small sector subtracted from a larger sector as shown below: You calculate the moment of inertia of the sector about the horizontal axis as follows: $$I=\frac{R^4}{24}(3\phi-3sin(\phi)-2sin(\phi)sin^2(\frac{\...


5

Since the friction is not known, the torque to overcome that friction can't be known either. You therefore can't do what you want since the problem is under-constrained.


5

It doesn't matter if you "guess right" whether each member is in tension and compression. The important thing is that the forces at the ends of each member are equal and opposite, and write all the equilibrium equations consistent with those assumptions. If you guess the member is in tension, and you find the force is negative, you have two options:...


4

Since you actually asked for the moment about the $x$ axis. Calculating the moment of inertia about the $x$ axis is a fair deal more complicated than calculating it about the $z$ axis as in my other answer. To start with, we will recognize that the symmetry about the $x$ axis lets us only work on the top half and then multiply by a factor of 2 in the end. ...


4

Without looking up all the formula's for you, the approach to this problem is rather simple. Find the polar moment of a solid cylinder, and subtract off the polar moment of the holes. For the off center holes you need to use the parallel axis therem, so you will have $$I_\text{assembly} = I_\text{solid cylinder} - I_\text{hexagon} - 6 (I_\text{small ...


4

This is a trigonometry question. Point B is a certain distance upwards from Point A. How much? You know the distance that it is to the left from Point A ($2\text{m}$) and the angle that it is from point A ($40°$), so using standard trigonometry $\tan(\theta)=\frac{opposite}{adjacent}$, you can calculate the vertical distance by rearranging to give: $...


3

I hope the following helps. This is from a book I used in college. Manufacturing Engineering and Technology, 5th Ed. by Kalpakjian and Schmid Minimum Bend Radius Where **r** is the tensile reduction of area.


3

A moment has a direction, which you can imagine when you look at the direction of the force and the axis you're calculating the moment from. Depending on which direction you're looking at it from, however, it can be seen as either clockwise or counter-clockwise. To remedy this confusion, we use the right-hand rule to identify whether a moment is positive or ...


3

The reason is that both the horizontal and the vertical component of $N_B$ generate a moment around $A$. Since the moment due to a force is equal to the product of the force and the perpendicular lever-arm between the force's line of action and the point being considered, this therefore means that the total moment around $A$ is equal to: $$\begin{align} N_{...


3

The fundamental beam equation is $$\dfrac{\text{d}^2}{\text{d}x^2}\left(EI\dfrac{\text{d}^2w}{\text{d}x^2}\right) = q$$ Which basically translates to "the fourth derivative of the deflection function is equal to the applied load". In fact the first derivative is the tangent of the deflection, which for small angles is approximately equal to the angle of ...


3

Imagine if AB didn't exist, so all we have is BD with the pinned support at D and the force at C. In this case, BD is hypostatic and becomes a mechanism, rotating around D. Obviously, we know that doesn't happen when AB is there. This tells us that the hinge is supporting BD, apparently generating an upwards vertical force at B. However, Newton's Third Law ...


3

consider an infinitesimally small section of a beam from coordinate $x_1$ to $x_1-dx$ loaded with a distributed load w. The moment of this load about point $x=0 \ is \\ M= w*x*dx.$ Now if we have the load area extended from $x_0 \ to\ x,$ we need to integrate the moment over the length of beam loaded with w: $$M= \frac{1}{2}wx^2$$ Or as you noted because ...


3

For every action there is an equal, but opposite, reaction. Never found a case that this is not true. Torque reaction on the P51 even caused uneven tire wear: https://www.aopa.org/news-and-media/all-news/2007/august/pilot/north-american-aviation-p-51d-mustang So, if you open the bonnet or hood of a car and run the engine with it in neutral, then blip the ...


2

I would add to the above answers: The beam could have sustained strain hardening from the manufacturing, cutting, preparation or any strain history that has left local changes in its elasticity modulus even if it has been straightened and looks within acceptable tolerances.


2

If the loads you apply are too big, you can get local crushing of the beam, or plastic deformation of the beam at the loading points instead of uniform elastic bending. But without more details of your experimental setup, it's a complete guess whether this is actually the cause. A basic check is whether the beam returns to its original state when the loads ...


2

What variation in value are you experiencing? Theoretically, the curvature or strain of the beam should be constant in between load points. Minor variation could be expected, but major probably indicates a problem. Assuming all strain gauges are calibrated thermo corrected etc then it could be a faulty strain gauge or poor adhesive/preparation.


2

'I' doesn't change for the different situations. 'I' is a property of the cross-section of the beam - a rectangle of width w and depth h. The Eb equations are correct for the situations set out, with the following caveats: The first situation is not for a beam that is fixed at both ends, it is for a beam that is supported at both ends but not 'fixed'. '...


2

The point where center of mass lies so the center of gravity. So, for center of mass For rod AC. B is the mid point of AC.So,let us take A as origin So coordinates of B($\frac{l}{2}$,0).Similarly for midpoint of BG We say there are four divisions which are equally space hence the length of BD,DE,EF,FG are$\frac{l}{4}$ So the mid point is E the mid point of ...


2

I have not repeated your maths, but the formulae are shown in the image. If you have the numbers correct the you have a shaft that will support nearly 5 times the load you are applying... Source : http://www.engineersedge.com/material_science/shear_stress_in_shafts_13117.htm where you can enter your values and have them confirmed.


2

Oscar, I'm not sure what your math background is, but there are a lot of different ways to get there. The most direct is to just reason from the properties of affine transforms, specifically shear mapping. During a horizontal shear, as in from triangle 1 to triangle 2, the distances moved by different points are proportional to their y coordinate. This ...


2

Area moment of inertia and mass moment of inertia are two different things. A) Mass moment of inertia (or moment of inertia): This is the resistance offered by a solid body when subjected to rotation (or application of torque). The mass moment of inertia is given by $$I = \int r^2 dM $$ The radius of gyration is the distance at which the mass can be ...


2

Bending moment is the area under the shear diagram which is definitely increasing by a slope of 5kn/m as it gets closer to support in a straight line, so it is maximum on the support. And this moment is balanced by the reaction of roller. If you remove the continuity in beam over the roller and allow a joint there thus removing the moment there, the ...


2

very close to 1E-3 in^4 More precisely: 997E-6 in^4 I wouldn't (didn't) use either of the other algebraic proposals. It's relatively easy to write an equation for the width of cut face at a defined offset from the centre: R.o = outer radius = 0.3125" R.i = inner radius = 0.25" G = gap = 0.375" width across the outer circle (whether it's actually there ...


2

Let's assume that in equilibrium the links 1 and 4 will be at angles $\theta_1$ and $\theta_4$ with the vertical. The relationship between these angles can be determined from the fact that the horizontal length between points A and B is $8+4=12$. $$ 8 \sin \theta _1+8 \cos \theta _1+4 \sin \theta _4+4 \cos \theta _4 =12 \ \ \ (1)$$ The external ...


2

"Hinges don't transfer moment" is just a way of saying that hinges do not resist bending, and therefore suffer no bending moment. You can basically "translate" that term as "Bending moment at hinges is always zero." Just note that in more complex structures such as frames there may be hinges which only affect some members but not others on the same node. ...


2

The resultant of distributed loads always acts on the centroid of the distributed load geometry, here the distributed load is uniform so its centroid lies half the way. If the distributed load varies linearly from zero at one end to a maximum value at the other end, then its centroid would lie at $\frac{1}{3} L$ from the "max load" end and $\frac{2}{3}L$ ...


2

You're on the right track, though there looks to be a typo in the last sentence of your question (you wrote "L1" and then typed the length for L2). In short, the "bottom" shear coefficient refers to the span extending toward the left. The "top" shear coefficient refers to the span extending toward the right. Perhaps a brief discussion of how to use ...


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