Hot answers tagged

8

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{...


7

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 ...


6

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.


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

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

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. ...


3

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 ...


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

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: $...


2

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 ...


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

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.


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

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

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 ...


2

You have done everything right! You calculated the polar moment of inertia $J$, which is the sum of second moment of areas $I_i$. But here you have a symmetric object so: $$J = I_x+I_y$$ and $$I_x=I_y=\frac{J}{2} = \pi r^3t$$.


2

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 ...


2

Two concepts are in play here: Calculation of moment of inertia for a composite section. Calculation of elastic beam stresses. First, the calculation of $I_{total}$ for the wide flange section. Your equation for $I_{total}$ via the parallel axis theorem is correct, but the execution went awry. $$I_{total} = \sum (I + A \cdot d^2)$$ Since the section 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

if you randomly put a moment at the center hinge, then the problem would still be indeterminate. Make the moment large the hinge will close up. Make the moment small the hinge will open up (if you consider weight). You would need to put a rotational constraint at the hinge point to make it stable. (I am not certain is 'statically determinate' would be an ...


2

If you think about it, a support is just an applied force or moment, right? If you have a simply supported beam with a force $F$ in the midspan, the supports will be two forces, each $F/2$ in the opposite direction. But then, what if, instead of supports, you simply had two forces $F/2$ applied at the ends? That'd be exactly the same, right? Wrong. The ...


1

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 ...


1

The short answer is that the same equations are applicable in both cases for $I_{xc}$. For $I_{yc}$, the equation you have is applicable only with a sign change if you assume that $t$ is positive in both cases. However it is possible to have an equation in $b$ and $s$ that is valid in both cases. A long math-based answer. I am going to call the vertex of ...


1

What you are after is a physically (material) non-linear analysis of a frame. From the SkyCiv documentation: The SkyCiv Non-Linear solver only takes into account geometric non-linearities. You can use the linear elastic moment distribution to determine the point at which the first plastic hinge will occur (the point with the largest bending moment). ...


1

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 ...


1

Let's start by simplifying this for ourselves. In such a simple beam, horizontal loads don't have any influence in anything other than the axial loads. You apparently need to solve for those as well, but you're only asking for the bending moment, so I'm going to ignore them, so let's see how the load looks considering just the vertical component. $$\begin{...


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