# Tag Info

22

As with all good things, it depends. If you can assume that your supports are totally stiff and that the loading on the shelf will be approximately uniform, then you basically have the following structure: A rectangular cross-section (such as a plank) will behave equally under positive or negative bending moment, so your objective should be to balance both....

10

For example, take a look at the following static systems. Assume they have the same length and the same (constant) cross-section. Thus an equal allowed bending moment $M_u$. The first system is statically determinate, as it is supported only by simple supports. The maximum moment developing within the beam is $M=\frac{QL}{4}$, thus the load under which the ...

9

Since the mass is 5k kg and the lever is 5m, this makes it quite easy to simplify because it is exactly 1k kg per m. The leftmost 2k kg (2m) of the mass has its center of mass exactly above the fulcrum so can be ignored as it provides no contribution to the moment. This leaves 3k kg (3m) spread from 1m to 4m on the right side. The center of mass will ...

9

The reason is that you assumed that the elements around node $\text{D}$ will be the first to fail. That is not the case. Indeed, it is the elements under compression ($\text{AB}$ and $\text{BC}$) that will fail first. Also, you assumed that all the members around $\text{D}$ will present the same axial force, which is untrue. To see this, here's your ...

9

The minimum force is in the direction the object would move without that force. Clearly the object would move along the inclined plain, down and right in your diagram. Put another way, its the component of the force along the direction of travel that matters. More mathematically, it's the dot product of the force with the acceleration unit vector, with ...

8

In any continuous situation, you simply use integration. The linear mass density of your block is $\lambda=\frac{m}{\ell}=$1000 kg/m. Now you can express the torque due to an infinitesimal slice of the rod of width $dx$ at position $x$ as $$d\tau=(\lambda dx) * x * g$$ where $x$ is measured from the fulcrum. Finally, you just sum up all of the little ...

8

The dimensions of the beam and magnitude of the deflection are important here. In most structural applications, it's reasonable to assume the length of a beam is unchanged by a small deformation. One of the basic assumptions of beam theory is that there is some internal surface of the beam called the neutral axis that holds no tension or compression stress, ...

8

This rule is typically applied when studying statics. Static means that your structure or object does not move. If the moments didn't all add up to zero, that would mean there was a net force action on the object, which would cause it to accelerate and move. Since it is static and not moving, we know that whatever forces are acting on the object, they must ...

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: $$I_x = \int _0^{r_0}\int _{-\frac{\alpha }{2}}^{\frac{\alpha }{2}}r^3\sin ^2(\theta) d \theta dr = \frac{... 7 Assumptions The angle between the wall and the strut is \theta a is the depth of the table top P is the weight on the table top, applied at the edge furthest from the wall The strut will fail when it buckles, which implies F_{\text{max}}=\frac{\pi^2EI}{L^2} where L, E and I are the length, the elastic modulus, and the moment of area, ... 7 I believe you have chosen a poor reference. Indeed, that link has absolutely nothing to do with structural analysis, but rather a specific implementation in steel structures. A pinned support is a boundary condition which restricts all displacements but allows the structure to rotate. A hinge (more formally, an internal hinge), on the ... 7 You're hanging it from a rod supported by both ends - and need to use the bending equations. For this case (case 7 in the link), the max weight is:$$W_{max} = \frac{\pi d^3 \sigma}{8L}$$\sigma is the tensile stress, L is the rod length, d is the rod diameter. The sag is:$$\delta = \frac{4W_{max}L^3}{3E \pi d^4}$$where W_{max} is the weight, ... 6 Adding to @Air's answer, there's also the issue of boundary conditions. A simple span where neither support allows for axial displacements will have a slight gain in length, including along the "neutral axis". This is because, in this case, the "neutral axis" will hold a tensile stress. Since the beam deforms, the neutral axis changes from a horizontal ... 6 To put it simply your loads are causing the beam/bar to bend right? So at the cut we need to consider the internal bending moment (M). Like you said: the professor has cut the bar to calculate the INTERNAL forces experienced by the member. In any 2D problem (say in the x-y plane) the first thing that you learn is that whenever you take a cut there's three ... 6 Play around with a simple version of this structure, made from a sheet of paper fixed in a slight curve, and see what happens when you apply a load to the mid point. If the first example, any deflection of the beam will increase its length, which creates more stiffness that is included in the nonlinear analysis but not in the linear one. In the second ... 6 If a bar is only supported by roller supports (which allow for horizontal displacements) and a horizontal force is applied, the bar will not suffer any internal forces because the entire structure (bar and supports) will suffer a rigid body motion sliding along the ground. At most, there'll be a slight force due to the friction of the rollers on the ground, ... 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{\...

6

No it doesn't, and it isn't meant for that either. A standard flat washer, that is. It's meant to give the nuts and bolts a better support, and it protects the material against the nut which may otherwise scratch it upon tightening. A washer may even reduce the friction on the nut, since the fastened material may in many cases have a higher friction against ...

5

Your general idea is correct. Your fingers apply a force which can be translated to a moment at the rotation axis. The reaction forces applied by the object to be cut must generate an equal moment at the axis. If the necessary reaction forces are greater than the object's shear resistance forces, then the object is sheared through. For this reason, it is ...

5

The answer to the headline question is yes, it is equal. It physically has to be equal, since you can't have two different values of a particular directional stress at one point. The quoted equation is for shear stress, where V = total shear force at the location in question; Q = first moment of area 'beyond' the point considered; t = thickness in the ...

5

To answer the new question, which it really rather different to the original question, you will require a 7500 g N downward force at the left-hand tip to balance forces. Taking moments about your support (which is now, indeed, a pivot): $$F_{\text{LHS free end}} * 1 = 5000*g * 1.5$$ $$F_{\text{LHS free end}} = 7500*g \text{ N}$$ In other words, yes, you ...

5

The force on the perimeter will depend on the boundary conditions at the edges of each plate. Determining the precise distribution of load at each point of the supported edges of the panels is going to be a headache if done by hand. I would determine the maximum force per unit length of the edges and use that as a uniform linear force on the supporting ...

5

It may be worth reconsidering your deduced zero force members. But for this question it is not really necessary to determine which are zero force members in advance as this will come out in the analysis. It is not important if the truss is determinate or not in order to get the member forces your are interested in (highlighted in red). Approach this ...

5

I understand your structure is as follows (where the circular version is just an example of a "curved" link): One thing that is clear is that this structure is isostatic, therefore the external reactions will be equal in both cases. However, the internal forces within the bars will obviously change due to the change in geometry. In this example, for ...

5

You've got your terms confused. The maximum shear stress at the midpoint is equal to $$\tau_{max} = 1.5\frac{V}{A} = 1.5\overline\tau$$ where $\dfrac{V}{A}=\overline\tau$, which is the average shear stress along the entire section. That is the only viable comparison to be made, stress to stress. And having a maximum stress greater than the average stress ...

5

We use this equation for a simply supported beam loaded at its center with 200lbs. M= P x L/4 =200 x 3 /4 = 150 lbs.ft for calculating the tube stress we convert this to 150 x 12 = 1800 lbs.inch For a hollow round tube we calculate section module which is $Z = 0.78 (R_o^4 -R_i^4)/R_o$ assuming you have 16 gauge tube with 0.065 inch walls, Z=0.1838 . ...

5

Consider a body to be in equilibrium under the action of forces $F_i$, $i=1,\ldots,n$ which act at locations $r_i$, $i=1,\ldots,n$. Equilibrium of the body implies $$\sum _{i=1}^n r_i \times F_i=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sum _{i=1}^n F_i=0$$ What about the moment about an arbitrary point p (that may or may not be on the body). That can be computed ...

5

Think about the centroid that way: If you have this curve $y(x)$ (first graphic), where its "mass" is its area $A$. The centroid $(x_s,y_s)$ is the place where you would have to put a point mass of equal mass to produce the same static moment about any arbitrary point, thus you could balance this thin plate on your finger if you put your finger right at the ...

5

The rear-driven car is able to climb higher inclination. Rational: Where is the engine and the weight has for sure an impact, but it is not related with rear or front driven car: You have rear-driven cars with engine in the front and reverse. Additionally, you specify that the center of gravity is in the same place (center), so the place of the engine has ...

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