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# Tag Info

## Hot answers tagged statics

41

More compact configuration that fully utilizes the area of the leaf. Avoid stresses concentrated on a single plane that is likely to cause the base material to progressively fail in shear or split. The staggered arrangement is more stable. It provides better strength in resisting the incidental bending resulted from the weight of the door and missing/loose ...

35

Because the screws go into wood and if the screws are in line then the wood will most often split between the screws in the grain direction and then the screws come loose.

24

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....

23

Different gears have different drivers for having holes. One way you can categorize gears is by whether they are used primarily for transferring: motion: transferring angular position and angular velocity (see clockwork) torque: when the gears are used in power transmission Motion Gears that transfer motion don't need to transfer power. So their strength ...

17

Staggering the screws will give a better chance for some of them to penetrate into solid grain. and discourage toilet paper perforation pattern. Provide for larger torque resistance. keeping the hinge from develoing a loose play, flapping out of door jamb plane. prhibiting door settling slanted out of its frame.

11

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

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: $$I_x = \int _0^{r_0}\int _{-\frac{\alpha }{2}}^{\frac{\alpha }{2}}r^3\sin ^2(\theta) d \theta dr = \frac{... 9 The minimum force is in the direction the object would move without that force. Clearly the object would move along the inclined plain, down and right in your diagram. Put another way, its the component of the force along the direction of travel that matters. More mathematically, it's the dot product of the force with the acceleration unit vector, with ... 8 Assumptions The angle between the wall and the strut is \theta a is the depth of the table top P is the weight on the table top, applied at the edge furthest from the wall The strut will fail when it buckles, which implies F_{\text{max}}=\frac{\pi^2EI}{L^2} where L, E and I are the length, the elastic modulus, and the moment of area, ... 8 In any continuous situation, you simply use integration. The linear mass density of your block is \lambda=\frac{m}{\ell}=1000 kg/m. Now you can express the torque due to an infinitesimal slice of the rod of width dx at position x as$$ d\tau=(\lambda dx) * x * g $$where x is measured from the fulcrum. Finally, you just sum up all of the little ... 8 The dimensions of the beam and magnitude of the deflection are important here. In most structural applications, it's reasonable to assume the length of a beam is unchanged by a small deformation. One of the basic assumptions of beam theory is that there is some internal surface of the beam called the neutral axis that holds no tension or compression stress, ... 8 This rule is typically applied when studying statics. Static means that your structure or object does not move. If the moments didn't all add up to zero, that would mean there was a net force action on the object, which would cause it to accelerate and move. Since it is static and not moving, we know that whatever forces are acting on the object, they must ... 8 In addition to the other answers such as weight reduction and inertia, there are other possibilities: Often there can be a precision machined hole for timing purposes. A common solution for some / many internal combustion engines to get camshafts timed to the crankshaft. Of course, dial gauges may also be used. Also, there can be threaded holes to help with ... 7 I believe you have chosen a poor reference. Indeed, that link has absolutely nothing to do with structural analysis, but rather a specific implementation in steel structures. A pinned support is a boundary condition which restricts all displacements but allows the structure to rotate. A hinge (more formally, an internal hinge), on the ... 7 You're hanging it from a rod supported by both ends - and need to use the bending equations. For this case (case 7 in the link), the max weight is:$$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}$$where W_{max} is the weight, ... 7 no. The change in moment is zero, as you can see on your plot. I think you can see if you imagine sectioning the beam slightly to the right of the support and constructing a free body diagram, the moment there is not zero. the support does not provide any concentrated moment, so the moment does not change there and so the moment at the support can not be ... 7 It could be for combination of reasons. lubrication: the holes will both pump and let the lubricant pass through. Reduced angular momentum reduces backlash and adds to gearbox responsiveness. Lighter gears need less shaft support, helping the averall compactness of the gearbox. 7 In a wing the normal situation is that the aerodynamic force is upwards (resisting gravity). You are right that there is some shear and a fair amount of torsion, but the result is that: the top side of the wing is in compression, while the bottom is in tension. So in essence what happens is the opposite from the picture below (the force is applied upwards, ... 6 Adding to @Air's answer, there's also the issue of boundary conditions. A simple span where neither support allows for axial displacements will have a slight gain in length, including along the "neutral axis". This is because, in this case, the "neutral axis" will hold a tensile stress. Since the beam deforms, the neutral axis changes from a horizontal ... 6 To put it simply your loads are causing the beam/bar to bend right? So at the cut we need to consider the internal bending moment (M). Like you said: the professor has cut the bar to calculate the INTERNAL forces experienced by the member. In any 2D problem (say in the x-y plane) the first thing that you learn is that whenever you take a cut there's three ... 6 You've got your terms confused. The maximum shear stress at the midpoint is equal to$$\tau_{max} = 1.5\frac{V}{A} = 1.5\overline\tau$$where \dfrac{V}{A}=\overline\tau, which is the average shear stress along the entire section. That is the only viable comparison to be made, stress to stress. And having a maximum stress greater than the average stress ... 6 Play around with a simple version of this structure, made from a sheet of paper fixed in a slight curve, and see what happens when you apply a load to the mid point. If the first example, any deflection of the beam will increase its length, which creates more stiffness that is included in the nonlinear analysis but not in the linear one. In the second ... 6 If a bar is only supported by roller supports (which allow for horizontal displacements) and a horizontal force is applied, the bar will not suffer any internal forces because the entire structure (bar and supports) will suffer a rigid body motion sliding along the ground. At most, there'll be a slight force due to the friction of the rollers on the ground, ... 6 Edit: There seems to be a much easier way I overlooked, which I'll explain. My first answer is kept below for reference. Your assembly consists of a small sector subtracted from a larger sector as shown below: You calculate the moment of inertia of the sector about the horizontal axis as follows:$$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 ...

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