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19

While not an ideal situation, it is common enough that this type of cut/reduction of the beam as it comes to its support actually has a name. This is more often referred to as a coped or dapped steel I beam. There are various ways to transition from the full depth of the beam to the depth you may require at your support. Some examples are: Sometimes the ...


16

This is a textbook example of what not to do. We don't get into stress concentration at the cut off of the corner of the beam, or the fact that the two very different stiffnesses of the beams are a constant cause of differential deflection and vibration. The thin edge of the web sitting on the column cap is an unstable mechanism waiting to either kick the CC ...


6

If the white beam on the left is adequate, the one on the right is much bigger than it needs to be, so hacking a piece out of it might not matter. In general this idea is a horrible example of engineering "design". Still, if you follow the sticker on the wall and "maintain social distancing," that will reduce the number of deaths when it ...


5

TL;DR: Since we can't see how the beam is supported on the other end, its not clear whether its structurally safe. Still, I don't believe this configuration can transfer safely any substantial structural load.. When I first saw the image, my first thought was that this is the one end of a simply supported beam. I thought is was an accident waiting to happen. ...


4

I would recommend you have your boys wear a helmet while playing. Also cover the landing area with 8 inch deep bed of mulch or some soft material like foam sheets covered by a mat. With all due respect to the enviably beautiful job your husband has done, 5 feet cantilever swing is the ideal unbalanced pendulum, overtime working its way to loosen up the nails ...


4

Let's start talking not about hinges, but supports. Specifically, why do supports generate the forces (including bending moments, if applicable) they do (or don't)? Think of a simply supported beam under a downward force. Its supports generate upward reaction forces. Why? Because if they didn't, the system would be unbalanced, and we wouldn't have a ...


4

For a rectangular simply supported plate with length $a$ and width $b$ (Roark's Formulas for stress and strain): $$\sigma_{max} = \sigma_b= \frac{\beta qb^2} {t^2} $$ $\beta$ ranges from 0.287 for a=b to 0.75 for a very long and narrow beam. 0.75 would make it equal to a beam. At a=5b ,$\beta$ is 0.74. Approximately at $a> 5b$ the plate behaves very ...


4

What is really proportional to the distance from the neutral axis (let's denote it $z$) is the strain. For pure bending and a symmetric cross-section the equation for strain is given by $$\epsilon(z) = \frac{z}{\rho}$$ where: $\rho$ or $R$ is the radius of curvature of the beam Because in the linear region the constitutive equation is $\sigma(z) = E\cdot \...


3

Shear buckling of beam web happens when the shear at a section of the beam under consideration surpasses the controlling combination of factored shear. The following three equations are the LFRD method for: no web instability Inelastic web buckling Elastic web buckling These depend on the ratio $h/t$ of the web. If this ratio is bigger than 260 web ...


3

For clarity, it's helpful to make the distinction between talking generally about "H-shaped" and "I-shaped" members, which is somewhat vague, versus the actual designations of standardized rolled shapes. The AISC Steel Construction Manual includes section information for standard W, S, HP, and M rolled shapes that will be fairly commonly ...


2

Usually continuity is to be maintained between all joints - wall/slab/beam, and side wall/side wall, or sidewalls/column, to provide water tightness. So, yes, moments due to the pool water (horizontally and vertically) need to be considered.


2

You seem to be mixing up the conditions on the forces and moments acting across the hinge, and the displacements and slopes there. There is no bending moment transmitted across the hinge, but there can be equal and opposite shear forces on each side, and the two slopes can be different non-zero values. The fact that the slopes can be non-zero should be ...


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

Here, $y'(L)$, i.e., the slope at $x=L$ cannot be determined beforehand. However, moment is prescribed at $x=L$. So, the boundary condition will be $EI\frac{d^2y}{dx^2}|_{x=L} = M$.


2

What you want to look for is Shear buckling or Web buckling. It is quite an advanced subject though. In order to determine mode of failure, critical loads etc, you need to specify explicity the boundary conditions, and then solve the problem for the dimensions. In case that you want to determine the permissible load of the member outside the context of a ...


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

The short answer would be no, or at least not reliably. In case of an actual static load there is no time dependency, hence no acceleration and thus no signal to integrate. You could try to use the transient response of the beam being loaded and try to integrate that twice in order to determine the deflection. Typically, it is difficult to accurately ...


2

The lower flanges resists a downward bending force that puts tensile stress on the lower flanges. Removing a short section of the lower flange lowers its bending strength but because of the very short lever arm at the cut , bending stresses are relatively very low ; So a clever solution. The shear strength is also reduced but from the visual sizes ,the black ...


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

Your best bet is to rig the planer with a couple of carpenter's horses or something on the sides. However, if we want to go your way need to check the following. Overturning moment on the fully extended setting with 44" cantilevered rollers. Torque on the joint between the two drawer slides. Let's say your machine and its base weigh 40# and the base ...


2

This is a case of a variable-cross-section beam. What you need to do in these cases is separate the cantilever into two parts: near the support it has one cross-section, near the free end, another. The analysis can then be done by hand, but it's a pain and I honestly wouldn't recommend it. Just use some analytical software to get the solution. You also ...


1

Call both the engineer and the builder, strong enough is one thing but meeting specifications ie building regs is another. If the engineer won’t sign it off then you have a problem.


1

Double integration of acceleration over time Double integration with an accelerometer would not yield any useful results under static loading. Alternative approach What you could do instead is the following (the longer the beam the better - a bridge would be ideal for this): place a few accelerometers along one side of the beam (if you can place on top and ...


1

Disclaimer: As a parent myself, I understand your anxiety to keep your boys safe. I have completed this calculation to the best of my knowledge, however I need to point out that I don't feel confident working with US units, nor do I have any experience in wooden structures. One thing to note is the loads on this swing. I'll just point out, outright that I'll ...


1

If we assume a single degree of freedom vibration of the beam as a spring and mass system and substitute beam with half of its mass concentrated at its center, which is not grossly wrong and we also assume no damping, the beam will vibrate under its natural frequency and a rough estimate of its vibration amplitude can be substituted for its deflection. We ...


1

Circular beam I , $I_{xx}=\pi/4(R^4-R_i^4)\rightarrow I_x= 1.1687in^4$ $M=300*10=3000 ft.lbs$ We use 2.5 as factor of safety due to dynamic gust loading. M=3000*2.5=7500lbs.ft $\sigma= My/I<25000 alum= (7500*1.5*12)/1.1687=115512.9psi $ We are by far underdesigned. testing for a 4in pipe. $1155512.9*3/4^2=64976.04psi>25000$ no good. It would be better ...


1

Using the properties you mentioned, the equation for bending stress is $$ \sigma = \frac{M}{I_{xx}}\cdot y_{max}$$ where : $M=F\frac{l}{2}=0.5kNm$ (that's a lot) $I_{xx} $= 1800mm4 $y_{max} = \frac{thickness}{2}=3[mm]$ The results is a whopping 833.[MPa], so it will fail. Assuming you use a solid beam (same material), you can estimate the minimum ...


1

It depends on the material and slack of the wire. A wire strong enough and pulled tight can impose large tension forces on the beam. Say the wire is pulled straight and under 50kg deflects only 1cm. The tension force or horizontal component of its reaction at beam will be Fx = 280/2(1)50 = 7000 kg. Which is impractically too large. And causes a huge moment ...


1

The floorboards functions are supporting and distributing loads on beams. Another function is to brace the beams. In the design of the beams we only consider the rectangle section beam, not I section. As you described in the question, the floorboards are nailed to the beam. Thus, if you cut the section of the beam, the connection between the boards and the ...


1

You are on the right track but your equation is wrong on some places and there's some misinterpretation. First error your drawing doesn't show that the midle piece thickness is 5mm (for this type of beam usualy the web and flange thickness is different). Just note some numbers aren't fully in cientific notation. I've done this to stick with your precision ...


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