New answers tagged

0

If you mean a cylinder with the diameter $Dcm$ and a square with a side $A=Dcm,$ The square is stronger, solid, or hollow. If they have the same moment of inertia, I, then they are equally strong only for bending moment, but then the square post has a bit less area and a bit weaker for the axial load. let's see what size square has equal I as a cylinder with ...


0

Of course you can replace pressure loads by point loads. That is exactly what almost every FEA program ever written does internally to analyze the model. However doing it correctly requires knowing exactly how the elements in the model were formulated. Of course the people who wrote the software do know everything about the element formulations. Trying to do ...


0

I don't recognize your equation, but, as you are interested in "boundary condition", at which only one face exist, so you shall eliminate one of the shear deformation terms as indicate below.


1

Besides the more elegant/classic method presented above, here is a practical, and simpler, method to derive the fixed end moments, using the concept of "consistent displacement" and "superposition". It can be used to verify the solutions obtained from the classic approach. 1) The assumptions and information from simply supported beam: 2) ...


0

if you want to be precise no. Lift is mostly generated by the uneven distribution of low pressure on top of the wing. that distribution varies with your wing design. In many profiles there is even downward pressure on parts of the top of the wing. you have to search For the wing that is similar to yours, the profile, aspect ratio and the sweep. many of the ...


1

You are on the right track. The way I'd solve it, is the following. I would break your problem in two subproblems right next to the application of the force F (at point K which is at a distance $x=a^+$). In that case, there compliance constraints that should be met. I.e. the point K is common at both problem 1 and problem 2, so: The shear forces at $x=a^+$...


0

The only reasonable possible concern is flatness of the metal top for the granite to sit on . Welds must be ground smooth at the top of the structure and should be as flat as possible as, no doubt ,the granite is flat. The traditional way to do this is to have the steel firmly held to a "strong back" ( welding table ) during welding. I have a 915 ...


0

The cross-section of the legs will be doing most of the supporting for the granite weigh in this case. Even the maximum of weight 238kg, is not going to be too taxing on the supporting frame. (For example if you had vertical legs a flat bar of almost any thickness more that 3 mm would be OK). Since you have this small inward angle, there will be some small ...


1

The equation of motion after the impulse is $$m\ddot x + c\dot x + kx = 0$$ with initial conditions $$x = 0, \dot x = \dot x_0 \text{ when } t = 0.$$ You have found the frequency and damping ratio already, so you can rewrite the EOM as $$m(\ddot x + \beta \omega\dot x + \omega^2 x) = 0$$ where you know $\beta$ and $\omega$. Scale your measured results to ...


3

Tl;DR: yes it is correct (disregarding rounding errors) assuming points at A: ( $x=0$) B: ( $x=\frac{1}{3}L$) C: ( $x=\frac{1}{2}L$) D: ($x=\frac{2}{3}L$) E: ( $x=L$) It is obvious that there symmetry in the problem. Therefor point C (in the middle will not move at all). For the half problem (ABC) you can write the equilibrium of forces. $$F_A + 14.4 +F_C=...


0

In classic structural theory/practice, a model beam is represent by a line element along its neutral axis (zero strain/stress), on which the load is placed on, and supports are attached to. From this assumption, all three beams in the example will yield the same internal forces (M, V), if loaded by an identical vertical load. However, in the real world, the ...


0

Since you want to challenge yourself with the design process, it would be good to use the force information not just for strength of the bar, but also the deflection. For example when you design a bow for archery you want deflection, and when you design a bridge, you don't. Olympic weightlifting bars are somewhere in between So it is worth considering how ...


1

I assume you modelled the structure in 3D structural analysis program, for such a case, if the two frames are pinned at the base of columns, but not connected on the top (two independent frames), the program shall issue a warning, because the frames are unstable in the transverse direction (pinned at the bottom, free on the top). You need to check your ...


2

Like alephzero already mentioned the pull up part is usually the less stresses part. The problem is with the deceleration forces when you are at the top of the pull up bar and suddenly you let your self down. Although that seldom happen (i.e. there is a constant force for deceleration), it is best to assume the worst case scenario. I'll make the following ...


2

I would have thought the nut and screw screw thread was more likely to be the failure point, but you are looking at the part and I'm only looking at a sketch of it. The yield strength of grade 5 steel is around 600 MPa. The area of the pin is about 47 mm^2. So to permanently bend the pin by shearing it would take about 28000 N or about 6300 lbf. (Or arguably ...


2

Yes, That is a good estimate. Another reasonable way is to measure how fast you can pull yourself up. Say if you can time your lift acceleration and it is $12m.s^2$ you are applying, $mg+1.2mg=2.2mg \ , $ 2.2 times your weight. Of course, If you have access to one of those benches with lift up weights, the ones that have rubber bars locking you in, you can ...


0

Your design is governed by the moment, the equivalent uniformly distributed load, $$ W = \frac{2*M}{L^2} = 17.6 lb/in$$ If you want to place multiple point loads on the beam, you should sum all moments due to the point loads, then compare it with the limit given by the manufacturer.


1

An easy way is to back-calculate the S, section modulus of the beam, and its bending strength then you can verify if it will support your set of loads or any other load. $$M = S*\sigma_{b .max} = wL^2/8=1960*2^2/8 = 980lbs.ft$$ Therefore you calculate the combined moment of say n point loads P1, P2, P3...Pn separately and add their moments to check if it ...


Top 50 recent answers are included