Such a calculation would be fantastic.
But what I was wondering was the vertical traction in the rear mount because I had found that anchor compound and an M20 threaded rod could handle about 3 tons.
Since I can't calculate the deflection myself, I'll have to test it out. Starting at 4 meters, if it sways too much, I'll reduce it to 3.8 meters, and so on.
But what I was wondering was the vertical traction in the rear mount because I had found that anchor compound and an M20 threaded rod could handle about 3 tons.
Since I can't calculate the deflection myself, I'll have to test it out. Starting at 4 meters, if it sways too much, I'll reduce it to 3.8 meters, and so on.
About 30 mm at the outer end for a beam of 90x200 with a static load of 1000 kg at the end, self-weight neglected, 8 GPa elasticity modulus assumed, linear model assumed (Euler-Bernoulli).D Derbyboy said:
More assumptions: 4 m assumed, vertical displacement set to 0 at a support right in the middle, vertical and horizontal displacement set to 0 at the inner support, i.e., the angle is never fixed at any part of the beam.
Also note dead load, in practice, you have something dynamic, reasonably some standard should be followed like Eurocode but it feels overkill for your bridge.
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Thank you
I believe it will be at most 200-300 kg at the outer edge. It will be a relatively narrow dock, 1.4-1.6, and will primarily be used for getting on and off a smaller motorboat, think of a larger Buster.
However, I was thinking of using sparre of 75x200
I believe it will be at most 200-300 kg at the outer edge. It will be a relatively narrow dock, 1.4-1.6, and will primarily be used for getting on and off a smaller motorboat, think of a larger Buster.
However, I was thinking of using sparre of 75x200
The results depend linearly on the width in the above model, so you get a factor of 90/75 more deflection. The same does NOT apply for the height, which has an exponential dependency.D Derbyboy said:
But then you have three of those, so it becomes a factor of 90/(3*75) instead. However, you have local deflection in what will constitute the floor, which is not included at all.
Disclaimer that all calculations were done in the loo using a mobile.
Found an error in the moment of inertia...
The deflection should be closer to 120 mm in my example with a 90*200 beam.
Now I'm going to do something else.
The deflection should be closer to 120 mm in my example with a 90*200 beam.
Now I'm going to do something else.
120 mm that's insane
The results are also linear against the load - half the load results in half the deflection.
In practice, this can be discussed, but this linear theory is an established engineering method taught at colleges and universities around the world. It applies to small deformations; however, the student is left to determine what constitutes a small deformation. A general rule of thumb can be that the deflection should be less than a factor of 10 of the beam's length, but this depends on the problem and how detailed an analysis one wants to conduct.
In practice, this can be discussed, but this linear theory is an established engineering method taught at colleges and universities around the world. It applies to small deformations; however, the student is left to determine what constitutes a small deformation. A general rule of thumb can be that the deflection should be less than a factor of 10 of the beam's length, but this depends on the problem and how detailed an analysis one wants to conduct.
Yes, but it's also a single wooden beam, quite loosely fixed, with 10 kN resting at the very end. You're not planning to hang ten people out there, right?D Derbyboy said:
Would have been a bit concerned about the bending stress in this beam...
I got lost in the calculations.Den ofrivillige klåparen said:Found an error in the moment of inertia...
Deflection should be closer to 120 mm in my example with a 90*200 beam.
Now I have to do something else.
Is the conclusion that if you hang 1000 kg at one (1) end of a 4m long 90x200 beam that is anchored in the middle and at one end, it will sink 120mm?
If there are 3 beams, the weight is distributed among them and the deflection reduces by a factor of 0.4 => 120mm*0.4 = 48mm?
So in plain English, you can place 10 well-built people at the end of the dock and it dips 5cm?
To clarify, I'm not questioning but asking if I interpreted it correctly
//Mirkan
Edit: clarification came while I was posting.
A quick practical test:
Prop up the beam at the outer ends and jump in the middle. Then you will have the same forces as if it were fixed at one outer end and in the middle.
An alternative to threaded rods on each side of the beam is beams on each side of the threaded rod.
Prop up the beam at the outer ends and jump in the middle. Then you will have the same forces as if it were fixed at one outer end and in the middle.
An alternative to threaded rods on each side of the beam is beams on each side of the threaded rod.