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Converting a trapezoidal load to point loads 2

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Structural noob

Structural noob

Structural
Jan 2, 2020
15
US
Hey all,

I wanted to know if there is any way I can convert a trapezoidal load to point loads at regular intervals so that I can demonstrate that as a moving load?

Thank you
 
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Well if you split it into enough loads, you could be close enough that the difference is negligible. I do the opposite with wind loads on taller structures, where there might be 10 or so girts applying point loads in reality but I model the wind as a trapezoid. As it's a moving load it's hard to say that one layout of point loads will always be more conservative.
 
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apologies is stating the obvious, but can you calculate a centroid ? Representing a distributed load with a point load (or a set of point loads) is pretty basic (1st year, 2nd year structures ?).

another day in paradise, or is paradise one day closer ?
 
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Yeah, that is what i want to know. How do i split it? Since it is a trapezoidal load ( Basic UDL + triangular load), the load distribution will not be uniform or equal for all point loads. The first point load, at the highest point of the trapezoid will be the largest point load and will keep decreasing as we move along the length of the load.

Had it been a simple UDL, converting it to a point load and then further converting it to a set of point loads over the distributed area would not have been a problem. I am trying to figure out a way to distribute a triangular load into various point loads.
 
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Lionel -

To me this seems pretty simple. If you've got a 10 ft long load, break it into 10 or 11 point loads each 1 ft apart based on the load tributary to that point.
 
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x measured postive to the right

S = slope = (q,right - q,left)/width of trapezoidal load
A = distance from 0 to start point of trapezoidal load

q,n = q,left + x,n*S
b,n = witdh of load chunk to represent as a point load = x,n - x,n-1
P,n = 0.5 * q,n-1 + q,n * (1/ b,n)
P,n application point = A + ((b,n * ((2 * q,n) + q,n-1)) / (3 * (q,n + q,n-1))



Open Source Structural Applications:
 
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Scan_20200117_wqt3id.png


BA
 
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I guess this is what the OP was looking for (equivalent concentrate loads).

Untitled_f2vitt.png
 
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You can also split a trapezoidal load into two triangular loads, then you only need to do the formulation once and apply it to both triangles.

Split trapazoid into a series of regularly spaced trapazoids, divide into two triangles, find centroid of each, find centroid of combined shape, find area and you have load and position.
 
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retired13 said:
I guess this is what the OP was looking for (equivalent concentrate loads).
Click to expand...

The OP wants point loads at regular intervals. Your way, the intervals are variable, not regular.
Edit: My error...retired13 was correct.

BA
 
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My fault, forgot to mark "N.T.S." :)
 
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Actually I found out a different way to do it,

I calculated the slope between the two extreme points of the trapezoid, and then used this equation.

P = Load on the higher side of trapezoid - x * (Slope) where, slope = (higher load - lower load)/total length(which is 13.25 ft) and x = my interval (1.325 ft)
 
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retired13 said:
My fault, forgot to mark "N.T.S." :)
Click to expand...

The scale doesn't matter. The intervals are still variable.
My error...retired13 was perfectly correct.

Lionel Messi said:
Actually I found out a different way to do it,

I calculated the slope between the two extreme points of the trapezoid, and then used this equation.

P = Load on the higher side of trapezoid - x * (Slope) where, slope = (higher load - lower load)/total length(which is 13.25 ft) and x = my interval (1.325 ft)
Click to expand...

Your equation for P is close but no cigar!
Your equation does not apply to the point load at each end of the trapezoid.



BA
 
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BA,

Sorry again, I misunderstood your graph, I thought it was calculating reactions. Isn't that we set interval to our pleasure, then take the average value of the boundaries? Or maybe I didn't get the question at all?
 
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retired13,
It is I who should apologize. Your method works perfectly well. I'm sorry.


BA
 
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BA,

Don't be apologetic, we basically misunderstood each other, because we're so used to our own way of doing business.

[ADD] Besides, after given a deeper thought, I found your method is more straightforward and attractive; as opposed to what I used to do, that is take the average load in an interval, then place the average/equivalent load in the centroid of the trapezoid using geometric formula.
 
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Thanks retired13. There is often more than one way to achieve the desired result.

BA
 
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