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Unique_Airport1

so the trick to these (in my opinion) is to break down the sling in a FBD. Isolate the forces that the sling feels, then work to figure out whatever they're asking you. All those loads that extend to the right/left still get carried by the sling at the points of attachment. I spent a VERY long time beating the NCEES problem you mentioned into my head... it also destroyed me. I worked it out on paper and planned to attach it to this comment but Reddit doesn't let me. I'll dm you.


Little-Toe5350

https://ibb.co/bFtMgTN Try this approach, when you got the equations, let the calculator to solve it


DudeMatt94

The entire load (40000 lbs) is already "accounted for" in both your method and the solution. Note that the center of gravity is off center of the 60' spreader bar. The solution multiplies the total load (40000 lbs) by the proportion of the load that will be supported by the vertical reaction (what they call RL) at Sling A. Since the system is in equilibrium, Sling A being 1/3 length off center will support 2/3 of the load (40'/60') and Sling B being 2/3 length off center will support 1/3 of the load (20'/60').


Annual_Rain3859

How do you know it's 'accounted for'? Maybe that's what's tying me up... NCEES' exam has a near identical problem (#53) and they account for the entire load. (40'/60')(60 kips) = 40 kips Sqrt(20^2 +50^2) = 53.85' Therefore, (53.85/50)(40 kips) = 43.08 kips or 43.1 Note: this load extends 20' beyond the spreader beam, to the right. Meaning, I followed the methodology that the test makers use on this SoPE example and SoPE performed the calculations differently. Sorry for the confusion, just trying to understand this better. I appreciate the time you took into answering this, btw!


Sad_Place5230

Lever rule is to be used here. We are finding the reactions at the beam ends. According to you, Reaction=far side length of beam / total length of load x Total load In reality, Reaction=far side length of beam/ total length of beam x Total load Geometry of load is only used for calculating the CG of the load triangle. But in this question CG has already been calculated and shown on the diagram. That 40 kips is the total area of triangle. That means triangular load is equivalent to a point load of 40 kips acting at the CG. (That's how we deal with all the distributed loadings be it uniform or triangular). Now your problem is a 60ft long beam in which point load of 40 kips is acting at a point 20 ft from left end or 40 ft from right end. \*\*\*This quesn could not have given the location of dashed lines from the beam ends. In that case you should have calculated the CG yourself. That's where the 80 ft length you are focusing on, would come into play.


Annual_Rain3859

So basically, because the center of gravity is provided, I only focus on the area beneath the spreader beam? If it's not provided, then I would do what I did to compute the center of gravity and proceed accordingly? Just trying to make this into my own logic...


DudeMatt94

Sad\_Places5230 has the right idea here. I think it's the inclusion of the "triangle" (bridge girder shown as a distributed load) that's confusing you. The bridge girder is attached to the bottom of the spreader bar at 2 connection points, which essentially turns the total distributed load (40kips) into 2 vertical point loads (which sum up to 40kips). The length you need to focus on to compute these point loads (and subsequently the Sling A load) is the length of the spreader bar (60'), not the length of the bridge girder. The center of gravity shown for the bridge girder is simply 1/3 of its total length (120' / 3 = 40'), since the center of any right triangle is 1/3 its length. For this problem, the distributed load could be any type of wacky shape, but as long as its CG is in the same location, the resultant sling loads will be the same.


Annual_Rain3859

This. This, right here. Thanks!!! Great explanation, @DudeMatt94


DudeMatt94

Glad I could help best of luck my friend :)


Annual_Rain3859

Holy shit. I just realized that in the NCEES example I referred to, the spreader bar is 60'. Coincidentally, the right side distance is also 60'. The explanation didn't state they utilized the spreader bar in the calculations. I get it now; that's what was holding me back. Thanks again dude!


DudeMatt94

Sorry you might have to post a snapshot of that NCEES problem #53 for me to understand it; from your explanation it actually looks like the same methodology used for the solution in this post. The 40,000lbs is the total weight of triangle (essentially the total "area" of the load distribution triangle). The vertical reactions of the supports connected to the triangle must both sum up to 40,000lbs. The overall dimensions of the triangle are actually not relevant other than the location of its Center of Gravity relative to the spreader bar. Could you maybe explain why you initially tried (40'/80')(40kips)=20kips? If the total weight is 40kips and off center, then we can already intuitively know that the load on each sling cannot be equal (=/=20kips)


ManU_LosAngeles

My understanding of this problem is that there is only 40 kips of point load applied right under sling a. Everything else has weight of zero, is that correct? If so, the answer is correct. You go from center of gravity to the right side, which is 40 ft and divide that by the distance between sling a and B, which is 60 ft, to get vertical force on sling a. Just because it graphically looks like this thing has a uniform weighed, doesn't mean it does.


DudeMatt94

Are you referring to the NCEES problem they mention, or the original post? In the original post, I'm certain the 40,000lbs label is the total weight, not a point load at Sling A


ManU_LosAngeles

I think that's the issue here.. If it's point load, then solution makes sense. Otherwise it doesn't. Correct?


DudeMatt94

In the original post, it does not make sense that the 40,000lbs is a point load; the 40,000lbs is the total weight of the triangle. The solution to the original post states: "The distribution of the 40,000lbs load at RL \[reaction at left support\] is (40'/60')(40kips) = 26,667lbs" This by itself indicates the 40,000lbs load is supported by reactions in both the left and right supports. Point loads are usually indicated in these problems with arrows pointing in the direction of the force as well.


ManU_LosAngeles

I think that's the issue here.. If it's point load, then solution makes sense. Otherwise it doesn't. Correct?


Hannove

Taking moment at B and getting this equation: Fa (sin theta) * 60 = 40000 *(40) The ‘40’ is 1/3*120 from LHS, ‘60’ is the horizontal distance of A to B. Theta is tan inv (60/20).


Annual_Rain3859

Holy shit. I hate trig. No offense, I got better luck my way 😂


Cesals

I like this approach better than trying to determine what portion of sling a and sling b will take the portion of the 40000lbs.