Maximizing Weight Capacity of Sawhorses with Evenly Distributed Loads on I-Beams

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SUMMARY

The discussion centers on calculating the weight capacity of sawhorses when supporting evenly distributed loads on I-beams. It is essential to determine the maximum load capacity of the I-beams, as they directly influence the load distribution across the sawhorses. Each sawhorse will bear a portion of the total load based on the number of beams and the load applied. For example, with a total load of 100 lbs evenly distributed across two I-beams, each beam would support 50 lbs, resulting in 25 lbs per sawhorse if two sawhorses are used.

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blake92
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In this type of setup, i was wondering what i needed to take into account to determine how much weight each sawhorse can hold. when an evenly distributed and uniform load is applied ontop of the I-beams.

Do i need to determine the max load of the I-beams going across?

How will the I-beams affect my answers?

Do i need to determine the forces in the saw horse legs?

These question and related material that you think is crucial is what I am trying to find out. Thanks for your help!
 

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What is it are you trying to find out?
 
SteamKing said:
What is it are you trying to find out?

I just want to know where all the forces are going and if i applied let's say 100lbs to the I-beams evenly would each take on half of that (50lbs) and then would each saw horse take half of that (25lbs) per beam?
 

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