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The best way is to add another HSS c channel back to back because this way you are quadrupling the I in the weak axis. Just make sure the bolts are tight. If the bolts are not tight fusing the two channels as one, you get at least something more than 2*I.


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What's acceptable will depend on your application. For a chair, 1mm of deflection will not be noticeable. For a CNC machine not so much. If you really care about optimizing the design, I would start by designing the object with narrow extrusions, and then see if/were the deflection will be too great, then bulk it up as necessary. This calculator ...


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In general the acceptable deflection has to do with perception. What I mean by that is, if it is possible to perceive the deflection of the structure. Therefore it is usually presented in building codes (usually that's where you most see it), as a percentage of the span of the beam. However, there is no universally accepted allowable deflection. For example: ...


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Let's label the upper beam as "A", the lower beam as "B". Assume rigid vertical studs, the deflection at beam A at points 1 & 2 must equal to beam B at the respective locations, that is $\Delta_{A1} = \Delta_{B1}$ and $\Delta_{A2} = \Delta_{B2}$. Now, release the studs and obtain the deflections at points 1 and 2 due to concentrate ...


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If I understood correctly from your post you started with the following cross-section (Lets say A) and you replaced it with the following (or the inverted) (let's call it B). There is one reason why you are getting larger deflections, the total second moment of area is smaller. Crosssection A The total second moment of area for Cross-section A along axis ...


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Basic rule of thumb would say, $$80* 3.5 =P*50 \quad P= 80*3.5/50 = 1.6*3.5= 5.6 kg$$ P is the light fixture weight but the fastners attaching the hanger to the ceiling need to be strong enough to support this load too.


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The formula is correct, the answer is likely wrong. Let's look at the two variables that can affect the outcome. The weight and the I. The schedule 80 I is. $$I = π (r_o^4 − r_i^4) / 4 = \pi ( 1.315^4-0.975^4)/4= 1.898"^4$$ $$\delta= \omega*120"^4/384EI $$ The schedule 40 I is. $$I = π (r_o^4 − r_i^4) / 4 = \pi ( 1.315^4-1.049^4)/4=1.397"^4$$ this gives a 1....


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