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So I've got a support system shown here in the attached image. Comparing configuration "A" to "B" it seems intuitive that "A" could support a larger load. The stubs are locking the two beams above and below the tube steel to eachother to increase resistance to deflection. How can I show this mathematically? A colleague suggested I would add the deflections of the beam above and below the tube steel to get a total deflection and you could show that this total deflection would be less than the deflection of the support beam in config. "B". This doesn't seem quite right to me, but I'm not sure. I should note I am assuming the taller base supports are 100% rigid.

beam loading

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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 $\Delta_{A2} = \Delta_{B2}$.

enter image description here

Now, release the studs and obtain the deflections at points 1 and 2 due to concentrate load $P_B$ at the mid-span of beam B.

$\Delta_{B1} = \Delta_{B2} = \frac {P_Ba}{48EI}(3L^2 - 4a^2)$

enter image description here

Next, apply a force $P_A$ at points 1 and 2 on beam A, and get the deflections at points 1 and 2.

enter image description here

$\Delta_{A1} = \Delta_{A2} = \frac {P_Aa}{6EI}(3aL - 4a^2)$

Now you have two equations and two unknowns. By equation $\Delta_{A1}$ and $\Delta_ {B1}$ you can get $P_A$ in proportion to $P_B$. And based on the equivalent forces concept, $2P_A = P_B = Applied Load$, you can get the exact/actual mangitude of $P_A$ and $P_B$.

Finally, compute the beam deflections using the actual $P_A and P_B$ using the appropriate equations as shown above. Note the difference in deflection in segments between points 1 and 2 of beams A and B, it implies the beams suffer different internal forces.

Note, this solution is valid for the case that the load is directly attached to beam B; beam A shares the load through force transfer from the rigid studs, not through connection to the load.

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  • $\begingroup$ I see the the logic here, but when I try this I don't think the equations are independent. All the deflections are the same so you get 1.) $ \frac{2P_Aa}{48EI}(3L^2-4a^2) = \frac{P_Aa}{6EI}(3aL-4a^2)$ so the loads just cancel out. What am I missing here? @r13 $\endgroup$
    – Ryan C
    Jun 24 at 16:45
  • $\begingroup$ If you equate the two equations correctly, the result should be P_A = P_B*(3L^2-4a^2)/{{8*(3aL-4a^2)]. If L = 10, a = 2,, P_A = 0.81*P_B. If the load = 100, then 2*P_A + P_B = 2*(0.81)*P_B + P_B = 100, that resulting in P_B = 38.2, and P_A = 0.81*38.2 = 30.9. $\endgroup$
    – r13
    Jun 24 at 17:06
  • $\begingroup$ You should use the equations exactly as shown. There are two mistakes, both on the left term 1) there is no 2 in the term for deflection, and 2) the term P_A should be P_B. $\endgroup$
    – r13
    Jun 24 at 17:15
  • $\begingroup$ BTW, how do you write equations in the "Add comment" block? $\endgroup$
    – r13
    Jun 24 at 17:32
  • $\begingroup$ engineering.meta.stackexchange.com/questions/307/… @r13 $\endgroup$
    – Ryan C
    Jun 24 at 17:59

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