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I'm making metal frame to support a wooden deck.

The yellow object is the concrete base where there is a 2" trough where my metal frame needs to support a 300-500lbs deck load.

The span between the unsupported trough is 15'. The length of the cross brace is 22"

The red frame is 2" welded steel tubing. Wall thickness 1/8".

I understand 2" will have huge deflection at 15' unsupported span but will adding cross beam like this reduce this deflection to be acceptable to support a ground-level deck?

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For HSS $2"$ x $2"$ x $1/8"$, $E = 29000 ksi, Ix = 0.486 in^4$

Assume uniformly distributed load. Let $w = 500 lbs/15 ft = 33.33 plf$, round up to $40 plf$ to include misc. weight.

$\Delta = 5*w*L^4/384*E*I = (5*40*15^4*12^3)/(2*384*29*10^6*0.486) = 1.616"$

$\Delta \approx L/111, NG(*)$

(*) Range of usual deflection limits:

  • Live Load: $L/180 - L/360$

  • Total Load: $L/120 - L/240$

As noted above, depending on the nature of the given load, the deflection limit varies. Since the deflection of your beam has exceeded the lowest limit, you need to determine a deflection limit according to the type of load, and that is suitable for your application.

(Note: Adding the cross bar does not add beam stiffness. You need a stronger beam)

As an example, assume $L/180$ governs, let's see what size of beam will satisfy it:

$I_{new} = I_{old}*(L/111)/(L/180) = 0.486*(180/111) = 0.788 in^4$

  • HSS $2"$ x $2"$ x $5/16", I = 0.815in^4, OK$

However, as the new tube is almost twice as heavy as the original tube, you need to re-check the deflection with the modified load.

Also, don't forget to check the flexural stress: $f_b = M/S <= 0.66fy$,

where $M = w*L^2/8$.

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  • $\begingroup$ I of HSS 2x2x1/8 is 0.303 inch^4. $\endgroup$
    – kamran
    Commented May 2, 2021 at 6:53
  • $\begingroup$ Double-check the moment of inertia. Depending on the radius of the corner and variation in the exact wall thickness, some producers may have a different value than that list on the AISC table. I got it from the AISC web table. $\endgroup$
    – r13
    Commented May 2, 2021 at 7:12
  • $\begingroup$ Note, if you are using cold-formed steel (which is not suggested), the properties will be very much different. $\endgroup$
    – r13
    Commented May 2, 2021 at 7:14
  • $\begingroup$ Thank you. Do you know where I can find a similar table for I beam to find the I value? $\endgroup$
    – user2727691
    Commented May 2, 2021 at 13:12
  • $\begingroup$ Based on this calculation, would it be better if I use a rectangular tube of 3" x 1" x 1/8" which weighs similar to 2x2 1/8 but has strong I value 0.817? On similar token, using I-beam/H-beam will even yield better weight to strength ratio, correct? Do you know where I can find similar chart for I beam? $\endgroup$
    – user2727691
    Commented May 2, 2021 at 13:20
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No, it won't work. It won't even support it's own weight at that span. regardless of lateral support or cross bracing.

It is easy to calculate what size steel tubing you neef. I estimate it needs to be at least 10 by 3 inches.

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