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Fundamentals of Beam Bracing: Understanding Joseph Yura's calculation of brace stiffness 2

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The cos2 is from the geometry. I hope my explanation makes sense.
For each inch the brace moves laterally, the brace elongates 1"/cosθ.
The horizontal component of the axial force to move the brace is F cosθ.
Stiffness is force/displacement = [F cosθ]/[1"/cosθ] = cos2θ F/displacement
 
I had the same question. I emailed Yura and got this response (and attachment):

You should view the relative stiffness and strength requirements as controlling the lateral displacement of a single panel. The stiffness and strength requirements are based controlling the shear displacement of the panel. If you make a cut through the panel, you slice through a single tension diagonal (the other diagonal is assumed to have no compressive strength). The required panel shear force is 2.97 kips, which is the horizontal component of the force in the diagonal.

See the attachment. When the diagonal extends a distance delta, the horizontal displacement is greater by a factor 1/ cosine. The shear stiffness of the panel is 6.45 k/in.
Joe
 
Thank you all for your responses.

nutte, your attachment was very helpful. I believe wannabeSE's response was incorrect now in light of nutte's attachment; what it should have said was that for every unit elongation in the brace, there is a 1/cos(theta) movement in the brace's lateral movement, (nutte had stated this in reverse order.)

Best regards,

jochav5280
 
Nutte's attachment is consistent with the intent of my explanation. In the attachment:
Fhorizontal = Fbrace cosθ
Δhorizontal = Δbrace/cosθ
 
Hello wannabeSE:

Your equations were correct, however, your following post was incorrect as you can see from Yura's quote: "For each inch the brace moves laterally, the brace elongates 1"/cosθ." It should have said, for each inch the brace elongates, the brace moves laterally by 1"/cosine(theta).

Best regards,

jochav5280
 
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