Seismic Design Manual EX. 3.14
Seismic Design Manual EX. 3.14
(OP)
I am having an issue understanding example 3.17 of the Seismic Design Manual. On page 3-137, when calculating the required column strength in accordance with section 15.8 (at bottom in page), we need to calculate the strength of the column to be greater than the sum of shear reactions from all links above the column due to the strain hardened expected yield strength of said link beams plus the factored gravity loads. When calculating this value, the manual uses 1.1x1.1x(318 + 205) = 633 kips for the summation on link beam shears. I don't understand where the 318kips comes from. Based on the givens, I would assume this value would be 1.1x1.1x205x4 = 992k since there are 4 levels above column C-1 and the only given beam size is the w16x77 from example 3.14.
Any guidance or clarification would be appreciated.
Any guidance or clarification would be appreciated.






RE: Seismic Design Manual EX. 3.14
Any chance you can scan the example?
JO
RE: Seismic Design Manual EX. 3.14
So they are simply saying that the second level is 205 kips (see Example 3.14 earlier in the text where they calculate this value) and they are saying that the two levels above (4th and roof) total 318 kips. They don't show the calculation but just ask you to assume that it is true.
So above column C-1 you have the "second" level at 205 and you have the two levels above that (4th and roof) that total 318. So the total is 205 + 318.
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RE: Seismic Design Manual EX. 3.14
I have attached Example 3.14, 3.15, 3.16 and 3.17 if that helps.
RE: Seismic Design Manual EX. 3.14
RE: Seismic Design Manual EX. 3.14
It seems to me that a leaning EBF frame would have the link at the level at the top of the column drive its shear into the column as well.
You might have to dig into the background of EBF's to figure it out.
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