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# Question about the TIA-222-G (urgent)

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 Jalmeida (Civil/Environmental) 15 Jun 12 16:01
 Good afternoon, someone can explain me the checks required by TIA-222-G for minimum bracing Resistance (page 69). I understand that the bracing must satisfy the condition of Ps, however I do not understand why Pr determine the minimum required design and Strengths of multiple connecting members. I wish I could explain the calculation procedure. Cumpz, thanks
 JoshPlum (Structural) 15 Jun 12 18:30
 Not sure that I understand the question. Where is Pr referred to in that section (4.4.1) of the G standard? The main provision is that the total brace strength in the direction braced should be between 1.5% and 2.5% of the maximum design axial force in the supported member. The idea that the supported member will require more support (Ps = 2.5%) when it is more slender seems rational to me.
 Jalmeida (Civil/Environmental) 15 Jun 12 18:57
 I'll put are a number of questions that would like to see clarified: 1)which is a support member in TIA? 2)I have to check the condition of PS in leg members? In my opinion as Pe is high is not necessary to check buckling problems, the geometric characteristics of own section. Thus, not understand the use of Pr, since it is a check for the leg members. 3)Someone can explain me the procedure of calculation I have to ralizar to check the conditions of 4.4.1, to better understand. It may be a simple concept, but I'm with difficulties in understand the elements of the verification procedure. Thanks.
 JoshPlum (Structural) 15 Jun 12 19:35
 Generally speaking, the leg is the primary member that will be braced by secondary horizontal or diagonal members. If you run an analysis by hand (or even using a computer) some of the forces in these secondary members may be very low. Therefore, there is a tendency to use the smallest possible members for these horizontal and diagonal braces. However, as the primary member approaches its buckling load, it will tend to push against its the secondary members at the brace points. These memberse must be strong enough to survive that loading. Therefore, the Ps value is for the design of these SECONDARY bracing members. If they cannot resist that Ps force (between 1.5% and 2.5% of the total axial force in the primary member) then they will fail before the primary leg member and the leg will fail before it reaches its calculated design loading.
 Jalmeida (Civil/Environmental) 16 Jun 12 4:36
 Thanks for the explanation, I understand. However, according to table 4.1 and 4.2, the diagonal members to be, for my case (1.15xPs) = Pr, in which the minimum strength required = 1.15xPsx0.5/cos (angle). This minimum resistance value is compared with anything? I do not understand what it affects, because we have that range between 1.5% and 2.5%.
 Jalmeida (Civil/Environmental) 18 Jun 12 8:40
 UP!
 Jalmeida (Civil/Environmental) 18 Jun 12 12:28
 I thank the web for the explanation of JoshPlum that somehow confirmed partially waht I was doing for the numerical design example in my master of science thesis in civil structural engineering. But a doubt still stands related to what control value should be used for the required design quantity of the diagonal minimum strength given by 1.15 x Ps x 0.5 / cos(orientation angle) ? I would appreciate a clear answer to this fact. Regards. JAlmeida (FEUP, Porto)
 JoshPlum (Structural) 18 Jun 12 13:59
 There are no additional requirements given in tables 4-1 and 4-2. Those tables are merely meant to provide examples on how to apply the provisions. Essentially, they're just helping you to figure out the basic trigonometry for those specific cases. The restraining force (Ps) needs to be perpendicular to the primary member. If you look at table 4-2, the cos(theta) term just resolves that force into a force in the direction of the brace (which is not perpendicular to the primary member). The restraining force need NOT be taken by a single member. If there are multiple secondary members restraining the primary member (which happens all the time for triagular towers) then each secondary member can be designed for a lower force. This is the basis of the 0.5 factor that you seen in table 4-2. When you are dealing with triagular tower, you know the horizontal angles are 60 degrees, right? Well, the 0.707 numbers in Table 4-1 are merely the square root of 2 and the 0.866 values are merely the sin of 60 degrees. That's where your 1/0.866 = 1.15 values comes from.... again basic trigonometry.
 Jalmeida (Civil/Environmental) 18 Jun 12 19:14
 Thank you for the explanation, you helped me a lot once again thank you. Regards, John
 Jalmeida (Civil/Environmental) 19 Jun 12 4:30
 however my triangular tower, presents a growth in the angle of the diagonals. What for angles greater than 60 (without affecting the coefficient 1.5), automatically 0.5xPr/cos value (angle) falls outside the range 1.5% to 2.5%. I can not have bars above 60 ยบ?
 Jalmeida (Civil/Environmental) 19 Jun 12 5:00
 Last post is not relevant, because I already knew. thank you
 Jalmeida (Civil/Environmental) 19 Jun 12 11:06
 My diagonals is not check, I'll let my interpretation here for your appreciation: If the tower is triangular: sin (alpha) = Ps / Pr <=> Pr = 1.15 x Ps As I have a brace in diamond, the diagonals have: cos (theta) = (0.5 x Pr) / Nresistant <=> Nresistant Pr = 0.5 x / cos (theta) Nresistant - resisting force of the diagonal member Nresistant Psx0.5x1.15/cos = (theta) Apart from my section to resist this effort, the Ps must be within the range 1.5 to 2.5 Fs/100. Example (extreme situation): angle: 60 Fs (effort of the leg member) = 1203 KN KL / r = 120 (threshold to be 2.5%) N = 46 KN (effort-diagonal) Nresitance = 34.6 KN therefore does not check the condition in question. This is an example of the checks that do not occur, which is almost all.

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