Question

Calculate the interaction diagram for the following 18" x 18" column with 4 #10 bars for the following 5 control points:

l8 1s

1) Po when Mn = 0.

2) Pn and Mn for ec = 0.003 and es = 0.00207

3) Pn and Mn when ec = 0.003 and es = 0.005

4) Mn when Pn = 0

5) Pn = Tn and Mn = 0

Then plot the diagram with the correct PHI applied and the 0.8 or 0.85 for Po, to develop and plot the acceptable column interaction diagram.

Then, take the actual loads and moments and plot the combinations for all 3 column load conditions in the given frame. (left column load, right column load, middle column load) Apply the SWAY moment magnifier for the column before you plot the points. State if "OK" or "NOT OK"

Beam is a 4,000 psi beam with bw=36" and an h=34", ln (clear span)= 50.5' on left and 48.5' on the right.

Column height to bottom of beam 7'-9" unsupported.

Answer 1

"F^1 c = 4000 psi fy = 60,000 Diagram 18^11 times 18^11 4 notequalto 10 Bar Design: Short Square Column Total Factored designed = Pu = 90 Skip Column size specified 18 in times 18 in F^1 c = 4000 psi, fy = 60,000 psi implies f^1 c = 4000 psi F^1 c = 4 ksi implies fy = 60,000 psi fy = 60 ksi Design of Column - longtudinal Rain forcement (a) phi_pn = phi times 0.80 [0.85 f^1 c (Ag - Ast) + fy Ast] Because Ag = 18 times 18 = 324 m^2 & phi_pu = 905 kips Therefore 905 = 0.65 times 0.80 [0.85 times 4 (324 - Ast) + 60 times Ast] 905 = 1.77 (324 - Ast) + 31.20 Ast \r\n 905 - 573.48 = 31.20 Ast - 1.77 Ast = 29.43 Ast Ast = 11.26 m^2 P_mide 8 # 11, Ast = 8 times 1.56 = 12.48 m^2 Therefore = 8 times 1.56 - 12.48 m^2 (b) Design of ties Assuming # 3 tires Spacing (1) 48 times die of tie = 48 times 3/8 = 18"" (2) 16 times"