Question

3. Determine the value of Pn for the short tied column if ex 500 mm. The concrete strength is 28 MPa and the steel is fy-413 MPa. Using Interaction Diagram Design a 300 mm wide rectangular short tied column to be used to support the following: Pu 2000 kN and Mu 450 kN-m. Place the bars in the two end faces. The concrete strength is 28 MPa and the steel with fy-413 MPa. Select a column with approximately 2 percent steel. Using Interaction Diagram 4.

Answer 1

Given the data is, e_x = 500 mm, f_c = 28 mpa, f_y = 413 mpa A ultimate Axial load capacity of the column As per Acl code: - A_g = 500 times 500 = 2500000 mm^2 A_sf = 11617.61 mm^2 e_g = A_sf/A_g = 11617.61/250000 = 0.0464 0.01 lessthanorequalto 0.0464 lessthanorequalto 0.08 as per Acl code 318 - 11 sec 10.9.1 For rectangular tie column, the design strength is pi p_n = 0.80 pi[0.85 f_e^1(Ag - A_sf) + A_s + fy] = 0.80 pi[0.85 times 28(250000 - 11617.61] + (11617.61 times 300)] = 69087.93 KN times pi = 69.087 times 10^3KN times 0.65 = 44.90 times 10^3 KN \r\n A ultimate the axial load as per BS 8110, 1: 1997: - p_n = 0.4 times f_cu times A_g + 0.75 times AS_c times A_y = 0.4 times 28 times [(250000 - 11617.61) + (0.75 times 11617.61 times 300)] = 31946259.9 N p_n = 3.194 times 10^3 KN