Question

Ql: design short tied column to carry dead load 1600 kN and live load 1400 kN. 28 N/mm and Ty -420 N/mm Q2: design circular column spirally reinforced subjected to the following loads P 1000 kN Use fc-28 N/mnn, and fy =420 N/m rm2 Q 3: The 300 x 500 mm column shown in Fig. Q3 is reinforced with 8 No. 29 bars arranged around the column perimeter, providing an area Ast 5160 mm2. A factored load Pu of 1 150 kN is to be applied with eccentricities ey-75 mm and ex 150 mm, as shown. Material strength are fe 28 MPa and fy -420 MPa. Check the adequacy of the design using the reciprocal load method. 75 mm 50 mm 75 mm 60 mm 75 mm 180 mm 300 150 mm 60 mm 500 mm Fig. Q3

Answer 1

Design of short hed column: - Dead load = (DL) = 1600 KN Live load = (LL) = 1400 KN f^prime c = 28 N/mm^2 f_y = 420 N/mm^2 Total load DL + LL = 1600 + 1400 = 3000 KN Factored load PU = 1.5 times 3000 pu = 4500 KN as per code load carrying capacity of short fied column PU = 0.4 fck Ac + 0.067 fy ASC assume steel area ASc = 0.8 % of Ag PU = 0.4 fck (Ag -Asc) + 0.67 fy (Asc) 4500 times 10^3 N = 0.4 fck (Ag - 0.8 % Ag) + 0.67 fy (0.8 % of Ag) = 0.4 times 28 (0.992 Ag) + 0.67 times 420 (0.8 % of Ag) 13.362 Ag = 4500 times 10^3 Ag = 336775.93 mm^2 = > side = squareroot 336775.9 = 580 mm provide a square column size of 600 mn times 600 mm actual area A = 600 times 600 = 36000