Question

20 A concrete column with hollow circular cross section is under external loads as shown in Figure 1. Horizontal distributed load is defined by w=1/8*x’ function and its value reaches to 8 kN at the top of the column. Also, 50 kN vertical load is applied from the top of the column and its application point is 400 mm away from the centroid of the column as shown in the figure. Concrete column has hollow circular cross section and outer radius is 400 mm, inner radius is 300 mm. Determine the maximum normal and shear stress for the column. Note: Unit weight for the column is y= 25 kN/m' and it should be considered in the calculation of the self-weight of the hollow circular column. 50 KN X Y=25 kN/m Zw=1/8*x’ kN/m 300 mm 8 m G 400 mm Section a-a a y<:- Results table Omax= Tmax Figure 1.

Answer 1

Page No - solution:- SOKN op Given figure - prohl spiti 400mm 222 in LW = 1/8x*² kN/m ZN 'R & Com portal /y.L to a 25 kn/m² P 300mm 400mm D=foomm d=600mm resultant of UUL R= area of bading diagram = // 2h = I x 8x8 = 64 kN 4 4 @ 22:38=6m from a-a Area compressive foorce @ (a-a] :50 kn + 1 - 60² XLXX = 50+ x 60.60 20:62) *8*25 =93.98 km @ 9-a = R= 64 KN * shear force @ * Bending moment @a-a -a - 64x6 +soxo 4 Scanner

Page No - 2 M= 148 kilom [hanging bending moment] * mare normal stress occars @ 'p' since Arcial 8 BM creates compressive stress smarne = 1 + 1 g Is moment of inertia = I loud") G4 (sooh 6004) 64 = 1.37481010 mma * max shear stress across @ Neutral axis shear stress @ Nort Tmore = Fay I b R.AT ID AJ =0,9,- 92Yz 8 x 800? x 4x 400 T Ex600 4x300 3XIT 8 3XIT = 24.667 x 100 mm2 Tmare = 64 x 103724.667x100 2 6 1.37481010 x 200 Tmare 0.191 mpa et poids C CamScanner erebrna