A reinforced concrete column cross-section is shown in Figure 1. Any applied design moment is to...
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A reinforced concrete column cross-section is shown in Figure 1. Any applied design moment is to be applied about the major principal axis shown. The asymmetrically arranged reinforcement is N-bars and for the concrete fc=32 MPa with y = 0.822 and Ec=28500 MPa. The nominal axial compression loads on the column are permanent action of NG=2200kN, imposed action of NQ = 1500kN, and M* = 650kNm. a) Calculate from first principles (ERSB) the value of the design pure axial compression capacity oNuo (in kN). b) Calculate the depth of the plastic centroid, dpc, from the indicated compression face and explain the significance of this point. c) Calculate the design bending capacity under pure bending oMuo. To do this calculate the effective depth, d, and the neutral axis depth parameter, ku, to check ductility. Hint: assume initially that all steel has yielded. d) Calculate from first principles (ERSB), the design capacities at the balanced condition oMub (in kKNm) and oNub (in kN). e) Approximating this column's design interaction diagram as a straight line between pure compression oNuo and the balanced' condition Mub and oNub, and another straight line between pure moment oMuo and the 'balanced' condition, is this proposed column cross-section adequate to carry the design actions N* and M*? Verify this by plotting these various values on the following page. 500 mm с N10 fitments 40. 2N20 || 800 mm 675 750 4N28 Figure 1 5N32 T 50 10000 9000 8000 7000 6000 5000 4000 3000 2000 1000 0 ØNu (kN) DESIGN INTERACTION DIAGRAM Student Name: 0 100 200 300 400 500, 600 700 Mu (kNm) Student No: 800 900 1000 1100 1200 A reinforced concrete column cross-section is shown in Figure 1. Any applied design moment is to be applied about the major principal axis shown. The asymmetrically arranged reinforcement is N-bars and for the concrete fc=32 MPa with y = 0.822 and Ec=28500 MPa. The nominal axial compression loads on the column are permanent action of NG=2200kN, imposed action of NQ = 1500kN, and M* = 650kNm. a) Calculate from first principles (ERSB) the value of the design pure axial compression capacity oNuo (in kN). b) Calculate the depth of the plastic centroid, dpc, from the indicated compression face and explain the significance of this point. c) Calculate the design bending capacity under pure bending oMuo. To do this calculate the effective depth, d, and the neutral axis depth parameter, ku, to check ductility. Hint: assume initially that all steel has yielded. d) Calculate from first principles (ERSB), the design capacities at the balanced condition oMub (in kKNm) and oNub (in kN). e) Approximating this column's design interaction diagram as a straight line between pure compression oNuo and the balanced' condition Mub and oNub, and another straight line between pure moment oMuo and the 'balanced' condition, is this proposed column cross-section adequate to carry the design actions N* and M*? Verify this by plotting these various values on the following page. 500 mm с N10 fitments 40. 2N20 || 800 mm 675 750 4N28 Figure 1 5N32 T 50 10000 9000 8000 7000 6000 5000 4000 3000 2000 1000 0 ØNu (kN) DESIGN INTERACTION DIAGRAM Student Name: 0 100 200 300 400 500, 600 700 Mu (kNm) Student No: 800 900 1000 1100 1200
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a To calculate the design pure axial compression capacity qNuo in kN from first principles using the Equivalent Rectangular Stress Block ERSB method we need to consider the concrete and steel contribu... View the full answer
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Fundamentals of Corporate Finance
ISBN: 978-0071051606
8th Canadian Edition
Authors: Stephen A. Ross, Randolph W. Westerfield
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