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Given: bxh = 400 mm x 600 mm Ast = 8 x 32 mm diameter bars Clear concrete cover to 12 mm diameter ties = 40 mm Concrete fc' = 28 MPa Steel fy = 415 MPa Reduction Factor for shear = 0.75 Due to reversal lateral force, the design axial load due to the combined effect of DL, LL and WL change are For WL to the +X direction: Mu = -420 kN-m; Vu = 370 kN; Nu = 580 kN (Compression) For WL to the -X direction: Mu = +420 kN-m; Vu = -370 kN; Nu = -450 kN (Tension)

Given: bxh = 400 mm x 600 mm Ast = 8 x 32 mm diameter bars Clear concrete cover to 12 mm diameter ties = 40 mm Concrete fc' = 28 MPa Steel fy = 415 MPa Reduction Factor for shear = 0.75 Due to reversal lateral force, the design axial load due to the combined effect of DL, LL and WL change are For WL to the +X direction: Mu = -420 kN-m; Vu = 370 kN; Nu = 580 kN (Compression) For WL to the -X direction: Mu = +420 kN-m; Vu = -370 kN; Nu = -450 kN (Tension)

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Given: bxh = 400 mm x 600 mm Ast=8 x 32 mm diameter bars Clear concrete cover to 12 mm diameter ties = 40 mm Concrete fc' = 28 MPa Steel fy = 415 MPa 400mm Mu = -420 kN-m; Vu = 370 kN; Nu = 580 KN (Compression) For WL to the -X direction: 600mm. Reduction Factor for shear = 0.75 Due to reversal lateral force, the design axial load due to the combined effect of DL, LL and WL change are follows: For WL to the +X direction: Mu = +420 kN-m; Vu = -370 kN; Nu = -450 kN (Tension) 1. Ans. 220 Determine the concrete shear strength (kN) for the positive x direction of WL by simplified calculation. 2. Ans. 0 Determine the concrete shear strength (kN) for the negative x direction of WL by simplified calculation. 3. Ans. 125 Calculate the required spacing (mm) of shear reinforcement. Apply code provisions on spacing limits of reinforcement where applicable.
Transcribed Image Text:Given: bxh = 400 mm x 600 mm Ast=8 x 32 mm diameter bars Clear concrete cover to 12 mm diameter ties = 40 mm Concrete fc' = 28 MPa Steel fy = 415 MPa 400mm Mu = -420 kN-m; Vu = 370 kN; Nu = 580 KN (Compression) For WL to the -X direction: 600mm. Reduction Factor for shear = 0.75 Due to reversal lateral force, the design axial load due to the combined effect of DL, LL and WL change are follows: For WL to the +X direction: Mu = +420 kN-m; Vu = -370 kN; Nu = -450 kN (Tension) 1. Ans. 220 Determine the concrete shear strength (kN) for the positive x direction of WL by simplified calculation. 2. Ans. 0 Determine the concrete shear strength (kN) for the negative x direction of WL by simplified calculation. 3. Ans. 125 Calculate the required spacing (mm) of shear reinforcement. Apply code provisions on spacing limits of reinforcement where applicable.
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