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
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4fbcdd5e-9854-4215-bf8d-33974a943929%2Fb36068e2-fec2-4f3f-837f-4dba06b949e1%2F58gn3p5e_processed.png&w=3840&q=75)
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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