Shear%20Design%20Example%20ACI%20v.1

Shear%20Design%20Example%20ACI%20v.1 - |V u | 454kN 5.32m...

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Unformatted text preview: |V u | 454kN 5.32m 9.14m W u Shear force envelope d- = 900 mm d + = 803 mm 25 414 075 ' . c yt f MPa f MPa = = = b w = 405 mm d = min(d- , d + ) = 803 mm Design for 3-stirrup zones 454kN 385.5kN 385.5kN d d Preliminary Calculations c c c c c 1 1 V 25 405 803 2710 6 6 V 271 075 1017 2 2 V 075 271 2033 3V 075 3 271 6098 5V 075 5 271 10163 ' . . . . . . . . . c w f b d kN kN kN kN kN = = = = = = = = = = = Cases 1 Case 2 Case 3 Case u V 1017 shear reinforcement is not needed . kN c u 2 v,min 2033 V 101 7 provide minimum shear reinforcement 600 401 5 , say 400 2 1 25 405 400 1 405 400 A 130 16 3 414 16 414 3 2 #3 stirrup: A ' . . min , . max , max , c w w yt yt kN kN d s mm mm s mm f b s b s mm f f legged = = = = = - 2 v v,min 2 71 142 A ok mm = = o c u 2 c 609 8 V 203 3 Calculate shear reinforcement using the 45 truss mechanism 600 401 5 , say 400...
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This note was uploaded on 04/09/2012 for the course SE 151A taught by Professor Restrepo during the Spring '12 term at UCSD.

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Shear%20Design%20Example%20ACI%20v.1 - |V u | 454kN 5.32m...

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