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

Shear%20Design%20Example%20ACI%20v.1 - Wu d = 900 mm Shear...

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|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 = = Φ = b w = 405 mm d = min(d - , d + ) = 803 mm Design for 3-stirrup zones 454kN 385.5kN 385.5kN d d

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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 = = × × = Φ = × = Φ = × = Φ = × × = Φ = × × =
Cases 1 Case 2 Case 3 Case u V 1017 shear reinforcement is not needed . kN c u 2 v,min 2033 V 1017 provide minimum shear reinforcement 600 4015 , 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 6098 V 2033 Calculate shear reinforcement using the 45 truss mechanism 600 4015 , say 400 2 1 385 500 400 1 V 271,000 292 4 075 803 414 2 4 2 . . min , . , . . # & u v yt kN kN d s mm mm s mm V s A mm d f legged l = = ÷ - = - = ÷ ÷ Φ - - 2 2 3 2 129 2 71 400 292 4 # : .

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