RC Columns subjected to shear forces – ACI 318
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Dear Artur / Rafal
Greetings
May I know the reason behind getting very narrow stirrups spacing when I verified it in both (Required / Provided Reinf.) for a simple RC Column subjected to very small shear forces according to ACI 318 ?
And which code formulas does RSA use to calculate (VC – Shear concrete strength & VS – Shear reinforcement)?
Manual Verification According to ACI318 -2002
Bar (6) in the attached model as example :
Fc = 24.10 Mpa ; bw =200 mm ; d = 600 – 48 = 552 mm ; Nu = 248.96 KN
Ag = 200*600 = 120000 mm2 reduction factor for shear strength = 0.75
11.3.1.1 Vc = 90.32 *0.75 = 67.74 KN > >> 5.31 KN
11.3.1.2 Vc = 103.70 *0.75 = 77.76 KN >>> 5.31 KN
While as RSA gave me very small stirrups spacing (75mm) as in the pictures
No need to use stirrups just using minimum shear reinforcement will be enough according to clause ACI 318 (7.10.5.2)
Thanks in advance
Refaat
