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Bulkhead Design <br />Page 3 <br />• Design for Shear <br />Cj, = 1.7 Yc = 1.7 x 62,064 lbs. = 105,509 lbs <br />= 105.5 Kips <br />~'/ = 1.7 W = 1.7 x 93,096 lbs. = 158,263 lbs. <br />= 158.3 Kips <br />= 15.8 Kips per ft. <br />'~'W = v.+ = 105,509 lbs = 287 psi <br />pbd 0.85 x 24 x 8 <br />287 psi ~ 110 psi = Vc = 2 f'c = 2 3,000 <br />.'. nominal shear carried by concrete, <br />web reinforcement is required. <br />Y ~ _ ( Y.,- V~) = 287 psi - 110 psi = 177 psi <br />177 psi < f'c = 3000 = 438 psi <br />• Cross section need not be increased <br />a = (L-d) (Y) _ (60 in - 18 in) x 177 psi <br />2 ~{u 287 psi <br />= 42.0 x .619 = 25.9" <br />d + a = 18" + 25.9" = 43.9" <br />Yu <br />_ (60 in - 18 in) 177 + 110/2 <br />287 <br />=42 in x.81-34 in. <br />287 ~ 4 f'c = 219 ~~ distance from supports to which <br />stirrups need to be placed = <br />a' d = 18" + 34" = 52" <br />.: two stirrups must intercept a potential diagonal crack <br />S = Avfy = 0.62) 60,000 = 8.757" ~ 8 3/4 in2 <br />Y'b 177 24 <br />• Av = 50 (24) 8.76 = 0.18 in2 <br />60,000 <br />0.18 in2` 0.40 in2 '• OK <br />