EX4.3 - CEE 415 Example 4.3 Design reinforcement for...

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CEE 415 1 2 20.0 k-in. 0.9 0.115 in. (60 ksi)(0.95)(3.4 in.) 2 u s y M A a fd ϕ ≥≅ =    2 2 0.115 in. (60 ksi) 0.169 in. (0.85) 0.85(4 ksi)(12 in.) 0.112 in. /ft 2 sy c u s y Af a fb M A a φ = = = ≥= Example 4.3: Design reinforcement for midspan and end sections of an interior span of the one-way slab shown in the floor system used in class discussion. Assume SDL = 25 psf and LL = 60 psf (no reduction). Use f c ' = 4000 psi and f y = 60 ksi. From Code Table 9.5(a), h (min) = /28 = 120 /28 = 4.3 , Use 4.5 in. Note: h (min) for an end span = ℓ/24 = 5″; Thus, we could either use 5 in. for all spans, or use 4.5 in. for all spans and do a deflection check in the end span. I will take the second option. w D = [(4.5/12)(ft) (150 lb/ft 3 ) + 25 psf] (1 ft.) = 81.3 lb/ft w L = 60 psf (1 ft) = 60 lb/ft w u = 1.2 w D + 1.6 w L = 194 lb/ft = 0.194 k/ft ( > 1.4 w D ) Start design at face of support (negative bending): From Code Sect. 8.3.3, moment coeff. = -(1/12) (slab with span 10 ft) n = - beam width = 10 ft – (8/12) ft = 9.33 ft (discuss) So, M u = (-1/12) (0.194 k/ft) (9.33 ft) 2 = 1.57 k-ft = 18.8 k-in. A pattern live loading in RISA gives, M u 20 k-in. (use this) Effective flexural depth, d = h – (1.1 in.) = 3.4 in.
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EX4.3 - CEE 415 Example 4.3 Design reinforcement for...

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