Tiêu đề PSAD 35 - RCD - T-Beams.pdf ✅

PSAD 35: T-Beams Video Transcript Problem 1. Refer to Fig. RCD-01. A concrete beam built monolithically into a slab of thickness 100 mm, is subjected to positive bending. The total depth of the beam is 600 mm, while the effective width is also 600 mm. It is reinforced with 5-20 mm diameter deformed bars arranged in two layers as shown. fc' = 21 MPa, fy = 420 MPa. 1. What is the effective depth of the beam, in mm? 2. At what depth, in mm, from the top of the flange is the neutral axis located? 3. What is the design moment capacity of the beam, in kNm? Solution: Solve for the centroid of the steel group. y = 2(32) 5 = 12.8 mm Therefore, the effective depth is d = 100 + 500 − 65 − 12.8 d = 522. 20 mm

Problem 2. The framing plan of a floor is consisting of parallel 400 mm-wide beams with simple span of 6 meters, spaced 3750 mm on centers, and built monolithically with a slab with thickness of 150 mm. The slab is to carry, aside from its own weight, a superimposed dead load of 2.4 kPa and live load of 6 kPa. The total depth of the beam is 800 mm from the top of the slab, while the effective depth is 710 mm. Assume that γconc =24 kN/m3, fc’ = 28 MPa, fy=420 MPa. Use NSCP 2015 provisions. 1. What is the value of the design moment, in kNm? 2. What is the effective width of the beam, to the nearest 10 mm? 3. What is the depth of the compression stress block, in mm? 4. What is the required tension steel area, in mm2? Solution: Solve for the total dead load, wDL = SDL + wSlab + wBeam wDL = (2.4)(3.75) + (24)[0.15(3.75) + 0.4(0.8 − 0.15)] wDL = 28.74 kN m Solve for the uniform live load, wLL = (6)(3.75) = 22.5 kN m Solve for the factored uniform load, wu = 1.2 wDL + 1.6wLL wu = 1.2(28.74) + 1.6(22.5) wu = 70.488 kN m Solve for the design moment Mu = wuL 2 8 Mu = (70.488)(6) 2 8 Mu = 317.196 kNm For the effective flange width, bf = min (bw + s1 + s2 2 , bw + 16tf, bw + ln 4 ) bf = min (400 + (3750 − 400), 400 + 16(150), 400 + 6000 4 ) bf = min(3750, 2800, 1900) bf = 1900 mm