Problem #2 (20 points) tlw 13 4 Design a simply suppornied, uniformly loaded. rc

Question

Problem #2 (20 points) tlw 13 4 Design a simply suppornied, uniformly loaded. rcinforced Tee beum for tensile steo the following conditions and specifications: Uniform live load: 3.0 kips/ Span of beam equals 20 feet fe 4 ksi dead load: 2.0 kips/ fy-60 ksi slab thickness equals 6 center to center spacing of beans equals 5'4" beam stem width equals 16 effective beam depth fiom top of slab to center of reinfoncing steel equals 28 inches Utilize the number of bars that will satisfy the spatial and minimum area #10 reinforcing bars Note: show all code checks. 2) 29613 (26'-3") 6 4060000 Flange width br 60 inches Maximum actual compressive stress in the concrete 32.03 10 Area ofensile steel in beamy squareinches Barconfiguration·

Explanation / Answer

Live load = 3 kip/ft Dead Load = 2 kip/ft Total factored load = 1.2D+1.6L = 7.2 kip/ft Mu = w x L x L /8 = 360 kip-ft Given Beam Width bw = 16 in Slab Width hf = 6 in Effective depth d = 28 in Span of beam s = 20 ft Center to center distance of beam c/c = 64 in Yield strength of steel fy = 60 ksi Concrete strength f'c = 4 ksi Ultimate Moment Mu = 360 kip-ft Effective Flange Width be = min(span/4, 16hf + bw, c/c distance) be = 60 in Calculate c c = 0.85*f'c*be*hf c = 1224 kips Moment Capacity Mn = c*(d-0.5hf) Mn = 2550 kip-ft a Mu/ Therefore, (d-0.5a) can be approximated to 0.9d As,req = Mu/(*(d-0.5a)*fy) As,req = 3.174603 in2 Nominal area of #10 bar is 1.27 in2 Total #10 bars required = 3.17/1.27 = 2.49968754 Use 3 #10 bars

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