(b) The modulus of rupture, f, of normal weight concrete with t, - 3000 psi is: 1, -7.5 - 7.5/103000 411 bin. The moment

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B The Modulus Of Rupture F Of Normal Weight Concrete With T 3000 Psi Is 1 7 5 7 5 103000 411 Bin The Moment 1 (42.28 KiB) Viewed 43 times

(b) The modulus of rupture, f, of normal weight concrete with t, - 3000 psi is: 1, -7.5 - 7.5/103000 411 bin. The moment that causes a stress equal to the modulus of rupture is Mo Hy (411 bin)(60,165 in) - 1167.344 in 1 - 97281-* (32 in.-10.61 in.) The uniformly distributed toad on a simple span that causes this much moment is: 8/97 28-) = 1.351 Wft = 1351 lb 24 fty BM hence: (c) If the beam is inverted, then the c term used to calculate M, 5 10.81 in. instead of 21.19 in.. M, 1411 bin. 160,185 in.) (10.87 in.) = 2,288.255 in-b=190.00 - The uniformly distributed load on a simple span that causes this much moment is -2.648 kft 2548 bytt 24 SM B/190.69 ft- A rectangular beam has the dimensions (see Fig. 3.2b) b = 250 mm, h - 650 mm, and d 600 mm and is reinforced with three No. 25 (No. 8) bars so that A, - 1530 mm. The concrete cylinder strength is 28 MPa, and the tensile strength in bending (modulus of rupture) is 3.27 MPa. The yield point of the steels, is 420 MPa, the stress-strain carves of the materials being those of Fig. 1.16. Determine the stresses caused by a bending moment M-61 kN-m. -250 mm 600 mm 650 mm 5355 5055 3 No.2 No. 8)