Question

The floor system shown below consists of normalweight concrete (150 pcf). Beams C2 ! LT 1 7.0 inch ! h=20 in G2! 12 in Girders BI 1 7.0 inch Slab 8 ft span h 24 in ----- - -- - --- 15 in 15 ft 15 ft Use 4000 psi concrete and Gr. 60 reinforcing steel. Assume 0.75 inch clear cover. Include concrete self-weight and a superimposed dead load of sp=25 psf. The design live load based on building occupancy is qlL=75 psf. Design the slab and beam B1 using LRFD. 1. Select a slab thickness, hs, based on the deflection criteria for the one end continuous condition. Round up according to the increments described in Lecture 11. 2. Compute the ultimate moment, M, and shear, V. in the slab based on self-weight of the floor system and the loads. 3. Design the slab to resist flexure and verify that OM, > M.. 4. Verify that the slab can resist shear, V. >V. 5. Specify the final slab design (bar size and spacing). 6. Compute the ultimate shear, V, in beam B1 based on self-weight of the floor system and the loads. 7. Compute the concrete contribution to design shear strength, 6V

Could you please help me on question 7.

Thank you very much.

Answer 1

7 concrete contribution to design shear Strength given that fo = 4000psi ty - 60,000 psi. As per ACI 318 code the shear strength of the concrete is given as Ve= 28 fé bud. where Ve= shear strength of concrete. bwa width of web d = depth of beam For beams select so bus 12 in. h = coin an effective cover of 1.5 in. effective depth de 2015 = 18.5 in Ve- 2x 54000 x 12x18.5. - 28081.02 lb = 28.08 kips - dve=0.75 x 28.08. = 21.06 Lips. shear strength of concrete of a beam 21.06kips

for griders bus 15in. h- zuin de 24-din – adin. Ve= 25té bud - 2x J4000 x 15x22. - 41742.06 lb - ul.74 kips OVc = 0.75x 41.74. = 31.305 kip