Question

Consider a floor plan and its overall dimensions as given in Assignment 2 and repeated in Figure 1 for convenience. The plan shows a series of spans in the Y-direction and one simple span in the X direction. The beam on axes 1, 2 and 3 could assumed to be simply supported. Concrete strength fc-30 MPa, maximum aggregate size is 20 mm, beams are of exterior exposure. For all beams, 10M stirrups are used. Reinforcement steel yield strength fy 400 MPa and its modulus of elasticity E 200 GPa. Column size is 400x400 mm2. For the given floor plan, the surface dead load is qu 6.0 kPa (including self-weight) and the surface live load is qu-7.0 kPa. Use the load combination of 1.25 DL+1.5 LL to find the ultimate factored load Ms Based on CSA A23.3 A) Design of tensile steel only for T-beam (10 points): For beam on axis-2, design the beam as T-beam. Determine beam cross section dimensions (b,and h) and the required tensile steel to resist the given factored load. Provide crack check z for this beam design. M, should be as close as possible to the factored moment M, (maximum over strength allowed is 15%). Iteration calculations will be needed to achieve Mr closest to Mr detailed hand calculations of the first iteration is required. Further iterations can be done with hand calculations, EXCEL spreadsheet or using SAFI B) Analysis of shear reinforcement (5 points): Validate if 10M two-legged stirrups at every 325 mm for the the whole length of the beam would be sufficient to resist the factored shear load for beam on axis-2. C) Shear design improvement (5 points) Improve the shear reinforcement for axis-2 beam previously analysed in B). There are two possible scenarios ) If in B) V,wasn't sufficient, provide a new design; i) If in B) V, was sufficient, create a more efficient design by using less total number of stirrups for the entire length of the beam. You can try two zones design or improve spacing. Show this by sufficient calculations and provide sketch of design summary (side view along beam length)

Answer 1

THe a) THE QjE - way FLOOR SPANS w TRIBUTARY w1DTH -10, = 65 Psf UDL 30 = 650 Pif TRe GiRDeR 18.975 (D) (-) 21.5 25