Answer Happy

QESTION IS
(3) Design and detailing of continuous Beam: Load Calculation: a) Lx = 4.50 m Ly= 8 10 m Area of slab (4.50*1) Nos = 4.50 m2 D=175mm d=150mm Self-weight of slab = 0.175 x 25 = 4.375 kN/m Floor finish = 1kN/m2 Self-weight of beam The depth of the beam Span/depth ratio=112 = 9100/12 = 758 mm - Width of the beam 250 mm (assume) Self-weight of beam= 0.25 x 0.775 x 25 (density of concrete) =4.8 KN/m Assume height of wal=3.5 m and thickness of wall =0.15 m Wall load = 3.5x0.15 x 18 (density of concrete) = 9.45 KN/m Dead load (self-weight of slab and floor finish) = 4.375 + 1 = 5.375kN/m Dead load (self-weight of slab and floor finish) in terms of N = 5.375 x 4.50 =24.2 KN/m Total dead load per m=(4.8+9.45+24.2)=38.45 KN/m Imposed Load = 3 kN/m
Total imposed load - 3 x 4.50 = 13.5 KN/m Design Load= (1.35gk + 1.50k) = (1.35*38.45+1.5*13,5) =72. 2KN/m Note: Calculation of +ve and -ve Bending moment for Beam Slab depth Depth of web End beam End beam L-Beam Intermediate T beams Dimensioning of T beam: Effective width of flange (bf): k=b1 =b2=(4500 - 25072 = 2125 mm. Assume bw = 250 mm = = Det = E(berri + bw) sb beff 1 = {0.20b1 + 0.10L) <0.20 x Lo (L. = 0.85 L = 0.85Ly) = [(0.2x2125+0.1x0.85x0100] > [0.2x0.85x9100] =1198.5 mm 3 1547 mm bf=1547 = 1600 mm
Effective depth of T beam (d) D = 775 mm d' = 50 mm, d = 725 mm 3.Leads moments and shear : Bf 160omm ht. 176 mm D, 775mm ds 725mm bwe a50mm T Cross section of T beam Maximum sagging moment = M.+ = + 0.088 x FL 2 M16+) = +0.088 x 72.2 x 9.102 = + 514.2 km Maximum Hogging moment = M11-) = -0.088 FL2=. - 514.2 kNm. = 4. Section verification: (comparison of K and K
K = ( [Equation 11... Page 8.... Data Book] fckbf d K= 545.2 x108 30 77000 7252 = 0.02 <0.167 K<k. Under reinforced section. Tension steel yields. No compression steel is required.
5. Steel calculations Steel for Sagging moment As[+) = M AS = 0.87 fyks = 1716.3 mm2 514.2x106 0.87 X 500 X U.Vb x 125 [Equation 11... Page 6-...Data Book] Lever arm = Z=d (0.50+ 3.60.25-K/1.134)) { Zmax..... [Equation 12...Page 8 – Data Book] mm. z =725[0.50 + (0.25 - (0.020/1.134); 50 ]2 = 711.9 mm Z max = 0.95d = (0.95 x 725) = 888.75 Z = 888.75 mm 514.2x106 = 1819.7 mm 2 0.87 x 500 x 0.95 x 125 M AS = 0.87 1 yks Aal+) = 1716.3 mm? Asi-) = 1716.3 mm? Provide 8 H 20 giving As = 1890 mm? > 1716.3 mm Refer Page No. 13 Minimum and maximum steel areas: As, min = (0.13/100)bwd = [(0.13/100) x 250 x 725)=235 mm As,max = [(4/100) x250 x 725] = 7250 mm? Check, As provided > As, min 1890 mm? > 235 mm”. Satisfied. (0) As provided < As maximum 1890 mm? <7250 mm Therefore, the steel areas provided are satisfactory.
Design for shear: (Table 8.10 - page 17, shear force coefficients) • Maximum shear force at support (V) = 0.40 F = 0.40 Wy = = 0.4 x72.2 x9.10 = 262.8 KN Maximum Shear force at Interior support (V2) = 0.8 FL = 0.8 x 72.2 x 9.10 = 394.2 KN • Shear Force at face of support (VEF) = WL-(W.bw/2)) • = 262.8- (72.2 x 0.25/2) = 253.78 k • Shear force at distance of effective d from the face of support Vxx = vef -(wx (bwx d)) = 253.78 - (72.2x0.25x0.725) = 240.7 KN Crushing strength of concrete at VRd max 22.8 VRd max 22.8 = 0.124 bwd (1 - fek /2501 fck = 0.124 x 250 x 725 (1 - 30/250). 30 = 593340 N = 593 KN Check VED <VRd max (22.8) 394.2 KN<593 KN (Therefore only nominal shear stirrups are required) Design of shear Links: AW 394.2 0.7x500x725x2.35x10- =0.53 0.87.Syk.d.con Table A.4 page 14 for (ASW/S) = 0.785 > 0.53Provide 2 legged H 10 vertical stirrups @200 mm clc
4. Analysis using software [Weightage 10%] Model a continuous T-beam using a standard commercially used software such as Etabs or Staad Pro or SAFE. Identify critical sections. Attach the data input and results output file showing only the necessary information. Prepare a table showing the moments at critical sections and the design results obtained from software analysis.