Combined Example

Example• Given Column 1 size 30x 30 Reinf. 422 Column 2 size 40x 40cm Reinf. 424 Ultimate soil bearing pressure , qult = 150kPa • • fyk = 300MPa fyd = 300/1.15 = 260.87 Mpa C25 fck= 20MPafctk = 1.5 MPa, Required:- Design of rectangular combined footing 1200k N 350m Property line 800k N b a . • SOLUTION • Proportioning of footing 800kN R X’ 15cm 350c m 1200k N a • R = 800 +1200 = 2000kN • RX’ = 1200 *350  X’ = (1200 *350)/2000 = 210cm • a = 2 ( X’ + 15) = 450 cm . 50*3.96m Take b =3. ok Shear force and bending moment diagrams .50 *b)  b =2.00) = 148..00m Actual contact pressure  = R/ab = 2000/ (4.50*b) 150 = 2000/(4.15kPa < qult ……….• • • • • qult = R/A = 2000/ (4. 00kNm 599.45kN/ m 66.65m 148.59kNm .15*3=444.67k N 377.76kN 733.30m 3.33k N 5.15m 800kN 0.24k d N 1.0.40m 0.99kNm 160.65m 1200kN 822. V = 444.085 k2 = 1.0017) =1.0017 Then.45d Take d= 0.6 – d =1.0*0.5 /300 = 0.488MN =488kN>V … OK! .085*1*3.45 (3.Thickness of the footing • Wide beam shear • The magnitude of the wide beam shear is read off from the shear force diagram at a distance of d from the face of the • • • • column.6 = 1 Vud = 0.25fctd k1k2 bwd (MN) k1 = ( 1+50) = (1 +50*0.6 -0. V =466.45 –d) -800 =733.60 =0.5/fyk = 0.25 *1*1.60m and  = min = 0.35-444.68kN The wide beam shear resistance according to EBCS-2 is given by: Vud = 0. 65=1.1 =4.30 =1.1m .95+2.95+1.95 450cm • Perimeters : Pr1 =1.3=2.30 =1.2 1.65=1.2+1.2 1.40 +0.2=6.5d +0.5d +0.2+2.40 +0.95 3d+0.2 3d+0.5m Pr2 =1.4=2.1 300cm 1.5d +0.5d +0.• Punching shear 1. 15* (1.38kN > V1…… ok! .15*(1.60m and  = min = 0.2*2.25fctd k1k2ud (MN) d= 0.5m .6 = 1 u1 = Pr1 =4.43kN • Punching shear resistance Vup = 0.66kN • Net shear force developed under column 2 V2 = 1200-148. u2 = Pr1 = 6.2)=564.1m • Punching shear resistance under column 1 • Vup = 0.5/fyk = 0.5 /300 = 0.1) =426.085 k2 = 1.6 -0.085*1*4.6 – d =1.6=732.5*0.0017 k1 = ( 1+50) = (1 +50*0.• Net shear force developed under column 1 V1= 800 -148.0017) =1.95*2.25 *1000* 1. 78kN > V2………… ok! M  0..25 *1000* 1.6  2  3915...99kN-m .65kN  m  M max .085*1*6.32 11.Punching shear resistance under column 2: • Vup = 0.6=992.00   0.1*0.32capacity  f cd  bd • Moment of concrete 2  0.Ok! • Calculation of reinforcement: – Long direction The reinforcement shall be calculated for the maximum moment Mmax = 599.33  103  3. use 13 bars Spacing = (300-2*5)/ (n-1) =24. n= 39.99  1  1   2 260.0   0.87  11.0022  300  60  39.f   cd f yd  2M   1 1 2  f bd cd   11.33  103  3.33  2  599.14cm2 No.6 . of bars .6     0.14 =12.6/3. as = 3.2cm Use 13  20 c/c 240mm .6cm 2 Use  = 20.0022   min  As  bd  0. 4+0.55 m l2 = (3-0.35m .0m Effective width at exterior and interior columns being a+d/2 and a+d. .35 m 0.30m a’+d= 0 . respectively.– Short direction l1 = (3-0.60m 2. a’+d/2= 0.3+0.3=0.3)/2 = 1.4)/2 = 1.6=1. • Contact pressure under columns 1 and 2 800 1   444.44kN / m 2 3.00 * 1.44 *1.35/2 = 243kN-m • M2 = 400 * 1.30 *1.00 X 0.30/2 = 338kN-m .60*1.00 X 1.35*0.60 1200 2   400kN / m 2 3.00 Bending moment: • M1 =444. 87  11.58 2  731.0049   min .33  103  0.58 As  bd  0.ok! .Ok! f   cd f yd  2M   1 1 2  f bd cd   11..32 11.• Calculation of Reinforcements – Under column 1 Moment capacity of concrete M  0.0049  60  58  17.33  103  0...05cm 2     0.79kN  m  M 1.33  2  243  1  1   2 260.32  f cd  bd 2  0....60   0.60   0. use 6 bars Spacing = (60-5)/ (n-1) =9.33  10 1.05/3. of bars ..Ok! 2 . n= 17.65kN  m  M 2 .Use  = 20.00   0.32 11.14cm No.58 3  1219.4 ..14 =5. as = 3.2cm Use 6  20 c/c 90mm – Under column 2 Moment capacity of concrete M  0.32  f cd  bd 2  0.. as = 3. n= 23.14 =7.20/3.33  2  338   1 1 2 260.14cm No.ok!  As  bd  0.14cm2 .. of bars .33  103 1   0..4 .f cd  f yd  2M   1 1 2  f bd cd   11.3cm Use 8  20 c/c 140mm • The reinforcement between the two strips will be nominal reinforcement to prevent shrinkage cracks • Short direction • Asmin = min bd = 0.0040   min .0017 *255 *58 =25.0040 100  58  23.58    0.20cm 2 Use  = 20..87  11. use 8 bars Spacing = (100)/ (n-1) =14. n= 25. n= 30. use 8 bars Spacing = (245)/ (7) =35cm < 400mm (smax for secondary bars)…….16/3.7 . of bars . as = 3.2cm < 400mm (smax for secondary bars)…….14 = 8 .14cm No.0017 *300 *60 =30.Use  = 20.14/3. of bars . ok Use 10  20 c/c 320mm . use 10 bars Spacing = (300-10)/ (9) =32. ok Use 8  20 c/c 350mm •Long direction Asmin = min bd = 0.14 = 9.60cm2 No. use 10 bars Spacing = (300-2*5)/ 9 =32.22cm Use 10  20 c/c 320mm .7 .33  2 160.87  11. long direction –Critical moment from bending moment diagram is M =160.60     0.•Cantilever portion –Bottom reinforcement .6/3.59kN-m f   cd f yd  2M   1 1 2  f bd cd   11.14 =9. of bars .59  1  1   2 260.0017  300  60  30.00057   min  As   min bd  0.33  103  3   0. n= 30.14cm No. as = 3.60cm 2 Use  = 20. 14 =1.87 ld    130.1 .0017 *35 *58 =3. of bars .44cm 4 f bd 4 1 .45/3.Short direction Provide minimum reinforcement • Asmin = min bd = 0.5cm Use 2  20 c/c 125mm • Development length • Short direction • Under column 1  f yd 2  260. n= 3.45cm2 • No. use 2 bars Spacing = (25)/ 2 =12. • Available development length.87 2: ld  4 f bd  4 1  130. f yd 2column • Under  260. la =130-5=125cm < ld Bend the bars upward with a minimum length of 10cm. .44cm • Available development length. la =135-5=130cm < ld Bend the bars upward with a minimum length of 10cm.