DESGIN OF SLAB SLAB 1 LL = 1.9 kPa Fc’ = 21 MPa Fy = 248 MPa Fs = 124 MPa 1. THICKNESS OF SLAB π‘= π 20 π‘= 1000 20 π = πππππ 2. Loads Carried Wt. of floor finishing = 0.75 kPa Wt. of concrete = 23.54(0.110) Wt. of concrete = 2.59 kPa LL = 1.9 kPa Total Load = 0.75 + 2.59 + 1.9 = 5.24 kPa < 7.9 kPa 3. Use SD 950 x 1.2 Wt. = 12.109 kg/mm2 A = 1560.9 mm2 Sx = 28.843x103 mm3 Ix = 757x103mm4 Load Capacity = 7.9 kPa 4. Check Load Capacity Total Load + Wt. of SD = 5.24 + 0.0792 Total Load + Wt. of SD = 5.32 kPa < 7.9 kPa (ok) 5. Location of SD Neutral axis from the bottom πΆ= πΌπ₯ ππ₯ πΆ= 757π₯103 28.843π₯103 πͺ = ππ. ππ ππ 6. Effective Depth 110 – 26.25 = 83.75 mm 7. Moment Capacity Considering (0.95) Strip π= πΉπππππ 2 2 π = √ππ2 + 2ππ − ππ ππ = π= π= π΄π π ππ πΈπ 4730√ππ ′ 200000 4730√21 = 9.23 ππ = π΄π π ππ ππ = 1560.9(9.23) 950(83.75) π·π = π. ππ π = √ππ2 + 2ππ − ππ π = √(0.18)2 + 2(0.18) − 0.18 π = π. ππ π=1− π 3 π=1− 0.44 3 π = π. ππ π= 0.45(21)(0.44)(0.85)(950)(83.75)2 2 π΄ = ππ. ππ ππ΅. π π= π€πΏ2 8 83.75(1)2 π= 8 M = 10.47 ππ. π < π = 47.21 ππ. π (ππ) 8. Moment Capacity of Steel deck M = Asfsjd M = 1560.9(124)(0.85)(83.75) M = 13.78kN.m > M = 10.49 kN.m (ok) 9. Steel area requirement π΄π = π ππ ππ 13.78π₯106 π΄π = 124(0.85)(83.75) π¨π = ππππ. ππ πππ 10. Spacing of reinforcing bars using 12 mm Ø π΄π = π (12)2 4 π¨π = πππ. π πππ π= π΄π (1000) π΄π π= 113.1(1000) 1561.08 S = 72.45 mm Say S = 30 mm 11. Spacing of temperature bars using 10 mm Ø As = 0.002bt As = 0.002(950)(110) As = 209 mm² π= π΄π (1000) π΄π π= 78.54(1000) 209 S = 375.79 mm Say S = 150 mm DESGIN OF SLAB SLAB 2 LL = 1.9 kPa Fc’ = 21 MPa Fy = 248 MPa Fs = 124 MPa 1. THICKNESS OF SLAB π‘= π 20 π‘= 2000 = 100 ππ 20 Say π = πππ ππ 2. Loads Carried Wt. of floor finishing = 0.75 kPa Wt. of concrete = 23.54(0.110) Wt. of concrete = 2.59 kPa LL = 3.8 kPa Total Load = 0.75 + 2.59 + 3.8 = 7.14 kPa < 7.9 kPa 3. Use SD 950 x 1.2 Wt. = 12.109 kg/mm2 A = 1560.9 mm2 Sx = 28.843x103 mm3 Ix = 757x103mm4 Load Capacity = 7.9 kPa 4. Check Load Capacity Total Load + Wt. of SD = 7.14 + 0.0792 Total Load + Wt. of SD = 7.22 kPa < 7.9 kPa (ok) 5. Location of SD Neutral axis from the bottom πΆ= πΌπ₯ ππ₯ πΆ= 57π₯103 28.843π₯103 πͺ = ππ. ππ ππ 6. Effective Depth 110 – 26.25 = 83.75 mm 7. Moment Capacity Considering (0.95) Strip π= πΉπππππ 2 2 π = √ππ2 + 2ππ − ππ ππ = π= π= π΄π π ππ πΈπ 4730√ππ ′ 200000 4730√21 = 9.23 ππ = π΄π π ππ ππ = 1560.9(9.23) 950(83.75) π·π = π. πππ π = √ππ2 + 2ππ − ππ π = √(0.181)2 + 2(0.181) − 0.181 π = π. πππ π=1− π 3 π=1− 0.447 3 π = π. πππ π= 0.45(21)(0.447)(0.851)(950)(83.75)2 2 π΄ = ππ. ππ ππ΅. π π= π€πΏ2 8 7.22(2)2 π= 8 M = π. ππ. π < π΄ = ππ. ππ ππ΅. π (ππ) 8. Moment Capacity of Steel deck M = Asfsjd M = 1560.9(124)(0.851)(83.75) M = 13.79 kN.m > M = 3.61 kN.m (ok) 9. Steel area requirement π΄π = π ππ ππ 3.61π₯106 π΄π = 124(0.851)(83.75) π¨π = πππ. ππ πππ 10. Spacing of reinforcing bars using 12 mm Ø π΄π = π (12)2 4 π¨π = πππ. π πππ π= π΄π (1000) π΄π π= 113.1(1000) 408.48 S = 276.88 mm Say S = 125 mm 11. Spacing of temperature bars using 10 mm Ø As = 0.002bt As = 0.002(950)(110) As = 209 mm² π= π΄π (1000) π΄π π= 78.54(1000) 209 S = 375.79 mm Say S = 175 mm DESGIN OF SLAB SLAB 3-1 Bed room LL = 1.9 kPa Fc’ = 21 MPa Fy = 248 MPa Fs = 124 MPa 1. THICKNESS OF SLAB π‘= π 20 π‘= 4000 20 π = πππ ππ 2. Loads Carried Wt. of floor finishing = 0.75 kPa Wt. of concrete = 23.54(0.200) Wt. of concrete = 4.71 kPa LL = 1.9 kPa Total Load = 0.75 + 4.71 + 1.9 = 7.36 kPa < 7.9 kPa 3. Use SD 950 x 1.2 Wt. = 12.109 kg/mm2 A = 1560.9 mm2 Sx = 28.843x103 mm3 Ix = 757x103mm4 Load Capacity = 7.9 kPa 4 Check Load Capacity Total Load + Wt. of SD = 7.358 + 0.0792 Total Load + Wt. of SD = 7.44 kPa < 7.9 kPa (ok) 5 Location of SD Neutral axis from the bottom πΆ= πΌπ₯ ππ₯ πΆ= 757π₯103 28.843π₯103 πͺ = ππ. ππ ππ 6 Effective Depth 200 – 26.25 = 173.75 mm 7 Moment Capacity Considering (0.95) Strip π= πΉπππππ 2 2 π = √ππ2 + 2ππ − ππ ππ = π= π= π΄π π ππ πΈπ 4730√ππ ′ 200000 4730√21 = 9.23 ππ = π΄π π ππ ππ = 1560.9(9.23) 950(173.75) π·π = π. ππ π = √ππ2 + 2ππ − ππ π = √(0.08)2 + 2(0.08) − 0.08 π = π. ππ π=1− π 3 π=1− 0.33 3 π = π. ππ π= 0.45(21)(0.33)(0.89)(950)(173.75)2 2 π΄ = ππ. ππ ππ΅. π π= π€πΏ2 8 7.44(4)2 π= 8 M = 14.88 ππ. π < π = 47.21 ππ. π (ππ) 8 Moment Capacity of Steel deck M = Asfsjd M = 1560.9(124)(0.89)(173.75) M = 30 kN.m > M = 10.49 kN.m (ok) 9 Steel area requirement π΄π = π ππ ππ 30π₯106 π΄π = 124(0.89)(173.75) π¨π = ππππ. ππ πππ 10 Spacing of reinforcing bars using 12 mm Ø π΄π = π (12)2 4 π¨π = πππ. π πππ π= π΄π (1000) π΄π π= 113.1(1000) 1564.53 S = 72.29 mm Say S = 40 mm 11 Spacing of temperature bars using 10 mm Ø As = 0.002bt As = 0.002(950)(200) As = 380 mm² π= π΄π (1000) π΄π π= 78.54(1000) 380 S = 206 mm Say S = 100 mm DESGIN OF SLAB SLAB 3-2 Bed room LL = 1.9 kPa Fc’ = 21 MPa Fy = 248 MPa Fs = 124 MPa 1. THICKNESS OF SLAB π‘= π 20 π‘= 4000 20 π = πππ ππ 2. Loads Carried Wt. of floor finishing = 0.75 kPa Wt. of concrete = 23.54(0.200) Wt. of concrete = 4.71 kPa LL = 1.9 kPa Total Load = 0.75 + 4.71 + 1.9 = 7.36 kPa < 7.9 kPa 3. Use SD 950 x 1.2 Wt. = 12.109 kg/mm2 A = 1560.9 mm2 Sx = 28.843x103 mm3 Ix = 757x103mm4 Load Capacity = 7.9 kPa 4. Check Load Capacity Total Load + Wt. of SD = 7.358 + 0.0792 Total Load + Wt. of SD = 7.44 kPa < 7.9 kPa (ok) 5. Location of SD Neutral axis from the bottom πΌπ₯ ππ₯ πΆ= 757π₯103 πΆ= 28.843π₯103 πͺ = ππ. ππ ππ 6. Effective Depth 200 – 26.25 = 173.75 mm 7. Moment Capacity Considering (0.95) Strip π= πΉπππππ 2 2 π = √ππ2 + 2ππ − ππ ππ = π= π= π΄π π ππ πΈπ 4730√ππ ′ 200000 4730√21 = 9.23 ππ = π΄π π ππ ππ = 1560.9(9.23) 950(173.75) π·π = π. ππ π = √ππ2 + 2ππ − ππ π = √(0.08)2 + 2(0.08) − 0.08 π = π. ππ π=1− π 3 π=1− 0.33 3 π = π. ππ π= 0.45(21)(0.33)(0.89)(950)(173.75)2 2 π΄ = ππ. ππ ππ΅. π π€πΏ2 π= 8 π= 7.44(4)2 8 M = 14.88 ππ. π < π = 47.21 ππ. π (ππ) 8. Moment Capacity of Steel deck M = Asfsjd M = 1560.9(124)(0.89)(173.75) M = 30 kN.m > M = 10.49 kN.m (ok) 9. Steel area requirement π΄π = π ππ ππ π΄π = 30π₯106 124(0.89)(173.75) π¨π = ππππ. ππ πππ 10. Spacing of reinforcing bars using 12 mm Ø π΄π = π (12)2 4 π¨π = πππ. π πππ π= π΄π (1000) π΄π π= 113.1(1000) 1564.53 S = 72.29 mm Say S = 40 mm 11. Spacing of temperature bars using 10 mm Ø As = 0.002bt As = 0.002(950)(200) As = 380 mm² π= π΄π (1000) π΄π π= 78.54(1000) 380 S = 206 mm Say S = 100 mm DESGIN OF SLAB SLAB 4 Bed room LL = 1.9 kPa Fc’ = 21 MPa Fy = 248 MPa Fs = 124 MPa 1. THICKNESS OF SLAB π‘= π 20 π‘= 4000 20 π = πππ ππ 2. Loads Carried Wt. of floor finishing = 0.75 kPa Wt. of concrete = 23.54(0.200) Wt. of concrete = 4.71 kPa LL = 1.9 kPa Total Load = 0.75 + 4.71 + 1.9 = 7.36 kPa < 7.9 kPa 3. Use SD 950 x 1.2 Wt. = 12.109 kg/mm2 A = 1560.9 mm2 Sx = 28.843x103 mm3 Ix = 757x103mm4 Load Capacity = 7.9 kPa 4. Check Load Capacity Total Load + Wt. of SD = 7.36 + 0.0792 Total Load + Wt. of SD = 7.44 kPa < 7.9 kPa (ok) 5. Location of SD Neutral axis from the bottom πΆ= πΌπ₯ ππ₯ πΆ= 757π₯103 28.843π₯103 πͺ = ππ. ππ ππ 6. Effective Depth 200 – 26.25 = 173.75 mm 7. Moment Capacity Considering (0.95) Strip π= πΉπππππ 2 2 π = √ππ2 + 2ππ − ππ ππ = π= π= π΄π π ππ πΈπ 4730√ππ ′ 200000 4730√21 = 9.23 ππ = π΄π π ππ ππ = 1560.9(9.23) 950(173.75) π·π = π. ππ π = √ππ2 + 2ππ − ππ π = √(0.08)2 + 2(0.08) − 0.08 π = π. ππ π=1− π 3 π=1− 0.33 3 π = π. ππ π= 0.45(21)(0.33)(0.89)(950)(173.75)2 2 π΄ = ππ. ππ ππ΅. π π= π€πΏ2 8 7.44(4)2 π= 8 M = 14.88 ππ. π < π = 47.21 ππ. π (ππ) 8. Moment Capacity of Steel deck M = Asfsjd M = 1560.9(124)(0.89)(173.75) M = 30 kN.m > M = 10.49 kN.m (ok) 9. Steel area requirement π΄π = π ππ ππ 30π₯106 π΄π = 124(0.89)(173.75) π¨π = ππππ. ππ πππ 10. Spacing of reinforcing bars using 12 mm Ø π΄π = π (12)2 4 π¨π = πππ. π πππ π= π΄π (1000) π΄π π= 113.1(1000) 1564.53 S = 72.29 mm Say S = 40 mm 11. Spacing of temperature bars using 10 mm Ø As = 0.002bt As = 0.002(950)(200) As = 380 mm² π= π΄π (1000) π΄π π= 78.54(1000) 380 S = 206 mm Say S = 100 mm