Marriage Hall.design

ANALYSIS ANDDESIGN OF A R.C.C. BUILDING FRAME ATRUCTURE Site location: DG Khan Building title: Marriage Hall Data used: ABCSOIL = 1ton per square foot Unit weights: Soil = 120pcf R.C.C. = 150pcf Brick ballast = 110pcf P.C.C+terrazzo = 144pcf Mud filling = 100pcf Brick tiles = 120pcf C/S mortar = 120pcf Wall loads: 1foot high 9inch wide = 90 lbs per foot Live loads: LL = 150psf (for slabs, beams etc.) LL = 200psf (for stairs) fc’ = 3ksi (for slabs and beams) Material strength: = 4ksi ( for basement and columns) fy = 60ksi (deformed steel bars) Page 1 of 85 Fig 1: Plan of building DESIGN OF SLABS Page 2 of 85 Page 3 of 85 Page 4 of 85 Page 5 of 85 . Page 6 of 85 . Page 7 of 85 . Page 8 of 85 . Page 9 of 85 . Page 10 of 85 . Page 11 of 85 . Page 12 of 85 . Page 13 of 85 . Page 14 of 85 . Page 15 of 85 . 27 k-ft Interior support Mu  = 136.5 = l 21 30  12 = 17.5  19. Width of web: bw = 9in d min for Mumax : d min hmin Thus.205  3  9 = 17.02 k-ft Exterior span Mu  = 10. h = 18in to keep the deflections in control.5in Page 16 of 85 .8 k-ft Interior span Mu  = 135.8 12  17.22 + 2.4 k-ft Minimum Depth of beam for deflection control: Exterior Span h= = Interior Span l 18.5 = 19.729in 18.22in 0.DESIGN OF BEAMS Roof Slab Beams Design Design of B1: Factored moments taken from ETABS: Exterior support Mu  = 35.5 h= 15 12 = 9.205 fc ' b = 136.72in h  22in d  22  2.143in 21 Say. = Mumax 0. 85 fc 'b = 1.35 k-ft b Mn  = 139.851   = 0.807  60 = 4. Sc 2 =9+ bf 13.35 k-ft Effective flange width for L-beam behavior: bf is lesser of the following: 15 12  15in for exterior span 12 30 12 =  30in for interior span 12 i) l 12 = ii) 6h f  bw = 6 (7)+9 = 51in iii) bw  Thus.5  = 12 4.807in 2 = a b Mn  AS 1 f y 0.0103  9 19.851   = 0.0103 7 fy = AS 1  max bd = 0.5  1.9(1.0135 8 fy =  max bd Page 17 of 85 .725in 0.5  12 = 90in 2 = 15in for exterior span = 30in for interior span Maximum capacity as singly reinforced section in spans:  max AS 1 3 fc ' = 0.725 ) 2 = 139.807)(60)(19.85  3  9 a = b AS 1 f y (d  ) 2 0.Maximum capacity as singly reinforced section at supports:  max 2 fc ' = 0. 369in 2 = a AS 1 f y 0.27 k-ft a = 3in As Mu = a 2 b f y (d  ) = Thus.4 k-ft a = 3in Page 18 of 85 . 2 . Mu  = 10.97 k-ft b Mn  = 197.85 fc 'b = 2.27 12 3 0.585in 2 200 200   0.#6.0135  9 19.85  3  30 a = b AS 1 f y (d  ) 2 b Mn  0. Design for positive moment in interior span: Assume. As 10.369  60 = 1.126in 2  Asmin = 0.369)(60)(19.97 k-ft Minimum steel ratio:  min = Asmin = 0.858in 0.858 ) 2 = 197.9(2.= 0.5  0.585in 2 Use.0033  9 19.5  = 12 1. Mu  = 135.9(60)(19.5  ) 2 = 0.0033 fy 60000 Design for positive moment in exterior span: Assume.5  2. #6. Mu 35.35 k-ft (OK) Flange is under tension: b = 9in Page 19 of 85 .5  ) 2 AS f y 0.85 fc 'b = = 1.As Mu = a 2 b f y (d  ) = a = 135.02 k-ft Flange is under tension: Thus. 2 .4 12 = 1.1227 bd 2 9(19.#8 + 1-#6 Design for negative moment at exterior support: Mu  = 35. Design for negative moment at interior support: Mu   = 136. b = 9in d = 19.8 k-ft < b Mn = 139.671in 2 1.31 0.671 60 = 1.585in 2 Use.5in R =  = 0.5  ) 2 Use.596in 2 1.85  3  30 (Assumption is correct) As = 135.31in 0.02 12 =  0.4 12 3 0.9(60)(19.5) 2  min 2 .0027 < As = 0.9(60)(19. 5  12 = 7.8 12 =  0.5 k-ft Minimum Depth of beam for deflection control: Exterior Span h= = Interior Span l 18.4797 bd 2 9(19. d = 19.35 k-ft Exterior span Mu  = 20. 3 ii) #3@9 " c/c in remaining middle portion.Thus. h = 9in to keep the deflections in control.5) 2 (OK) 2 .14in 21 Say.0102 >  min As = 0.#6 Design for Shear: Provide: i) #3@6 " c/c utpo l from both supports.1 k-ft Interior support Mu  = 20. Mu 136. Design of B2: Factored moments taken from ETABS: Exterior support Mu  = 27.5 12 = 8.5 h= 13.0102  9 19.5  1.#8 + 1.34 k-ft Interior span Mu  = 15.756in 18.1 k-ft First Interior support Mu  = 17. Width of web: bw = 9in d min for Mumax : Page 20 of 85 .5in R =  = 0.79in 2 Use.5 = l 21 13. 2in Thus.0103  9 13.268in 0.5in 12 i) l 12 = ii) 6h f  bw = 6 (7)+9 = 51in Page 21 of 85 .25)(60)(13.d min hmin = Mumax 0.5in Maximum capacity as singly reinforced section at supports:  max 2 fc ' = 0.0103 7 fy = AS 1  max bd = 0.9(1.5  1.205  3  9 = 7.85 fc 'b = 1.5  = 12 3.5  13.7in 0.205 fc ' b = 27.5 12  13.25  60 = 3. h  16in d  16  2.74 k-ft Effective flange width for L-beam behavior: bf is lesser of the following: 13.85  3  9 a = b AS 1 f y (d  ) 2 0.27 ) 2 = 66.7 + 2.35 12  7.851   = 0.25in 2 = a b Mn  AS 1 f y 0.74 k-ft b Mn  = 66.5 = 10. iii) bw  Sc 2 =9+ bf Thus.5  1.86in 0.5  12 = 90in 2 = 13.5  0.1 k-ft Minimum steel ratio:  min = Asmin = 0.0135 8 fy = AS 1  max bd = 0.85 fc 'b = 1.64  60 = 2.1 k-ft a = 3in As = Mu a 2 b f y (d  ) Page 22 of 85 .9(1.0033 fy 60000 Design for positive moment in exterior span: Assume.0033  9 13.5 a = b AS 1 f y (d  ) 2 b Mn  0. 13.0135  9 13.851   = 0.85  3 13.86 ) 2 = 89.401in 2 200 200   0.5  = 12 2.5in Maximum capacity as singly reinforced section in spans:  max 3 fc ' = 0.64in 2 = a AS 1 f y 0. Mu  = 20.64)(60)(13.1 k-ft b Mn  = 89. 0039 >  min As = 0.35 k-ft < b Mn = 66. b = 9in d = 13.= Thus.9(60)(13. Page 23 of 85 .5  .474in 2 Use. 2 .5) 2 (OK) 2 . Mu 27.5in R =  = 0.#6. Design for positive moment in interior span: Mu  = 15.0039  9 13. 3 ii) #3@10 " c/c in remaining middle portion. 20.#6 Design for negative moment at interior supports: Mu  Use.372in 2  Asmin = 0.74 k-ft (OK) Flange is under tension: Thus.401in 2 As Use.5 k-ft Use. 2 .34 k-ft < b Mn = 66.5  ) 2 = 0.20 bd 2 9(13.#6 Design for negative moment at exterior support: Mu   = 27.35 12 =  0.1 12 3 0.  = 20.#6 Design for Shear: Provide: i) #3@7 " c/c utpo l from both supports.74 k-ft (OK) 2 . Width of web: bw = 12in d min for Mumax : d min hmin Thus.Design of B3: Factored moments taken from ETABS: Exterior support Mu  = 65.5in Maximum capacity as singly reinforced section at supports: Page 24 of 85 .143in 21 Say.205 fc ' b = 221.44 12  18.5  19.729in 18. h = 18in to keep the deflections in control.5 h= 15 12 = 9.205  3 12 = 18. = Mumax 0.18 k-ft Interior support Mu  = 187.5 = l 21 30  12 = 17.47in h  22in d  22  2.975in 0.5 = 21.2 k-ft Interior span Mu  = 221.1 k-ft Exterior span Mu  = 9.97 + 2.44 k-ft Minimum Depth of beam for deflection control: Exterior Span h= = Interior Span l 18. 5 12 = 162in bf = 45in for exterior span = 90in for interior span Maximum capacity as singly reinforced section in interior spans:  max AS 1 3 fc ' = 0.85 fc 'b = 2. max 2 fc ' = 0.41in 2 = a AS 1 f y 0. = 13.725 ) 2 = 185.86 k-ft Effective flange width for T-beam behavior: bf is lesser of the following: 15 12  45in for exterior span 4 30 12 =  90in for interior span 4 i) l 4 = ii) 16h f  bw = 16 (7)+12 = 124in iii) S Thus.41 60 = 4.851   = 0.5  = 12 4.5  3.0103 12 19.16in 2 Page 25 of 85 .851   = 0.9(2.41)(60)(19.85  3 12 a = b AS 1 f y (d  ) 2 b Mn  0.5  2.725in 0.0135 8 fy =  max bd = 0.86 k-ft b Mn  = 185.0103 7 fy = AS 1  max bd = 0.0135 12 19. Mu  = 221.18 12 3 0.5  = 12 0.85  3  90 a = b AS 1 f y (d  ) 2 0.0033 12 19.16  60 = 0.44 k-ft a = 3in As = Mu a 2 b f y (d  ) Page 26 of 85 .5  0. 2 .5  ) 2 = 0.0033 fy 60000 Design for positive moment in exterior span: Assume.9(60)(19. As 9.85 fc 'b = 3.= a b Mn  AS 1 f y 0.826 ) 2 = 271. Design for positive moment in interior span: Assume.42 k-ft Minimum steel ratio:  min = Asmin = 0.16)(60)(19.42 k-ft b Mn  = 271.18 k-ft a = 3in As Mu = a 2 b f y (d  ) = Thus.826in 0.9(3.#6.772in 2 200 200   0.772in 2 Use.113in 2  Asmin = 0. Mu  = 9. 715 0.1 k-ft Flange is under tension: Thus.734  60 = 0.112 Mu =  0. d' = 2.5in d = 19.#8 Design for negative moment at exterior support: Mu  = 65.715in 0.9(60)(19.5in M2 = Mu   b Mn = 1.5  ) 2 Use.85  3  90 (Assumption is correct) As = 221.734in 2 2. Design for negative moment at interior support: Mu   = 187.5in R =  = 0.6k-ft  Page 27 of 85 . 65.171 2 bd 12(19.57in 2 0. b = 12in d = 19.0033 = As = 0.86 k-ft Design as doubly reinforced section of size 12in×22in.= a = 221.85 fc 'b = = 2.772in 2 Use.44 12 = 2.5  ) 2 AS f y 0.5) 2  min 2 .9(60)(19.2 k-ft > b Mn = 185. 4 .44 12 3 0.#6. 87 ksi < f y As2revised Thus.#8 + 2 .87 2 . 60 = 0.75 k-ft Exterior span Mu  = 43.725 = 47.Assuming compression steel is yielding: M2 1.#6 ( Bottom bars) Design for Shear: Provide: i) #3@5 " c/c utpo l from both supports.44 in 2 = As1 + Use.6 12 = = 0.5  2. As = As2 fy fs ' = 0.01 k-ft Interior span Mu  = 170. 3 ii) #3@8 " c/c in remaining middle portion.5) AS 2 = fs ' = As2 fy = fs ' 87(4.026 = 2.5) 4.41+ 0.9 f y ( d  d ') 0. Design of B4: Factored moments taken from ETABS: Exterior support Mu  = 15.9(60)(19.1 k-ft Interior support Mu  = 197.026 in 2 47.021in 2 0.#6 (Top bars) 2 .85  2.6 k-ft Minimum Depth of beam for deflection control: Exterior Span h= l 18.725  0.5 Interior Span h= l 21 Page 28 of 85 .021 As2revised = 2. 398in Thus.0112  17.85  3 12 a = b AS 1 f y (d  ) 2 0.898in 0.41)(60)(19. h = 18in to keep the deflections in control.5 = 30  12 = 17.85 fc 'b = 2.205 fc ' b = 197.41 60 = 4.86 k-ft b Mn  = 185. Width of web: bw = 12in d min for Mumax : d min hmin = Mumax 0.86 k-ft Page 29 of 85 .5in Maximum capacity as singly reinforced section at supports:  max AS 1 2 fc ' = 0.0103 7 fy =  max bd = 0.5  2. h  22in d  22  2.5  19.0103 12 19.851   = 0.725in 0.205  3 12 = 17.143in 21 Say.9(2.725 ) 2 = 185.5  = 12 4.= 15 12 = 9.729in 18.5 = 20.898 + 2.41in 2 a b Mn  = AS 1 f y 0. 0135 12 19.42 k-ft Minimum steel ratio:  min = Asmin = 0.85 fc 'b = 3.5  3.42 k-ft b Mn  = 271. bf = 45in for exterior span = 90in for interior span Maximum capacity as singly reinforced section in interior spans:  max AS 1 3 fc ' = 0.0033 12 19.851   = 0.9(3.5  = 12 0.826in 0.0135 8 fy =  max bd = 0.16)(60)(19.Effective flange width for T-beam behavior: bf is lesser of the following: 15 12  45in for exterior span 4 30 12 =  90in for interior span 4 i) l 4 = ii) 16h f  bw = 16 (7)+12 = 124in iii) S = 13.826 ) 2 = 271.5 12 = 162in Thus.772in 2 200 200   0.5  0.0033 fy 60000 Page 30 of 85 .85  3  90 a = b AS 1 f y (d  ) 2 0.16in 2 a b Mn  = AS 1 f y 0.16  60 = 0. 9(60)(19.1 k-ft a = 3in As Mu = a 2 b f y (d  ) = Thus.772in 2 Use.551in 0.85  3  90 (Assumption is correct) As = Use.5  ) 2 AS f y 0. As 43.6 12 = 1.#6.6 12 3 0. Mu  = 43.9(60)(19.106  60 = 0.5  ) 2 3 .Design for positive moment in exterior span: Assume.1 12 3 0.5  ) 2 = 0. Design for positive moment in interior span: Assume. 2 .85 fc 'b = = 2.75 k-ft Flange is under tension: Page 31 of 85 .551 0.6 k-ft a = 3in As = Mu a 2 b f y (d  ) = a = 170.532in 2  Asmin = 0. 170. Mu  = 170.106in 2 2.972in 2 0.#8 Design for negative moment at exterior support: Mu  = 15.9(60)(19. 9(60)(19. Mu 15.725 = 47.5in R =  < Thus.#8 (Top bars) Use.85  2.75 12 =  0.041 bd 2 12(19.146  60 = 0.41+ 0.5in M2 = Mu   b Mn = 11.5in d = 19.183 in 2 47.#6.772in 2 Use.5) AS 2 = fs ' = As2 fy fs ' = 87(4.725  0. d' = 2.87 As2revised = 2.5  2.b = 12in d = 19.#6 (Bottom bars) Design for Shear: Page 32 of 85 .15k-ft  Assuming compression steel is yielding: M2 11.5) 2  min As = 0.87 ksi < f y As2revised Thus.01 k-ft > b Mn = 185. Design for negative moment at interior support: Mu   = 197.9 f y ( d  d ') 0.146in 2 0.593 in 2 4 . As = As2 fy fs ' = As1 + = 0. 2 .183 = 2.5) 4. 2 .86 k-ft Design as doubly reinforced section of size 12in×22in.15 12 = = 0. 3 ii) #3@8 " c/c in remaining middle portion.56 12  20.1 k-ft Interior span Mu  = 281.56 k-ft Interior support Mu  = 306.92in h  22in d  22  2.5in Page 33 of 85 .5  19. Design of B5: Factored moments taken from ETABS: Exterior support Mu  = 214.5 = l 21 27 12 = 15.205  3 15 = 20. say = Mumax 0.42in 0. Width of web: bw = 15in d min for Mumax : d min hmin Thus.205 fc ' b = 320.42in 21 Say.36 k-ft Exterior span Mu  = 320.51in 18.Provide: i) #3@5 " c/c utpo l from both supports.5 h= 27 12 = 17. h = 18in to keep the deflections in control.9 k-ft Minimum Depth of beam for deflection control: Exterior Span h= = Interior Span l 18.5 = 22.42 + 2. 013in 2 a = AS 1 f y 0.948  60 = 1.726 ) 2 = 232.013  60 = 4.851   = 0.0103 15 19.1467in 0.0103 7 fy = AS 1  max bd = 0.9(3.0135 15 19. = 13.5  = 12 4.948in 2 a = AS 1 f y 0.5  3.5 12 = 162in bf = 81in Maximum capacity as singly reinforced section in interior spans:  max AS 1 3 fc ' = 0.013)(60)(19.Maximum capacity as singly reinforced section at supports:  max 2 fc ' = 0.35 k-ft Effective flange width for T-beam behavior: b f is lesser of the following: 27 12  81in 4 i) l 4 = ii) 16h f  bw = 16 (7)+12 = 124in iii) S Thus.726in 0.851   = 0.35 k-ft b Mn  = 232.85 fc 'b = 3.85  3 15 a = b AS 1 f y (d  ) 2 b Mn  0.85 fc 'b = 3.5  3.0135 8 fy =  max bd = 0.85  3  81 Page 34 of 85 . 147 ) 2 = 336.9(60)(19.9(3.9572in2  Asmin (OK) 3.56 12 3 0.5  ) 2 As = As = 3.149in 0.#8.9(60)(19. 320. Design for positive moment in interior span: Page 35 of 85 .5  ) 2 AS f y 0.248 k-ft b Mn  = 336.0033 fy 60000 Design for positive moment in exterior span: Assume.5  = 12 1.56 k-ft a = 3in As = = a = Mu a 2 b f y (d  ) 320.0033 15 19.a = b AS 1 f y (d  ) 2 b Mn  0.85  3  81 (Assumption is correct) Thus.965in 2 200 200   0.248 k-ft Minimum steel ratio:  min = Asmin = 0.56 12 1.5  0.957  60 = 1.764in 2 Use. 5. Mu  = 320.85 fc 'b = = 3.948)(60)(19.149 0. 5  ) 2 AS f y 0. Mu  = 281.01 0.5  2.36 12 =  0.451 bd 2 15(19.35 k-ft  Page 36 of 85 .5in R =  = 0.#8 Design for negative moment at exterior support: Mu  = 214.#8 Design for negative moment at interior support: Mu  = 306.85 fc 'b = = 3.9(60)(19.298in 2 1.1k-ft > b Mn = 232.5  ) 2 Use.0094 >  min As =  0. b = 15in d = 19.9(60)(19.85  3  81 (Assumption is correct) As = 281.9 12 3 0.48  60 = 1. 5 .9 12 = 3.48in 2 3.75in 2 Mu 214.Assume. Use.36 k-ft Flange is under tension: Thus.9 k-ft a = 3in As = = a = Mu a 2 b f y (d  ) 281.5) 2 (OK) 4 .01in 0.0094 15 19. Design as doubly reinforced section of size 15in×22in. d' = 2.5in d = 19.5in M2 = Mu   b Mn = 73.75k-ft  Assuming compression steel is yielding: M2 73.75 12 = = 0.964in 2 0.9 f y ( d  d ') 0.9(60)(19.5  2.5) AS 2 = fs ' = As2 fy fs ' = 87(4.726  0.85  2.5) 4.726 = 47.88 ksi < f y As2revised Thus, = As2 fy fs ' = As1 + As = 0.964  60 = 1.21 in 2 47.88 As2revised = 3.013+ 1.21 = 4.22 in 2 4 - #8 + 3 - #6 (Top bars) Use, 2 - #8 (Bottom bars) Design for Shear: Provide: i) #3@4 " c/c utpo l from both supports. 3 ii) #3@7 " c/c in remaining middle portion. Ground,1st & 2nd Slab Beams Design Design of B1: Factored moments taken from ETABS: Exterior support Mu  = 64.56 k-ft Exterior span Mu  = 28.73 k-ft Interior support Mu  = 248.41 k-ft Page 37 of 85 Interior span Mu  = 203.66 k-ft Minimum Depth of beam for deflection control: Exterior Span h= = Interior Span l 18.5 h= 15 12 = 9.729in 18.5 = l 21 30  12 = 17.143in 21 Say, h = 18in to keep the deflections in control. Width of web: bw = 9in d min for Mumax : d min hmin = Mumax 0.205 fc ' b = 248.4112  23.2in 0.205  3  9 = 23.2 + 2.5 = 25.71in Thus, h  25in d  25  2.5  22.5in Maximum capacity as singly reinforced section at supports:  max AS 1 2 fc ' = 0.851   = 0.0103 7 fy =  max bd = 0.0103  9  22.5  2.086in 2 a = AS 1 f y 0.85 fc 'b = 2.086  60 = 5.454in 0.85  3  9 Page 38 of 85 b Mn  a = b AS 1 f y (d  ) 2 0.9(2.086)(60)(22.5  = 12 5.454 ) 2 = 185.61 k-ft b Mn  = 185.61 k-ft Effective flange width for L-beam behavior: bf is lesser of the following: 15 12  15in for exterior span 12 30 12 =  30in for interior span 12 iv) l 12 = v) 6h f  bw = 6 (7)+9 = 51in vi) bw  Thus, Sc 2 =9+ bf 13.5  12 = 90in 2 = 15in for exterior span = 30in for interior span Maximum capacity as singly reinforced section in spans:  max AS 1 3 fc ' = 0.851   = 0.0135 8 fy =  max bd = 0.0135  9  22.5  2.734in 2 a b Mn  = AS 1 f y 0.85 fc 'b = 2.734  60 = 2.144in 0.85  3  30 a = b AS 1 f y (d  ) 2 0.9(2.734)(60)(22.5  = 12 2.144 ) 2 = 263.63 k-ft Page 39 of 85 b Mn  = 263.63 k-ft Minimum steel ratio:  min = Asmin = 0.0033  9  22.5  0.668in 2 200 200   0.0033 fy 60000 Design for positive moment in exterior span:` Assume, Mu  = 28.73 k-ft a = 3in As Mu = a 2 b f y (d  ) = Thus, As 28.73 12 3 0.9(60)(22.5  ) 2 = 0.304in 2  Asmin = 0.668in 2 Use, 2 - #6. Design for positive moment in interior span: Assume, Mu  = 203.66 k-ft a = 3in As = Mu a 2 b f y (d  ) = a = 203.66 12 3 0.9(60)(22.5  ) 2 AS f y 0.85 fc 'b = = 2.155in 2 2.155  60 = 1.69in 0.85  3  30 (Assumption is correct) Page 40 of 85 5) 5.#8 Design for negative moment at exterior support: Mu  = 64.5) AS 2 = fs ' = As2 fy fs ' = 87(5.698in 2 0.5in M2 = Mu   b Mn = 62. Design for negative moment at interior support: Mu  = 248. b = 9in d = 22.454  0.69 0.#6.9(60)(22.5) 2  min 2 .41 k-ft > b Mn = 185.5  ) 2 Use.85  2.9 f y ( d  d ') 0.5in R =  = 0. 3 .454 = 53.56 12 =  0.17 bd 2 9(22.9(60)(22.1in 2 1. d' = 2. Mu 64.As = 203.61 k-ft  Design as doubly reinforced section of size 9in×25in.56 k-ft Flange is under tension: Thus.0033 = As = 0.8k-ft  Assuming compression steel is yielding: M2 62.5  2.8 12 = = 0.14 ksi < f y Page 41 of 85 .668in 2 Use.66 12 = 2.5in d = 22. 698  = As1 + As Use. 60 = 0.14 As2revised = 2. Width of web: bw = 9in d min for Mumax : Page 42 of 85 . 3 ii) #3@9 " c/c in remaining middle portion.874 in 2 4 .5 h= 13.756in 18.5  12 = 7.086+ 0. h = 9in to keep the deflections in control.18 k-ft First Interior support Mu  = 36.83 k-ft Interior span Mu  = 22.788 in 2 53.04 k-ft Exterior span Mu  = 27. Design of B2: Factored moments taken from ETABS: Exterior support Mu  = 43.#8 (Top bars) 2 . = As2 fy fs ' = 0.9 k-ft Interior support Mu  = 39.51 k-ft Minimum Depth of beam for deflection control: Exterior Span h= = Interior Span l 18.5 = l 21 13.14in 21 Say.788 = 2.#6 ( Bottom bars) Design for Shear: Provide: i) #3@6 " c/c utpo l from both supports.5 12 = 8.As2revised Thus. 74 k-ft b Mn  = 66.d min hmin = Mumax 0.0103  9 13.5 12  13.66in 0.268in 0.16in Thus.0103 7 fy =  max bd = 0.205  3  9 = 9.25)(60)(13.25  60 = 3.85  3  9 a = b AS 1 f y (d  ) 2 0.66 + 2.5  1.9(1.04 12  9.851   = 0.74 k-ft Effective flange width for L-beam behavior: bf is lesser of the following: 13.5in Maximum capacity as singly reinforced section at supports:  max AS 1 2 fc ' = 0.27 ) 2 = 66.5 = 12.5in 12 iv) l 12 = v) 6h f  bw = 6 (7)+9 = 51in Page 43 of 85 .25in 2 = a b Mn  AS 1 f y 0.205 fc ' b = 43.85 fc 'b = 1. h  16in d  16  2.5  13.5  = 12 3. 86 ) 2 = 89.64)(60)(13.9(1.85  3 13.5  1.64  60 = 2.5 a = b AS 1 f y (d  ) 2 b Mn  0.64in 2 = a AS 1 f y 0.0135  9 13.851   = 0.0033 fy 60000 Design for positive moment in exterior span: Assume. Mu  = 27.5  12 = 90in 2 = 13.85 fc 'b = 1.0135 8 fy =  max bd = 0.18 k-ft a = 3in As = Mu a 2 b f y (d  ) Page 44 of 85 .5  = 12 2.1 k-ft Minimum steel ratio:  min = Asmin = 0.5  0. =9+ 13.1 k-ft b Mn  = 89.86in 0.vi) bw  Sc 2 bf Thus.0033  9 13.401in 2 200 200   0.5in Maximum capacity as singly reinforced section in spans:  max AS 1 3 fc ' = 0. Design for positive moment in interior span: Mu  = 22.18 12 3 0. Page 45 of 85 . b = 9in d = 13. Mu 43.503in 2 As Use.74 k-ft  (OK) 2 .#6 Design for negative moment at exterior support: Mu  = 43.74 k-ft  (OK) Flange is under tension: Thus. 2 .= Thus. 2 . 3 ii) #3@10 " c/c in remaining middle portion.315 = bd 2 9(13.#6 Design for negative moment at interior supports: Mu  Use. 27.0064  9 13.9 k-ft < b Mn = 66.5) 2 (OK) 2 .5  .#6 Design for Shear: Provide: i) #3@7 " c/c utpo l from both supports.51 k-ft Use.0064 >  min As = 0.5in R =  = 0.04 k-ft < b Mn = 66.#6.9(60)(13.503in 2  Asmin (OK) = 0. = 36.777in 2 Use.04 12  0.5  ) 2 = 0. 7in 0.5 h= 15 12 = 9.143in 21 Say.0103 7 fy Page 46 of 85 . h = 18in to keep the deflections in control. Width of web: bw = 12in d min for Mumax : d min hmin Thus.Design of B3: Factored moments taken from ETABS: Exterior support Mu  = 129.1 k-ft Exterior span Mu  = 3.851   = 0.2in h  25in d  25  2.79 k-ft Interior span Mu  = 345.5in Maximum capacity as singly reinforced section at supports:  max 2 fc ' = 0.205  3 12 = 23.35 12  23.35 k-ft Minimum Depth of beam for deflection control: Exterior Span h= = Interior Span l 18.5 = 26.5  22.729in 18.5 = l 21 30  12 = 17.205 fc ' b = 345. = Mumax 0.73 k-ft Interior support Mu  = 303.7 + 2. 0135 12  22.5  2.AS 1 =  max bd = 0.85 fc 'b = 3.9(2.851   = 0.46 k-ft b Mn  = 247.953in 0.5  3.453in 0.781in 2 = a AS 1 f y 0.5  = 12 5.645  60 = 0.0135 8 fy =  max bd = 0.85  3 12 a = b AS 1 f y (d  ) 2 b Mn  0.85 fc 'b = 2.5 12 = 162in bf = 45in for exterior span = 90in for interior span Maximum capacity as singly reinforced section in interior spans:  max AS 1 3 fc ' = 0.85  3  90 Page 47 of 85 . = 13.781)(60)(22.645in 2 a = AS 1 f y 0.46 k-ft Effective flange width for T-beam behavior: bf is lesser of the following: 15 12  45in for exterior span 4 30 12  90in for interior span = 4 iv) l 4 = v) 16h f  bw = 16 (7)+12 = 124in vi) S Thus.0103 12  22.453 ) 2 = 247.781 60 = 5. 2 .953 ) 2 = 361. As 3.#6.0033 fy 60000 Design for positive moment in exterior span: Assume. Design for positive moment in interior span: Assume.645)(60)(22.24 k-ft Minimum steel ratio:  min = Asmin = 0.039in 2  Asmin = 0.9(60)(22.73 12 3 0. Mu  = 3.24 k-ft b Mn  = 361.73 k-ft a = 3in As Mu = a 2 b f y (d  ) = Thus.35 k-ft a = 3in As = Mu a 2 b f y (d  ) Page 48 of 85 .5  = 12 0.772in 2 Use.b Mn  a = b AS 1 f y (d  ) 2 0.5  ) 2 = 0.5  0.772in 2 200 200   0.9(3. Mu  = 345.0033 12 19. 955 0. Design for negative moment at interior support: Mu  = 303.33k-ft  Page 49 of 85 .9(60)(22.5) 2  min (OK) 2 .1 k-ft Flange is under tension: Thus.255 = bd 2 12(22. Mu 129.484in 2 0.5  1.9(60)(22.5in R =  = 0.955in 0.5in M2 = Mu   b Mn = 56.5in d = 22.35 12 3 0.0051 > As = 0.85 fc 'b = = 3.0051 12  22.#8.46 k-ft  Design as doubly reinforced section of size 12in×25in.654in 2 3.= a = 345.112  0. 3 .85  3  90 (Assumption is correct) As = 345.5  ) 2 AS f y 0. d' = 2.#6 Design for negative moment at exterior support: Mu  = 129. b = 12in d = 22.654  60 = 0.377in 2 Use.#8 + 3 .5  ) 2 Use.35 12 = 3.79 k-ft > b Mn = 247. 488 in 2 = As1 + Use. Design of B4: Factored moments taken from ETABS: Exterior support Mu  = 50 k-ft Exterior span Mu  = 70 k-ft Interior support Mu  = 340.626in 2 0. = 0.5 Interior Span h= l 21 Page 50 of 85 .9(60)(22.21 k-ft Interior span Mu  = 258. 3 ii) #3@8 " c/c in remaining middle portion. As = As2 fy fs ' 60 = 0.8 k-ft Minimum Depth of beam for deflection control: Exterior Span h= l 18.453  0.33 12 = = 0.#6 ( Bottom bars) Design for Shear: Provide: i) #3@5 " c/c utpo l from both supports.5) AS 2 = fs ' = As2 fy fs ' = 87(5.85  2.626  3 .096 As2revised = 2.09 ksi < f y As2revised Thus.5) 5.781+ 0.5  2.#8 + 3 .453 = 53.707 = 3.Assuming compression steel is yielding: M2 56.#6 (Top bars) 2 .9 f y ( d  d ') 0.707 in 2 53. 453in 0.781in 2 a b Mn  = AS 1 f y 0.453 ) 2 = 247.5 + 2.84 12  23.9(2.5in 0.781)(60)(22.0103 7 fy =  max bd = 0.205  3 12 = 23.729in 18.5  = 12 5.46 k-ft b Mn  = 247.0103 12  22.85  3 12 a = b AS 1 f y (d  ) 2 0.5in Maximum capacity as singly reinforced section at supports:  max AS 1 2 fc ' = 0.5  2.5  22. h = 18in to keep the deflections in control.= 15 12 = 9. h  25in d  25  2. Width of web: bw = 12in d min for Mumax : d min hmin = Mumax 0.781 60 = 5.5 = 30  12 = 17.851   = 0.205 fc ' b = 339.46 k-ft Page 51 of 85 .5 = 26in Thus.85 fc 'b = 2.143in 21 Say. 5 12 = 162in bf = 45in for exterior span = 90in for interior span Maximum capacity as singly reinforced section in interior spans:  max AS 1 3 fc ' = 0.851   = 0.5  = 12 0. = 13.0033 12 19.0135 12  22.85 fc 'b = 3.24 k-ft Minimum steel ratio:  min = Asmin = 0.5  0.645  60 = 0.645in 2 = a b Mn  AS 1 f y 0.772in 2 200 200   0.24 k-ft b Mn  = 361.0135 8 fy =  max bd = 0.5  3.0033 fy 60000 Page 52 of 85 .953 ) 2 = 361.9(3.Effective flange width for T-beam behavior: bf is lesser of the following: 15 12  45in for exterior span 4 30 12 =  90in for interior span 4 i) l 4 = ii) 16h f  bw = 16 (7)+12 = 124in iii) S Thus.953in 0.645)(60)(22.85  3  90 a = b AS 1 f y (d  ) 2 0. Mu  = 258.9(60)(22.#8 Design for negative moment at exterior support: Mu  = 50 k-ft Page 53 of 85 .5  ) 2 AS f y 0.741in2  Asmin = 0.74in 2 2. 4 .85 fc 'b = = 2.8 12 = 2.772in 2 Use.74  60 = 0.716in 0.9(60)(22.85  3  90 (Assumption is correct) As = 258. As 70 12 3 0.Design for positive moment in exterior span: Assume. Design for positive moment in interior span: Assume.#6.5  ) 2 = 0.8 k-ft a = 3in As = Mu a 2 b f y (d  ) = a = 258.716 0.5  ) 2 Use.8 12 3 0.6in 2 0. 2 .9(60)(22. Mu  = 70 k-ft a = 3in As Mu = a 2 b f y (d  ) = Thus. Flange is under tension: b = 12in d = 22.031in 2 0.453  0.165 = 3.85  2.5in M2 = Mu   b Mn = 92.031 60 = 1.5in d = 22.21 k-ft > b Mn = 247. 2 . As 50 12 Mu =  0. Design for negative moment at interior support: Mu  = 340. d' = 2.096 As2revised = 2.949 in 2 4 .5  2.46 k-ft  Design as doubly reinforced section of size 12in×25in.781+ 1.#6.#6 (Top bars) 3 .9(60)(22.75k-ft  Assuming compression steel is yielding: M2 92.099 2 bd 12(22.5) 5.09 ksi < f y As2revised Thus.5) 2  min = 0.#6 ( Bottom bars) Page 54 of 85 .9 f y ( d  d ') 0.#8 + 2 . = As2 fy fs ' = As1 + = 1.772in 2 Use.165 in 2 53. As Use.5) AS 2 = fs ' = As2 fy fs ' = 87(5.75 12 = = 1.453 = 53.5in R = < Thus. 77 k-ft Interior support Mu  = 537.63in h  27in d  27  2.5 h= 27 12 = 17.205  3 18 = 24.5 = 26.46 k-ft Exterior span Mu  = 504.Design for Shear: Provide: i) #3@5 " c/c utpo l from both supports.13 12  24. say = Mumax 0. Width of web: bw = 18in d min for Mumax : d min hmin Thus.51in 18.205 fc ' b = 537.13 k-ft Interior span Mu  = 462.5  24.42in 21 Say.03 k-ft Minimum Depth of beam for deflection control: Exterior Span h= = Interior Span l 18. 3 ii) #3@8 " c/c in remaining middle portion. Design of B5: Factored moments taken from ETABS: Exterior support Mu  = 384.5 = l 21 27 12 = 15.5in Page 55 of 85 .13 + 2. h = 18in to keep the deflections in control.13in 0. 0103 18  24.0135 8 fy =  max bd = 0.5  = 12 5.85 fc 'b = 5.Maximum capacity as singly reinforced section at supports:  max AS 1 2 fc ' = 0.54  60 = 5.96 k-ft b Mn  = 439.85  3  81 Page 56 of 85 .85  3 18 a = b AS 1 f y (d  ) 2 b Mn  0.0103 7 fy =  max bd = 0.93in 0.851   = 0.96 k-ft Effective flange width for T-beam behavior: bf is lesser of the following: 27 12  81in 4 iv) l 4 = v) 16h f  bw = 16 (7)+18 = 130in vi) S Thus.9(4. = 13.95in 2 a = AS 1 f y 0.54in 2 = a AS 1 f y 0.851   = 0.54)(60)(24.0135 18  24.5 12 = 162in bf = 81in Maximum capacity as singly reinforced section in interior spans:  max AS 1 3 fc ' = 0.93 ) 2 = 439.5  4.7in 0.85 fc 'b = 4.5  5.95  60 = 1. 5  1.455in 2 200 200   0.5  ) 2 AS f y 0.22 k-ft Minimum steel ratio:  min = Asmin = 0.03 k-ft Page 57 of 85 .42 0.5  = 12 1.77 12 3 0.9(60)(24. Design for positive moment in interior span: Mu  = 462.0033 18  24.77 12 1.72in 2 Use.9(60)(24.42in 0. Mu  = 504.a = b AS 1 f y (d  ) 2 b Mn  0.5  ) 2 As = As = 4.877in 2  Asmin (OK) 4.95)(60)(24.#8.7 ) 2 = 633.877  60 = 1.77 k-ft a = 3in As = Mu a 2 b f y (d  ) = a = 504.22 k-ft b Mn  = 633.0033 fy 60000 Design for positive moment in exterior span: Assume.9(5.85 fc 'b = = 4. 504. 6.85  3  81 (Assumption is correct) Thus. 5  ) 2 As = As = 4.3in 0.5) 2 (OK) 5 .03 12 3 0.46 12 =  0. b = 18in d = 24.0089 18  24.96 k-ft  Page 58 of 85 .3in 2 Use. 462. a As = 3in Mu = a 2 b f y (d  ) = a = 462.464  60 = 1.#8.9(60)(24. 6.3 0.5in R =  = 0. Use.427 bd 2 18(24.5  3.85  3  81 (Assumption is correct) Thus.464in 2  Asmin (OK) 4.5  ) 2 AS f y 0.13k-ft > b Mn = 439.03 12 1.9(60)(24.46 k-ft Flange is under tension: Thus.0089 >  min As =  0.85 fc 'b = = 4. Design for negative moment at exterior support: Mu  = 384.92in 2 Mu 384.Assume.#8 Design for negative moment at interior support: Mu  = 537. 9 f y ( d  d ') 0.#8 + 2 . 3 ii) #3@7 " c/c in remaining middle portion.93 = 55. d' = 2.05 = 5. As = As2 fy fs ' = As1 + = 0.5in d = 24.9(60)(24.Design as doubly reinforced section of size 18in×27in.5  2.5) AS 2 = fs ' = As2 fy fs ' = 87(5. Page 59 of 85 .6 in 2 6 . 3 .85  2.82 As2revised = 4.5) 5.982in 2 0.93  0.05 in 2 55.#6 (Bottom bars) Design for Shear: Provide: i) #3@4 " c/c utpo l from both supports.982  60 = 1.#6 (Top bars) Use.54+ 1.82 ksi < f y As2revised Thus.17 12 = = 0.17k-ft  Assuming compression steel is yielding: M2 97.5in M2 = Mu   b Mn = 97. 81in 2 0.27  65.8 0.03 Ast = 0.6(91.67  0.8k Mu x = 91.03  314  9.01 f y 1176.7 ex = ex h = Mu y Pu = 65. 15  0.6Mux  0.5 fc ' 0.56)  488.67in 1176.6Mu y 0. D = 20 in Let.56 12  0.DESIGN OF COLUMNS Design of C3: Basement & GF: Pu = 1176.81  24.0335 20 Page 60 of 85 .14 = D = 20in Let assume.56k-ft Let’s design the column by Reciprocal Load Method: Ag trail = = D Pu  0.5  4  0.95in 3.8  0.01  60 4A =  4  488.27k-ft Mu y = 65.75 20  = Ag =  = 0.42in  4  202  314in 2  2 D = 20-5=15 Bending about Y-axis:  = 0. 1 Pno = 1.6k Pnxo  = 1.6k Pn yo  = 1.26 fc ' Ag  = 1.26  4  314  1582. Diameter of spiral d sp Concrete cover = 3 in 8 = 2.6 1582.27 12  0.56k Pno 1 1 1 1    Pn Pnxo Pnyo Pno 2 1  = 8. #3.56k Pno Bending about X-axis:  = 0.93in = 1176.26  4  314  1582.8 Pu = 0. Provide 12 .7 ey = Mu x 91.1 4  314  1381.Pnyo fc ' Ag  1.26 fc ' Ag  = 1.1 fc ' Ag Pno = 1.5 = 15in Diameter of chore Ag = 314 in 2 Ach =  4 d ch2   4  152  176. (OK).625in 2 Page 61 of 85 .046 20 ey h Pnxo  1.157 104 k 1 1381.#8 longitudinal bars.92k > Pu.93  0.56 1 Pn = Pn = 1225.1 4  314  1381. Spiral Reinforcement: Let the bar used.5in = 20  2  2. 6(187.14 D = 20in Let assume.43)  281.75  18.01  60 4A =  = 4  281. 1stF & 2ndF: Pu = 527.75in 2 0.45d ch fc '(  1) 0.03  314  9.7 Page 62 of 85 .18k-ft Mu y = 154.59k Mu x = 187. D = 20 in Let.03 Ast = 0.5  4  0.3in = Ag 314  1) 0.01 f y 527.18 154.59  0.6Mu y 0.43k-ft Let’s design the column by Reciprocal Load Method: Ag trail = = D Pu  0.75 20  = Ag =  = 0.14     60 2  d sp f y 8   1.2 Smax 3 3.45 15  4( 176.6Mux  0.42in 2  4  202  314in 2  D = 20-5=15 Bending about Y-axis:  = 0.625 Ach Provide #3 spiral at a pitch of 1inch c/c.94in 3. 15  0.5 fc ' 0. 2k Pn yo  Mu y = 1.2 791.59 Pu = 4.26in = 527.56k Pno 1 1 1 1    Pn Pnxo Pnyo Pno 1 Pn = Pn = 565.195 18 = 0. #3.Pnyo fc ' Ag  0.63 4  314  791. Diameter of spiral d sp Concrete cover = 3 in 8 = 2.7  4  314  879.59 3.2k > Pu.51  0.63 fc ' Ag Pno = 1.26  4  314  1582. Provide 12 .56k Pno Bending about X-axis:  = 0.34 fc ' Ag   = 0. Spiral Reinforcement: Let the bar used.7 Pno = 1.769 10 3 k 1 879.18 12  4.26  0.28k Pnxo = 1.5in Page 63 of 85 .26  4  314  1582.26 fc ' Ag  ex = ex h = Pu = 154.28 1582.56 (OK).#8 longitudinal bars.237 18 ey h Pnxo  0.43  12  3.51in 527.7 ey = Mu x 187. 1 1 1   = 1. 01 f y 231.604 0.625 Ach  4 d ch2   4  152  176.62in 231.37 9 12in Page 64 of 85 .53k-ft Mu y = 69.37  0.87)  113.5 fc ' 0.45d ch fc '(  1) 0.87 12  3.6 12 Mu y Pu = 69.6Mu y 0.5 = 15in Diameter of chore Ag = 314 in 2 Ach = Smax 3 3.6Mux  0.025 Ast = 0.87k-ft Let’s design the column by Reciprocal Load Method: Ag trail = = Pu  0.025 180  4.5in 2 y 15in x Bending about Y-axis:  = ex = 12  2(2.45 15  4( 176.6(34.625in 2 2 Provide #3 spiral at a pitch of 1inch c/c.375)  0.= 20  2  2. Ag = 12 15  180in 2  = 0. The selected section is 12''15'' Let.37k Mu x = 34.5  4  0.53  69.1in 2 0.3in = Ag 314  1) 0. Design of C2: Pu = 231.01 60 Let.14     60 2  d sp f y 8   1. 6  4 180  432k Pno = 1.ex h Pnyo = 3.8k 1 1 1 1    Pn Pnxo Pnyo Pno 1 1 1   = 3.55  Pn yo = 0.55  4 180  396k Pno = 1.8 1 Pn = Pn = 272.19 fc ' Ag  Pno = 1.6 fc ' Ag  Pnxo = 0.37 Pu = 3.62in = 231. Diameter of spiral d sp = 3 in 8 Spacing of the ties = least of the: i) 16 dlb = 16 1  16in Page 65 of 85 .87 12  3.30 12  0.67 0.5)  0.6 15 ey = Mu x 69.62  0.62  0.#8 longitudinal bars. Lateral ties Reinforcement: Let the bar used.19 fc ' Ag  Pno = 1.19  4 180  856.241 15 ey h Pnxo  0. #3.19  4 180  856.8k fc ' Ag Bending about X-axis:  = 15  2(2.673 103 k 1 396 432 856. Provide 6 . (OK).26k > Pu. 58in 867.39 114.58  0.01  60 Let.5)  0. Ag = 15  24  360in 2  = 0.18k-ft Let’s design the column by Reciprocal Load Method: Ag trail = = Pu  0.79 0.5 fc ' 0.01 f y 867. Design of C1: Basement & GF: Pu = 867.065 24 Page 66 of 85 .ii) 48 dtb iii) iv) Least dimension 12in 3 = 48   18in 8 = 12in = 12in Provide 2 .7 24 Mu y Pu = 114.#3 spiral at 10inch c/c.6Mux  0.6(122.025  360  9in 2 24in Bending about Y-axis: x 15in  = ex = ex h = y x X 24  2(2.18  12  1.39k-ft Mu y = 114.6Mu y 0.46in 2 0.67  0.67k Mu x = 122.025 Ast = 0. The selected section is 15'' 24'' Let.5  4  0.81)  388.67 1. 6 15 ey = Mu x 122.94 Pno = 1.2k 1 1 1 1    Pn Pnxo Pnyo Pno 1 1 1   = 9.6 1699.071 15 ey h Pnxo  0.92 fc ' Ag  Pnxo = 0.2  4  360  1728k Bending about X-axis:  = 15  2(2.05110 4 k 1 1324.2 1 Pn = Pn = 1104.8k Pno = 1.2 fc ' Ag  Pn yo = 0. Provide 12 . Diameter of spiral d sp = 3 in 8 Spacing of the ties = least of the: i) 16 dlb = 16 1  16in ii) 48 dtb 3 = 48   18in 8 Page 67 of 85 .8 1353.693in = 867.6k  Pno = 1.18  4  360  1699.67 0. (OK). #3.Pnyo fc ' Ag  0.86k > Pu.67 Pu = 1.94  4  360  1353.39 12  1.693  0. Lateral ties Reinforcement: Let the bar used.92  4  360  1324.18 fc ' Ag  Pno = 1.#8 longitudinal bars.5)  0. 8 Pno = 1.7 24 Mu y Pu = 117.9 12  3. 1stF & 2ndF: Pu = 383.#3 spiral at 10inch c/c.5)  0.01 f y 383.9k-ft Let’s design the column by Reciprocal Load Method: Ag trail = = Pu  0.01 60 Let.69  0.5 fc ' 0.5 3.iii) iv) Least dimension 12in = 12in = 12in Provide 2 .6Mux  0.025  360  9in 2 24in Bending about Y-axis: Pnyo fc ' Ag  0.3 117.5  0.6Mu y 0. Ag = 15  24  360in 2  = 0.5  4  0.154 24 Pn yo = 0.6(194.8  4  360  1152k Pno = 1.2  4  360  1728k Page 68 of 85 . The selected section is 15'' 24'' Let.791 0.025 Ast = 0.2 fc ' Ag   x 15in  = ex = ex h = y x 24  2(2.5k Mu x = 194.9)  219.55in 2 0.69in 383.3k-ft Mu y = 117. #8 longitudinal bars.8k Pno = 1.#3 spiral at 10inch c/c.5 = 6.3 12  6.1  0.Bending about X-axis:  = 15  2(2.46 10 3 k 1 1152 460.5k > Pu.32  4  360  460. Lateral ties Reinforcement: Let the bar used.6 15 ey = Mu x 194. Provide 12 . Diameter of spiral d sp = 3 in 8 Spacing of the ties = least of the: i) 16 dlb ii) 48 dtb iii) iv) Least dimension 12in = 16 1  16in 3 = 48   18in 8 = 12in = 12in Provide 2 .67 0. #3.2  4  360  1728k 1 1 1 1    Pn Pnxo Pnyo Pno 1 Pn = Pn = 406.2 fc ' Ag  Pno = 1. Page 69 of 85 .32 fc ' Ag  Pnxo = 0. 1 1 1   = 2.5)  0.1in = Pu 383.508 12 ey h Pnxo  0.8 1728 (OK). Design strips B A FigA: Column forces at base Page 70 of 85 . 5 = 31.6)     9918 y ' = 54(215.7  643.9 153.5 1147.5   M x' 313963.9     60    223.23 10 3 ft 4 12 Iy = 1  83  623  1648.4  219.14  81   0.356 2 Mu x = Pu  ey = 9918   0.7  221.65 9918 x' = ex = 31.9  1163.7  1147.4)  40.8  930.8)   67.5(223.7  1176.65  16364.9  231.9)  27(95.7  226  221.5(219.8  221.14 9918 y' = ey = 40.4  867.Total factored load at the base Pu = 9918k Foot print area of the base A = 62' 83'  5146 ft 2 Ix = 1  62  833  2954.4 10 3 ft 4 12 Eccentricities of the load Pu : M y' 0     15(648.9  1176.65 ft 2 0 13.356=  3532.5(29.4  223.6  231.5  715.ft Page 71 of 85 .13  266)     81(146  648.7)  398146.7 k.7  221.5  930.2 k.8 1163.6)    153.5)     9918x ' = 45(643.65  60  1.725 =40.ft Mu y = Pu  ex = 9918 1. 23 103 = 1.72 1.25 40.08 2.5 40.02 2.99 2.5 -13.14 2.17 2.q = Pu Mu y x Mux y   A Iy Ix = 9918 16364. ft 2 Thickness of MAT: Let’s check the thickness of MAT for critical columns: C1.61 1.12 1.7 3532.5 -40.81 2.68 1.90  0.C6 & C11 shown in FigA.204ksf .22 1.65 1.5 0 -13.2  x y 3 62  84 1648 10 2954.5 -27 27 13.19 2.60 1.20 The soil pressure under the all columns is less than ABC of soil = 1.25 -13.63 1.5 40.5 -40.75 2.0012 y Column# x (ft) y (ft) q (ksf) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 -15 15 15 -15 -30 -15 15 30 -30 -15 15 30 -30 -30 -30 -30 -30 30 30 30 30 30 13.58 1.5 27 13.5 13.78 1.5 -27 1.15 2.009 x  0. Page 72 of 85 .05 1.5 0 -13.5 40.5 -40.5 -40.66 2.0 ton = 2. 675d (20  d ) = 13.14(20  d ) d 1000 = 0.8k bo =  (20  d )  3.675d 2  13.8 = 0 d = 32.85  4 4000  3.5d -1176.85  4 fc 'bo d = 0.675d 2 0.5d  0.14(20  d ) vVc  Pu vVc = 0.93in 33in Page 73 of 85 .C1 Critical Perimeter: d 2 d 2 (20  d ) in Pu = 1176. 3225d 2  10.6k bo d 3 = (27  )  (24  d)  51  d 2 2 vVc  Pu vVc = 0.965d -648.215(51  d ) d = 0.7 = 0 d = 30.85  4 fc 'bo d 3 0.96in 31in Page 74 of 85 .85  4 4000  (51  d ) d 2 = 1000 3 = 0.3225d 2  10.C6 12in 24 15 d 2 d 2 d 2 Pu = 867.965 d 2 0. 215(39  2 d ) d = 0.385d 0.385d -648.43d 2  8.43d 2  8.C11 d 2 24 d 15 2 d 2 d 2 Pu = 648.296in 31in Thickness of MAT = 33inches Page 75 of 85 .7 = 0 d Thus. = 30.85  4 4000  (39  2 d ) d 1000 = 0.7k bo = (15  d )  (24  d)  39  2 d vVc  Pu vVc = 0.85  4 fc 'bo d = 0. 5  30. positive moment at bottom of foundation = 1466.29k  ft  1466.5in  C =T 0.85  4 12a = 60 As As = 0.Reinforcement Calculation: BOTTOM STEEL a) Along the shorter direction of MAT (Strip-A): Load Diagram Bending Moment Diagram Bottom Steel: Max.68a Mu a =  As f y (d  ) 2 Page 76 of 85 .85 fc ' ab = As f y 0.2k  ft / ft 12 b = 12in d = 33  2.29  122. positive moment at bottom of foundation  = 1769.0033 12  30. = 1.77  147.68a  60(30.5  1.911in 2 = 200 200   0.86  0 a = 1.34  0.21in 2 Thus. As Use.2 12 a 2  61a  79.a = 0.9  0.5  30.77 k  ft  1769.5in C =T Page 77 of 85 .0033 fy 60000 Asmin = 0.68 = 0.34 As  min = 1.5  ) 2 122.48 k  ft / ft 12 b = 12in d = 33  2.21in 2 #8@7inch c/c b) Along the longer direction of MAT(Strip-B): Load Diagram Bending Moment Diagram Max. 48 12 a 2  61a  96.85  4 12a = 60 As Page 78 of 85 .21in 2 Thus.0.5  30.5in C =T 0.5  1.68a  60(30.85  4 12a = 60 As As = 0.68a Mu a =  As f y (d  ) 2 a = 0.85 fc ' ab = As f y 0.68 = 1. negative moment at top of foundation = 1114.23k  ft   1114.0033 12  30.39  0 a = 1.85k  ft / ft 12 b = 12in d = 33  2.9  0.5  ) 2 147.62  0.62 As  min = 1.23  92.1in 2 = 200 200   0.85 fc ' ab = As f y 0.21in 2 #8@7inch c/c TOP STEEL a) Along the shorter direction of MAT (Strip-A): Max.0033 fy 60000 Asmin = 0. = 1. As Use. 01 As  min = 1.21in 2 #8@7inch c/c b) Along the longer direction of MAT (Strip-B): Max.68a Mu a =  As f y (d  ) 2 Page 79 of 85 .As = 0.0033 12  30.68 = 0.68a  60(30.21in 2 Thus.5  1.42k  ft / ft 12 b = 12in d = 33  2.9  0. = 1.85 12 a 2  61a  60.68a Mu a =  As f y (d  ) 2 a = 0. As Use.69  0 a = 1.69in 2 = 200 200   0.85  4 12a = 60 As As = 0.5  ) 2 92.5in  C =T 0.85 fc ' ab = As f y 0.0033 fy 60000 Asmin = 0. negative moment at top of foundation = 1445k  ft  1445  120.5  30.01 0. 5  1.5  ) 2 120.0033 12  30. = 1.68a  60(30.68 = 0.21in 2 Thus. As Use.9  0.9in 2 = 200 200   0.42 12 a 2  61a  78.21in 2 #8@7inch c/c Page 80 of 85 .0033 fy 60000 Asmin = 0.32  0.32 As  min = 1.7  0 a = 1.a = 0. 33 1  sin  1  sin 30 Backfill Loading (active pressure): 1 1  K a H 2  120  0. Surcharge Loading: S = 500psf Ka = PS =  S K a H  500  0. 9in NSL surcharge = 500psf   30 H = 6feet Ps Pb 3’   120 pcf 2’ O Base Slab Lateral Loads Calculation: Let’s assume 1foot depth into the page.8lbs 2 2 Page 81 of 85 .33  6  990lbs Pb = 1  sin  1  sin 30   0. As the wall is monolithically connected to a very rigid thick base slab. only.33  62  712. so it’s safe against sliding and overturning.DESIGN OF RETAINING WALL Wall need to be designed against shear and bending. 14  2 12  27.14k Factored Bending Moments about point O: Mu S = 1.757in  9in 0.205 fc ' b = 74.6  990  1.81k-in d min = Mu 0.36 = 74. Mus d = 57 (3  0.36 k  in 2 2 Mubd d distance from O.14 = 2. (OK) Thickness of wall = 9in Shear Check: Critical shear at d-distance from O.45+23. 2 2 Mu = 51.205  4 12 Thus.45k  in 3 = 27.36 (2  0.96k  Vu 1000 (OK) Page 82 of 85 .Factored Loads: PuS = 1.36k-in Let thickness of the wall d = 9in = 92  7 Value of moment at critical section.724k vVc = v 2 fc 'bd = 0.292)  51.81  2.584+1.584  3 12  57k-in Mub = 1.58K Pub = 1.6  712. Vu = 1.8  1.292)  23.75  2 4000 12  7  7. 0025 12  7  0.81   0.Flexural Design of Stem: Mu 74.1271 bd 2 12  7 2 R =  = 0.0018 12  9  0. (OK) #3 @ 6inch c/c Page 83 of 85 .21in 2 Asmin = 0.0025 As = 0.194in 2 As > Asmin Use. 36 lb / ft  ft Live Loads Factored slab loads: Factored Bending Moment: Page 84 of 85 .3psf qL = 200psf qu = 1.55 psf = 12 T 10 Loads Calculation: Dead Loads Weight of steps = R 150 7 150   43.2(145.6(200)  494.42inch  6inch 24 d = 5inch R. slab 6 R2  T 2 72  102 150  75   91.75 psf =  2 12 2 12 1inch floor finish = 1  120  10 psf 12 qD = 145.DESIGN OF STAIRS 10@ 7inch = 5'10" 5foot UP 5foot 5foot UP = 5' 5'10"  10'10" l hmin considering one end continuous  l 24 = 10 12  10  5.3)  1.C. 0046 12  5  0.0018 12  6  0.07in = 0. = 0.Mu = d min for singly reinforced section = Thus.0018 As = 0.62k  in 10 10 12 69.129in 2 #3 @ 10inch c/c Page 85 of 85 .276in 2 (OK) #6 @ 8inch c/c Temperature Reinforcement: As Use. 69.36 qul 2   1302  1000  69.0046 > min  0.205 fc ' b h = 6inch Reinforcement Calculation: Use. 1 1 494.205  3 12 0.62 Mu  3.62  0.232 12  52 R =  = 0.