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					     Department of Civil Engineering N-W.F.P UET, Peshawar                    Reinforced Concrete Design



(1) ISOLATED FOOTING:
     Given Data:
              Service D.L = 250 k
              Service L.L = 200 k
              Allowable bearing capacity (qa) = 4 k/ft2
              Depth of footing below ground level (z) = 4'
              Column size = 18? x 18?
              fc ' = 3 ksi
              fy = 40 ksi




                                          Figure 1.1: Isolated footing.


Solution: -
     Step No 1: Sizes.
              Assume h =18?, d = h – 4 = 14?
              bo = 4(c + d) = 4 x (32) = 128
              Since the space between the bottom of footing and the ground level is occupied
              partly by concrete and partly by soil, therefore, the pressure (W) of this material at
              4 ft depth is:
              W = ?avgz
                 =125 x 4
                 = 500 psf




     Prof Dr. Qaisar Ali                                                                Page 1 of 13
Department of Civil Engineering N-W.F.P UET, Peshawar                    Reinforced Concrete Design



      qe (Bearing pressure effective to carry column’s service load):
      qe = qa –W
         = 4 – 0.500 = 3.5 ksf
      Areq = Service load on column /qe
      Areq = (250 + 200) /3.5 =128.57 ft2
      B x B = Areq
      B x B =128.57
      B = 11.34' ˜ 11.5'
      Areq = 11.5' x 11.5'


Step No 2: Loads.
      qu (Bearing pressure for strength design of footing):
      qu = Factored load on column/ Areq
      qu = (1.2 x 250 +1.6 x 200) / (11.5' x 11.5') = 4.68 ksf

Step No 3: Analysis.
       (1) Punching Shear:
           Vup = qu [B2 – {(c + d)/12}2 ]
               = 4.68 x [11.52 – {(18 + 14)/12} 2 ] = 585.65 k




                           Figure 1.2: Critical Perimeter for punching shear.


Prof Dr. Qaisar Ali                                                                Page 2 of 13
Department of Civil Engineering N-W.F.P UET, Peshawar                  Reinforced Concrete Design



      (2) Beam Shear:
           Vub = qu {(B – c)/2 – d}B
           Vub = 4.68 x [{11.5 – (18/12)}/2 – (14/12)] x 11.5 = 206.31 k




               Figure 1.3: Section at distance “d” from column face for Beam shear.

      (3) Moment:
           Mu = qu B[{(B – c)/2} 2 ]/2
               = 4.68 x 11.5 x [{11.5 - (18/12)}/2]2 /2
               = 672.75 ft-k = 8073 in-k




                              Figure 1.4: Section for Moment Calculation.


Prof Dr. Qaisar Ali                                                              Page 3 of 13
Department of Civil Engineering N-W.F.P UET, Peshawar                   Reinforced Concrete Design



Step No 4: Design.
      (1) Design for Punching Shear:
           Vup = 585.65 k
           F Vc = F4v (fc')bo d
           bo = 4(c + d) = 128
           F Vc = 0.75 x 4 x v (3000) x 128 x 14/1000 = 294.45 k < 585.65 k
           Since FVc < Vup , therefore increase the depth of footing.
           Let,
           h = 24?, d = h – 4 = 20?
           Vup = 572 k
           F Vc = 499.52 k < 572 k (Again increase the depth)
           Let,
           h = 27?, d = 23?
           Vup = 564.29 k
           F Vc = 619.80 k > 564.29 k, O.K.
           Therefore the depth d = 23? is adequate for punching shear.

      (2) Design for Beam Shear:
           Vub = 165.95 k
           F Vc = F2v (fc')B*d
           F Vc = 0.75 x 2 x v (3000) x (11.5 x 12) x 23/1000 = 260.77 k > 165.95 k
           Therefore d = 23? is also adequate for beam shear.

      (3) Design for Flexure:
           Mu = 672.75 ft-k = 8073 in-k
       (i) Let a = 0.2d = 0.2 x 23 = 4.6?
            As = Mu / { F fy (d – a/2)}
            As = 8073/ {0.9 x 40 x (23 – 4.6/2)} = 10.83 in2
      (ii) a = As fy / (0.85fc'B)
            a = 10.60 x 40/ (0.85 x 3 x 11.5 x 12) = 1.23?
            As = 8073/ {0.9 x 40 x (23 – 1.23/2)} = 10.02 in2
      (iii) a = 10.02 x 40/ (0.85 x 3 x 11.5 x 12) = 1.138?


Prof Dr. Qaisar Ali                                                               Page 4 of 13
Department of Civil Engineering N-W.F.P UET, Peshawar                 Reinforced Concrete Design



           As = 8073/ {0.9 x 40 x (23 – 1.138/2)} = 9.99 in2
      (iv) a = 9.99 x 40/ (0.85 x 3 x 11.5 x 12) = 1.135?
           As = 8073/ {0.9 x 40 x (23 – 1.135/2)} = 9.99 in2 , O.K.

      •      Check the minimum reinforcement:
      Asmin = {3v (fc ')/fy }B*d; but not less than (200/fy )B*d
      {3v (fc')/fy }B*d = {3v (3000)/40000} x (11.5 x 12) x 23 =13.03 in2
      (200/fy )B*d = (200/40000) x (11.5 x 12) x 23 = 15.87 in2 .
      So Asmin = 15.87 in2 . As Asmin > As, thus Asmin governs.
      As= 15.87 in2
      Using 1? F (#8) {#25, 25 mm}, with bar area Ab = 0.79 in2
      No of bars = BAb /As
                      = (11.5 x 12) x 0.79/15.87 = 6.86 ˜ 6? c/c
      Finally use #8 @ 6? c/c {#25 @ 150 mm c/c} both ways.




Prof Dr. Qaisar Ali                                                             Page 5 of 13
Department of Civil Engineering N-W.F.P UET, Peshawar                                 Reinforced Concrete Design




              GL


                                                                     18" RCC Square column

                                                                          #8 @ 6" c/c (both ways)
      4'-0"                                                                                           3" (1:3:6) PCC Pad


              27"                                                         3" clear cover



                                                11'-6"
                                                12'-0"




                                           # 8 @ 6" c/c (bothways)




                                                                                             11'-6"




                                                          18" RCC
                                                          Square column




                                                11'-6"


                                Figure 1.5: Drafting for Isolated footing.




Prof Dr. Qaisar Ali                                                                             Page 6 of 13
    Department of Civil Engineering N-W.F.P UET, Peshawar                          Reinforced Concrete Design



(2) COMBINED FOOTING (Ex 16.3, Nelson pg 562):
    Given Data:
          External column C1 (24? x 18?): D.L = 170 k
                                                  L.L = 130 k
          Internal column C2 (24? x 24?): D.L = 250 k
                                                 L.L = 200 k
          Footing depth = 6' below G.L
          fc ' = 3000 psi
          fy = 60000 psi
          Surcharge = 100 lb/ft2
          qa = 6 k/ft 2


    9"                                           18'



                 18" x 24" column                           24" x 24" column




                                                                                                                h



                                                Length of footing




                 18" x 24" column                               24" x 24" column                                B




                                          Figure 1.6: Combined footing.


    Prof Dr. Qaisar Ali                                                                      Page 7 of 13
     Department of Civil Engineering N-W.F.P UET, Peshawar                         Reinforced Concrete Design



Solution: -
     Step No 1: Sizes.
              Service load for column C1: P1 = D + L = 170 + 130 = 300 k
              Service load for column C2: P2 = D + L = 250 + 200 = 450 k


              Net bearing capacity, qe = qa – {(weight of fill and concrete) + (surcharge)}
              An average unit weight of 125 pcf can be assumed for fill and concrete.
                                       qe = 6 – (0.125 x 6 + 0.1) = 5.15 k/ft2
              Area of footing = (Sum of column loads)/qe = (300 + 450)/5.15 = 145.63 ft2


              Location of resultant:




                     P1                                                              P2




                                                             R
                                        χ
                                                      18'
                                     Figure 1.7: Calculation for resultant load.


              R = P1 + P2 = 750 k
              Take moment about exterior column centerline.
              750 x ? = 450 x 18
              ? = 450 x 18/750 = 10.8 ft
              Therefore, length of footing = 2(10.8 + 0.75) = 23.1 ft = 23'-3?
              Width of footing (B) = 145.63/23.25 = 6.3' ˜ 6'-6?
              Assume depth of footing, h = 3'-5? = 41?, d = h – 3.5 = 37.5?



     Prof Dr. Qaisar Ali                                                                     Page 8 of 13
Department of Civil Engineering N-W.F.P UET, Peshawar                     Reinforced Concrete Design



Step No 2: Loads.
      Pu, ext = 1.2 x 170 + 1.6 x 130 = 412 k
      Pu, int = 1.2 x 250 + 1.6 x 200 = 620 k
      Total factored load (P) = 1032 k
      The net upward pressure (qu ) caused by the factored column load is:
      qu = P/ (Footing area) = 1032/ (23.25 x 6.5) = 6.83 k/ft 2
      Net upward linear pressure/ foot in the longitudinal direction= 6.83 x 6.5= 44.4 k/ft.


Step # 3: Analysis.
      (i) Punching Shear:




                              Figure 1.8: Critical Perimeter for punching shear.

           Critical perimeter for column C1: bo = 2 x 36.75 + 61.5 = 135?
           Vu, ext = Pu, ext – qu{(36.75 x 61.5)/144} = 304.80 k
           Critical perimeter for column C2: bo = 4 x 61.3 = 246?
           Vu, int = Pu, int – qu{(61.5)/12}2 = 440.60 k
      (ii) Moment:
           The maximum negative moment between the columns occurs at a section of
           zero shears. Let “?” be the distance from the outer edge of the exterior column
           to this section,
           Vu = 44400? – 412000 = 0
           ? = 9.28 ft (See Shear force diagram)




Prof Dr. Qaisar Ali                                                                 Page 9 of 13
Department of Civil Engineering N-W.F.P UET, Peshawar                                          Reinforced Concrete Design


            412000lb                                                               620000 lb




                                                         44400 lb/ft

              1'-6"                             16'-3"                                2'-0"        3'-6"
                                                                       376000 lb



                              9.28'



                                                                                     135000 lb



                       345000 lb
                                                                                   3250000 lb-in




                      3100000 lb-in




                                      19230000 lb-in
                           Figure 1.9: Shear force & bending moment Diagram.

           The moment at 9.28 ft is:
           M (9.28) = {44400 x (9.282 /2) – 412000 x (9.28 – 0.75)} x 12 = –19230000 in- lb
           The moment at the right edge of the interior column is:
           Mint = 44400 x (3.52 /2) x 12 = 3263400 in- lb

Step # 4: Design.
      (i) Design for punching shear:
           For column C1:
           Vu, ext = 304.80 k
           F Vc = 0.75 x 4 x v (3000) x 135 x 37.5/1000 = 831.85 k > Vu, ext , O.K.
           For column C2:
           Vu, int = 440.60 k
           F Vc = 0.75 x 4 x v (3000) x 246 x 37.5/1000 = 1515.82 k > Vu, int, O.K.


Prof Dr. Qaisar Ali                                                                                        Page 10 of 13
Department of Civil Engineering N-W.F.P UET, Peshawar                         Reinforced Concrete Design



      (ii) Design for Moment:
      •    Check the minimum reinforcement ratio:
           Asmin = {3v (fc ')/fy }B*d = (200/fy )B*d
                 = (3 x v (3000) x 78 x 37.5/60000) = (200/40000) x 78 x 37.5
                  = 8.01 in2 < 9.75 in2 , Asmin = 9.75 in2
           For M (9.28) = 19230 in-k
           Try 11, #9, As = 11.0 in2
           a = As fy / (0.85fc 'B) = 11 x 60/ (0.85 x 3 x 78) = 3.31 in
           Md = F As fy (d – a/2)
               = 0.9 x 11 x 60 x (37.5 – 3.31/2) = 21309 in-k > M u, O.K.
           For Mint = 3263.4 in-k
           Try 16, #7, As = 9.62 in2
           a = As fy / (0.85fc 'B) = 9.62 x 60/ (0.85 x 3 x 78) = 2.90 in
           Md = F As fy (d – a/2)
               = 0.9 x 9.62 x 60 x (37.5 – 2.90/2) = 18727 in-k > M u, O.K.

      Design of transverse beam under interior column:
           Width of transverse beam = 24 + 2(d/2) = 24 + d = 24 + 37.5 = 61.5?
           Net upward load per linear foot for transverse beam= 620000/6.5 = 95400 lb/ft
           Mu = 95400 x [{(6.5)-(24/12)}/2]2 x 12/2 = 2900000 in- lb = 2900 in-k
           Since the transverse bars are placed on top of the longitudinal bars, the actual
           value of “d t1 ” furnished is:
           dt1 = 37.5 – 1 = 36.5?
           Asmin = 0.0033 x 61.5 x 36.5 = 7.48 in2
           Try 13, # 7, As = 7.80 in2
           a = As fy / (0.85fc 'b) = 7.80 x 60/ (0.85 x 3 x 61.5) = 2.98 in
           Md = F As fy (dt1 – a/2)
               = 0.9 x 7.80 x 60 x (36.5 – 2.98/2) = 14746 in-k > M u, O.K.




Prof Dr. Qaisar Ali                                                                    Page 11 of 13
Department of Civil Engineering N-W.F.P UET, Peshawar                         Reinforced Concrete Design



      Design of transverse beam under exterior column:
           Width of transverse beam (b) = 18 + d/2 = 18 + 37.5/2 = 36.75?
           Net upward load per linear foot for transverse beam=412000/6.5=63384.61 lb/ft
           Mu = 63384.61 x [{(6.5)-(24/12)}/2]2 x 12/2 = 1925305 in- lb = 1925.305 in-k
           dt2 = 36.5?
           Asmin = 0.0033bdt2
           Asmin = 0.0033 x 36.75 x 36.5 = 4.43 in2
           Try 8, # 7, As = 4.80 in2
           a = As fy / (0.85fc 'b) = 4.80 x 60/ (0.85 x 3 x 36.75) =1.44 in
           Md = F As fy (dt2 – a/2)
               = 0.9 x 4.80 x 60 x (36.5 – 1.44/2) = 9274 in-k > M u, O.K.




Prof Dr. Qaisar Ali                                                                    Page 12 of 13
        Department of Civil Engineering N-W.F.P UET, Peshawar                                 Reinforced Concrete Design



              9"                                           18'                                          4'-6"




                              18" x 24" column                     24" x 24" column




     6'-0"
                                            3"
                                           clear
Supporting bars                                                                                                                       3'-5"
                                                                           Supporting bars
                                                                                                                           3" clear

                         3'                                                           2'-6"           2'-6"
                                                                  23'-3"



                                  8 # 7 (6'-0")
                                                                                         13 # 7 (6'-0")

    18" x 24" column
                                          11 # 9 {19'-6" (top)}                                                       6'-6"


                                                                                  24" x 24" column


                              A

                                        Figure 1.9: Drafting for Combined footing.




        Prof Dr. Qaisar Ali                                                                            Page 13 of 13