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Chapter 12.1 Footing by SD

VIEWS: 33 PAGES: 29

									                     CHAPTER 12.1
            DESIGN OF FOOTING
       (STRENGTH DESIGN METOD : SD)




12.1 DESIGN RC. FOOTING BY SD
Footing is a structure which is used :
      a) to transmit the load of the structure to the soil stratum of
         sufficient strength.
      b) to spread the load over a large sufficient area of that stratum
         to minimize bearing pressure.

The two essential requirements in the design of foundations are that the
total settlement of the structure be limited to a tolerably small amount
and that differential settlement of the various parts of the structure be
eliminated as nearly as possible, i.e. differential settlement of the same
structure is as small as possible.




                                                                             1
12.1 DESIGN RC. FOOTING BY SD (Cont.)
Two types of foundation are classified as the resisting load :
   • If adequate soil is not found immediately below the structure, it
     become necessary to use deep foundation such as piles (เสาเข็ม ) or
     caissons (ฐานรากปลอง) to transmit the load to deeper, firmer
     layers. This type of foundation is widely used in Bangkok area
     since the Bangkok clay is a very soft soil.
   • If satisfactory soil directly underlies the structure, it merely
     necessary to spread the load, by footing or other means. Such
     structures are known as spread foundation.




12.1 DESIGN RC. FOOTING BY SD (Cont.)
Footing may be classified as :
   • Wall footing (ฐานรากตอเนื่องรับกําแพง) is simply a strip of
     reinforced concrete, wider than the wall, that distributed its
     pressure.
   • Single-column footings (ฐานรากเดี่ยว) are usually square,
     sometimes rectangular, and represent the simplest and most
     economical type.




  Wall Footing (ฐานรากตอเนื่องรับกําแพง) Single-column Footing (ฐานรากเดี่ยว)




                                                                                 2
12.1 DESIGN RC. FOOTING BY SD (Cont.)
  • Combined footing (ฐานรากรวม) is a footing under two or more
    columns are also used under closed spaced and heavily loaded
    interior columns.
  • Strap footing (ฐานชนิดมีคานรัด) is used when the property rights
    prevent the use of footing projecting beyond the exterior wall. In
    this case strap footings (ฐานรากตีนเปด) are used that enable one to
    design a footing that will not project beyond the wall column.




    Combined Footing (ฐานรากรวม)         Strap Footing (ฐานรากตีนเปด)




12.1 DESIGN RC. FOOTING BY SD (Cont.)
  • Raft footing (ฐานรากแบบแพ) is a very large footing like a large
    concrete slab reinforced in both direction. It consists of a solid
    reinforced concrete slab which extends under the entire building
    and which consequently distributes the load of the structure over
    the maximum area. This foundation has a high rigidity to minimize
    differential settlement.




                        Raft Footing (ฐานรากแบบแพ)




                                                                           3
12.1 DESIGN RC. FOOTING BY SD (Cont.)
          Behavior of Single Footing subjected to Loads

                P               P              P




             Fig. 12.1A      Fig. 12.1B      Fig. 12.1C

 If the load is symmetrical with respect to the bearing area, the bearing
 pressure is assumed to be uniformly distributed (as shown in Figure
 12.1A), it is known that this is approximately true.




12.1 DESIGN RC. FOOTING BY SD (Cont.)
• For footing resting on coarse-grained soils the pressure is
  larger at the center of the footing and decreases toward the
  perimeter (Figure 12.1B). This is so because the individual
  grains in such soils are somewhat mobile, so that the soil
  located close to the perimeter can shift very slightly outward
  in the direction of lower soil stresses.
• For footing resting on clay soils, the pressure is higher near
  the edge than at the center of the footing, since in such clay
  soils the load produces a shear resistance around the
  perimeter which adds to the upward pressure (Figure
  12.1C).




                                                                            4
12.1 DESIGN RC. FOOTING BY SD (Cont.)
• It is generally not economical in footing to use shear
  reinforcement. For the rare cases where the thickness is
  restricted so that shear reinforcement must be used. Singly
  reinforcement is used to resist flexure in design footing
  since the footing is under the soil and does not have the
  limitation on the thickness.




12.1 DESIGN RC. FOOTING BY SD (Cont.)
 • Consider footing under loads in Figure 12.2, the
   reinforcement is as follows :




   Fig. 12.2A Steel Reinforcement for Flexure   Fig. 12.2B Steel Reinforcement for Shear
              (เหล็กเสริมรับโมเมนต)                        (เหล็กรับแรงเฉือน)




                                                                                           5
12.1 DESIGN RC. FOOTING BY SD (Cont.)
 • Weights of footing are generally 4 to 8% of the column load. For a
   high column load, a lower value is assumed while for a low column
   load, a high value of footing weight is assumed.

 • For spread footing (footing on soil), to compute bending moments
   and shears, only the upward pressure “qu” that is caused by the
   factored column loads is considered. The weight of the footing does
   not cause moments and shears, just as, obviously, no moments or
   shears are present in a book lying flat on a table.

 • However, for footing on piles, the analysis and design of moments
   and shears must include the weight of the footing.




12.1 DESIGN RC. FOOTING BY SD (Cont.)
Loads, Bearing Pressures, and Footing Size
 • Allowable bearing pressures are established from principles of soil
   mechanics, on the basis of load tests and other experimental
   determinations.

 • Allowable pressure “qa” under service loads are usually based on a
   safety factor of 2-3 against exceeding the ultimate bearing capacity
   of the particular soil and to keep settlements within tolerable limits.




                                                                             6
12.1 DESIGN RC. FOOTING BY SD (Cont.)
For concentrically loaded footings the required area is :

                                           D+ L
                                  Areq =
                                            qa


In addition, mode code permit a 33% increase in allowable pressure when
the effect of wind “W” or earthquake “E” are included, in which case:

                             D+L+ W      D+L+E
                   A req =            or
                              1.33q a     1.33q a


Footing sizes are determined for UNFACTORED SERVICE




12.1 DESIGN RC. FOOTING BY SD (Cont.)
LOADS and SOIL PRESSURE.
 • A footing is eccentrically loaded if the supported column is not
   concentric with the footing area or if the column transmits at its
   juncture with the footing, not only a vertical load but also a bending
   moment (as shown in Figure 12.3).
 • If the resulting eccentricity e = M/P does not exceed the kern
   distance “k” of the footing area, the pressure on soil can be
   determined as (see Figure 12.3 (a)) :

                                  P Mc
                   q max/ min =     ±   --- (12.1)
                                  A   I




                                                                            7
12.1 DESIGN RC. FOOTING BY SD (Cont.)
The footing area is found by trial and error from the condition
that qmax ≤ qa
 • If the eccentricity falls outside the kern “k”, Equation 12.1 gives a
   negative value (tension) for soil pressure “q” along one edge of the
   footing. Because no tension can be transmitted at the contact area
   between soil and footing, Equation 12.1 is no longer valid and
   bearing pressures are distributed as in Figure 12.3b.
 • For rectangular footings of size l x b the maximum pressure can be
   found from :
                             2P
                     q max =          --- (12.2)
                             3bm

 • qmax must be less than the allowable pressure “qa”




12.1 DESIGN RC. FOOTING BY SD (Cont.)
Bending Moments, Bond, and Reinforcement
(แรงดัด, แรงยึดเหนี่ยว, และการเสริมเหล็ก)
 • In spreading footing, the critical section for bending moment of the
   footing is shown in Figure 12.1




            Fig. 12.1 Critical section for Consider Moment
                  and Development Length in Footing




                                                                           8
12.1 DESIGN RC. FOOTING BY SD (Cont.)
• Section “cd” has the highest bending moment. The moment about
  “cd” is caused by the upward pressure qu on the area to one side of
  the section (area “abcd”). The reinforcement perpendicular to that
  section, i.e., the bar running in the long direction, is calculated from
  this bending moment.
• Likewise, the moment about section “ef” is caused by the pressure qu
  on the area “befg”, and the reinforcement is in the short direction, i.
  e., perpendicular to “ef”, is calculated for this bending moment.
• In square footing, it is customary to determine As based on the
  average depth “d” and to use the same arrangement of reinforcement
  for both layers.




12.1 DESIGN RC. FOOTING BY SD (Cont.)
 • In rectangular footing, the reinforcement in the long direction is
   uniformly distributed over the pertinent (shorter) width. In locating
   the bars in the short direction, one has to consider that the support
   provided to the footing by the column is concentrated near the
   middle. Consequently, the curvature of the footing is sharpest, i.e.,
   the moment per meter largest, immediately under the column, and it
   decreases in the long direction with increasing distance from the
   column.

For this reason, a larger steel area per longitudinal meter is needed in
the central portion than near the far ends of the footing. E.I.T. Code
provides the following :




                                                                             9
12.1 DESIGN RC. FOOTING BY SD (Cont.)
For reinforcement in the short direction, a portion of the total
reinforcement (given in Eq. 12.3) shall be distributed uniformly over a
band width (centered on the centerline of the column or pedestal) equal
to the length of the short side of the footing. The remainder of the
reinforcement required in the short direction shall be distributed
uniformly outside the center band width of the footing

         Reinforcem ent in Band Width                 2
                                               =             --- (12.3)
   Total Reinforcem ent in Short Direction         (S + 1)


In Equation 12.3, “S” is the ratio of the long side to the short side of
the footing.




12.1 DESIGN RC. FOOTING BY SD (Cont.)
Shears on Footing
• In single footings the effective depth “d” is mostly governed
  by shear.
• There are 2 different types of shear acting on footing :
       1. Two-way or punching shear and
       2. One-way or beam shear.




                                                                           10
12.1 DESIGN RC. FOOTING BY SD (Cont.)
Punching Shear :
• If the footing is failed by punching shear, the fracture takes the form
  of the truncated pyramid shown in Figure 12.2. (or of a truncated
  cone for a round column), with sides sloping outward at an angle
  approaching 45 degree.
• The average shear stress in the concrete that fails in this manner can
  be taken as that acting on vertical planes laid through the footing
  around the column on a perimeter a distance “d/2” from the face of
  the column (vertical section through “abcd” in Figure 12.3).




12.1 DESIGN RC. FOOTING BY SD (Cont.)




  Fig. 12.2 Punching shear failure in single footing   Fig. 12.3 Critical sections for shear

E.I.T. 1008-38 (วสท. 1008-38) allows the nominal punching shear of
concrete on the perimeter “abcd” as :
                              V c = φ 1 . 06 f c′ b 0 d
Except for column of very elongated cross section, for which
                                         ⎛     4 ⎞
                            V c = 0 . 27 ⎜ 2 +
                                         ⎜        ⎟ f c′ b 0 d
                                         ⎝     βc ⎟
                                                  ⎠




                                                                                               11
12.1 DESIGN RC. FOOTING BY SD (Cont.)
For cases in which the ratio of critical perimeter to slab depth, bo/d, is
very large :
                                    ⎛α d    ⎞
                       V c = 0 . 27 ⎜ s + 2 ⎟ f c′ b 0 d
                                    ⎜ b     ⎟
                                    ⎝ 0     ⎠

Where :      Vc   Nominal shear strength of concrete in (kg)
                   =
             fc′  Ultimate strength of concrete (ksc)
                   =
             b0   The perimeter “abcd” at distance (cm)
                   =
             d    Effective depth of footing (cm)
                   =
             βc   Ratio of the long to short sides of the column cross
                   =
                  section
             αs = 40 for interior loading, 30 for edge loading, and
                  20 for corner loading of a footing.




12.1 DESIGN RC. FOOTING BY SD (Cont.)
Beam Shear or One-Way Shear
 • This kind of shear is occurred at the distance “d” from face of
   column or section “ef” in Figure 12.3.
 • The nominal shear strength of concrete is equal to :
            Vc = φ 0.53 f c′ bd (approximate method) or
                   ⎛         ′ 176 ρ w Vu d ⎞
             Vc = φ⎜ 0.50 f c +
                   ⎜                        ⎟bd ≤ φ 0.93 f c′ bd
                                            ⎟
                   ⎝              Mu        ⎠

 Vu =   Total factored shear force at that section
    =   qu times footing area outside that section
  b =   width of footing at distance “d” from face of column
 Mu =   moment of Vu about “ef” and Vud/Mu must equal or less than 1.0.




                                                                             12
12.1 DESIGN RC. FOOTING BY SD (Cont.)
Footing on Piles
 • If pile is used, the spacing of each pile shall be at least 3 times the
   diameter of the pile.

                                3D min.


                           D

 • The depth of pile cap is usually governed by shear. In design footing
   on piles, both punching and beam shears and flexural moment need
   to be considered.




12.1 DESIGN RC. FOOTING BY SD (Cont.)
 • The CRITICAL SECTION of footing on pile is the same as that spread
   footing. The difference is that shears on caps are caused by
   concentrated pile reaction rather than by distributed bearing
   pressures.
 • This poses the question of how to calculate shear if the critical
   section intersects the circumference of one or more piles.
 • For this reason, ACI code 15.5.3 accounts for the fact that pile
   reaction is not really a point load, but rather distributed over the
   pile-bearing area. Correspondingly, for piles with diameters, “dp”, it
   stipulates as follows : (See Figure 12.4)




                                                                             13
12.1 DESIGN RC. FOOTING BY SD (Cont.)




                      Fig. 12.4 Reaction from Piles




12.1 DESIGN RC. FOOTING BY SD (Cont.)
(a) The entire reaction from any pile whose center is located dp/2 or
    more outside this section shall be considered as producing shear on
    that section.
(b) The reaction from any pile whose center is located dp/2 or more
    inside the section shall be considered as producing no shear on that
    section.
(c) For intermediate positions of the pile center, the portion of the pile
    reaction to be considered as producing shear on the section shall be
    based on straight-line interpolation between the full value at dp/2
    outside the section and zero at dp/2 inside the section.




                                                                             14
12.1 DESIGN RC. FOOTING BY SD (Cont.)
 • Punching shear must also be investigated for the individual pile.
   Particularly in caps on a small number of heavily loaded piles, it is
   this possibility of a pile punching upward through the cap that may
   govern the required depth. The critical perimeter for this action is
   located at a distance “d/2” outside the upper edge of the pile.
 • For relatively deep caps and closely spaced piles, critical perimeters
   around adjacent piles may overlap. In this case, the critical perimeter
   is also located that its length is a minimum, as shown in Fig. 12.5.




12.1 DESIGN RC. FOOTING BY SD (Cont.)
Some Requirements for Design Footing
 • Critical section of footing for punching shear is at “d/2” from face
   of column.
 • Critical section of footing for beam shear is at “d” from face of
   column.
 • Shear stress of concrete is calculated as :
                                      Vu
                               vu =
                                      b 0d

        when : Vu and b0 are considered at critical section.




                                                                             15
12.1 DESIGN RC. FOOTING BY SD (Cont.)
 • Shear stress of concrete (punching shear) shall not exceed :

                              1 . 06 φ f c′

 • If shear reinforcement is used, the shear stress of concrete plus
   shear reinforcement shall not exceed

                     1 . 06 φ f c′ ; φ = 0.85

 • Shear stress of concrete (one way or beam shear) shall not exceed

                     v c = φ 0 . 53 f c′   ; φ = 0.85




12.1 DESIGN RC. FOOTING BY SD (Cont.)
• Minimum Thickness of Footing : Minimum thickness of
  concrete above the steel reinforcement shall be at least 15
  cm. for spread footing and shall be at least 30 cm. for
  footing on piles.
• Covering of concrete in footing shall be at least 7.5 cm if the
  concrete is exposed to earth or increased to be 10 cm. in
  case of immersed in water at all time.




                                                                       16
12.1 DESIGN RC. FOOTING BY SD (Cont.)
 • Strength of Bangkok Clay : If no data for the strength of
   soil is available, the strength of Bangkok clay can be used as
   follows :
   (เทศบัญญัติ กทม. พ.ศ. 2522 -Bangkok Code 2522) :
   - If the depth of the soil is less than 7.0 m, the
     allowable friction of soil shall not exceed 600 kg/m2.
   - For the depth of soil higher than 7 m, the allowable
     friction of soil shall be used as :
              Friction (kg/m2) = 800 + 200L
       where L (m) is the length of pile that is longer than 7 m.




Example 12.1 Spread Footing
Given fc′ = 160 ksc and fy = 3000 ksc, allowable bearing pressure of soil
=10,000 ksc (with safety factor about 2-3). A column 0.35 x 0.35 m2
is used to support dead load of 16,000 kg. and live load of 20,000 kg.,
design spread footing by using strength design method.

Solution : Dead weight of footing is about 4-8%, in this case use 8% :
    Weight of footing = (L+D) × 0.08 = (20000 + 16000)(0.08)
                       = 2880 kg
       ∴Dead Load = 16000 + 2880 = 18,880 kg
           Live Load = 20,000 kg




                                                                            17
Note : The size of footing is determined based on working stress
method.
                                          D+L
         Size of footing             =                     (No factored was applied)
                                         Pr essure
                                         18800 + 20000
                                     =                 = 3 .88 m 2
                                             10000
                                     use footing 2 x 2 m2
 In design, factor load is applied to find pressure :
               1.4D + 1.7L 1.4 × 16000 + 1.7 × 20000 56400
  Pnet =                  =                         =      = 14100 kg/m 2
                   A                 2× 2              4

Note : In calculating moment or shear for spread footing, the weight
of footing is not included (use dead load of 16000 kg instead of
18880 kg.)




Critical Section of moment at face of column :
                  2.0
                                           M = bP net
                                                        (0 .825 )2
                                                            2
                  0.35
 2.0       0.35
                                              = 2.0(14100 )
                                                              (0.825 )2   = 9596 .8 kg-m
                                                                 2

                  0.35
                             0.825
                                           From M = Rnbd2
       0.825             0.825
                                                9596.8
                                                  0.9
                                                       × 100 = R n (200) d 2  ( )
                                     Try thickness of 30 cm, d = 30 – 8 = 22 cm.
               14100 kg/m2
                                              9596 .8 × 100
                                     Rn =                      = 11 .01 ksc
                                             0 .9 × 200 × 22 2




                                                                                           18
fc′ = 160 ksc and fy = 3000 ksc                                ρ = 0.004                      Rn = 11.47ksc
                 As = ρbd = 0.004 x 200 x 22 = 17.6 cm2
     Use 9 – DB16                  ∑As = 18.09 cm2                                      ∑0 = 45.26 cm

 In general, footing is governed by shears (punching or beam
 shears), this example is designed moment at the beginning which is
 not a good practice. In design footing, it is recommended to design
 shears first and followed by moment.
    Punching Shear: Critical section at d/2 from column
                                    d                              d   22
                                      = 11           35              =    = 11
                                    2                              2    2

                                                              57
                                   200          d
                                   cm.
                                                2                     Vdiagonal
                                                          d
                                                    200
                                         VPunching cm




Punching Shear : Critical section at d/2 from column
                                         35 cm
                      d                                        d 22
                        = 11                                     =   = 11
                      2                                        2   2



                200 cm                                                57 cm
                               d
                               2
                                                                                  Vdiagonal
                                         d
                  VPunching         200 cm


                ( 5600 - 0 . 57 × 0 . 57 × 14100
 v punching =                                    = 10 . 33 ksc
                          4 × 57 × 22
     v allow = φ1.06 fc′ = 0.85 × 1.06 160 = 11.39 ksc > 10.33 ksc                                      OK




                                                                                                              19
Beam Shear : Critical section at “d” from column
                         14100 × 2 × (0 . 825 − 0 . 22 )
        v diagonal =                                     = 3 . 878 ksc
                                 22 × 200
             v allow = φ 0 . 53 f c′ = 0 . 85 × 0 . 53 160

                      = 5 .698 ksc > 3 .878 ksc            OK

Development Length (ระยะฝงเพิ่ม) (See E.I.T. 4502)
 • For bar with diameter smaller than 36 mm :
                                     ⎛ π (1 .6 2 ) ⎞
                               0 .06 ⎜
                                     ⎜             ⎟ (3000 )
               0 .06 A b f y         ⎝     4 ⎟     ⎠           0 .06 (2 .01 )(3000 )
      l db   =               =                               =
                     f c′                   160                          160

      l db = 28 .61 cm < 30 cm                   Use 30 cm




 • For tension and has covering more than 2(db) = 2(1.6)
   = 3.2 cm., spacing more than 3db = 3(1.6) = 4.8 cm.,
   multiply factor = 1.0
 • But the development length must not less than
         0.11d b fy 0.11(1.6 )(3000 )
                   =                  = 41 .74 cm > 30 cm
             f c′          160

         Thus             use ld = 42 cm

 • In construction, ld (measured from face of column)
   ≈ 82.5 – 8 ≈ 74.5 cm > 42 cm OK.




                                                                                       20
 Check Dead Weight of Footing
          Weight of Footing = 2 x 2 x 0.3 x 2400 = 1880 kg
               (which is equal to the assumed dead load)
                                             P

                           0.35
                                                 c
                                                                    9 – DB 16 ตะแกรง
                                                                       1 – DB 16 รัดรอบ
           0.30                                                        0.22


                                        2.00
                    0.08




Example 12.2 :
Given a column of 0.50 x 0.50 m2, dead load = 150,000 kg, live load =
90,000 kg, 5-pile of 0.35 x 0.35 x 21 m is used to resist the specified
loads (for only this example). Design a footing to carry the loads
if fc′ =160 ksc and fy = 3000 ksc.
 Solution :                                                     b0= h+d = 0.5+0.8 = 1.3
                 0.50
                                                                                      0.35
                                                              0.5
                                                        0.5                           1.60
                                    d
                                     0.10
                                                                                      0.35
 0.375
                                                 0.35         1.60         0.35
                        0.35 x 0.35 x 21 m
     0.35 0.55                                                2.30
        0.90




                                                                                             21
 DL + LL = 150,000 + 90,000 = 240,000 kg
 Weight of footing 4% = 0.04 × 240,000 = 9600 kg
 Weight of footing + (DL + LL) = 9600 + 240000 = 249600 kg
 Total 5 piles are used, each pile will resist a safe load of
                              249600
                         =           = 49920 kg/pile
                                 5
 • A pile shall resist a safe load not less than 50000 kg (with a
   safety factor about 2-3)
 • Total loads : Pu = 1.4D + 1.7L
                    = 1.4(150,000) + 1.7(90,000) + 1.4(9600)
                    = 376,400 kg
                                 376440
       Pressure/pile         =          = 75288 kg
                                    5




• Use footing of 2.30 x 2.30 m2 , spacing of each pile is not less
  than 3D = 3(0.35) = 1.05 m       see figure :
• Try t = 100 cm                    d = 90 cm.
Punching Shear at “d/2” from face of column
                                 critical section

                              27.5 cm       45 cm = d/2


                         0%                 100 %

                                 35 cm          37.5 cm      face of column


                  pile                         55 cm

                                                          27.5
   Pressure of column will reduce to =                         xP = 0.786x75288
                                                           35
                                                                  = 59155 kg/pile




                                                                                    22
          4 × 59155
   vp =                        b0 = 90 + 50 (width of column) = 140 cm
             4b0d
           4 × 59155
      =                = 4 . 69 ksc
          4 × 140 × 90
                                         (value of d)
 v allow = 1.06 φ f c′ = 1.06 × 0.85 × 160 = 11.39 ksc > 4.69 ksc


• vd at distance 90 cm from face of column is longer than the
  distance from column to outer side of pile (55 + 17.5 = 72.5
  cm). Thus, beam shear is not occurred in the footing, i.e., no
  need to check beam shear.
• The result suggested that the trial thickness of “t” = 100 cm is
  too thick and can be reduced. Try t = 75 cm and use d = 65 cm




Punching shear: Critical section = 65/2 = 32.5 cm from face of
column.
   b0 = 50 + 65 = 115 cm          d = 65 cm   P = 75288 kg
                      4 × 75288
               vp =                 = 10 .07 ksc < 11 .39 ksc            OK
                      4 (115 )(65 )

Beam Shear : Critical section at d= 65 cm. from face of column.
                      critical section

                                         65 cm = d
                 7.5 cm
                               27.5 cm
                      0%                   100 %
                             35 cm           37.5 cm    face of column

                                             55 cm
                  pile




                                                                              23
                                   7 .5
                       Pmod =           × 75288 = 16133 kg
                                   35
                                   2 × 16133   2 × 16133
                            vd =             =           = 2 . 16 ksc
                                       bd       230 × 65
                        ′
  Vd-allow = 0 .53 φ f c = 0 .53 × 0 .85 × 160 = 5 .69 ksc > 2 .16 ksc                 OK

 Although vallow = 11.39 ksc is somewhat higher than vp = 10.07 ksc,
 it is not practical to design value of vp to close to vallow. In this case,
 use t = 75 cm and d = 65 cm.




 AS for Bending Moment                                2 columns
                                                                  Pressure of column
                                                                         (distance)

                                   Bending Moment = 2 × 75288 × 0.55
                                                  = 82816 kg-m
                                    M = Rnbd2 = Rn (230)(65)2
                                   82816 × 100
                   0.55 m                      = R n × 230 × 65 2
75288 kg x 2 columns                   0.9
                                            Rn = 9.47 ksc

     f c′ = 160 ksc ⎫
                    ⎬         ρ = 0.0035          R n ≈ 10 ksc > 9.47 ksc              OK
    f y = 3000 ksc ⎭




                                                                                            24
  As = ρbd = 0.0035(230)(65) = 52.3 cm2
  Use 11-DB25        ∑As = 54.01 cm2                             ∑0 = 86.4 cm

Development Length
                                           0.06
                                                  (π(2.5) ) (3000)
                                                         2

                          0.06A bf y                 4
                 l db =                =
                               f c′                  160
                     = 69.85 cm > 30 cm
           Use ldb = 70 cm                  OK
If the covering is not less than 2db = 2(2.5) = 5 cm and spacing of
bars is not less than 3db = 3(2.5) = 7.5 cm, the multiply factor is
1.0. Thus, ldb = 70 cm




But the development length shall be larger than
                      0 .11d b f y         0 .11(2 .5 )(3000 )
                                       =
                             f c′                  160
                                       = 65.22 cm < 70 cm
                          Thus, ld = 70 cm
In practice, it is found that the development length is
           l d = 115 − 25 − 10 = 80 cm             OK

Check Dead Load of Footing
          = 2 . 3 × 2 . 3 × 0 .75 × 2400 = 9522 kg < 9600 kg              OK




                                                                                25
            Check for Punching around the perimeter of Column
 - Critical section is at d/2 from the edge of pile
          Check by yourself !
                                         0.50


          115 – 25 - 10 = 80 cm
                                                       11 – DB 25#
                 0.10
                                                                     DB 25 รัดรอบ

                                                                0.65

                                                                0.10
  Safety load 50 tons
                                                              0.10
                        0.35      0.80          0.80   0.35




Example 12.3 :
Design footing to carry loads of 14,000 kg. (60% is live load and 40% is
dead load). The footing is constructed on Bangkok clay which has
C = 600 kg/m2. If a hollow pile of φ15 cm x 6.00 m long is used. Given
column section of 0.20 x 0.20 m, fc′ = 140 ksc and fy = 3000 ksc.
 Solution :
 Weight of footing ≈ 8%    0.08 x 14000 = 1120 kg
 Live load + Dead Load = 14000 + 1120 = 15120 kg

 A hollow pile of φ 0.15 x 6 m can resist a safe load of :
                          = πd x l x C = π(0.15)(6) x 600
                          = 1696 kg/pile
            ∴Total piles = 15120/1696 = 8.9             Use 9 piles




                                                                                    26
                                  d
                                          b0= 20+17.5 = 37.5 cm


                1.30             0.2
                                        0.2

                                0.40
                         0.15    0.50      0.50   0.15
                                        1.30

Total load = 1.4D + 1.7L
           = 1.4(0.4)(14000) + 1.7(0.6)(14000)+1.4(1120)
           = 23,688 kg
Pressure on each pile = 23688/9 = 2632 kg
Spacing of each pile = 3d = 3 x 15 = 45 cm                        Use = 50 cm
(Also See Figure)




Try footing of 1.30 x 1.30 m2
Thickness of 25 cm          d= 17.5 cm. (Assume that footing is not
serious contact to earth, however, for new code the footing on pile
shall have “d” at least 30 cm.

Bending Moment = 2632 x 3(piles) x 0.4 = 3158 kg-m

Punching Shear : Critical section at distance d/2 from column
   bo = 17.5 + 20 = 37.5 cm
          8(piles ) × 2632     8 × 2632
   vp =                    =
             4 × b0 × d      4(37 .5 )(17 .5 )
       = 8 .02 ksc < v c = φ 1 .06 140 = 10 .66           OK




                                                                                27
Beam Shear or Diagonal Shear :
             3(piles) × 2632 3 × 2632
       vd =                  =
                   bd          130 × 17.5
       v d = 3.47 ksc < v all = φ 0.53 140 = 5.33 ksc                 OK

**** The footing with t = 25 cm can resist punching shear and beam shear. ****

Find As for bending moment of 3158 kg-m
                 From                Mn = Rnbd2
                                   3158
                                         = R n (1.3 )(17 .5 )
                                                             2
                                                                          R n = 8.81 ksc
                                    0 .9
  Choose ρ = 0.004      Rn = 11.39 ksc > 8.81 ksc
  As = ρbd = 0.004(130)(17.5) = 9.1 cm2
       Use 9 - DB12     ∑As = 10.17 cm2      ∑0= 33.94 cm




Development Length :
                                              0.06
                                                     (π(1.2 ) ) (3000 )
                                                             2

                           0.06 A b f y                  4
                 l db =                   =
                                 f c′                    140
                        = 17 .2 cm < 30 cm             Use          l db = 30 cm

Multiply factor to include effect of covering, spacing, reinforcement
in the perpendicular direction (see ว.ส.ท. ขอ 4502) Use 1.0, Thus,
ld = ldb x 1.0 = 30 cm.
But ld must be at least :
              0.11d bf y       0.11(1.2)(3000)
                           =                   = 33.46 cm > 30 cm
                  fc′                 140

Choose ld = 33.46 cm       use 35 cm and the development length in
the footing is ≈ 65 – 10 –7.5 + 15 ≈ 62.5 cm. OK




                                                                                           28
 Check weight of footing :
        1.3 × 1.3 × 0.25 × 2400 = 1014 kg < 1120 kg OK
                                           0.20 x 0.20


                                                9 – DB 12 ตะแกรง
                                                                  DB 12 รัดรอบ

                                                   0.175 0.25


                                                         9 - เสาหกเหลี่ยมกลวง


              0.15   0.50           0.50    0.15
                             1.30




HOME WORK FOR CHAPTER 12 (SD)
   Please see details in handout.

Do not forget to submit home work for
              chapter 10




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