1.8 Spacing of Reinforcement and Concrete Cover

      The spacing of reinforcement and the concrete cover  should be sufficient to make concreting more easier; consequently, the concrete surrounding the reinforcement can be efficiently vibrated, resulting in a dense concrete cover which provides suitable protection of the reinforcement against corrosion.

1.8.1 Spacing of Reinforcement

      Figure 1.19 shows two reinforced concrete sections. The bars are placed such that the clear spacing s shall be at least equal to the maximum diameter of the bars, or 25 mm, or 1.50 times maximum size of aggregate, whichever is greatest, according to the Egyptian Code. Vertical clear spacing  between bars, in more than one layer, shall not be less than 25 mm.

            FIGURE 1.19  Spacing of steel bars (a) in one row or (b) in two rows.

1.8.2 Concrete Cover

      The specified minimum concrete cover for different structural members, according to their degree of exposure, is given in the Egyptian Code, Table 4-13. Concrete cover for beams is equal to 25 mm for main bars and 20 mm for stirrups and that for slabs is equal to 15 mm, when concrete is not exposed to weather or in contact with ground.

1.8.3 Number of  Steel Layers and Overall depth of Concrete Section

      The general equation for the required width of a concrete section  is as follows:

   

(1.55)

            The total depth t is equal to the effective depth d plus the distance from the centroid of the tension reinforcement to the extreme tension concrete fibers, which depends on the number of layers of the steel bars. In application to the section shown in Fig. 1.19a,

 

    (1.56)

 

      for one row of steel bars and

     (1.57)

 

      for two layers of steel bars, Fig. 1.19b. The overall depth t shall be increased to the nearest 5 cm. If No. 8  (25 mm) or smaller bars are used, a practical estimate of the overall depth can be made as follows:

 

               t = d + 50 mm, for one  layer  of steel bars

               t = d + 75 mm, for two layers of steel bars

Example 1.4:

      For the cantilever beam shown in Fig. 1.20, if DL = 13.5 kN/m' (including own weight) and LL = 35 kN, it is required to:

      a.   Design the beam section for a minimum depth when b = 250 mm.

      b.   Design the beam section for a minimum depth when b = 120 mm.

      c.   Design the beam section for an effective depth d = 450 m when b = 250 mm.

      d.   Design the beam section for an overall depth t = 700 m when b = 120 mm.

            Given:  = 25 N/mm2 and  = 280 N/mm2.

 

            FIGURE 1.20  Example 1.4.

 

Solution:

      The ultimate moment  as specified by the Egyptian Code (where   and  are the dead and live service moments, respectively) is to be:

    150 kNm.

 

Part a:

         Enter Table A.1 with fcu = 25 N/mmand  = 280 N/mm2  and obtain

        

 = 0.534 and   = 196.7

         Then, calculate  and  as follows:

 

        414 mm

 

       mm2

 

      For  = 1842 mm2, different choices of steel bars may be selected as follows:

Steel Bars

Area of steel, mm2

f  20

1884

f  25 1960
f  22 1960
f 16 1800
2 f 25 + 2 f 22 1740

 

      The area of steel bars must be closest to the required steel area. If  2 f 25 plus 2 f 22 are chosen, As = 1740 mm2, which is 102 mm2 less than the required area of 1842 mm2. But since the overall depth t may be increased a fraction of 50 mm, the actual effective depth will be a little greater than the calculated dmin, consequently reducing the required As.

            The 2 f 25 plus 2 f 22  would have to be placed in  one  row  as 250 mm width is sufficient. Calculating the required width to place 2 f 25 plus 2 f 22 in one layer:

= 2 (f25 + 22f22) + 3s + 2fstr + 2c

                              = 2 (25 + 22) + 3 25 + 2 8 + 2 25 = 235 mm

      which is less than b = 250 mm. The overall depth t, is then computed from:

 

t = d + 0.5f25 +f str + c

                          = 414 + 0.5 25 + 8 + 25 = 461.5 mm;  say 500 mm

      The actual effective depth d = 500 -  50 = 450 mm

 

      which is greater than the calculated d of 414 mm. Because of the small variation, reduction in the required steel area can be approximated by the ratio of the calculated d to the actual d.         As     actually needed is as follows

    1693 mm2

 

      which is less than 1740 mm2 (2 f 25 plus 2 f 22) provided, Fig. 1.21.

  FIGURE 1.21  Example 1.4, part a.

Part b:

 

      The minimum effective depth that correspond to b = 120 mm equals 597.5 mm. The area of steel As required equals Asmax  or 1276 mm2. If  4 f 20 is chosen, As = 1256 mm2, which is 20 mm2 less than 1276 mm2. If the steel bars are placed in one row:

    = 4 20 + 3 25 + 2 8 + 2 25 = 211 mm

 

      which is greater than b = 120 mm, therefore, the steel bars have to be placed in two rows as 120 mm width is not sufficient. The overall depth t is thus,

                        t = 597.5 + 25 + 8 + 20 + 0.5 25 = 663 cm ; say 700 mm

      The actual  d = 700 - 75 = 625 mm

 

  FIGURE 1.22  Example 1.4, part b.

 

Part c:

      First calculate K1  from:

 

   

which results in K1 = 0.581.

 

      Enter Table B.2 with K1 = 0.581, then traverse horizontally to  = 25, and finally obtain

 

    K2 = 208. Then, 1630 mm2  (Use 3 f 22 + 3 f 19)

Asmin = the least of  :                                             

    442 mm2 

 

    mm2

                        1.3  = 1.3   1630 = 2119  mm2 

      But not less than                

     mm2 

Asmin  = 442 mm2                                        Asmax = 1968.75 mm2 

 

Here,  1630 mm2 which is greater than Asmin   and less than .

       163 mm2  (Use 2 f 12)

Part d:

If steel is assumed to be placed in two layers as 120 mm width is not sufficient.

      d = 600 -  75  = 525 mm

      First calculate K1  from:

 

    which results in K1 = 0.468.

 

              Enter Table B.2 with K1 = 0.468, the first value of K1  (that correspond to  = 25) is 0.534 which is greater than 0.468. This implies that  is required. Enter Table C.3 (where  0.10) and obtain K2 = 201.8  and a = 0.225.

    1416 mm2 

 

    319 mm2    

      Since  is less than 0.20,

 

          \  .

 

1.9 Flanged Sections

      Concrete floor slabs and beams are normally tied together by means of stirrups and bent-up bars if any and then are cast form one mass of concrete. Such a monolithic system will act integrally i.e., it is allowed to assume that part of the slab acts with the beam and they form what is known as a flanged beam, Fig. 1.23.

     FIGURE 1.23  Slab-beam floor system.

      The part of the slab acting with the beam is called the flange, and it is indicated in Fig. 1.24a by the area Bts. The rest of the section confining the area (t-ts)b is called the stem or web. As Fig. 1.24b indicates, in an I-section there are two flanges, a compression flange, which is actually effective, and a tension flange, which is ineffective as it lies below the neutral axis and is thus neglected completely. Therefore, the design of an I-section is similar to that of a T-section.

                                                                               FIGURE 1.24  (a) T-section and (b) I-section.

1.9.1 Effective Flange Width, B

      As Fig. 1.25 indicates, the compressive stresses, in a T-section,  are at a maximum value at points adjacent to the beam and decrease approximately in a parabolic form to zero at a distance x from the face of the beam. Stresses also vary vertically from a maximum at the top fibers of the flange to a minimum at the lower fibers of the flange.

                                                                                FIGURE 1.25  Effective flange width, B.

      As a means of simplification, rather than varying with distance from the web, an effective width B of uniform stress may be assumed. The effective width B is a function of span length of the beam and depends on:

      1.   Spacing of beams

      2.   Width of web of beam

      3.   The ratio of the slab thickness to the total beam depth

      4.   End conditions of the beam (simply supported or continuous)

      5.   The way in which the load is applied (distributed load or point load)

      6.   The ratio of the length of beam between points of zero moment to the width of the web and the distance between webs.

      1.   Spacing of beams

      2.   Width of web of beam

      3.   The ratio of the slab thickness to the total beam depth

      4.   End conditions of the beam (simply supported or continuous)

      5.   The way in which the load is applied (distributed load or point load)

      6.   The ratio of the length of beam between points of zero moment to the width of the web and the distance between webs.

 

      The Egyptian Code prescribes that the effective flange width B of a T-section, as in Fig. 1.26, shall be taken as the web width b plus the effective overhanging flange sides  x1    and x2. Thus,  

 

B = b + (x1   + x2)                                                       (1.58)

      where x1  + x2  equal the least of:

   

     (1.59a)

    or when ts1    = ts2

     (1.59b)

   

     (1.59c)

        where L2 is the distance between the points of zero moments. For a simply supported beam, the distance L2 referred to above is just the span distance between centers of supports. For beams continuos from one end and simply supported from the other end, the distance L2  may be taken as 0.80 times the span distance between centers of supports.  For beams continuos from both ends, the distance L2  may be taken as 0.70 times the span distance between centers of supports. ts1 and ts2  are the thicknesses of the right and left slabs and S1  and S2  are the clear distances to the next right and left beams.

            FIGURE 1.26  Effective flange width of T-beams.

 

      To increase the compression force capacity of isolated rectangular beams, concrete overhanging flange sides are added, Fig. 1.27. This isolated T-shaped section is most commonly used  as prefabricated units. The Egyptian Code specifies the size of isolated T-sections as:

                   

   

and    

 (1.60)

            FIGURE 1.27  Isolated T-shaped sections.

 

 

      The end beam of a slab-beam girder floor is called a spandrel beam. The beam joins the slab from only one side.

     FIGURE 1.28  Effective flange width of L-beams.

      The Egyptian Code specifies that the effective flange width B shall be taken as the web width b plus the effective overhanging flange width x1. Thus,

                       B = b + x1                                                      (1.61)

      where x1   equals the least of:

                    

    (1.62a)
    (1.62b)
    (1.62c)

                                                                                                                         

            The design of inverted L-shaped sections may approximately follow the same procedure of T- and I-shaped sections but with employing the respective effective width B.

1.10 Flexural Design of Reinforced Concrete Flanged Sections

      In flanged sections, it can be seen that a large area of the compression flange, forming a part of the slab, is effective in resisting a great part or all of the compressive force due to bending. If the section is designed on this basis, the depth of the web will be small; consequently the moment arm yct  is small, resulting in a large amount of tension steel which is not favorable.

            Because of the large area of the compression flange, the design of a T-section does not need, in most practical cases, to consider a doubly reinforced section. But, in case of precast units, when the width of the flange is small and the effective depth is limited, compression steel may be added.

1.10.1 Effective Depth d

      In many cases, the effective depth d can be known based on the flexural design of the section at the support in a continuous beam, e.g. section 2-2 in Fig. 1.29a.  The section at the support is subjected to a negative moment, the slab being under tension and ignored, and the beam width is that of the web b.

            FIGURE 1.29 Slab and beam systems

      Iýýýf the effective depth d of section 1-1 in Fig. 1.29b is not known, an approximate effective depth can be obtained by considering a rectangular section with a reduced width , Fig. 1.30. The reduced width  is greater than the width of the web b and less than the effective flange width B. A reasonable choice of  ratio varies between  and  depending on the applied moment and shear requirements. If shear is high or a small amount of  is required, a greater depth is needed; i.e.  approaches . For shallow sections, a higher ratio is used; i.e. the ratio  may approach . After determining the ratio , the next step is to estimate the effective depth using equation

       (1.63)

 

            FIGURE 1.30 Reduced width of T-section.

      It is also possible to estimate the effective depth d using

                                                                                                                             (1.64)

      Table D.1 gives values for K1min for all grades of steel and a range of commonly used concrete strengths.

1.10.2 Design of T- and I-Sections

      As already stated in Section 1.9, the design of an I-section is similar to that of a T-section. When the depth of the equivalent stress block a lies within the flange; i.e. a ts, the section behaves as a rectangular section with the beam width equal to the flange width. Otherwise, if a is greater than ts, a T-section design is a must.

      If a ts , the section may be designed as a rectangular section of width B, Fig. 1.31.

            FIGURE 1.31. Rectangular section behavior.

      The design may be commenced by assuming that a   ts. Taking moments of forces about the tension steel, we have

                                                                      (1.65)

      solution of the quadratic equation yields a. If a ts as assumed, the tension steel  can be found using

                                                                                                                   (1.66)

 When the depth of the equivalent stress block is greater than the flange thickness, i.e. a > ts, the section may be designed using the equations for a doubly reinforced beam, as follows. As Fig. 1.32 indicates, the tension steel As may be considered to be divided into an area As1, which resists the compression in the concrete over the web, and an area As2  or Asf, which resists the compression in the concrete in the overhanging of the flange.

    FIGURE 1.32. Design of a T-Section when a amax.

      Assuming that the tension steel is yielding, considering equation T2 = C2, then

                                                                                                            (1.67)

or

                                                                                                                (4.68)

      The ultimate moment of the section is the sum of the two moments Mu1 and Mu2:

                                                                                                                           (1.69)

      where

                                                                               (1.70)

      and

                                                                       (1.71)

      solving the quadratic equation yields a.

 

    If a amax 

      This implies that the section is adequate without . Considering equation, T1 = C1, then

                                                                                                                      (1.72)

      or                          

                                                                                                                          (1.73)

      The total steel used in the T-section is

                                                                                                             (1.74)

If a > amax 

    This implies that  is necessary, Fig. 1.33. Here also,

                                                                                                                (1.73)

      The ultimate moment of the section is the sum of the three moments Mu1, Mu2    and Mu3:

                                                                                                                (1.75)

      where

                                                                                (1.70)

                                                                  (1.76)

      and

                                                                                                                     (1.77)

      and

                                                             (1.78)

      The total steel used in the T-section is

                                                                                         (1.79)

      If , then

                                                                                                                    (1.80)

            FIGURE 1.33. Design of a T-Section, a > amax.

Egyptian Code Solution

      The Egyptian Code allows another approach to determine As when a > ts. Ignoring the compression in the web part below the flange as shown in Fig 1.34, the tension steel can be obtained from:

                                                                                                                       (1.81)

      giving

                                                                                                                           (1.82)

 

      FIGURE 1.34. Design of a T-section.

Example 1.5

      A T-beam section with B = 1000 mm, b = 250 mm and ts = 100 mm is to have a design flexural strength Mu of 450 kNm. If  fcu = 25 N/mm2 and steel 360/520, calculate the required steel area when:

      

      a.   d = 550 mm

      b.   d = 440 mm

      c.   d = 400 mm

  FIGURE 1.35. Example 1.5.

 

Solution:

      Assume a ts. Then,

     

      giving :

     

Part a: d = 550 mm

     

      solution of the quadratic equation gives a = 79 mm which is less than  ts. Therefore, the section will behave as a rectangular section. For equilibrium, C = T, we have

     

      As = 2818 mm2; use 6  25.       

       mm2; use 3  12.

Part b: d = 440 mm

     

      solution of the quadratic equation gives a = 104 mm which is greater than ts. Therefore, a T-section design is required. With reference to Fig. 1.32, for equilibrium, , hence from

     

      we have

      kNm

      giving

      kNm

     

      Solving the quadratic equation yields a = 116 mm and c = 145 mm.

      But cmax = 0.44 d = 0.44 x 440 = 193.6 mm  which  is  greater  than c.  This  implies  that the section is adequate without .

      For equilibrium, T = C1 + C2, hence from

       kN

      we have

     

      giving As = 3710 mm2

Another Solution

      For equilibrium, C1 = T1, we can put

       which results in As1 = 1034 mm2

      Also, for equilibrium, C2 = T2, we can put

     

      Asf = As2  = 2675 mm2

      A= As1 + As2 = As1 + Asf = 1034 + 2675 = 3709 mm2

      As1 = As - Asf = 1034 mm2      which is less than

      Asmax =  m max b d = 5 x 10-4 x 25 x 250 x 440 = 1375 mm2 

      and greater than Asmin =  mmin b d =  x 250 x 440 = 336.11 mm2

Egyptian Code Solution

      Upon neglecting the compression in the web part below the neutral axis, we have

       mm2

 

Part c: d = 400 mm

     

      solution of the quadratic equation gives a = 118.5 mm which is greater than ts. Therefore, a T-section design is required. For equilibrium, , hence from

       kN

      we have

      kNm

      giving

      kNm

     

      Solving the quadratic equation yields a = 182 mm and c = 227.5 mm.

      But cmax = 0.44 d = 0.44 x 400 = 176 mm which is less than c. This implies that compression steel is required, Fig. 1.33. 

      Here, also kNm

     

      Asf = As= 2675 mm2

      amax = 0.80 cmax = 140 mm

     

      kNm

      As1 = Asmax = mmax b d = 5 x 10-4 x 25 x 250 x 400 = 1250 mm2 

      kNm

      Since  which is less than 0.15 for steel 360/520, this implies that .

     

       = 255.56 mm2 

      As = As1 + As2 + As3 = Asmax + Asf +  = 1250 + 2675 + 255.56 = 4180.6 mm2

Egyptian Code Solution

      Upon neglecting the compression in the web part below the neutral axis, we have

     mm2

 

 

             = 0.10 As = 410.7 mm2 or more.

 

Example 1.6

      In a slab-beam floor system, the smallest effective flange width B was found to be 1450 mm, the web width b was 250 mm and the slab thickness was 120 mm, Fig. 1.36a. Design a T-section to resist an ultimate external moment Mu of 240 kNm. Given: fcu = 20 N/mm2 and steel 240/350.

            FIGURE 1.36. Example 1.6.

Solution:

      Since the effective depth is not given, a reduced flange width is assumed; say . Thus,  mm.

      That is, an equivalent rectangular section, Fig. 1.36b, can be chosen with Br = 580 mm and

     

      which results in d = 380 mm. Assume two rows of steel bars (to be checked later).

      t = 380.8 + 75 =  455.8 mm; say t = 500 mm

      actual d = 500 - 75 = 425 mm

      Proceed as in the previous example to calculate As.

      Assume a ts

     

     

      a = 46 mm which is less than ts

            For equilibrium, T = C, we have

     

       mm2 choose 6 f 25 (2950 mm2)

       should not be less than 0.10 As, use 3 f 12.

 

            FIGURE 1.37. Example 1.6.

1.11 Design of T and I Sections using Design Aids

      Once b and d are known, the design of a T-section simulates that of a rectangular section when    a , with b equals B, Fig. 1.38a. Otherwise, if a >  as in Fig. 1.38b, the code allows the neglecting of compression in the web part below the flange as shown in Fig 1.38c.

      First  calculate the ratio   and K1  from 

                                                                                                                                (1.67)

      Then,  with  the  known  value  of  , determine the design table that corresponds (Tables E.1 through E.5). Traverse vertically to the  value and  also to  value, then  horizontally  to  the K1  value,  and  finally  obtain the value K2 to be used. Then, calculate  As  from

                                                                                                                                     (1.68)

      If a >  take the value K2  that correspond to a = .

      FIGURE 1.38. Design of T and I sections

Example 1.7:

      In a slab-beam floor system, the smallest effective flange width B was found to be 1450 mm, the web width b was 250 mm and the slab thickness was 120 mm. Design a T-section to resist an ultimate external moment Mu of 240 kNm. Given: fcu = 20 N/mm2 and steel 240/350.

Solution:

      = 580 mm

 

       = 380 mm

 

      Assume two rows of steel bars (to be checked later)

 

      t = 380 + 75 = 455 mm; say t = 500 mm and therefore, actual d = 500 -75 = 425 mm

       =  and  which results in K1 = 1.0446

      Enter Table E.1 and obtain K2 = 197.3  and a = 0.40 ts = 48 mm. Then,

       2862 mm2  (Use 6 f 25) and  = 286.2 mm2     (Use 3 f 12)

Example 1.8:

         A T-beam section with B = 1000 mm, b = 250 mm and ts = 100 mm is to have a design flexural strength Mu of 450 kNm. Using fcu = 25 N/mm2 and steel 360/520, calculate the required steel area when d = 550, 440 and 400 mm.

Solution:

a.   d = 550 mm

       =  and  which results in K1 = 0.8198 Enter Table E.3 and obtain

       K2 = 291.2  and a = 0.80 ts = 80 mm. Then,

       = 2810 mm2  (Use 6  25) and  = 281 mm2     (Use 3  12)

 

b.   d = 440 mm

       =  and  which results in K1 = 0.6558

      Enter Table E.3, since a > ts, take K2 = 277.8  at a = ts. Then,

       = 3682 mm2  and  = 368.2 mm2

 

c.   d = 400 mm

       =  and  which results in K1 = 0.596

      Enter Table E.3, since a > ts, take K2 = 273.9 at a = ts. Then,

       = 4107 mm2  and  = 410.7 mm2

 

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