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.
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/mm2 and = 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 |
6 f 20 |
1884 |
4 f 25 | 1960 |
5 f 22 | 1960 |
9 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 + 2f_{str} + 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 |
A_{smin} = 442 mm2 A_{smax} = 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,
\ . |
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.
T- and I -Shaped Sections
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 t_{s}_{1} = t_{s2} |
(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.
Isolated T- Shaped Sections
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.
Inverted L-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.
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.
T-Section Behaves as a Rectangular Section
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)
T-Section Behavior
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
A_{sf} = A_{s2} = 2675 mm2
A_{s }= A_{s1} + A_{s2 }= A_{s1} + A_{sf}_{ }= 1034 + 2675 = 3709 mm2
A_{s1} = A_{s} - A_{sf} = 1034 mm2 which is less than
A_{smax} = m _{max} b d = 5 x 10-4 x 25 x 250 x 440 = 1375 mm2
and greater than A_{smin} = m_{min} 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
A_{sf} = A_{s}_{2 }= 2675 mm2
a_{max} = 0.80 c_{max} = 140 mm
kNm
A_{s1 }= A_{smax} = m_{max} 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
A_{s} = A_{s1} + A_{s2} + A_{s3} = A_{smax} + A_{sf} + = 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