Chapter 13 Reinforced Concrete Beams Concrete is a material strong in its resistance to compression, but very weak indeed in tension. good concrete will safely take a stress upwards of 7 N/mm 2 in compression, but the safe stress in tension is usually limited to no more than one-tenth of its compressive stress. It will be remembered that in a homogenous beam the stress distribution is as shown in figure 1a, and in the case of a section symmetrical about the X-X axis, the actual stress in tension equals the actual stress in compression. If such a beam was of ordinary unreinforced concrete, then the stresses in tension and in compression would of necessity have to be limited to avoid overstressing in the tension face, while the compressive fibres would be taking only one-tenth of their safe allowable stress. The result would be most uneconomical since the compression concrete would be understressed. In order to increase the safe load-carrying capacity of the beam, and allow the compression concrete to use to the full its compressive resistance, steel bars are introduced in the tension zone of the beam to carry the whole of the tensile forces. This may, at first, seem to be a little unreasonable for the concrete is capable of carrying some tension, even though its safe tensile stress is very low. In fact it will be seen in figure 1b that the strain in the steel and in the concrete at are the same. b f d N X X d 1 st f (a) (b) Strain in steel and in concrete at level - Figure 1: Concrete beams: (a) homogeneous beam: (b) beam with steel bars Chapter 13: Reinforced concrete beams Page 1
Therefore if the stress in steel is 140 N/mm 2, the concrete would be apparently stressed to 1/m times this and, since m, the ratio of the elastic moduli of steel and concrete, is usually taken as 15, the tensile stress in concrete would be in excess of 9 N/mm 2. The concrete would crack through failure in tension at a stress very much lower than this and thus its resistance to tension is disregarded. 12.1 Basis of Design Elastic theory Method In the design of reinforced concrete beams the following assumptions are made: Plane sections through the beam before bending remain plane after bending. The concrete above the neutral axis carries all the compression. The tensile steel carries all the tension. Concrete is an elastic material, and therefore stress is proportional to strain. The concrete shrinks in and thus grips the steel bars firmly so that there is no slip between the steel and the surrounding concrete. s the concrete below the neutral axis (N) is ignored from the point of view of its tensile strength, the actual beam shown in Fig. 2(a) is, in fact, represented by an area of concrete above the neutral axis carrying compression, and a relatively small area of steel taking tension as in Fig. 2(b). Thus the position of the neutral axis will not normally be at d 1 /2 from the top of the beam, but will vary in position according to the relation between the amount of concrete above the N and the area of steel. The distance from the top face of the beam to the N which indicates the depth of concrete in compression in a beam is denoted by. b b N N d 1 d 1 st st d 1 (a) (b) Figure 2: Elastic Theory Method: (a) beam cross-section; (b) representation of tensile strength Chapter 13: Reinforced concrete beams Page 2
12.2 Critical rea of Steel In a beam of a given size, say 200mm wide with a 300mm effective depth, the area of steel could be varied it could be 1000mm 2 in area, or 2000mm 2, etc. and each addition of steel would make for a stronger beam, but not necessarily for an economical one. If, for example, only a very small steel area is included, as in Fig. 3(a), and this small steel content is stressed up to its safe allowable value, then the concrete will probably be very much understressed, so that the concrete is not used to its full advantage. From this point of view, an uneconomical beam results. (a) b (b) Strain in concrete in compression N d 1 d 1 (c) Strain in steel (d) Figure 3: Critical area of steel: (a) small steel area; (b) large steel area; (c) normal reinforced concrete beam; (d) normal strain diagram If on the other hand, an overgenerous amount of steel is included, as in Fig. 3(b), then stressing this steel to its safe allowable value will probably result in overstressing the concrete. Thus, to keep the concrete stress to its safe amount, the stress in the steel will have to be reduced, and again the result is an uneconomical section. Chapter 13: Reinforced concrete beams Page 3
Given that m= 15, there will be only one critical area of steel for a given size of beam which will allow the stress in the concrete to reach its permissible value at the same time as the stress in the steel reaches its maximum allowable value. In reinforced concrete beam calculations the designer is nearly always seeking to choose a depth of beam and an area of steel which will allow these stresses to be attained together thus producing what is called a balanced or economical section. The breadth of the beam is normally chosen from practical considerations. If the two given safe stresses are reached together, then the required area of steel will always remain a constant fraction of the beam s area (bd 1 ) and the factor /d 1 will also have a constant value. Figure 3(c) shows a normal reinforced concrete (RC) beam and Fig. 3(d) the strain diagram for the beam. From similar triangles it will be seen that Equation 1 But stress/strain for any material is E (Young s modulus) for the material. Therefore strain = stress/e. Let p cb represent the permissible compressive stress for concrete in bending in N/mm 2 and let p st represent the permissible tensile stress in the steel in N/mm 2. Thus from equation (1) But E s /E c = m, modular ratio, therefore Equation 2 Equation (2) shows that and d 1 will always be proportional to each other when m, p cb,and p st are constant. From equation (2) Chapter 13: Reinforced concrete beams Page 4
So Equation 3 Note: is the distance down to the neutral axis from the compression face of the beam, and is expressed in millimetres. s is, however, seen to be a constant fraction of d 1 for fixed values of m, p cb and p st, it is frequently useful to know the value of the constant /d 1, and this constant is usually denoted by n, i.e. s is equal to from equation(3), then Equation 4 For the values m = 15, p cb = 7 N/mm 2, p st = 140 N/mm 2 and = 0.43d 1. Figure 4 shows that the stress in the concrete from the top of the section to the N follows the normal distribution for ordinary homogeneous sections, and the total compression force an average rate of N/mm 2. i.e. an area b of concrete, taking stress at The tension, however, will now all be taken by the steel and will be, i.e. an area of steel st taking stress at p st N/mm 2. These two internal forces T and C are equal and opposite in their direction, and they form an internal couple acting at apart, where, is the lever arm, denoted by l 2. Chapter 13: Reinforced concrete beams Page 5
p cb 1 / 3 C p cb b d 1 l a = lever arm = d T = st p st Figure 4: Stress within concrete This internal couple or moment of resistance may thus be written as follows: In terms of concrete Equation 5 In terms of steel Equation 6 But just as /d 1 is a constant n, similarly is also a constant, say j, and so the lever arm l a may be expressed as jd 1. For the values m = 15, p cb = 7 N/mm 2, p st = 140 N/mm 2, and l a = 0.86d 1 Substituting the constants n and j into equation (5), Chapter 13: Reinforced concrete beams Page 6
s, n and j are all constants, then has a constant value which may be called R, and therefore Equation 7 Using the values m = 15, p cb = 7 N/mm 2, p st = 140 N/mm 2 and Hence and Equation 8 s then the required area of steel is Equation 9 Note: s the area of steel is a constant fraction of the area b x d 1 (i.e. st = bd 1 constant), this relationship may be specified as the percentage area of steel, p, where or From equation (9) But Therefore Chapter 13: Reinforced concrete beams Page 7
So p = percentage of steel Equation 10 Using the values m = 15, p cb = 7 N/mm 2, p st = 140 N/mm 2 Hence It must be remembered, however, that there is a wide range of concrete strengths, hence it is not advisable to memorize the numerical values of these design constants. They may be derived, quite easily, from From all the above it will be seen that in designing simple reinforced concrete beams the procedure is as follows: 1. Calculate maximum bending moment, M max, in Nmm. 2. Choose a suitable breadth of beam, b. 3. Calculate effective depth to centre of steel 4. Determine area of steel required from or Chapter 13: Reinforced concrete beams Page 8
EXMPLE 1: Design a simple reinforced concrete beam 150 mm wide to withstand a maximum bending moment of 20 x 10 6 N mm using the following permissible stresses: p cb = 7 N/mm 2, p st = 140 N/mm 2 and m = 15. nswer Solution by Design Constants lternative Solution From equation (3) From equation (5) and substituting 0.43d 1 for From equation (6) Chapter 13: Reinforced concrete beams Page 9
Required d1 = 321mm Unit 48: Structural Behaviour and Detailing for Construction Use five 12mm diameter bars (area = 5 113.1 = 566mm 2 ). The overall depth d is determined as follows. 150mm Overall Depth = 321 + ½ 12 + 25 = (say) 325mm ½ 12mm 25mm Figure 5: Example 1 The effective depth d 1 is measured from the top face to the centre of the steel bars. The bars themselves need an effective cover of at least 25mm, so to determine the overall depth d it is necessary to add (half the bar diameter + 25mm) to the effective depth d 1 (see Fig. 5) In this case overall depth d = 321 + 6 + 25 = 352 mm. Say 150 mm x 355 mm overall with five 12 mm diameter bars. BS 5328: 1991, in its Guide to specifying concrete, lists four types of concrete mix (see Part 1, Clause 7.2) and gives detailed methods for specifying concrete mixes in Part 2. For the purpose of exercises in this chapter the permissible stresses in compression due to bending, p cb, may be taken as follows: 10.0 N/mm 2 for 1:1:2 mix (or C30 grade) 8.5 N/mm 2 for 1:1.5:3 mix (or C25 grade) 7.0 N/mm 2 for 1:2:4 (or C20 grade) Chapter 13: Reinforced concrete beams Page 10
The permissible tensile stress, p st, for steel reinforcement may be taken as 0.55f y, which for hot rolled mild steel bars is generally accepted to be 140 N/mm 2. EXMPLE 2: simple supported reinforced concrete beam is to span 5m carrying a total uniform load of 60 kn inclusive of self-weight, and a point load of 90 kn from a secondary beam as shown in Fig. 6. The beam is to be 200mm wide. Choose a suitable overall depth and area of tensile steel reinforcement for the maximum bending moment. ssume: p cb = 8.5 N/mm 2, p st = 140 N/mm 2, m =15. 90kN 60 kn 1 m 4 m 5 metres Figure 6 nswer See Lecturer s solution in class EXMPLE 3: simply supported beam is to span 3.6 m carrying a uniform load of 60 kn inclusive of self -weight. The beam is to be 150mm wide, and the stresses in steel and concrete respectively are to be 140 N/mm 2 and 10 N/mm 2 (m = 15). Determine the constants n, j, R and p, and choose a suitable effective depth, overall depth, area of steel anumber of steel bars for the beam. nswer See Lecturer s solution in class Chapter 13: Reinforced concrete beams Page 11