©COMPUTERS AND STRUCTURES, INC., BERKELEY, CALIFORNIA AUGUST 2002 CONCRETE FRAME DESIGN BS 8110-97 Technical Note Beam Design This Technical Note describes how this program completes beam design when the BS 8110-97 code is selected. The program calculates and reports the required areas of steel for flexure and shear based upon the beam moments, shears, load combination factors and other criteria described herein. Overview In the design of concrete beams, the program calculates and reports the required areas of steel for flexure and shear based upon the beam moments, and shears, load combination factors, and other criteria described herein. The reinforcement requirements are calculated at a user-defined number of check stations along the beam span. All beams are designed for major direction flexure and shear only. Effects resulting from axial forces, minor direction bending, and torsion that may exist in the beams must be investigated independently by the user. The beam design procedure involves the following steps: Design beam flexural reinforcement Design beam shear reinforcement Design Beam Flexural Reinforcement The beam top and bottom flexural steel areas are designed at a user-defined number of check stations along the beam span. The following steps are involved in designing the flexural reinforcement for the major moment for a particular beam at a particular section: Determine the maximum factored moments Determine the reinforcing steel Overview Page 1 of 10 Concrete Frame Design BS 8110-97 Beam Design Determine Factored Moments In the design of flexural reinforcement of concrete frame beams, the factored moments for each load combination at a particular beam station are obtained by factoring the corresponding moments for different load cases with the corresponding load factors. The beam section is then designed for the maximum positive and maximum negative factored moments obtained from all of the load combinations at that section. Negative beam moments produce top steel. In such cases, the beam is always designed as a rectangular section. Positive beam moments produce bottom steel. In such cases, the beam may be designed as a rectangular section, or T-Beam effects may be included. Determine Required Flexural Reinforcement In the flexural reinforcement design process, the program calculates both the tension and compression reinforcement. Compression reinforcement is added when the applied design moment exceeds the maximum moment capacity of a singly reinforced section. The user has the option of avoiding the compression reinforcement by increasing the effective depth, the width, or the grade of concrete. The design procedure is based on the simplified rectangular stress block as shown in Figure 1 (BS 3.4.4.1). It is assumed that moment redistribution in the member does not exceed 10% (i.e., βb ≥ 0.9) (BS 3.4.4.4). The code also places a limitation on the neutral axis depth, x/d ≤ 0.5, to safeguard against non-ductile failures (BS 3.4.4.4). In addition, the area of compression reinforcement is calculated assuming that the neutral axis depth remains at the maximum permitted value. The design procedure used by the program for both rectangular and flanged sections (L- and T-beams) is summarized in the next section. It is assumed that the design ultimate axial force does not exceed 0.1fcu Ag (BS 3.4.4.1); hence, all the beams are designed for major direction flexure and shear only. Design of a Rectangular beam For rectangular beams, the moment capacity as a singly reinforced beam, Msingle, is obtained first for a section. The reinforcing steel area is determined Design Beam Flexural Reinforcement Page 2 of 10 Concrete Frame Design BS 8110-97 Beam Design based on whether M is greater than, less than, or equal to Msingle. See Figure 1. 0.67fcu/γc Figure 1: Design of a Rectangular Beam Section Calculate the ultimate moment of resistance of the section as a singly reinforced beam. Msingle = K'fcubd2, where (BS 3.4.4.4) K' = 0.156. If M ≤ Msingle, no compression reinforcement is required. The area of tension reinforcement, As, is obtained from As = M , where ( 0.95 fy )z K z = d 0.5 + 0.25 − ≤ 0.95d, and 0.9 Design Beam Flexural Reinforcement (BS 3.4.4.4) (BS 3.4.4.4) Page 3 of 10 Concrete Frame Design BS 8110-97 K= M f cu bd 2 Beam Design . (BS 3.4.4.4) This is the top steel if the section is under negative moment and the bottom steel if the section is under positive moment. • If M > Msingle, the area of compression reinforcement, As' , is given by As' = ( f s' M − M single ) − 0.67 f cu γ c ( d − d' ) , (BS 3.4.4.4) where d' is the depth of the compression steel from the concrete compression face, and 2d' ≤ 0.95 fy. f s' = 700 1 − d (BS 3.4.4.4) This is the bottom steel if the section is under negative moment. From equilibrium, the area of tension reinforcement is calculated as: As = M single ( 0.95 f y ) z + M − M single (0.95 f y ) ( d − d' ) , where K ' z = d 0.5 + 0.25 − = 0.776887 d. 0.9 (BS 3.4.4.4) (BS 3.4.4.4) As is to be placed at the bottom of the beam and As' at the top for positive bending and vice versa for negative bending. Design as a T-Beam (i) Flanged beam under negative moment The contribution of the flange to the strength of the beam is ignored. The design procedure is therefore identical to the one used for rectangular beams, except that in the corresponding equations, b is replaced by bw. See Figure 2. (ii) Flanged beam under positive moment With the flange in compression, the program analyzes the section by considering alternative locations of the neutral axis. Initially the neutral axis is as- Design Beam Flexural Reinforcement Page 4 of 10 Concrete Frame Design BS 8110-97 Beam Design sumed to be located in the flange. On the basis of this assumption, the program calculates the depth of the neutral axis. If the stress block does not extend beyond the flange thickness, the section is designed as a rectangular beam of width bf. If the stress block extends beyond the flange width, the contribution of the web to the flexural strength of the beam is taken into account. See Figure 2. 0.67fcu/γc Figure 2: Design of a T-Beam Section The T-beam requires only tension reinforcement when the moment is positive, the flange is in compression, the moment is less than βf fcu bd2 and the flange depth is less than 0.45d. In those conditions, the tension reinforcing steel area of the T-beam is calculated as follows (BS 3.4.4.5): As = M + 0.1 fcu bw d (0.45 d − hf ) 0.95 fy (d − 0.5 hf ) (BS 3.4.4.5) where, β f = 0.45 hf d 1 − bw 1 − hf + 0.15 bw bf 2d b Design Beam Flexural Reinforcement (BS 3.4.4.5) Page 5 of 10 Concrete Frame Design BS 8110-97 Beam Design If the above conditions are not met, the T-beam is designed using the general principle of the BS 8110 code (BS 3.4.4.4, BS 3.4.4.5), which is as follows: Assuming that the neutral axis lies in the flange, the normalized moment is computed as K= M f cu bf d 2 . (BS 3.4.4.4) Then the moment arm is computed as K z = d 0.5 + 0.25 − ≤ 0.95d, 0.9 (BS 3.4.4.4) the depth of neutral axis is computed as x= 1 (d − z), and 0.45 (BS 3.4.4.4) the depth of compression block is given by a = 0.9 x. (BS 3.4.4.4) If a ≤ hf, the subsequent calculations for As are exactly the same as previously defined for the rectangular section design. However, in that case the width of the compression block is taken to be equal to the width of the compression flange, bf for design. Compression reinforcement is required if K > K'. If a > hf, the subsequent calculations for As are performed in two parts. The first part is for balancing the compressive force from the flange, Cf, and the second part is for balancing the compressive force from the web, Cw, as shown in Figure 2. In this case, the ultimate resistance moment of the flange is given by Mf = 0.67 (fcu /γc) (bf − bw) hf (d − 0.5 hf), (BS 3.4.4.1) the balance of moment taken by the web is computed as Mw = M − Mf, and Design Beam Flexural Reinforcement Page 6 of 10 Concrete Frame Design BS 8110-97 Beam Design the normalized moment resisted by the web is given by Kw = − Mw f cu bw d 2 (BS 3.4.4.1) If Kw ≤ K', the beam is designed as a singly reinforced concrete beam. The area of steel is calculated as the sum of two parts: one to balance compression in the flange and one to balance compression in the web. As = − . Mf Mw , where + 0.95 fy ( d − 0.5 hf ) 0.95 fy z (BS 3.4.4.1) K z = d 0.5 + 0.25 − w ≤ 0.95d. 0.9 (BS 3.4.4.1) If Kw > K', compression reinforcement is required. The compression reinforcing steel area is calculated using the following procedure: The ultimate moment of resistance of the web only is given by Muw = K' fcu bw d2. (BS 3.4.4.4) The compression reinforcement is required to resist a moment of magnitude Mw Mlw. The compression reinforcement is computed as A's = ( f s' Mw − M uw ) − 0.67 f cu / γ c ( d − d' ) , (BS 3.4.4.1) where d' is the depth of the compression steel from the concrete compression face, and 2d' ≤ 0.95 fy. f s' = 700 1 − d (BS 3.4.4.1) The area of tension reinforcement is obtained from equilibrium As = M uw M − M uw Mf 1 + + w 0.95 f y d − 0.5 hf 0.777 d d − d' Design Beam Flexural Reinforcement . (BS 3.4.4.1) Page 7 of 10 Concrete Frame Design BS 8110-97 Beam Design Determination of the Required Minimum Flexural Reinforcing The minimum flexural tensile reinforcing steel required for a beam section is given by the following table, which is taken from BS Table 3.25 (BS 3.12.5.3) with interpolation for reinforcement of intermediate strength: Table 1 Minimum Percentage of Tensile Reinforcing Section Rectangular T-Beam with web in tension T-Beam with web in compression Situation Definition of percentage bw < 0.4 bf Minimum percentage fy= 250 MPa fy = 460 MPa As bh A 100 s bw h 100 0.24 0.13 0.32 0.18 bw ≥ 0.4 bf 100 As bw h 0.24 0.13 100 As bw h 0.48 0.26 The minimum flexural compression steel, if it is required, provided in a rectangular or T-beam section is given by the following table, which is taken from BS Table 3.25 (BS 3.12.5.3) with interpolation for reinforcement of intermediate strength: Table 2 Minimum Percentage of Compression Reinforcing (if required) Section Situation Definition of percentage Minimum percentage ' Rectangular Web in tension T-Beam As bh ' A 100 s bf hf 100 0.20 0.40 ' Web in compression 100 As bw h 0.20 In addition, an upper limit on both tension reinforcement and compression reinforcement has been imposed to be 0.04 times the gross cross-sectional Design Beam Flexural Reinforcement Page 8 of 10 Concrete Frame Design BS 8110-97 Beam Design area (BS 3.12.6.1). The program reports an overstress when the ratio exceed 4 percent. Design Beam Shear Reinforcement The shear reinforcement is designed for each load combination in the major and minor directions of the column. The following steps are involved in designing the shear reinforcement for a particular beam for a particular load combination resulting from shear forces in a particular direction (BS 3.4.5): Calculate the design shear stress and maximum allowable shear stress as v= V , where Acv (BS 3.4.5.2) v ≤ 0.8 RLW f cu , (BS 3.4.5.2, BS 3.4.5.12) v ≤ N/mm2 , (BS 3.4.5.2, BS 3.4.5.12) vmax = min {0.8RLW f cu , 5.0 MPa}, (BS 3.4.5.2, BS 3.4.5.12) Acv = bw d, and RLW is a shear strength reduction factor that applies to light-weight concrete. It is equal to 1 for normal weight concrete. The factor is specified in the concrete material properties. If v exceeds 0.8RLW f cu or 5 MPa, the program reports an overstress. In that case, the concrete shear area should be increased. Note The program reports an overstress message when the shear stress exceed 0.8RLW f cu or 5 MPa (BS 3.4.5.2, BS 3.4.5.12). Calculate the design concrete shear stress from 0.79 k 1k 2 vc = RLW γm Design Beam Shear Reinforcement 100 As bd 1 3 400 d 1 4 , (BS 3.4.5.4, Table 3.8) Page 9 of 10 Concrete Frame Design BS 8110-97 Beam Design where, k1 is the enhancement factor for support compression, and is conservatively taken as 1, f k2 = cu 25 1 3 ≥ 1, γm = 1.25, and (BS 3.4.5.8) (BS 3.4.5.4, Table 3.8) (BS 2.4.4.1) As is the area of tensile steel. However, the following limitations also apply: 0.15 ≤ 100 As ≤ 3, bd (BS 3.4.5.4, Table 3.8) 400 ≥ 1, and d (BS 3.4.5.4, Table 3.8) fcu ≤ 40 N/mm2 (for calculation purpose only). (BS 3.4.5.4, Table 3.8) If v ≤ vc +0.4, provide minimum links defined by Asv 0.4 b ≥ , sv 0.95 f yv (BS 3.4.5.3) else if vc +0.4 < v < vmax, provide links given by Asv (v − v c ) b , ≥ sv 0.95 fyv (BS 3.4.5.3) else if v ≥ vmax, a failure condition is declared. (BS 3.4.5.2, 3.4.5.12) In shear design, fyv cannot be greater than 460 MPa (BS 3.4.5.1). If fyv is defined as greater than 460 MPa, the program designs shear reinforcing assuming that fyv equals 460 MPa. Design Beam Shear Reinforcement Page 10 of 10