COPUTERS AND STRUCTURES, INC., BERKELEY, CALIFORNIA DECEBER 2001 CONCRETE FRAE DESIGN ACI-318-99 Technical Note This Technical Note describes how this program completes beam design when the ACI 318-99 code is selected. The program calculates and reports the required areas of steel for flexure and shear based on 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 on the beam moments, shears, load combination factors, and other criteria described below. The reinforcement requirements are calculated at a user-defined number of check/design stations along the beam span. All beams are designed for major direction flexure and shear onl. Effects resulting from an axial forces, minor direction bending, and torsion that ma exist in the beams must be investigated independentl b 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 is designed at check/design stations along the beam span. The following steps are involved in designing the flexural reinforcement for the major moment for a particular beam for a particular section:! Determine the maximum factored moments! Determine the reinforcing steel Overview Page 1 of 13
Determine Factored oments In the design of flexural reinforcement of Special, Intermediate, or Ordinar moment resisting concrete frame beams, the factored moments for each load combination at a particular beam section are obtained b factoring the corresponding moments for different load cases with the corresponding load factors. u and maxi- The beam section is then designed for the maximum positive mum negative tions. u factored moments obtained from all of the load combina- Negative beam moments produce top steel. In such cases, the beam is alwas designed as a rectangular section. Positive beam moments produce bottom steel. In such cases, the beam ma be designed as a Rectangular- or a T-beam. 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 capacit of a singl reinforced section. The user has the option of avoiding the compression reinforcement b 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 (ACI 10.2). It is assumed that the compression carried b concrete is less than 0.75 times that which can be carried at the balanced condition (ACI 10.3.3). When the applied moment exceeds the moment capacit at this designed balanced condition, the area of compression reinforcement is calculated assuming that the additional moment will be carried b compression and additional tension reinforcement. The design procedure used b this program for both rectangular and flanged sections (L- and T-beams) is summarized below. It is assumed that the design ultimate axial force does not exceed 0.1 f c A g (ACI 10.3.3); hence, all the beams are designed for major direction flexure and shear onl. Design Beam Flexural Reinforcement Page 2 of 13
Figure 1 Design of Rectangular Beam Section Design for Rectangular Beam In designing for a factored negative or positive moment, u (i.e., designing top or bottom steel), the depth of the compression block is given b a (see Figure 1), where, a = d - d 2 2 0.85f u 1 c, (ACI 10.2.7.1) ϕb where, the value of ϕ is 0.90 (ACI 9.3.2.1) in the above and the following equations. Also β 1 and c b are calculated as follows: β 1 = 0.85-0.05 f c 4,000, 0.65 β 1,000 1 0.85, (ACI 10.2.7.3) c b = ε c ε E c s E s f d = 87,000 87,000 f d. (ACI 10.2.3, 10.2.4) Design Beam Flexural Reinforcement Page 3 of 13
The maximum allowed depth of the compression block is given b a max = 0.75β 1 c b. (ACI 10.2.7.1, 10.3.3)! If a a max, the area of tensile steel reinforcement is then given b A s = ϕf u. a d 2 This steel is to be placed at the bottom if u is positive, or at the top if u is negative.! If a > a max, compression reinforcement is required (ACI 10.3.3) and is calculated as follows: The compressive force developed in concrete alone is given b c C = 0.85 f ba max, and (ACI 10.2.7.1) the moment resisted b concrete compression and tensile steel is a uc = C d max ϕ. 2 Therefore the moment resisted b compression steel and tensile steel is us = u - uc. So the required compression steel is given b A s = us, where f ( d d )ϕ s c f s = 0.003E s c d. (ACI 10.2.4) The required tensile steel for balancing the compression in concrete is Design Beam Flexural Reinforcement Page 4 of 13
A s1 = f d uc a max 2 ϕ, and the tensile steel for balancing the compression in steel is given b A s2 = us. f ( d d )ϕ Therefore, the total tensile reinforcement, A s = A s1 A s2, and total compression reinforcement is A s. A s is to be placed at bottom and A s is to be placed at top if u is positive, and vice versa if u is negative. Design for T-Beam In designing for a factored negative moment, u (i.e., designing top steel), the calculation of the steel area is exactl the same as above, i.e., no T-Beam data is to be used. See Figure 2. If u > 0, the depth of the compression block is given b 2 2u a = d - d. 0.85fc ϕbf The maximum allowed depth of compression block is given b a max = 0.75β 1 c b. (ACI 10.2.7.1, 10.3.3) If a d s, the subsequent calculations for A s are exactl the same as previousl defined for the rectangular section design. However, in this case the width of the compression flange is taken as the width of the beam for analsis. Compression reinforcement is required if a > a max. If a > d s, calculation for A s is performed in two parts. The first part is for balancing the compressive force from the flange, C f, and the second part is for balancing the compressive force from the web, C w, as shown in Figure 2. C f is given b C f = 0.85 f c (b f - b w )d s. Design Beam Flexural Reinforcement Page 5 of 13
Figure 2 Design of a T-Beam Section Therefore, A s1 = given b C f f and the portion of u that is resisted b the flange is uf = C f ds d ϕ. 2 Again, the value for ϕ is ϕ(flexure), which is 0.90 b default. Therefore, the balance of the moment, u to be carried b the web is given b uw = u - uf. The web is a rectangular section of dimensions b w and d, for which the design depth of the compression block is recalculated as a 1 = d - d 2 2 0.85f uw i c ϕ b w.! If a 1 a max, the area of tensile steel reinforcement is then given b A s2 = ϕf uw, and a d 1 2 Design Beam Flexural Reinforcement Page 6 of 13
A s = A s1 A s2. This steel is to be placed at the bottom of the T-beam.! If a 1 > a max, compression reinforcement is required (ACI 10.3.3) and is calculated as follows: The compressive force in web concrete alone is given b C = 0.85 fc ba max. (ACI 10.2.7.1) Therefore, the moment resisted b concrete web and tensile steel is a uc = C d max ϕ, and 2 the moment resisted b compression steel and tensile steel is us = uw - uc. Therefore, the compression steel is computed as A s = s us f ( d d )ϕ, where c f s = 0.003E s c d. (ACI 10.2.4) The tensile steel for balancing compression in web concrete is A s2 = f ( d uc a max )ϕ 2, and the tensile steel for balancing compression in steel is A s3 = us. f ( d d )ϕ Design Beam Flexural Reinforcement Page 7 of 13
The total tensile reinforcement, A s = A s1 A s2 A s3, and total compression reinforcement is A s. A s is to be placed at bottom and A s is to be placed at top. inimum Tensile Reinforcement The minimum flexural tensile steel provided in a rectangular section in an Ordinar moment resisting frame is given b the minimum of the two following limits: 3 f c 200 A s max bw d and bw d f f or (ACI 10.5.1) A s (4/3)A s(required). (ACI 10.5.3) Special Consideration for Seismic Design For Special moment resisting concrete frames (seismic design), the beam design satisfies the following additional conditions (see also Table 1 ):! The minimum longitudinal reinforcement shall be provided at both the top and bottom. An of the top and bottom reinforcement shall not be less than A s(min) (ACI 21.3.2.1). A s(min) max 3 f c 200 bw d and bw d or (ACI 10.5.1) f f A s(min) 3 4 As(required). (ACI 10.5.3)! The beam flexural steel is limited to a maximum given b A s 0.025 b w d. (ACI 21.3.2.1)! At an end (support) of the beam, the beam positive moment capacit (i.e., associated with the bottom steel) would not be less than 1/2 of the beam negative moment capacit (i.e., associated with the top steel) at that end (ACI 21.3.2.2).! Neither the negative moment capacit nor the positive moment capacit at an of the sections within the beam would be less than 1/4 of the Design Beam Flexural Reinforcement Page 8 of 13
maximum of positive or negative moment capacities of an of the beam end (support) stations (ACI 21.3.2.2). For Intermediate moment resisting concrete frames (i.e., seismic design), the beam design would satisf the following conditions:! At an support of the beam, the beam positive moment capacit would not be less than 1/3 of the beam negative moment capacit at that end (ACI 21.10.4.1).! Neither the negative moment capacit nor the positive moment capacit at an of the sections within the beam would be less than 1/5 of the maximum of positive or negative moment capacities of an of the beam end (support) stations (ACI 21.10.4.1). Design Beam Shear Reinforcement The shear reinforcement is designed for each load combination at a user defined number of stations along the beam span. The following steps are involved in designing the shear reinforcement for a particular beam for a particular load combination at a particular station due to the beam major shear: Determine the factored shear force, V u. Determine the shear force, V c, that can be resisted b the concrete. Determine the reinforcement steel required to carr the balance. For Special and Intermediate moment resisting frames (ductile frames), the shear design of the beams is also based upon the probable and nominal moment capacities of the members, respectivel, in addition to the factored load design. The following three sections describe in detail the algorithms associated with this process. Design Beam Shear Reinforcement Page 9 of 13
Table 1 Design Criteria Table Tpe of Check/ Design Column Check (interaction) Ordinar oment Resisting Frames (non-seismic) Intermediate oment Resisting Frames (Seismic) Special oment Resisting Frames (Seismic) Column Design (interaction) 1% < ρ < 8% 1% < ρ < 8% α = 1.0 1% < ρ < 6% Column Shears odified (earthquake loads doubled) Column capacit ϕ = 1.0 and α = 1.0 Column shear capacit ϕ = 1.0 and α = 1.25 Beam Design Flexure ρ 0.025 3 ρ f f c, ρ 200 f Beam in. oment Override Check No Requirement 1 uend u 3 END uspan uspan { u u } END 1 max, 5 1 max u, 5 { u } END 1 uend u 2 END uspan uspan 1 max 4, 1 max 4, { u u } END { u u } END Beam Design Shear odified (earthquake loads doubled) Beam Capacit Shear (Vp) with α = 1.0 and ϕ = 1.0 plus VDL Beam Capacit Shear (Vp) with α = 1.25 and ϕ = 1.0 plus VDL Vc = 0 Joint Design No Requirement No Requirement Checked for shear Beam/Column Capacit Ratio No Requirement No Requirement Reported in output file NLD a = Number of specified loading Design Beam Shear Reinforcement Page 10 of 13
Determine Shear Force and oment In the design of the beam shear reinforcement of an Ordinar moment resisting concrete frame, the shear forces and moments for a particular load combination at a particular beam section are obtained b factoring the associated shear forces and moments with the corresponding load combination factors. In the design of Special moment resisting concrete frames (i.e., seismic design), the shear capacit of the beam is also checked for the capacit shear resulting from the probable moment capacities at the ends and the factored gravit load. This check is performed in addition to the design check required for Ordinar moment resisting frames. The capacit shear force, V p, is calculated from the probable moment capacities of each end of the beam and the gravit shear forces. The procedure for calculating the design shear force in a beam from probable moment capacit is the same as that described for a column in section Design Column Shear Reinforcement of Technical Note Column Design Concrete Frame Design ACI318-99. See also Table 1 for details. The design shear force V u is then given b (ACI 21.3.4.1) V u = V p V DL (ACI 21.3.4.1) where V p is the capacit shear force obtained b appling the calculated probable ultimate moment capacities at the two ends of the beams acting in two opposite directions. Therefore, V p is the maximum of V and V, where P 1 P 2 V P 1 = V P 2 = I L L I J J, and, where I = oment capacit at end I, with top steel in tension, using a steel ield stress value of αf and no ϕ factors (ϕ = 1.0), J = oment capacit at end J, with bottom steel in tension, using a steel ield stress value of αf and no ϕ factors (ϕ = 1.0), Design Beam Shear Reinforcement Page 11 of 13
I = oment capacit at end I, with bottom steel in tension, using a steel ield stress value of αf and no ϕ factors (ϕ = 1.0), J = oment capacit at end J, with top steel in tension, using a steel ield stress value of αf and no ϕ factors (ϕ = 1.0), and L = Clear span of beam. For Special moment resisting frames α is taken as 1.25 (ACI 21.0, R21.3.4.1). V DL is the contribution of shear force from the in-span distribution of gravit loads. For Intermediate moment resisting frames, the shear capacit of the beam is also checked for the capacit shear based on the nominal moment capacities at the ends and the factored gravit loads, in addition to the check required for Ordinar moment resisting frames. The design shear force in beams is taken to be the minimum of that based on the nominal moment capacit and modified factored shear force. The procedure for calculating nominal (ϕ = 1.0) moment capacit is the same as that for computing the probable moment capacit for Special moment resisting frames, except that α is taken equal to 1 rather than 1.25 (ACI 21.10.3.a, R21.10). The modified factored shear forces are based on the specified load factors, except the earthquake load factors are doubled (ACI 21.10.3.b). The computation of the design shear force in a beam of an Intermediate moment resisting frame is the same as described for columns in section Determine Section Forces of Technical Note Column Design Concrete Frame Design ACI318-99. See also Table 1 for details. Determine Concrete Shear Capacit The allowable concrete shear capacit is given b V c = 2 fc b w d. (ACI 11.3.1.1) For Special moment resisting frame concrete design, V c is set to zero if both the factored axial compressive force, including the earthquake effect P u, is less than f c A g /20 and the shear force contribution from earthquake V E is more than half of the total maximum shear force over the length of the member V u (i.e., V E 0.5V u ) (ACI 21.3.4.2). Design Beam Shear Reinforcement Page 12 of 13
Determine Required Shear Reinforcement Given V u and V c, the required shear reinforcement in area/unit length is calculated as A v = ( Vu / ϕ Vc ) s. (ACI 11.5.6.1, 11.5.6.2) f d s The shear force resisted b steel is limited b (V u / ϕ - V c ) 8 fc bd. (ACI 11.5.6.9) Otherwise, redimensioning of the concrete section is required. Here, ϕ, the strength reduction factor for shear, is 0.85 b default (ACI 9.3.2.3). The maximum of all the calculated A v values, obtained from each load combination, is reported along with the controlling shear force and associated load combination number. The beam shear reinforcement requirements displaed b the program are based purel on shear strength considerations. An minimum stirrup requirements to satisf spacing and volumetric considerations must be investigated independentl of the program b the user. Design Beam Shear Reinforcement Page 13 of 13