Beam Design RC BEAM DESIGN Based on BS

Beam Design

RC BEAM DESIGN Based on BS 8110

Partial Safety Factor γm v Cl. 2. 4. 4. 1 v Given in Table 2. 2

Durability v. Cl. 2. 2. 4 and 3. 1. 5 v. Exposure conditions (Integrity of rc concrete – ability to prevent corrosion) : moderate, severe, very severe, most severe and abrasive (Table 3. 2) v. Protection a agaist corrosion of steel (Table 3. 3) v. Fire resistance requirement (Table 3. 4)

Nominal cover v. Cl. 3. 3. 1. 1 v. Bar Size (Cl. 3. 3. 1. 2) § Single Bars: nominal cover ≥ main bar diameter, d 1 § Paired bars: nominal cover ≥ √ 2 d 1 § Bundles bars: 2 √(Aequivalent/π)

Example: Using the data given, determine: (i) the nominal cover required to the underside of the beam, and (ii) the minimum width of beam required. Data: v Exposure condition mild v Characteristic strength of concrete (fcu) 40 N/mm 2 v Nominal maximum aggregate size (hagg) 20 mm v Diameter of main tension steel 25 mm v Diameter of shear links 8 mm v Minimum required fire resistance 1. 5 hours

SOLUTION v Clause 3. 3. 1. 2 Nominal cover ≥ (main bar diameter − link diameter) ≥ (25 − 8) = 17 mm v Clause 3. 3. 1. 3 Nominal cover ≥ nominal maximum aggregate size > 20 mm v Clause 3. 3. 3 Exposure condition is mild v Grade of concrete is C 40 v Table 3. 3 Nominal cover ≥ 20 mm* v Clause 3. 3. 6 Minimum fire resistance = 1. 5 hr v The beam is simply supported v Table 3. 4 Nominal cover ≥ 20 mm The required nominal cover = 20 mm

Flexural Strength of Sections v. The flexural strength (i. e. the ultimate moment of resistance of a cross-section) is determined assuming the following conditions as given in Clause 3. 4. 4. 1, BS 8110: Part 1

Concrete Capacity v The maximum compressive force which can be resisted by the concrete corresponds to the maximum depth permitted for the neutral axis, as shown in Figure 5. 28 (i. e. x = d/2).

Concrete Capacity v Consider the moment of the compressive force about the line of action of Ft : v Mult, concrete = (Fc × z) where Fc = compressive force = (stress × area) = [0. 45 fcu × (b × 0. 9 x )] (for the maximum concrete force x = d/2) = [0. 45 fcu × (b × 0. 45 d)] = 0. 2 bdfcu z = lever arm = [d − (0. 5 × 0. 45 d)] = 0. 775 d {Note: In general z = [d − (0. 5 × 0. 9 x)] } Mult, concrete = (0. 2 bdfcu )×(0. 775 d) =0. 156 bd 2 fcu v This equation can be rewritten as 0. 156 =M/bd 2 fcu

Steel Capacity v Mult, steel = (Fs × z) v where Fs = tensile force = (stress × area) = (0. 95 fy × As) z = lever arm v Mult, steel = 0. 95 fy. As z v As = M/0. 95 fyz v Consider z, the lever arm: v z = [d − (0. 5 × 0. 9 x)] , v 0. 9 x = 2(d − z) v Mult, concrete = [0. 45 fcu × (b × 0. 9 x)] × z (substitute for 0. 9 x) = 0. 9 fcu b (d − z) z

Steel Capacity Mult, concrete = K bd 2 fcu = 0. 9 fcu b (d − z) z Kd 2 = 0. 9 fcu dz − 0. 9 z 2 − dz + Kd 2/0. 9 = 0 The solution of this quadratic equation (ax 2 + bx + c = 0) gives an expression which can be used to determine the lever arm, z.

Summary *The smaller of these two values being the critical case When K ≤ K′ for a singly-reinforced section the maximum moment permitted, based on the concrete strength, is equal to 0. 156 bd 2 fcu When K > K′ a section requires compression reinforcement.

Area of steel reinforcement, As

Areas/metre width for various pitches of bars

Example 1: v A rectangular beam section is shown in Figure 5. 29. Using the data given, determine the maximum ultimate moment which can be applied to the section assuming it to be singly reinforced. v Data: § Characteristic strength of concrete (fcu) 40 N/mm 2 § Characteristic strength of steel (fy) 460 N/mm 2

Solution: v Strength based on concrete: Mult = 0. 156 bd 2 fcu = (0. 156 × 250 × 4202 × 40)/106 = 275. 2 k. Nm v Strength based on steel: As = 1260 mm 2 z = 0. 775 d = 0. 775 x 420 = 325. 5 Mult = 0. 95 fy. As z = (0. 95 × 460 × 1260 × 325. 5)/106 = 179. 5 k. Nm v The maximum design moment which can be applied is: Mult = 179. 5 k. Nm

v The cross-section of a simply Example 2: supported rectangular beam is shown in Figure 5. 30. Using the data given, determine the Data: maximum ultimate moment v Characteristic strength of which can be applied to the concrete (fcu) 30 N/mm 2 section assuming it to be v Characteristic strength of steel singly-reinforced. (fy) 460 N/mm 2 v Nominal maximum aggregate size (hagg) 20 mm v Diameter of main tension steel 32 mm v Diameter of shear links 8 mm v Exposure condition mild v Minimum required fire resistance 1. 0 hour

Solution: Clause 3. 3. 1 Nominal cover to all steel Clause 3. 3. 1. 2 ≥ bar size cover ≥ (32 − 8) = 24 mm Clause 3. 3. 1. 3 ≥ Nominal maximum aggregate size cover ≥ 20 mm Clause 3. 3. 3 Exposure condition: mild Table 3. 3 and fcu = 30 N/mm 2 cover ≥ 25 mm Clause 3. 3. 6 Min. fire resistance: 1 hr beam is simply supported Table 3. 4 cover ≥ 20 mm v The required nominal cover to all steel = 25 mm

Solution v Figure 3. 2 Minimum width b for 1 hour fire resistance = 200 mm adequate v Effective depth d = (h − cover − link diameter − bar diameter/2) = (475 − 25 − 8 − 16) = 426 mm v Strength based on concrete: Mult = 0. 156 bd 2 fcu = (0. 156 × 250 × 4262 × 30)/106 = 212. 3 k. Nm v Strength based on steel: As = 2410 mm 2 z = 0. 775 d Mult = 0. 95 fy. As z = (0. 95 × 460 × 2410 × 0. 775 × 426)/106 = 349. 9 k. Nm v The maximum design moment which can be applied is: Mult = 212. 3 k. Nm

Example 3: v The cross-section of a simply supported rectangular beam is shown in Figure 5. 31. Using the data given, and assuming the section to be singly-reinforced, determine the area of tension reinforcement required to resist an applied ultimate bending moment of 150 k. Nm. v Data: § § § § Characteristic strength of concrete (fcu) 40 N/mm 2 Characteristic strength of steel (fy) 460 N/mm 2 Nominal maximum aggregate size (hagg) 20 mm Diameter of main tension steel Assume 25 mm Diameter of shear links 10 mm Exposure condition moderate Minimum required fire resistance 2. 0 hours

Solution: • Clause 3. 3. 1 Nominal cover to all steel • Clause 3. 3. 1. 2 ≥ bar size cover = (25 − 10) = 15 mm • Clause 3. 3. 1. 3 ≥ Nominal maximum • aggregate size cover = 20 mm • Clause 3. 3. 3 Exposure condition: moderate • Table 3. 3 fcu = 40 N/mm 2 cover ≥ 30 mm • Clause 3. 3. 6 Min. fire resistance: 2. 0 hr • beam is simply supported • Table 3. 4 cover ≥ 40 mm The required nominal cover to the main steel = 40 mm • Figure 3. 2 Minimum width b for 2 hours, fire resistance = 200 mm adequate Figure 5. 31

Solution: v Effective depth d = (h − cover − link diameter − bar diameter/2) = (450 − 40 − 13) = 387 mm v Clause 3. 4. 4. 4 v Check that the section is singly-reinforced: v K = M/bd 2 fcu = 0. 125 < 0. 156 v Since K < K ′ the section is singly-reinforced v As = M/0. 95 fyz = [(150 × 106) / (0. 95 × 460 × 0. 83 × 387)] = 1068 mm 2 v Adopt 4/ 20 mm diameter HYS bars providing 1260 mm 2.

Shear Strength of Sections v The equation in BS 8110 (Cl 3. 4. 5. 2) v Where § § v = design shear stress V = design shear force due to ultimate loads bv=breadth of the section d =effective depth v v ≤ 0. 8 √(fcu) or ≤ 5 N/mm 2

Shear Strength of Sections v This is the equation given in Table 3. 7 of the code for the cross-sectional area of designed links when the design shear stress v at a crosssection is greater than (vc + 0. 4). It is recognised that the truss analogy produces conservative results and the code specifies that: designed links are required when: v (vc + 0. 4) < v < 0. 8 (fcu)1/2 or 5 N/mm 2

Shear Strength of Sections v It is important to provide minimum areas of steel in concrete to minimize thermal and shrinkage cracking, etc. v Minimum links are specified in the code which provide a shear resistance of 0. 4 N/mm 2 in addition to the design concrete shear stress vc, and consequently designed links are required when v > (vc + 0. 4) as indicated. v In Table 3. 7: minimum links are required when 0. 5 vc < v < (vc + 0. 4) v The cross-sectional area of the links required is given by:

Example 3: v A concrete beam is simply supported over a 5. 0 m span as shown in Figure 5. 48. Using the data given, determine suitable shear reinforcement. v Data: § Characteristic strength of concrete (fcu) 40 N/mm 2 § Characteristic strength of mild steel (fyv) 250 N/mm 2 § Maximum shear force at the support 40 k. N

Solution: v Consider the end of the beam at the support where the shear force is a maximum = 40 k. N v Clause 3. 4. 5. 2

Solution: v Table 3. 7 To determine the required reinforcement, evaluate vc and either adopt minimum links throughout or use designed links. v As = area of 2/16 mm diameter bars = 402 mm 2 From Table 3. 8

Solution: v Since 0. 5 vc < v < (vc + 0. 4), minimum links are required for the whole length of the beam. v The cross-sectional area of the links required is given by Option 1: v Assume 2 -legged / 6 mm diameter mild steel links. ∴Asv = 56. 6 mm 2 v Table 3. 7 The spacing required , v Clause 3. 4. 5. 5 v The maximum spacing of the links ≤ 0. 75 d = (0. 75 x 225) = 169 mm v Adopt 6 mm diameter mild steel links @ 150 mm centres throughout the length of the beam.

Solution: Option 2: v. Clause 3. 4. 5. 5 v. The maximum spacing of the links ≤ 0. 75 d v= (0. 75 × 225) = 169 mm v. Assume sv = 150 mm. *Adopt 6 mm diameter mild steel links (56. 6 mm 2) @ 150 mm centres throughout the length of the beam as in option 1.

Example 4: v A concrete beam is simply supported over a 7. 0 m span as shown in Figure 5. 49. Using the data given, determine suitable shear reinforcement. v Data: § § Characteristic strength of concrete (fcu) 40 N/mm 2 Characteristic strength of mild steel (fyv) 250 N/mm 2 Characteristic dead load (gk) 5. 0 k. N/m Characteristic imposed load (qk) 30. 0 k. N/m

Solution: v Ultimate design load = [(1. 4 × 5. 0) + (1. 6 × 30. 0)] = 55. 0 k. N/m v Ultimate design shear force at the support V = (55. 0 × 3. 5) = 192. 5 k. N

Solution: • Table 3. 7 : To determine the required reinforcement evaluate vc and to determine the requirement for either minimum links or designed links.

Solution: v Table 3. 7 Since (vc + 0. 4) < v < 0. 8 √(fcu) < 5 N/mm 2 designed links are required v The maximum shear force which can be resisted by the provision of minimum links is given by: v V = (vc + 0. 4)bvd = (1. 06 × 240 × 570)/103 = 145 k. N

Solution: v Provide designed links for the first metre from each end and minimum links elsewhere. v The cross-sectional area of the designed links required is given by v Assuming 8 mm diameter mild steel links, mm 2 and fyv = 250 N/mm 2. Asv = 101

Solution: v Clause 3. 4. 5. 5 The maximum spacing of the links ≤ 0. 75 d = (0. 75 × 570) = 428 mm v Adopt 8 mm diameter mild steel links @ 125 mm centres for 1. 0 m from each end of the beam. v The cross-sectional area of the minimum links required is given by Asv = v Adopt 8 mm diameter mild steel links @ 250 mm centres throughout the central 5. 0 m of the beam.

Deflections (Cl. 3. 4. 6) v Allowances for these factors to reflect the actual behaviour are made in the BS 8110 in terms of: ♦ using a basic ‘span/effective depth’ ratio (span/d) rather than a ‘span/h ’ ratio (Table 3. 9), ♦ applying a modifying factor which is dependent on the steel strength (steel percentage and steel strength are directly related through the factor K = M/bd 2) and the steel stress (fs) under service conditions (Table 3. 10), ♦ applying a modifying factor which is dependent on the percentage of compression steel provided (Table 3. 11).

Deflections v. The basic ratios given in Table 3. 9 relate to three support conditions: § cantilevers, § simply supported beams, § continuous beams. *Note: Continuous beams are considered to be any beam in which at least one end of the beam is continuous, i. e. this includes propped cantilevers at the end of a series of continuous beams.

Example 5: v A rectangular concrete beam 250 mm wide × 475 mm overall depth is simply supported over a 7. 0 m span. Using the data given, check the suitability of the beam with respect to deflection. v Data: § Characteristic strength of concrete (fcu) 40 N/mm 2 § Characteristic strength of main steel (fy) 460 N/mm 2 § Design ultimate bending moment at mid-span (M) 150. 0 k. Nm § Assume the distance to the centre of the main steel from the tension face is 50 mm

Solution: v Effective depth d = (475 − 50) = 425 mm v Check that section is singly-reinforced: Adopt 3/20 mm diameter HYS bars providing 943 mm 2

Solution: v Clause 3. 4. 6 There is no redistribution of moments β b = 1. 0 v Table 3. 9 The beam is simply supported: v L/d = 20 v Clause 3. 4. 6. 5 M/bd 2 = (K × fcu ) = (0. 082 × 40) = 3. 32 N/mm 2 Interpolate from Table 3. 10 0. 92

Solution: v. L/d x value from Table 3. 10 = L/d x 0. 92 v. Allowable deflection = 20 x 0. 92 = 18. 4 v. Actual deflection = L/d = 6000/425 = 14. 12 So, the deflection is acceptable.

Minimum Areas of Steel (Cl. 3. 12. 5) v Minimum areas of steel are required in structural elements to ensure that any unnecessary cracking due to thermal/shrinkage effects or tension induced by accidental loading can be minimized. v The required minimum steel areas for different situations, i. e. tension or compression reinforcement, rectangular beams/slabs, flanged beams and columns/walls, are given in Table 3. 25

Example 5: Minimum Areas of Steel v Determine the minimum areas of steel required for the cross-sections shown in Figure 5. 57; in all cases fy = 460 N/mm 2.

(a) Rectangular beam The required area of steel is given by: 100 As /Ac = 0. 13 vwhere: § As is the minimum recommended area of reinforcement, § Ac is the total area of concrete. v. Ac = (550 × 320) = 176 × 103 mm 2 v. As = (0. 13× 176× 103)/100= 229 mm 2

(b) Flanged beam (see Cl 3. 4. 1. 5) v where: § b is the breadth of the section bw is the breadth or effective breadth of the rib; for a box, T or I section, bw is § taken as the average breadth of the concrete below the flange, § h is the overall depth of the cross-section of a reinforced member. v bw / b = (300 / 1200) = 0. 25 < 0. 4 the required area of steel is given by: v 100 As /bwh = 0. 18 v bw h = (300 × 600) = 180 × 103 mm 2 v As = 0. 18× 180× 103/100 = 324 mm 2

Maximum Areas of Steel (Cl. 3. 12. 6) v. The maximum percentages of steel given in the code are based on the physical requirements for placing the concrete and fixing the steel. They are: v. In beams (Clause 3. 12. 6. 1): § As ≤ 4% Ac § As′ ≤ 4% Ac

Minimum Spacing of Bars (Cl. 3. 12. 11. 1) v Guidance is given in the code for minimum bar spacing to ensure that members can be constructed achieving adequate penetration and compaction of the concrete to enable the reinforcement to perform as designed. v It is important that reinforcing bars are surrounded by concrete for two main reasons: § (i) to develop sufficient bond between the concrete and the bars such that the required forces are transferred between the steel and the concrete, and § (ii) to provide protection to the steel against corrosion, fire, etc.

Maximum Spacing of Bars (Clause 3. 12. 11. 2) v The requirement to limit the maximum spacing of reinforcement is to minimize surface cracking. v Using Table 3. 28. v Since very small bars when mixed with larger bars would invalidate the assumptions on which the Table 3. 28 values are based, the code specifies that any bar in a section with a diameter < (0. 45 × the largest bar in the section) should be ignored except when considering those in the side faces of deep beams.

Maximum Spacing of Bars v. The minimum size of bar to be used in the side faces is given in Clause 3. 12. 5. 4 as: vbar diameter ≥ vwhere: sb is the bar spacing (e. g. 250 mm) vb is the breadth of the section ≤ 500 mm