Beam Design

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    22-Dec-2015

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beam design

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  • Beam Design

  • RC BEAM DESIGNBased on BS 8110

  • Partial Safety Factor mCl. 2.4.4.1

    Given in Table 2.2

  • DurabilityCl. 2.2.4 and 3.1.5Exposure conditions (Integrity of rc concrete ability to prevent corrosion) : moderate, severe, very severe, most severe and abrasive (Table 3.2)Protection a agaist corrosion of steel (Table 3.3)Fire resistance requirement (Table 3.4)

  • Nominal coverCl. 3.3.1.1Bar Size (Cl. 3.3.1.2)Single Bars: nominal cover main bar diameter, d1Paired bars: nominal cover 2 d1Bundles bars: 2 (Aequivalent/)

  • 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:Exposure condition mildCharacteristic strength of concrete (fcu) 40 N/mm2Nominal maximum aggregate size (hagg) 20 mmDiameter of main tension steel 25 mmDiameter of shear links 8 mmMinimum required fire resistance 1.5 hours

  • SOLUTIONClause 3.3.1.2 Nominal cover (main bar diameter link diameter)

    (25 8) = 17 mmClause 3.3.1.3 Nominal cover nominal maximum aggregate size > 20 mmClause 3.3.3 Exposure condition is mildGrade of concrete is C40Table 3.3 Nominal cover 20 mm*Clause 3.3.6 Minimum fire resistance = 1.5 hrThe beam is simply supportedTable 3.4 Nominal cover 20 mm

    The required nominal cover = 20 mm

  • Flexural Strength of SectionsThe 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, BS8110:Part 1

  • Concrete CapacityThe 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 CapacityConsider the moment of the compressive force about the line of action of Ft :Mult,concrete = (Fc z)

    where Fc = compressive force = (stress area) = [0.45fcu (b 0.9x )] (for the maximum concrete force x = d/2) = [0.45fcu (b 0.45d)] = 0.2bdfcu z = lever arm = [d (0.5 0.45d)] = 0.775d{Note: In general z = [d (0.5 0.9x)] }Mult,concrete = (0.2bdfcu )(0.775d) =0.156bd2fcuThis equation can be rewritten as 0.156 =M/bd2fcu

  • Steel CapacityMult,steel = (Fs z)where Fs = tensile force = (stress area)

    = (0.95fy As) z = lever armMult,steel = 0.95fyAs zAs = M/0.95fyzConsider z, the lever arm:z = [d (0.5 0.9x)] , 0.9x = 2(d z)Mult,concrete = [0.45fcu (b 0.9x)] z (substitute for 0.9x)

    = 0.9fcu b (d z) z

  • Steel CapacityMult,concrete = K bd 2 fcuK bd 2 fcu = 0.9fcu b (d z) z Kd 2 = 0.9fcu dz 0.9z 2z2 dz + Kd2/0.9 = 0The solution of this quadratic equation (ax2 + 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 caseWhen K K for a singly-reinforced section the maximum moment permitted, based on the concrete strength, is equal to 0.156 bd2fcu

    When K > K a section requires compression reinforcement.

  • Area of steel reinforcement, As

  • Areas/metre width for various pitches of bars

  • Example 1: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.Data:Characteristic strength of concrete (fcu) 40 N/mm2Characteristic strength of steel (fy) 460 N/mm2

  • Solution:Strength based on concrete:

    Mult = 0.156bd2fcu = (0.156 250 4202 40)/106 = 275.2 kNmStrength based on steel:

    As = 1260 mm2 z = 0.775 d = 0.775 x 420 = 325.5Mult = 0.95fyAs z = (0.95 460 1260 325.5)/106 = 179.5 kNmThe maximum design moment which can be applied is: Mult = 179.5 kNm

  • Example 2:The cross-section of a simply supported rectangular beam is shown in Figure 5.30. Using the data given, determine the maximum ultimate moment which can be applied to the section assuming it to be singly-reinforced.

    Data:Characteristic strength of concrete (fcu) 30 N/mm2Characteristic strength of steel (fy) 460 N/mm2Nominal maximum aggregate size (hagg) 20 mmDiameter of main tension steel 32 mmDiameter of shear links 8 mmExposure condition mildMinimum required fire resistance 1.0 hour

  • Solution:Clause 3.3.1 Nominal cover to all steelClause 3.3.1.2 bar sizecover (32 8) = 24 mmClause 3.3.1.3 Nominal maximum aggregate sizecover 20 mmClause 3.3.3 Exposure condition: mildTable 3.3 and fcu = 30 N/mm2cover 25 mmClause 3.3.6 Min. fire resistance: 1 hrbeam is simply supportedTable 3.4 cover 20 mmThe required nominal cover to all steel = 25 mm

  • SolutionFigure 3.2 Minimum width b for 1 hour fire resistance = 200 mm adequateEffective depth

    d = (h cover link diameter bar diameter/2) = (475 25 8 16) = 426 mmStrength based on concrete:

    Mult = 0.156bd2fcu = (0.156 250 4262 30)/106 = 212.3 kNmStrength based on steel:

    As = 2410 mm2 z = 0.775 dMult = 0.95fyAs z = (0.95 460 2410 0.775 426)/106 = 349.9 kNmThe maximum design moment which can be applied is: Mult = 212.3 kNm

  • Example 3: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 kNm.Data:Characteristic strength of concrete (fcu) 40 N/mm2Characteristic strength of steel (fy) 460 N/mm2Nominal maximum aggregate size (hagg) 20 mmDiameter of main tension steel Assume 25 mmDiameter of shear links 10 mmExposure condition moderateMinimum required fire resistance 2.0 hours

  • Solution:Clause 3.3.1 Nominal cover to all steelClause 3.3.1.2 bar sizecover = (25 10) = 15 mmClause 3.3.1.3 Nominal maximumaggregate sizecover = 20 mmClause 3.3.3 Exposure condition: moderateTable 3.3 fcu = 40 N/mm2cover 30 mmClause 3.3.6 Min. fire resistance: 2.0 hrbeam is simply supportedTable 3.4 cover 40 mmThe required nominal cover to the main steel = 40 mmFigure 3.2 Minimum width b for 2 hours, fire resistance = 200 mm adequateFigure 5.31

  • Solution:Effective depth d = (h cover link diameter bar diameter/2)

    = (450 40 10 13) = 387 mmClause 3.4.4.4Check that the section is singly-reinforced:K = M/bd2fcu = 0.125 < 0.156Since K < K the section is singly-reinforced

    As = M/0.95fyz = [(150 106) / (0.95 460 0.83 387)]

    = 1068 mm2Adopt 4/ 20 mm diameter HYS bars providing 1260 mm2.

  • Shear Strength of SectionsThe equation in BS 8110 (Cl 3.4.5.2)

    Wherev = design shear stressV = design shear force due to ultimate loadsbv=breadth of the sectiond =effective depthv 0.8 (fcu) or 5 N/mm2

  • Shear Strength of SectionsThis 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 cross-section 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:(vc + 0.4) < v < 0.8 (fcu)1/2 or 5 N/mm2

  • Shear Strength of SectionsIt is important to provide minimum areas of steel in concrete to minimize thermal and shrinkage cracking, etc. Minimum links are specified in the code which provide a shear resistance of 0.4N/mm2 in addition to the design concrete shear stress vc, and consequently designed links are required when v > (vc + 0.4) as indicated. In Table 3.7: minimum links are required when 0.5vc < v < (vc + 0.4)The cross-sectional area of the links required is given by:

  • Example 3: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.Data:Characteristic strength of concrete (fcu) 40 N/mm2Characteristic strength of mild steel (fyv) 250 N/mm2Maximum shear force at the support 40 kN

  • Solution:Solution:Consider the end of the beam at the support where the shear force is a maximum = 40 kNClause 3.4.5.2

  • Solution:Table 3.7 To determine the required reinforcement, evaluate vc and either adopt minimum links throughout or use designed links.As = area of 2/16 mm diameter bars = 402 mm2

    From Table 3.8

  • Solution:Since 0.5vc < v < (vc + 0.4), minimum links are required for the whole length of the beam.The cross-sectional area of the links required is given by

    Option 1:Assume 2-legged / 6 mm diameter mild steel links. Asv = 56.6 mm2Table 3.7 The spacing required , Clause 3.4.5.5The maximum spacing of the links 0.75d

    = (0.75 x 225) = 169 mmAdopt 6 mm diameter mild steel links @ 150 mm centres throughout the length of the beam.

  • Solution:Option 2:Clause 3.4.5.5The maximum spacing of the links 0.75d= (0.75 225) = 169 mmAssume sv = 150 mm.

    *Adopt 6 mm diameter mild steel links (56.6 mm2) @ 150 mm centres throughout the length of the beam as in option 1.

  • Example 4: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.Data:Characteristic strength of concrete (fcu) 40 N/mm2Characteristic strength of mild steel (fyv) 250 N/mm2Characteristic dead load (gk) 5.0 kN/mCharacteristic imposed load (qk) 30.0 kN/m

  • Solution:Ultimate design load = [(1.4 5.0) + (1.6 30.0)] = 55.0 kN/mUltimate design shear force at the support V = (55.0 3.5) = 192.5 kN

  • Solution:Table 3.7 : To determine the required reinforcement evalua