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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