reinforced concrete beam design - deflections

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Computation Sheet NoClient Job NoProject/Job BySubject Date:

Title : Reinforced Concrete Beam Design

General Description :

Limitations :

AS3600 - 2009 (Incorporating Amendment 1, 2010 )

Nomenclature : Symbols and notation as generally used in AS3600.

Input : Yellow cells require data input by designer

For bending in rectangular beams including T and L beams

Section geometry including width and depth and for T and L beams dimensions of flange and cover Material propertiesBending moment M* at a sectionArea of reinforcement based on bar diameter and number of barsFor shear in beamsSection geometry including width and depth, cover and material propertiesFitment diameterArea of tensile reinforcement at section where shear is to be checkedUltimate design shear force at cross section being checkedInsert value of β1Cross sectional area of shear reinforcementFor torsion in beamsSection geometry including width and depth, cover and material propertiesUltimate design actions including bending moment, shear force and torsional moment

Cross sectional area of torsional reinforcementInsert value of β1For deemed to comply ratios for deflection for beamsSection geometry including width and depth, cover and material propertiesArea longitudinal tensile reinforcement in compression reinforcement

Factor k1 for deflectionMean value of the modulus of elasticity of the concrete at 28 daysDeflection limitations from table 2.3.2 of AS 3600

Where pink fill is used, it alerts designer to options or information

Output : Boxed cells with green background calculated automatically using formulae.For flexure in rectangular beams including T and L beamsArea of reinforcement required

Minimum tensile reinforcement requiredFor shear in beams

Spacing of fitments required and the maximum spacing of fitments allowedChecks for minimum reinforcement and spacingFor torsion in beams

Requirements or torsional reinforcement and additional longitudinal reinforcementChecks that the diameter and the spacing of fitment is appropriateFor deemed to comply ratios for deflection for beamsCalculation on the ratio of compression reinforcement to tension reinforcementReinforcement ratio at the midspanThe total deflection ratio The incremental deflection ratio

Provides summary of the results

Feedback : For comments, corrections, suggestions or other feedback regarding this spread sheet, please contact the CCAA

Version 1.1

Date 7/1/2012

Calculates tensile reinforcement requirements for a singly reinforced rectangular concrete beam in flexure. Calculates tensile reinforcement requirements for a singly reinforced T or L concrete beam in flexure. Calculates shear reinforcement requirements for a singly reinforced rectangular concrete beam in shear. Calculates torsional reinforcement requirements for a singly reinforced rectangular concrete beam in torsion. Calculates deemed-to-comply deflections for a reinforced rectangular concrete beam

For rectangular sections or T or L beams only . Fitments are perpendicular only

Codes / Theoretical Basis : Warner, Rangan, Hall & Faulkes, Concrete Structures, Longman, Melbourne, 1999

Foster, Kilpatrick and Warner, Reinforced Concrete Basics 2E, Pearson Australia, 2010

Diameter and spacing of fitments, area and diameter of diameter of longitudinal flexural reinforcement

Applied actions

Maximum moment at kuo

Various values for shear parameters including the Vumax, Vuc, Vu.min, and Vuc

Torsional modulus Jt

Various values for torsion parameters including the Tu.max, Tuc, Tu.min, and Tuc

Various values for shear parameters including the Vu.max, Vuc, Vu.min, and Vuc


Computation Sheet No

Client Job No

Project/Job By

Subject Date:

Reinforced Rectangular Concrete Beam Design - BendingDesign for beams that are under reinforced

Data InputCapacity Reduction Factor

Geometry & Material PropertiesWidth of rectangular section b 3408 mm f 0.8

Overall depth D 350 mm 25 MPa

Cover to fitments 30 mm 500 MPa

Effective depth d 300 mm 0.36 (Effective depth usually = D - cover to fitment - fitment size - distance Fitment diameter 10 mm to the centroid of the tensile reo, rounded down to nearest 5mm )

Applied actionsUltimate strength moment at section M* 303 kNm

Initial 2,971 This figure is a first approximation of the required reoIterate as required to get the area of reo required

1670

Proposed reinforcement arrangement (use minimum of 2 bars)

Bar size 12 16 20 24 28 32 36 40 Area of bar mm2 113 201 314 452 616 804 1020 1260 Theoretical No. of bars 26.29 14.78 9.46 6.57 4.82 3.69 2.91 2.36Number of bars required 27.0 15.0 10.0 7.0 5.0 4.0 3.0 3.0 (Note number of bars rounded up)

Area of bars mm2 Total 3051 3015 3140 3164 3080 3216 3060 3780 (Note areas of bars rounded up)

Number of bars required 26.0 14.0 9.0 6.0 4.0 3.0 2.0 2.0 (Note bars of number rounded down)

Area of bars mm2 Total 2938 2814 2826 2712 2464 2412 2040 2520 (Note areas of bars rounded down)

3140 Choose no of bars and area from above table and input it into the yellow box left. Use a minimum of 2 bars

2,620 (This area of reinforcement is based on interpolation)

13,296

Design Calculations

Percentage

1351 kNm OK 446%

13,296

Check Moment using stress blocks

363.2 kNm OK 120%

363.2 kNm OK 120%

0.09 OK 24%

Ø = = 0.80 limits 0.6 ≤ Ø ≤ 0.8

Calculations (Cl 8.1.3)

= 0.85

= 0.85 limits 0.67 ≤ α2 ≤ 0.85

Initial strength requirements Requirements based on ultimate strength moment M* and d

Initial calculation of reo = 2971

Minimum strength requirements (Cl 8.1.6.1)

= 3.00 MPa

= 1670 OK 188%

Results based on tensile reinforcement sizes chosen Ultimate Design Moment M* 303.0 kNm

Depth to neutral axis 25.5 mm Actual bending capacity 363.2 kNm

Neutral axis parameter 0.085 Reinforcement ratio p 0.003

Depth of compression block 21.7 mm Area of steel chosen 3140

f'c

fsy

kuo

Ast required for M* < fMu Nominal Ast.required mm2

Minimum Ast. mm2

Suggested bar numbers & sizes for limit state requirement M* < fM u

Ast.provided mm2

Minimum Ast for flexure mm2

Maximum Ast for flexure kuo=0.36 mm2

Maximum Moment Muo at kuo = 0.36

Check ØMuo = Ø α2 f'c γ kuo (1-0.5 γ kuo) b d2 fMuo M* < fMu

Ast max= 0.306 γ f'c/fsy b d Ast max= mm2

Check ØMu = Ast fsy d (1- 0.5/α2 (Ast fsy / (b d f'c)) = M* with Ast =Ast.provided fMu M* < fMu

Check ØMu = Ø f'c γ ku (1-0.5 γ ku) bd2 for concrete stress block fMu M* < fMu

Check ku calculated using provided Ast.provided ku ku0 ≤ 0.36

Calculations (Table 2.2.2) with kuo=0.36

(1.19 − 13kuo/12)

g = 1.05 - 0.007 f'c limits 0.67 ≤ g ≤ 0.85

α2 = 1.0 − 0.003 f’c

Ast = M* / (Ø *fsy * 0.85 *d) mm2

f'ctf = 0.6 ( f'c )0.5

Deemed to comply A st.min = 0.2 ( D/d )2 f'ctf / fsy bw d mm2 Astmin

< Ast provided

dn f Mu

ku

g ku d Ast mm2

d

b

D



Client Job No

Project/Job By

Subject Date:

Reinforced Concrete T- or L- Beam Design - BendingFor beams that are under reinforced

Data Input Capacity Reduction Factor

Geometry & Material Properties

340 mm f 0.8

1330 mm 15 MPa

270 mm 200 MPa

Overall depth D 930 mm 0.36

Bottom cover to fitments 30 mm Fitment diameter 12 mmEffective depth d 875 mm

(Effective depth usually = D - cover to fitment - fitment size - distance WARNING fsy not to AS 3600

to the centroid of the tensile reo, rounded down to nearest 5mm ) WARNING f'c not to AS 3600

Applied actionsUltimate strength moment at section M* 359 kNm

Initial 3,017 This figure is a first approximation of the required reo.For a T or L beam the reo required will probably be lower.

Approximate Strength requirements Iterate as required to get the area of reo required. Requirements based on ultimate strength moment M* and d

3,017

Proposed reinforcement arrangement (use minimum of 2 bars)

Bar size 12 16 20 24 28 32 36 40 Area of bar mm2 113 201 314 452 616 804 1020 1260 Theoretical No. of bars 26.70 15.01 9.61 6.67 4.90 3.75 2.96 2.39Number of bars required 27.0 16.0 10.0 7.0 5.0 4.0 3.0 3.0 (Note number of bars rounded up)

Area of bars mm2 Total 3,051 3,216 3,140 3,164 3,080 3,216 3,060 3,780 (Note areas of bars rounded up)

Number of bars required 26.0 15.0 9.0 6.0 4.0 3.0 2.0 2.0 (Note bars of number rounded down)

Area of bars mm2 Total 2,938 3,015 2,826 2,712 2,464 2,412 2,040 2,520 (Note areas of bars rounded down)

4,050

2,636 For T or L beam with the Stress Block in the Flange Only

-3,430

Design Calculations

Depth to neutral axis 48 mm

Check if

Calculations as a Rectangular Beam

688 kNm

OK 192%

Calculations as T or L Beam with Compression Stress Block in the Flange Only

Check Moment using stress blocks assuming a rectangular beam with part of the flange in compression

810 kN

Tension force in reinforcing assuming it is at yield T 810 kN

T 56 mm

Calculate Ultimate Moment 551.5 kNm

0.06

Width of web bw

Effective width of compression flange bef f'c

Thickness of flange, tf fsy

kuo

Ast required for M* < ØMu Ast.required mm2

Initial calculation of reo Ast =M* / (Ø *fsy * 0.85 *d) mm2

Suggested bar numbers & sizes for limit state requirement M* < fM u

Ast.provided mm2 Choose no of bars and area from above table and input it into the yellow box left. Use a minimum of 2 bars

Minimum Ast for flexure mm2

Minimum Ast for flexure mm2 For T or L beam with the Stress Block in the Flange and Web. If negative do not use this value

Check t ≤ Ast * fsy / (α2 * f'c * bw) dn notional=

dn notional< tf

DESIGN AS RECTANGULAR BEAM WITH THE COMPRESSION STRESS BLOCK IN THE FLANGE

ONLY

USE THIS CALCULATIONS FOR T OR L BEAM WITH THE COMPRESSION STRESS BLOCK IN

THE FLANGE

Maximum Moment Muo at kuo=0.36 assuming a rectangular beam

Check ØMuo = Ø α2 f'c γ kuo (1-0.5 γ kuo) bw d2 ØMuo

M* < ØMuo

USE THIS CALCULATIONS FOR T OR L BEAM WITH THE COMPRESSION STRESS

BLOCK IN THE FLANGE

Compression force in flange Cf Cf = α2 f'c Ύ dn bef

T = Ast fsy

dn =

α2 f'c γ bw

ØMu= Ø*T* (d-0.5 γ dn)

Check ku calculated using Ast.provided ku

Dd

bw

bef

tf

Check OK 154%

Check OK 561%

Calculations as T or L Beam with Compression Stress Block in the Flange & Web

Compression force in flange outstand 3408.075 kN

Tension force in reinforcing assuming it is at yield 810 kN

Compression force in web and flange equal to web width -2598.075 kN

-705 mm

Calculate Ultimate Moment -423.9 kNm

-0.81

FAIL -118%

OK -224%

Calculations for both types of beam

Ø = = 0.80 limits 0.6 ≤ Ø ≤ 0.8

Calculations (Cl 8.1.3)

= 0.85

= 0.85 limits 0.67 ≤ α2 ≤ 0.85

Minimum strength requirements (Cl 8.1.6.1)

= 2.32 MPa

= 781

OK 519%

Results based on reinforcement arrangement for a T or L Beam when the compression stress block is in the flange onlyDepth to neutral axis 56.2 mm Actual bending capacity 551.5 kNm

Neutral axis parameter 0.064 Reinforcement ratio p 0.014

Depth of compression block 47.8 mm Area of steel chosen 4050

Depth to neutral axis -705.1 mm Actual bending capacity -423.9 kNm

Neutral axis parameter -0.806 Reinforcement ratio p 0.014

Depth of compression block -599.3 mm Area of steel chosen 4050

M* < fMu

ku < 0.36

USE THIS CALCULATIONS FOR T OR L BEAM WITH THE COMPRESSION STRESS

BLOCK IN THE FLANGE

Cf = α2 f'c tf (bef - bw)

T =Ast fsy

Cw =( T - Cf )

dn = T-Cf

α2 f'c γ bw

ØMu= Ø (Cf (d-0.5tf) + Cw (d-0.5 γ dn))

Check ku calculated using Ast.provided ku

M* < fMu

ku< 0.36

Calculations (Table 2.2.2) with kuo=0.36

(1.19 − 13kuo/12)

g = 1.05 - 0.007 f'c limits 0.67 ≤ g ≤ 0.85

α2 = 1.0 − 0.003 f’c

f'ctf = 0.6 ( f'c )0.5

Deemed to comply A st.min = 0.2 ( D/d )2 f'ctf / fsy bw d mm2

Astmin < Ast provided

dn Ø Mu

ku

g dn Ast mm2

Results based on reinforcement arrangement for T or L beam when the compression stress block is in the flange and webdn Ø Mu

ku

g dn Ast mm2



Client Job No

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Reinforced Concrete Beam Design - Shear Cl 8.2 (Design shear force at a cross section)

Data Input Capacity Reduction Factor


500 mm f 0.7

Depth, D 700 mm 25 MPa

Bottom cover to fitments 50 mm 500 MPa

615 mm

Fitment diameter 12 mm

to the centroid of the outermost layer of tensile reo, rounded down to nearest 5mm )

1,808

Applied actionsDesign shear force at a cross section V* 360 kN

Maximum Transverse Spacing of Fitments = 600 mmCl 8.2.1.2. Single fitments OK

Design Calculations

= 1,538 kN OK 299%

= 1,076 kN

Calculate shear strength of a beam without shear reinforcement Cl 8.2.7

Max Spacing of fitments 300 mm

105 mm2

β1

1.100 β1 ≥ 1.1

OTHERWISE 1.084 β1 ≥ 0.8 For beams with no shear reo

Insert value for β1 β1 = 1.100

β2 =

β2 = 1

β3 =

β3 = 1

2.92 MPa

Shear strength of a beam without shear reinforcement Cl 8.2.7 178.5 kN

125.0 kN

Calculate shear strength of a beam with minimum shear reinforcement Cl 8.2.9

= 363.0 kN 254.1 kN

= 332.3 kN

232.6 kN therefore = 363.0 kN therefore = 254.1 kN

Requirements for shear reinforcing Cl 8.2.5 V* = 360.0 kN

= 62.5 kN Shear Reo may be Required

D > 750 mm No

= 62.5 kN Shear Reo Required



Shallow Wide Beams Check if D does not exceed the greater of 250mm and half the width of the web Cl 8.2.5 (i)

= 125 kN Shear Reo Required

Check if D does not exceed the greater of 250mm D = 700 mm D > 250 D > 250

or half the width of the web = 250 mm Shear Reo Required

V* = 360 kN

Spacing of Fitments Cl 8.2.12.2 Maximum spacing 0.75 D = 525 mm Lesser of 0.75D or 500 500 mm

of fitments 0.5 D = 350 mm Lesser of 0.5D or 300 300 mm

= 254.1 kN 300 mm

Minimum shear reinforcement Cl 8.2.8 Max Spacing = 105 mm2

= 90 mm2

= 105 mm2

Check shear capacity of the beam with minimum reinforcement at minimum spacing

(for perpendicular shear reinforcement)

Width, bv = effective width for shear Cl 8.2.6

f'c

fsy.f

Depth do to centroid of outermost layer of reo

(Depth do usually = D - cover to fitment - fitment size - distance

Ast.provided mm2

Calculate web crushing limit Vu.max CL 8.2.6

Vu.max = 0.2 fc′ bv do Vu.max V* < f Vu.max

ØVu.max

Vuc= β1 β2 β3 bv do fcv { Ast }1/3

( bv do)

Asv.min

For members where the cross-sectional area of shear reinforcement provided (Asv) is equal to or greater than the minimum area specified in Clause 8.2.8

β1 = 1.1(1.6 − do/1000) ≥ 1.1

β1 = 1.1(1.6 − do/1000) ≥ 0.8

1, for members subject to pure bending; or 1 − (N*/3.5Ag) ≥ 0 for members subject to axial tension; or 1 + (N*/14Ag) for members subject to axial compression

1, or may be taken as 2do/av but not greater than 2, provided the applied loads and the support are orientated so as to create diagonal compression over the length (a v)

fcv = f ′c 1/3 ≤ 4MPa fcv =

Vuc =

ØVuc =

Vu.min = Vuc + 0.10√ fc′ bvdo) ≥ Vuc + 0.6 bv do Vuc + 0.6 bv do

Ø(Vuc + 0.6 bv do) =

Vuc + 0.10√ fc′ bvdo)

Ø(Vuc + 0.10√ fc′ bvdo) =

Vu.min

ØVu.min

Check if V* ≤ 0.5ØVuc 0.5 Ø Vuc V* ≤ 0.5 Ø Vuc

Check if 0.5 Ø Vuc < V * ≤ ØVu.min 0.5 Ø Vuc 0.5Ø Vuc < V*

Ø Vu.min V* < Ø Vu.min

Check if V * > Ø Vu.min Ø Vu.min V* > Ø Vu.min

Check if V* ≤ Ø Vuc Ø Vuc V* < Ø Vuc

D > Bw/2

Check for shallow wide beams if V* ≤ Ø Vuc and D does not exceed the greater of 250mm and half the width of the web Cl 8.2.5 (i) then minimum shear reinforcement is waived

Where V* > Ø Vu.min however minimum reinforcement shall be provided in accordance with Clause 8.2.10

Ø Vu.min V* < Ø Vu.min

Asv.min = 0.06 √ f′c bv s / fsy.f ≥0.35 bv s / fsy.f 0.35 bv s / fsy.f

0.06 √ f′c bv s / fsy.f

Asv.min

Ø Vus min = Ø (Asv min fsy.f do / s) cot θv Adopt θv = 45 deg

do

bv

D

4df or 70mm whichever is the greater

H41

Choose β1 best suited

= 75.3 kN (1)

= 235.0 kN (2) If figure is negative then concrete will take all the shear

= 235 kN 105 mm2

= 360 kN Design shear force at a cross section V* = 360 kN

= 328 mm2 at maximum spacing = 300 mm

Warning Grade 250 Fitment diameters and areas for 2 legs

Fitment Dia (nominal) R6 R10 L6 L8 L10 N12 N16 N20Grade of Fitment 250 250 500 500 500 500 500 500 Area of one bar mm2 30 80 28 45 71 113 201 314 Area of 2 legs mm2 60 160 57 91 142 226 402 628 Single fitments

Area of 4 legs mm2 120 320 113 182 284 452 804 1256 Fitments in pairs

Area of 6 legs mm2 180 480 170 272 425 678 1206 1884 Fitments in threes

Area of 8 legs mm2 240 640 226 363 567 904 1608 2512 Fitments in fours

Area of 10 legs mm2 300 800 283 454 709 1130 2010 3140 Fitments in fives

Maximum Transverse Spacing of Fitments = 600 mm Single fitments OKIf the beam width is greater than maximum spacing then will need to have fitments in groups

= 500 MPaFitment Dia assumed = 12 mm

Because of the various combinations, some iteration will be required for shear reinforcementSee table above for various areas and diameters of fitments and different numbers of legsCheck input of fitment size and grade at top RH part of this sheet is consistent.If spacing of fitments greater than the maximum spacing, reduce area of shear reinforcementNote it is usual to rationalise the spacing down to the nearest 25 mm

= 142 mm2

= 105 mm2

Spacing of fitments required = 130 mm

Maximum spacing of fitments allowed = 130 mm

Results based on fitment arrangement with 2 legs Design shear force at the cross section V* 360.0 kN

1076.3 kN

Fitment diameter 12 mm 125.0 kN

Maximum Spacing 130 mm 254.1 kN

360.0 kN

Ø Vus min

Contribution of shear reinforcing Vus Cl 8.2.2

Vus = V*/Ø - Vuc Ø Vus

with a minimum reinforcement. Use Vus min

Take the greater value of Ø Vus and Ø Vus min ie (1) or (2) Ø Vus Design but area must exceed Asv.min

Ø Vuc + Ø Vus Design = Ø Vu

Therefore Asv = Vus/( (fsy.f do / s) cot θv) Asv

fsy.f

Input Asv = cross sectional area of shear reinforcement

Asv.min

Vu.max = 0.2 fc′ bv do Ø Vu.max

Vuc= β1 β2 β3 bv do fcv { A st }1/3 Ø Vuc

bv do

Vu.min = Vuc + 0.10√ fc′ bvdo) ≥ Vuc + 0.6b bvdoØ Vu.min

Ø Vu = Ø Vuc + Ø Vus Ø Vu



Client Job No

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Subject Date:

Reinforced Concrete Beam Design - Torsion Cl 8.3 (Design torsion at a cross section)

Data InputCapacity Reduction Factor


500 mm f 0.7

Overall depth, D 700 mm 20 MPa

Bottom cover to fitments 40 mm 500 MPa

615 mm 500 MPa

Fitment diameter 12 mm

to the centroid of the outermost layer of tensile reo, rounded down to nearest 5mm ) Spacing of fitment s 200 mmDiameter of longitudinal reo d 24 mm

Longitudinal Tensile Reinforcement (fully anchored) Flexural Reo 1,808

1,808

Applied actionsUltimate strength moment at section M* 380 kNmDesign shear force at a cross section V* 360 kN

Design torsional moment at a cross section T* 50 kNm

Design CalculationsCalculate torsional modulus Jt CL 8.3.3 x = 500 mm y = 700 mm

5.775E+07 mm3

= 231.0 kN OK 323%

= 161.7 kN

= 1230.0 kN OK 239%= 861.0 kN = 0.73 Web Crushing OK

Calculate torsional strength of a beam Cl 8.3.5 Details of closed fitmentsWithout closed fitments and no prestress

= 77.5 kNm

= 54.2 kNm

With closed fitments= 212,784 mm2

= 1,888 mm

Input cross sectional area of bar forming fitment from the diameter chosen above top RHS of spread sheet

See table below for various areas of bar of the fitments of various diameters

Fitment Dia (nominal) R6 R10 L6 L8 L10 N12 N16 N20

Grade 250 250 500 500 500 500 500 500 Area of one bar mm2 30 80 28 45 71 113 201 314 Single fitments

Area of two bars mm2 60 160 57 91 142 226 402 628 Fitments in pairs

Area of three bars mm2 120 320 113 182 284 452 804 1256 Fitments in threes

= 113 mm2 at spacing = 200 mm

(Cross sectional area of bar forming the fitment)Adopt θv = 45 deg and cot θv = 1.0 120.2 kN

84.2 kN

Calculate shear strength of a beam without shear reinforcement Cl 8.2.7

β1

1.100 β1 ≥ 1.1

OTHERWISE 1.084 β1 ≥ 0.8 For beams with no shear reo

Insert value for β1 β1 = 1.100

β2 = β2 = 1

β3 = β3 = 1

2.71 MPa

165.7 kN

116.0 kN

Requirements for torsional reinforcing Cl 8.3.4 (a)

= 13.6 kNm Torsional Reo Required

Check if = 4.0 Torsional Reo Required

The overall depth does not exceed the greater of 250 mm Depth = 700 If D < 250 Torsional Reo may be Requiredand half the width of the web

Half the Width = 250 D < b/2 Torsional Reo may be Required

Check if = 4.0 Torsional Reo Required

Requirements for torsional reinforcing Cl 8.3.4 (b)Check if = 0.59 Torsional reo is required

OK Complies with Cl 8.3.4 (b)If NOT OK then either revise size and spacing or both in top right-hand corner of spread sheet

Requirements for additional longitudinal reinforcing (Cl 8.3.6)= 533 mm2

Requirements for additional longitudinal reinforcing (Cl 8.3.6)= 533 mm2

Width, bv = effective width for shear Cl 8.2.6

f'c

fsy.f

Depth do to centroid of outermost reo fsy

(Depth do usually = D - cover to fitment - fitment size - distance

Ast. mm2

Ast.provided mm2

Jt =

Calculate web crushing limit CL 8.3.3 T* + V* < 1.0

ØTu.max ØVu.max

Tu.max = 0.2Jtf’c Tu.max T* < Ø Tu.max

ØTu.max

Vu.max = 0.2 fc′ bv do Vu.max V* <Ø Vu.max

ØVu.max

T* + V*

ØTu.max ØVu.max

Tuc = 0.3Jt √f’c Tuc

ØTuc

Tus = Asw fsy.f 2At cot Ɵv/s At = area of a polygon

ut = perimeter of the polygon defined for At

Fitment diameters and cross sectional areas of bar forming the closed fitment asw

Asw

Tus = Asw fsy.f 2At cot Ɵv/s Tus =

ØTus =

Vuc= β1 β2 β3 bv do fcv { Ast }1/3 /(bv do)

For members where the cross-sectional area of shear reinforcement provided (Asv) is equal to or greater than the minimum area specified in Clause 8.2.8

β1 = 1.1(1.6 − do/1000) ≥ 1.1

β1 = 1.1(1.6 − do/1000) ≥ 0.8

1, for members subject to pure bending; or 1 − (N*/3.5Ag) ≥ 0 for members subject to axial tension; or 1 + (N*/14Ag) for members subject to axial compression

1, or may be taken as 2do/av but not greater than 2, provided the applied loads and the support are orientated so as to create diagonal compression over the length (av)

fcv = f ′c 1/3 ≤ 4MPa fcv =

Vuc =

Ø Vuc =

Check if T* ≤ 0.25ØTuc 0.25 Ø Tuc T* < 0.25 Ø Tuc

T* + V* ≤ 0.5 T* + V* T* + V* ≤ 0.5

ØTuc ØVuc ØTuc ØVuc ØTuc ØVuc

T* + V* ≤1.0 T* + V* T* + V* ≤1.0

ØTuc ØVuc ØTuc ØVuc ØTuc ØVuc

T* ≤1.0 T* T* ≤ 1.0

ØTus ØTus ØTus

Alt = (0.5 fsy.f / fsy)(Asw / s)( ut cot2 θv) (in the tensile zone).

Alt = (0.5 fsy.f / fsy)(Asw / s)( ut cot2 θv) (in the compression zone).

do

bv

D

4df or 70mm whichever is the greater

Proposed additional reinforcement arrangement

Bar size 12 16 20 24 28 32 36 40

Area of bar mm2 113 201 314 452 616 804 1020 1260 Theoretical No. of bars 4.72 2.65 2.65 1.18 0.87 0.66 0.52 0.42Actual bars required 5.0 3.0 3.0 2.0 1.0 1.0 1.0 1.0

Area of bars mm2 Total 565 603 942 904 616 804 1020 1260

Maximum spacing of fitments allowed = 230 mm Fitment spacing okIf spacing NOT OK then revise in top right-hand corner of spread sheet

Requirements for torsional reinforcing (Cl 8.3.7)

Minimum transverse reinforcement shall be the greater of(1) Minimum shear reinforcement required by Clause 8.2.8 in the form of closed ties or fitments and

= 70 mm2

= 54 mm2

= 70 mm2 OK area less than chosen fitment areaWith closed fitments If area NOT OK then revise in top right-hand corner of spread sheet

= 212,784 mm2

= 1,912 mm= 26 mm2 OK area less than chosen fitment area

If area NOT OK then revise in top right-hand corner of spread sheet

Results based on fitment arrangement for Torsion Design torsion force at a cross section T* 50.0 kNm

Based on 2 legs with closed fitments Design shear force at a cross section V* 360.0 kNm

500 MPa 231.0 kNm

Fitment diameter 12 mm 1230.0 kN

Maximum Spacing 230 mm 77.5 kNm

Additional tension Reo 533 mm2 116.0 kN

Additional compression Reo 533 mm2

Reminder: Do not forget to add the fitments for torsion to those required shear and the additional longitudinal reo to the reo required for flexure

Suggested bar numbers & sizes for limit state requirement M* < fMu

(11) Torsional capacity equal to 0.25 TUC

Asv.min = 0.06 √ f′c bv s / fsy.f ≥0.35 bv s / fsy.f 0.35 bv s / fsy.f

0.06 √ f′c bv s / fsy.f

Asv.min

Asw = s * 0.25 Tuc/ (fsy.f 2At cot Ɵv )/ ØAt = area of a polygon

ut = perimeter of the polygon defined for At

Asw

Fitment Yield Strength fsy.f Tu.max = 0.2Jtf’c Tu.max

Vu.max = 0.2 fc′ bv do Vu.max

Tuc = 0.3Jt √f’c Tuc

Vuc= β1 β2 β3 bv do fcv { Ast }1/3 Vuc =

bv do

J128

John Woodside:



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Subject Date:

Reinforced Concrete Beam Design - Deflections - Deemed to comply ratios for beams Cl 8.5.4(Please note results may be conservative. Refer to discussion in RCDH)Data Input

Capacity Reduction Factor


400 mm f 0.8

990 mm 20 MPa

Depth, D 700 mm 500 MPa

Cover to fitments 30 mm

Effective depth d 640 mm

(Effective depth usually = D - cover to fitment - fitment size - distance to the centroid of the tensile reo, rounded down to nearest 5mm )

Longitudinal Tensile Reinforcement (fully anchored) 628

Longitudinal Compression Reinforcement (fully anchored) 628

Applied actions Imposed action Short-term factor Long-term factor

Dead load kN/m g 26.9 kNm

Live load kN/m q 11.4 kNm0.7 0.4

Short term and long term actions from AS 1170.0 Table 4.1 Offices 0.7 0.4

Short term factor 0.7 Parking 0.7 0.4

Long term factor 0.4 Retail 0.7 0.4

Storage 1 0.6

Other 1 0.6

8400 mm Roof actions

and L for a beam where Ln is the clear span 0.7 0.4

Other roofs 0.7 0

13.1

Design Calculations= 1.000

= 0.80

= 0.0010

= 2.48

= 0.0015

= 0.6681

For k1 factor use k1 (2)

= 0.0062 WARNING Do not use this value for k1!

= 0.0320 OK Use this value for k1

= 0.032 For k1 factor use k1 (2)

(b) for continuous beams, where the ratio of the longer to the shorter of two adjacent spans does not exceed 1.2 and where no end span is longer than an interior span—

= 0.00391

= 60.05 kNm

= 33.15 kNm

CONCRETE PROPERTIES AT 28 DAYS

20 25 32 40 50 65 80 100

24 000 26 700 30 100 32 800 34 800 37 400 39 600 42 200

= 26,700 MPa From Table 2.3.2 of AS 3600. Limits for calculated vertical deflections of beams and slabs

See options above for values

Type of member

All members The total deflection 1/250 = 0.004 1/125 = 0.002

1/800 = 0.00125

Transfer members Total deflection 1/250 = 0.004

Input ∆/Lef to be used = 0.004

See options above and to the right, for value TO INPUT

Calculated

Width, bw = width of rectangular beam

Width, bef = effective width of compressive width or flange f'c

fsy

Ast.provided mm2

Asc.provided mm2

Short-term, long-term and combination factors ys and yl (after AS 1170.0 )

(ys) (yl)

Distributed imposed actions, QResidential and domestic structures

ys

yl

See table to the right for values of ψ

Lef is the effective span of beam which is the lesser of Ln +D Lef

Roofs used for floor-type activities

Actual span to depth ratio Lef/d Lef/d

Calculate Asc/Ast Asc/Ast

kcs

Reinforcing ratio p = Ast/bef/d at the midspan

β = bef / bw > 1.0

Factor = 0.001 (f'c)1/3 / β 2/3 Ratio of p/( 0.001 (f'c)1/3 β 2/3)

0.1/β2/3

Ratio of p/( 0.001 (f'c)1/3 β 2/3) 0.06/β2/3

k1(1) Deflection coefficient (5 − 0.04 f′c)p + 0.002 ≤ 0.1/ β2/3 When p ≥0.001 (f'c)1/3/β2/3

k1(2) Deflection coefficient 0.055(f′c)1/3/ β2/3 − 50 p ≤ 0.06 /β2/3 When p < 0.001 (f'c)1/3/β2/3

Insert factor k1

Factor k2

k2 = deflection constant, taken as—

(a) for simply supported beams, 5/384 = 0.01302; or

(i) 2.4/384 = 0.00625 in an end span; or

(ii) 1.5/384 = 0.00391 in interior spans.

Insert factor k2

See options above for values

Effective design load for total deflection Fd.ef

Effective design load for incremental deflection Fd.ef

f'c (MPa)

Ec (MPa)

Input Ec Mean value of the modulus of elasticity at 28 day

Deflection to be considered

Deflection limitation (∆/Lef) for spans (Notes

1 and 2)

Deflection limitation (∆/Lef) for cantilevers

(Note 4)

Members supporting masonry partitions

The deflection that occurs after the addition

or attachment of the partitions

1/500 = 0.002 where provision is made to

minimize the effect of movement, otherwise

1/1000 = 0.001

1/250 = 0.004 where provision is made to

minimize the effect of movement, otherwise

1/500 = 0.002

Members supporting other brittle finishes

The deflection that occurs after the addition

or attachment of the finish

Manufacturer’s specification but not more than

1/500 = 0.002

Manufacturer’s specification but not

more than 1/250 = 0.004

Members subjected to vehicular or pedestrian

traffic

The imposed action (live load and dynamic impact) deflection

1/400 = 0.0025

1/500 = 0.002 where provision is made to minimize the effect of deflection of the transfer

member on the supported structure, otherwise

1/1000 = 0.001

Actual Lef/d

d

bw

D

= OK COMPLIES 24.33 13.13

= OK COMPLIES 29.66 13.13

Results Deflections Beam D 700 mm

b 400 mm

Calculated Span L 8400 mm

= OK COMPLIES 24.33 13.1 Ast provided 628 mm2

Asc provided 628 mm2

= OK COMPLIES 29.66 13.1 =1/ 250

[k1(D/Lef)(befEc)/(k2Fd.ef)]1/3 Total deflection

[k1(D/Lef)(befEc)/(k2Fd.ef)]1/3 Incremental deflection

Actual Lef/d

[k1(D/Lef)(befEc)/(k2Fd.ef)]1/3 Total deflection

[k1(D/Lef)(befEc)/(k2Fd.ef)]1/3 Incremental deflection ∆/Lef used

reinforced concrete beam design - deflections

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