Mathcad in Concrete Structures ACI 318-05 6th Edition

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CONSTRUCTION MANAGEMENT

Concept-Develop-Execute-Finish

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Seun Sambath, PhD

Concrete Structures

(ACI 318-05)

in Mathcad

Sixth Edition

Phnom Penh 2010

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Address: #M41, St.308, Sangkat Tonle Basac, Khan Chamkarmon,

Phnom Penh, Cambodia.

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1. Unit Conversion

Length

in = inch ft = foot yd = yard mi = mile 1in25.4 mm 1cm0.394 in 1ft0.305 m 1m3.281 ft 1ft12 in 1yd 0.914 m 1m1.094 yd 1yd 3 ft 1mi 1.609 km 1km0.621 mi 1mi 1760 yd 1mi 5280 ft 1m100mm1.1 m 1mi 200m 1.124 mi

Size of standard concrete cylinder D 6in 15.24 cm H 12in 30.48 cm

Force

lbf = pound force kip = kilopound force

1lbf 4.448 N 1lbf 0.454 kgf 1kgf 9.807 N 1N0.225 lbf 1kgf 2.205 lbf 1N0.102 kgf

1kip4.448 kN 1kip0.454 tonnef 1tonf 0.907 tonnef

1kN0.225 kip 1tonnef  2.205 kip 1tonnef  1.102 tonf

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Stress

psi = pound per square inch 1psi 1 lbf in2  

ksi = kilopound per square inch 1ksi 1 kip in2  

psf = pound per square foot 1psf 1 lbf ft2  

1psi6.895 kPa 1kPa 0.145 psi 1Pa 1 N m2  

1ksi6.895 MPa 1MPa0.145 ksi 1kPa 1 kN m2   1MPa 1 N mm2   1psf 0.048 kN m2   1 kN m2 20.885 psf  1psf 4.882 kgf m2   1 kgf m2 0.205 psf 

Concrete compression strength

3000psi20.7 MPa 20MPa 2900.8 psi 4000psi27.6 MPa 25MPa 3625.9 psi 5000psi34.5 MPa 35MPa 5076.3 psi 8000psi55.2 MPa 55MPa 7977.1 psi

Steel yield strength

60ksi 413.7 MPa 390MPa 56.6 ksi 75ksi 517.1 MPa 490MPa 71.1 ksi Live loads 40psf 1.915 kN m2   200kgf m2 40.963 psf  100psf 4.788 kN m2   4.80kN m2 100.25 psf 

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Density

1pcf 1 lbf ft3   125pcf 19.636 kN m3   145pcf 22.778 kN m3  

Moments

1ft kip  1.356 kN m 

User setting

Riels 1 USD 4165Riels 124USD 516460 Riels 200000Riels48.019 USD CostSteel 665 USD 1tonnef  670kgf CostSteel  445.55 USD

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2. Simple Calculation

1 3 2 3  

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3 23.472  a 2 b 4 c 1 Δ b2 4 a c x1 b_{2 a}_ Δ 0.225 x2 b_{2 a}_ Δ 2.225

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3. Materials

1. Concrete

Standard concrete cylinder D 6in 15.24 cm D 150mm H 12in 30.48 cm H 300mm

Concrete compression strength

f'c 3000psi20.7 MPa _f'c 20MPa 2900.8 psi f'c 4000psi27.6 MPa _f'c 25MPa 3625.9 psi f'c 5000psi34.5 MPa _f'c 35MPa 5076.3 psi Concrete ultimate strain

εu 0.003

Cubic and cylinder compression strength

fcube= _0.85f'c f'c 20MPa _fcube f'c 0.85  23.529 MPa  f'c 25MPa _fcube f'c 0.85  29.412 MPa  f'c 35MPa _fcube f'c 0.85  41.176 MPa 

Modulus of rupture (tensile strength)

fr 7.5 f'c=  (in psi) Metric coefficient 7.5psi f'c

psi  7.5MPa psi MPa  f'c MPa MPa psi   = C MPa f'c MPa  = C 7.5 psi MPa MPa psi   0.623  fr 0.623 f'c=  (in MPa)

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Modulus of elasticity

Ec 33 wc=  1.5 f'c (in psi) wc is a unit weight of concrete (in pcf)

Metric coefficient 33psi

kN m3 pcf

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1.5  MPa psi  44.011 MPa Ec 44 wc=  1.5 f'c (in MPa) wc in kN/m3

Example 3.1

Concrete compression strength _f'c 25MPa Unit weight of concrete _wc 24kN m3  145pcf 22.778 kN m3   Modulus of rupture fr 7.5psi f'c psi  3.114 MPa  fr 0.623MPa f'c MPa   3.115 MPa  Modulus of elasticity Ec 33psi wc pcf

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1.5  f'c psi   2.587 104MPa  Ec 44MPa wc kN m3

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1.5  f'c MPa   2.587 104MPa 

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2. Steel Reinforcements

Steel yield strength of deformed bar (DB) _fy 390MPa 56.565 ksi Steel yield strength of round bar (RB) _fy 235MPa 34.084 ksi

Modulus of elasticity _Es 29000000psi 1.999105MPa

Es 2 10 5MPa Steel yield strength _εy fy

Es =

US Steel Reinforcements

Bar No. (#) Bar diameter

no 3 4 5 6 7 8 9 10 11 12 13 14

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 D no 8 in  D 9.5 12.7 15.9 19 22.2 25.4 28.6 31.8 34.9 38.1 41.3 44.4

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mm  

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Steel area A π D 2  4  A 0.71 1.27 1.98 2.85 3.88 5.07 6.41 7.92 9.58 11.4 13.38 15.52

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cm2  

Weight of steel reinforcements

W A 7850 kgf m3  W 0.559 0.994 1.554 2.237 3.045 3.978 5.034 6.215 7.52 8.95 10.503 12.182

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kgf m   ORIGIN 1 n 1 9 Area n  A n

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1 2 3 4 5 6 7 8 9 3 9.5 0.71 1.43 2.14 2.85 3.56 4.28 4.99 5.70 6.41 0.559 4 12.7 1.27 2.53 3.80 5.07 6.33 7.60 8.87 10.13 11.40 0.994 5 15.9 1.98 3.96 5.94 7.92 9.90 11.88 13.86 15.83 17.81 1.554 6 19.1 2.85 5.70 8.55 11.40 14.25 17.10 19.95 22.80 25.65 2.237 7 22.2 3.88 7.76 11.64 15.52 19.40 23.28 27.16 31.04 34.92 3.045 8 25.4 5.07 10.13 15.20 20.27 25.34 30.40 35.47 40.54 45.60 3.978 9 28.6 6.41 12.83 19.24 25.65 32.07 38.48 44.89 51.30 57.72 5.034 10 31.8 7.92 15.83 23.75 31.67 39.59 47.50 55.42 63.34 71.26 6.215 11 34.9 9.58 19.16 28.74 38.32 47.90 57.48 67.06 76.64 86.22 7.520 12 38.1 11.40 22.80 34.20 45.60 57.00 68.41 79.81 91.21 102.61 8.950 13 41.3 13.38 26.76 40.14 53.52 66.90 80.28 93.66 107.04 120.42 10.503 14 44.5 15.52 31.04 46.55 62.07 77.59 93.11 108.63 124.14 139.66 12.182 Bar # Diameter_(mm) Area of cross section (cm2) for the number of bars is equal to Weight (kgf/m)

Metric Steel Reinforcements

D (6 8 10 12 14 16 18 20 22 25 28 32 36 40)Tmm A π D 2  4  W A 7850 kgf m3  n 1 9 Area n  A n 1 2 3 4 5 6 7 8 9 6 0.28 0.57 0.85 1.13 1.41 1.70 1.98 2.26 2.54 0.222 8 0.50 1.01 1.51 2.01 2.51 3.02 3.52 4.02 4.52 0.395 10 0.79 1.57 2.36 3.14 3.93 4.71 5.50 6.28 7.07 0.617 12 1.13 2.26 3.39 4.52 5.65 6.79 7.92 9.05 10.18 0.888 14 1.54 3.08 4.62 6.16 7.70 9.24 10.78 12.32 13.85 1.208 16 2.01 4.02 6.03 8.04 10.05 12.06 14.07 16.08 18.10 1.578 18 2.54 5.09 7.63 10.18 12.72 15.27 17.81 20.36 22.90 1.998 20 3.14 6.28 9.42 12.57 15.71 18.85 21.99 25.13 28.27 2.466 22 3.80 7.60 11.40 15.21 19.01 22.81 26.61 30.41 34.21 2.984 25 4.91 9.82 14.73 19.63 24.54 29.45 34.36 39.27 44.18 3.853 28 6.16 12.32 18.47 24.63 30.79 36.95 43.10 49.26 55.42 4.834 32 8.04 16.08 24.13 32.17 40.21 48.25 56.30 64.34 72.38 6.313

36 10.18 20.36 30.54 40.72 50.89 61.07 71.25 81.43 91.61

7.990

40 12.57 25.13 37.70 50.27 62.83 75.40 87.96 100.53 113.10

9.865

Diameter (mm)

Area of cross section (cm2) for the number of bars is equal to _Weight (kgf/m)

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4. Safety Provision

Required_Strength Design_Strength

U_ϕSn where

U = required strength (factored loads) ϕSn = design strength

Sn = nominal strength

ϕ = strength reduction factor

a. Load Combinations

Basic combination U=1.2 D 1.6 L Roof combination U=1.2 D 1.6 L  _{1.0 Lr} Wind combination U=1.2 D 1.6 W 1.0 L  _{0.5 Lr} where D = dead load L = live load Lr = roof live load W = wind load

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b. Strength Reduction Factor

Strength Condition Strength reduction factor ϕ

Tension-controlled members ( εt 0.005 ) ϕ=0.9 Compression-controlled ( εt 0.002 )

Spirally reinforced ϕ=0.70

Other ϕ=0.65

Shear and torsion ϕ=0.75

where

εt = net tensile strain

For spirally reinforced members

ϕ 0.9 if _{εt 0.005} 0.70 if _{εt 0.002} 1.7 _{200 εt} 3 otherwise = 0.7 0.90.7 0.0050.002

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εt 0.002

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 0.7 200 3 

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εt 0.002

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 1.7200 εt 3 =

For other members

ϕ 0.9 if _{εt 0.005} 0.65 if _{εt 0.002}

1.45 _{250 εt}

3 otherwise =

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5. Loads on Structures (Case of Two-Way Slabs)

Slab dimension

Short side _La 4m Long side _Lb 6m

A. Preliminary Design

Thickness of two-way slab Perimeter

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_{La Lb}

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2 tmin Perimeter₁₈₀ 111.111 mm t 1 30 1 50

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La (133.333 80) mm  t 110mm Section of beam B1 L 6m h 1 10 1 15

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L (600 400) mm  h 500mm b (0.3 0.6) h  (150 300) mm b 250mm For girders h 1 8 1 10

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L =

For two-way slab beams h 1 10 1 15

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L =

For floor beams h 1

15 1 20

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L = Section of beam B2 L 4m h 1 10 1 15

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L (400 266.667) mm  h 300mm b (0.3 0.6) h  (90 180) mm b 200mm

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B. Loads on Slab

Floor cover Cover 50mm 22 kN m3 1.1 kN m2    RC slab Slab t 25 kN m3 2.75 kN m2    Ceiling Ceiling 0.40kN m2  M & E Mechanical 0.20kN m2  Partition Partition 1.00kN m2 

Dead load DL Cover Slab CeilingMechanical Partition 5.45 kN m2   

Live load for Lab LL 60psf 2.873 kN m2    Factored load _wu 1.2 DL 1.6 LL 11.137 kN m2   

C. Loads of Wall

Void 30mm 30 mm 190 mm 4 wbrick.hollow (90mm 90 mm 190 mm Void) 20 kN m3 1.744 kgf   wbrick.solid 45mm 90 mm 190 mm 20 kN m3 1.569 kgf   ρbrick.hollow _{90mm 90}wbrick.hollow_{ mm 190}_ _mm 11.111 kN m3    Brickhollow.10 120mm Void 55 1m2  

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20  kN m3 1.648 kN m2    Brickhollow.20 220mm Void 110 1m2  

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20  kN m3 2.895 kN m2   

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D. Loads on Beam B1

Self weight SW 25cm 50cm(  110mm)25kN m3 2.438 kN m   

Wall _wwall _{Brickhollow.10 3.5m 50cm}(  ) 4.943 kN m    Slab α 4m 2 6m  0.333  k 12 α 2α30.815 wD.slab DL 4m 2   k2 17.763 kN m    wL.slab LL 4m 2   k2 9.363 kN m   

Dead load _wD SW_wwall_wD.slab 25.143 kN m   

Live load _wL _{wL.slab 9.363} kN m    Factored load _wu _{1.2 wD} _{1.6 wL} 45.153 kN m   

E. Loads on Beam B2

Self weight SW 20cm 30cm(  110mm)25kN m3 0.95 kN m   

Wall _wwall _{Brickhollow.10 3.5m 30cm}(  ) 5.272 kN m    Slab α 4m 2 4m  0.5  k 12 α 2α30.625 wD.slab DL 4m 2   k2 13.625 kN m    wL.slab LL 4m 2   k2 7.182 kN m   

Dead load _wD SW_wwall_wD.slab 19.847 kN m   

Live load _wL _{wL.slab 7.182} kN m    Factored load _wu _{1.2 wD} _{1.6 wL} 35.308 kN m   

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F. Loads on Column

Tributary area B 4m L 6m Slab loads _PD.slab DL B L  130.8 kN

PL.slab LL B L  68.948 kN Beam loads _PB1 25cm 50cm(  110mm)25kN m3 L   14.625 kN  PB2 20cm 30cm(  110mm)25kN m3 B   3.8 kN 

Wall loads _Pwall.1 _{Brickhollow.10 3.5m 50cm}(  )L 29.657 kN

Pwall.2 Brickhollow.10 3.5m 30cm(  )B 21.089 kN SW of column SW=(5% 7% _{) PD}

Total loads for number of floors n 7

PD

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PD.slab PB1 PB2 Pwall.1 Pwall.2

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1.05n 1469.787 kN PL PL.slab n  482.633 kN Pu 1.2 PD  1.6 PL 2535.958 kN Pu PD PL  1.299 Control PD PL n B L 11.622 kN m2   (Ref. 10kN m2 18kN m2  ) PL PD PL  24.72 % (Ref. 15% 35% )

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06 . Loads on Structures (Case of One-Way Slabs)

A. Preliminary Design

Thickness of one-way slab (both ends continue) L 6m 2 3 m  tmin ₂₈L  107.143 mm t 1 25 1 35

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L(120 85.714) mm  t 120mm

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Section of floor beam B1 L 8m h 1 15 1 20

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L (533.333 400) mm  h 500mm b (0.3 0.6) h  (150 300) mm b 250mm Section of girder B2 L 6m h 1 8 1 10

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L (750 600) mm  h 600mm b (0.3 0.6) h  (180 360) mm b 300mm

B. Loads on Slab

Cover 50mm 22 kN m3 1.1 kN m2    Slab t 25 kN m3 3 kN m2    Ceiling 0.40kN m2  Mechanical 0.20kN m2  Partition 1.00kN m2 

DL Cover Slab CeilingMechanical Partition 5.7 kN m2    LL 60psf 2.873 kN m2    wu 1.2 DL 1.6 LL 11.437 kN m 1m   

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C. Loads on Beam B1

Void 30mm 30 mm 190 mm 4 Brickhollow.10 120mm Void 55 1m2  

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20  kN m3 1.648 kN m2    SW 25cm 50cm(  120mm)25kN m3 2.375 kN m    wwall Brickhollow.10 3.5m 50cm(  ) 4.943 kN m    wD.slab DL 3 m 17.1 kN m    wL.slab LL 3 m 8.618 kN m    wD SW_wwall_wD.slab 24.418 kN m    wL wL.slab 8.618 kN_m wu 1.2 wD 1.6 wL 43.091 kN m   

D. Loads on Girder B2

SW 30cm 60cm(  120mm)25kN m3 3.6 kN m    wwall Brickhollow.10 3.5m 60cm(  ) 4.778 kN m    PB1 25cm 50cm(  120mm)25kN m3 8m4m 2  14.25 kN  Pwall Brickhollow.10 3.5m 50cm(  ) 8m 4m 2  29.657 kN  PD.slab DL 3 m 8m4m 2   102.6 kN  PL.slab LL 3 m 8m 4m 2  51.711 kN 

(21)

Factored loads wD SW_wwall 8.378 kN m    wL 0 wu 1.2 wD 1.6 wL 10.054 kN m    PD PB1 Pwall  PD.slab146.507 kN PL PL.slab 51.711 kN  Pu 1.2 PD  1.6 PL 258.545 kN

(22)

7. Loads on Staircase

Run and rise of step G 3.5m

12  291.667 mm 

H 3.8m

24  158.333 mm

 G2 H  60.833 cm

Slope angle α atan H

G









 28.496 deg



Loads on Waist Slab

Thickness of waist slab t 120mm

Step cover Cover 50mm H(  G)22kN m3 1m G 1 m  1.697 kN m2   

(23)

SW of step Step G H 2 24 kN m3 1m G 1 m  1.9 kN m2   

SW of waist slab Slab t 25 kN m3 1m2 1m2cos α( )  3.414 kN m2    Renderring Renderring 0.40kN m2 1m2 1m2cos α( )  0.455 kN m2    Handrail Handrail 0.50kN m2 

Total dead load DL Cover Step Slab Renderring Handrail

DL 7.966 kN m2  

Live load for public staircase LL 100psf 4.788 kN m2    Factored load _wu 1.2 DL 1.6 LL 17.22 kN m2   

Loads on Landing Slab

Cover 50mm 22 kN m3 1.1 kN m2    Slab 150mm 25 kN m3 3.75 kN m2    Renderring 0.40kN m2  Handrail 0.50kN m2 

DL Cover Slab RenderringHandrail 5.75 kN m2    LL 100psf 4.788 kN m2    wu 1.2 DL 1.6 LL 14.561 kN m2   

(24)

8. Loads on Roof

Slope angle α atan 3m 10m 2





















30.964 deg  

Srokalinh tile Tile 30mm 20 kN m3 0.6 kN m2    Purlins _w20x20x1.0 (20mm 20 mm 18mm 18  mm) 7850 kgf m3 0.597 kgf 1m    Purlin _w20x20x1.0 1m 1m 100 mm cos α ( )  0.068 kN m2    Rafters _w40x80x1.6 (40mm 80 mm 36.8mm 76.8  mm) 7850 kgf m3  w40x80x1.6 2.934 kgf 1m   Rafter _w40x80x1.6 1m 750mm 1 m cos α ( )  0.045kN m2   

(25)

Roof beam Beam 20cm 30 cm 25 kN m3 1m 1m 10m 4   0.6 kN m2   

Roof column Column 20cm 20 cm 3m 2  25kN m3 1 10m 4  m4









 0.15 kN m2   

Total dead load _wD Tile PurlinRafterBeamColumn 1.463 kN m2    Live load _wL 1.00kN m2  Factored load _wu _{1.2 wD} _{1.6 wL} 3.356 kN m2   

(26)

9. ASCE Wind Loads

Basic wind speed V 120km hr

 V 33.333m

s

 V74.565 mph

Exposure category Expoure=C

Importance factor I 1.15 Topograpic factor _Kzt 1.0 Gust factor G 0.85 Wind directionality factor _Kd  0.85

Static wind pressure _qs 0.613N m2 V m s













2  0.681kN m2   

Velocity pressure coefficients

zg 274m α 9.5 (For exposure C) Kz z( ) 2.01 max z 4.6m(  ) zg













2 α   _{Kz 10m}( ) 1.001

Velocity wind pressure

qz z( ) _{qs Kz z} ( )_Kzt KI _d _{qz 10m}( ) 0.667kN m2  

Design wind pressure

pz z Cp

 

  qz z( ) G C _p

Dimension of building in plan

B 6m 3 18 m L 4m 5 20 m

λ L

B 1.111 

External pressure coefficients

(27)

Cp.leeward linterp 0 1 2 4 40













0.5  0.5  0.3  0.2  0.2 













 λ













0.478    Cp.side 0.7 Floor heights H 3.5m 3.5m 3.5m 3.5m 3.5m 3.5m 3.5m













 H reverse H( ) h



H 24.5 m ORIGIN 1 n rows H( )7 Wind forces i 1 n a i 1 i k H k



 H i 2   a n 1 



H reverse a( ) 24.5 22.75 19.25 15.75 12.25 8.75 5.25 1.75













m  Bwindward 6m _Bleeward 6m Bside 4m Pwindward_i a_i a_{i 1}_ z pz z Cp.windward







   d Bwindward  Pleeward_i pz h Cp.leeward







a i 1  ai





 _Bleeward  Pside_i pz h Cp.side







a i 1  ai





 _Bside 

(28)

reverse augment Pwindward Pleeward



 Pside



5.70 11.13 10.71 10.21 9.61 8.81 8.10 3.43  6.86  6.86  6.86  6.86  6.86  6.86  3.35  6.71  6.71  6.71  6.71  6.71  6.71 













kN   Alternative ways i 1 n b i 1 i k H k



 

Prectangle_i pz b



iCp.windward







ai 1  ai



Bwindward

Ptrapezium_i pz ai Cp.windward 





pz a



i 1 Cp.windward



2 



ai 1  ai



Bwindward 

reverse augment Pwindward Prectangle



 Ptrapezium



5.70 11.13 10.71 10.21 9.61 8.81 8.10 5.75 11.13 10.71 10.22 9.62 8.83 8.08 5.70 11.12 10.70 10.20 9.59 8.78 8.20













kN 

(29)

10. Design of Singly Reinforced Beams

A. Concrete Stress Distribution

In actual distribution Resultant C=_{α f'c}  bc Location β c In equivalent distribution Location β c a 2 = Resultant C=_{α f'c}  bc =_{γ f'c}  ba Thus, a=2 β c =_{β1 c} where _{β1 2 β}=  γ α c a  = α β1 =

f'c 4000psi 5000psi 6000psi 7000psi 8000psi

α 0.72 0.68 0.64 0.60 0.56 β 0.425 0.400 0.375 0.350 0.325 β1 2 β=  0.85 0.80 0.75 0.70 0.65 γ α β1 = 0.72 0.85  0.847 0.68 0.80  0.85 0.64 0.75  0.853 0.60 0.70  0.857 0.56 0.65  0.862

(30)

Conclusion: γ=0.85 β1 0.85 if _{f'c 4000psi} 0.65 if _{f'c 8000psi} 0.85 0.05 f'c 4000psi  1000psi   otherwise = 4000psi27.6 MPa 8000psi55.2 MPa 1000psi6.9 MPa

B. Strength Analysis

Equilibrium in forces X



=0 C=T 0.85 f'c  ba =_{As fs} (1) Equilibrium in moments M



=0 Mn C d= 



_

 a₂



_

T d a 2 









 = Mn 0.85 f'c  ba d a 2 









 = (2.1) Mn As fs d a 2 









 = (2.2)

Conditions of strain compatibility εs εu d c c = εs εu= d _c c or _{εt εu} dt c  c  = (3.1) c d εu εu εs  = or c _dt εu εu εt  = (3.2) Unknowns = 3 _{a As} _fs Equations = 2



X=0



M=0

(31)

C. Steel Ratios

ρ As b d = As fy  b d f_y = 0.85 f'c   ba b d f_y = _{0.85 β1} f'c fy  c d  = _{0.85 β1} f'c fy  c dt  dt d  = ρ _{0.85 β1} f'c fy  εu εu εs  = _{0.85 β1} f'c fy  εu εu εt  dt d  =

Balanced steel ratio

fc f'c= _{fs fy}= _{εs εy}= fy Es = ρb 0.85 β1 f'c fy  εu εu εy  = _{0.85 β1} f'c fy  600MPa 600MPa_fy  =

εu 0.003 _Es 2 10 5MPa _{εu Es} 600 MPa Maximum steel ratio

ACI 318-99 _{ρmax 0.75 ρb}=  ACI 318-02 and later _{ρmax 0.85 β1} f'c

fy  εu εu εt  = _{with εt 0.004} For fy 390MPa _εs fy Es 0.002  For εt 0.004 ρmax ρb εu εy εu 0.004 = 5 7 = =0.714 For εt 0.005 ρmax ρb εu εy εu 0.005 = 5 8 = =0.625

Minimum steel ratio

ρmin 3 _f'c fy 200 fy  = (in psi) ρmin 0.249 _f'c fy 1.379 fy  = (in MPa)

(32)

D. Determination of Flexural Strength

Given: b d_As_f'c_fy

Find: _ϕMn

Step 1. Checking for steel ratio

ρ As b d =

ρ_ρmin : Steel reinforcement is not enough ρmin ρ ρmax : the beam is singly reinforced ρ_ρmax : the beam is doubly reinforced

ρ=_ρmax _{As ρ b}=  d Step 2. Calculation of flexural strength

a As fy  0.85 f'c b = c a β1 = Mn As fy d a 2 









 = εt εu dt c  c  = ϕ=_{ϕ εt}

 

The design flexural strength is ϕ Mn

(33)

Concrete dimension b 200mm h 350mm Steel reinforcements _As 5 π 16mm( ) 2  4  10.053 cm 2  d h 30mm6mm 16mm 40mm 2 









 278 mm  dt h 30mm 6mm 16mm 2 









  306 mm 

Materials _f'c 25MPa _fy 390MPa

Solution

Checking for steel ratios

β1 0.65 max 0.85 0.05 f'c 27.6MPa  6.9MPa  













min 0.85













 0.85  εu 0.003 ρmax 0.85 β1 f'c fy  εu εu 0.004  0.02  ρmin max 0.249MPa f'c MPa  fy 1.379MPa fy 













0.00354  ρ As b d 0.018 

Steel_Reinforcement "is Enough" if ρ_ρmin "is not Enough" otherwise 

Steel_Reinforcement "is Enough"

As min ρ ρmax







 db 10.053 cm 2 Calculation of flexural strength

a As fy  0.85 f'c b 92.252 mm  c a β1  108.532 mm  Mn As fy d a 2 









 90.911 kN m  

(34)

εt εu dt c  c   0.00546  ϕ 0.65 max 1.45 _{250 εt} 3













min 0.9













 0.9 

The design flexural strength is ϕ Mn 81.82 kN m 

E. Determination of Steel Area

Given: _{Mu b d}  _f'c_fy

Find: _As

Relative depth of compression concrete

w a d = 0.85 f'c   ba 0.85 f'c  db = As fy  0.85 f'c  db = ρ fy  0.85 f'c 1 =

Flexural resistance factor

R Mn b d 2 = As fy d a 2 









 b d 2 = As b d fy d a 2  d  = _{ρ fy} 1 1 2w 









 = R _{ρ fy} 1 ρ fy  1.7 f'c 













 = _{0.85 f'c} w 1 1 2w 









 =

Quadratic equation relative w R 0.85 f'c w 1 1 2w 









 = w22 w 2 R 0.85 f'c   =0 w1 1 1 2 R 0.85 f'c    1 = _{w2 1} 1 2 R 0.85 f'c    1 = w 1 1 2 R 0.85 f'c    = ρ 0.85 f'c fy  w = 0.85 f'c fy  1 1 2 R 0.85 f'c   













 =

(35)

Step 1. Assume ϕ=0.9

Mn= Mu_ϕ Step 2. Calculation of steel area

R Mn b d 2 = ρ 0.85 f'c fy  1 1 2 R 0.85 f'c   













 =

ρ_ρmax : the beam is doubly reinforced (concrete is not enough) ρ_ρmax : the beam is singly reinforced

As max ρ ρmin=







 db (this is a required steel area)

Step 3. Checking for flexural strength

a As fy  0.85 f'c b

= _{( As is a provided steel area)} Mn As fy d a 2 









 = c a β1 = _{εt εu} dt c  c  = ϕ=_{ϕ εt}

 

FS Mu ϕ Mn = (usage percentage)

FS 1 : the beam is safe FS 1 : the beam is not safe

Example 10.2

Required strength _Mu 153kN m Concrete section b 200mm h 500mm d h 30mm8mm 18mm 40mm 2 









 424 mm 

(36)

dt h 30mm 8mm 18mm 2 









  453 mm 

Solution Steel ratios β1 0.65 max 0.85 0.05 f'c 27.6MPa  6.9MPa  













min 0.85

























0.00354  Assume ϕ 0.9 Mn Mu_ϕ  170 kN m  Steel area R Mn b d 2 4.728 MPa   ρ 0.85 f'c fy  1 1 2 R 0.85 f'c   













  0.014  ρmin ρ ρmax 1 As ρ b d 11.783 cm 2 As 6 π 16mm( ) 2  4  12.064 cm 2 

Checking for flexural strength









 173.444 kN m  

(37)

εt εu dt c  c   0.00743  ϕ 0.65 max 1.45 _{250 εt} 3













min 0.9













 0.9  FS Mu ϕ Mn  0.98 

The_beam "is safe" if FS 1 "is not safe" otherwise

 The_beam"is safe"

F. Determination of Concrete Dimension and Steel Area

Given: _{Mu f'c} _fy Find: b d_As

Step 1. Determination of concrete dimension

Assume εt 0.004 (Usually εt 0.005 ) ρ _{0.85 β1} f'c fy  εu εu εt  = R _{ρ fy} 1 ρ fy  1.7 f'c 













 = ϕ=_{ϕ εt}

 

_Mn Mu ϕ = bd2 Mn R = Option 1: b Mn R d2 = Option 2: d Mn R b = Option 3: k b d = d 3 Mn R k = b =k d

Step 2. Calculation of steel area

R Mn b d 2 =

(38)

ρ 0.85 f'c fy  1 1 2 R 0.85 f'c   













 = As max ρ ρmin=







 db

Step 3. Checking for flexural strength

a As fy  0.85 f'c b = c a β1 = Mn As fy d a 2 









 = εt εu dt c  c  = ϕ=_{ϕ εt}

 

FS Mu ϕ Mn =

Example 10.3

Required strength _Mu 700kN m













min 0.85

























0.00354  Assume _εt 0.007 ρ _{0.85 β1} f'c fy  εu εu εt  0.014  R _{ρ fy} 1 ρ fy  1.7 f'c 













 4.728 MPa 

(39)

ϕ 0.65 max 1.45_{250 εt} 3













min 0.9













 0.9  _Mn Mu ϕ  777.778 kN m   Concrete dimension k b d = k 400 600  Cover 30mm 10mm 25mm 40mm 2   Cover85 mm d 3 Mn R k 627.231 mm  b k d  418.154 mm

h Round d( Cover50mm) 700 mm b Round b 50mm(  ) 400 mm

d hCover615 mm b h













400 700













mm  Steel area R Mn b d 2 5.141 MPa   ρ 0.85 f'c fy  1 1 2 R 0.85 f'c   













  0.015  As max ρ ρmin







 db 37.741 cm 2 As 8 π 25mm( ) 2  4   39.27 cm 2  _dt h 30mm 10mm 25mm 2 









  dt 647.5 mm  Checking for flexural strength









 803.914 kN m   εt εu dt c  c   0.00616  ϕ 0.65 max 1.45 _{250 εt} 3













min 0.9













 0.9  FS Mu ϕ Mn  96.749 % 

(40)

11. Design of Doubly Reinforced Beams

ρ_ρmax : the beam is singly reinforced (with tensile reinforcements only) ρ_ρmax : the beam is doubly reinforced

(with tensile and compression reinforcements)

A. Strength Analysis

Equilibrium in forces X



=0 T=C_Cs (1) T=_{As fs} =_{As fy} C=_{0.85 f'c}  ba Cs A's f's=  Equilibrium in moments M



=0 Mn Mn1 Mn2=  (2) Mn1 T d d'= (  ) =_{A's f's} (d d') Mn2 C d= 



_

 a₂



_

_{0.85 f'c}  ba d a 2 









 = Mn2



T _Cs



d a 2 









 =



_{As fy} _{A's f's}



d a 2 









 =

(41)

Conditions of strain compatibility εs εu d c c = (3.1) εs εu= d _c c or _{εt εu} dt c  c  = c d εu εu εs  = or c _dt εu εu εt  = ε's εu c d' c = (3.2) ε's εu= c _cd' c d' εu εu ε's  =

B. Steel Ratios

Compression steel ratio ρ' A's

b d =

Tensile steel ratio

ρ As b d = As fy  b d f_y = 0.85 f'c   ba _{A's f's} b d f_y = _{0.85 β1} f'c fy  c d  ρ' f's fy   =

Maximum tensile steel ratio

ρt.max ρmax ρ' f's fy   =

ρ_ρt.max : concrete is enough ρ_ρt.max : concrete is not enough

Minimum tensile steel ratio

ρ _{0.85 β1} f'c fy  c d  ρ' f's fy   = _{0.85 β1} f'c fy  εu εu ε's  d' d  ρ'f's fy   = f's fy= _{ε's εy}= fy Es =

(42)

ρcy 0.85 β1 f'c fy  εu εu εy  d' d   ρ' =

ρ_ρcy : compression steel will yield _{f's fy}= ρ_ρcy : compression steel will not yield _{f's fy}

C. Determination of Flexural Strength

Given: b d d'_dt A _s_A's_f'c_fy

Find: _ϕMn

Step 1. Checking for singly reinforced beam

ρ As b d =

ρ_ρmax : the beam is singly reinforced ρ_ρmax : the beam is doubly reinforced

ρ' A's b d

= _{ρt.max ρmax ρ'}= 

ρ=_ρt.max _{As ρ b}=  d

Step 2. Determination of compression parameters 2.1. Assume f's fy= 2.2. Calculate a As fy   _{A's f's} 0.85 f'c b = c a β1 = ε's εu= c _cd' _{f's.revised Es ε's}=  _fy If _{f's.revised f's} then _{f's f's.revised}=

(43)

Direct calculation Case f's fy= a As fy   _{A's fy} 0.85 f'c b = Case f's fy

As fy =0.85 f'c  ba _{A's f's} _{0.85 f'c}  ba _{A's Es} _εu c d' c   = As fy 0.85 f'c  ba _{A's Es} _εu β1 c   _{β1 d'} β1 c   = _{0.85 f'c}  ba _{A's f1} a_{β1 d'} a   =

where f1 Es εu=  =600MPa

0.85 f'c a2b



_{A's f1} _{As fy}



a _{A's f1} β ₁d'=0 0.85 f'c a2b



_{A's f1} _{As fy}



a _{A's f1} β ₁d' 0.85 f'c d2b 0 = a d









2 _{ρ' f1}_ __{ρ fy}_ 0.85 f'c a d   ρ' f1 β 1 0.85 f'c d' d   =0 w22 p w  q=0 p 1 2 ρ' f1 ρ fy 0.85 f'c  = q ρ' f1  β ₁ 0.85 f'c d' d  = w1 =p  p2 q0 _w2 =p  p2 q0 a =d



p p2 q



c a β1 = ε's εu= c _cd' _{f's Es ε's}=  _fy Step 3. Calculation of flexural strength

Mn1 A's f's=  (dd') Mn2



As fy  A's f's



d a 2 









 = εt εu dt c  c  = ϕ=_{ϕ εt}

 

ϕMn ϕ Mn1 Mn2= 







(44)

Example 11.1

Concrete dimension b 300mm h 550mm Steel reinforcements _As 8 π 20mm( ) 2  4  25.133 cm 2  d h 30mm10mm16mm 20mm 40mm 2 









  d 454 mm dt h 30mm 10mm 16mm 20mm 2 









  484 mm  A's 4 π 20mm( ) 2  4  12.566 cm 2  d' 30mm 10mm 20mm 2   50 mm 

Solution

Steel ratios β1 0.65 max 0.85 0.05 f'c 27.6MPa  6.9MPa  













min 0.85

























0.00354 

Checking for singly reinforced beam

ρ As

b d 0.0185 

The_beam "is singly reinforced" if ρ_ρmax "is doubly reinforced" otherwise 

The_beam"is doubly reinforced"

ρ' A's b d 9.226 10 3     _ρt.max _{ρmax ρ'}  0.027

(45)

Concrete "is enough" if ρ_ρt.max "is not enough" otherwise 

Concrete"is enough" ρ _{min ρ ρt.max}







0.0185 As ρ b d 25.133 cm 2

Determination of compression parameters

Es 2 10 5MPa _f1 _{Es εu}  600 MPa εy fy Es 1.95 10 3     ρcy 0.85 β1 f'c fy  εu εu εy  d' d  ρ'0.024 

Compression_steel "will yield" if ρ_ρcy "will not yield" otherwise 

Compression_steel "will not yield"

a As fy

 _{A's fy}

0.85 f'c b if Compression_steel="will yield"

p 1 2 ρ' f1  ρ fy 0.85 f'c   q ρ' f1  β ₁ 0.85 f'c d' d   d



p p2 q



otherwise  a 90.825 mm c a β1  106.853 mm 

f's fy if Compression_steel="will yield"

ε'sεuc_cd' min Es ε's



 fy



otherwise  f's 319.24 MPa  Flexural strength

(46)

Mn1 A's f's (d d') 162.072 kN m  Mn2



As fy A's f's



d a 2 









 236.576 kN m   εt εu dt c  c   0.011  ϕ 0.65 max 1.45 _{250 εt} 3













min 0.9













 0.9  ϕMn ϕ Mn1 Mn2







358.783 kN m 

D. Design of Doubly Reinforced Beam

Given: _{Mu b d}   d'_dt f'_c_fs Find: _{As A's} Step 1. Assume εt ρ _{0.85 β1} f'c fy  εu εu εt  = ϕ=_{ϕ εt}

 

Step 2. Cheching for singly reinforced beam

As ρ b=  d a As fy  0.85 f'c b = Mn As fy d a 2 









 =

Mu ϕMn : the beam is singly reinforced Mu ϕMn : the beam is doubly reinforced Step 3. Case of doubly reinforced beam

Mn2 Mn=

(47)

c a β1 = _{f's Es εu} c d' c  _fy = A's Mn1 f's d d'(  ) = ρ' A's b d = _{ρt.max ρmax ρ'}=  As 0.85 f'c   ba  _{A's f's} fy = ρ As b d =

Example 11.2

Required strength _Mu 1350kN m Concrete dimension b 400mm h 800mm d h 30mm12mm25mm 25mm 40mm 2 









  d 688 mm dt h 30mm 12mm 25mm 25mm 2 









  720.5 mm  d' 30mm 12mm 25mm 2   54.5 mm 













min 0.85

























0.00354  Assume _εt 0.0092 ρ _{0.85 β1} f'c fy  εu εu εt  0.011 

(48)

ϕ 0.65 max 1.45_{250 εt} 3













min 0.9













 0.9 

Checking for singly reinforced beam As ρ b d 31.342 cm 2 a As fy  0.85 f'c b 143.803 mm  c a β1  169.18 mm  Mn2 As fy d a 2 









  753.074 kN m  

The_beam "is singly reinforced" if _{Mu ϕ Mn2}  "is doubly reinforced" otherwise 

The_beam"is doubly reinforced"

Case of doubly reinforced beam

Mn1 Mu_ϕ  Mn2 746.926 kN m  f's min Es εu



_

 c_c d'fy



_

390 MPa A's Mn1 f's d d'(  ) 30.232 cm 2    ρ' A's b d  0.011  _ρt.max _{ρmax ρ'}  0.028 As 0.85 f'c   ba _{A's f's} fy 61.574 cm 2    ρ As b d 0.022 

Concrete "is enough" if ρ_ρt.max "is not enough" otherwise

 Concrete"is enough"

Compression steel _{A's 30.232 cm}  2 5 π 28mm( )

2  4   30.788 cm 2 Tensile steel _{As 61.574 cm}  2 10 π 28mm( ) 2  4   61.575 cm 2 400mm (12mm30mm) 2 28mm 5 4 44 mm

(49)

E. Determination of Tensile Steel Area

Given: _{Mu b d}   d'_dt A' _s_f'c_fy

Find: _As

Step 1. Calculation of compression parameters 1.1. Assume _{f's fy}= ϕ=0.9 1.2. Calculate Mn= Mu_ϕ Mn1 A's f's=  (dd') Mn2 Mn Mn1=  R Mn2 b d 2 = ρ 0.85 f'c fy  1 1 2 R 0.85 f'c   













 = As ρ b=  d a As fy  0.85 f'c b = c a β1 = f's.revised Es εu=  c _cd' fy εt εu dt c  c  = ϕ=_{ϕ εt}

 

f's.revised f's : Goto 1.2 Mu ϕ Mn  : Goto 1.2 Step 2. Calculation of tensile steel area

As 0.85 f'c   ba  _{A's f's} fy = ρ As b d = ρ' A's b d = _{ρt.max ρmax ρ'}= 

(50)

Example 11.3

Required strength _Mu 1350kN m Concrete dimension b 400mm h 800mm d h 30mm12mm25mm 28mm 40mm 2 









  d 685 mm dt h 30mm 12mm 25mm 28mm 2 









  719 mm  d' 30mm 12mm 28mm 2   56 mm 

Compression reinforcements _A's 5 π 28mm( )

2

 4

 30.788 cm 2 













min 0.85

























0.00354  Compression parameters

(51)

Compression ε( ) _f's_fy ϕ0.9 Mn Mu_ϕ Mn1A's f's (d d') Mn2Mn Mn1 R Mn2 b d 2  ρ 0.85 f'c fy  1 1 2 R 0.85 f'c   













  Asρ b d a As fy  0.85 f'c b  c a β1 

f's.revisedmin Es εu



_

 c _cd'fy



_

Z i f's fy ϕ a d f's.revised fy













 εt εu dt c  c   ϕ 0.65 max 1.45_{250 εt} 3













min 0.9













 break ( ) f's.revised f's  f's ε

















Mu ϕ Mn 



if f'sf's.revised i0 99 for reverse Z

 

T  Z Compression 0.000001( )

(52)

a Z 0 2 d 142.792 mm  c a β1  167.99 mm  f's Z 0 0fy390 MPa  Tensile steel area

ρ' A's b d  0.011  _ρt.max _{ρmax ρ'}  0.029 As 0.85 f'c   ba _{A's f's} fy 61.909 cm 2    ρ As b d 0.023 

Concrete "is enough" if ρ_ρt.max "is not enough" otherwise

Tensile steel _{As 61.909 cm}  2 10 π 28mm( )

2

 4

(53)

12. Design of T Beams

12.1. Effective Flange Width

For symmetrical T beam:

b L 4

 b _{bw 16hf} b s

where

L = span length of beam s = spacing of beam

12.2. Strength Analysis

Design as rectangular section Design as T section

a_hf a_hf or _{Mu ϕMnf} or _{Mu ϕMnf} where _Mnf _{0.85 f'c} _hfb d hf 2 













 =

(54)

Equilibrium in forces



X=0 T=_{C1 C2} T=_{As fs} =_{As fy} C1 0.85 f'c=  hf



b _bw



=_{Asf fy} C2 0.85 f'c=   ba _w=T _C1=_{As fy} _{Asf fy} Equilibrium in moments



M=0 Mn Mn1 Mn2=  Mn1 C1 d= 





_

 hf₂





_

_{Asf fy} d hf 2 













 = Mn2 C2 d= 



_

 ₂a



_

_{0.85 f'c}  ba _w d a 2 









 = Mn2



T _C1



d a 2 









 =



_{As fy} _{Asf fy}



d a 2 









 =

Condition of strain compatibility εs εu d c c = or εt εs dt c c = εs εu= d _c c _{εt εu} dt c  c  = c d εu εu εs  = c _dt εu εu εt  =

12.3. Steel Ratios

ρw As bw d = As fy  bw d fy = 0.85 f'c   ba _w _{Asf fy} bw d fy = ρw 0.85 β1 f'c fy  c d   _ρf = where _ρf Asf bw d =

(55)

Maximum steel ratio

ρw.max ρmax ρf= 

ρw ρw.max : concrete is enough ρw ρw.max : concrete is not enough

12.4. Determination of Moment Capacity

Given: _{bw d}  db h_t _f_As_f'c_fy

Find: _ϕMn

Step 1. Checking for rectangular beam

a As fy  0.85 f'c b =

a_hf : the beam is rectangular a_hf : the beam is tee

Step 2. Case of T beam

Asf 0.85 f'c  _hf



b _bw



fy = Mn1 Asf fy d hf 2 













 = a As fy   _{Asf fy} 0.85 f'c bw = c a β1 = Mn2



As fy  Asf fy



d a 2 









 = εt εu dt c  c  = ϕ=_{ϕ εt}

 

ϕMn ϕ Mn1 Mn2= 







(56)

Example 12.1

Concrete dimension b 28in711.2 mm hf 6in152.4 mm bw 10in 254 mm h 30in762 mm d 26in660.4 mm dt 27.5in698.5 mm Steel reinforcements _As 6 π 10 8 in









2  4   _{As 7.363 in}  2

Materials _f'c 3000psi20.684 MPa fy 60ksi413.685 MPa

Solution

Steel ratios β1 0.65 max 0.85 0.05 f'c 4000psi  1000psi  













min 0.85













 0.85  εu 0.003 ρmax 0.85 β1 f'c fy  εu εu 0.005  0.014  ρmin max 3psi f'c psi  fy 200psi fy 













0.00333 

(57)

a As fy 

0.85 f'c b 157.162 mm 

The_beam "is rectangular" if a_hf "is T" otherwise  The_beam"is T" Case of T beam Asf 0.85 f'c  _hf



b_bw



fy 29.613 cm 2    ρf Asf bw d 0.018  _ρw.max _{ρmax ρf} 0.031 ρw As bw d  0.028 

Concrete "is enough" if _{ρw ρw.max} "is not enough" otherwise

As min ρw ρw.max







bwd _{As 7.363 in}  2 Mn1 Asf fy d hf 2 













 715.669 kN m   a As fy  _{Asf fy} 0.85 f'c bw  165.734 mm  c a β1  194.981 mm  Mn2



As fy Asf fy



d a 2 









  427.446 kN m   εt εu dt c  c   0.008  ϕ 0.65 max 1.45 _{250 εt} 3













min 0.9













 0.9  ϕMn ϕ Mn1 Mn2







1028.803 kN m 

(58)

12.5. Determination of Steel Area

Given: _{Mu bw d}   b_dt h _f_f'c_fy

Find: _As

Step 1. Checking for rectangular beam Mnf 0.85 f'c hfb d hf 2 













 = ϕ=0.9

Mu ϕMn : the beam is rectangular Mu ϕMn : the beam is tee

Step 2. Case of T beam

Asf 0.85 f'c  _hf



b _bw



fy = _ρf Asf bw d = Mn1 Asf fy d hf 2 













 = Mn2= Mu_ϕ Mn1 R Mn2 bw d 2 = ρ 0.85 f'c fy  1 1 2 R 0.85 f'c   













 = As2 ρ bw=  d a As2 fy  0.85 f'c bw = As 0.85 f'c   ba _w _{Asf fy} fy = _ρw As bw d = ρw.max ρmax ρf= 

ρw ρw.max : concrete is enough ρw ρw.max : concrete is not enough

(59)

Example 12.2

Concrete dimension _hf 3in76.2 mm

L 24ft7.315 m s 47in1.194 m bw 11in 279.4 mm d 20in508 mm Required strength _Mu 6400in kip 723.103 kN m 

Materials _f'c 3000psi20.684 MPa _fy 60ksi413.685 MPa

Solution

Effective flange width b min L 4 bw 16 hf  s









 b 1193.8 mm Steel ratios β1 0.65 max 0.85 0.05 f'c 4000psi  1000psi  













min 0.85













 0.85  εu 0.003 ρmax 0.85 β1 f'c fy  εu εu 0.005  0.014  ρmin max 3psi f'c psi  fy 200psi fy 













0.00333 

Checking for rectangular beam ϕ 0.9 Mnf 0.85 f'c hfb d hf 2 













 751.538 kN m  

The_beam "is rectangular" if _{Mu ϕ Mnf}  "is tee" otherwise

 The_beam"is tee"

(60)

Asf 0.85 f'c  _hf



b_bw



fy 29.613 cm 2    _ρf Asf bw d 0.021  Mn1 Asf fy d hf 2 













 575.646 kN m   Mn2 Mu_ϕ  Mn1 227.801 kN m  R Mn2 bw d 2 3.159 MPa   ρ 0.85 f'c fy  1 1 2 R 0.85 f'c   













  8.484103  As2 ρ bw d 12.042 cm 2 a As2 fy  0.85 f'c bw 101.408 mm  c a β1  119.304 mm  As 0.85 f'c   ba _w_{Asf fy} fy 41.655 cm 2    _ρw As bw d  0.029  ρw.max ρmax ρf 0.034

Concrete "is enough" if _{ρw ρw.max} "is not enough" otherwise

As.min ρmin bw d As max As As.min







 41.655 cm 2 6 π 32mm( ) 2  4   48.255 cm 2

(61)

13. Shear Design

Safety provision

Vu ϕVn

where _Vu = required shear strength Vn = nominal shear strength

ϕ=0.75 is a strength reduction factor for shear

ϕVn = design shear strength

(62)

Nominal shear strength

Vn Vc Vs=  where

Vc = concrete shear strength Vs = steel shear strength Concrete shear strength

Vc 2 f'c=  bwd (in psi) Vc 0.166 f'c=  bwd (in MPa) Steel shear strength

Vs Av fy  d s = where Av = area of stirrup

fy = yield strength of stirrup s = spacing of stirrup No required stirrups Vu  ϕVc₂ : no stirrup is required ϕVc 2 VuϕVc : stirrup is minimum Vu ϕVc : stirrup is required Minimum stirrups Av.min 0.75 f'c bw s  fy  50 bw s  fy   = (in psi) Av.min 0.062 f'c bw s  fy  0.345 bw s  fy   = (in MPa)

(63)

Maximum spacing of stirrup smax Av fy  0.75 _f'c_bw Av fy 50 bw  = (in psi) smax Av fy  0.062 _f'c_bw Av fy 0.345 bw  = (in MPa) Case _{Vs 2 Vc}  smax= d₂ 24in=600mm Case _{2 Vc} _Vs_{4 Vc} smax= d₄ 12in=300mm Case _{Vs 4 Vc} 

Concrete is not enough

Example 13.1

(64)

Live load for garage LL 6.00kN m2  Loads on slab Hardener 8mm 24 kN m3 0.192 kN m2    Slab 200mm 25 kN m3 5 kN m2    Mechanical 0.30kN m2 

DL Hardener Slab Mechanical 5.492 kN m2    LL 6 kN m2   Loads on beam wbeam 30cm 60cm(  200mm)25kN m3 3 kN m    wD.slab DL 3.5 m 19.222 kN m    wL.slab LL 3.5 m 21 kN m    wD wbeam wD.slab 22.222 kN m    wL wL.slab 21 kN_m wu 1.2 wD 1.6 wL 60.266 kN m    Shear L 8m V0 wu L  2  241.066 kN  V x( ) _{V0 wu x} 

Concrete shear strength

bw 300mm d 600mm 40mm 10mm 20mm 2 









 540 mm  Vc 0.166MPa f'c MPa  _bwd 134.46 kN  ϕ 0.75

(65)

Location of no stirrup zone V0 wu x  = ϕVc₂ x V0 ϕ Vc  2  wu  3.163 m  Minimum stirrup Av 2 π 10mm( ) 2  4  1.571 cm 2  _fy 390MPa smax min Av fy  0.062MPa f'c MPa  _bw Av fy 0.345MPa bw 













591.894 mm  

smax Floor smax 50mm







550 mm Vs.min Av fy

 d

smax  60.147 kN 

Location of minimum stirrup zone

V0 wu x  =ϕ Vc Vs.min







x V0 ϕ Vc Vs.min 





  wu 1.578 m 

Required spacing of stirrup

Vu V0 wu 



_

400mm₂



_

229.012 kN Vs Vu_ϕ  Vc 170.89 kN

Concrete "is enough" if _{Vs 4 Vc}  "is not enough" otherwise

s Av fy  d Vs 193.581 mm  smax.1 smax 550 mm  smax.2 min d 2600mm









if Vs 2 Vc  min d 4300mm









otherwise  _{smax.2 270 mm} 

(66)

Example 13.2

Design of shear in support and midspan zones.

Stirrups in Support Zone

Required shear strength _Vu _{V0 wu} 400mm 2 

  229.012 kN 

Concrete shear strength Vc 0.166MPa f'c

MPa

 _bwd 134.46 kN 

ϕ 0.75

Stirrup "is minimum" if _{Vu ϕ Vc}  "is required" otherwise

 Stirrup "is required"

Required steel shear strength Vs Vu_ϕ  Vc 170.89 kN

Concrete "is enough" if _{Vs 4 Vc}  "is not enough" otherwise

Spacing of stirrup Av 2 π 10mm( ) 2  4  1.571 cm 2  _fy 390MPa s Av fy  d Vs 193.581 mm  smax.1 min Av fy  0.062MPa f'c MPa  _bw Av fy 0.345MPa bw 

Mathcad in Concrete Structures ACI 318-05 6th Edition

CONSTRUCTION MANAGEMENT

Concept-Develop-Execute-Finish

Seun Sambath, PhD

Concrete Structures

(ACI 318-05)

in Mathcad

Sixth Edition

Phnom Penh 2010

Address: #M41, St.308, Sangkat Tonle Basac, Khan Chamkarmon,

Phnom Penh, Cambodia.

Table of Contents

1. Unit Conversion

Length

Force

Stress

Density

Moments

User setting

2. Simple Calculation













3. Materials

1. Concrete

















Example 3.1

























2. Steel Reinforcements



























































