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5>1> karKNnaF wm E2-B11. Drawing for calculation1> KMnrbg;srab;KNna

bg;F wmCan;TI2-10

B1

1. Drawing for calculation

2.Unit use in calculated2> xateRbIk gkarKNnakN 1000N MPa 10

6Pa KPa 10

3Pa

K 1000lb lb 4.448N K 4.448kN

psi 0.689 Ncm

2 psi 0.00689MPa psi 6.895 kNm

2

Ksi 1000psi Ksi 6.895MPa Ksi 6.895 103

KPa

K ft 1.356kN m K in 0.113kN m

psf 0.04788kN

m2

psf 47.88N

m2

psf 47.88Pa

Ksf 1000psf Ksf 47.88KPa Ksf 0.048MPa

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pcf 16.02kgf

m3

pcf 0.157kN

m3

2.Unit use in calculated

3. Data apply for design

3> TnnyTUeTA nig sm tikm er:sIusg;sgt;rbs;ebtugenAGayu28f f'c 25MPa f'c 3.626Ksi

er:sIusg;Tajrbs;EdkeFVIkar fy 400MPa fy 58.015Ksi

er:sIusg;Tajrbs;Edkkg fy.s 235MPa fy.s 34.084Ksi

m:UDuleGLasicrbs;ebtug Ec 24000MPa Ec 3.481 103 Ksi

m:UDuleGLasicrbs;Edk

Es 200000MPa Es 2.901 104

Ksi

TMgn;maDrbs;ebtug c 150pcf c 24kN

m3

TMgn;maDrbs;Edk s 500pcf s 78.55kN

m3

Strength-reduction factor for bending moment : 0.9

bnkGefrelIkMralxN LL 50psf LL 2.4KPa

kMras;rbs;kMralxN hf 19cm hf 7.48in

muxkat;rbs;ssr Cx 80cm Cx 31.496in

Cy 80cm Cy 31.496in

muxkat;ssEdkeFVIkar db 20mm db 0.787in

muxkat;ssEdkkg ds 10mm ds 0.394in

RsTab;karBarebtug Cc 30mm Cc 1.181in

RbEvgRbelaHElVg L 6m L 19.685ft

3. Data apply for design

4. M & V Value

4> DdaRkam ng TinnytMl2

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6000 6000 6000

-242.57kN.m -271.72kN.m -278.98kN.m

123.17kN.m 125.33kN.m 135.47kN.m

-239.55kN -235.20kN -213.34kN

199.47kN 212.30kN 210.17kN

-3.50kN.m -3.17kN.m -8.64kN.m

2.33kN.m2.72kN.m

Diagram Moment

1.18kN.m

Diagram Shear Force

Diagram Torson

Moment value:

On support : Mu.s 278.98kN m Mu.s 206K ft

At mid span : Mu.mid 135.47kN m Mu.mid 100K ft

Shear value:

On support : Vu.s 239.55kN Vu.s 54K

At mid span : Vu.mid 76.24kN Vu.mid 17K

Torsion value:

Maximum torsion : Tu 8.64kN m Tu 6.373K ft

4. M & V Value

5. Section Design

5> karKNnamuxkat;rbsF wm

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b

t

bf

h

- karKNnakMBs;rbs;Fwm

hminL

10 hmin 23.622in hmin 60cm

h ceilhmin

5cm

5 cm h 23.622in h 60 cm

t hf t 7.48 in t 19 cm

- karKNnaTTwgrbs;Fwm

b 0.5 h b 11.811in b 30 cm

bw b bw 11.811in bw 30cm

- kMBs;eFVIkarrbs;ssEdkenAkgFwm

d h 2.5in d 21.122in d 53.65cm

d' 2.5ind' 6.35cm

5. Section Design

6.1. Rectangular beam design

6> karKNnasrsrEdkk gmuxkat;6>1 > karKNnasrsrEdkenAelITr

4

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b

h

As

A'sd'

d

Maximum bending moment on support is :

Mu K ftMu Mu kN m

Calculate the steel ratio:

Equivalent constant :Mu Mu.s

fc 4000psi( ) f'c 4000psiif

8000psi( ) f'c 8000psiif

f'c otherwise

fc 27.579MPa

fc 4Ksi

1 0.85 0.05

fc 4000psi

1000psi 1 0.85

min max3 f'c psi

fy

200psi

fy

min 0.003

max 0.6375 1f'c

fy

87Ksi

87Ksi fy

max 0.02

Calculate R:

Ru.max max fy 1max fy

1.7 f'c

Ru.max 858.013psi

Ru.max 5.916MPa

Calculate the maximum moment:

Mu1 Ru.max b d2

Mu1 376.785K ft

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Mu1 510.826kN m

Check Compression steel is required or not?

Compression "steel is required" Mu Mu1if

"steel is not required" otherwise

Compression "steel is not required"

Calculate reinforcement bars:

Tension Steel:

As As' RuMu

b d2

0.85 f'c

fy1 1

2 Ru

0.85 f'c

As b d( ) min maxif

0 otherwise

Mu Mu1if

As'' As1 max b d

Mu2 Mu Mu1

As2Mu2

fy d d'( )

'As2

b d

As.max b d max '

As min As1 As2 As.max

Mu Mu1if

As 2.469in2

As 15.928cm2

Compression Steel:

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A's A''s As1 max b d

Mu2 Mu Mu1

As2Mu2

fy d d'( )

As.max max b d

As min As1 As2 As.max

a As1fy

0.85 f'c b

ca

1

f's min 87 Ksic d'

c

fy

A's As2 f's fy=if

As2fy

f's

f's fyif

Mu Mu1if

0 otherwise

A's 0 in2

A's 0cm2

Choose Number of reinforcement bars:

Area of diameter bar:

Ab db

2

4 Ab 0.487in

2 Ab 3.142cm

2

Number of compression reinforcement bars:

Nc.s ceilA's

Ab

Nc.s 0

Number of tensile reinforcement bars:

Nt.s ceilAs

Ab

Nt.s 6

Actual Area of reinforcement bars in use:

Compression steel:

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A's Ab Nc.s A's 0in2

A's 0cm2

'A's

b d ' 0

Tensile steel:

As Ab Nt.s As 2.922in2

As 18.85cm2

As

b d 0.012

Total reinforcement ratio is:

total ' total 0.012

Verification of nominal moment:

k 0.85 1f'c

fy

d'

d

87Ksi

87Ksi fy

k 0.003

a aAs fy

0.85 f'c b

a'As A's fy0.85 f'c b

a'' A1 0.85 1 f'c bA2 A's 87 Ksi 0.85 f'c As fy

A3 87 Ksi A's d'

c c'1

2 A1A2 A2

24 A1 A3

c''1

2 A1A2 A2

24 A1 A3

c c' c' 0if

c'' otherwise

a'' 1 c

a a Mu Mu1if

a a' '( ) kif

a'' otherwise

otherwise

a 4.656in a 11.827cm

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b

t

bf

h

As

d

Maximum bending moment at mid span is :

Mu Mu.mid Mu 100K ft Mu 135.47kN m

Calculate ultimate moment of total flange is:

bf min

L

4 16t bw

L

bf 59.055in

bf 150cm

Cft 0.85 f'c bf t Cft 1362K Cft 6056kN

Muft Cft dt

2

Muft 1775K ft Muft 2406kN m

Check if the section acts as a rectangular or T-section:

Equivalent constant :

fc 4000psi( ) f'c 4000psiif

8000psi( ) f'c 8000psiif

f'c otherwise

fc 27.579MPa

fc 4Ksi

1 0.85 0.05fc 4000psi

1000psi 1 0.85

min max3 f'c psi

fy

200psi

fy

min 0.003

b 0.85 1f'c

fy

87Ksi

87Ksi fy b 0.027

max 0.75 b max 0.02

Check "T-section design." Mu Muftif

"Rectangular section design." otherwise

Check "Rectangular section design."

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b bw Check "Rectangular section design."=if

bf Check "T-section design."=if

b 30 cm

b 11.811in

Cf 0.85 f'c t b bw Cf 0K Cf 0kN

AsfCf

fy Asf 0 in

2 Asf 0cm

2

Muw Muf Cf dt

2

Muw Mu Muf

Muw 99.923K ft Muw 135kN m

Ru RuMu

bw d2

Ruw Muw

bw d2

Ru Ru Mu Muftif

Ruw otherwise

Ru 1.569MPa

Ru 0.228Ksi

0.85f'c

fy

1 1

2 Ru

0.85 f'c

0.005

As As' bw d

As'' As' Asf

As As' Mu Muftif

As'' otherwise

As 7.328cm2

As 1.136in2

wAs

b d w 0.005

Control the maximum reinforcement area use in section:

As.max 0.0319 bf t bwd

2t

f'c 3Ksiif

0.0425 bf t bwd

2t

f'c 4Ksiif

As.max 20.321in2 As.max 131.102cm

2

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maxAs.max

b d max 0.081

Control "Section is OK!!!" As.max Asif

"Section is too hight reinforcement ratio." otherwise

Control "Section is OK!!!"

Number of reinforcement bars in used :

Nt.mid Av db

2

4

N ceilAs

Av

Nt.mid 3

As.new Nt.mid db

2

4 As.new 1.461in

2 As.new 9.425cm

2

newAs.new

bw d new 0.006

Verification with nominal moment :

a'As.new fy

0.85 f'c b a' 2.328in a' 5.914cm

Check "Rectangular analysis." a' tif

"T-section analysis." otherwise

Check "Rectangular analysis."

a Asf0.85 f'c t b bw

fy

aAs.new Asf fy

0.85 f'c bw

a a' Check "Rectangular analysis."=if

a Check "T-section analysis."=if

a 5.914cm

a 2.328in

Mn Mn' As.new fy da

2

Mn'' As.new Asf fy da

2

Asf fy dt

2

Mn Mn' Check "Rectangular analysis."=if

Mn'' Check "T-section analysis."=if

Mn 127K ft Mn 172kN m

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Check "Section is Ok!!!" Mu Mnif

"Section is damage" otherwise

Check "Section is Ok!!!"

6.2. T-beam design

7. V & T Design

7> karKNnaEdkkg7>1> karKNnaEdkkgTb;ng kMlaMgkat;TTwg

Maximum shear force on support is :

Vu Vu.s Vu 53.856K Vu 239.55kN

Shear value at distance d from support :

Ln L Cx Ln 17.06ft Ln 5.2m

Vu.d Vud

0.5 LnVu Vu.mid Vu.d 46K Vu.d 206kN

Vc 2 f'c psi bw d Vc 30K Vc 134kN

Vc 2 f'c psi bw d Vc 27K Vc 120kN

VsVu.d Vc

Vs 21K Vs 95kN

Vs Vu.d Vc Vs 19K Vs 86kN

Av 2 ds

2

4 Av 0.243in

2 Av 0.243in

2

Sv.max Smax1 mind

2

Av fy.s

50psi bw

24in

Smax2 mind

412in

S' floorSmax1

1cm

1 cm Vs 2Vcif

floorSmax2

1cm

1 cm Vs 2Vcif

0 otherwise

Sv.max 26cm

Sv.max 10.236in

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Sv SAv fy d

Vs

S min S Sv.max

S floorS

1cm

1 cm

Vs 4Vcif

0 otherwise

Sv 26cm

Sv 10.236in

7>2> karKNnaEdkkgTb;ng kMlaMgrmYlMaximum torsion force is : Tu 6.373K ft Tu 8.64kN m

Determine sectional properties :

A oh

X o

Y o Y 1

X 1

xo b yo h

x1 b 3.5in x1 8.311in x1 21.11cm

y1 h 3.5in y1 20.122in y1 51.11cm

Aoh x1 y1 Aoh 167.235in2

Aoh 1079cm2

Ao 0.85 Aoh Ao 142.15in2

Ao 917cm2

Acp xo yo Acp 279.001in

2

Acp 1800cm

2

Ph 2 x1 y1 Ph 56.866in Ph 144.44cm

Pcp 2 xo yo Pcp 70.866in Pcp 180cm

Ta f'c psi Acp

2

Pcp Ta 5K f t Ta 7kN m

4Ta 20K ft 4Ta 27kN m

Tu 4Ta Tu 4TaifTu otherwise

Tu 6K f t Tu 8.64kN m

Check the adequacy of the size of the section :

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LiftVu.d

bw d

2Tu Ph

1.7 Aoh2

2

Lift 161.384psi

Right Vc

bw

d

8 f'c psi

Right 541.942psi

Check "The section is adequate." Lift Rightif

"Increase section." otherwise

Check "The section is adequate."

Determine the required closed stirrups due to torsion :

45deg cot ( ) 1

fyv fy.s fyl fyv

AtTu

2 Ao fy.s cot ( ) At 0.009

in2

in At 0.022

cm2

cm

- KMlatEdkkgTb;Tl;CamYynigkMlaMgrmYl

St AvVs

2 fy.s d

Avt max At Av 50 psibw

fy.s

Ast ds

2

4

S floor

minAst

Avt

Ph

8

1 cm

1 cm

S S Tu Ta Lift Rightif

0 otherwise

St 13cm

St 5.118in

To find the distribution of longitudinal bars is :

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Al Al At Phfyv

fyl

cot ( )

2

Al.min5 f'c psi Acp

fylAt Ph

fyv

fyl

Al max Al Al.min Tu Ta Lift Rightif

0 otherwise

Al 12.683cm2

Al 1.966in2

As.top1

3Al As.top 0.655in

2 As.top 4.228cm

2

As.bottom1

3Al As.bottom 0.655in

2 As.bottom 4.228cm

2

As.mid1

3Al As.mid 0.655in

2 As.mid 4.228cm

2

dt1

2

4As.mid

dt 0.457in dt 12mm

7>3> kardMerobEdkkgenAkgF wmd

Vu

Vu.d

cV s+V

Maximum

spacing

1x

x

c1/2V

u.midV

Ln/2

CL

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S S So x1 x

Ln/2

Stirrup spacing is:

S Sv Tu Taif

St otherwise

S 5.118in S 13 cm

First distance from support which can start placing the first stirrups is :

x0S

2 x0 2.559in x0 6.5cm

S0 x0 S0 2.559in S0 6.5cm

Calculate Vs for maximum stirrup spacing:

VsAv fy.s d

Sv.maxTu Taif

0 otherwise

Vs 0K Vs 0kN

Vs Vs Vs 0K Vs 0kN

The distance x1 at which maximum stirrups spacing can be used is:

x1Vu Vc Vs

Vu

Ln

2

Tu Taif

0 otherwise

x1 0cm

x1 0 in

The distance x at which no shear reinforcement is needed is:

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x

VuVc

2

Vu

Ln

2

Tu Taif

Ln

2x0 otherwise

x 253.5cm

x 99.803in

Number of stirrups placing at distance x1 is :

Nx1 ceilx1 x0

S

Nx1 0

Sx1 S Nx1 0if

0 otherwise

Sx1 0 in Sx1 0cm

Number of stirrups placing at distance x is :

Nx ceilx x1

S

Tu Taif

ceilx x1

Sv.max

otherwise

Nx 20

Sx Sv.max Nx1 0if

S otherwise

Sx 5.118in Sx 13cm

Total number of stirrups is:

Nst Nx1 Nx Nst 20

Total distance placing total stirrups is:

X x0 Sv Nx1 Sv.max Nx Tu Taif

x0 S Nst otherwise

X 266.5cm

X 104.921in

Free distance at midspan is:

x'Ln

2X

Ln

2X 0if

0 otherwise

x' 0 cm

x' 0 in

7. V & T Design

8. Deflection Control

8> karepgpat;PaBdabMaximum moment in member at stage for which deflection is being computed:

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MaMu.mid

1.5 Ma 67K ft Ma 90kN m

w8 Ma

L2

w 1.375K

ft w 20

kN

m

Moment of inertia of gross concrete section aboutthe centroidal axis:

Igb h

3

12 Ig 12974in

4 Ig 540000cm

4

Modulus of rupture of concrete:

fr 7.5 f'c psi fr 452psi fr 3MPa

Distance from centroidal axis of concrete section:

Yt h2

Yt 11.811in Yt 30cm

Cracking moment:

Mcrfr Ig

Yt Mcr 496K in Mcr 56kN m

Distance to the neutral axis :

nEs

Ec

n 8

b

2x

2 n As x n As d 0=

x a' 0.5 b

b' n As

c' n As d

' b'2

4 a' c'

x' b' '2 a'

x''b' '

2 a'

x x' x' 0if

x'' otherwise

x 5.072in x 12.883cm

Moment of inertia of cracked transformed section:

Icr b x3

3

n As d x( )2 Icr 2952in4 Icr 122873cm4

The effective moment of inertia is:

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IeMcr

Ma

3

Ig 1Mcr

Ma

3

Icr Ie 222575cm

4

Ie 5347in4

Calculate the deflections from the differerent loads:

5

384

w L4

Ec Ie 0.25in 0.634cm

Compare the calculated values with the allowable deflection:

allL

360 all 0.656in all 1.667cm

The immediate deflection due to a uniform live load is:

ll 0.6 ll 0.15in ll 0.38cm

Deflection " is OK!!! " ll allif

" is not allowed. " otherwise

Deflection " is OK!!! "

8. Deflection Control

9. Cracking Control

9> karepgpat;sameRbHCrack width limiting:

The ratio of distance from the netural axis to the tension face and to the steel centroid:

- For beam : 1.2

Stress in reinforcement at service load:

fs 0.6 fy fs 34.809Ksi fs 240MPa

dc Cc dsdb

2 dc 1.969in dc 5cm

m Nt.midNumber of reinforcement bars:

A2 dc b

m A 15.5 in

2 A 100 cm

2

W 0.076

in2

lb fs

3

A dc 106

W 0.25198mmW 0.00992in

Tolerable crack widths for reinforced concrete:

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Exposure condition: Tolerable crack width

1. Dry air or protective membrane: W1 0.016in

2. Humidity, moist air, soil: W2 0.012in

3. Deicing chemicals: W3 0.007in

4. Seawater and Seawater spray, wetting and drying: W4 0.006in

5. Water-retaining structures,excluding nonpressure pipes: W5 0.004in

Cracking "is OK!!! for Condition No.1" W W1if

"is OK!!! for Condition No.2" W W2if

"is OK!!! for Condition No.3" W W3if

"is OK!!! for Condition No.4" W W4if

"is OK!!! for Condition No.5" W W5if

"Cracking is not allowable, thus redesign the section." otherwise

Cracking "is OK!!! for Condition No.2"

ACI code provisions for crack control:

Factor related to width of crack:

z fs3

dc A z 109K

in z 19

kN

mm

Check "Cracking is OK!!! for exterior exspoure." z 145K

inif

"Cracking is OK!!! for interior exspoure." z 175K

inif

"Redesign or reducing the value of reinforcement bars" otherwise

Check "Cracking is OK!!! for interior exspoure."

9. Cracking Control

10. Summary Calculation

10> segbTinnyKNna- kMBs;rbs;Fwm h 60 cm

- TTwgrbs;Fwm b 30 cm

- m uxkat;ssEdkeFVIkarrbs;Fwm db 20mm

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- cMnYnssEdkxageRkamenAelITMrrbs;Fwm Nc.s 0

- cMnYnssEdkxagelIenAelITMrrbs;Fwm Nt.s 6

- cMnYnssEdkxageRkamenAkNalElVgrbs;Fwm Nt.mid 3

- m uxkat;ssEdkkgrbs;Fwm ds 10mm- cMnYnssEdkkgsrubrbs;Fwm Nst 20

- KMlatssEdkkgdMbUgecjBIEKmrbs;ssr S0 65mm

- cMnYnssEdkkgerobenACMhanTI1rbs;Fwm Nx1 0 KMlat Sx1 0mm

- cMnYnssEdkkgerobenACMhanTI2rbs;Fwm Nx 20 KMlat Sx 130mm

- PaBdabrbs;Fwm Deflection " is OK!!! " ll 4mm- sameRbHrbs;Fwm W 0.3mm

10. Summary Calculation