DOCUMENT No

CALCULATIONSOFFICE PROJECT TITLE

MANILA MU000749

SUBJECT SHEET No

MANUAL VERIFICATION OF PROKON - LONG TERM DEFLECTION

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DESIGN BASIS STATEMENT (Inc. sources of info/data, assumptions made, standards, etc.)

DOCUMENT No. (Project/ Doc)

SHEET No.1 OF

PROJECT TITLE

DOCUMENT No SHEET

MU000749 OF

SUBJECTPROKON and SAFE - long term deflection verification

SUBJECT CALCULATIONS OUTPUT

Reference : BS8110 - Part 2 1985

Calculation of curvatures :

slab mark :

grid : tolevel :

Mp = 181 kN .m Applied design moment due to permanent load.

Mt = 28 kN .m Applied design moment due transitory load.

Es = 200 kN / mm² modulus of elasticity of reinforcement

fc = 45 N / mm² design service stress in concrete

fy = 460 N / mm² yield strength of steel

h = 350 mm total height of section

b = 1000 mm width of section

d = 308 mm effective depth of section

As = 2946 mm² reinforcement proivided

Ø = 2.00 creep coefficient

3.00E-04 free shrinkage strain

curvature can be obtained from the following equation.

a ) strain - stress diagram for cracked section :

b ) strain - stress diagram for uncracked section :

εcs =

gerry

DOCUMENT No SHEET

MU000749 OF

SUBJECTPROKON and SAFE - long term deflection verification

SUBJECT CALCULATIONS OUTPUT

Calculation of short term curvature - uncracked section - (total load)

=

= 2.02E-06 /mm

Calculation of short term curvature - cracked section - (total load)

x = 102 mm depth of neutral axis (assumed)

from fig. a (strain diagram)

M =

eq - 01

fs = 215.67 N / mm²

from fig. a (strain diagram)

eq - 02

fc = 15.43 N / mm²

from fig. a (stress diagram)

fct / (d - x ) = ŝ / (d - x) where : ŝ is tensile stress of concrete eq - 03

fct =

fct = 1.20 N / mm²

by force equilibrium :

fc = eq - 04

fc = 15.43 N / mm²

eq - 02 and eq - 04 should be more or less equal fc/fc =1.00

Correct assumption of neutral axis

Therefore use x = 102 mm

5.23E-06 /mm

Asfs(d - x /3) + 1/3 bh fct (h-x)

fc / Ecx =

fs /Esd - x

fs =M - 1/3 bh fct (h - x)

As (d - x / 3)

fc =x fs Ec

Es ( d - x )

ŝ ( h - x )

( d - x )

fs As + 1/2 b (h - x ) fct

1/2 b x

1

rb=

DOCUMENT No SHEET

MU000749 OF

SUBJECTPROKON and SAFE - long term deflection verification

SUBJECT CALCULATIONS OUTPUT

Calculation of short term curvature - cracked section - (permanent load)

x = 103 mm depth of neutral axis (assumed)

from fig. a (strain diagram)

M =

eq - 05

fs = 181.49 N / mm²

from fig. a (strain diagram)

eq - 06

fc = 13.25 N / mm²

from fig. a (stress diagram)

fct / (d - x ) = ŝ / (d - x) where : ŝ is tensile stress of concrete eq - 07

fct =

fct = 1.21 N / mm²

by force equilibrium :

fc = eq - 08

fc = 13.25 N / mm²

eq - 06 and eq - 08 should be more or less equal fc/fc =1.00

Correct assumption of neutral axis

Therefore use x = 103 mm

4.43E-06 /mm

Asfs(d - x /3) + 1/3 bh fct (h-x)

1

rb=

fs =M - 1/3 bh fct (h - x)

As (d - x / 3)

fc / Ecx

=fs /Es

d - x

fc =x fs Ec

Es ( d - x )

ŝ ( h - x )

( d - x )

fs As + 1/2 b (h - x ) fct

1/2 b x

DOCUMENT No SHEET

MU000749 OF

SUBJECTPROKON and SAFE - long term deflection verification

SUBJECT CALCULATIONS OUTPUT

Calculation of longt term curvature - cracked section - (permanent load)

x = 147 mm depth of neutral axis (assumed)

from fig. a (strain diagram)

M =

eq - 09

fs = 215.85 N / mm²

from fig. a (strain diagram)

eq - 10

fc = 9.58 N / mm²

from fig. a (stress diagram)

fct / (d - x ) = ŝ / (d - x) where : ŝ is tensile stress of concrete eq - 11

fct =

fct = 0.69 N / mm²

by force equilibrium :

fc = eq - 12

fc = 9.58 N / mm²

eq - 10 and eq - 12 should be more or less equal fc/fc =1.00

Correct assumption of neutral axis

Therefore use x = 147 mm

6.72E-06 /mm

Asfs(d - x /3) + 1/3 bh fct (h-x)

1

rb=

fs =M - 1/3 bh fct (h - x)

As (d - x / 3)

fc / Ecx

=fs /Es

d - x

fc =x fs Ec

Es ( d - x )

ŝ ( h - x )

( d - x )

fs As + 1/2 b (h - x ) fct

1/2 b x

DOCUMENT No SHEET

MU000749 OF

SUBJECTPROKON and SAFE - long term deflection verification

SUBJECT CALCULATIONS OUTPUT

Calculation of shrinkage curvature

where :

shrinkage curvature

modular ratio Es / Eeff

free shrinkage strain

Eeff effective modulus of elasticity of concrete

Ec short term modulus of concrete

Ø creep coefficient

I second moment of area of either the cracked

or the gross, depending on whether the curvature

due to loading is derived.

Ss first moment of area of the reinforcement about

the centroid of the cracked section, which ever

is appropriate.

Es / Eeff

20.690

I =

I = 2.64E+09

Ss = As ( d - x )

Ss = 4.73E+05 mm³

1.1E-06

long term curvaturre permanent + shrinkage curvature+ short term curvature total loads

- short term curvature permanent loads

8.63E-06

Δy =

where :

l = 10000 mm effective span of member

K = 0.104 constant depends on the shape of bending

moment diagram.

Δy = 89.77 mm

1/ rcs

άe

εcs

άe =

άe =

bx³/12 + bx (x/2)² + άe As (d - x )²

mm4

1/ rcs =

Calculation of long term deflection (Δy)

1/ rx =

1/ rx =

K l² (1/ rx)

1000 kN.m

K = 0.092

1.125541

300 kN.m

1155 kN.m

MA =

MB =

MC =

b = 800 mmh = 600 mm

effective thickness =

effective thickness = 600

2 x cross section of concrete

exposed peremeter

2 x cross section of concrete

exposed peremeter

DOCUMENT No SHEET

MU000749 OF

SUBJECTManual - Long Term Deflection

SUBJECT CALCULATIONS OUTPUT

Reference : BS8110 - Part 2 1985

Calculation of curvatures :

slab mark :

grid : tolevel :

Mp = 181 kN .m Applied design moment due to permanent load.

Mt = 28 kN .m Applied design moment due transitory load.

Es = 200 kN / mm² modulus of elasticity of reinforcement

fc = 45 N / mm² design service stress in concrete

fy = 460 N / mm² yield strength of steel

h = 350 mm total height of section

b = 1000 mm width of section

d = 308 mm effective depth of section

As = 2946 mm² reinforcement proivided

Ø = 2.00 creep coefficient

3.00E-04 free shrinkage strain

curvature can be obtained from the following equation.

a ) strain - stress diagram for cracked section :

b ) strain - stress diagram for uncracked section :

εcs =

gerry

DOCUMENT No SHEET

MU000749 OF

SUBJECTManual - Long Term Deflection

SUBJECT CALCULATIONS OUTPUT

Calculation of short term curvature - uncracked section - (total load)

=

= 2.02E-06 /mm

Calculation of short term curvature - cracked section - (total load)

x = 102 mm depth of neutral axis (assumed)

from fig. a (strain diagram)

M =

eq - 01

fs = 215.67 N / mm²

from fig. a (strain diagram)

eq - 02

fc = 15.43 N / mm²

from fig. a (stress diagram)

fct / (d - x ) = ŝ / (d - x) where : ŝ is tensile stress of concrete eq - 03

fct =

fct = 1.20 N / mm²

by force equilibrium :

fc = eq - 04

fc = 15.43 N / mm²

eq - 02 and eq - 04 should be more or less equal fc/fc =1.00

Correct assumption of neutral axis

Therefore use x = 102 mm

5.23E-06 /mm

Asfs(d - x /3) + 1/3 bh fct (h-x)

fc / Ecx =

fs /Esd - x

fs =M - 1/3 bh fct (h - x)

As (d - x / 3)

fc =x fs Ec

Es ( d - x )

ŝ ( h - x )

( d - x )

fs As + 1/2 b (h - x ) fct

1/2 b x

1

rb=

DOCUMENT No SHEET

MU000749 OF

SUBJECTManual - Long Term Deflection

SUBJECT CALCULATIONS OUTPUT

Calculation of short term curvature - cracked section - (permanent load)

x = 103 mm depth of neutral axis (assumed)

from fig. a (strain diagram)

M =

eq - 05

fs = 181.49 N / mm²

from fig. a (strain diagram)

eq - 06

fc = 13.25 N / mm²

from fig. a (stress diagram)

fct / (d - x ) = ŝ / (d - x) where : ŝ is tensile stress of concrete eq - 07

fct =

fct = 1.21 N / mm²

by force equilibrium :

fc = eq - 08

fc = 13.25 N / mm²

eq - 06 and eq - 08 should be more or less equal fc/fc =1.00

Correct assumption of neutral axis

Therefore use x = 103 mm

4.43E-06 /mm

Asfs(d - x /3) + 1/3 bh fct (h-x)

1

rb=

fs =M - 1/3 bh fct (h - x)

As (d - x / 3)

fc / Ecx

=fs /Es

d - x

fc =x fs Ec

Es ( d - x )

ŝ ( h - x )

( d - x )

fs As + 1/2 b (h - x ) fct

1/2 b x

DOCUMENT No SHEET

MU000749 OF

SUBJECTManual - Long Term Deflection

SUBJECT CALCULATIONS OUTPUT

Calculation of longt term curvature - cracked section - (permanent load)

x = 147 mm depth of neutral axis (assumed)

from fig. a (strain diagram)

M =

eq - 09

fs = 215.85 N / mm²

from fig. a (strain diagram)

eq - 10

fc = 9.58 N / mm²

from fig. a (stress diagram)

fct / (d - x ) = ŝ / (d - x) where : ŝ is tensile stress of concrete eq - 11

fct =

fct = 0.69 N / mm²

by force equilibrium :

fc = eq - 12

fc = 9.58 N / mm²

eq - 10 and eq - 12 should be more or less equal fc/fc =1.00

Correct assumption of neutral axis

Therefore use x = 147 mm

6.72E-06 /mm

Asfs(d - x /3) + 1/3 bh fct (h-x)

1

rb=

fs =M - 1/3 bh fct (h - x)

As (d - x / 3)

fc / Ecx

=fs /Es

d - x

fc =x fs Ec

Es ( d - x )

ŝ ( h - x )

( d - x )

fs As + 1/2 b (h - x ) fct

1/2 b x

DOCUMENT No SHEET

MU000749 OF

SUBJECTManual - Long Term Deflection

SUBJECT CALCULATIONS OUTPUT

Calculation of shrinkage curvature

where :

shrinkage curvature

modular ratio Es / Eeff

free shrinkage strain

Eeff effective modulus of elasticity of concrete

Ec short term modulus of concrete

Ø creep coefficient

I second moment of area of either the cracked

or the gross, depending on whether the curvature

due to loading is derived.

Ss first moment of area of the reinforcement about

the centroid of the cracked section, which ever

is appropriate.

Es / Eeff

20.690

I =

I = 2.64E+09

Ss = As ( d - x )

Ss = 4.73E+05 mm³

1.1E-06

long term curvaturre permanent + shrinkage curvature+ short term curvature total loads

- short term curvature permanent loads

8.63E-06

Δy =

where :

l = 10000 mm effective span of member

K = 0.104 constant depends on the shape of bending

moment diagram.

Δy = 89.77 mm

1/ rcs

άe

εcs

άe =

άe =

bx³/12 + bx (x/2)² + άe As (d - x )²

mm4

1/ rcs =

Calculation of long term deflection (Δy)

1/ rx =

1/ rx =

K l² (1/ rx)

L 10 10000 b 1000 As 6fy 460 h 350fc' 55 d 307.5Ec 34856.132889 4700*SQRT(B3)Es 200000s/w 8.75sdl 5TDL 13.75LL 3

applieD momentTL 209.375 Kn.MPL 171.875 Kn.M

n 5.7378711699Ig 3.573E+09fr 4.598043062Mcr 93.876712516

nAs 16899.425545nAsd 5196573.3551 c -5196573NAs(-x) -16899.42554 x b 16899.426 xbxx/2 500 xx a 500 xxx 86.438550504 -2.444E-05

Icr 1041122661

Mcr/Ma 0.4483663881

Ie 1269328899

Ydl 49.294699192 49.29469919186

FACTORD0.8208955224FActorLL 0.1791044776LTDD 80.931595688LTDLL 8.8289013478

89.760497036

T 25 2945.243

X 520.256 520.256

H 1450 1450

b 300 300

d 1350 1350

M 2992.45 Kn.M 2992450000 N.mm 2.99E+09

As' 6434As 6434fc 40dt 100Es 200000Ecstd 28000Creep 2 6434Ecltd 9333.333FCTstd 1.120519 N/MM2

FCTltd 0.616286

bfs 375.3425 N/MM2

fc1 #NAME?

fc2 #NAME? N/mm2

fc1-fc2 #NAME?

X 793.1140472H 1450b 300d 1350M 2992.45 Kn.M 2992450000 N.mmAs' 6434As 6434fc 40dt 100Es 200000Ecstd 28000Creep 2.6Ecltd 7777.7777778FCTstd 1.1795699811 N/MM2

FCTltd 0.6487634896

bfsltd 412.32994713 N/MM2

fc1 22.83703892

fc2 22.83703025 N/mm2

fc1-fc2 8.66989E-06

n 25.714285714

Icr 180678887156