DOCUMENT No
CALCULATIONSOFFICE PROJECT TITLE
MANILA MU000749
SUBJECT SHEET No
MANUAL VERIFICATION OF PROKON - LONG TERM DEFLECTION
ISSUE AUTHOR DATE CHECKED BY DATE APPROVED BY DATE COMMENTS
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SUPERSEDES DOC DATE
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TOTAL SHEETS
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