Flexural Crack width Calculation for Rectangular RC
Slab
According ACI350-06 &ACI224R-01, Rev0
Project Building
Project :Building :Element:Location:-
Designed by:- M. Abu Shady Checked by:- M. Abu Shady 15-Nov-13 Date:-
Element Location
M.A.S.
General Input :Width b
1000
mm
Depth h
300
mm
cover
50
fy
420
mm N/mm2
fc'
32
N/mm2
Ma service
2.9
KN.m
Section RFT. Top Rft.
No of bars
N/mm2
Es εc
200000 0.003
mm
εy
0.0021
mm
Ec
26587
N/mm2
d
244
mm
d'
56
mm
n= Es /Ec =
7.52
ACI318 clause 8.5.1
AS mm2
rebar Dia 8 T 12 Top
S spacing Top = 125 mm S spacing Bot. = 125 mm
= 912 mm2
Bot. Rft.
8 T 12 Bot. Exposure condition Water-retaining structures
= 912 mm2
ACI224R-01,Table 4.1
1- Locate the Neutral axis & Calculate setcion properties:X-X
εc= 0.000019
fc
εs'
d'= 56 As Top
(n-1)As Top N.A.
N.A. h= 300
Cc = 0.5fc*y*b
y= 51.83 N.A.
d= 244 As Bot.
nAs Bot.
A Tranformed Section
εs= 0.000070 Strain
T= fs * As Stress
b= 1000 Taking moment of areas @ X-X, [by+nAs Bot.+ (n-1)As Top]*y = by*y/2 + nAs Bot.*d + (n-1)As Top*d' and simplify in form of Quadratic equation (ay2+by+c=0), (b-b/2) * y2 + (nAs Bot.+ (n-1)As Top) * y + (-nAs Bot.*d - (n-1)As Top*d') = 0 Thus, a =
500.00 b= 12808.88 c = -2007060.08
Ig = bh3/12
Ag = b*h
= 2,250,000,000 mm4
Acr =by+ nAs Bot.+ (n-1)As Top
= 51.83 mm
Thus,
= 300,000 mm2
= 64,639 mm2
Icr =by3/12+by*(y/2)2+ nAs Bot.*(d-y)2+ (n-1)As Top*(y-d')2
= 299,865,909 mm4
2- Check of Concrete & Steel Reinforcement stresses:-
fc a = Ma*y/Icr
= 3.51
N/mm2
ACI 318M-11, Eq 9-10
= 52.61
KN.m
ACI 318M-11, Eq 9-9
= 0.50
N/mm2
= 0.016 fc'
2
Ok < f s max = 13.98 = 250.00 N/mm N/mm2 , ACI350-06, Eq 10-4 & 10-5 = 0.033 fy Ok < f s max= 320N/mm2 allowed for rebar Dia 12 mm used ,According ACI224R-04, Table4.2 Ok < f s max= 240N/mm2 allowed for spacing 125 mm used ,According ACI224R-04, Table4.3
fs a=Ma(y-d)/(Icr/n)
3- Calculation of crack width := 0.018 mm
Crack width,
ACI224R-01, Eq 4-2a
ok Crack Width <= wcr all= 0.100 mm, ACI224R-01,Table 4.1 Where, dc = 56.000 mm
ᵝ
= 1.291 mm
A = 14,000 mm2
ACI224R-01, 4.2.1 ACI224R-01, 4.2.1 & ACI350-06, Eq 10-6 ACI224R-01, 4.2.2
Page 1 of 5
Rebar Area DIA (mm) 6 8 10 12 14 16 18
Area (mm2) 29 51 79 114 154 201 255
20
314
22 25 32
380 491 804
Element Type Beam Slab
ACI224R-01,Table 4.1, reasonable* crack widths, reinforced concrete under service loads Exposure condition Dry air or protective membrane Humidity, moist air, soil Deicing chemicals Seawater and seawater spray, wetting and drying Water-retaining structures
Crack width mm
0.41 0.3 0.18 0.15 0.1
Table 4.2—Maximum bar diameter for high bond bars bar size, mm 32 25 20 16 12 10 8 6 Table 4.3—Maximum bar spacing for high bond bars Maximum bar spacing, mm Pure flexure 300 250 200 150 100 50
Steel stress, MPa Maximum 160 200 240 280 320 360 400 450
Pure tension 200 150 125 75 — —
Steel stress, Mpa 160 200 240 280 320 360
References 1-ACI350-06 &ACI224R-01, 2- Design of Reinforced Concrete by Jack C. McCormac_9th Ed-2013-1118129849_ACI 318-11.pdf pages 43 (Example 2.3), 6.10 Control of Flexural Cracks and excel sheets for chapter 6 http://www.mediafire.com/download/k42shb458odynac/Sheets_used_in_Design_of_Reinforced_Concrete_by_Jack_C._McC 3- RCM ACI Builder v 5.3.0.0 by Eng. Hussein Rida http://www.structural-experts.com/showthread.php?tid=742 4- Moment of Inertia of Cracked Section Transformed to Concrete http://www.concrete.org/technical/ckc/314/314_design_aid_J_1-14_001.pdf 5-Quadratic eq http://www.sosmath.com/algebra/quadraticeq/quadraformula/quadraformula.html
6- Equivalent Stiffness , Is =Ic/n http://www.bgstructuralengineering.com/BGSMA/BGSMA_Itr/BGSMA_ITR02.htm for axial force, εc = εc fc/Ec = fs/Es ==> F/(Ac Ec) = F/(As Es) ==> thus As Es = Ac Ec ==> As = Ac * Ec/Es = Ac /n and for moment, εc = εc fc/Ec = fs/Es ==> M*y/(Is Es) = M*y/(Ic Ec) == > Is = Ic Ec/Es = Ic/n
ncrete_by_Jack_C._McCormac_9th_Ed-2013-1118129849_ACI_318-11.rar
s = Ac /n
