Aci - Crack Calculations.pdf

Steel Properties (For steel reinforcement, ASTM A615M Grade 420 MPa) 5

fy ≔ 420 MPa = 60.916 ksi

Es ≔ 2 ⋅ 10 MPa

Concrete Properties 3 kN ⎛ kgf γconcrete ≔ 25 ―― = ⎝2.549 ⋅ 10 ⎞⎠ ―― 3 3 m m

f'c ≔ 32 ⋅ MPa

λ ≔ 1.0

7 Ec ≔ 4700 ⋅ ‾‾‾‾‾‾‾ f'c ⋅ MPa = ⎛⎝2.659 ⋅ 10 ⎞⎠ kPa

Es n ≔ ― = 7.522 Ec

fr ≔ 0.62 ⋅ λ ⋅ ‾‾‾‾‾‾‾ f'c ⋅ MPa = 3.507 MPa

: modulus of rupture of concrete (ACI 318­08, eq. 9­10)

Geometric Properties

b ≔ 1000 mm

h ≔ 900 mm

cc ≔ 75 mm

wlimit ≔ 0.1 mm

T

Dia ≔ [ 10 mm 12 mm 16 mm 20 mm 25 mm 32 mm ] T 2 2 2 2 2 2 Area ≔ ⎡⎣ 0.7854 cm 1.131 cm 2.011 cm 3.142 cm 4.909 cm 8.042 cm ⎤⎦

: Rebar Selec on D10=1, D12=2, D16=3, D20=4, D25=5, D32=6

Dn ≔ 6

s ≔ 150 mm

ds ≔ Dia

3

4 b⋅h Ig ≔ ―― = 0.061 m 12

Dn

= 32 mm

As ≔ Area

h yt ≔ ―= 0.45 m 2

Dn

3 2 b ⋅ ―= ⎛⎝5.361 ⋅ 10 ⎞⎠ mm s

d ≔ h − cc = 0.825 m d ≔ 809 mm

ds dc ≔ cc + ―= 91 mm 2

Cracking Moment and Moment at Service Loads

Msrv ≔ 250 kN ⋅ m

fr ⋅ Ig Mcr ≔ ――= 473.479 kN ⋅ m yt

: cracking moment (ACI 318­08, eq. 9­9)

Calculation of Tensile Stress in Reinforcement at Service Loads

‾‾‾‾‾‾‾‾‾ 2⋅B⋅d+1 −1 c ≔ ―――――― = 218.283 mm B

−1 b B ≔ ――= 0.025 mm n ⋅ As

: neutral axis depth

3

2 10 4 b⋅c Icr ≔ ――+ n ⋅ As ⋅ (d − c)) = ⎛⎝1.754 ⋅ 10 ⎞⎠ mm 3

Msrv fs ≔ ――――= 63.336 MPa ⎛ c⎞ As ⋅ d − ― ⎜⎝ 3 ⎟⎠

: moment of iner a of cracked sec on transformed to concrete

Msrv fs ≔ ――n ⋅ (d − c)) = 63.336 MPa Icr

Msrv fc ≔ ――c = 3.111 MPa Icr

: steel stress

: concrete stress

­ verificaton ­ T ≔ As ⋅ fs = 339.564 kN

C ≔ 0.5 ⋅ b ⋅ c ⋅ fc = 339.564 kN

⎛ c⎞ M ≔ T ⋅ d − ― = 250 kN ⋅ m ⎜⎝ 3 ⎟⎠

Msrv = 250 kN ⋅ m

: internal couple T ＝ C

: internal couple mement M ＝ Msrv

­ if there is a compression axial force, the tensile stress minus compress can be applied to calcua lte the crack width.

P ≔ 0 kN

Ac ≔ b ⋅ h − As

P ⋅ Es σs ≔ ―――――= 0 MPa Es ⋅ As + Ec ⋅ Ac

4 fs ≔ fs − σs = ⎛⎝6.334 ⋅ 10 ⎞⎠ kPa

P σs ≔ ――――= 0 MPa 1 As + ―⋅ Ac n : steel stress considering compression

1. Calculation of Crack Width (ACI) Step 1. Calculation of Spacing and Crack Width (ACI 318-08) 3 280 smax ≔ 380 ⋅ ―― ⋅ MPa ⋅ mm − 2.5 ⋅ cc = ⎛⎝1.492 ⋅ 10 ⎞⎠ mm fs

: maxium spacing (ACI 318­08, eq. 10­4)

Judge ≔ if ⎛⎝smax > s , “OK” , “NG”⎞⎠ = “OK”

Step 2. Comparison of Cracking Moment and Moment at Service Loads

Mcr = 473.479 kN ⋅ m

Judge ≔ if ⎛⎝Mcr > Msrv , “OK” , “NG”⎞⎠ = “OK”

Msrv = 250 kN ⋅ m

Step 3. Calculation of Crack Width (ACI 224.1R) h−c β ≔ ――= 1.154 dc = 91 mm d−c 2 ‾‾‾‾‾‾‾‾‾ fs 2 ⎛s⎞ w ≔ 2 ⋅ ― ⋅ β ⋅ dc + ― = 0.086 mm ⎜⎝ 2 ⎟⎠ Es

4 2 A ≔ 2 dc ⋅ s = ⎛⎝2.73 ⋅ 10 ⎞⎠ mm

: crack width (ACI 224.1R­07, eq. 1­1)

w ≔ ‖ if cc > 50 mm = 0.073 mm ‖ ‖ c ← 50 mm ‖ c ‖ ‖ ‖ dc ← cc + 0.5 ⋅ ds ‖ 2 ‾‾‾‾‾‾‾‾‾ fs ‖ 2 ⎛s⎞ ⋅ β ⋅ dc + ― ‖w←2⋅― ⎜ ⎟ ⎝2⎠ Es ‖ w ‖

Judge ≔ ‖ if Msrv ≤ Mcr ‖ ‖ ‖ ‖ “Not cracked” ‖ else if w ≤ wlimit ‖ ‖ ‖ ‖ “Lower than limit” ‖ else ‖ ‖ “NG” ‖ ‖

= “Not cracked”

: crack width calculated with reduced cover depth, 50 mm (ACI 350­06, Clause 10.6.4)