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)