Structural Concrete - Hassoun

2.13 Models for Predicting Shrinkage and Creep of Concrete 65
φ 0 = φ RH β(f cm28
)β(t 0 )=(1.596)(2.49)(0.488) =1.939
β H = 1.5h e [1 +(0.012H) 18 ]+250 = 1.5(76)[1 +(0.012 × 75) 18 ]+250
= 381 ≤ 1500 days
( ) t − 0.3 t0
(
)
35 − 28 0.3
β c (t, t 0 )=
=
= 0.3
β H + t − t 0 381 + 35 − 28
φ(t, t 0 )=φ 0 β c (t, t 0 )=1.939 × 0.3 = 0.582
J(t, t 0 )= 1 + φ(t, t 0 ) 1
=
E cmt0 E cm28
35,548 + 0.582
35,548 = 44.5 × 10−6 MPa −1
Example 2.12 (SI Units)
Use the CEB MC 90–99 model to calculate the shrinkage strain and creep function for the specimen
given in Example 2.8.
Solution
Shrinkage Calculation
ε s (t, t c )=ε as (t)+ε ds (t, t c )
Calculation of ε as (t):
α as = 600 for rapidly hardening high − strength cements
(
) 2.5
fcm28 ∕10
ε as0 (f cm28
)=−α as × 10 −6
6 + f cm28
∕10
( ) 2.5 45.2∕10
=−600
× 10 −6 =−72.6 × 10 −6 mm∕mm
6 + 45.2∕10
β as (t) =1 − exp(−0.2(t) 0.5 )=1 − exp(−0.2(35) 0.5 )=0.694
ε as (t) =ε as0 (f cm28
)β as (t) =(−72.6 × 10 −6 )(0.694) =−50.4 × 10 −6 mm∕mm
Calculation of ε ds (t, t c ):
α ds1 = 6 for rapidly hardening high − strength cements
α ds2 = 0.12 for rapidly hardening high − strength cements
ε ds0 (f cm28
)=[(220 + 110α ds1 )exp(−α ds2 f cm28
∕10)] × 10 −6
=[(220 + 110 × 6)exp(−0.12 × 45.2∕10)] × 10 −6 = 511.6 × 10 −6 mm∕mm
( ) 0.1
( )
35 35 0.1
β s1 =
= = 0.97 ≤ 1.0
f cm28
45.2
For 40% < H = 75% < 99% (0.97) = 96.5%,
[ ( ) H 3
] [ ( ) 75 3
]
β RH =−1.55 1 − =−1.55 1 − =−0.896
100
100