Structural Concrete - Hassoun

2.13 Models for Predicting Shrinkage and Creep of Concrete 61
E cm607 =(1.167) 1 ∕ 2
E cm28
=(1.167) 1 ∕ 2
(31833.9) =34389.5 MPa
(
) 1∕ 2 (
)
E cm(tc +τ sh ) = t c + τ
1∕ 2
sh
4 + 0.85 ( 8 + 160.3
) E cm28 = (31833.9)
t c + τ sh
4 + 0.85 (8 + 160.3)
= 34055.9 MPa
ε shu =−ε su
E cm607
E cm(tc +τ sh )
= −(−827 × 10 −6 ) 34389.5
34055.9 = 835.1 × 10−6 mm∕mm
Determination of K h :
According to the Table 2.10, for H = 75%
( ) H 3 ( ) 75 3
K h = 1 − = 1 − = 0.578
100 100
Determination of S(t):
√
√
t − t
S(t) =tanh c 35 − 8
= tanh
T sh
160.3 = 0.389
ε s (t) =(ε shu )(K h )S(t) =(835.1 × 10 −6 )(0.578)(0.389) =187.8 × 10 −6 mm∕mm
Creep Calculation:
Determination of q 1 :
Calculation of C 0 (t,t 0 ):
c =
J(t, t 0 )=q 1 + C 0 (t, t 0 )+C d (t, t 0 , t c )
q 1 = 0.6 0.6
=
E cm28
31833.9 = 18.85 × 10−6 MPa −1
w
w∕c = 207.92 = 452 kg∕m3
0.46
q 2 = 185.4(c) 0.5 (f cm28
) −0.9 × 10 −6 = 185.4(452) 0.5 (45.2) −0.9 × 10 −6
Q f (t 0 )=
= 127.7 × 10 −6
1
0.086(t 0 ) 2∕9 + 1.21(t 0 ) = 1
= 0.182
4∕9 0.086(28) 2∕9 + 1.21(28)
4∕9
Z(t, t 0 )= ln[1 +(t − t 0 )0.1 ]
√
t0
= ln[1 +(35 − 28)0.1 ]
√
28
= 0.150
r(t 0 )=1.7(t 0 ) 0.12 + 8 = 1.7(28) 0.12 + 8 = 10.54
⎡
Q(t, t 0 )=Q f (t 0 ) ⎢
⎢
1 + Q ( ) −1∕r(t r(t0 )
f t0
⎤
0 )
[
] −1∕10.54
⎥ = 0.182 1 + 0.18210.54
= 0.148
Z(t, t
⎣
0 ) r(t 0) ⎥
⎦
0.150 10.54
( ) w 4
q 3 = 0.29q 2 = 0.29(127.7 × 10 −6 )(0.46) 4 = 1.66 × 10 −6
c