Structural Concrete - Hassoun

2.13 Models for Predicting Shrinkage and Creep of Concrete 49
Creep Calculation
Determination of E cmt0 :
Determination of C c (t):
C cu = 2.35
J(t, t 0 )= 1 + C c(t)
E cmt0
a = 2.30 b = 0.92 (Table 2.5)
f c ′ (t 0 )=f t 0
28
cm 28
= 6556
= 6542 psi
a + bt 0 2.3 + 0.92 × 28
E cmt0 = 33(γ) 3∕2 √f ′ c (t 0 )=33(146) 3∕2√ 6542 = 4,708,673 psi
K ch = 1.27 − 0.0067(H) =1.27 − 0.0067(75) =0.768
K ca = 1.25(t 0 ) −0.118 = 1.25(28) −0.118 = 0.844
( ) V
K cs = 1.14 − 0.092 = 1.14 − 0.138 = 1.002
S
C c (t) =
(t − t 0 ) 0.60
10 +(t − t 0 ) C 0.60 cu K ch K ca K cs = (35 − 28) 0.60
2.35 × 0.768 × 0.844 × 1.002 = 0.37
10 +(35 − 28)
0.60
J(t, t 0 )= 1 + C c (t)
E cmt0
= 1 + 0.37
4,708,673 = 0.29 × 10−6 psi −1
Example 2.2 (in.−lb Units)
Using the B3 model, calculate the shrinkage strain and creep function for the specimen given in Example
2.1.
Solution
Shrinkage Calculation
Determination of ε shu :
α 1 = 1.10 (Table 2.8)
α 2 = 1.0 (Table 2.9)
ε s (t) =(ε shu )(K h )S(t)
ε shu =−ε su
E cm607
E cm(tc +τ sh )
ε su =−α 1 α 2 [0.025(w) 2.1 (f cm28
) −0.28 + 270]×10 −6
=−(1.10)(1.0)[0.025(345) 2.1 (6556) −0.28 + 270]×10 −6 =−798 × 10 −6 in.∕in.
E cm28 = 57000 √ √
f cm28 = 57,000 6556 = 4,615,240 psi
k s = 1.0