Structural Concrete - Hassoun

64 Chapter 2 Properties of Reinforced Concrete
Solution
Shrinkage Calculation
Calculation of ε cs0 :
β sc = 8
ε s (f cm28
)=
=
ε s (t, t c )=(ε cs0 )β s (t, t c )
[
160 + 10 ( ) ( β sc 9 − f cm 28
10
[
(
160 + 10 (8)
9 − 45.2
10
For H = 75%,
β RH =−1.55β arh
( ) H 3 ( ) 75 3
β arh = 1 − = 1 − = 0.578
100 100
β RH =−1.55β arh =−1.55 × 0.578 =−0.896
ε cs0
Calculation of β s (t − t c ):
)]
× 10 −6
)]
× 10 −6 = 518.4 × 10 −6 mm∕mm
= ε s (f cm28
)(β RH )=(518.4 × 10 −6 )(−0.896) =−464.5 × 10 −6 mm∕mm
h e = 2A c
= 76 mm
u
√
√
t − t
β s (t − t c )=
c
0.56(h e ∕4) 2 +(t − t c ) = 35 − 8
0.56(76∕4) 2 +(35 − 8) = 0.343
ε s (t, t c )=(ε cs0 )β s (t − t c )=(−464.5 × 10 −6 )(0.343) =−159.3 × 10 −6 mm∕mm
Creep Calculation
Calculation of E cmt0 and E cm28
:
J(t, t 0 )= 1
E cmt0
+ φ 28 (t, t 0 )
E cm28
t 0 = 28 days ⇒ E cmt0 = E cm28
√
√
f
E cm28
= 21,500 3 cm28
45.2
10 = 21,500 3 = 35,548 MPa
10
Calculation of φ(t, t 0 ):
φ RH = 1 +
1 − √ H∕100
0.46 3 he ∕100 = 1 + 1 − √ 75∕100 = 1.596
0.46 3 76∕100
5.3 5.3
β(f cm28
)= √ = √ = 2.49
f cm28
∕10 45.2∕10
β(t 0 )=
1
1
=
= 0.488
0.1 + t 0.02
0.2
0
0.1 + 28