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Structural Concrete - Hassoun

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60 Chapter 2 Properties of Reinforced <strong>Concrete</strong><br />

Determination of E cmt0 :<br />

Determination of C c (t):<br />

C cu = 2.35<br />

a = 2.30 b = 0.92 (Table 2.5)<br />

f c ′(t 0 )=f t 0<br />

28<br />

cm 28<br />

= 45.2<br />

= 45.1 MPa<br />

a + bt 0 2.3 + 0.92 × 28<br />

E cmt0 = 0.043(γ) 3∕2 √f ′ c (t 0 )=0.043(2405) 3∕2√ 45.1 = 34058.8MPa<br />

K ch = 1.27 − 0.0067(H) =1.27 − 0.0067(75) =0.768<br />

K ca = 1.25(t 0 ) −0.118 = 1.25(28) −0.118 = 0.844<br />

( ) V<br />

K cs = 1.14 − 0.0035 = 1.14 − 0.0035(38) =1.007<br />

S<br />

C c (t) =<br />

(t − t 0 ) 0.60<br />

10 +(t − t 0 ) C 0.60 cu K ch K ca K cs = (35 − 28) 0.60<br />

2.35 × 0.768 × 0.844 × 1.00 = 0.37<br />

10 +(35 − 28) 0.60<br />

J(t, t 0 )= 1 + C c (t)<br />

E cmt0<br />

= 1 + 0.37<br />

34058.8 = 40.2 × 10−6 MPa −1<br />

Example 2.9 (SI Units)<br />

Using the B3 model, calculate the shrinkage strain and creep function for the specimen given in<br />

Example 2.8.<br />

Solution<br />

Shrinkage Calculation<br />

ε s (t) =(ε shu )(K h )S(t)<br />

Determination of ε shu :<br />

α 1 = 1.10 (Table 2.8)<br />

α 2 = 1.0 (Table 2.9)<br />

ε shu =−ε su<br />

E cm607<br />

E cm(tc +τ sh )<br />

ε su =−α 1 α 2 [0.019(w) 2.1 (f cm28<br />

) −0.28 + 270]×10 −6<br />

=−(1.10)(1.0)[0.019(207.92) 2.1 (45.2) −0.28 + 270]×10 −6 =−827 × 10 −6 mm∕mm<br />

E cm28 = 4735 √ √<br />

f cm28 = 4735 45.2 = 31833.9MPa<br />

k s = 1.0<br />

(Since the type of member is not defined)<br />

[ ( )]<br />

T sh = 0.085(t c ) −0.08 (f cm28 ) −0.25 V 2<br />

2k s<br />

S<br />

T sh = 0.085(8) −0.08 (45.2) −0.25 [2(1)(38)] 2<br />

= 160.3days

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