Crack Paths 2009

mortar matrix is

17 .7 — 3.5 0

(19)

[am]: —3.5 17.7 0 (GPa)

0

0 7.1

I Thepredicted tensile strength ofmortarmatrix

When vm :38%, vm :1_vm :62%, the strain of mortar matrix is computed from Eq. (11):

-s

_ 422.7><10i6 _

(20)

{ e m } _ —96.1>

0

Accordingto Eq. (4), the m a x i m utemnsile stress of mortar matrix is

cs,

17.7 - 3.5 0

422 .7

7.8

(21)

{cm}: 6,. =[q,,]{em}= —3.5 17.7 0 > < 1 0 " - - % . 1 X 1 0=7 6— 3 2 (MPa)

1 x,

0

0

7.1

0

0

Therefore, the mortar matrix should be designed with the following parameters:

E m: l 7 G P ,Lvlm : 0 . 2 76 1 9 m I

-

2.2.2 Step 11 * Calculations on the primary material parameters of hardened cement

paste

I The equivalent stificness offine aggregate

W h e nEfa :78GPa,vfa I ()_1, the equivalent stiffness of fine aggregate can be computed

according to Eq. (8) and Eq. (9).

78 .8 — 7.9

0

(22)

[q,,]: —7.9 78.8

0 (GPa)

0

0

35 .5

I The m a x i m usmtrain offine aggregate

W h e n oLfi, : 10MPa, the strain of fine aggregate can be computedfrom Eq. (6).

0.1

0

78

78

10 ><106

128.2 >< 10 *6

(23)

,

0.1

1

,

,

{e,,}:[q,,] ‘ { c m } :— XE 0 ><10 9~

0

= -12.8><10 6

0

0 0.03

0

0

I The predicted material parameters of hardened cementpaste

W h e nvfa I 42%,“ :1_vfa : 58%, the strain of hardened cement paste is computed from

Eq. (13), (20) and (23);

{a }: _6f56~‘_’4:1f0:

.

(24)

0

Equation (12) can be rewritten:

v, [as ] {8C } = {o m } — vf, [qfa ] {8 fa}

(25)

Based on Eqs (2), (4), (6), (7) and (12), the elastic modulus of the hardened cement

1138