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2 Theoretical Background

2.7.10.6 Shrinkage influence

Shrinkage influence

The influence of shrinkage is directly introduced in the calculation with the defined value of the free shrinkage εsh. Thus, the influence of structural restraints or redistributions of the shrinkage forces is not taken into account.

In our example, the shrinkage strain is applied with the following value:

εsh = -0.5 · 10-3 

The free shrinkage strain causes additional forces in the cross-section:

nsh,ϕ1 = -Es · εsh · as1,ϕ1+ as2,ϕ1 = -200 · 109 · -0.5 + 10-3 · 1000 + 15 · 10-6 =            = 101.5 kN/m 

The forces act for both crack states c (cracked and uncracked) with the eccentricity to the centroid of the ideal cross-section:

esh,c,ϕ1 = as1,ϕ1 · d1,ϕ1 + as2,ϕ1 · d2,ϕ1as1,ϕ1 + as2,ϕ1 - zc,ϕ1 

  • uncracked state:

esh,c,ϕ1 = 1000 · 150 + 15 ·501000 + 15 - 101.4 = 47.1 mm 

  • cracked state:

esh,c,ϕ1 = 1000 · 150 + 15 ·501000 + 15 -58.5 = 90.0 mm 

The bending moment caused by the axial force nsh,φ1 for both crack states c is:

msh,c,ϕ1 = nsh,ϕ1 · esh,c,ϕ1 

  • uncracked state:

msh,l,ϕ1 = 101.5 · 103 · 0.047 = 4.8 kNm/m 

  • cracked state:

msh,lI,ϕ1 = 101.5 · 103 · 0.090 = 9.1 kNm/m 

When determining the coefficient ksh,c,d for both crack states c, we have to distinguish:

  • for mφ1 ≠ 0:

ksh,c,ϕ1 = msh,c,ϕ1 + mϕ1 - nϕ1 · ec,ϕ1mϕ1 - nϕ1· ec,ϕ1 

  • for mφ1 = 0:

ksh,c,ϕ1 = 1        where ksh,c,ϕ1  1,100

In this example: mφ1 ≠ 0

  • uncracked state:

ksh,l,ϕ1 = 4.771 · 103 + 30 · 103 - -100 · 103 · 1.4 · 10330 · 103 -  -100 · 103 · 1.4 · 103 = 1.159 

  • cracked state:

ksh,Il,ϕ1 = 9.135 · 103 + 30 · 103 - -100 · 103 · -41.5 · 10330 · 103 - -100 · 103 · -41.5 · 103 = 1.354 

Image 2.122 Shrinkage influence
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