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Overview

1 Introduction

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 ε_sh. 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:

$ε_{s h} = - 0.5 \cdot 10^{- 3}$

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

$n_{s h, ϕ_{1}} =_{} - E_{s} \cdot ε_{s h} \cdot (a_{s_{1}, ϕ_{1}} +_{} a_{s_{2}, ϕ_{1}}) =_{} - 200 \cdot 10^{9} \cdot (- 0 . 5_{} +_{} 10^{- 3}) \cdot (1000_{} +_{} 15) \cdot {10^{- 6}}_{} = = 101.5 kN/m$

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

$e_{s h, c, ϕ_{1}} = \frac{a_{s_{1}, ϕ_{1}} \cdot d_{1, ϕ_{1}} + a_{s_{2}, ϕ_{1}} \cdot d_{2, ϕ_{1}}}{a_{s_{1}, ϕ_{1}} + a_{s_{2}, ϕ_{1}}} - z_{c, ϕ_{1}}$

uncracked state:

$e_{s h, c, ϕ_{1}} = \frac{1000 \cdot 150 + 15 \cdot 50}{1000 + 15} - 101.4 = 47.1 mm$

cracked state:

$e_{s h, c, ϕ_{1}} = \frac{1000 \cdot 150 + 15 \cdot 50}{1000 + 15} - 58.5 = 90.0 mm$

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

$m_{s h, c, ϕ_{1}} = n_{s h, ϕ_{1}} \cdot e_{s h, c, ϕ_{1}}$

uncracked state:

$m_{s h, l, ϕ_{1}} = 101.5 \cdot 10^{3} \cdot 0.047 = 4.8 kNm/m$

cracked state:

$m_{s h, l I, ϕ_{1}} = 101.5 \cdot 10^{3} \cdot 0.090 = 9.1 kNm/m$

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

for m_φ₁ ≠ 0:

$k_{s h, c, ϕ_{1}} = \frac{m_{s h, c, ϕ_{1}} + m_{ϕ_{1}} - n_{ϕ_{1}} \cdot e_{c, ϕ_{1}}}{m_{ϕ_{1}} - n_{ϕ_{1}} \cdot e_{c, ϕ_{1}}}$

for m_φ₁ = 0:

$k_{s h, c, ϕ_{1}} = 1 where k_{s h, c, ϕ_{1}} \in \{1, 100\}$

In this example: m_φ₁ ≠ 0

uncracked state:

$k_{s h, l, ϕ_{1}} = \frac{4.771 \cdot 10^{3} + 30 \cdot 10^{3} - (- 100 \cdot 10^{3}) \cdot 1.4 \cdot 10^{3}}{30 \cdot 10^{3} - (- 100 \cdot 10^{3}) \cdot 1.4 \cdot 10^{3}} = 1.159$

cracked state:

$k_{s h, I l, ϕ_{1}} = \frac{9.135 \cdot 10^{3} + 30 \cdot 10^{3} - (- 100 \cdot 10^{3}) \cdot (- 41.5 \cdot 10^{3})}{30 \cdot 10^{3} - (- 100 \cdot 10^{3}) \cdot (- 41.5 \cdot 10^{3})} = 1.354$

Parent section

2.7.10 Example