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Overview

1 Introduction

2 Theoretical Background

3 Input Data

4 Calculation

5 Results

6 Result Evaluation

7 Printout

8 General Functions

9 Literature

2.7.10.8 Final cross-section properties

Final cross-section properties

The curvature for both crack states c (cracked/uncracked) is calculated as follows:

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

uncracked state:

$κ_{l, ϕ_{1}} = 1.158 \cdot \frac{30 \cdot 10^{3} - (- 100 \cdot 10^{3}) \cdot 1.4 \cdot 10^{- 3}}{11 \cdot 10^{9} \cdot 6.816 \cdot 10^{- 4}} = 4.655 \cdot 10^{- 3}$

cracked state:

$κ_{I I, ϕ_{1}} = 1.353 \cdot \frac{30 \cdot 10^{3} - (- 100 \cdot 10^{3}) \cdot (- 41.5 \cdot 10^{- 3})}{11 \cdot 10^{9} \cdot 2 . 193^{- 4}} = 14.499 \cdot 10^{- 3}$

The strain for both crack states is determined as follows:

$ε_{c, ϕ_{1}} = \frac{n_{ϕ_{1}}}{E \cdot A_{c, ϕ_{1}}}$

uncracked state:

$ε_{l, ϕ_{1}} = \frac{- 100 \cdot 10^{3}}{11 \cdot 10^{9} \cdot 0.206} = - 4.413 \cdot 10^{- 5}$

cracked state:

$ε_{I l, ϕ_{1}} = \frac{- 100 \cdot 10^{3}}{11 \cdot 10^{9} \cdot 0.087} = - 10.449 \cdot 10^{- 5}$

Thus, it is possible to determine the mean strain.

$ε_{ϕ_{1}} = ζ_{ϕ_{1}} \cdot ε_{I I, ϕ_{1}} + (1 - ζ_{ϕ_{1}}) \cdot ε_{I, ϕ_{1}} = = 0.835 \cdot (- 10.449 \cdot 10^{- 5}) + (1 - 0.835) \cdot (- 4.413 \cdot 10^{- 5}) = - 9.459 \cdot 10^{- 5}$

The mean curvature is determined as follows:

$κ_{ϕ_{1}} = ζ_{ϕ_{1}} \cdot κ_{I I, ϕ_{1}} + (1 - ζ_{ϕ_{1}}) \cdot κ_{I, ϕ_{1}} = = 0.835 \cdot (14.449 \cdot 10^{- 3}) + (1 - 0.835) \cdot 4.655 \cdot 10^{3} = 12.885 \cdot 10^{- 3}$

With the mean curvature and the longitudinal strain, you can calculate the final cross-section properties while taking account of shrinkage, creep, and tension stiffening.

Ideal cross-section area

$A_{ϕ_{1}} = \frac{n_{ϕ_{1}}}{E \cdot ε_{ϕ_{1}}} = \frac{- 100 \cdot 10^{3}}{11 \cdot 10^{9} \cdot (- 9.459 \cdot 10^{- 5})} = 958.59 {cm}^{2}$

Ideal moment of inertia to the ideal center of the cross-section

$I_{ϕ_{1}} = \frac{I_{I, ϕ_{1}} \cdot I_{I I, ϕ_{1}}}{ζ_{ϕ_{1}} \cdot I_{I, ϕ_{1}} \cdot k_{d h, I I, ϕ_{1}} + (1 - ζ_{ϕ_{1}}) \cdot I_{I I, ϕ_{1}} \cdot k_{d h, I, ϕ_{1}}} = = \frac{6.816 \cdot 10^{- 4} \cdot 2.193 \cdot 10^{- 4}}{0.836 \cdot 6.816 \cdot 10^{- 4} \cdot 1.353 + (1_{} -_{} 0.836) \cdot 2.193 \cdot 10^{- 4} \cdot 1.158} = 18 391.50 {cm}^{4}$

Eccentricity of centroid

$e_{ϕ_{1}} = \frac{m_{ϕ_{1}} - κ_{ϕ_{1}} \cdot E \cdot I_{ϕ_{1}}}{n_{ϕ_{1}}} = \frac{30 \cdot 10^{3} - 12.855 \cdot 10^{- 3} \cdot 11 \cdot 10^{9} \cdot 1.839 \cdot 10^{- 4}}{- 100 \cdot 10^{3}} =_{} - 39 mm$

Ideal moment of inertia to the geometric center of the cross-section

$I_{0, ϕ_{1}} = I_{ϕ_{1}} + A_{ϕ_{1}} \cdot {e_{ϕ_{1}}}^{2} = 1.839 \cdot 10^{- 4} + 0.096 \cdot {(- 0.0393)}^{2} = 33 207 {cm}^{4}$

Poisson's ratio is determined as follows:

$ν = (1 - \underset{d \in {1, 2}}{m a x} (ζ_{d})) \cdot ν_{i n i t} = (1 - m a x (0.0836)) \cdot 0.2 = 0.0328$

Image 2.124 Final cross-section properties

Parent Chapter

2.7.10 Example