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Why is the crack moment M cr smaller for a double bend than for a single bend?

Answer

The crack moment of a concrete cross-section is calculated from the mean tensile strength of the concrete and the ideal section modulus. The crack moment describes the internal force that occurs when the tension stress f ctm is reached in the outermost fiber of the cross-section and crack formation occurs.

For simple bending, it is possible to calculate the crack moment analytically. For double bending, the introduction of a weighting factor k is helpful in order to determine from the components M cr, y, and M cr, z M cr .

Calculation for the attached example:

Bending moment M y = 20 kNm
Bending moment M z = 20 kNm

Ultimate modulus of resistivity W y = 3081 cm 3
Ideal section modulus W z = 3081 cm 3

Mean tensile strength of concrete f ctm = 0.290 kN / cm 2

Member 1: Single bending M y :

$\begin{array}{l}M_{cr\;}=f_{ctm}\times W_y\\M_{cr\;}=0,29\;\frac{kN}{cm^2}\times3081\;cm^3\\M_{cr\;}=893\;kNcm\;=\;8,9\;kNm\end{array}$

Member 2: Simple bending M z :

$\begin{array}{l}M_{cr\;}=f_{ctm}\times W_z\\M_{cr\;}=0,29\;\frac{kN}{cm^2}\times3081\;cm^3\\M_{cr\;}=893\;kNcm\;=\;8,9\;kNm\end{array}$

Member 3: Double bending M y and M z :

$\begin{array}{l}M_{cr\;}=\sqrt{M_{cr,y}^2+M_{cr,z}^2}\\M_{cr,y\;}=k\times My\\k=\frac{f_{ctm}}{\sigma_M}\\\sigma_M=\frac{M_y}{W_y}+\frac{M_z}{W_z}=\\\end{array}$


Keywords

Crack moment Mean tensile strength concrete

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