Exponential Traction Law
The Traction Law
This traction law assumes a linear elastic response up to σc followed by a scaled exponential decrease to reach zero at critical crack opening displacement (COD) of δc. The exponential decay in the post peak region (i.e., δ>δe) is given by
[math]\displaystyle{ S(\delta) = \sigma_c\frac{e^{-\alpha\frac{\delta-\delta_e}{\delta_c-\delta_e}}-e^{-\alpha}}{1-e^{-\alpha}} }[/math]
This cohesive law defines traction as a function of crack opening displacement (COD) during uniaxial, monotonic loading. There are separate and uncoupled cohesive laws for opening displacement (mode I) and sliding displacement (mode II).
Notice how the shape changes with α. In the limit of α=0, this law is identical to the Triangular Traction Law (and that one should be used instead because it is more efficient. Increasing α causes traction to drop faster, which might simulate increasingly brittle response. The toughness of this cohesive law is the area under the curve that can be cast as
[math]\displaystyle{ J_c = {1\over 2} \sigma_c\delta_c\left(\delta_{peak} + \bigl(1-\delta_{peak}\bigr)f(\alpha)\right) \quad{\rm where}\quad f(\alpha) = 2\left(\frac{1}{\alpha}-\frac{e^{-\alpha}}{1-e^{-\alpha}}\right) }[/math]
The function f(α) is 1 when α=0 and decays to zero as α increases. The toughness is thus bracketed by
[math]\displaystyle{ {1\over 2} \sigma_c\delta_e \le J_c \le {1\over 2} \sigma_c\delta_c }[/math]
The minimum corresponds large α with an instant drop to zero traction at δe while the maximum corresponds to α=0 and is area under the corresponding Triangular Traction Law.
When creating this traction law, you have to enter α, exactly two of σc, δp, and k and either Jc or δc. These four required properties must be entered for both mode I and mode II. If you enter δc, then the missing one of σc, δp, and k are calculated from one of
[math]\displaystyle{ k = {\sigma_c\over \delta_{peak}\delta_c} \qquad {\rm or} \qquad \delta_{peak} = {\sigma_c\over k\delta_c} \qquad {\rm or} \qquad \sigma_c = k \delta_{peak}\delta_c }[/math]
Finally Jc is found from the expression above. If instead you enter Jc, the calculation of δc depends on which two of σc, δp, and k are provided:
[math]\displaystyle{ {\rm from\ }\sigma_c\ {\rm and}\ \delta_p:\quad \delta_c = \frac{2J_c}{\sigma_c\left(\delta_{peak} + \bigl(1-\delta_{peak}\bigr)f(\alpha)\right)} }[/math]