# Viscosity of Clayey Soils: Experimental Studies
## 3 Results
### Effect of Shear Rate on Ultimate Shear Resistance of Clay Soil
The total cohesion (\(c\)), in turn, significantly depends on the shear rate (\(\dot{\gamma}\)) and as the shear rate increases, the ultimate shear resistance of clay soil increases proportionally to (\(ln\dot{\gamma}\)) (Figure 9). Figure 8: Graph of the dependence of the peak and residual strength of Quaternary clay with undisturbed structure and disturbed structure on the vertical stress (\(\sigma_{n_{1}}=0.1\) MPa; \(\sigma_{n_{2}}=0.2\) MPa and \(\sigma_{n_{3}}=0.3\) MPa), acting on the sample at different shear rates: 1—\(\dot{\gamma}_{1}=0.066\) min\({}^{-1}\); 2—\(\dot{\gamma}_{2}=0.0041\) min\({}^{-1}\); 3—\(\dot{\gamma}_{3}=0.00026\) min\({}^{-1}\); 4—\(\dot{\gamma}_{4}=0.00002\) min\({}^{-1}\)[33]. Figure 9: Graph of dependence of peak strength of disturbed structure Quaternary clay on the deformation rate [33]. Since, according to the experimental results, it was obtained that the structural cohesion (\(c_{c}\)) and the angle of internal friction (\(\varphi\)) are independent of the shear rate (\(\dot{\gamma}\)), we present the viscous resistance (\(c_{\dot{\gamma}}\)) as the following relationship:  
\[c_{\dot{\gamma}}=c_{\dot{\gamma}_{0}}+\alpha ln(\dot{\gamma}/\dot{\gamma}_{1}), \tag{4}\]  
where \(\alpha\) is the experimental parameter. Based on the analysis of the experimental research results (performed on clay soil samples) and also on the rheological model of a Maxwell viscoelastic body, we present the rheological equation for kinematic shear as applied to a soil medium in the following form:  
\[\dot{\gamma}=\frac{\tau-\tau^{*}}{\eta}+\frac{\dot{\tau}}{G}, \tag{5}\]  
where \(\tau^{*}=\sigma tg\varphi+c_{0}\) (with \(c_{0}=c_{c}+c_{w}\)) [kPa]; \(\eta\) is the overall viscosity of the soil [kPa-s]; \(\dot{\tau}\) is the rate of change in shear stress [kPa/s]; and G is the shear modulus of the soil [kPa]. By performing a certain grouping in Equation (5), we obtain a differential equation of the following form:  
\[\dot{\tau}+\tau\cdot\frac{G}{\eta}=\frac{G}{\eta}\cdot\left(c_{\dot{\gamma}}+ \tau^{*}\right), \tag{6}\]  
The general solution of Equation (6) is known [34] and is as follows:  
\[\tau=e^{-\int\frac{G}{\eta}dt}\cdot\left\{\int\frac{G}{\eta}\cdot\left(c_{ \dot{\gamma}}+\tau^{*}\right)\cdot e^{\int\frac{G}{\eta}dt}dt+C\right\}, \tag{7}\]  
Figure 3:0  
We will determine the constant of integration \(C\) from the initial Condition (8) at \(t=0\):  
Figure 3:1  
where \(\tau^{*}=\sigma tg\varphi+c_{0}\) [kPa]. By substituting (9) into (8), we obtain the solution of the equation for constant shear rate (\(\dot{\gamma}=const\)):  
Figure 3:2  
when \(t\rightarrow\infty\)\(\tau(\infty)=\sigma tg\varphi+c_{c}+c_{w}+c_{\dot{\gamma}}\), with \(c_{\dot{\gamma}}=\dot{\gamma}\eta\). The obtained result corresponds to our understanding of the ultimate shear resistance (3). However, in this case, the viscous resistance (\(c_{\dot{\gamma}}\)) increases in direct proportion to the shear rate \(\left(\dot{\gamma}\right)\), which does not correspond to the experimental results (4). This dependence can be obtained if the following relationship is taken as a rheological model:  
Figure 3:3  
where \(\tau^{*}=\sigma tg\varphi+c_{c}+c_{w}\) [kPa]. Equation (5) can also be represented in the following form:  
Figure 3:4  
The solution of this equation is known [34] and is given by:  
Figure 3:5  
when \(t\rightarrow\infty\)\(\tau(\infty)=\sigma tg\varphi+c_{c}+c_{w}+\alpha ln(\dot{\gamma}/\dot{ \gamma}_{1})\). It follows from Equation (13) that \(c_{\gamma}=\alpha ln(\dot{\gamma}/\dot{\gamma}_{1})\). The obtained result coincides with the result of experiment (4), and using Equation (13), it is possible to construct the \((\tau^{*}-ln\dot{\gamma})\) diagram (Figure 9), where the asymptotic lines at different shear rates are parallel and offset from each other according to the law \(\alpha ln(\dot{\gamma}/\dot{\gamma}_{1})\).