Granite crushed is a material produced from granite massifs mechanically crushed into different sizes [1] [10] - [12] . The process produces sand, gravel and pebbles in a variety of angular and sharp-edged shapes. Due to its natural origin, granite crushed stone offers exceptional durability and strength.

In this study, we used 0/31.5 mm granite crushed stone from the Dan quarry in the Republic of Benin. The village of Dan is located in the commune of Djidja (Zou department), around 30 km from the commune of Bohicon in the Republic of Benin . The commune of Djidja lies to the north between latitude 7˚10' and 7˚40', and to the east between longitude 1˚04' and 2˚10'. The Dan quarry is located at latitude 7˚21'44'' to the north and longitude 2˚6'38'' to the east. At this quarry, crushed materials are screened and stockpiled by granular class [4] [13] . Figure 1 shows the location of the Dan quarry.

The equipment used for the characterization tests complies with the requirements of applicable standards in the field.

This section includes geotechnical testing equipment such as the CBR test, the shear test and the odometer test.

Samples are taken in accordance with ISO 22475-1 [26] .

The various geotechnical tests are carried out in accordance with the standards cited in §2.1.2.

Determination of the friction angle and internal cohesion by the Casagrande box shear test goes through the calibration of the raw material from the initial condition through the values obtained from the Modified Proctor test, then:

Step 1: Determination of the optimal water content and dry density on the material from the quarry (initial state).

Step 2: Determination of the optimal water content and dry density on the class 0/5 test sample.

Step 3: Carrying out the Casagrande box shears the test.

Step 4: Determination of the optimum water content and dry density on the test sample after the test.

Previous studies have shown that the elastic behavior of soils is never linear in reality . Therefore, it is important to focus studies on the nonlinear behavior of soils used in road construction. To do this, several mathematical models, both hyperelastic and hypoelastic, can be used to describe these nonlinear behaviors of soils. However, it has been proven that hypoelastic models are the most recommended when it comes to small deformation studies. Two types of hypoelastic models exist, namely hyperbolic models and variable modulus models, as reported by Babaliyè in 2020. In the context of this study, hyperbolic models mathematically based on a representation of the stress-strain relationship using a hyperbolic or parabolic curve [28] are best suited to describe the nonlinear elastic behavior of soils .

According to Hardin and Drnevich [29] , the hypoelastic behavior of a material is given by Equation (1).

$\tau =\frac{\gamma }{\frac{1}{G_{\max }}+\frac{\gamma }{{\tau }_{\max }}}$ (1)

where ${\tau }_{\max }$ represents the maximum shear stress, $G_{\max }$ the maximum shear modulus, $\tau$ the shear stress and $\gamma$ the shear strain.

To determine the parameters ${\tau }_{\max }$ and $G_{\max }$, Equation (1) was reformulated by setting: $a=1/G_{\max }$ and $b=1/{\tau }_{\max }$. This gives the following Equation (2):

$\tau =\phi \left(\gamma ,a,b\right)=\frac{\gamma }{a+b\gamma },a,b\in ℝ$ (2)

Using a nonlinear least fit method, the parameters a and b are evaluated. This method consists of fitting the experimental data $y_i$ to the function $\phi$ by minimizing the distance ${\phi }_i$ between $y_i$ and $\phi \left(\gamma ,a,b\right)$:

${\phi }_i=\underset{i=1}{\overset{n}{{\displaystyle \sum }}}{\left[y_i-\phi \left(\gamma ,a,b\right)\right]}^2$ (3)

The implementation of nonlinear regression follows the following steps:

1st Step: Linearization $\phi \left(\gamma ,a,b\right)$ of around $\left(a_0,b_0\right)$

$\{\begin{array}{l}\frac{\partial \phi }{\partial a}\left(\gamma ,a,b\right)=-\frac{\gamma }{{\left(a+b\gamma \right)}^2}\\ \frac{\partial \phi }{\partial b}\left(\gamma ,a,b\right)=-\frac{{\gamma }^2}{{\left(a+b\gamma \right)}^2}\end{array}$ (4)

$\phi \left(\gamma ,a,b\right)=\frac{\gamma }{a_0+\gamma b_0}-\frac{\gamma }{{\left(a_0+b_0\gamma \right)}^2}\left(a-a_0\right)-\frac{{\gamma }^2}{{\left(a_0+b_0\gamma \right)}^2}\left(b-b_0\right)$ (5)

Let us set: $A=\frac{\gamma }{{\left(a_0+b_0\gamma \right)}^2}$; $B=\frac{{\gamma }^2}{{\left(a_0+b_0\gamma \right)}^2}$ and $C=\frac{\gamma }{a_0+b_0\gamma }$.

Equation (5) becomes:

$\phi \left(\gamma ,a,b\right)=C-A\left(a-a_0\right)-B\left(b-b_0\right)$ (6)

2nd Step: Determination of a and b.

The minimization of $\phi$ consists of canceling its first derivative with respect to the unknowns a and b. Let:

$\{\begin{array}{l}\frac{\partial \phi }{\partial a}=0\\ \frac{\partial \phi }{\partial b}=0\end{array}$ (7)

with $\phi \left(\gamma ,a,b\right)=C-A\left(a-a_0\right)-B\left(b-b_0\right)$.

Development of the terms of the system of Equation (7).

Case of the first equation:

$\frac{\partial {\phi }_i}{\partial a}=0\iff \frac{\partial }{\partial a}\left[{\displaystyle \sum {\left(y_i-\phi \left(\gamma ,a,b\right)\right)}^2}\right]=0$ (8)

$\iff {\displaystyle \sum \frac{\partial }{\partial a}\left(y_i-\phi \left(\gamma ,a,b\right)\right)\cdot \left(y_i-\phi \left(\gamma ,a,b\right)\right)}=0$

$\iff -{\displaystyle \sum \frac{\partial }{\partial a}\phi \left(\gamma ,a,b\right)\cdot \left(y_i-\phi \left(\gamma ,a,b\right)\right)}=0$

$\iff -{\displaystyle \sum \left(-A\right)\cdot \left(y_i-\phi \left(\gamma ,a,b\right)\right)}=0$

$\iff {\displaystyle \sum A\left(y_i-C+A\left(a-a_0\right)+B\left(b-b_0\right)\right)}=0$

$\iff {\displaystyle \sum A{y}_i}-{\displaystyle \sum A\cdot C}+{\displaystyle \sum A^2\left(a-a_0\right)}+{\displaystyle \sum A\cdot B}\left(b-b_0\right)=0$

$\iff \left(a-a_0\right){\displaystyle \sum A^2}+\left(b-b_0\right){\displaystyle \sum A\cdot B}={\displaystyle \sum A\cdot C}-{\displaystyle \sum A{y}_i}$

$\iff \left(a-a_0\right){\displaystyle \sum A^2}+\left(b-b_0\right){\displaystyle \sum A\cdot B}={\displaystyle \sum A\left(C-y_i\right)}$ (9)

Case of the second equation:

$\frac{\partial {\phi }_i}{\partial b}=0\iff \frac{\partial }{\partial b}\left[{\displaystyle \sum {\left(y_i-\phi \left(\gamma ,a,b\right)\right)}^2}\right]=0$ (10)

$\iff {\displaystyle \sum \frac{\partial }{\partial b}\left(y_i-\phi \left(\gamma ,a,b\right)\right)\left(y_i-\phi \left(\gamma ,a,b\right)\right)}=0$

$\iff {\displaystyle \sum \left(-\frac{\partial }{\partial b}\phi \left(\gamma ,a,b\right)\right)\cdot \left(y_i-C+A\left(a-a_0\right)+B\left(b-b_0\right)\right)}=0$

$\iff {\displaystyle \sum -\left(-B\right)\left(y_i-C+A\left(a-a_0\right)+B\left(b-b_0\right)\right)}=0$

$\iff {\displaystyle \sum B\left(y_i-C+A\left(a-a_0\right)+B\left(b-b_0\right)\right)}=0$

$\iff {\displaystyle \sum \left(B{y}_i-B\cdot C+A\cdot B\left(a-a_0\right)+B^2\left(b-b_0\right)\right)}=0$

$\iff {\displaystyle \sum B\left(y_i-C\right)}+\left(a-a_0\right){\displaystyle \sum A\cdot B}+\left(b-b_0\right){\displaystyle \sum B^2}=0$

$\iff \left(a-a_0\right){\displaystyle \sum A\cdot B}+\left(b-b_0\right){\displaystyle \sum B^2}={\displaystyle \sum B\left(C-y_i\right)}$ (11)

So we have the following system:

$\{\begin{array}{l}\left(a-a_0\right){\displaystyle \sum A^2}+\left(b-b_0\right){\displaystyle \sum A\cdot B}={\displaystyle \sum A\left(C-y_i\right)}\\ \left(a-a_0\right){\displaystyle \sum A\cdot B}+\left(b-b_0\right){\displaystyle \sum B^2}={\displaystyle \sum B\left(C-y_i\right)}\end{array}$ (12)

Put into matrix form, the system of Equation (12) becomes:

$\left(\begin{array}{cc}{\displaystyle \sum A^2}& {\displaystyle \sum A\cdot B}\\ {\displaystyle \sum A\cdot B}& {\displaystyle \sum B^2}\end{array}\right)\left(\begin{array}{c}a-a_0\\ b-b_0\end{array}\right)=\left(\begin{array}{c}{\displaystyle \sum A\left(C-y_i\right)}\\ {\displaystyle \sum B\left(C-y_i\right)}\end{array}\right)$ (13)

Thus, the determinant (det M) of this system of equations is:

$\det M=\left|\begin{array}{cc}{\displaystyle \sum A^2}& {\displaystyle \sum A\cdot B}\\ {\displaystyle \sum A\cdot B}& {\displaystyle \sum B^2}\end{array}\right|$ (14)

$\iff \det M={\displaystyle \sum A^2}{\displaystyle \sum B^2}-{\displaystyle \sum A\cdot B}{\displaystyle \sum A\cdot B}$ (15)

Similarly, the determinants associated with a and b are det(a) and det(b) respectively.

Either:

$\det \left(a\right)=\left|\begin{array}{cc}{\displaystyle \sum A\left(C-y_i\right)}& {\displaystyle \sum A\cdot B}\\ {\displaystyle \sum B\left(C-y_i\right)}& {\displaystyle \sum B^2}\end{array}\right|$ (16)

Which is worth: $\det \left(a\right)={\displaystyle \sum A\left(C-y_i\right)}{\displaystyle \sum B^2}-{\displaystyle \sum B\left(C-y_i\right)}{\displaystyle \sum A\cdot B}$

So,

$a-a_0=\frac{\det \left(a\right)}{\det M}$ (17)

That is:

$a=a_0+\frac{{\displaystyle \sum A\left(C-y_i\right)}{\displaystyle \sum B^2}-{\displaystyle \sum B\left(C-y_i\right)}{\displaystyle \sum A\cdot B}}{{\displaystyle \sum A^2}{\displaystyle \sum B^2}-{\displaystyle \sum A\cdot B}{\displaystyle \sum A\cdot B}}$ (18)

Also,

$\det \left(b\right)=\left|\begin{array}{cc}{\displaystyle \sum A^2}& {\displaystyle \sum A\left(C-y_i\right)}\\ {\displaystyle \sum A\cdot B}& {\displaystyle \sum B\left(C-y_i\right)}\end{array}\right|$ (19)

which gives: $\det \left(b\right)={\displaystyle \sum A^2}{\displaystyle \sum B\left(C-y_i\right)}-{\displaystyle \sum A\cdot B}{\displaystyle \sum A\left(C-y_i\right)}$

So,

$b-b_0=\frac{\det \left(b\right)}{\det M}$ (20)

Consequently,

$b=b_0+\frac{{\displaystyle \sum A^2}{\displaystyle \sum B\left(C-y_i\right)}-{\displaystyle \sum A\cdot B}{\displaystyle \sum A\left(C-y_i\right)}}{{\displaystyle \sum A^2}{\displaystyle \sum B^2}-{\displaystyle \sum A\cdot B}{\displaystyle \sum A\cdot B}}$ (21)

The following system is made up:

$\{\begin{array}{l}a=a_0+\frac{{\displaystyle \sum A\left(C-y_i\right)}{\displaystyle \sum B^2}-{\displaystyle \sum B\left(C-y_i\right)}{\displaystyle \sum A\cdot B}}{{\displaystyle \sum A^2}{\displaystyle \sum B^2}-{\displaystyle \sum A\cdot B}{\displaystyle \sum A\cdot B}}\\ b=b_0+\frac{{\displaystyle \sum A^2}{\displaystyle \sum B\left(C-y_i\right)}-{\displaystyle \sum A\cdot B}{\displaystyle \sum A\left(C-y_i\right)}}{{\displaystyle \sum A^2}{\displaystyle \sum B^2}-{\displaystyle \sum A\cdot B}{\displaystyle \sum A\cdot B}}\end{array}$ (22)

From a Python program, the optimal value of each parameter a and b of Equation (22) is determined by respecting the stopping criterion defined by Equation (23) (Montgomery and Runger [30] ; Houanou [31] ; Babaliye [2] ).

$\frac{a-a_0}{a_0}<{10}^{-6}$ (23)

According to Gérard Degoutte and Paul Royet [32] and Leipholz [33] , the calculation of Young’s modulus (E) and Poisson’s ratio (υ) can be done from oedometric and shear tests. Thus, the following Equation (24) and Equation (25) are used:

$E=2G\left(1+\upsilon \right)$ (24)

$E=E_{oed}\frac{\left(1+\upsilon \right)\left(1-2\upsilon \right)}{1-\upsilon }$ (25)

where we denote by:

$G$, the shear modulus;

$E$, the Young’s modulus;

$\upsilon$, Poisson’s ratio;

$E_{oed}$, the oedometric module.

Equation (24) and Equation (25) allowed us to obtain the following Equation (26):

$2G\left(1+\upsilon \right)=E_{œd}\frac{\left(1+\upsilon \right)\left(1-2\upsilon \right)}{1-\upsilon }$ (26)

Thus, the transformation of Equation (26) becomes:

$\upsilon =\frac{E_{œd}-2G}{2\left(E_{œd}-G\right)}$ (27)

Furthermore, the determination of the oedometric modulus is obtained by Equation (28):

$E_{œd}=\frac{1+e_0}{C_c}\frac{{{\sigma }^{\prime }}_{\text{final}}-{{\sigma }^{\prime }}_{\text{initial}}}{\log \left(\frac{{{\sigma }^{\prime }}_{\text{final}}}{{{\sigma }^{\prime }}_{\text{initial}}}\right)}$ (28)

with

$e_0$, Index of voids in the soil in place;

$C_c$, Compression index of the soil in place;

${{\sigma }^{\prime }}_{\text{initial}}$, Initial normal stress;

${{\sigma }^{\prime }}_{\text{final}}$, Final normal stress.

The results of experimental tests carried out on a series of samples of Dan granite crushed rock are as follows.

Samples from the Dan quarry were sieved for particle size analysis. The following Figure 11 and Figure 12 show the various particle size curves.

Key information from Figure 11 and Figure 12 is shown in Table 1 below.

We note that the curves of all three samples are within the CEBTP grading range for the sub-base layer, but not within the CEBTP grading range for the crushed materials to be used in the base layer. We note that the 0.08 sieve pass of the samples studied is less than 35%, which complies with CEBTP requirements.

The material can therefore be used as a base course. These results are similar to those found by Elenga et al. [34] , Babaliyè [2] , Houanou et al. [4] and Dossou [1] .

The results of sand equivalence values carried out on three samples made from the fine 0/2-part of 0/31.5 granitic crushed materials are presented in Table 2 below:

According to these values, the sand equivalent of granitic crushed stone 0/2 is 57.33%; this sand is clean and this value is higher than the minimum 40% recommended for T3 - T4 traffic [35] [36] .

The results of the Micro Deval test on the three 0/31.5 granite crushed sand samples are presented in Table 3 below:

From the values recorded in Table 3 , it can be seen that the average value of the Micro-Deval coefficient, 7.72%, is less than 10%. This material is therefore rated as very good to good according to NF P 18-572 [37] and EN 1097-1 [38] . In accordance with the CEBTP guide [35] , a mean MD coefficient, i.e. 7.72%, is less than 12%, so it can be concluded that the material can be used for the construction of pavements designed for T3 - T4 traffic.

The results of the Los Angeles test carried out on three samples of 0/31.5 granitic crushed stone are presented in Table 4 below:

Analysis of Table 4 shows that, with an average Los Angeles coefficient of 23.4%, granite crushed stone offers good resistance to wear and mechanical impact. This value is less than 25% of the maximum value recommended for high-traffic wearing courses -. It follows that this material can be used in other parts of the pavement -.

The results of the methylene blue test are shown in Table 5 below:

Analysis of Table 5 shows that the methylene blue value for granitic crushed stone 0/31.5 is 0.16%, less than 0.2%. Granitic crushed stone is therefore sandy according to standard NF P 94-068 [16] .

The results of the organic matter test are presented in Table 6 below:

Analysis of Table 6 shows that the organic matter content of granite crushed stone is 0.14%, less than 1%, making the material low organic [17] .

The results obtained for the density of the crushed materials are shown in Table 7 below:

Analysis of Table 7 shows that the results for the absolute density of granitic crushed stone vary slightly between the different samples studied. The average is 2.67 g/cm³. This value indicates that the material tends to provide higher densities, which may contribute to greater strength and durability.

1) Modified Proctor test

The results obtained on the Modified Proctor test samples are shown in Table 8 below:

Table 8 shows the results of the Modified Proctor test carried out on 0/31.5 granitic crushed stone. The values obtained at the Modified Proctor Optimum are 2.27 t/m³ for maximum dry density and 6.47% for water content. This value shows that granitic crushed stone 0/31.5 is more compact and potentially more stable when used as backfill or pavement material [21] [36] [43] [44] . Moreover, this value, similar to those obtained by Dossou [1] , Houanou et al. [4] and Babaliyè [2] , is higher than 2 t/cm³. Granite crushed stone can therefore be used in road construction to CEBTP specifications in sub-base and base courses.

2) CBR test

The results of the CBR test on granitic crushed stone are shown in Table 9 below:

Analysis of Table 9 shows that the CBR value for granite crushed rock at 95% OPM after immersion is 96.29%. This value is higher than 30%, which is the minimum value required by CEBTP , AGEROUTE-Sénégal for granular materials suitable for use in sub-base courses. This material can also be used as a base course for pavements, as its CBR is higher than 80% [1] [43] . For example, Dan’s granite crushed stone can be used for both sub-base and base courses.

The relative linear swelling of this material, at 0.068%, is less than 0.5%. As a result, the material is not very sensitive to water and will exhibit very good volumetric behavior under prolonged humidity conditions [45] . It can be used as a base course, as its linear swelling is less than 1% [35] [43] .

Table 10 gives the average values for dry density and optimum moisture content for Dan’s granite crushed stone (all-material) and for the graded material before and after the shear test. These values are determined to specify the test conditions.

The values obtained enable us to plot tangential stress as a function of displacement ( Figures 13(a)-(c) ), and shear stress as a function of normal stress ( Figures 14(a)-(c) ).

By identification, the equations of the straight line derived from the tests ( Figure 14 ) gave the values of c and were recorded in Table 11 .

Analysis of Figure 13 shows that shear stress evolves with increasing applied load, whatever the normal load applied.

Figure 14 shows that tangential stress evolves in the same direction as normal stress. The slope reflecting this increase is of the order of 0.60. The equation of the Coulomb line typical of a shear test is of the form:

$\tau =c+\sigma \tan \phi$ (29)

where τ is shear stress, c cohesion, σ normal stress and φ angle of internal friction [23] .

Table 11 shows that the angle of internal friction of Dan’s granite crushed stone is 31.90˚, compared with 0.77 kPa for internal cohesion.

The results of the odometer test were used to draw the following odometer curves ( Figure 15 ):

Table 12 shows the parameters derived from the odometer test and the corresponding curve ( Figure 15 ).

Analysis of Table 12 shows that the void index is 0.496, while the pre-consolidation stress is 28 kPa with a density of 1.875 g/m³. Furthermore, the coefficient of compressibility is 0.093 while the coefficient of swelling is 0.008. It can be deduced that the material is not very compressible and does not swell [25] .

Evaluation of the ratio shows that Dan’s granitic crushed stone has low compressibility [46] . This may be due to its low fine particle content of 6.66%.

Also, the oedometric modulus evaluated is 295.248 MPa.

Successive iterations, based on the equation system, led to the determination of the optimal value of the Hardin Drnevich numerical model parameters. These values are presented in Table 13 below. They concern normal stress, shear modulus and shear stress.

Figure 16(a) and Figure 16(b) below show the Hardin and Drnevich hyperbolic behavior curves for Dan’s granitic crushed stone.

It can be seen that the stress-strain curves of the model are very close to those of the observations. This means that the model fits the observations well.

The fit between the observations and the model is reflected by the coefficient of determination of each of the curves ( Figure 16(a) and Figure 16(b) ). Table 14 shows the different values of the calculated coefficient of determination.

According to Table 14 , the coefficients of determination of the model according to the different compaction energies are close to 100% (ranging from 99.03 to 99.82%). This clearly shows that the Hardin and Drnevich model is adequate [1] [2] [27] [31] .

Table 15 below gives the Poisson’s ratios and Young’s moduli derived respectively from Equations (24)-(27) and Equation (24) or Equation (25).

Analysis of Table 15 shows that the shear modulus varies from 12.858 kPa to 172.653 kPa and the Poisson’s ratio varies from 0.477 to 0.204, while Young’s modulus varies from 37.988 MPa to 274.808 MPa. Poisson’s ratio decreases with increasing normal stress. In addition, shear modulus and Young’s modulus increase with normal stress at different load applications.

Analysis of the data in Table 16 shows that the material contains little water, as its average water content of 3.5% is less than 4%. According to standard NF P 94-093 [47] , this material is suitable for compaction.

Table 17 shows the result of the analysis of the characteristics of crushed granite for its use in the foundation layer according to CBTP 1984 amended 2019.

According to Table 17 above, Dan’s granite crushed meets all the criteria for use as a base course for flexible pavements [35] [36] [43] [44] .

Similarly, Table 18 shows the result of the analysis of the characteristics of crushed granite for its use in the foundation layer according to CBTP 1984 amended 2019.

According to Table 18 , Dan’s granite crushed meets all the criteria for use as a base course for flexible pavements [36] [44] . In addition, it can be used as a reinforcement material to improve the characteristics of other materials with poor qualities [1] .