## 3 Assessment of Shear Force by Means of Diagonal Compression Test  
As previously discussed, diagonal compression tests are simpler to realize in comparison with shear-compressive tests, but their interpretation is more uncertain and, indeed, different approaches are available in literature ([PERSON] et al., 2008 [17], [PERSON] et al., 2012 [44], [PERSON] et al., 2015 [45]) to calculate significant mechanical parameters of masonry, usually the tensile strength or the pure shear strength, based on the maximum applied compressive diagonal load. In this paper, an attempt to convert the maximum diagonal force in a shear force was done, since many diagonal compression tests are available in literature, while the design formulations, see Equations (1) and (3), are usually expressed in term of shear force capacity. According to the American standard (ASTM E519 2010 [46]), the diagonal compression force applied during the test can be decomposed along the directions parallel to the wall sides (i.e., direction inclined of 45\({}^{\circ}\) with respect to the applied load). Such an approach assumes, indeed, a uniform shear stress distribution within the wall and, thus, the following stress state at its center (pure shear state, \(\sigma=0\), see the Mohr's circle in Figure 1b) when applied the maximum load compression load \(P\):  
\[\sigma=0\ \ \ \ \ \tau_{0}=\frac{P}{A\sqrt{2}}, \tag{4}\]  
Figure 1: (**a**) American standard (ASTM) interpretation of the diagonal compression test; (**b**) Mohr’s circle corresponding to ASTM and RILEM interpretation of diagonal compression test. in which \(\tau_{o}\) is the shear strength corresponding to the normal stress \(\sigma=0\), and \(A\) is the area of the cross section of the wall, i.e., \(A=t\cdot l\). Under these assumptions, the tensile strength of masonry, \(f_{t}\), and the shear force, \(V_{t}\), can be calculated as follows:  
\[\sigma_{I}=\sigma_{II}=f_{t}=\tau_{0}, \tag{5}\]  
\[\tau_{0}=\frac{P}{A\sqrt{2}}, \tag{6}\]  
\[V_{t}=\tau_{0}\cdot A=\frac{P}{\sqrt{2}}\approx 0.707P, \tag{7}\]  
with \(\sigma_{\rm I}\) and \(\sigma_{\rm II}\) being the tensile and compressive principal stresses, respectively. On the other hand, RILEM guidelines (RILEM LUMB6 1994 [47]) also provide a criterion for the assessment of the stress state at the center of the panel on the basis of several experimental results obtained by [PERSON] (1931) [48]. Assuming the masonry wall as isotropic and homogeneous material and under linear elastic analysis, [PERSON] (1931) [48] found that the stress state is not uniform along the diagonal of the panel and is characterized by both shear and normal stresses as also supposed by ([PERSON] et al., 2008 [17], [PERSON] et al., 2012 [44], [PERSON] et al., 2015 [45]). This means that the Mohr's circle relative to the center of the panel is no longer centerd in the origin (see Figure 1b). Therefore, for the tensile strength of masonry, \(f_{t}\), and the pure shear strength, \(\tau_{o}\), i.e., the shear stress at zero normal stress, evaluated at the maximum compression force \(P\), RILEM proposes a different interpretation, which leads to a further evaluation of the shear force \(V_{t}\):  
\[f_{t}=\sigma_{I}=0.5\frac{P}{A}, \tag{8}\]  
\[\tau_{0}=0.88\frac{P}{A}, \tag{9}\]  
\[V_{t}=\tau_{0}\cdot A=0.88P. \tag{10}\]  
Thus, according to RILEM interpretation, the tensile strength of masonry is lower than that provided by the ASTM standard, while the pure shear strength is higher. Moreover, the Italian code (IBC, 2018 [35]) suggests that, in the case of diagonal compression tests, the tensile strength of masonry for diagonal cracking, \(f_{t}\), can be assumed equal to the maximum diagonal applied load divided by twice the mean cross section area of the tested wall, \(A\), i.e., it adopts Equation (5a), but it does not give any indication about the evaluation of the shear strength. Thus, two possibilities were examined in the following, starting from Equation (5a).