## 4 Maximum Strength of Masonry-Infilled Frames
### Frame Strength Accounting for Interaction with the Masonry Infill Wall  
The strength prediction in Table 6 underestimates the strength of specimens with double wythe (IF-1.0 and IFS-1.0). It should be noted that strength prediction was made based on the flexural failure strength of the columns, although the actual failure mode of the two specimens was the shear failure of columns rather than flexural failure. The reason for such change in failure mode is presumably the existence of an interaction between the masonry infills and the surrounding frame. It is generally assumed that, in a masonry-infilled frame, the infill walls and frame resist lateral loads by formulating a diagonal compression strut, as shown in Figure 10b. The horizontal component of this strut force will be applied to the upper part of the windward column and the lower part of the leeward \begin{table}
\begin{tabular}{c c c c c c c c c} \hline \hline \multirow{2}{*}{**Specimen ID**} & **Drift at Max** & \multirow{2}{*}{\(\tau_{m}\) **Equation (4)**} & \multirow{2}{*}{\(V_{m}\) **Equation (6)**} & \multirow{2}{*}{\(2V_{p}\) **Equation (1)**} & \multirow{2}{*}{\(P_{u,pred}\) (Prediction)**} & \multirow{2}{*}{\(P_{u}^{*}\) (Test)} & \multirow{2}{*}{\(P_{u,pred}\)/\(P_{u}^{*}\)} & \multirow{2}{*}{\(P_{u,pred}\)/\(P_{u}^{*}\)} \\  & & & & & & & & \\ \hline IF0.5 & 1.0\% & 0.781 MPa & 154.6 kN & 178.8 kN & 333.4 kN & 360.5 kN & \(-\)317.2 kN & 0.92 & 1.05 \\ \hline IFS0.5 & - & 0.462 MPa & 91.5 kN & 178.8 kN & 270.3 kN & 288.3 kN & \(-\)253.8 kN & 0.94 & 1.07 \\ \hline IF1.0 & 0.89\% & 0.733 MPa & 306.3 kN & 178.8 kN & 485.1 kN & 590.1 kN & \(-\)530.6 kN & 0.82 & 0.91 \\ \hline IFS1.0 & - & 0.462 MPa & 193.1 kN & 178.8 kN & 371.9 kN & 468.7 kN & \(-\)399.9 kN & 0.79 & 0.93 \\ \hline \hline \end{tabular}
\end{table} 
Table 6: Prediction of maximum force based on masonry shear force and flexural yielding of the frame. column. Figure 10a is the free body diagram of the windward column when it was assumed that both ends of the column yield by flexure. Figure 11 shows the shear force diagram and bending moment diagram for the free body diagram in Figure 10a. It was assumed that plastic hinges were formulated at both ends, and the reactive force from the masonry wall acts as a triangular distributed load at the top of the column. If \(H_{top}\), which represents the shear force at the top of the column, is lower than the shear strength of the column section \(V_{n}\), flexural yielding will occur as assumed. However, as shown in the figure, if \(H_{top}\) exceeds the shear strength of column \(V_{n}\), shear failure will occur before flexural yielding at both ends of the column. Figure 11: Shear force and bending moment diagrams of the windward column subjected to reaction from masonry infill. (**a**) Shear force diagram. (**b**) Bending moment diagram. Figure 10: Free body diagram of the windward column subjected to reaction from masonry infill. (**a**) Free body diagram. (**b**) Horizontal component of strut force. In the IF-1.0 and IFS-1.0 specimens, in which the double wythe masonry wall was used, shear failure occurred. Therefore, the contribution of the lateral-load capacity of a frame part was adjusted considering the difference in the failure mode as follows. First, as shown in Figure 10a, when flexural yielding occurs at both ends, the equilibrium moment equation centered on the lower part of the column can be established as shown in Equation (7). \[H_{top}h_{col}-V_{m}\left(h_{col}-\frac{z}{3}\right)-2M_{p}=0 \tag{7}\]  
where \(h_{col}\) is the clear height of the column, \(M_{p}\) is the plastic moment capacity of a column section, and \(z\) is the contact length between the masonry infill and the frame. In this study, \(z\) was estimated according to ASCE 41 [25] as \(z=w/2 cos\theta\), with \(w\) being the equivalent strut width.