The cross-sectional area of a $16-\mathrm{mm}$ diameter bolt is $201.06 {\mathrm{mm}}^2$. When used in a double shear connection, two cross-sectional surfaces of the bolt are subjected to shear stress. The allowable bolt shear force (denoted $\mathrm{V}$) is therefore:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{allow}}={\mathrm{\tau }}_{\mathrm{allow}} 2{\mathrm{A}}_{\mathrm{bolt}} \\ =(140 \mathrm{MPa})(2 \mathrm{surfaces})(201.06 {\mathrm{mm}}^2 \mathrm{per} \mathrm{surface}) \\ =56,297 \mathrm{N}=56.3 \mathrm{kN} \end{gathered}$$

The shear force $\mathrm{V}$ acting on bolts $\mathrm{A}$ and $\mathrm{B}$ is equal to the axial force in member $\left(1\right)$. As shown previously, the force in member $\left(1\right)$ can be expressed in terms of $\mathrm{w}$ as:

${\mathrm{F}}_1=(-2.2595 \mathrm{m})(\mathrm{w} \mathrm{kN/}\mathrm{m})$