Question 1.1: Load Resultants at a Section of a Piping An L-shaped pipe as......

See Figure 1.3 and Equation (1.5).

$\begin{array}{rccc} \Sigma F_x=0 & \Sigma F_y=0 & \Sigma F_z=0 \\ \Sigma M_x=0 & \Sigma M_y=0 & \Sigma M_z=0 \end{array}$          (1.5)

Using the conversion factor from Table A.1, $w$=5.79 (N/m)/(0.0685) =84.53 N/m. Thus, the weights of the pipes $AB$ and $BO$ are equal to

$W_{A B}=(84.53)(0.6)=50.72  N , \quad W_{B O}=(84.53)(0.48)=40.57  N$

Free-body: Part $ABO$. We have six equations of equilibrium for the 3D force system of six unknowns (Figure 1.3b). The first three from Equation (1.5) results in the internal forces on the pipe at point $O$ as follows:

$\begin{array}{ll} \Sigma F_x=0: & F=0 \\ \Sigma F_y=0: & V_y-50.72-40.57-100=0: \quad V_y=191.3  N \\ \Sigma F_z=0: & V_z=0 \end{array}$

Applying the last three from Equation (1.5), the moments about point $O$ are found to be

$\begin{array}{ll} \Sigma M_x=0: & T+(50.72)(0.3)+100(0.6)=0, \quad T=-75.2  N \cdot m \\ \Sigma M_y=0: & M_y=0 \\ \Sigma M_z=0: & M_z-25-100(0.48)-(50.72)(0.48)-(40.57)(0.24)=0, \\ & M_z=107.1  N \cdot m \end{array}$

Comment: The negative value calculated for $T$ means that the torque vector is directed opposite to that indicated in Figure 1.3b.