Question 7.130: Draw the shear and moment diagrams for the beam ABC.

Support Reactions: The 6 kN load can be replacde by an equivalent force and couple moment at B as shown on FBD (a).

+↑\Sigma F_y=0 ; \quad A_y+6.364 \sin 45^{\circ}-6=0 \quad A_y=1.50  \mathrm{kN}

Shear and Moment Functions: \text { For } \mathbf{0} \leq \boldsymbol{x}<\mathbf{3}  \mathbf{m} [\mathrm{FBD}(\mathrm{b})] \text {, }

+↑ \Sigma F_y=0 ; \quad 1.50-V=0 \quad V=1.50  \mathrm{kN}

For 3 m < x ≤ 6 m [FBD (c)],

+↑ \Sigma F_y=0 ; \quad V+6.364 \sin 45^{\circ}=0 \quad V=-4.50  \mathrm{kN}

M=\{27.0-4.50 x\}  \mathrm{kN} \cdot \mathrm{m}