Frame AFD, SFD, BMD (10 kN/m, 60 kN, hinge)

Assumptions (taken from the figure):

• Left column AB = 2 m, with the internal hinge at its top B (the top-left corner); A(0,6) hinged; the 60 kN horizontal force acts at the support A and is resisted directly by it.
• Beam B–C = 6 m with 10 kN/m downward; right column C–D = 8 m, D(6,0) hinged; 2 kN vertical (down) at 5 m below C on CD.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B-------------------C
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A                   |
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                    D

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 60.00 kN ←
• $V_A$ = 30.00 kN ↑
• $H_D$ = 0.00 kN ←
• $V_D$ = 32.00 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V constant (steps at point loads), M linear
• BC: V linear, M parabolic
• CD: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 45.00 kN·m at 3.00 m from B on BC, tension on the bottom
• V = 0 at 3.00 m from B along BC: M = 45.00 (bottom) kN·m (local maximum)
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).