Frame AFD, SFD, BMD (t/m loads, hinge)

Assumptions (taken from the figure):

• Units: tonne (t), t/m, t·m. A(0,0) hinged; AB = 4 m with 5 t/m horizontal (→) over the full height (20 t).
• Beam B–C–D = 3 m + 3 m with internal hinge at C; triangular load increasing uniformly from 0 at B to 1 t/m at D (total 3 t).
• Right column DE = 3 m, E hinged at (6, 1); 3 t/m horizontal (←) over the top 1 m of DE (3 t).

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------------------D
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A

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.

• $H_A$ = 12.57 t ←
• $V_A$ = 3.18 t ↓
• $H_E$ = 4.43 t ←
• $V_E$ = 6.18 t ↑

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 linear, M parabolic
• BC: V parabolic, M cubic
• CD: V parabolic, M cubic
• DE: V linear, M parabolic
• 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 = 15.80 t·m at 2.51 m from A on AB, tension on the right face
• V = 0 at 2.51 m from A along AB: M = 15.80 (right face) t·m (local maximum)
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).