Frame AFD, SFD, BMD (5 and 10 kN/m, hinge)

Assumptions (taken from the figure):

• A(0,0) hinged; AB = 5 m; horizontal triangular load 10 kN/m at A to 0 at B (→). Beam B–C–D–E: BC = CD = 3 m, DE = 2 m; hinge at C.
• 5 kN/m downward over B to D; 10 kN downward at E (overhang, equivalent to 10 kN plus a 20 kN·m hogging moment at D). Right column DF = 5 m, F hinged; 10 kN horizontal (←) at 2 m below D, i.e. 3 m above F (the 2 m and 3 m dimensions on the right column).

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---------E
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A                              F

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.

• $\Sigma M_A = 0$: $6V_F = 25\left(\tfrac{5}{3}\right) + 30(3) + 10(8) - 10(3) = 181.67$, so $V_F = 30.28$ kN ↑
• $\Sigma F_y = 0$: $V_A = 40 - 30.28 = 9.72$ kN ↑
• $M_C = 0$ (left part, $H_A$ taken positive to the right): $5H_A = 3V_A - 25\left(\tfrac{10}{3}\right) - 15(1.5) = -76.67$, so $H_A = -15.33$, i.e. $H_A$ = 15.33 kN ←
• $\Sigma F_x = 0$: $H_F$ = 0.33 kN →

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

Joint D check: the beam moment 38.33 kN·m on CD is balanced by 20.00 kN·m in the overhang DE plus 18.33 kN·m in the column DF.

Shape of the diagrams

• AB: V parabolic, M cubic
• BC: V linear, M parabolic
• CD: V linear, M parabolic
• DE: V constant (10 kN), M linear from 20 kN·m (top) at D to 0 at E
• DF: 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 = 38.33 kN·m at D (on CD), tension on the top
• V = 0 at 1.89 m from A along AB: M = 13.37 kN·m (right face), a local maximum
• Point of contraflexure (M = 0) at 4.30 m from A along AB
• V = 0 at 1.94 m from B along BC: M = 2.79 kN·m (bottom), a local maximum
• Point of contraflexure (M = 0) at 0.89 m from B along BC
• Point of contraflexure (M = 0) at 1.90 m from D along DF
• Bending moment is zero at the internal hinge C (value read from both sides: 0.00).