Frame AFD, SFD, BMD (60 kNm, 50 kN)

Assumptions (taken from the figure):

• A(0,0) hinged; inclined member A–B–C at 45°: B = (2,2), C = (5,5); 50 kN downward and 60 kN·m clockwise couple at B.
• Beam C–D = 6 m with 5 kN/m downward; column D–E = 4 m with E on a roller (vertical reaction only).

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.

                   /C------------------------D
                  /                          |
               ///                           |
              /                              |
           ///                               |
          /                                  |
       /B/                                   |
      /                                      |
   ///                                       E
  /
A/

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$.

• $H_A$ = 0.00 kN →
• $V_A$ = 43.64 kN ↑
• $V_E$ = 36.36 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 constant (steps at point loads), M linear
• CD: V linear, M parabolic
• DE: 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 = 147.27 kN·m at B, tension on the right face