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Frame AFD, SFD, BMD (100 kN, 25 kN/m, hinge)

Short answer

After computing support reactions, axial force, shear force and bending moment diagrams are plotted. Maximum bending moment is 433.33 kN·m at E (tension top in DE, right face of EF); zero at hinge D; maximum sagging moment 116.67 kN·m under the 100 kN load on CD. AB has N=-41.67 kN, V=26.67 kN constant.

Draw axial force, shear force and bending moment diagrams for the frame. Indicate numerical values at salient points. [Figure: left column A-B-C with A hinged at the base, a 50 kN horizontal force at B (3 m above A), and C a further 2 m above B; top member C-D-E with a 100 kN point load at 2 m from C, an internal hinge at D (4 m from C) and a 25 kN/m UDL over D to E (4 m); an inclined member from E down to a hinged support F located 2 m inside the horizontal projection of E.]

Assumptions (taken from the figure):

  • A(0,0) hinged; AB = 3 m, BC = 2 m; 50 kN horizontal (→) at B. Top member C–D–E: CD = 4 m (hinge at D), DE = 4 m.
  • 100 kN downward at 2 m from C; 25 kN/m downward over DE (total 100 kN).
  • The inclined member EF goes from E (8, 5) down to the hinged support F (6, 0) at ground level, 2 m inside E.

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--------------------E
|                                        /
|                                       //
|                                      /
|                                     //
B                                     /
|                                   //
|                                   /
|                                  //
|                                 /
|                                //
|                                /
|                              //
A                              F

Support reactions

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

  • HAH_A = 26.67 kN ←
  • VAV_A = 41.67 kN ↑
  • HFH_F = 23.33 kN ←
  • VFV_F = 158.33 kN ↑

Check: the three equilibrium equations of the complete frame and the hinge condition are satisfied to rounding.

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

SectionN (kN)V (kN)M (kN·m)
AB at A-41.6726.670.00
AB at B-41.6726.6780.00 (right face)
BC at B-41.67-23.3380.00 (right face)
BC at C-41.67-23.3333.33 (right face)
CD at C-23.3341.6733.33 (bottom)
CD at 100 kN load (2 m from C)-23.3341.67 / -58.33116.67 (bottom)
CD at D-23.33-58.330.00
DE at D-23.33-58.330.00
DE at E-23.33-158.33-433.33 (top)
EF at E-138.3480.47-433.33 (right face)
EF at F-138.3480.470.00

Shape of the diagrams

  • AB: V constant, M linear
  • BC: V constant, M linear
  • CD: V constant on each side of the 100 kN load (it steps by 100 kN there), M linear on each side
  • DE: V linear, M parabolic
  • EF: V constant, M linear
  • AFD: constant between loads on a member, changing only where a load component acts along the member.
  • The 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 = 433.33 kN·m at joint E (top of DE, right face of EF)
  • Maximum sagging moment = 116.67 kN·m under the 100 kN load on CD (2 m from C), tension on the bottom
  • The bending moment is zero at the internal hinge D (from both sides).

✓ Answer checked by Claude Opus 5.5 (high effort)

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