Skip to main content

Frame AFD, SFD, BMD (80 kN, 30 kNm, hinge)

Short answer

Reactions: Ha=12.19→, Va=31.04↑, Hf=32.19←, Vf=86.46↑. AB: M linear 0→-73.13 kN·m left face. BC: M 18.96 bottom at C. Hinge D: M=0. DE: triangular load, M cubic, max -373.13 top at E. EF: horizontal distributed load, M parabolic from -373.13 to 0. Max BM=373.13 kN·m at E.

Draw axial force, shear force and bending moment diagram of the frame loaded as shown in the figure below. [Figure: left column A-B, with A hinged at the base and B at 6 m above A (2 m and 6 m marked on the left); an 80 kN horizontal force acting on the left column at B; a 30 kNm couple at the top-left corner; a 50 kN point load on the beam; beam B-D-E with an internal hinge at D and a triangular load rising to 15 kN/m on the right part ending at E; right column E-F with a 10 kN/m distributed horizontal load and F hinged at the base; horizontal dimensions 2 m, 1 m and 9 m as marked.]

Assumptions (taken from the figure):

  • A(0,0) hinged; AB = 6 m; 80 kN horizontal (→) and 30 kN·m clockwise couple at B.
  • Beam B–C–D–E: BC = 2 m (50 kN downward at C), CD = 1 m, internal hinge at D, DE = 9 m with triangular load from 0 at D to 15 kN/m at E (67.5 kN).
  • Right column EF = 6 m, F hinged, with 10 kN/m horizontal (←) over its full height.

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
|                                            |
|                                            |
|                                            |
|                                            |
|                                            |
|                                            |
|                                            |
|                                            |
|                                            |
|                                            |
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 = 12.19 kN →
  • VAV_A = 31.04 kN ↑
  • HFH_F = 32.19 kN ←
  • VFV_F = 86.46 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below 10−910^{-9}).

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

SectionN (kN)V (kN)M (kN·m)
AB at A-31.04-12.190.00
AB at B-31.04-12.19-73.13 (left face)
BC at B-92.1931.04-43.13 (top)
BC at C-92.1931.0418.96 (bottom)
CD at C-92.19-18.9618.96 (bottom)
CD at D-92.19-18.960.00
DE at D-92.19-18.960.00
DE at E-92.19-86.46-373.13 (top)
EF at E-86.4692.19-373.13 (right face)
EF at F-86.4632.190.00

Shape of the diagrams

  • AB: V constant (steps at point loads), M linear
  • BC: V constant (steps at point loads), M linear
  • CD: V constant (steps at point loads), M linear
  • DE: V parabolic, M cubic
  • EF: 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 = 373.13 kN·m at E, tension on the right face
  • Point of contraflexure (M = 0) at 1.39 m from B along BC
  • Bending moment is zero at the internal hinge (value read from both sides: 0.00).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Strength of Materials chapter 1 questions →