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

Short answer

By equilibrium with hinge at C, reactions are found (H_A=21.67 kN left, V_A=1.94 kN up, H_F=8.33 kN left, V_F=38.06 kN up). Diagrams show max bending moment 61.67 kN·m at D on CD (tension top); zero at hinge C; local maxima 27.32 kN·m on AB and 17.04 kN·m on BC where shear vanishes. Overhang DE has linear moment to 20 kN·m hogging at D.

Draw axial force, shear force and bending moment diagrams for the frame shown in figure below, indicating the principal numerical values at salient points. [Figure: left column A-B with A hinged at the base, carrying a triangular horizontal load from 10 kN/m at A to 2 kN/m at B; horizontal member B-C-D-E with a 5 kN/m UDL over B to D, an internal hinge at C, a 10 kN downward point load at E; right column D-F with F hinged at the base; horizontal dimensions 3 m, 3 m, 2 m; vertical dimensions 2 m, 3 m, 1 m on the right.]

Assumptions (taken from the figure):

  • A(0,0) hinged; AB = 5 m; horizontal triangular load from 10 kN/m at A to 2 kN/m 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). Right column DF = 5 m, F hinged at ground level.
  • The overhang DE is replaced at D by 10 kN downward plus a hogging moment 10×2=2010\times2 = 20 kN·m.

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                              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 = 21.67 kN ←
  • VAV_A = 1.94 kN ↑
  • HFH_F = 8.33 kN ←
  • VFV_F = 38.06 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-1.9421.670.00
AB at B-1.94-8.3316.67 (right face)
BC at B-8.331.9416.67 (bottom)
BC at C-8.33-13.060.00
CD at C-8.33-13.060.00
CD at D-8.33-28.06-61.67 (top)
DF at D-38.068.33-41.67 (right face)
DF at F-38.068.330.00

Shape of the diagrams

  • AB: V parabolic, M cubic
  • BC: V linear, M parabolic
  • CD: V linear, M parabolic
  • 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 = 61.67 kN·m at D (on CD), tension on the top
  • V = 0 at 2.79 m from A along AB: M = 27.32 (right face) kN·m (local maximum)
  • V = 0 at 0.39 m from B along BC: M = 17.04 (bottom) kN·m (local maximum)
  • Bending moment is zero at the internal hinge (value read from both sides: 0.00).
  • Overhang DE: V = 10 kN (constant), M = 0 at E varying linearly to 20 kN·m (hogging, tension on top) at D.

✓ Answer checked by Claude Opus 5.5 (high effort)

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