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Frame AFD, SFD, BMD (30 kN, 20 kN/m)

Short answer

Support reactions: H_A=25.0 kN right, V_A=117.5 kN up, H_D=101.0 kN left, V_D=45.0 kN up. Diagrams show max bending moment 250.0 kN·m at B (tension left face of AB); local peak 77.94 kN·m (bottom) at 5.20 m from C on CD where shear zero; zero at hinge C.

Draw axial force, shear force and bending moment diagram for the given frame. Indicate numerical values at salient points. [Figure: left column A-B of 5 m (2.5 m + 2.5 m) with A hinged at the base and a 50 kN horizontal force at the mid-height; top member B-C-D with a 30 kN force at 30∘30^\circ at B, a 20 kN/m UDL over B to C (4 m), an internal hinge at C, then a triangular load increasing from zero at C to 15 kN/m at D over 9 m; D is a hinged support.]

Assumptions (taken from the figure):

  • A(0,0) hinged; AB = 5 m; 50 kN horizontal (→) at mid-height (2.5 m). Top member B–C–D: BC = 4 m, CD = 9 m, hinge at C, D hinged support at beam level.
  • 30 kN at B acts at 30° to the horizontal, to the right and downward; 20 kN/m downward on BC; triangular load on CD from 0 at C to 15 kN/m at D (total 67.5 kN).

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

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 = 25.00 kN →
  • VAV_A = 117.50 kN ↑
  • HDH_D = 100.98 kN ←
  • VDV_D = 45.00 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-117.50-25.000.00
AB at B-117.50-75.00-250.00 (left face)
BC at B-100.98102.50-250.00 (top)
BC at C-100.9822.500.00
CD at C-100.9822.500.00
CD at D-100.98-45.000.00

Shape of the diagrams

  • AB: V constant (steps at point loads), M linear
  • BC: V linear, M parabolic
  • CD: V parabolic, M cubic
  • 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 = 250.00 kN·m at B (on AB), tension on the left face
  • V = 0 at 5.20 m from C along CD: M = 77.94 (bottom) kN·m (local maximum)
  • Bending moment is zero at the internal hinge (value read from both sides: 0.00).

✓ Answer checked by Claude Opus 5.5 (high effort)

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