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

Short answer

Determinate by assumng a hinge at the crown. Reactions: H_A=7.78 kN ←, V_A=21.67 kN ↑, H_E=2.22 kN ←, V_E=18.33 kN ↑. Maximum bending moment is 46.67 kN·m at the top of left column (tension on left face).

Draw axial force, shear force and bending moment diagrams for the frame shown indicating salient points with their values. [Figure: portal frame; left column hinged at the base, 4 m + 2 m high, carrying a triangular horizontal load increasing to 10 kN/m at the base over 4 m; top beam carrying 5 kN/m UDL; a 10 kN horizontal force on the right column at 4 m above the right base (2 m + 4 m heights marked on the right); right base hinged; horizontal dimensions 2 m, 4 m, 2 m.]

Assumptions (taken from the figure):

  • A(0,0) hinged; AB = 6 m (4 m + 2 m); horizontal triangular load 10 kN/m at A to 0 at 4 m above A (→).
  • Beam B–C–D: 8 m with 5 kN/m downward. Right column DE = 6 m, E hinged; 10 kN horizontal (←) at 4 m above E.
  • With two hinged supports a hinge is assumed at the crown (mid-span of BD) to make the frame determinate.

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                                  E

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 = 7.78 kN ←
  • VAV_A = 21.67 kN ↑
  • HEH_E = 2.22 kN ←
  • VEV_E = 18.33 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-21.677.780.00
AB at B-21.67-12.22-46.67 (left face)
BC at B-12.2221.67-46.67 (top)
BC at C-12.221.670.00
CD at C-12.221.670.00
CD at D-12.22-18.33-33.33 (top)
DE at D-18.3312.22-33.33 (right face)
DE at E-18.332.220.00

Shape of the diagrams

  • AB: V parabolic, M cubic
  • BC: V linear, M parabolic
  • 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 = 46.67 kN·m at B (on AB), tension on the left face
  • V = 0 at 0.87 m from A along AB: M = 3.26 (right face) kN·m (local maximum)
  • Point of contraflexure (M = 0) at 1.84 m from A along AB
  • V = 0 at 0.33 m from C along CD: M = 0.28 (bottom) kN·m (local maximum)
  • Point of contraflexure (M = 0) at 0.67 m from C along CD
  • 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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