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Frame AFD, SFD, BMD (20 kNm, 15 kN)

Short answer

Reactions: H_A=39.67 kN←, V_A=39.00 kN↓; H_G=14.00 kN←, V_G=79.00 kN↑. Maximum bending moment 107.99 kN·m at C (top tension). On AB, local moment maximum 90.69 kN·m at 5.01 m from A. Contraflexure point on BC at 2 m from B. Axial forces constant on each member.

Draw axial force, shear force and bending moment diagram of given loaded frame. Also show the salient feature. [Figure: left column A-B, A hinged at the base, carrying a horizontal distributed load varying from 10 kN/m at A to 5 kN/m at B; beam B-C with a 5 kN/m UDL; a 20 kNm couple at C; right column from C down to a hinged support G; a 10 kN force at 30∘30^\circ acting on the right column; a 15 kN downward force at the end E of the beam extension; horizontal dimensions 4 m and 2 m; vertical dimensions 3 m and 3 m on the right.]

Assumptions (taken from the figure):

  • A(0,0) hinged; AB = 6 m; beam B–C = 4 m; column C–G = 6 m with G hinged; overhang C–E = 2 m carrying 15 kN downward at E.
  • Load on AB: horizontal, varying from 10 kN/m at A to 5 kN/m at B (acting →); 5 kN/m downward on BC; 20 kN·m couple at C (clockwise); 10 kN at 30° to the horizontal acts at mid-height of CG, to the right and downward.
  • Two hinged supports with no internal hinge are indeterminate, so a hinge is assumed at the mid-span of BC.

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
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A                G

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 = 39.67 kN ←
  • VAV_A = 39.00 kN ↓
  • HGH_G = 14.00 kN ←
  • VGV_G = 79.00 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 A39.0039.670.00
AB at B39.00-5.3387.99 (right face)
BC at B-5.33-39.0087.99 (bottom)
BC at C-5.33-59.00-107.99 (top)
CG at C-74.005.33-57.99 (right face)
CG at G-79.0014.000.00

Shape of the diagrams

  • AB: V parabolic, M cubic
  • BC: V linear, M parabolic
  • CG: 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 = 107.99 kN·m at C (on BC), tension on the top
  • V = 0 at 5.01 m from A along AB: M = 90.69 (right face) kN·m (local maximum)
  • Point of contraflexure (M = 0) at 2.00 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)

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