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AFD, SFD, BMD for frame with UDL and couple

Short answer

Reactions: A_x = 25.05 kN ←, A_y = 2.95 kN ↓, H_y = 42.02 kN ↑. Maximum bending moment is 129.10 kN·m at knee C (tension inside). Contraflexure in DE at 0.97 m from D.

Draw AFD, SFD and BMD of the given frame loaded as shown in figure below. Indicate also the salient features if any. [Figure: frame with pin support at A (bottom of left column) and roller/hinge at H; left column A-B-C 3 m + 3 m with 10 kN at 45° at B and 30 kN at 30° at C; beam C-D-E-F with 4 kN/m UDL, 20 kN·m moment at E, 5 kN downward at F; right column from E/F down to G (8 kN horizontal) and H with 2 m + 2 m; horizontal dimensions 3 m, 1 m, 1 m.]

Reading of the figure

  • Origin at A, x to the right, y upward: A(0,0) pin, B(0,3), C(0,6), D(3,6), E(4,6), F(5,6), G(4,4), H(4,2) roller (vertical reaction only).
  • Loads: 30 kN at C (30° below horizontal, towards right) = 25.98 kN → and 15.00 kN ↓; 10 kN at B (45°) = 7.07 kN → and 7.07 kN ↓; UDL 4 kN/m on CD (12 kN at 1.5 m from C); couple 20 kN·m at E (taken anticlockwise); 5 kN ↓ at F; 8 kN ← at G.
  • r = 3, so the frame is statically determinate.

Reactions

 30kN\_  4 kN/m   20kN·m  5kN
      C=======D==E====F ↓
      |          |
 10kN\_B         G←8kN
      |          |
      |          H (roller)
      A (pin)
  • ΣFx = 0: Aₓ + 7.07 + 25.98 − 8 = 0, so Aₓ = 25.05 kN (←).
  • ΣM_A = 0 (anticlockwise +): 4H_y − 7.07(3) − 25.98(6) − 12(1.5) + 20 − 5(5) + 8(4) = 0, so 4H_y = 168.10 and H_y = 42.02 kN (↑).
  • ΣFy = 0: A_y + 42.02 − 7.07 − 15 − 12 − 5 = 0, so A_y = 2.95 kN (↓).

Member forces and diagrams

Sign notes: N is + for tension and − for compression; V follows the member axis in the order of the letters (e.g. A→B); for M the side in brackets is the tension side, on which the BMD is drawn.

  • AB (A to B, 3.00 m): N = 2.95 kN (tension); V = 25.05 to 25.05 kN; M = 0.00 (–) to 75.16 (right) kN·m, linear.
  • BC (B to C, 3.00 m): N = 10.02 kN (tension); V = 17.98 to 17.98 kN; M = 75.16 (right) to 129.10 (right) kN·m, linear.
  • CD (C to D, 3.00 m): N = 8.00 kN (compression); V = −25.02 to −37.02 kN; M = 129.10 (bottom) to 36.02 (bottom) kN·m, parabolic.
  • DE (D to E, 1.00 m): N = 8.00 kN (compression); V = −37.02 to −37.02 kN; M = 36.02 (bottom) to 1.00 (top) kN·m, linear. M changes sign (contraflexure) at 0.97 m from D.
  • EF (E to F, 1.00 m): N = 0.00 kN (zero); V = 5.00 to 5.00 kN; M = 5.00 (top) to 0.00 (–) kN·m, linear.
  • EG (E to G, 2.00 m): N = 42.02 kN (compression); V = 8.00 to 8.00 kN; M = 16.00 (right) to 0.00 (–) kN·m, linear.
  • GH (G to H, 2.00 m): N = 42.02 kN (compression); V = 0.00 to 0.00 kN; M = 0.00 (–) to 0.00 (–) kN·m, constant.

Salient features

  • Maximum bending moment = 129.10 kN·m at the knee C (tension on the inside of the frame). Maximum axial force = 42.02 kN compression in the right column.
  • Shear in CD never becomes zero, so the moment falls steadily from C to D; in DE the moment changes sign 0.97 m from D (point of contraflexure).
  • The roller at H gives no horizontal reaction, so the whole horizontal load is resisted at A. The 8 kN load is carried by compression in beam CDE.
  • Moment is carried round the rigid corner C unchanged (129.10 kN·m on both members, tension on the inner fibres). Joint E check: 1.00 + 20 (couple) = 5.00 + 16.00.

✓ Answer checked by Claude Opus 5.5 (high effort)

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