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AFD, SFD, BMD of frame with 60 kN/m UDL

Short answer

Reactions: Ax=3 kN left, Ay=136.6 kN up, Hy=113.4 kN up. Maximum bending moment 128.50 kN·m (tension bottom) at 2.28 m from C in CD. Contraflexure points: 0.75 m above B in BC, 0.21 m from C in CD, 0.38 m from D in DE. Axial forces: AB-BC 136.6 kN compression, EG-GH 113.4 kN compression, CD-DE 12 kN compression.

Draw AFD, SFD and BMD of the given frame loaded as shown in figure below. Indicate the salient feature if any. [Figure: frame with pin support at A and at H; left column A-B-C 3 m + 3 m with 15 kN horizontal at B; top beam C-D-E with 60 kN/m UDL; 20 kN·m moment at E, 10 kN downward load at F; right part G with 12 kN horizontal, heights 2 m + 2 m; horizontal dimensions 4 m, 1 m, 2 m.]

Reading of the figure

  • Origin at A: A(0,0) pin, B(0,3), C(0,6); beam C–D(4,6)–E(5,6)–F(7,6); right column from E down: G(5,4), H(5,2).
  • Loads: 15 kN → at B; UDL 60 kN/m on CD (240 kN, 2 m from C); couple 20 kN·m at E (taken clockwise); 10 kN ↓ at F; 12 kN ← at G.
  • With pins at both A and H and no internal hinge the frame would have r = 4 and be indeterminate, so H is taken as a roller (vertical reaction only). Then r = 3 with no hinge, and the frame is determinate.

Reactions

 60 kN/m          20kNm   10kN
 C========D=======E=======F ↓
 |                |
 B→15kN           G←12kN
 |                |
 A (pin)          H (roller)
  • ΣFx: Aₓ + 15 − 12 = 0, so Aₓ = 3.00 kN (←).
  • ΣM_A: 5H_y − 15(3) − 240(2) − 20 − 10(7) + 12(4) = 0, so H_y = 113.40 kN (↑).
  • ΣFy: A_y + H_y − 240 − 10 = 0, so A_y = 136.60 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 = 136.60 kN (compression); V = 3.00 to 3.00 kN; M = 0.00 (–) to 9.00 (right) kN·m, linear.
  • BC (B to C, 3.00 m): N = 136.60 kN (compression); V = −12.00 to −12.00 kN; M = 9.00 (right) to 27.00 (left) kN·m, linear. M changes sign (contraflexure) at 0.75 m above B, from M = 3y − 15(y − 3) = 0 at y = 3.75 m.
  • CD (C to D, 4.00 m): N = 12.00 kN (compression); V = 136.60 to −103.40 kN; M = 27.00 (top) to 39.40 (bottom) kN·m, parabolic. V = 0 at 2.28 m from C, where M = 128.50 kN·m. M changes sign (contraflexure) at 0.21 m from C.
  • DE (D to E, 1.00 m): N = 12.00 kN (compression); V = −103.40 to −103.40 kN; M = 39.40 (bottom) to 64.00 (top) kN·m, linear. M changes sign (contraflexure) at 0.38 m from D.
  • EF (E to F, 2.00 m): N = 0.00 kN (zero); V = 10.00 to 10.00 kN; M = 20.00 (top) to 0.00 (–) kN·m, linear.
  • EG (E to G, 2.00 m): N = 113.40 kN (compression); V = 12.00 to 12.00 kN; M = 24.00 (right) to 0.00 (–) kN·m, linear.
  • GH (G to H, 2.00 m): N = 113.40 kN (compression); V = 0.00 to 0.00 kN; M = 0.00 (–) to 0.00 (–) kN·m, constant.

Salient features

  • Maximum bending moment = 128.50 kN·m (tension at the bottom) at 2.28 m from C in the beam, where the shear is zero.
  • Hogging moments occur at C (27.00 kN·m, tension on the top/outside) and at E (64.00 kN·m, tension on top).
  • Points of contraflexure: 0.75 m above B in the column BC, 0.21 m from C in CD and 0.38 m from D in DE.
  • Joint E check: 64.00 (DE) = 20.00 (EF) + 24.00 (EG) + 20.00 (clockwise couple).
  • Axial force: AB–BC 136.60 kN compression; EG–GH 113.40 kN compression (= H_y); CD–DE 12.00 kN compression (the net horizontal force of 12 kN passed to the right column).

✓ Answer checked by Claude Opus 5.5 (high effort)

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