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AFD, SFD, BMD of frame with UDL and triangular load

Short answer

Reactions: Ax=5 kN right, Ay=40 kN up, MA=56.67 kN·m anticlockwise. Maximum bending moment 86.67 kN·m at knee C (tension outside). Beam CD moment varies parabolically from 86.67 to 6.67 kN·m; DE moment cubic from 6.67 to 0. Axial forces: AC 40 kN compression, CD 10 kN compression, DE zero.

Draw bending moment diagram, shear force diagram and axial force diagram for the given figure below. And also indicate the salient points if any. [Figure: frame with fixed/pin support at A; column A-B-C of 4 m total height with 5 kN horizontal load at B 2 m above A; top beam C-D of 4 m carrying 10 kN/m UDL; right column D-E of 2 m with a triangular load.]

Reading of the figure

  • Origin at A: A(0,0) fixed support (the column ends in a ground hatch with no free end), B(0,2), C(0,4), D(4,4), E(4,2). Column AC is 4 m, beam CD 4 m, and the right column DE 2 m with a free end at E.
  • Loads: 5 kN → at B; UDL 10 kN/m on CD (40 kN, 2 m from C); triangular load on DE acting to the left, 10 kN/m at D and zero at E (total 10 kN, acting 0.67 m below D).
  • E has no support, so the frame is a cantilever frame fixed at A: r = 3, determinate.

Reactions

 10 kN/m
 C================D
 |                 |<- 10 kN/m
 |                 | triangular
 B→5kN             E (free)
 |
 A (fixed)
  • ΣFx: Aₓ + 5 − 10 = 0, so Aₓ = 5.00 kN (→).
  • ΣFy: A_y − 40 = 0, so A_y = 40.00 kN (↑).
  • ΣM_A: M_A − 5(2) − 40(2) + 10(3.33) = 0, so M_A = 56.67 kN·m (anticlockwise).

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, 2.00 m): N = 40.00 kN (compression); V = −5.00 to −5.00 kN; M = 56.67 (left) to 66.67 (left) kN·m, linear.
  • BC (B to C, 2.00 m): N = 40.00 kN (compression); V = −10.00 to −10.00 kN; M = 66.67 (left) to 86.67 (left) kN·m, linear.
  • CD (C to D, 4.00 m): N = 10.00 kN (compression); V = 40.00 to 0.00 kN; M = 86.67 (top) to 6.67 (top) kN·m, parabolic.
  • DE (D to E, 2.00 m): N = 0.00 kN (zero); V = 10.00 to 0.00 kN; M = 6.67 (right) to 0.00 (–) kN·m, cubic.

Salient features

  • Maximum bending moment = 86.67 kN·m at the knee C (tension on the outside, i.e. the left face of the column and the top of the beam). The moment at the fixed base is 56.67 kN·m.
  • The shear in the beam falls linearly from 40.00 kN at C to zero at the corner D (no vertical load acts on DE), so the beam moment varies as a parabola from 86.67 kN·m at C to 6.67 kN·m at D, with zero slope at D.
  • DE acts as a cantilever under the triangular load: the moment falls from 6.67 kN·m at D to zero at E along a cubic curve, and the shear falls from 10 kN at D to zero at E.
  • Axial force: AC 40.00 kN compression; CD 10.00 kN compression (equal to the 10 kN resultant of the triangular load); DE zero.

✓ Answer checked by Claude Opus 5.5 (high effort)

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