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Frame AFD, SFD, BMD (inclined column, triangular)

Short answer

Assuming hinges at B and C for determinacy, reactions: H_A=0, V_A=50 kN up, H_D=20 kN left, V_D=100 kN up, M_D=200 kN·m. Diagrams show max bending moment 200.0 kN·m at D (tension left face of CD); zero at hinges B, C; AB peaks 50.0 kN·m under 20 kN load; BC local max 192.45 kN·m at 5.77 m from B.

Draw axial force, shear force and bending moment diagram of the frame shown in figure below. [Figure: frame with an inclined left leg A-B (A hinged at the base, 5 m of rise to the level of the 20 kN horizontal force, total height 10 m to B), a 20 kN horizontal force acting on the inclined leg at 5 m height, top member B-C of 10 m carrying a triangular load increasing from zero at B to 30 kN/m at C, right column C-D 10 m high with D fixed; horizontal dimensions 2 m and 10 m.]

Assumptions (taken from the figure):

  • Inclined leg A(0,0) to B(2,10) (length 10.2 m), A hinged; 20 kN horizontal (→) at mid-length (5 m height).
  • Beam B–C = 10 m, triangular load 0 at B to 30 kN/m at C (total 150 kN downward). Column CD = 10 m, D fixed.
  • With A hinged, D fixed and no internal hinge, the frame is indeterminate (5 reactions), so hinges are assumed at B and C to make it 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.

     Bo------------------------Co
     |                         |
    |                          |
    |                          |
   ||                          |
   |                           |
  ||                           |
  |                            |
  |                            |
  |                            |
 |                             |
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A                              D

Support reactions

Hinge at B (forces on AB): −2VA+10HA+20(5)=0-2V_A + 10H_A + 20(5) = 0. Hinge at C (forces on AB and BC): −12VA+10HA+20(5)+150(3.33)=0-12V_A + 10H_A + 20(5) + 150(3.33) = 0. Together with the equilibrium of the whole frame:

  • HAH_A = 0.00 kN
  • VAV_A = 50.00 kN ↑
  • HDH_D = 20.00 kN ←
  • VDV_D = 100.00 kN ↑
  • MDM_D = 200.00 kN·m (anticlockwise)

Check: the three equilibrium equations of the complete frame and both hinge conditions are satisfied.

Axial force, shear force and bending moment at the key sections

SectionN (kN)V (kN)M (kN·m)
AB at A-49.039.810.00
AB at 20 kN load (mid-length)-49.03 / -52.959.81 / -9.8150.00 (right face)
AB at B-52.95-9.810.00
BC at B-20.0050.000.00
BC at C-20.00-100.000.00
CD at C-100.0020.000.00
CD at D-100.0020.00200.00 (left face)

Shape of the diagrams

  • AB: V constant on each half (it changes sign at the 20 kN load), M linear on each half (triangular, peak 50 kN·m at mid-length)
  • BC: V parabolic, M cubic (a simply supported span between the hinges B and C)
  • CD: V constant, M linear
  • AFD: constant between loads on a member, changing only where a load component acts along the member.
  • The BMD is drawn on the tension side of each member and is zero at hinges and at free ends.

Salient features

  • Maximum bending moment = 200.00 kN·m at D (on CD), tension on the left face
  • V = 0 at 5.77 m from B along BC: M = 192.45 kN·m (bottom), the local maximum (=wL2/93= wL^2/9\sqrt3)
  • On AB the moment rises from 0 at A to 9.81×5.10=50.09.81\times5.10 = 50.0 kN·m under the 20 kN load and falls back to 0 at B (tension on the right, inner face)
  • The bending moment is zero at the hinges B and C.

✓ Answer checked by Claude Opus 5.5 (high effort)

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