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AFD, SFD, BMD for frame with internal hinge

Short answer

Reactions: A_x = 6.50 kN ←, A_y = 7.29 kN ↓, F_x = 6.50 kN ←, F_y = 34.79 kN ↑. Maximum hogging moment is 72.48 kN·m at D. Moment is zero at hinge C, supports A and F, and free end E.

Draw the axial force, shear force and bending moment diagram for the given frame shown in figure below. Also show the salient features. [Figure: frame with pin support at A and roller support at F; internal hinge at C; column A-B-C of 3 m + 3 m with 15 kN load at 30° at B; beam C-D-E with triangular load rising to 5 kN/m, 20 kN·m moment at D and 10 kN downward at E; column D/E-F with 1 m dimension; horizontal dimensions 4 m and 2 m.]

Reading of the figure: A is a hinge support at (0, 0). Column A–B–C is 6 m high (3 + 3 m) with an internal hinge at C (top). The beam C–D–E has C–D = 4 m and D–E = 2 m. The right column D–F is 5 m long (F is 1 m above the level of A, at (4, 1)). Loads: 15 kN at B at 30° (down and to the right), a triangular load from zero at C to 5 kN/m at D, a 20 kN·m couple at D (clockwise), and 10 kN downward at E.

Support at F: the figure description calls F a roller, but with the internal hinge at C a roller at F would give only 3 reactions for 4 conditions (3 equilibrium equations + M_C = 0). The column AC could then rotate about A, so the frame would be a mechanism. F is therefore taken as a hinge (pin) support, giving 4 reactions.

Reactions (4 unknowns, 3 equations + hinge condition M_C = 0)

Components: 15 kN at B = (12.99 →, 7.50 ↓). Triangular load = ½ × 5 × 4 = 10 kN at 2.667 m from C. Hinge at C (left part, moments about C): A_x × 6 + 12.99 × 3 = 0 → A_x = −6.50 kN (6.50 kN ←)

ΣFx = 0:   A_x + F_x + 12.99 = 0   ->  F_x = −6.50 kN (6.50 kN ←)
ΣM_A = 0 (anticlockwise +):
   −12.99 × 3  (15 kN, horizontal part at B)  = −38.97
   −10 × 2.667 (triangular load)               = −26.67
   −10 × 6     (10 kN at E)                    = −60.00
   −20         (clockwise couple at D)         = −20.00
   +4 F_y − 1 × F_x = 4 F_y + 6.50             (F is at (4, 1))
   =>  4 F_y = 139.14,  F_y = 34.79 kN ↑
ΣFy = 0:   A_y + 34.79 − 7.50 − 10 − 10 = 0  ->  A_y = −7.29 kN (7.29 kN ↓)

Reactions: A: 6.50 kN ← and 7.29 kN ↓; F: 6.50 kN ← and 34.79 kN ↑.

Member forces

MemberN (kN)V (kN)M (kN·m)
AB+7.29 (tension)6.500 at A → 19.49 at B
BC+14.79 (tension)−6.5019.49 at B → 0 at C (hinge)
CD (triangular load)−6.50 (compression)−14.79 at C → −24.79 at D0 at C → 72.48 at D (hogging); 31.24 at 2 m from C
DE (overhang)0+1020 at D → 0 at E (hogging)
FD (column)−34.79 (compression)6.500 at F → 32.48 at D
  • Hinge forces at C: 6.50 kN horizontal and 14.79 kN vertical (the beam is held by the hinge).
  • Shear in CD: V(x) = −14.79 − 0.625x² (x from C); M(x) = 14.79x + 0.2083x³ (hogging).
  • Joint D check: 72.48 (beam) = 32.48 (column) + 20 (overhang) + 20 (couple) ✓.
BMD (kN·m)
   0 C ______ 31.2 ____ 72.5 D --20-- E 0
   |  (hinge)                 |
 19.5 .. 0                   32.5 
   |                          |
 B 19.5                       |
   |                          |
   0 A                       0 F

Salient features: the bending moment is zero at the internal hinge C, at the supports A and F, and at the free end E; the maximum moment is 72.48 kN·m, hogging at D; the maximum moment in the column AC is 19.49 kN·m at B, where the 15 kN load acts; the shear in the beam does not change sign between C and D; the column AC is in tension (the reaction at A is downward); the 20 kN·m couple at D causes no jump in the shear, only a jump of 20 kN·m in the BMD at D.

✓ Answer checked by Claude Opus 5.5 (high effort)

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