- Asked 2/15
- 2071 Shrawan · 14 marks
AFD, SFD, BMD of frame with hinge at C
Short answer
Reactions: A_x = 4.33 kN ←, A_y = 28.22 kN ↑, E_x = 25.67 kN →, E_y = 24.10 kN ↑. Maximum beam moment is 53.94 kN·m at 4.64 m from hinge C. Moment at knee D is 49.35 kN·m (tension inside).
Draw axial force, shear force and bending moment diagram for the given frame. Also indicate salient features if any. [Figure: frame with pin at A and support at E; hinge at C; left column A-B-C, 3 m + 3 m, 10 kN at 60° at B; top beam C-D of 6 m with 5 kN/m UDL; 20 kN at 30° at D; right column D-E of 4 m with triangular load up to 10 kN/m.]
Reading of the figure
- Origin at A: A(0,0) pin, B(0,3), C(0,6) hinge at the top of the left column, D(6,6), E(6,2) pin. Left column 6 m, beam CD 6 m, right column DE 4 m.
- Loads: 10 kN at B making 60° with the column, i.e. 30° below the horizontal pointing right: 8.66 kN → and 5.00 kN ↓; UDL 5 kN/m on CD (30 kN); 20 kN at D making 30° with the vertical, pointing down-left: 10.00 kN ← and 17.32 kN ↓; triangular load on DE, zero at D and 10 kN/m at E, acting to the left (total 20 kN, acting 2.67 m below D).
- r = 4 and the hinge at C gives one extra equation, so the frame is determinate.
Reactions
5 kN/m \ 20 kN
C(hinge)==========D
| |
| | triangular load
B --> 10 kN | (0 at D, 10 kN/m at E)
| (30° below |
| horizontal) |
A (pin) E (pin)
- Hinge C (left part, M_C = 0): 6Aₓ + 8.66(3) = 0, so Aₓ = 4.33 kN (←).
- ΣFx: Aₓ + Eₓ + 8.66 − 10 − 20 = 0, so Eₓ = 25.67 kN (→).
- ΣM_A: 6E_y − 2Eₓ − 8.66(3) − 30(3) − 17.32(6) + 10(6) + 20(3.33) = 0, so E_y = 24.10 kN (↑).
- ΣFy: A_y + E_y − 5 − 30 − 17.32 = 0, so A_y = 28.22 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 = 28.22 kN (compression); V = 4.33 to 4.33 kN; M = 0.00 (–) to 12.99 (right) kN·m, linear.
- BC (B to C, 3.00 m): N = 23.22 kN (compression); V = −4.33 to −4.33 kN; M = 12.99 (right) to 0.00 (–) kN·m, linear.
- CD (C to D, 6.00 m): N = 4.33 kN (compression); V = 23.22 to −6.78 kN; M = 0.00 (–) to 49.35 (bottom) kN·m, parabolic. V = 0 at 4.64 m from C, where M = 53.94 kN·m.
- DE (D to E, 4.00 m): N = 24.10 kN (compression); V = −5.67 to −25.67 kN; M = 49.35 (left) to 0.00 (–) kN·m, cubic.
Salient features
- The maximum moment in the beam is 53.94 kN·m at 4.64 m from the hinge C, where the shear is zero. At C the moment is zero (hinge) and the beam shear there is 23.22 kN.
- The moment at the knee D is 49.35 kN·m (tension on the inside); the right column has its largest moment at D and zero at the pin E.
- The left column has small moments (maximum 12.99 kN·m at B), because the hinge at C makes AC act like a member pinned at both ends (A and C) carrying the 10 kN load.
- The axial forces are large in DE (24.10 kN compression) and AB (28.22 kN compression).
✓ Answer checked by Claude Opus 5.5 (high effort)
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