- 2070 Chaitra · 12 marks
Frame AFD, SFD, BMD (30 and 20 kN/m, hinge)
Short answer
Reactions: Ha=47.25→, Va=93.00↑, He=77.25←, Ve=117.00↑. AB: N=-93, V=-47.25, M linear 0→-378 kN·m left face. BC: triangular load, M cubic to 0 at hinge C. CD: UDL, M parabolic from 0 to -342 top at D. DE: N=-137.71, V=26.31, M linear -342→0. Max BM=378 kN·m at B.
Draw axial force, shear force and bending moment diagram for the frame shown in figure below. [Figure: left column A-B 8 m high with A hinged and a 30 kN horizontal force at B; top member B-C-D with a 30 kN/m triangular load over B to C (6 m), an internal hinge at C, and 20 kN/m UDL over C to D (6 m); inclined member from D down to E, a hinged support 12 m below, 5 m to the right horizontally; horizontal dimensions 6 m, 6 m, 5 m.]
Assumptions (taken from the figure):
- A(0,0) hinged; AB = 8 m; 30 kN horizontal (→) at B. Beam B–C–D: BC = CD = 6 m; hinge at C.
- Triangular load on BC from 0 at B to 30 kN/m at C; 20 kN/m on CD (both downward).
- Inclined member DE: horizontal 5 m, vertical 12 m downward, so E = (17, −4) is the hinged support (length 13 m).
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.
B------------C------------D
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A \
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E
Support reactions
Equilibrium of the whole frame gives , and ; the internal hinge adds the condition (moments of all forces on one side of the hinge), which makes the reactions determinate.
- = 47.25 kN →
- = 93.00 kN ↑
- = 77.25 kN ←
- = 117.00 kN ↑
Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below ).
Axial force, shear force and bending moment at the key sections
| Section | N (kN) | V (kN) | M (kN·m) |
|---|---|---|---|
| AB at A | -93.00 | -47.25 | 0.00 |
| AB at B | -93.00 | -47.25 | -378.00 (left face) |
| BC at B | -77.25 | 93.00 | -378.00 (top) |
| BC at C | -77.25 | 3.00 | 0.00 |
| CD at C | -77.25 | 3.00 | 0.00 |
| CD at D | -77.25 | -117.00 | -342.00 (top) |
| DE at D | -137.71 | 26.31 | -342.00 (right face) |
| DE at E | -137.71 | 26.31 | 0.00 |
Shape of the diagrams
- AB: V constant (steps at point loads), M linear
- BC: V parabolic, M cubic
- CD: V linear, M parabolic
- DE: V constant (steps at point loads), M linear
- AFD: constant between loads on a member, changing only where a load component acts along the member.
- BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.
Salient features
- Maximum bending moment = 378.00 kN·m at B (on AB), tension on the left face
- V = 0 at 0.15 m from C along CD: M = 0.23 (bottom) kN·m (local maximum)
- Point of contraflexure (M = 0) at 0.30 m from C along CD
- Bending moment is zero at the internal hinge (value read from both sides: 0.00).
✓ Answer checked by Claude Opus 5.5 (high effort)
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