- 2078 Bhadra · 16 marks
Frame AFD, SFD, BMD (20 kNm, 15 kN)
Short answer
Reactions: H_A=39.67 kN←, V_A=39.00 kN↓; H_G=14.00 kN←, V_G=79.00 kN↑. Maximum bending moment 107.99 kN·m at C (top tension). On AB, local moment maximum 90.69 kN·m at 5.01 m from A. Contraflexure point on BC at 2 m from B. Axial forces constant on each member.
Draw axial force, shear force and bending moment diagram of given loaded frame. Also show the salient feature. [Figure: left column A-B, A hinged at the base, carrying a horizontal distributed load varying from 10 kN/m at A to 5 kN/m at B; beam B-C with a 5 kN/m UDL; a 20 kNm couple at C; right column from C down to a hinged support G; a 10 kN force at acting on the right column; a 15 kN downward force at the end E of the beam extension; horizontal dimensions 4 m and 2 m; vertical dimensions 3 m and 3 m on the right.]
Assumptions (taken from the figure):
- A(0,0) hinged; AB = 6 m; beam B–C = 4 m; column C–G = 6 m with G hinged; overhang C–E = 2 m carrying 15 kN downward at E.
- Load on AB: horizontal, varying from 10 kN/m at A to 5 kN/m at B (acting →); 5 kN/m downward on BC; 20 kN·m couple at C (clockwise); 10 kN at 30° to the horizontal acts at mid-height of CG, to the right and downward.
- Two hinged supports with no internal hinge are indeterminate, so a hinge is assumed at the mid-span of BC.
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
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A G
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.
- = 39.67 kN ←
- = 39.00 kN ↓
- = 14.00 kN ←
- = 79.00 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 | 39.00 | 39.67 | 0.00 |
| AB at B | 39.00 | -5.33 | 87.99 (right face) |
| BC at B | -5.33 | -39.00 | 87.99 (bottom) |
| BC at C | -5.33 | -59.00 | -107.99 (top) |
| CG at C | -74.00 | 5.33 | -57.99 (right face) |
| CG at G | -79.00 | 14.00 | 0.00 |
Shape of the diagrams
- AB: V parabolic, M cubic
- BC: V linear, M parabolic
- CG: 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 = 107.99 kN·m at C (on BC), tension on the top
- V = 0 at 5.01 m from A along AB: M = 90.69 (right face) kN·m (local maximum)
- Point of contraflexure (M = 0) at 2.00 m from B along BC
- 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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