- 2068 Chaitra · 8 marks
Stiffness matrix: frame with 40 kN/m, 60 kN
Short answer
Use coordinates theta_B, theta_C, sway Delta; axial deformation neglected. Member stiffnesses from EI/L give K = EI [[2.333, 0.5, -0.667], [0.5, 2.0, -0.375], [-0.667, -0.375, 0.6319]] for (theta_B, theta_C, Delta). Solving gives EI theta_B=26.34, EI theta_C=-9.85, EI Delta=44.80.
Determine the stiffness matrix for the frame shown in figure below. [Figure: portal frame, EI constant; beam of 4 m carrying 40 kN/m UDL; left column 2 m + 1 m (60 kN horizontal at the 1 m level), right column 2 m + 2 m (60 kN horizontal at mid-height, pointing left); both bases fixed.]
Assumptions (figure partly illegible): A(0,0), B(0,3), C(4,3), D(4,-1): the left column is 3 m high (2 m + 1 m) and the right column is 4 m high (2 m + 2 m), with the beam BC = 4 m horizontal at the top; EI is constant; the bases A and D are fixed. Loads: 40 kN/m on BC (downward); 60 kN horizontal (to the right) on AB at 2 m above A; 60 kN horizontal (to the left) at mid-height of CD. Coordinates: 1 = , 2 = (clockwise), 3 = horizontal sway of the beam (right +). Axial deformation is neglected.
Chord rotations
and .
Stiffness matrix (column by column)
Member stiffnesses: AB (): , , , ; CD (): , , , ; BC (): , .
- , ,
- ,
Load vector (for completeness)
Fixed-end moments: BC ; AB (60 kN at 2 m from A, ): , ; CD (60 kN to the left at 2 m from C): , kNm. The equivalent loads are
and gives
End moments (kNm, clockwise +)
| End | (kNm) | End | (kNm) |
|---|---|---|---|
| AB | -25.64 | BA | 31.92 |
| BC | -31.92 | CB | 56.65 |
| CD | -56.65 | DC | 8.28 |
Checks:
- Joint B: (31.92) + (-31.92) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (56.65) + (-56.65) = 0.00 kNm (applied clockwise moment 0.00) ✓
Answer: (coordinates ).
✓ Answer checked by Claude Opus 5.5 (high effort)
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