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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 = θB\theta_B, 2 = θC\theta_C (clockwise), 3 = horizontal sway Δ\Delta of the beam (right +). Axial deformation is neglected.

Chord rotations

ψAB=Δ/3\psi_{AB}=\Delta/3 and ψCD=Δ/4\psi_{CD}=\Delta/4.

Stiffness matrix (column by column)

Member stiffnesses: AB (L=3L=3): 4EI3=1.3333EI\dfrac{4EI}{3}=1.3333EI, 2EI3=0.6667EI\dfrac{2EI}{3}=0.6667EI, 6EI9=0.6667EI\dfrac{6EI}{9}=0.6667EI, 12EI27=0.4444EI\dfrac{12EI}{27}=0.4444EI; CD (L=4L=4): EIEI, 0.5EI0.5EI, 0.375EI0.375EI, 0.1875EI0.1875EI; BC (L=4L=4): EIEI, 0.5EI0.5EI.

  • k11=4EI3+4EI4=2.3333EIk_{11}=\dfrac{4EI}{3}+\dfrac{4EI}{4}=2.3333EI, k12=2EI4=0.5000EIk_{12}=\dfrac{2EI}{4}=0.5000EI, k13=−6EI32=−0.6667EIk_{13}=-\dfrac{6EI}{3^2}=-0.6667EI
  • k22=4EI4+4EI4=2.0000EIk_{22}=\dfrac{4EI}{4}+\dfrac{4EI}{4}=2.0000EI, k23=−6EI42=−0.3750EIk_{23}=-\dfrac{6EI}{4^2}=-0.3750EI
  • k33=12EI33+12EI43=0.6319EIk_{33}=\dfrac{12EI}{3^3}+\dfrac{12EI}{4^3}=0.6319EI
[K]=EI[2.33330.5000−0.66670.50002.0000−0.3750−0.6667−0.37500.6319][K]=EI\begin{bmatrix}2.3333 & 0.5000 & -0.6667 \\ 0.5000 & 2.0000 & -0.3750 \\ -0.6667 & -0.3750 & 0.6319\end{bmatrix}

Load vector (for completeness)

Fixed-end moments: BC ∓40×4212=∓53.33\mp\dfrac{40\times4^2}{12}=\mp53.33; AB (60 kN at 2 m from A, L=3L=3): FEMAB=−13.33FEM_{AB}=-13.33, FEMBA=26.67FEM_{BA}=26.67; CD (60 kN to the left at 2 m from C): FEMCD=−30.00FEM_{CD}=-30.00, FEMDC=30.00FEM_{DC}=30.00 kNm. The equivalent loads are

{P}={26.667−23.33314.444}\{P\}=\begin{Bmatrix}26.667\\-23.333\\14.444\end{Bmatrix}

and [K]{d}={P}[K]\{d\}=\{P\} gives

EIθB=26.3388,EIθC=−9.8519,EIΔ1=44.7970EI\theta_{B}=26.3388,\quad EI\theta_{C}=-9.8519,\quad EI\Delta_{1}=44.7970

End moments (kNm, clockwise +)

EndMM (kNm)EndMM (kNm)
AB-25.64BA31.92
BC-31.92CB56.65
CD-56.65DC8.28

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: [K]=EI[2.3330.500−0.6670.5002.000−0.375−0.667−0.3750.6319][K]=EI\begin{bmatrix}2.333&0.500&-0.667\\0.500&2.000&-0.375\\-0.667&-0.375&0.6319\end{bmatrix} (coordinates θB,θC,Δ\theta_B,\theta_C,\Delta).

✓ Answer checked by Claude Opus 5.5 (high effort)

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