Stiffness matrix: element matrices, member forces, BMD

Data and assumptions: $E=2\times10^5$ MPa $=2\times10^8$ kN/m², $A=200$ cm² $=0.02$ m². The printed $I=2.5\times10^5$ mm⁴ would give $EI=50$ kNm² (deflections of several metres), so $I=2.5\times10^5$ cm⁴ $=2.5\times10^{-3}$ m⁴ is used: $EA=4\times10^6$ kN, $EI=5\times10^5$ kNm². Geometry: beam AC = 4 m (2 m + 2 m) with the 600 kN load downward at B (mid-span); A is a hinge (translations fixed, rotation free); column CD = 5 m, fixed at D. Coordinates at C: 1 = horizontal (right +), 2 = vertical (up +), 3 = rotation (anticlockwise +). B is only a load point, so AC is one element and the load is replaced by fixed-end actions.

Step 1: Element stiffness matrices (global axes, kN and m)

Member constants: $EA/L$, $12EI/L^3$, $6EI/L^2$, $4EI/L$, $2EI/L$.

Element 1 (A: dofs $u,v,\theta$; then C: $u,v,\theta$), member axis along global $x$:

$$[k_1]=\begin{bmatrix}1000000 & 0 & 0 & -1000000 & 0 & 0 \\ 0 & 93750 & 187500 & 0 & -93750 & 187500 \\ 0 & 187500 & 500000 & 0 & -187500 & 250000 \\ -1000000 & 0 & 0 & 1000000 & 0 & 0 \\ 0 & -93750 & -187500 & 0 & 93750 & -187500 \\ 0 & 187500 & 250000 & 0 & -187500 & 500000\end{bmatrix}$$

Element 2 (C: $u,v,\theta$; then D: $u,v,\theta$), member axis along global $-y$ (axial terms act in $v$, bending terms in $u$):

$$[k_2]=\begin{bmatrix}48000 & 0 & 120000 & -48000 & 0 & 120000 \\ 0 & 800000 & 0 & 0 & -800000 & 0 \\ 120000 & 0 & 400000 & -120000 & 0 & 200000 \\ -48000 & 0 & -120000 & 48000 & 0 & -120000 \\ 0 & -800000 & 0 & 0 & 800000 & 0 \\ 120000 & 0 & 200000 & -120000 & 0 & 400000\end{bmatrix}$$

Step 2: Fixed-end actions of element 1

600 kN at mid-span: $FEM=\mp\dfrac{PL}{8}=\mp300$ kNm and end shears 300 kN. The fixed-end actions are reversed to give the equivalent joint loads.

Step 3: Structure stiffness matrix at C (hinge A condensed) and solution

Assembling $[k_1]$ and $[k_2]$ and eliminating the free rotation at A ($\theta_A$, using $M_A=0$):

$$[K]_{CC}=\begin{bmatrix}1048000 & 0 & 120000 \\ 0 & 823438 & -93750 \\ 120000 & -93750 & 775000\end{bmatrix},\qquad \{P\}=\begin{Bmatrix}0.00\\-412.50\\450.00\end{Bmatrix}\ (\text{kN, kN, kNm})$$ $$\{d\}=[K]^{-1}\{P\}:\quad u_C=-0.06148\ \text{mm},\ v_C=-0.4398\ \text{mm},\ \theta_C=0.000537\ \text{rad},\ \theta_A=-0.001033\ \text{rad}$$

Step 4: Member end forces (local axes, kN, kNm)

• Element 1 (A to C): axial -61.48 kN (tension +), shear at A 248.15 kN, $M_A=0$; at C: shear 351.85 kN, $M_C=207.41$ kNm (clockwise on the member end).
• Element 2 (C to D): axial -351.85 kN (tension +), shear at C 61.48 kN, $M_C=-207.41$ kNm; at D shear -61.48 kN, $M_D=-100.01$ kNm.

Joint C equilibrium of moments: $207.41+(-207.41)=0$ ✓.

Step 5: Bending moment diagram (kNm)

Reactions: A: $H=61.48$, $V=248.15$, $M=0.00$; D: $H=-61.48$, $V=351.85$, $M=100.01$.

Answer: $u_C=-0.0615$ mm, $v_C=-0.440$ mm, $\theta_C=0.000537$ rad; $M_C=207.41$ kNm, $M_D=-100.01$ kNm, BM under the load 496.30 kNm.