Truss: displacement method (stiffness matrix)

Setup: A is on the wall at the top and C is on the wall 3 m below A; B is 4 m from the wall, so AB is horizontal (4 m) and CB is inclined (3-4-5 triangle, length 5 m). The loads at B are 100 kN horizontal (to the right) and 50 kN vertical (downward). The axial stiffness $EA/L$ of each member is 400 kN/cm $=40\,000$ kN/m. The only free joint is B, with two displacements: $u$ (horizontal) and $v$ (vertical).

Member stiffness in global coordinates

For a bar with direction cosines $(c,s)$ taken from B towards the support, its contribution to the joint stiffness is $k\begin{bmatrix}c^2&cs\\cs&s^2\end{bmatrix}$.

Structure stiffness matrix and displacements

$$[K]=40000\begin{bmatrix}1.64&0.48\\0.48&0.36\end{bmatrix}=\begin{bmatrix}65600&19200\\19200&14400\end{bmatrix}$$ $$\begin{bmatrix}65600&19200\\19200&14400\end{bmatrix}\begin{Bmatrix}u\\v\end{Bmatrix}=\begin{Bmatrix}100\\-50\end{Bmatrix} \;\Rightarrow\; u=0.4167\ \text{cm},\quad v=-0.9028\ \text{cm}$$

Member forces

Elongation $\delta=-(uc+vs)$, force $F=k\delta$:

• AB: $\delta=u=0.4167$ cm $\Rightarrow F_{AB}=400\times0.4167=166.67$ kN (tension)
• CB: $\delta=-(u(-0.8)+v(-0.6))=-0.2083$ cm $\Rightarrow F_{CB}=-83.33$ kN (compression)

Equilibrium of joint B (AB in tension pulls B towards A; CB in compression pushes B away from C): horizontal $100-166.67+0.8\times83.33=0.000$ ✓; vertical $-50+0.6\times83.33=0.000$ ✓.

Answer: $F_{AB}=166.67$ kN (tension), $F_{CB}=-83.33$ kN (compression); joint B displaces 0.4167 cm horizontally and -0.9028 cm vertically.