Truss: nodal displacements, reactions, member forces

Data: node 1 $(0,0)$ pinned, node 2 $(4,0)$ loaded, node 3 $(0,3)$ pinned. $A=2\times10^{-3}$ m$^2$, $E=2\times10^5$ MPa $=2\times10^8$ kN/m$^2$, so $EA=4\times10^5$ kN. Loads at node 2: $F_x=100$ kN, $F_y=50$ kN. Units: kN, m.

   3 (0,3)
   |\
   | \ 5 m
 3 m  \
   |   \    50 kN
   1----2 -> 100 kN
     4 m

Element data

Direction cosines $c=\dfrac{x_j-x_i}{L}$, $s=\dfrac{y_j-y_i}{L}$; axial stiffness $EA/L$.

Element stiffness in global axes:

$$[k]=\frac{EA}{L}\begin{bmatrix}c^2&cs&-c^2&-cs\\cs&s^2&-cs&-s^2\\-c^2&-cs&c^2&cs\\-cs&-s^2&cs&s^2\end{bmatrix}$$

• Member 1-2: $10^5\begin{bmatrix}1&0&-1&0\\0&0&0&0\\-1&0&1&0\\0&0&0&0\end{bmatrix}$ (DOFs $u_1,v_1,u_2,v_2$)
• Member 2-3: $10^4\begin{bmatrix}5.12&-3.84&-5.12&3.84\\-3.84&2.88&3.84&-2.88\\-5.12&3.84&5.12&-3.84\\3.84&-2.88&-3.84&2.88\end{bmatrix}$ (DOFs $u_2,v_2,u_3,v_3$)
• Member 1-3: $1.3333\times10^5$ in the $v_1,v_3$ terms only.

Assembly and solution

Nodes 1 and 3 are pinned ($u_1=v_1=u_3=v_3=0$); free DOFs are $u_2,v_2$. Adding the node-2 terms of members 1-2 and 2-3:

$$\begin{bmatrix}151\,200&-38\,400\\-38\,400&28\,800\end{bmatrix}\begin{Bmatrix}u_2\\v_2\end{Bmatrix}=\begin{Bmatrix}100\\50\end{Bmatrix}$$ $$u_2=1.6667\ \text{mm}\ (\rightarrow),\qquad v_2=3.9583\ \text{mm}\ (\uparrow)$$

Member forces

$F=\dfrac{EA}{L}\,[-c\ \ -s\ \ c\ \ s]\{d\}$:

• Member 1-2: $166.67$ kN (tension)
• Member 2-3: $-83.33$ kN (compression)
• Member 1-3: $0.00$ kN (zero force, both ends fixed in the vertical direction)

Reactions $\{R\}=[K]\{d\}-\{F\}$

• Node 1: $R_x=-166.67$ kN, $R_y=0.00$ kN
• Node 3: $R_x=66.67$ kN, $R_y=-50.00$ kN

Check: $\Sigma F_x=-166.67+66.67+100=0$; $\Sigma F_y=0-50+50=0$; $\Sigma M_1=50\times4-66.67\times3=0$.

Answer: $u_2=1.667$ mm, $v_2=3.958$ mm; member forces 166.67 kN (T), 83.33 kN (C), 0; reactions $(-166.67,\,0)$ kN at node 1 and $(66.67,\,-50)$ kN at node 3.