Truss Element (Bar Element in 2D and 3D)
Truss Element in 2D
A typical plane truss element is shown in local and global co-ordinate systems as shown below. In the local co-ordinate system, two nodes are numbered as 1 & 2. The local co-ordinate system will be denoted by $x'$ & $y'$ and the global co-ordinate system as $x$ & $y$ respectively.
In global co-ordinate, each node has two degrees of freedom. Let $u_1, v_1$ and $u_2, v_2$ be the nodal displacements at node 1 & 2 respectively along $x$ & $y$ directions. Let the truss member be inclined with angle $\theta$ with x-axis, as shown in figure.
Clearly, we have
$$u_1' = u_1 \cos\theta + v_1 \sin\theta$$
and
$$u_2' = u_2 \cos\theta + v_2 \sin\theta$$
Displacement Transformation Matrix
Introducing direction cosines: $l = \cos\theta$ & $m = \cos(90^\circ - \theta) = \sin\theta$
The above equations can be written as:
$$u_1' = u_1 \cdot l + v_1 \cdot m$$
$$u_2' = u_2 \cdot l + v_2 \cdot m$$
In matrix form:
$$\begin{Bmatrix} u_1' \\ u_2' \end{Bmatrix} = \begin{bmatrix} l & m & 0 & 0 \\ 0 & 0 & l & m \end{bmatrix} \begin{Bmatrix} u_1 \\ v_1 \\ u_2 \\ v_2 \end{Bmatrix}$$
Displacement Transformation Matrix
$$\Rightarrow \{u'\} = [T]\{u\}$$
where
$$[T] = \begin{bmatrix} l & m & 0 & 0 \\ 0 & 0 & l & m \end{bmatrix}$$
is the displacement transformation matrix, which transforms the four displacement components in global co-ordinate to only two displacement components in local co-ordinate along the length of the bar element/truss element only, i.e., one at each node.
How to Calculate $l$ & $m$
$$l_e = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
$$l = \cos\theta = \frac{x_2 - x_1}{l_e}$$
$$m = \cos(90^\circ - \theta) = \sin\theta = \frac{y_2 - y_1}{l_e}$$
Force Transformation Matrix
Let $P_1'$ & $P_2'$ be the forces in truss member at node 1 & 2 respectively producing displacements $u_1'$ & $u_2'$ in local co-ordinate system, and $P_{1x}, P_{1y}, P_{2x}$ & $P_{2y}$ are the forces in global co-ordinate system at node 1 and 2 producing displacements $u_1, v_1, u_2$ & $v_2$ respectively.
Clearly,
$$P_{1x} = P_1' \cos\theta = P_1' \cdot l$$
$$P_{1y} = P_1' \sin\theta = P_1' \cdot \cos(90^\circ - \theta) = P_1' \cdot m$$
Similarly,
$$P_{2x} = P_2' \cdot l \qquad \text{&} \qquad P_{2y} = P_2' \cdot m$$
In matrix form:
$$\begin{Bmatrix} P_{1x} \\ P_{1y} \\ P_{2x} \\ P_{2y} \end{Bmatrix} = \begin{bmatrix} l & 0 \\ m & 0 \\ 0 & l \\ 0 & m \end{bmatrix} \begin{Bmatrix} P_1' \\ P_2' \end{Bmatrix}$$
Force Transformation Matrix
$$\therefore \{P\} = [T]^T \{P'\}$$
where,
$\{P\}$ = global components of forces at nodes 1 & 2
$\{P'\}$ = local components of forces
$[T]^T$ is the force transformation matrix to find component of global component of forces from local component of forces.
Note: Force transformation matrix is transpose of displacement transformation matrix $[T]$ which transforms global displacement components to local displacement components.
Global Stiffness Matrix for Truss Element in 2D
We know,
$$\{P'\} = [K']\{u'\} \quad --- (i)$$
where,
$\{P'\}$ = column matrix/vector for local component of forces.
$\{u'\}$ = column matrix/vector for local component of displacements.
$[K'] = k \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \frac{AE}{l_e} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$ (element stiffness matrix in local co-ordinate system)
Also,
$$\{u'\} = [T]\{u\} \quad --- (ii)$$
and
$$\{P\} = [T]^T \{P'\} \quad --- (iii)$$
Putting (i) into (iii):
$$\{P\} = [T]^T [K'] \{u'\}$$
$$\Rightarrow \{P\} = [T]^T [K'] [T] \{u\} \quad (\text{from ii})$$
$$\therefore \{P\} = [K] \{u\} \quad --- (iv)$$
Eq. (iv) represents the force-displacement relation in global co-ordinate and hence
$$[K] = [T]^T [K'] [T]$$
represents the stiffness matrix in global co-ordinates.
Derivation of $[K]$
$$[K] = [T]^T [K'] [T]$$
$$= \begin{bmatrix} l & 0 \\ m & 0 \\ 0 & l \\ 0 & m \end{bmatrix}_{4 \times 2} \frac{AE}{l_e} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}_{2 \times 2} \begin{bmatrix} l & m & 0 & 0 \\ 0 & 0 & l & m \end{bmatrix}_{2 \times 4}$$
$$= \frac{AE}{l_e} \begin{bmatrix} l & 0 \\ m & 0 \\ 0 & l \\ 0 & m \end{bmatrix} \begin{bmatrix} l & m & -l & -m \\ -l & -m & l & m \end{bmatrix}$$
$$[K] = \frac{AE}{l_e} \begin{bmatrix} l^2 & lm & -l^2 & -lm \\ lm & m^2 & -lm & -m^2 \\ -l^2 & -lm & l^2 & lm \\ -lm & -m^2 & lm & m^2 \end{bmatrix}$$
Each element of stiffness matrix $[K]$ i.e. $K_{ij}$ in global coordinates represents the forces in x- and y-direction at node i required due to unit displacement at node j.
Stress & Strain Calculation
$$\sigma = \frac{E}{l_e} \begin{bmatrix} -1 & 1 \end{bmatrix} \{u'\}$$
$$= \frac{E}{l_e} \begin{bmatrix} -1 & 1 \end{bmatrix} [T] \{u\}$$
$$= \frac{E}{l_e} \begin{bmatrix} -1 & 1 \end{bmatrix} \begin{bmatrix} l & m & 0 & 0 \\ 0 & 0 & l & m \end{bmatrix} \{u\}$$
$$\therefore \sigma = \frac{E}{l_e} \begin{bmatrix} -l & -m & l & m \end{bmatrix} \{u\}$$
i.e., $\sigma = E[B]\{u\}$
and Strain,
$$\epsilon = \frac{\sigma}{E} = \frac{1}{l_e} \begin{bmatrix} -l & -m & l & m \end{bmatrix} \{u\} = [B]\{u\}$$
Strain-Displacement Matrix
$$[B] = \frac{1}{l_e} \begin{bmatrix} -l & -m & l & m \end{bmatrix}$$
Temperature Effect
Element load vector due to temperature change in local co-ordinate is given by
$$\{f_t'\} = \begin{Bmatrix} f_{t1}' \\ f_{t2}' \end{Bmatrix} = AE\alpha \cdot \Delta t \begin{Bmatrix} -1 \\ 1 \end{Bmatrix}$$
But in global co-ordinate, element load vector is given by
$$\{f_t^e\} = \begin{Bmatrix} f_{t1x} \\ f_{t1y} \\ f_{t2x} \\ f_{t2y} \end{Bmatrix} = AE\alpha \cdot \Delta t \begin{Bmatrix} -\cos\theta \\ -\sin\theta \\ \cos\theta \\ \sin\theta \end{Bmatrix} = AE\alpha \cdot \Delta t \begin{Bmatrix} -l \\ -m \\ l \\ m \end{Bmatrix}$$
Stress in element (with temperature):
$$\sigma = \frac{E}{l_e} \begin{bmatrix} -l & -m & l & m \end{bmatrix} \{u\} - E\alpha \cdot \Delta t$$
$$\therefore \sigma = E[B]\{u\} - E\alpha \cdot \Delta t$$
where, $[B] = \frac{1}{l_e} \begin{bmatrix} -l & -m & l & m \end{bmatrix}$ and $\{u\} = \begin{Bmatrix} u_1 \\ v_1 \\ u_2 \\ v_2 \end{Bmatrix}$
Strain in element (with temperature):
$$\epsilon = \frac{1}{l_e} \begin{bmatrix} -l & -m & l & m \end{bmatrix} \begin{Bmatrix} u_1 \\ v_1 \\ u_2 \\ v_2 \end{Bmatrix} - \alpha \cdot \Delta t$$
$$\therefore \epsilon = [B]\{u\} - \alpha \cdot \Delta t$$
Effect of Lack of Fit
$$\{f_\delta^e\} = \begin{Bmatrix} f_{\delta1x} \\ f_{\delta1y} \\ f_{\delta2x} \\ f_{\delta2y} \end{Bmatrix} = \frac{AE\delta}{l_e} \begin{Bmatrix} -l \\ -m \\ l \\ m \end{Bmatrix}$$
$\delta$ is +ve when element is longer and −ve if element is shorter.
Stress (with lack of fit):
$$\sigma = \frac{E}{l_e} \begin{bmatrix} -l & -m & l & m \end{bmatrix} \{u\} - E \cdot \frac{\delta}{l_e}$$
$$\therefore \sigma = E[B]\{u\} - E \cdot \frac{\delta}{l_e}$$
Strain (with lack of fit):
$$\epsilon = \frac{1}{l_e} \begin{bmatrix} -l & -m & l & m \end{bmatrix} \{u\} - \frac{\delta}{l_e}$$
$$\therefore \epsilon = [B]\{u\} - \frac{\delta}{l_e}$$
Example 1
Two element truss in figure shown below is subjected to loading as shown. Determine the displacement component at node 3, the reaction force components at nodes 1 and 2, and element displacements, stresses and forces. Take $A_1 = A_2 = 1.5 \text{ in}^2$ and $E_1 = E_2 = 10 \times 10^6 \text{ lb/in}^2$.
Solution
For element 1 (1 to 3):
- $\theta = \tan^{-1}\left(\frac{40}{40}\right) = 45^\circ$
- $l = \cos 45^\circ = 0.707$
- $m = \sin 45^\circ = 0.707$
- $l_e = \sqrt{40^2 + 40^2} = 56.568 \text{ inch}$
For element 2 (2 to 3):
- $\theta = 0^\circ$
- $l = \cos 0^\circ = 1$
- $m = \sin 0^\circ = 0$
- $l_e = 40 \text{ inch}$
Force vector:
$$\{F\} = \begin{Bmatrix} f_{1x} \\ f_{1y} \\ f_{2x} \\ f_{2y} \\ f_{3x} \\ f_{3y} \end{Bmatrix} = \begin{Bmatrix} R_{1x} \\ R_{1y} \\ R_{2x} \\ R_{2y} \\ 500 \\ 300 \end{Bmatrix} \text{ lb}$$
Displacement vector:
$$\{u\} = \begin{Bmatrix} u_1 \\ v_1 \\ u_2 \\ v_2 \\ u_3 \\ v_3 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ u_3 \\ v_3 \end{Bmatrix}$$
Stiffness Matrix for Element 1:
$$[K^1] = \frac{A_1 E_1}{l_{e1}} \begin{bmatrix} l^2 & lm & -l^2 & -lm \\ lm & m^2 & -lm & -m^2 \\ -l^2 & -lm & l^2 & lm \\ -lm & -m^2 & lm & m^2 \end{bmatrix} \begin{matrix} u_1 \\ v_1 \\ u_3 \\ v_3 \end{matrix}$$
$$= \frac{1.5 \times 10 \times 10^6}{56.568} \begin{bmatrix} (0.707)^2 & (0.707)^2 & -(0.707)^2 & -(0.707)^2 \\ (0.707)^2 & (0.707)^2 & -(0.707)^2 & -(0.707)^2 \\ -(0.707)^2 & -(0.707)^2 & (0.707)^2 & (0.707)^2 \\ -(0.707)^2 & -(0.707)^2 & (0.707)^2 & (0.707)^2 \end{bmatrix}$$
$$= 2.65165 \times 10^5 \begin{bmatrix} 0.5 & 0.5 & -0.5 & -0.5 \\ 0.5 & 0.5 & -0.5 & -0.5 \\ -0.5 & -0.5 & 0.5 & 0.5 \\ -0.5 & -0.5 & 0.5 & 0.5 \end{bmatrix}$$
$$[K^1] = 10^5 \begin{bmatrix} 1.326 & 1.326 & -1.326 & -1.326 \\ 1.326 & 1.326 & -1.326 & -1.326 \\ -1.326 & -1.326 & 1.326 & 1.326 \\ -1.326 & -1.326 & 1.326 & 1.326 \end{bmatrix} \begin{matrix} u_1 \\ v_1 \\ u_3 \\ v_3 \end{matrix}$$
Stiffness Matrix for Element 2:
$$[K^2] = \frac{A_2 E_2}{l_{e2}} \begin{bmatrix} l^2 & lm & -l^2 & -lm \\ lm & m^2 & -lm & -m^2 \\ -l^2 & -lm & l^2 & lm \\ -lm & -m^2 & lm & m^2 \end{bmatrix} \begin{matrix} u_2 \\ v_2 \\ u_3 \\ v_3 \end{matrix}$$
$$= \frac{1.5 \times 10 \times 10^6}{40} \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
$$[K^2] = 10^5 \begin{bmatrix} 3.75 & 0 & -3.75 & 0 \\ 0 & 0 & 0 & 0 \\ -3.75 & 0 & 3.75 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{matrix} u_2 \\ v_2 \\ u_3 \\ v_3 \end{matrix}$$
Global Stiffness Matrix:
$[K] = [K^1] + [K^2]$
$$[K] = 10^5 \begin{bmatrix} 1.326 & 1.326 & 0 & 0 & -1.326 & -1.326 \\ 1.326 & 1.326 & 0 & 0 & -1.326 & -1.326 \\ 0 & 0 & 3.75 & 0 & -3.75 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ -1.326 & -1.326 & -3.75 & 0 & 5.076 & 1.326 \\ -1.326 & -1.326 & 0 & 0 & 1.326 & 1.326 \end{bmatrix} \begin{matrix} u_1 \\ v_1 \\ u_2 \\ v_2 \\ u_3 \\ v_3 \end{matrix}$$
Note: $K_{55} = 1.326 + 3.75 = 5.076$
Force-Displacement Relationship:
$\{F\} = [K]\{u\}$
$$\begin{Bmatrix} R_{1x} \\ R_{1y} \\ R_{2x} \\ R_{2y} \\ 500 \\ 300 \end{Bmatrix} = 10^5 \begin{bmatrix} 1.326 & 1.326 & 0 & 0 & -1.326 & -1.326 \\ 1.326 & 1.326 & 0 & 0 & -1.326 & -1.326 \\ 0 & 0 & 3.75 & 0 & -3.75 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ -1.326 & -1.326 & -3.75 & 0 & 5.076 & 1.326 \\ -1.326 & -1.326 & 0 & 0 & 1.326 & 1.326 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ u_3 \\ v_3 \end{Bmatrix}$$
After elimination (since $u_1 = v_1 = u_2 = v_2 = 0$, only rows/columns 5 & 6 remain); reduced matrix relation is:
$$\begin{Bmatrix} 500 \\ 300 \end{Bmatrix} = 10^5 \begin{bmatrix} 5.076 & 1.326 \\ 1.326 & 1.326 \end{bmatrix} \begin{Bmatrix} u_3 \\ v_3 \end{Bmatrix}$$
Solving the 2×2 system using matrix inversion: $\{u\} = [K_r]^{-1}\{F_r\}$.
Determinant: $|K_r| = (10^5)^2 \times (5.076 \times 1.326 - 1.326 \times 1.326) = 10^{10} \times (6.7308 - 1.7583) = 4.9725 \times 10^{10}$
$$[K_r]^{-1} = \frac{1}{4.9725 \times 10^{10}} \begin{bmatrix} 1.326 \times 10^5 & -1.326 \times 10^5 \\ -1.326 \times 10^5 & 5.076 \times 10^5 \end{bmatrix}$$
$$u_3 = \frac{1}{4.9725 \times 10^{10}} \left(1.326 \times 10^5 \times 500 + (-1.326 \times 10^5) \times 300\right)$$
$$= \frac{1.326 \times 10^5 \times 200}{4.9725 \times 10^{10}} = \frac{2.652 \times 10^7}{4.9725 \times 10^{10}} = 5.333 \times 10^{-4} \text{ in.}$$
$$v_3 = \frac{1}{4.9725 \times 10^{10}} \left(-1.326 \times 10^5 \times 500 + 5.076 \times 10^5 \times 300\right)$$
$$= \frac{-6.63 \times 10^7 + 15.228 \times 10^7}{4.9725 \times 10^{10}} = \frac{8.598 \times 10^7}{4.9725 \times 10^{10}} = 1.729 \times 10^{-3} \text{ in.}$$
$$\begin{Bmatrix} u_3 \\ v_3 \end{Bmatrix} = \begin{Bmatrix} 5.333 \times 10^{-4} \\ 1.729 \times 10^{-3} \end{Bmatrix} \text{ in.}$$
Reaction Forces:
Back-substituting the computed displacements into the full equation $\{F\} = [K]\{u\}$:
Row 1 ($R_{1x}$): $R_{1x} = 10^5 \times (-1.326 \times u_3 + (-1.326) \times v_3)$
$$= 10^5 \times (-1.326)(5.333 \times 10^{-4} + 1.729 \times 10^{-3}) = 10^5 \times (-1.326)(2.2623 \times 10^{-3}) = -300.0 \text{ lb}$$
Row 2 ($R_{1y}$): $R_{1y} = 10^5 \times (-1.326 \times u_3 + (-1.326) \times v_3)$
$$= 10^5 \times (-1.326)(2.2623 \times 10^{-3}) = -300.0 \text{ lb}$$
Row 3 ($R_{2x}$): $R_{2x} = 10^5 \times (-3.75 \times u_3 + 0 \times v_3)$
$$= 10^5 \times (-3.75)(5.333 \times 10^{-4}) = -200.0 \text{ lb}$$
Row 4 ($R_{2y}$): $R_{2y} = 10^5 \times (0 \times u_3 + 0 \times v_3) = 0$
$$R_{1x} = -300 \text{ lb}, \quad R_{1y} = -300 \text{ lb}, \quad R_{2x} = -200 \text{ lb}, \quad R_{2y} = 0$$
Equilibrium check:
$\sum F_x = R_{1x} + R_{2x} + 500 = -300 - 200 + 500 = 0$ $\checkmark$
$\sum F_y = R_{1y} + R_{2y} + 300 = -300 + 0 + 300 = 0$ $\checkmark$
Element Displacement in Local Co-ordinates:
$\{u'\} = [T]\{u\}$
For element 1:
$$\begin{Bmatrix} u_1' \\ u_3' \end{Bmatrix} = \begin{bmatrix} 0.707 & 0.707 & 0 & 0 \\ 0 & 0 & 0.707 & 0.707 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 5.333 \times 10^{-4} \\ 1.729 \times 10^{-3} \end{Bmatrix} = \begin{Bmatrix} 0 \\ 1.6 \times 10^{-3} \end{Bmatrix} \text{ inch}$$
For element 2:
$$\begin{Bmatrix} u_2' \\ u_3' \end{Bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 5.333 \times 10^{-4} \\ 1.729 \times 10^{-3} \end{Bmatrix} = \begin{Bmatrix} 0 \\ 5.333 \times 10^{-4} \end{Bmatrix} \text{ inch}$$
Element Stress:
Using $\sigma = \frac{E}{l_e} \begin{bmatrix} -l & -m & l & m \end{bmatrix} \{u\}$:
For element 1 ($l = m = 0.707$, $l_e = 56.568$):
$$\sigma_1 = \frac{10 \times 10^6}{56.568} \begin{bmatrix} -0.707 & -0.707 & 0.707 & 0.707 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 5.333 \times 10^{-4} \\ 1.729 \times 10^{-3} \end{Bmatrix}$$
$$= 1.7678 \times 10^5 \times [0 + 0 + 0.707 \times 5.333 \times 10^{-4} + 0.707 \times 1.729 \times 10^{-3}]$$
$$= 1.7678 \times 10^5 \times [3.7704 \times 10^{-4} + 1.2224 \times 10^{-3}]$$
$$= 1.7678 \times 10^5 \times 1.5994 \times 10^{-3}$$
$$\sigma_1 = 282.75 \text{ lb/in}^2 \text{ (Tensile)}$$
For element 2 ($l = 1, m = 0$, $l_e = 40$):
$$\sigma_2 = \frac{10 \times 10^6}{40} \begin{bmatrix} -1 & 0 & 1 & 0 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 5.333 \times 10^{-4} \\ 1.729 \times 10^{-3} \end{Bmatrix}$$
$$= 2.5 \times 10^5 \times [0 + 0 + 1 \times 5.333 \times 10^{-4} + 0] = 2.5 \times 10^5 \times 5.333 \times 10^{-4}$$
$$\sigma_2 = 133.325 \text{ lb/in}^2 \text{ (Tensile)}$$
Element Forces:
$$F_1 = A_1 \times \sigma_1 = 1.5 \times 282.75 = 424.125 \text{ lb}$$
$$F_2 = A_2 \times \sigma_2 = 1.5 \times 133.325 = 200 \text{ lb}$$
Example 2
Determine the nodal displacements, reaction forces and member forces of the given truss structure. Given that $A = 2 \times 10^{-3} \text{ m}^2$ & $E = 2 \times 10^5 \text{ MPa}$ for each member.
Solution
Force Vector:
$$\{F\} = \begin{Bmatrix} f_{1x} \\ f_{1y} \\ f_{2x} \\ f_{2y} \\ f_{3x} \\ f_{3y} \end{Bmatrix} = \begin{Bmatrix} R_{1x} \\ R_{1y} \\ R_{2x} \\ R_{2y} \\ 100 \\ 50 \end{Bmatrix} \text{ kN}$$
Displacement vector:
$$\{u\} = \begin{Bmatrix} u_1 \\ v_1 \\ u_2 \\ v_2 \\ u_3 \\ v_3 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ u_3 \\ v_3 \end{Bmatrix}$$
Element 1 (2 to 1):
$\theta = 90^\circ$, $l = \cos 90^\circ = 0$, $m = \sin 90^\circ = 1$, $l_e = 3\text{m}$
$AE = 2 \times 10^{-3} \text{ m}^2 \times 2 \times 10^5 \times 10^6 \text{ N/m}^2 = 4 \times 10^8 \text{ N}$
$$[K^1] = \frac{4 \times 10^8}{3} \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 1 \end{bmatrix}$$
$$[K^1] = 4 \times 10^8 \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0.333 & 0 & -0.333 \\ 0 & 0 & 0 & 0 \\ 0 & -0.333 & 0 & 0.333 \end{bmatrix} \text{ N/m} \quad \begin{matrix} u_2 \\ v_2 \\ u_1 \\ v_1 \end{matrix}$$
Element 2 (2 to 3):
$\theta = 0^\circ$, $l = \cos 0^\circ = 1$, $m = \sin 0^\circ = 0$, $l_e = 4\text{m}$
$$[K^2] = 4 \times 10^8 \begin{bmatrix} 0.25 & 0 & -0.25 & 0 \\ 0 & 0 & 0 & 0 \\ -0.25 & 0 & 0.25 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \text{ N/m} \quad \begin{matrix} u_2 \\ v_2 \\ u_3 \\ v_3 \end{matrix}$$
Element 3 (1 to 3):
$\theta = 360^\circ - \tan^{-1}(3/4) = 323.13^\circ$
$l = \cos(323.13^\circ) = 0.8$, $m = \sin(323.13^\circ) = -0.6$, $l_e = \sqrt{3^2 + 4^2} = 5\text{m}$
$$[K^3] = \frac{4 \times 10^8}{5} \begin{bmatrix} 0.64 & -0.48 & -0.64 & 0.48 \\ -0.48 & 0.36 & 0.48 & -0.36 \\ -0.64 & 0.48 & 0.64 & -0.48 \\ 0.48 & -0.36 & -0.48 & 0.36 \end{bmatrix}$$
$$[K^3] = 4 \times 10^8 \begin{bmatrix} 0.128 & -0.096 & -0.128 & 0.096 \\ -0.096 & 0.072 & 0.096 & -0.072 \\ -0.128 & 0.096 & 0.128 & -0.096 \\ 0.096 & -0.072 & -0.096 & 0.072 \end{bmatrix} \text{ N/m} \quad \begin{matrix} u_1 \\ v_1 \\ u_3 \\ v_3 \end{matrix}$$
Global Stiffness Matrix:
$[K] = [K^1] + [K^2] + [K^3]$
$$[K] = 4 \times 10^8 \begin{bmatrix} 0.128 & -0.096 & 0 & 0 & -0.128 & 0.096 \\ -0.096 & 0.405 & 0 & -0.333 & 0.096 & -0.072 \\ 0 & 0 & 0.25 & 0 & -0.25 & 0 \\ 0 & -0.333 & 0 & 0.333 & 0 & 0 \\ -0.128 & 0.096 & -0.25 & 0 & 0.378 & -0.096 \\ 0.096 & -0.072 & 0 & 0 & -0.096 & 0.072 \end{bmatrix} \begin{matrix} u_1 \\ v_1 \\ u_2 \\ v_2 \\ u_3 \\ v_3 \end{matrix} \text{ N/m}$$
Force-Displacement Relationship:
$\{F\} = [K]\{u\}$
$$\begin{Bmatrix} R_{1x} \\ R_{1y} \\ R_{2x} \\ R_{2y} \\ 100 \\ 50 \end{Bmatrix} = 4 \times 10^8 \begin{bmatrix} 0.128 & -0.096 & 0 & 0 & -0.128 & 0.096 \\ -0.096 & 0.405 & 0 & -0.333 & 0.096 & -0.072 \\ 0 & 0 & 0.25 & 0 & -0.25 & 0 \\ 0 & -0.333 & 0 & 0.333 & 0 & 0 \\ -0.128 & 0.096 & -0.25 & 0 & 0.378 & -0.096 \\ 0.096 & -0.072 & 0 & 0 & -0.096 & 0.072 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ u_3 \\ v_3 \end{Bmatrix}$$
After elimination (since $u_1 = v_1 = u_2 = v_2 = 0$, only rows/columns for $u_3$ & $v_3$ remain):
$$\begin{Bmatrix} 100 \\ 50 \end{Bmatrix} = 4 \times 10^5 \begin{bmatrix} 0.378 & -0.096 \\ -0.096 & 0.072 \end{bmatrix} \begin{Bmatrix} u_3 \\ v_3 \end{Bmatrix} \text{ (Converting } 4\times 10^8 \text{ N/m to } 4\times 10^5 \text{ kN/m)}$$
Solving the 2×2 system using $\{u\} = [K_r]^{-1}\{F_r\}$:
Determinant: $|K_r| = (4 \times 10^5)^2 (0.378 \times 0.072 - (-0.096)(-0.096))$
$$= 16 \times 10^{10} \times (0.027216 - 0.009216) = 16 \times 10^{10} \times 0.018 = 2.88 \times 10^9$$
$$[K_r]^{-1} = \frac{1}{2.88 \times 10^9} \begin{bmatrix} 0.072 \times 4 \times 10^5 & 0.096 \times 4 \times 10^5 \\ 0.096 \times 4 \times 10^5 & 0.378 \times 4 \times 10^5 \end{bmatrix} = \frac{1}{2.88 \times 10^9} \begin{bmatrix} 28800 & 38400 \\ 38400 & 151200 \end{bmatrix}$$
Therefore:
$$u_3 = \frac{28800 \times 100 + 38400 \times 50}{2.88 \times 10^9} = \frac{2{,}880{,}000 + 1{,}920{,}000}{2.88 \times 10^9} = \frac{4{,}800{,}000}{2.88 \times 10^9} = 1.6667 \times 10^{-3} \text{ m}$$
$$v_3 = \frac{38400 \times 100 + 151200 \times 50}{2.88 \times 10^9} = \frac{3{,}840{,}000 + 7{,}560{,}000}{2.88 \times 10^9} = \frac{11{,}400{,}000}{2.88 \times 10^9} = 3.9583 \times 10^{-3} \text{ m}$$
$$\begin{Bmatrix} u_3 \\ v_3 \end{Bmatrix} = \begin{Bmatrix} 1.6667 \times 10^{-3} \\ 3.9583 \times 10^{-3} \end{Bmatrix} \text{ m} = \begin{Bmatrix} 1.667 \\ 3.958 \end{Bmatrix} \text{ mm}$$
Reaction Forces:
Back-substituting into the full equation $\{F\} = [K]\{u\}$. Since only $u_3$ and $v_3$ are non-zero, we use the corresponding columns 5 & 6 of $[K]$:
Row 1 ($R_{1x}$): $R_{1x} = 4 \times 10^5 \times (-0.128 \times u_3 + 0.096 \times v_3)$
$$= 4 \times 10^5 \times \{-0.128 \times 1.6667 \times 10^{-3} + 0.096 \times 3.9583 \times 10^{-3}\}$$
$$= 4 \times 10^5 \times \{-2.1333 \times 10^{-4} + 3.8 \times 10^{-4}\} = 4 \times 10^5 \times 1.6667 \times 10^{-4} = 66.667 \text{ kN}$$
Row 2 ($R_{1y}$): $R_{1y} = 4 \times 10^5 \times (0.096 \times u_3 + (-0.072) \times v_3)$
$$= 4 \times 10^5 \times \{0.096 \times 1.6667 \times 10^{-3} - 0.072 \times 3.9583 \times 10^{-3}\}$$
$$= 4 \times 10^5 \times \{1.6 \times 10^{-4} - 2.85 \times 10^{-4}\} = 4 \times 10^5 \times (-1.25 \times 10^{-4}) = -50 \text{ kN}$$
Row 3 ($R_{2x}$): $R_{2x} = 4 \times 10^5 \times (-0.25 \times u_3 + 0 \times v_3)$
$$= 4 \times 10^5 \times (-0.25 \times 1.6667 \times 10^{-3}) = 4 \times 10^5 \times (-4.1667 \times 10^{-4}) = -166.667 \text{ kN}$$
Row 4 ($R_{2y}$): $R_{2y} = 4 \times 10^5 \times (-0.333 \times v_1 + 0.333 \times v_2)$. Since $v_1 = v_2 = 0$:
$$R_{2y} = 0 \text{ kN}$$
$$R_{1x} = 66.667 \text{ kN}, \quad R_{1y} = -50 \text{ kN}, \quad R_{2x} = -166.667 \text{ kN}, \quad R_{2y} = 0 \text{ kN}$$
Equilibrium check:
$\sum F_x = R_{1x} + R_{2x} + 100 = 66.667 - 166.667 + 100 = 0$ $\checkmark$
$\sum F_y = R_{1y} + R_{2y} + 50 = -50 + 0 + 50 = 0$ $\checkmark$
Member Forces:
Using the formula: $f_e = \frac{EA}{l_e} \begin{bmatrix} -l & -m & l & m \end{bmatrix} \{u\}$ (positive = tension, negative = compression).
Element 1 (nodes 2→1): $l=0, m=1, l_e=3$ m, $\frac{EA}{l_e} = \frac{4\times 10^5}{3}$ kN/m
Node displacements: $u_2 = 0, v_2 = 0, u_1 = 0, v_1 = 0$
$$f_1 = \frac{4 \times 10^5}{3} \begin{bmatrix} 0 & -1 & 0 & 1 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{Bmatrix} = 0$$
$$f_1 = 0 \text{ kN}$$ (No force — nodes 1 and 2 have zero displacement)
Element 2 (nodes 2→3): $l=1, m=0, l_e=4$ m, $\frac{EA}{l_e} = \frac{4\times 10^5}{4} = 1 \times 10^5$ kN/m
Node displacements: $u_2 = 0, v_2 = 0, u_3 = 1.6667\times10^{-3}, v_3 = 3.9583\times10^{-3}$
$$f_2 = 1 \times 10^5 \begin{bmatrix} -1 & 0 & 1 & 0 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 1.6667 \times 10^{-3} \\ 3.9583 \times 10^{-3} \end{Bmatrix}$$
$$= 10^5 \times (0 + 0 + 1.6667 \times 10^{-3} + 0) = 10^5 \times 1.6667 \times 10^{-3}$$
$$f_2 = 166.667 \text{ kN (Tensile)}$$
Element 3 (nodes 1→3): $l=0.8, m=-0.6, l_e=5$ m, $\frac{EA}{l_e} = \frac{4\times 10^5}{5} = 0.8 \times 10^5$ kN/m
Node displacements: $u_1 = 0, v_1 = 0, u_3 = 1.6667\times10^{-3}, v_3 = 3.9583\times10^{-3}$
$$f_3 = 0.8 \times 10^5 \begin{bmatrix} -0.8 & 0.6 & 0.8 & -0.6 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 1.6667 \times 10^{-3} \\ 3.9583 \times 10^{-3} \end{Bmatrix}$$
$$= 0.8 \times 10^5 \times \{0 + 0 + 0.8 \times 1.6667 \times 10^{-3} + (-0.6) \times 3.9583 \times 10^{-3}\}$$
$$= 0.8 \times 10^5 \times \{1.3333 \times 10^{-3} - 2.375 \times 10^{-3}\} = 0.8 \times 10^5 \times (-1.0417 \times 10^{-3})$$
$$f_3 = -83.333 \text{ kN} = 83.333 \text{ kN (Compression)}$$