Chapter 14, Problem 14.1P

To determine

The structure stiffness matrix for the truss.

Answer to Problem 14.1P

The structure stiffness matrix for the truss is shown below.

$k=AE\left[\begin{array}{cccccccc}0.3536& -0.3536& -0.3536& 0.3536& 0& 0& 0& 0\\ -0.3536& 0.8536& 0.3536& -0.3536& 0& 0& 0& -0.5\\ -0.3536& 0.3536& 1.0608& -0.3536& -0.3536& 0.3536& -0.3536& -0.3536\\ 0.3536& -0.3536& -0.3536& 1.0608& 0.3536& -0.3536& -0.3536& -0.3536\\ 0& 0& -0.3536& 0.3536& 0.8536& -0.3536& -0.5& 0\\ 0& 0& 0.3536& -0.3536& -0.3536& 0.3536& 0& 0\\ 0& 0& -0.3536& -0.3536& -0.5& 0& 0.8536& 0.3536\\ 0& -0.5& -0.3536& -0.3536& 0& 0& 0.3536& 0.8536\end{array}\right]$

Explanation of Solution

Concept Used:

Write the expression for direction cosine in x-direction.

${\lambda }_x=\frac{x_F-x_N}{L}$          ...... (I)

Here, coordinate of $x$ at near end of the member is $x_F$ and coordinate of $x$ at far end of the member is $x_N$.

Write the expression for direction cosine in x-direction.

${\lambda }_y=\frac{y_F-y_N}{L}$          ...... (II)

Here, coordinate of $y$ at near end of the member is $y_F$ and coordinate of $y$ at far end of the member is $y_N$.

Write the stiffness matrix for the member.

$\frac{AE}{L}\left[\begin{array}{cccc}{\lambda }_x{}^2& {\lambda }_x{\lambda }_y& -{\lambda }_x{}^2& -{\lambda }_x{\lambda }_y\\ {\lambda }_x{\lambda }_y& {\lambda }_y{}^2& -{\lambda }_x{\lambda }_y& -{\lambda }_y{}^2\\ -{\lambda }_x{}^2& -{\lambda }_x{\lambda }_y& {\lambda }_x{}^2& {\lambda }_x{\lambda }_y\\ -{\lambda }_x{\lambda }_y& -{\lambda }_y{}^2& {\lambda }_x{\lambda }_y& {\lambda }_y{}^2\end{array}\right]$

Here, cross-sectional area of the member is $A$, length of the member is $L$, modulus of elasticity is $E$ and moment of inertia is $I$.

Write the expression for total structural stiffness matrix.

$k=k_1+k_2+k_3+k_4+k_5$          ...... (III)

Here, the structural stiffness matrix is $k$.

Calculation:

The free body diagram of the truss is shown below.

          Figure (1)

Consider the member-1.

Calculate the length of member $1$.

$\begin{array}{c}L=\sqrt{{\left(1\text{ m}\right)}^2+{\left(1\text{ m}\right)}^2}\\ =\sqrt{2}\text{ m}\end{array}$

Calculate direction cosine in x-direction.

Substitute $1\text{ m}$ for $x_F$, $\text{0 m}$ for $x_N$ and $\sqrt{2}\text{ m}$ for $L$ in Equation (I).

$\begin{array}{c}{\lambda }_x=\frac{1\text{ m}-0\text{ m}}{\sqrt{2}\text{ m}}\\ =0.70711\end{array}$

Calculate direction cosine in y-direction.

Substitute $1\text{ m}$ for $y_F$, $\text{2 m}$ for $y_N$ and $\sqrt{2}\text{ m}$ for $L$ in Equation (II).

$\begin{array}{c}{\lambda }_y=\frac{1\text{ m}-2\text{ m}}{\sqrt{2}\text{ m}}\\ =-0.70711\end{array}$

The stiffness matrix for member-1 is shown below.

Substitute $\sqrt{2}\text{m}$ for $L$, $0.70711$ for ${\lambda }_x$ and $-0.70711$ for ${\lambda }_y$ in the stiffness matrix.

   $k_1=\frac{AE}{\sqrt{2}}\left[\begin{array}{cccc}{\left(0.70711\right)}^2& \left(0.70711\right)\left(-0.70711\right)& -{\left(0.70711\right)}^2& -\left(0.70711\right)\left(-0.70711\right)\\ \left(0.70711\right)\left(-0.70711\right)& {\left(-0.70711\right)}^2& -\left(0.70711\right)\left(-0.70711\right)& -{\left(-0.70711\right)}^2\\ -{\left(0.70711\right)}^2& -\left(0.70711\right)\left(-0.70711\right)& {\left(0.70711\right)}^2& \left(0.70711\right)\left(-0.70711\right)\\ -\left(0.70711\right)\left(-0.70711\right)& -{\left(-0.70711\right)}^2& \left(0.70711\right)\left(-0.70711\right)& {\left(-0.70711\right)}^2\end{array}\right]$

$k_1=AE\left[\begin{array}{cccc}0.3536& -0.3536& -0.3536& 0.3536\\ -0.3536& 0.3536& 0.3536& -0.3536\\ -0.3536& 0.3536& 0.3536& -0.3536\\ 0.3536& -0.3536& -0.3536& 0.3536\end{array}\right]\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}$          ...... (IV)

Consider the member- $2$.

Calculate the length of member $2$.

$\begin{array}{c}L=\sqrt{{\left(1\text{ m}\right)}^2+{\left(1\text{ m}\right)}^2}\\ =\sqrt{2}\text{ m}\end{array}$

Calculate direction cosine in x-direction.

Substitute $\text{2 m}$ for $x_F$, $\text{1 m}$ for $x_N$ and $\sqrt{2}\text{ m}$ for $L$ in Equation (I).

$\begin{array}{c}{\lambda }_x=\frac{\text{2 m}-1\text{ m}}{\sqrt{2}\text{ m}}\\ =0.70711\end{array}$

Calculate direction cosine in y-direction.

Substitute $\text{0 m}$ for $y_F$, $\text{1 m}$ for $y_N$ and $\sqrt{2}\text{ m}$ for $L$ in Equation (II).

$\begin{array}{c}{\lambda }_y=\frac{\text{0 m}-1\text{ m}}{\sqrt{2}\text{ m}}\\ =-0.70711\end{array}$

The stiffness matrix for member-2 is shown below.

Substitute $\sqrt{2}\text{m}$ for $L$, $0.70711$ for ${\lambda }_x$ and $-0.70711$ for ${\lambda }_y$ in stiffness matrix.

   $k_2=\frac{AE}{\sqrt{2}}\left[\begin{array}{cccc}{\left(0.70711\right)}^2& \left(0.70711\right)\left(-0.70711\right)& -{\left(0.70711\right)}^2& -\left(0.70711\right)\left(-0.70711\right)\\ \left(0.70711\right)\left(-0.70711\right)& {\left(-0.70711\right)}^2& -\left(0.70711\right)\left(-0.70711\right)& -{\left(-0.70711\right)}^2\\ -{\left(0.70711\right)}^2& -\left(0.70711\right)\left(-0.70711\right)& {\left(0.70711\right)}^2& \left(0.70711\right)\left(-0.70711\right)\\ -\left(0.70711\right)\left(-0.70711\right)& -{\left(-0.70711\right)}^2& \left(0.70711\right)\left(-0.70711\right)& {\left(-0.70711\right)}^2\end{array}\right]$

$k_2=AE\left[\begin{array}{cccc}0.3536& -0.3536& -0.3536& 0.3536\\ -0.3536& 0.3536& 0.3536& -0.3536\\ -0.3536& 0.3536& 0.3536& -0.3536\\ 0.3536& -0.3536& -0.3536& 0.3536\end{array}\right]\begin{array}{c}3\\ 4\\ 5\\ 6\end{array}$          ...... (V)

Consider the member- $3$.

Calculate direction cosine in x-direction.

Substitute $\text{0 m}$ for $x_F$, $\text{2 m}$ for $x_N$ and $\text{2 m}$ for $L$ in Equation (I).

$\begin{array}{c}{\lambda }_x=\frac{\text{0 m}-2\text{ m}}{\text{2 m}}\\ =-1\end{array}$

Calculate direction cosine in y-direction.

Substitute $\text{0 m}$ for $y_F$, $\text{0 m}$ for $y_N$ and $2\text{ m}$ for $L$ in Equation (II).

$\begin{array}{c}{\lambda }_y=\frac{\text{0 m}-0\text{ m}}{2\text{ m}}\\ =0\end{array}$

The stiffness matrix for member-3 is shown below.

Substitute $2\text{m}$ for $L$, $-1$ for ${\lambda }_x$ and $0$ for ${\lambda }_y$ in stiffness matrix.

$\begin{array}{c}{k}_3=\frac{AE}{2}\left[\begin{array}{cccc}{\left(-1\right)}^2& \left(-1\right)\left(0\right)& -{\left(-1\right)}^2& -\left(-1\right)\left(0\right)\\ \left(-1\right)\left(0\right)& {\left(0\right)}^2& -\left(-1\right)\left(0\right)& -{\left(0\right)}^2\\ -{\left(-1\right)}^2& -\left(-1\right)\left(0\right)& {\left(-1\right)}^2& \left(-1\right)\left(0\right)\\ -\left(-1\right)\left(0\right)& -{\left(0\right)}^2& \left(-1\right)\left(0\right)& {\left(0\right)}^2\end{array}\right]\\ =AE\left[\begin{array}{cccc}0.5& 0& -0.5& 0\\ 0& 0& 0& 0\\ -0.5& 0& 0.5& 0\\ 0& 0& 0& 0\end{array}\right]\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\end{array}$          ...... (VI)

Consider the member- $4$.

Calculate the length of member $4$.

$\begin{array}{c}L=\sqrt{{\left(1\text{ m}\right)}^2+{\left(1\text{ m}\right)}^2}\\ =\sqrt{2}\text{ m}\end{array}$

Calculate direction cosine in x-direction.

Substitute $\text{0 m}$ for $x_F$, $\text{1 m}$ for $x_N$ and $\sqrt{\text{2}}\text{ m}$ for $L$ in Equation (I).

$\begin{array}{c}{\lambda }_x=\frac{\text{0 m}-1\text{ m}}{\sqrt{\text{2}}\text{ m}}\\ =-0.70711\end{array}$

Calculate direction cosine in y-direction.

Substitute $\text{0 m}$ for $y_F$, $\text{1 m}$ for $y_N$ and $\sqrt{\text{2}}\text{ m}$ for $L$ in Equation (II).

$\begin{array}{c}{\lambda }_y=\frac{\text{0 m}-1\text{ m}}{\sqrt{2}\text{ m}}\\ =-0.70711\end{array}$

The stiffness matrix for member-4 is shown below.

Substitute $\sqrt{2}\text{m}$ for $L$, $-0.70711$ for ${\lambda }_x$ and $-0.70711$ for ${\lambda }_y$ in stiffness matrix.

   $k_4=\frac{AE}{\sqrt{2}}\left[\begin{array}{cccc}{\left(-0.70711\right)}^2& \left(-0.70711\right)\left(-0.70711\right)& -{\left(-0.70711\right)}^2& -\left(-0.70711\right)\left(-0.70711\right)\\ \left(-0.70711\right)\left(-0.70711\right)& {\left(-0.70711\right)}^2& -\left(-0.70711\right)\left(-0.70711\right)& -{\left(-0.70711\right)}^2\\ -{\left(-0.70711\right)}^2& -\left(-0.70711\right)\left(-0.70711\right)& {\left(-0.70711\right)}^2& \left(-0.70711\right)\left(-0.70711\right)\\ -\left(-0.70711\right)\left(-0.70711\right)& -{\left(-0.70711\right)}^2& \left(-0.70711\right)\left(-0.70711\right)& {\left(-0.70711\right)}^2\end{array}\right]\begin{array}{c}3\\ 4\\ 7\\ 8\end{array}$

$k_4=AE\left[\begin{array}{cccc}0.3536& 0.3536& -0.3536& -0.3536\\ 0.3536& 0.3536& -0.3536& -0.3536\\ -0.3536& -0.3536& 0.3536& 0.3536\\ -0.3536& -0.3536& 0.3536& 0.3536\end{array}\right]\begin{array}{c}3\\ 4\\ 7\\ 8\end{array}$          ...... (VII)

Consider the member- $5$.

Calculate direction cosine in x-direction.

Substitute $\text{0 m}$ for $x_F$, $\text{0 m}$ for $x_N$ and $\text{2 m}$ for $L$ in Equation (I).

$\begin{array}{c}{\lambda }_x=\frac{\text{0 m}-0\text{ m}}{2\text{ m}}\\ =0\end{array}$

Calculate direction cosine in y-direction.

Substitute $\text{0 m}$ for $y_F$, $\text{0 m}$ for $y_N$ and $2\text{ m}$ for $L$ in Equation (II).

$\begin{array}{c}{\lambda }_y=\frac{\text{0 m}-2\text{ m}}{2\text{ m}}\\ =-1\end{array}$

The stiffness matrix for member-5 is shown below.

Substitute $2\text{m}$ for $L$, $0$ for ${\lambda }_x$ and $-1$ for ${\lambda }_y$ in stiffness matrix.

$\begin{array}{c}{k}_5=\frac{AE}{2}\left[\begin{array}{cccc}{\left(0\right)}^2& \left(-1\right)\left(0\right)& -{\left(0\right)}^2& -\left(-1\right)\left(0\right)\\ \left(-1\right)\left(0\right)& {\left(-1\right)}^2& -\left(-1\right)\left(0\right)& -{\left(-1\right)}^2\\ -{\left(0\right)}^2& -\left(-1\right)\left(0\right)& {\left(0\right)}^2& \left(-1\right)\left(0\right)\\ -\left(-1\right)\left(0\right)& -{\left(-1\right)}^2& \left(-1\right)\left(0\right)& {\left(-1\right)}^2\end{array}\right]\\ =AE\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0.5& 0& -0.5\\ 0& 0& 0& 0\\ 0& -0.5& 0& 0.5\end{array}\right]\begin{array}{c}4\\ 2\\ 7\\ 8\end{array}\end{array}$          ...... (VIII)

Calculate total structural stiffness matrix.

Substitute the values of $k_1$, $k_2$, $k_3$, $k_4$ and $k_5$ in Equation (III).

$k=AE\left[\begin{array}{cccccccc}0.3536& -0.3536& -0.3536& 0.3536& 0& 0& 0& 0\\ -0.3536& 0.8536& 0.3536& -0.3536& 0& 0& 0& -0.5\\ -0.3536& 0.3536& 1.0608& -0.3536& -0.3536& 0.3536& -0.3536& -0.3536\\ 0.3536& -0.3536& -0.3536& 1.0608& 0.3536& -0.3536& -0.3536& -0.3536\\ 0& 0& -0.3536& 0.3536& 0.8536& -0.3536& -0.5& 0\\ 0& 0& 0.3536& -0.3536& -0.3536& 0.3536& 0& 0\\ 0& 0& -0.3536& -0.3536& -0.5& 0& 0.8536& 0.3536\\ 0& -0.5& -0.3536& -0.3536& 0& 0& 0.3536& 0.8536\end{array}\right]$

Conclusion:

The structure stiffness matrix for the truss is shown below.

$k=AE\left[\begin{array}{cccccccc}0.3536& -0.3536& -0.3536& 0.3536& 0& 0& 0& 0\\ -0.3536& 0.8536& 0.3536& -0.3536& 0& 0& 0& -0.5\\ -0.3536& 0.3536& 1.0608& -0.3536& -0.3536& 0.3536& -0.3536& -0.3536\\ 0.3536& -0.3536& -0.3536& 1.0608& 0.3536& -0.3536& -0.3536& -0.3536\\ 0& 0& -0.3536& 0.3536& 0.8536& -0.3536& -0.5& 0\\ 0& 0& 0.3536& -0.3536& -0.3536& 0.3536& 0& 0\\ 0& 0& -0.3536& -0.3536& -0.5& 0& 0.8536& 0.3536\\ 0& -0.5& -0.3536& -0.3536& 0& 0& 0.3536& 0.8536\end{array}\right]$