Chapter 15, Problem 15.1P

To determine

The reactions at supports

Answer to Problem 15.1P

The support reaction $Q_4$ is $17.495\text{kN}\uparrow$.

The support reaction $Q_5$ is $7.5\text{kN}\downarrow$.

The support reaction $Q_6$ is $5\text{kN-m}$.

Explanation of Solution

Given:

The flexural rigidity, $EI$ of the beam is constant.

Concept Used:

Write the expression for fixed end moments in a beam with uniformly distributed load.

   $FEM=\frac{w{l}^2}{12}$ ...... (I).

Here, the fixed end moment is FEM, the load on the beam is $w$ and the length of the beam is $l$.

Write the expression for fixed end moments in a beam with point load at mid-span.

   $FEM=\frac{wl}{8}$ ...... (II)

Write the expression for load matrix.

   $Q=kD$ ...... (III)

Here, $Q$ is the load matrix, $k$ is the stiffness and $D$ is the displacement.

Calculations:

Calculate the fixed end moment at $1$ using Equation (I).

Substitute $5\text{kN}/\text{m}$ for $w$ and $2\text{m}$ for $l$ in Equation (I).

   $\begin{array}{c}FE{M}_1=\frac{\left(5\text{kN}/\text{m}\right)\times {\left(2\text{m}\right)}^2}{12}\\ =\frac{\left(20\text{kN-m}\right)}{12}\\ =1.67\text{kN-m}\end{array}$

Write the force matrix from the known force.

   $Q=\left[\begin{array}{c}-5\\ -1.667\\ 1.667\\ Q_4-5\\ Q_5\\ Q_6\end{array}\right]$

Write the displacement matrix for the known displacement.

   $D=\left[\begin{array}{c}{D}_1\\ D_2\\ D_3\\ 0\\ 0\\ 0\end{array}\right]$

Develop the stiffness matrix.

   $k=EI\left[\begin{array}{cccc}\frac{12}{L^3}& \frac{6}{L^2}& -\frac{12}{L^3}& \frac{6}{L^2}\\ \frac{6}{L^2}& \frac{4}{L}& -\frac{6}{L^2}& \frac{2}{L}\\ -\frac{12}{L^3}& -\frac{6}{L^2}& \frac{12}{L^3}& -\frac{6}{L^2}\\ \frac{6}{L^2}& \frac{2}{L}& -\frac{6}{L^2}& \frac{4}{L}\end{array}\right]$

Calculate the member stiffness matrix for member 1.

   $k_{\text{1}}=EI\left[\begin{array}{cccc}\frac{12}{L^3}& \frac{6}{L^2}& -\frac{12}{L^3}& \frac{6}{L^2}\\ \frac{6}{L^2}& \frac{4}{L}& -\frac{6}{L^2}& \frac{2}{L}\\ -\frac{12}{L^3}& -\frac{6}{L^2}& \frac{12}{L^3}& -\frac{6}{L^2}\\ \frac{6}{L^2}& \frac{2}{L}& -\frac{6}{L^2}& \frac{4}{L}\end{array}\right]\begin{array}{c}1\\ 2\\ 4\\ 3\end{array}$ ...... (IV)

Substitute $2\text{m}$ for $L$ in Equation (IV).

   $\begin{array}{c}{k}_1=EI\left[\begin{array}{cccc}\frac{12}{2^3}& \frac{6}{2^2}& -\frac{12}{2^3}& \frac{6}{2^2}\\ \frac{6}{2^2}& \frac{4}{2}& -\frac{6}{2^2}& \frac{2}{2}\\ -\frac{12}{2^3}& -\frac{6}{2^2}& \frac{12}{2^3}& -\frac{6}{2^2}\\ \frac{6}{2^2}& \frac{2}{2}& -\frac{6}{2^2}& \frac{4}{2}\end{array}\right]\begin{array}{c}1\\ 2\\ 4\\ 3\end{array}\\ =EI\left[\begin{array}{cccc}1.5& 1.5& -1.5& 1.5\\ 1.5& 2& -1.5& 1\\ -1.5& -1.5& 1.5& -1.5\\ 1.5& 1& -1.5& 2\end{array}\right]\begin{array}{c}1\\ 2\\ 4\\ 3\end{array}\end{array}$

Calculate the member stiffness matrix for member 2.

   $k_{\text{2}}=EI\left[\begin{array}{cccc}\frac{12}{L^3}& \frac{6}{L^2}& -\frac{12}{L^3}& \frac{6}{L^2}\\ \frac{6}{L^2}& \frac{4}{L}& -\frac{6}{L^2}& \frac{2}{L}\\ -\frac{12}{L^3}& -\frac{6}{L^2}& \frac{12}{L^3}& -\frac{6}{L^2}\\ \frac{6}{L^2}& \frac{2}{L}& -\frac{6}{L^2}& \frac{4}{L}\end{array}\right]\begin{array}{c}4\\ 3\\ 5\\ 6\end{array}$ ...... (V)

Substitute $2\text{m}$ for $L$ in Equation (V).

   $\begin{array}{c}{k}_2=EI\left[\begin{array}{cccc}\frac{12}{2^3}& \frac{6}{2^2}& -\frac{12}{2^3}& \frac{6}{2^2}\\ \frac{6}{2^2}& \frac{4}{2}& -\frac{6}{2^2}& \frac{2}{2}\\ -\frac{12}{2^3}& -\frac{6}{2^2}& \frac{12}{2^3}& -\frac{6}{2^2}\\ \frac{6}{2^2}& \frac{2}{2}& -\frac{6}{2^2}& \frac{4}{2}\end{array}\right]\begin{array}{c}4\\ 3\\ 5\\ 6\end{array}\\ =EI\left[\begin{array}{cccc}1.5& 1.5& -1.5& 1.5\\ 1.5& 2& -1.5& 1\\ -1.5& -1.5& 1.5& -1.5\\ 1.5& 1& -1.5& 2\end{array}\right]\begin{array}{c}4\\ 3\\ 5\\ 6\end{array}\end{array}$

Calculate the stiffness matrix for the beam.

   $K=k_1+k_2$ ...... (VII)

Substitute the values of $k_{\text{1}}$ and $k_2$ in Equation (VII).

   $\begin{array}{l}k=EI\left[\begin{array}{cccc}1.5& 1.5& -1.5& 1.5\\ 1.5& 2& -1.5& 1\\ -1.5& -1.5& 1.5& -1.5\\ 1.5& 1& -1.5& 2\end{array}\right]\begin{array}{c}1\\ 2\\ 4\\ 3\end{array}+EI\left[\begin{array}{cccc}1.5& 1.5& -1.5& 1.5\\ 1.5& 2& -1.5& 1\\ -1.5& -1.5& 1.5& -1.5\\ 1.5& 1& -1.5& 2\end{array}\right]\begin{array}{c}4\\ 3\\ 5\\ 6\end{array}\\ k=EI\left[\begin{array}{cccccc}1.5& 1.5& 1.5& -1.5& 0& 0\\ 1.5& 2& 1& -1.5& 0& 0\\ 1.5& 1& 4& 0& -1.5& 1\\ -1.5& -1.5& 0& 3& -1.5& 1.5\\ 0& 0& -1.5& -1.5& 1.5& -1.5\\ 0& 0& 1& -1.5& -1.5& 2\end{array}\right]\end{array}$

Substitute the values of stiffness matrix and displacement matrix in Equation (III).

   $\left[\begin{array}{c}-5\\ -1.667\\ 1.667\\ Q_4-5\\ Q_5\\ Q_6\end{array}\right]=EI\left[\begin{array}{cccccc}1.5& 1.5& 1.5& -1.5& 0& 0\\ 1.5& 2& 1& -1.5& 0& 0\\ 1.5& 1& 4& 0& -1.5& 1\\ -1.5& -1.5& 0& 3& -1.5& 1.5\\ 0& 0& -1.5& -1.5& 1.5& -1.5\\ 0& 0& 1& -1.5& -1.5& 2\end{array}\right]\left[\begin{array}{c}{D}_1\\ D_2\\ D_3\\ 0\\ 0\\ 0\end{array}\right]$

Solve the above matrix to get the values of displacement.

   $-5=EI\left(1.5D_1+1.5D_2+1.5D_3\right)$ ...... (VIII)

   $-1.667=EI\left(1.5D_1+2D_2+D_3\right)$ ...... (IX)

   $1.667=EI\left(1.5D_1+D_2+4D_3\right)$ ...... (X)

Solve Equations (VIII), (IX) and (X).

   $\begin{array}{l}{D}_1=\frac{-19.99}{EI}\\ D_2=\frac{11.6}{EI}\\ D_3=\frac{5}{EI}\end{array}$

Calculate the value of $Q_4$.

   $Q_4-5=EI\left(-1.5D_1-1.5D_2\right)$ ...... (XI)

Substitute $\frac{-19.99}{EI}$ for $D_1$ and $\frac{11.6}{EI}$ for $D_2$ in Equation (XI).

   $\begin{array}{l}{Q}_4-5=EI\left(\left(-1.5\times \frac{-19.99}{EI}\right)-\left(1.5\times \frac{11.6}{EI}\right)\right)\\ Q_4=\left(12.495+5\right)\text{kN}\\ Q_4=17.495\text{kN}\end{array}$

Calculate the value of $Q_5$.

   $Q_5=EI\left(-1.5D_3\right)$ ...... (XII)

Substitute $\frac{5}{EI}$ for $D_3$ in Equation (XII).

   $\begin{array}{l}{Q}_5=\left(EI\left(-1.5\times \frac{5}{EI}\right)\right)\text{kN}\\ Q_5=-7.5\text{kN}\end{array}$

Calculate the value of $Q_6$.

   $Q_6=EI\left(D_3\right)$ ....... (XIII)

Substitute $\frac{5}{EI}$ for $D_3$ in Equation (XIII).

   $\begin{array}{l}{Q}_6=\left(EI\left(\frac{5}{EI}\right)\right)\text{kN-m}\\ Q_6=5\text{kN-m}\end{array}$

Conclusion:

The support reaction $Q_4$ is $17.495\text{kN}\uparrow$.

The support reaction $Q_5$ is $7.5\text{kN}\downarrow$.

The support reaction $Q_6$ is $5\text{kN-m}$.