Chapter 4, Problem 4.5.39P

To determine

The support reaction at point A and C and plot the shear, moment, and axial force diagram.

Answer to Problem 4.5.39P

Explanation of Solution

Given: .

The given figure.

AB column and BC beam forms the plane frame that carries a load that is distributed in the triangular shape. At C there is roller support and support A is fixed. Below the B joint there is a moment release at column AB

Concept Used: $$.

  $$.

  $$.

Vertical force equilibrium is given as,

${\displaystyle \sum F_V}=0$.

$A_y+C_y=\frac{q_{0}L}{2}$.

Horizontal force equilibrium is given as,

  $\begin{array}{l}{\displaystyle \sum F_H}=0\\ A_x=0\end{array}$.

Calculation: .

Vertical force equilibrium is given as,

${\displaystyle \sum F_V}=0$.

$A_y+C_y-\frac{q_{0}L}{2}=0$.

$A_y+C_y=\frac{q_{0}L}{2}$.

Horizontal force equilibrium is given as,

  $\begin{array}{l}{\displaystyle \sum F_H}=0\\ A_x=0\end{array}$.

At the top of the moment release the moment is given as,

$\begin{array}{l}{\displaystyle \sum M_B}=0\\ \frac{q_{0}L}{2}(2L/3)-C_{y}L=0\\ \frac{q_{0}L}{2}(2L/3)=C_{y}L\\ C_y=\frac{q_{0}L}{3}\end{array}$.

In equation (1),$C_y$ is substituted,

$A_y=\frac{q_{0}L}{2}-\frac{q_{0}L}{3}\Rightarrow \frac{q_{0}L}{6}$.

Below B there is moment release at point A,

$M_A=0$.

At x the shear force is given and equated to 0,

  $\begin{array}{l}\frac{q_{0}L}{6}=\left(1/2\right)x\left(q_0/L\right)x\\ \frac{q_{0}L}{6}=\frac{q_0}{2L}{x}^2\\ x^2=\frac{q_{0}L}{6}(2L/q_0)\end{array}$.

  $=\frac{L^2}{3}$.

  $\Rightarrow 0.57735L$.

At point x bending moment is maximum,

  $M_x=\frac{q_{0}L}{6}\left(L/\sqrt{3}\right)-\frac{q_0}{2L}\frac{L^2}{3}\left(L/3\sqrt{3}\right)$.

  $\begin{array}{l}=\frac{q_0{L}^2}{6\sqrt{3}}-\frac{q_0{L}^2}{18\sqrt{3}}\\ =\frac{3q_0{L}^2-q_0{L}^2}{18\sqrt{3}}\end{array}$.

  $=\frac{2q_0{L}^2}{18\sqrt{3}}$.

  $M_x=\frac{q_0{L}^2}{9\sqrt{3}}\Rightarrow 0.0641q_0{L}^2$.

Axial force critical values$N_{\max }=\frac{-q_{0}L}{6}$.

Shear force critical values$V_{\max }=\frac{-q_{0}L}{3}$.

Moment critical values$M_{\max }=0.06415q_0{L}^2$.

Conclusion: .

Thus, the support reaction at point A and C and plot the shear, moment and axial force diagram.

To determine

For the load that is parabolic, lateral acts to the right added to AB column and part (a) is repeated.

Answer to Problem 4.5.39P

Explanation of Solution

Given: .

The given figure:.

AB column and BC beam forms the plane frame that carries a load that is distributed in the triangular shape. At C there is roller support and support A is fixed. Below the B joint, there is a moment release at column AB.

Concept Used: $$.

  $$.

Vertical force equilibrium is given as,

${\displaystyle \sum F_V}=0$.

$A_y+C_y=\frac{q_{0}L}{2}$.

Horizontal force equilibrium is given as,

  $\begin{array}{l}{\displaystyle \sum F_H}=0\\ A_x=0\end{array}$.

Calculation: .

With the equilibrium force being vertical,

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y+C_y=\frac{q_{0}L}{2}\to (2)\end{array}$.

Moment at point B above moment release,

  $\begin{array}{l}{\displaystyle \sum M_B}=0\\ C_{y}L-\frac{q_{0}L}{2}(2L/3)=0\\ C_{y}L=\frac{q_{0}L}{2}(2L/3)\\ C_y=\frac{q_{0}L}{3}\end{array}$.

In (2) substitute$C_y$.

  $A_y=\frac{q_{0}L}{2}-\frac{q_{0}L}{3}$.

  $\begin{array}{l}=\frac{3q_{0}L-3q_{0}L}{6}\\ =\frac{q_{0}L}{6}\end{array}$.

With the equilibrium force being horizontal,

  ${\displaystyle \sum F_H}=0$.

As left of x axis is negative,$f(x)\le 0$.

  $A_x={\displaystyle \underset{0}{\overset{2L}{\int }}-(\sqrt{1-y/2L}})q_0{d}_y$.

  $=-q_0{\displaystyle \underset{0}{\overset{2L}{\int }}(\sqrt{1-y/2L}})d_y$.

  $\begin{array}{l}=-q_0{\left(\frac{-2L{(1-y/2L)}^{3/2}}{3/2}\right)}_0^{2L}\\ =\frac{4q_{0}L}{3}\left((1-2L/2L)-(1-0)\right)\\ =\frac{4q_{0}L}{3}(0-1)\end{array}$.

  $A_x=\frac{-4q_{0}L}{3}$.

At A moment below the moment release,

  $M_A={\displaystyle \underset{0}{\overset{2L}{\int }}(q_{0}y\sqrt{1-y/2L}})d_y+\frac{1}{2}{q}_{0}L(2L/3)-\frac{q_{0}L}{3}(L)$.

  $=I_1+\frac{q_0{L}^2}{3}-\frac{q_0{L}^2}{3}$.

  $\begin{array}{l}{I}_1={\displaystyle \underset{0}{\overset{2L}{\int }}(q_{0}y\sqrt{1-y/2L}})dy\\ u=y,dv={\left(1-y/2L\right)}^{1/2}\\ du=dy,v=\frac{{\left(1-y/2L\right)}^{3/2}-2L}{3/2}\end{array}$.

  $=\frac{-4L}{3}{\left(1-y/2L\right)}^{3/2}$.

  ${\displaystyle \int d(uv)=uv-{\displaystyle \int vdu}}$.

  $\begin{array}{l}=q_0\left\{{(y(-4L/3){(1-y/2L)}^{3/2})}_0^{2L}-{\displaystyle \underset{0}{\overset{2L}{\int }}((-4L/3){(1-y/2L)}^{3/2})dy}\right\}\\ =q_0(0+(4L/3){\left(\frac{{(1-y/2L)}^{5/2}}{5/2(-1/2L)}\right)}_0^{2L}\\ =q_0\left(\frac{4L}{3}(0-(-4L/5)\right)\\ =q_0\left((4L/3)(4L/5)\right)\end{array}$.

  $M_A=\frac{16q_0{L}^2}{15}$.

At x shear force is calculated and equated to 0,

  $\begin{array}{l}\frac{q_{0}L}{6}=\left(1/2\right)x\left(q_0/L\right)x\\ \frac{q_{0}L}{6}=\frac{q_0}{2L}{x}^2\end{array}$.

  $x^2=\frac{q_{0}L}{6}\left(2L/q_0\right)$.

  $\begin{array}{l}=\frac{L^2}{3}\\ x=0.57735L\end{array}$.

At x bending moment is calculated

  $M_x=\frac{q_{0}L}{6}(L/\sqrt{3})-\frac{q_0}{2L}\frac{L^2}{3}(L/3\sqrt{3})$.

  $\begin{array}{l}=\frac{q_0{L}^2}{6\sqrt{3}}-\frac{q_0{L}^2}{18\sqrt{3}}\\ =\frac{3q_0{L}^2-q_0{L}^2}{18\sqrt{3}}\\ =\frac{2q_0{L}^2}{18\sqrt{3}}\Rightarrow \frac{q_0{L}^2}{9\sqrt{3}}\\ M_x=0.06415q_0{L}^2\end{array}$.

For plane frame and critical values of N, V, M, the axial, shear and moment is given.

Axial force critical values$N_{\max }=\frac{-q_{0}L}{6}$.

Shear force critical values in beam$V_{\max }=\frac{-q_{0}L}{3}$.

Shear force critical values in column$V_{\max }=\frac{4q_{0}L}{3}$.

Moment critical values in beam$M_{\max }=0.06415q_0{L}^2$.

Moment critical values in column$M_{\max }=\frac{-16q_0{L}^2}{15}$.

Conclusion: .

Thus, for the load that is parabolic lateral acts to the right added to AB column and part (a) is repeated.