Chapter 1, Problem 1.3.32P

A soccer goal is subjected to gravity loads (in the - z direction, w = 73 N/m for DG, BG, and BC; w = 29 N/m for all other members; see figure) and a force F = 200 N applied eccentrically at the mid-height of member DG. Find reactions at sup ports C, D, and H.

  

To determine

The reactions at support $C$.

The reaction at support $D$.

The reaction at support $H$.

Answer to Problem 1.3.32P

The reactions at support $C$in x, y and z axes are $120\text{N}$, $160\text{N}$, and $506\text{N}$

The reactions at support $H$in x and y axis are $320\text{N}$, $\text{499}\text{N}$.

The reactions at support $D$in z axis is $466\text{N}$.

Explanation of Solution

The weight per unit length in the members $DG$, $BC$, and $BG$is $w=73\text{N}/\text{m}$, the weight per unit length for all other members is $w=29\text{N}/\text{m}$, the value of force is $F=200\text{N}$.

The following figure shows the forces on the goal post.

Figure-(1)

Write the expression for the length of the members $PQ$and $RS$

$\begin{array}{l}{L}_{PQ}=L_{RS}\\ L_{PQ}=\sqrt{{\left(\frac{1}{2}{L}_{DQ}\right)}^2+L^2{}_{DQ}}\end{array}$…… (I)

Here, the length of member $PQ$is $L_{PQ}$, the length of member $RS$is $L_{RS}$, and the length of member $DQ$is $L_{DQ}$.

Write the expression for horizontal component of force on member $GD$

$F_H=F_{GD}\times \sin \left({\tan }^{-1}\left(\frac{4}{\sqrt{3^2+4^2}}\right)\right)$…… (II)

Here the horizontal component of force is $F_H$, and the force of $200\text{N}$is $F_{GD}$.

Write the expression for the vertical component of force on member $GD$

$F_V=F_{GD}\times \cos \left({\tan }^{-1}\left(\frac{4}{\sqrt{3^2+4^2}}\right)\right)$…… (III)

Here, the vertical component of force is $F_V$.

Write the expression for moment about x axis.

$\begin{array}{l}{\displaystyle \sum M_x}=0\\ \left[\begin{array}{l}\left(H_z\times L_{DQ}\right)-\left(F_H\times L_{GP}\right)\\ -\left(2\times \left(F_{DQ}\times \frac{L_{DQ}}{2}\right)\right)\\ -\left(2\times \left(F_{PQ}\left(L_{GP}+\frac{L_{GP}}{2}\right)\right)\right)\\ -\left(F_{QS}\times L_{CS}\right)-2\times \left(F_{GP}\times \frac{L_{GP}}{2}\right)\end{array}\right]=0\end{array}$…… (VI)

Here, the force along the member $PQ$is $F_{PQ}$, the length of the member $GP$is $L_{GP}$, the force along the member $QS$is $F_{QS}$.

Write the expression for moment about Y axis.

$\begin{array}{l}{\displaystyle \sum M_y}0\\ \left[\begin{array}{l}-\left(D_z\times L_{DC}\right)-\left(H_z\times L_{QH}\right)+\left(F_{GB}\times \frac{L_{GB}}{2}\right)-\left(F_V\times \frac{L_{DQ}}{2}\right)\\ +\left(F_{QS}\times L_{HS}\right)+\left(F_{DQ}\times L_{QS}\right)+\left(F_{PQ}\times L_{QS}\right)\\ +\left(F_{GD}\times L_{GB}\right)+\left(F_{GP}\times L_{GB}\right)\end{array}\right]=0\end{array}$…… (VII)

Here, the length of member $DC$is $L_{DC}$, the length of member $QH$is $L_{QH}$, the force along the member $GB$is $F_{GB}$, the length of the member $GB$is $L_{GB}$, the length of member $HS$is $L_{HS}$, the length of member $QS$is $L_{QS}$, the force along the member $GD$is $F_{GD}$.

Write the expression for moment about Z axis and equate it to zero.

$H_y\times L_{HS}-\left(F_{DQ}\cos 36.86\text{N}\times L_{QS}\right)=0$…… (VIII)

Here, the reaction of Y axis about $C$is $H_y$.

Write the expression for the sum of forces about x axis.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ C_x-F_{DQ}\sin 36.86°=0\end{array}$…… (IX)

Write the expression for the sum of forces about Y axis.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ C_y+H_y-F_{DQ}\cos 36.86°=0\end{array}$…… (X)

Write the expression for the sum of forces about Z axis.

$\begin{array}{l}{\displaystyle \sum F_z}=0\\ D_z+H_z+C_z=\left[\begin{array}{l}{F}_{GP}+F_{BR}+F_{PQ}+F_{RS}+F_{DQ}\\ +F_{CS}+F_{GB}+F_{DG}+F_{BC}+F_{QS}\end{array}\right]\end{array}$…… (XI)

Calculation:

Substitute $2.44\text{m}$for $L_{DQ}$in Equation (I)

$\begin{array}{c}{L}_{PQ}=L_{RS}\\ =\sqrt{{\left(\frac{1}{2}2.44\text{m}\right)}^2+{\left(2.44\text{m}\right)}^2}\\ =2.728\text{m}\end{array}$

Substitute $200\text{N}$for $F_{GD}$in equation (II)

$\begin{array}{c}{F}_H=\left(200\text{N}\right)\times \sin \left({\tan }^{-1}\left(\frac{4}{\sqrt{3^2+4^2}}\right)\right)\\ =\left(200\text{N}\right)\sin \left(53.13°\right)\\ =159.99\text{N}\\ \approx \text{160}\text{N}\end{array}$

Substitute $200\text{N}$for $F_{GD}$in equation (III)

$\begin{array}{c}{F}_V=\left(200\text{N}\right)\times \cos \left({\tan }^{-1}\left(\frac{4}{\sqrt{3^2+4^2}}\right)\right)\\ =\left(200\text{N}\right)\cos \left(53.13°\right)\\ =120\text{N}\end{array}$

Substitute $2.44\text{m}$for $L_{DQ}$, $160\text{N}$for $F_H$, $70.76\text{N}$for $F_{DQ}$, $1.22\text{m}$for $L_{GP}$, $79.17\text{N}$for $F_{PQ}$, $211.7\text{N}$for $F_{QS}$, $2.44\text{m}$for $L_{CS}$and $35.38\text{N}$for $F_{GP}$in Equation (VI).

$\begin{array}{l}\left[\begin{array}{l}\left(H_z\times 2.44\text{m}\right)-\left(160\text{N}\times 1.22\text{m}\right)-\left(2\times \left(70.76\text{N}\times \frac{2.44\text{m}}{2}\right)\right)\\ -\left(2\times \left(79.17\text{N}\left(1.22\text{m}+\frac{1.22\text{m}}{2}\right)\right)\right)-\left(211.7\text{N}\times 2.44\text{m}\right)\\ -2\times \left(35.38\text{N}\times \frac{1.22\text{m}}{2}\right)\end{array}\right]=0\\ H_z=\frac{1217.33\text{N}\cdot \text{m}}{2.44\text{m}}\\ H_z=498.91\text{N}\end{array}$

Substitute $7.3\text{m}$for $L_{DC}$, $7.3\text{m}$for $L_{GB}$, $7.3\text{m}$for $L_{GB}$

$498.91\text{N}$for $H_z$, $3.65\text{m}$for $L_{QH}$and $L_{HS}$, $532.9\text{N}$for $F_{GB}$, $120\text{N}$for $F_V$, $2.44\text{m}$for $L_{DQ}$, $79.17\text{N}$for $F_{GP}$, $211.7\text{N}$for $F_{QS}$, $79.17\text{N}$for $F_{PQ}$, $70.76\text{N}$for $F_{DQ}$, $178.12\text{N}$for $F_{GD}$in Equation (VII)

$\begin{array}{l}\left[\begin{array}{l}-\left(D_z\times 7.3\text{m}\right)-\left(498.91\text{N}\times 3.65\text{m}\right)+\left(532.9\text{N}\times \frac{7.3\text{m}}{2}\right)\\ -\left(120\text{N}\times \frac{2.44\text{m}}{2}\right)+\left(211.7\text{N}\times 3.65\text{m}\right)\\ +\left(70.76\text{N}\times 7.3\text{m}\right)+\left(79.17\text{N}\times 7.3\text{m}\right)\\ +\left(178.12\text{N}\times 7.3\text{m}\right)+\left(79.17\text{N}\times 7.3\text{m}\right)\end{array}\right]=0\\ D_z=\frac{3399.429\text{N}\cdot \text{m}}{7.3\text{m}}\\ D_z=465.675\text{N}\end{array}$

The following figure shows the free body diagram of forces along y and z axes.

Figure-(2)

Substitute $3.65\text{m}$for $L_{HS}$, $200\text{N}$for $F_{DQ}$and $7.3\text{m}$for $L_{QS}$in Equation (VIII)

$\begin{array}{l}{H}_y\times 3.65\text{m}-\left(200\text{N}\times \cos 36.86\text{N}\times 7.3\text{m}\right)=0\\ H_y=\frac{1168.15}{3.65}\\ H_y=320\text{N}\end{array}$

Substitute $200\text{N}$in Equation (IX)

$\begin{array}{l}{C}_x-200\text{N}\sin 36.86°=0\\ C_x-120\text{N=0}\\ {\text{C}}_x=120\text{N}\end{array}$

Substitute $200\text{N}$for $F_{DQ}$and $320\text{N}$for $H_y$in Equation (X)

$\begin{array}{l}{C}_y+H_y-200\text{Ncos36}\text{.86}°\text{=0}\\ {\text{C}}_y+320\text{N}-\text{200cos36}\text{.86}°\text{=0}\\ {\text{C}}_y=-160\text{N}\end{array}$

Substitute $466\text{N}$for $D_z$, $499\text{N}$for $H_z$

$35.38\text{N}$for $F_{GP}$, $79.17\text{N}$for $F_{BR}$, $79.17\text{N}$for $F_{PQ}$, $79.17\text{N}$for $F_{RS}$

$70.76\text{N}$for $F_{DQ}$, $70.76\text{N}$for $F_{CS}$

$532.9\text{N}$for $F_{GB}$, $178.12\text{N}$for $F_{DG},F_{BC}$, $211.7\text{N}$for $F_{QS}$in Equation (XI).

$\begin{array}{l}{D}_z+H_z+C_z=\left[\begin{array}{l}35.38\text{N}+79.17\text{N}+79.17\text{N}+79.17\text{N}+70.76\text{N}\\ +70.76\text{N}+532.9\text{N}+178.12\text{N}+178.12\text{N}+211.7\text{N}\end{array}\right]\\ 466\text{N}+499\text{N}+C_z=1471.46\text{N}\\ C_z=505.46\text{N}\end{array}$

Conclusion:

The reactions at support $C$in x, y and z axes are $120\text{N}$, $160\text{N}$, and $506\text{N}$

The reactions at support $H$in x and y axis are $320\text{N}$, $\text{499}\text{N}$.

The reactions at support $D$in z axis is $466\text{N}$.