Chapter 11, Problem 96P

A 2-zn-high, 4-zn-wide rectangular advertisement panel is attached to a 4-rn-wide. 0.15-rn-high rectangular concrete block (density = 2300 kg/m³) by two 5-cm-diameter. 4-m-high (exposed part) poles, as shown in Fig. 11-96. If the sign is to withstand 150 kin/h winds from any direction, determine (a) the maximum drag force on the panel. (b) the drag force acting on the poles, and (c) the minimum length L of the concrete block for the panel to resist the winds. Take the density of air to be 1.30 kg/m³.

To determine

(a)

The maximum drag force on the panel.

Answer to Problem 96P

The maximum drag force on the panel is $18055.55\text{N}$.

Explanation of Solution

Given information:

The height of the rectangular advertisement panel is $2\text{m}$, the width of the rectangular advertisement panel is $4\text{m}$, the height of the rectangular concrete block is $0.15\text{m}$, the width of the rectangular concrete block is $4\text{m}$, the diameter of the poles $5\text{cm}$, the height of the poles is $4\text{m}$, the velocity of the wind is $150\text{km}/\text{h}$, the density of the concrete block is $2300\text{kg}/{\text{m}}^{\text{3}}$.

Write the expression for the drag force for the panel.

  $F_{Dpanel}=C_D\frac{1}{2}{\rho }_a{V}^{2}A$........ (I)

Here, the drag force coefficient is $F_D$, the velocity of the airplane is $V$ and the density of air is $\rho$.

Write the expression for the frontal area of the panel.

  $A=h_1\times w_1$........ (II)

Here, the height of the rectangular advertisement panel is $h_1$ and the width of the rectangular advertisement panel is $w_1$.

Calculation:

The drag coefficient for the turbulent flow for the thin rectangular plate is $2$.

Substitute $2\text{m}$ for $h_1$ and $4\text{m}$ for $w_1$ in Equation (II).

  $\begin{array}{c}A=2\text{m}\times 4\text{m}\\ =\text{8}{\text{m}}^2\end{array}$

Substitute $2$ for $C_D$, $1.30\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $8{\text{m}}^2$ for $A$ and $150\text{km}/\text{h}$ for $V$ in Equation (I).

  $\begin{array}{c}{F}_{Dpanel}=\left(2\right)\frac{1}{2}\left(1.30\text{kg}/{\text{m}}^{\text{3}}\right){\left(150\text{km}/\text{h}\right)}^2\left(8{\text{m}}^2\right)\\ =\left(1.30\text{kg}/{\text{m}}^{\text{3}}\right){\left(150\text{km}/\text{h}\left(\frac{5\text{m}/\text{s}}{18\text{km}/\text{h}}\right)\right)}^2\left(8{\text{m}}^2\right)\\ =18055.55\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =18055.55\text{N}\end{array}$

Conclusion:

The maximum drag force on the panel is $18055.55\text{N}$.

To determine

(b)

The drag force acting on the pole.

Answer to Problem 96P

The drag force acting on the pole is $67.70\text{N}$.

Explanation of Solution

Write the expression for the frontal area of the pole.

  $A_p=D\times h_3$........ (III).

Here, the diameter of the pole is $D$ and the height of the pole is $h_3$.

Write the expression for the drag force for the pole.

  $F_{Dpole}=C_D\frac{1}{2}{\rho }_a{V}^{2}A$........ (IV)

Calculation:

The drag coefficient for the turbulent flow for circular rod is $0.3$.

Substitute $4\text{m}$ for $h_3$ and $5\text{cm}$ for $D$ in Equation (III).

  $\begin{array}{c}{A}_p=4\text{m}\times 5\text{cm}\\ =4\text{m}\times 5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.2{\text{m}}^2\end{array}$

Substitute $0.3$ for $C_D$, $1.30\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.2{\text{m}}^2$ for $A_p$ and $150\text{km}/\text{h}$ for $V$ in Equation (IV).

  $\begin{array}{c}{F}_{Dpole}=\left(0.3\right)\frac{1}{2}\left(1.30\text{kg}/{\text{m}}^{\text{3}}\right){\left(150\text{km}/\text{h}\right)}^2\left(0.2{\text{m}}^2\right)\\ =0.15\left(1.30\text{kg}/{\text{m}}^{\text{3}}\right){\left(150\text{km}/\text{h}\left(\frac{5\text{m}/\text{s}}{18\text{km}/\text{h}}\right)\right)}^2\left(0.2{\text{m}}^2\right)\\ =67.70\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =67.70\text{N}\end{array}$

Conclusion:

The drag force acting on the pole is $67.70\text{N}$.

To determine

(c)

The minimum length $L$ of the concrete block for the panel to resist the winds.

Answer to Problem 96P

The minimum length $L$ of the concrete block for the panel to resist the winds is $3.70\text{m}$.

Explanation of Solution

Draw the diagram for the side view of the advertisement panel.

Figure-(1)

Write the expression for the moment about point A.

  $\begin{array}{l}{\displaystyle \sum M_A}=0\\ F_{Dpanel}\left(5.15\text{m}\right)+F_{Dpole}\left(2.15\text{m}\right)-W\left(\frac{L}{2}\right)=0\end{array}$........ (V)

Here, the minimum length of the concrete block for the panel to resist the winds is $L$ and the weight of the concrete block is $W$.

Write the expression for the weight of the concrete block.

  $W=mg$........ (VI)

Here, the mass of the block is $m$ and the acceleration due to gravity is $g$.

Write the expression for the volume of the block.

  $V=L\times h_2\times w_2$........ (VII)

Here, the height of the rectangular concrete block is $h_2$ and the width of the rectangular concrete block is $w_2$.

Write the expression for the mass of the block.

  $m={\rho }_c\times V$

Here, the density of the concrete block is ${\rho }_c$.

Substitute ${\rho }_c\times V$ for $m$ in Equation (VI).

  $W=\left({\rho }_c\times V\right)g$........ (VIII)

Substitute $L\times h_2\times w_2$ for $V$ in Equation (VIII).

  $W=\left({\rho }_c\times L\times h_2\times w_2\right)g$........ (IX)

Calculation:

Substitute $2300\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_c$, $0.15\text{m}$ for $h_2$, $4\text{m}$ for $w_2$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (IX).

  $\begin{array}{c}W=\left(2300\text{kg}/{\text{m}}^{\text{3}}\times L\times 0.15\text{m}\times 4\text{m}\right)9.81\text{m}/{\text{s}}^{\text{2}}\\ =\left(1380\text{kg}/\text{m}\right)L\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =13537.8\text{kg}/{\text{s}}^{\text{2}}\left(\frac{1\text{N}\cdot \text{m}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)L\\ =\left(13537.8\text{N}\cdot \text{m}\right)L\end{array}$

Substitute $\left(13537.8\text{N}\cdot \text{m}\right)L$ for $W$, $67.70\text{N}$ for $F_{Dpole}$ and $18055.55\text{N}$ for $F_{Dpanel}$ in Equation (V).

  $\begin{array}{l}18055.55\text{N}\left(5.15\text{m}\right)+67.70\text{N}\left(2.15\text{m}\right)-\left(13537.8\text{N}\cdot \text{m}\right)L\text{m}\left(\frac{L\text{m}}{2}\right)=0\\ L^2=\left(\frac{92986.082\text{N}\cdot \text{m}+145.55\text{N}\cdot \text{m}}{6768.9\text{N}\cdot \text{m}}\right){\text{m}}^{\text{2}}\\ L=\sqrt{13.7587{\text{m}}^{\text{2}}}\\ L=3.70\text{m}\end{array}$

Conclusion:

The minimum length $L$ of the concrete block for the panel to resist the winds is $3.70\text{m}$.