Chapter 11, Problem 43P

During major windstorms, high vehicles such as RVs and semis may be thrown off the road and boxcars off their tracks, especially when they are empty and in open areas. Consider a 6000-kg semi that is
10 m long, 2.5 m high, and 2 in wide. The distance between the bottom of the truck and the road is 0.8 in. Now the truck is exposed to winds from its side surface. Determine the wind velocity that will tip the truck over to its side. Take the air density to be 1.1 kg/m³ and assume the weight to be uniformly distributed.

To determine

The wind velocity that will tip the truck over to its side.

Answer to Problem 43P

The wind velocity that will tip the truck over to its side is $41.98\text{m}/\text{s}$.

Explanation of Solution

Given information:

The mass of the truck is $6000\text{kg}$, the length of the truck is $10\text{m}$,the height of the truck is $2.5\text{m}$, the width of the truck is $2\text{m}$, the distance between the bottom of the truck and the road is $0.80\text{m}$ and the air density is $1.1\text{kg}/{\text{m}}^{\text{3}}$.

Write the expression of weight of the truck.

  $W=mg$...... (I)

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

Write the expression of frontal area.

  $A=L\times h$...... (II)

Here, the length of the truck is $L$ and the height of the truck is $h$.

The below figure represent the forces acting on the truck.

Figure-(1)

Write the expression of equilibrium of momentum about point $P$ as shown in Figure-(1).

  $F_D\left(\frac{h}{2}+b\right)=W\left(\frac{a}{2}\right)$...... (III)

Here, the width of the truck is $a$, the distance between the bottom of the truck and the road is $b,$ and the drag force is $F_D$.

Refer table 11.2 "Representative drag coefficient $C_D$ for various three dimensional bodies based on the frontal area for $\mathrm{Re}>{10}^4$ unless stated otherwise" to obtain the drag coefficient for rectangular plate as.

  $C_D=1.10+0.02\left(\frac{L}{h}+\frac{h}{L}\right)$...... (IV)

Write the expression of drag force acting on the truck.

  $F_D=\frac{1}{2}{C}_D\rho A{V}^2$...... (V)

Here, the density of the air is $\rho$ and the velocity of the wind is $V$.

Write the expression of characteristics length of the truck.

  $L_C=\frac{A}{P}$...... (VI)

Here, the perimeter of the truck is $P$.

Write the expression of perimeter of the truck.

  $P=2\left(L+h\right)$...... (VII)

Write the expression of Reynolds number.

  $\mathrm{Re}=\frac{V{L}_C}{\nu }$...... (VIII)

Here, the kinematic viscosity is $\nu$.

Calculation:

Substitute $6000\text{kg}$ for $m$ and $9.81{\text{m}/\text{s}}^2$ for $g$ in Equation (I).

  $\begin{array}{c}W=\left(6000\text{kg}\right)\times \left(9.81{\text{m}/\text{s}}^2\right)\\ =\left(58860{\text{kg}\cdot \text{m}/\text{s}}^2\right)\times \left(\frac{1\text{N}}{1{\text{kg}\cdot \text{m}/\text{s}}^2}\right)\\ =58860\text{N}\end{array}$

Substitute $10\text{m}$ for $L$ and $2.5\text{m}$ for $h$ in Equation (II).

  $\begin{array}{c}A=\left(10\text{m}\right)\times \left(2.5\text{m}\right)\\ =25{\text{m}}^{\text{2}}\end{array}$

Substitute $2.5\text{m}$ for $h$, $0.8\text{m}$ for $b$, $58860\text{N}$ for $W$ and $2\text{m}$ for $a$ in Equation (III).

  $\begin{array}{l}{F}_D\left(\frac{2.5\text{m}}{2}+0.8\text{m}\right)=\left(58860\text{N}\right)\left(\frac{2\text{m}}{2}\right)\\ F_D\times 2.05\text{m}=\left(58860\text{N}\right)\times \left(1\text{m}\right)\\ F_D=\frac{\left(58860\text{N}\right)\times \left(1\text{m}\right)}{\left(2.05\text{m}\right)}\\ F_D=28712.195\text{N}\end{array}$

Substitute $10\text{m}$ for $L$ and $2.5\text{m}$ for $h$ in Equation (IV).

  $\begin{array}{c}{C}_D=1.10+0.02\left(\frac{10\text{m}}{2.5\text{m}}+\frac{2.5\text{m}}{10\text{m}}\right)\\ =1.10+0.02\times 4.25\\ =1.185\end{array}$

Substitute $1.185$ for $C_D$, $1.1\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $25{\text{m}}^{\text{2}}$ for $A$ and $28712.195\text{N}$ for $F_D$ in Equation (V).

  $\begin{array}{l}\left(28712.195\text{N}\right)=\frac{1}{2}\times \left(1.185\right)\times \left(1.1\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(25{\text{m}}^{\text{2}}\right)V^2\\ V^2=\frac{2\times \left(28712.195\text{N}\right)}{\left(1.185\right)\times \left(1.1\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(25{\text{m}}^{\text{2}}\right)}\\ V=\sqrt{\frac{2\times \left(28712.195\text{N}\right)\times \left(\frac{\left(1{\text{kg}\cdot \text{m}/\text{s}}^2\right)}{1\text{N}}\right)}{\left(1.185\right)\times \left(1.1\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(25{\text{m}}^{\text{2}}\right)}}\\ V=41.98\text{m}/\text{s}\end{array}$

Substitute $10\text{m}$ for $L$ and $2.5\text{m}$ for $h$ in Equation (VII).

  $\begin{array}{c}P=2\left(10\text{m}+2.5\text{m}\right)\\ =2\times 12.5\text{m}\\ =\text{25}\text{m}\end{array}$

Substitute $\text{25}\text{m}$ for $P$ and $25{\text{m}}^{\text{2}}$ for $A$ in Equation (VI).

  $\begin{array}{c}{L}_C=\frac{\left(25{\text{m}}^{\text{2}}\right)}{\left(\text{25}\text{m}\right)}\\ =5\text{m}\end{array}$

Refer table A-9 "Properties of air" to find the value of kinematic viscosity of air as $1.75\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}$ corresponding to density of air is $1.1\text{kg}/{\text{m}}^{\text{3}}$.

Substitute $1.75\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}$ for $\nu$, $5\text{m}$ for $L_C$ and $41.98\text{m}/\text{s}$ for $V$ in Equation (VIII).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(41.98\text{m}/\text{s}\right)\times \left(5\text{m}\right)}{\left(1.75\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}\\ =\frac{209.9{\text{m}}^{\text{2}}/\text{s}}{1.75\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}}\\ =119.94\times {10}^5\end{array}$

So, the Equation used for $C_D$ is correct because the obtained Reynolds number is greater than ${10}^4$.

Conclusion:

The wind velocity that will tip the truck over to its side is $41.98\text{m}/\text{s}$.