# Physics Exercises/Strength of Materials

## Exercises

### The Too-Tall Tower

#### Part 1. (Easy)

1.) Consider a tall block of material with constant cross sectional area. Since the block has mass, it exerts some amount of pressure on the material below it. Ultimately at some height the object's own mass must cause the tower to crumble. What is the height for a material with a density of ${\displaystyle \rho }$ and an ultimate tensile strength of ${\displaystyle \sigma _{t}}$ on a planet with gravity ${\displaystyle g}$?

Compute the maximum heights in meters of the following substances:

• Iron
• Concrete

#### Part 2. (Medium)

2.) Woah, that's pretty tall. Over the course of several kilometers, obviously, the air pressure would weaken, so the tower could be a tad bit taller, considering that not as much pressure is pushing on the tower. Since the tower is in the couple of mile tall range, the air pressure needs to be determined from tables. w:Atmospheric Pressure lists the following equation:

There are two different equations for computing pressure at various height regimes below 86 km (or 278,400 feet). Equation 1 is used when the value of standard temperature lapse rate is not equal to zero and equation 2 is used when standard temperature lapse rate equals zero.

Equation 1:

${\displaystyle {P}=P_{b}\cdot \left[{\frac {T_{b}}{T_{b}+L_{b}\cdot (h-h_{b})}}\right]^{\frac {g_{0}\cdot M}{R^{*}\cdot L_{b}}}}$

Equation 2:

${\displaystyle {P}=P_{b}\cdot \exp \left[{\frac {-g_{0}\cdot M\cdot (h-h_{b})}{R^{*}\cdot T_{b}}}\right]}$

where

${\displaystyle P}$ = Static pressure (pascals)
${\displaystyle T}$ = Standard temperature (kelvins)
${\displaystyle L}$ = Standard temperature lapse rate (kelvins per meter)
${\displaystyle h}$ = Height above sea level (meters)
${\displaystyle R^{*}}$ = Universal gas constant for air: 8.31432×103 N·m / (kmol·K)
${\displaystyle g_{0}}$ = Gravitational constant (9.80665 m/s²)
${\displaystyle M}$ = Molar mass of Earth's air (28.9644 g/mol)

The value of subscript b ranges from 0 to 6 in accordance with each of seven successive layers of the atmosphere shown in the table below. In these equations, g0, M and R* are each single-valued constants, while P, L, T, and h are multivalued constants in accordance with the table below. It should be noted that the values used for M, g0, and ${\displaystyle R^{*}}$ are in accordance with the U.S. Standard Atmosphere, 1976, and that the value for ${\displaystyle R^{*}}$ in particular does not agree with standard values for this constant.[1] The reference value for Pb for b = 0 is the defined sea level value, P0 = 101325 pascals or 29.92126 inHg. Values of Pb of b = 1 through b = 6 are obtained from the application of the appropriate member of the pair equations 1 and 2 for the case when ${\displaystyle h=h_{b+1}}$.:[1]

Subscript b Height Above Sea Level Static Pressure Standard Temperature
(K)
Temperature Lapse Rate
(m) (ft) (pascals) (inHg) (K/m) (K/ft)
0 0 0 101325 29.92126 288.15 -0.0065 -0.0019812
1 11,000 36,089 22632.1 6.683245 216.65 0.0 0.0
2 20,000 65,617 5474.89 1.616734 216.65 0.001 0.0003048
3 32,000 104,987 868.019 0.2563258 228.65 0.0028 0.00085344
4 47,000 154,199 110.906 0.0327506 270.65 0.0 0.0
5 51,000 167,323 66.9389 0.01976704 270.65 -0.0028 -0.00085344
6 71,000 232,940 3.95642 0.00116833 214.65 -0.002 -0.0006096

Assuming this table is accurate:

Find a general value of the atmospheric pressure using the height you found previously with this formula. Considering that the ultimate compressive strength was probably determined in a sea-level atmosphere, what's the new maximum height of these materials?

#### Part 3. (Medium Hard)

3.) That added a few meters. Perhaps more interesting is the change of the earth's gravity. Considering the earth's gravity changes as follows:

${\displaystyle g{\frac {R^{2}}{(R+h)^{2}}}\,}$

Where:

• g is the gravity at a radius of R (i.e. the base of the tower)
• h is the height of a particle above the earth's surface
• R is the radius of the earth

Find the new maximum height in terms of the given quantities. And toss on the atmospheric pressure as well.

#### Part 4. (Hard)

4.) Well, that added a few more meteres as well. But to truly increase the height, one would need a non-constant cross sectional area. So, what if the tower was, in fact, a cone? Find an equation relating the maximum height of the tower to the cone angle of the tower using the quantities given. If the base of the tower could only be ${\displaystyle D}$ meters in diameter, what's the maximum height the tower can be (ignore the atmospheric pressure and gravity change, otherwise the integral is impossible to evaluate).

#### Part 5. (Very Hard)

5.) So while a cone could make for a very useful amount of height, what if you could pick any function for the area as a function of height? Is it possible, using the weakening of the earth's gravitational field, to make a tower that could reach an indefinite height? If not, what shape gives the maximum height of the tower, assuming that the diameter of the tower can't be greater then ${\displaystyle D}$. (Note: No solution found so far).

I'll post the solutions later. I did this problem as a mental exercise at one point so I know the towers generally hit a couple miles to almost a dozen miles for some of the lighter materials out there.

## References

1. a b U.S. Standard Atmosphere, 1976, U.S. Government Printing Office, Washington, D.C., 1976. (Linked file is very large.)