Chapter 17, Problem 29P

(a)

To determine

Find the average strength-to-weight ratio for the aluminum alloy.

Answer to Problem 29P

The average strength-to-weight ratio for the aluminum alloy is $1372\frac{\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{\frac{\text{kN}}{{\text{m}}^3}}$.

Explanation of Solution

Given data:

Use Table 17.3 in the textbook.

Specific weight of aluminum alloy is $25.5\frac{\text{kN}}{{\text{m}}^3}$.

Use Table 17.4 in the textbook.

Yield strength of aluminum alloy is $35\text{MPa}$.

Calculation:

For Aluminum alloy:

Given,

$\text{Ratio}=\frac{\text{Yield Strength}}{\text{Specific Weight}}$ (1)

Substitute $25.5\frac{\text{kN}}{{\text{m}}^3}$ for $\text{Specific Weight}$ and $35\text{MPa}$ for $\text{Yield Strength}$ in equation (1).

$\begin{array}{c}\text{Ratio}=\frac{35\text{MPa}}{25.5\frac{\text{kN}}{{\text{m}}^3}}\\ =\frac{35\times {10}^6\text{Pa}}{25.5\frac{\text{kN}}{{\text{m}}^3}}\left[\therefore 1\text{MPa}={10}^6\text{Pa}\right]\end{array}$ (2)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (2).

$\begin{array}{c}\text{Ratio}=\frac{35\times {10}^6\frac{\text{N}}{{\text{m}}^{\text{2}}}}{25.5\frac{\text{kN}}{{\text{m}}^3}}\\ =\frac{35\times {10}^6\left({10}^{-3}\frac{\text{kN}}{{\text{m}}^{\text{2}}}\right)}{25.5\frac{\text{kN}}{{\text{m}}^3}}\left[\therefore 1\frac{\text{N}}{{\text{m}}^{\text{2}}}={10}^{-3}\frac{\text{kN}}{{\text{m}}^{\text{2}}}\right]\\ =\frac{35000\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{25.5\frac{\text{kN}}{{\text{m}}^3}}\\ =1372\frac{\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{\frac{\text{kN}}{{\text{m}}^3}}\end{array}$

Thus, the average strength-to-weight ratio for the aluminum alloy is $1372\frac{\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{\frac{\text{kN}}{{\text{m}}^3}}$.

Conclusion:

Hence, the average strength-to-weight ratio for the aluminum alloy is $1372\frac{\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{\frac{\text{kN}}{{\text{m}}^3}}$.

(b)

To determine

Find the average strength-to-weight ratio for the titanium alloy.

Answer to Problem 29P

The average strength-to-weight ratio for the titanium alloy is $17233\frac{\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{\frac{\text{kN}}{{\text{m}}^3}}$.

Explanation of Solution

Given data:

Use Table 17.3 in the textbook.

Specific weight of titanium alloy is $44.1\frac{\text{kN}}{{\text{m}}^3}$.

Use Table 17.4 in the textbook.

Yield strength of titanium alloy is $760\text{MPa}$.

Calculation:

For titanium alloy:

Given,

$\text{Ratio}=\frac{\text{Yield Strength}}{\text{Specific Weight}}$ (3)

Substitute $44.1\frac{\text{kN}}{{\text{m}}^3}$ for $\text{Specific Weight}$ and $760\text{MPa}$ for $\text{Yield Strength}$ in equation (3).

$\begin{array}{c}\text{Ratio}=\frac{760\text{MPa}}{44.1\frac{\text{kN}}{{\text{m}}^3}}\\ =\frac{760\times {10}^6\text{Pa}}{44.1\frac{\text{kN}}{{\text{m}}^3}}\left[\therefore 1\text{MPa}={10}^6\text{Pa}\right]\end{array}$ (4)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (4).

$\begin{array}{c}\text{Ratio}=\frac{760\times {10}^6\frac{\text{N}}{{\text{m}}^{\text{2}}}}{44.1\frac{\text{kN}}{{\text{m}}^3}}\\ =\frac{760\times {10}^6\left({10}^{-3}\frac{\text{kN}}{{\text{m}}^{\text{2}}}\right)}{44.1\frac{\text{kN}}{{\text{m}}^3}}\left[\therefore 1\frac{\text{N}}{{\text{m}}^{\text{2}}}={10}^{-3}\frac{\text{kN}}{{\text{m}}^{\text{2}}}\right]\\ =\frac{760000\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{44.1\frac{\text{kN}}{{\text{m}}^3}}\\ =17233\frac{\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{\frac{\text{kN}}{{\text{m}}^3}}\end{array}$

Thus, the average strength-to-weight ratio for the titanium alloy is $17233\frac{\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{\frac{\text{kN}}{{\text{m}}^3}}$.

Conclusion:

Hence, the average strength-to-weight ratio for the titanium alloy is $17233\frac{\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{\frac{\text{kN}}{{\text{m}}^3}}$.

(c)

To determine

Find the average strength-to-weight ratio for the steel.

Answer to Problem 29P

The average strength-to-weight ratio for the steel is $2597\frac{\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{\frac{\text{kN}}{{\text{m}}^3}}$.

Explanation of Solution

Given data:

Use Table 17.3 in the textbook.

Specific weight of steel is $77.0\frac{\text{kN}}{{\text{m}}^3}$.

Use Table 17.4 in the textbook.

Yield strength of steel (structural) is $200\text{MPa}$.

Calculation:

For steel:

Given,

$\text{Ratio}=\frac{\text{Yield Strength}}{\text{Specific Weight}}$ (5)

Substitute $77.0\frac{\text{kN}}{{\text{m}}^3}$ for $\text{Specific Weight}$ and $200\text{MPa}$ for $\text{Yield Strength}$ in equation (5).

$\begin{array}{c}\text{Ratio}=\frac{200\text{MPa}}{77.0\frac{\text{kN}}{{\text{m}}^3}}\\ =\frac{200\times {10}^6\text{Pa}}{77.0\frac{\text{kN}}{{\text{m}}^3}}\left[\therefore 1\text{MPa}={10}^6\text{Pa}\right]\end{array}$ (6)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (6).

$\begin{array}{c}\text{Ratio}=\frac{200\times {10}^6\frac{\text{N}}{{\text{m}}^{\text{2}}}}{77.0\frac{\text{kN}}{{\text{m}}^3}}\\ =\frac{200\times {10}^6\times {10}^{-3}\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{77.0\frac{\text{kN}}{{\text{m}}^3}}\left[\therefore 1\frac{\text{N}}{{\text{m}}^{\text{2}}}={10}^{-3}\frac{\text{kN}}{{\text{m}}^{\text{2}}}\right]\\ =\frac{200000\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{77.0\frac{\text{kN}}{{\text{m}}^3}}\\ =2597\frac{\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{\frac{\text{kN}}{{\text{m}}^3}}\end{array}$

Thus, the average strength-to-weight ratio for the steel is $2597\frac{\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{\frac{\text{kN}}{{\text{m}}^3}}$.

Conclusion:

Hence, the average strength-to-weight ratio for the steel is $2597\frac{\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{\frac{\text{kN}}{{\text{m}}^3}}$.