Chapter 1, Problem 29P

Convert the following to appropriate ips units:

1. (a)   A stress, σ = 150 MPa.
2. (b)   A force, F = 2 kN.
3. (c)   A moment, M = 150 N · m.
4. (d)   An area, A = 1 500 mm².
5. (e)   A second moment of area, I = 750 cm⁴.
6. (f)     A modulus of elasticity, E = 145 GPa.
7. (g)   A speed, v = 15 km/h.
8. (h)   A volume, V = 1 liter.

To determine

The stress in ips unit.

Answer to Problem 29P

The stress in ips unit is $21.8\text{kpsi}$.

Explanation of Solution

To convert the units from SI unit to ips unit, divide the quantity by the conversion factor.

Calculate the stress in ips unit.

${\sigma }_{\text{kpsi}}=\frac{{\sigma }_{\text{MPa}}}{A}$ (I)

Here, stress in ips unit is ${\sigma }_{\text{kpsi}}$, stress in the given unit is ${\sigma }_{\text{MPa}}$ and the conversion factor is $A$.

Conclusion:

Refer Table A-2 “Conversion factors $A$ to convert input $X$ to output $Y$ using the formula $Y=AX$”.

The conversion factor for conversion of $\text{MPa}$ to $\text{kpsi}$ is $6.89\text{MPa}/\text{kpsi}$.

Substitute $150\text{MPa}$ for ${\sigma }_{\text{MPa}}$ and $6.89\text{MPa}/\text{kpsi}$ for $A$ in Equation (I).

$\begin{array}{c}{\sigma }_{\text{kpsi}}=\frac{150\text{MPa}}{6.89\text{MPa}/\text{kpsi}}\\ =21.8\text{kpsi}\end{array}$

Thus, the stress in ips unit is $21.8\text{kpsi}$.

To determine

The force in ips unit.

Answer to Problem 29P

The force in ips unit is $449\text{lbf}$.

Explanation of Solution

To convert the units from SI unit to ips unit, divide the quantity by the conversion factor.

Calculate the force in ips unit.

$F_{\text{lbf}}=\frac{F_{\text{kN}}}{A}$ (II)

Here, force in ips unit is $F_{\text{lbf}}$, force in the given unit is $F_{\text{kN}}$ and the conversion factor is $A$.

Conclusion:

Refer Table A-2 “Conversion factors $A$ to convert input $X$ to output $Y$ using the formula $Y=AX$”.

The conversion factor for conversion of $\text{kN}$ to $\text{lbf}$ is $4.45\text{kN}/\text{lbf}$.

Substitute $2\text{kN}$ for $F_{\text{kN}}$ and $4.45\text{kN}/\text{lbf}$ for $A$ in Equation (II).

$\begin{array}{c}{F}_{\text{lbf}}=\frac{2\text{kN}}{4.45\text{kN}/\text{lbf}}\\ =449\text{lbf}\end{array}$

Thus, the force in ips unit is $449\text{lbf}$.

To determine

The moment in ips unit.

Answer to Problem 29P

The moment in ips unit is $\text{1}\text{.33}\text{kip}\cdot \text{in}$.

Explanation of Solution

To convert the units from SI unit to ips unit, divide the quantity by the conversion factor.

Calculate the moment in ips unit.

$M_{\text{kip}\cdot \text{in}}=\frac{M_{\text{N}\cdot \text{m}}}{A}$ (III)

Here, moment in ips unit is $M_{\text{lbf}}$, moment in the given unit is $M_{\text{kN}}$ and the conversion factor is $A$.

Conclusion:

Refer Table A-2 “Conversion factors $A$ to convert input $X$ to output $Y$ using the formula $Y=AX$”.

The conversion factor for conversion of $\text{N}\cdot \text{m}$ to $\text{lbf}\cdot \text{in}$ is $0.113\text{N}\cdot \text{m}/\text{lbf}\cdot \text{in}$.

Substitute $150\text{N}\cdot \text{m}$ for $M_{\text{N}\cdot \text{m}}$ and $0.113\text{N}\cdot \text{m}/\text{lbf}\cdot \text{in}$ for $A$ in the Equation (III).

$\begin{array}{c}{M}_{\text{kip}\cdot \text{in}}=\frac{150\text{N}\cdot \text{m}}{0.113\text{N}\cdot \text{m}/\text{lbf}\cdot \text{in}}\\ =\left(1327\text{lb}\cdot \text{in}\right)\left(\frac{1\text{kip}}{1000\text{lbf}}\right)\\ \approx \text{1}\text{.33}\text{kip}\cdot \text{in}\end{array}$

Thus, the moment in ips unit is $\text{1}\text{.33}\text{kip}\cdot \text{in}$.

To determine

The area in ips unit.

Answer to Problem 29P

The area in ips unit is $2.33{\text{in}}^2$.

Explanation of Solution

To convert the units from SI unit to ips unit, divide the quantity by the conversion factor.

Calculate the area in ips unit.

$A_{{\text{in}}^2}=\frac{A_{{\text{mm}}^{\text{2}}}}{A}$ (IV)

Here, area in ips unit is $A_{{\text{in}}^2}$, area in the given unit is $A_{{\text{mm}}^{\text{2}}}$ and the conversion factor is $A$.

Conclusion:

Refer Table A-2 “Conversion factors $A$ to convert input $X$ to output $Y$ using the formula $Y=AX$”.

The conversion factor for conversion of ${\text{mm}}^2$ to ${\text{in}}^2$ is $645{\text{mm}}^2/{\text{in}}^2$.

Substitute $1500{\text{mm}}^2$ for $A_{{\text{mm}}^2}$ and $645{\text{mm}}^2/{\text{in}}^2$ for $A$ in Equation (IV).

$\begin{array}{c}{A}_{{\text{in}}^2}=\frac{1500{\text{mm}}^2}{645{\text{mm}}^2/{\text{in}}^2}\\ =2.33{\text{in}}^2\end{array}$

Thus, the moment in ips unit is $2.33{\text{in}}^2$.

To determine

The second moment of area in ips unit.

Answer to Problem 29P

The second moment of area in ips unit is $18.0{\text{in}}^4$.

Explanation of Solution

To convert the units from SI unit to ips unit, divide the quantity by the conversion factor.

Calculate the second moment of area in ips unit.

$I_{\text{in}4}=\frac{I_{{\text{cm}}^4}}{A}$ (V)

Here, second moment of area in ips unit is $I_{{\text{in}}^4}$, second moment of area in the given unit is $I_{{\text{Cm}}^4}$ and the conversion factor is $A$.

Conclusion:

Refer Table A-2 “Conversion factors $A$ to convert input $X$ to output $Y$ using the formula $Y=AX$”.

The conversion factor for the conversion of $\text{cm}$ to $\text{in}$ is $2.54\text{cm}/\text{in}$. So, the conversion factor for the conversion of ${\text{mm}}^4$ to ${\text{in}}^{\text{4}}$ is ${\left(2.54\text{cm}/\text{in}\right)}^4$.

Substitute $750{\text{cm}}^4$ for $I_{{\text{Cm}}^4}$ and ${\left(2.54\text{cm}/\text{in}\right)}^4$ for $A$ in Equation (V).

$\begin{array}{c}{I}_{{\text{in}}^4}=\frac{750{\text{cm}}^4}{{\left(2.54\text{cm}/\text{in}\right)}^4}\\ =18.0{\text{in}}^4\end{array}$

Thus, the second moment of area in ips unit is $18.0{\text{in}}^4$.

To determine

The modulus of elasticity in ips unit.

Answer to Problem 29P

The modulus of elasticity in ips unit is $21.0\text{Mpsi}$.

Explanation of Solution

To convert the units from SI unit to ips unit, divide the quantity by the conversion factor.

Calculate the modulus of elasticity in ips unit.

$E_{\text{Mpsi}}=\frac{E_{\text{GPa}}}{A}$ (VI)

Here, modulus of elasticity in ips unit is $E_{\text{Mpsi}}$, modulus of elasticity in the given unit is $E_{\text{GPa}}$ and the conversion factor is $A$.

Conclusion:

Refer Table A-2 “Conversion factors $A$ to convert input $X$ to output $Y$ using the formula $Y=AX$”.

The conversion factor for conversion of $\text{GPa}$ to $\text{Mpsi}$ is $6.89\text{GPa}/\text{Mpsi}$.

Substitute $6.89\text{GPa}/\text{Mpsi}$ for $A$ and $145\text{GPa}$ for $E_{\text{GPa}}$ in Equation (VI).

$\begin{array}{c}{E}_{\text{Mpsi}}=\frac{145\text{GPa}}{6.89\text{GPa}/\text{Mpsi}}\\ =21.0\text{Mpsi}\end{array}$

Thus, the modulus of elasticity in ips unit is $21.0\text{Mpsi}$.

To determine

The velocity in ips unit.

Answer to Problem 29P

The velocity in ips unit is $46.6\text{mi}/\text{h}$.

Explanation of Solution

To convert the units from SI unit to ips unit, divide the quantity by the conversion factor.

Calculate the velocity in ips unit.

$V_{\text{mi}/\text{h}}=\frac{V_{\text{km}/\text{h}}}{A}$ (VII)

Here, velocity in ips unit is $V_{\text{mi}/\text{h}}$, velocity in the given unit is $V_{\text{km}/\text{h}}$ and the conversion factor is $A$.

Conclusion:

Refer Table A-2 “Conversion factors $A$ to convert input $X$ to output $Y$ using the formula $Y=AX$”.

The conversion factor for conversion of $\text{km}/\text{h}$ to $\text{mi}/\text{h}$ is $1.61\left(\text{km}/\text{h}\right)/\left(\text{mi}/\text{h}\right)$.

Substitute $1.61\left(\text{km}/\text{h}\right)/\left(\text{mi}/\text{h}\right)$ for $A$ and $75\text{km}/\text{h}$ for $V_{\text{km}/\text{h}}$ in Equation (VII).

$\begin{array}{c}{V}_{\text{mi}/\text{h}}=\frac{75\text{km}/\text{h}}{1.61\left(\text{km}/\text{h}\right)/\left(\text{mi}/\text{h}\right)}\\ =46.6\text{mi}/\text{h}\end{array}$

Thus, the velocity in ips unit is $46.6\text{mi}/\text{h}$.

To determine

The volume in ips unit.

Answer to Problem 29P

The volume in ips unit is $1.06\text{qt}$.

Explanation of Solution

The given volume $v$ in SI unit is $1\text{litre}$.

To convert the units from SI unit to ips unit, divide the quantity by the conversion factor.

Calculate the volume in ips unit.

$v_{\text{qt}}=\frac{v_{\text{mililiter}}}{A}$ (VII)

Here, volume in ips unit is $V_{\text{mi}/\text{h}}$, volume in the given unit is $V_{\text{km}/\text{h}}$ and the conversion factor is $A$.

Conclusion:

Refer Table A-2 “Conversion factors $A$ to convert input $X$ to output $Y$ using the formula $Y=AX$”.

The conversion factor for conversion of milliliter to quart $\left(\text{qt}\right)$ is $946\text{mililiter}/\text{qt}$.

Substitute $946\text{mililiter}/\text{qt}$ for $A$ and $1000\text{militer}$ for $v_{\text{mililetr}}$ in Equation /(VIII).

$\begin{array}{c}{v}_{\text{qt}}=\frac{1000\text{militer}}{946\text{mililiter}/\text{qt}}\\ =1.06\text{qt}\end{array}$

Thus, the volume in ips unit is $1.06\text{qt}$.