Chapter 1, Problem 30P

Convert the following to appropriate SI units:

1. (a)   A length, l = 5 ft.
2. (b)   A stress, σ = 90 kpsi.
3. (c)   A pressure, p = 25 psi.
4. (d)   A section modulus. Z = 12 in³.
5. (e)   A unit weight, w = 0.208 Ibf/in.
6. (f)     A deflection. δ = 0.001 89 in.
7. (g)   A velocity, v = 1 200 ft/min.
8. (h)   A unit strain. ∈ = 0.002 15 in/in.
9. (i)     A volume. V = 1830 in³.

To determine

The length in $\text{SI}$ unit.

Answer to Problem 30P

The length in $\text{SI}$ unit is $1.524\text{m}$.

Explanation of Solution

Write the expression for length.

$l_1=l\left(n_1\right)$ (I)

Here, the length in $\text{SI}$ unit is $l_1$, the length in $\text{ft}$ is $l$ and conversion factor is $n_1$.

Write the expression for conversion factor.

$n_1=\left(\frac{0.3048\text{m}}{1\text{ft}}\right)$ (II)

Conclusion:

Substitute $\left(\frac{0.3048\text{m}}{1\text{ft}}\right)$ for $n_1$ in Equation (I).

$l_1=l\left(\frac{0.3048\text{m}}{1\text{ft}}\right)$ (III)

Substitute $5\text{ft}$ for $l$ in Equation (III).

$\begin{array}{c}{l}_1=\left(5\text{ft}\right)\left(\frac{0.3048\text{m}}{1\text{ft}}\right)\\ =1.524\text{m}\end{array}$

Thus, the length in $\text{SI}$ unit is $1.524\text{m}$.

To determine

The stress in $\text{SI}$ unit.

Answer to Problem 30P

The stress in $\text{SI}$ unit is $620.55\text{MPa}$.

Explanation of Solution

Write the expression for stress.

${\sigma }_1=\sigma \left(n_2\right)$ (IV)

Here, the stress in $\text{SI}$ unit is ${\sigma }_1$, the stress in $\text{kpsi}$ is $\sigma$ and conversion factor for stress is $n_2$

Write the expression for conversion factor.

$n_2=\left(\frac{6.895\text{MPa}}{1\text{kpsi}}\right)$ (V)

Conclusion:

Substitute $\left(\frac{6.895\text{MPa}}{1\text{kpsi}}\right)$ for $n_2$ in Equation (IV).

${\sigma }_1=\sigma \left(\frac{6.895\text{MPa}}{1\text{kpsi}}\right)$ (VI)

Substitute $90\text{kpsi}$ for $\sigma$ in Equation (VI).

$\begin{array}{c}{\sigma }_1=\left(90\text{kpsi}\right)\left(\frac{6.895\text{MPa}}{1\text{kpsi}}\right)\\ =620.55\text{MPa}\end{array}$

Thus, the stress in $\text{SI}$ unit is $620.55\text{MPa}$.

To determine

The pressure in $\text{SI}$ unit.

Answer to Problem 30P

The pressure in $\text{SI}$ unit is $172.375\text{kPa}$.

Explanation of Solution

Write the expression for pressure.

$p_1=p\left(n_3\right)$ (VII)

Here, the pressure in $\text{SI}$ unit is $p_1$, the pressure in $\text{psi}$ is $p$ and conversion factor for pressure is $n_3$

Write the expression for conversion factor.

$n_3=\left(\frac{6.895\text{kPa}}{1\text{psi}}\right)$ (VIII)

Conclusion:

Substitute $\left(\frac{6.895\text{kPa}}{1\text{psi}}\right)$ for $n_3$ in Equation (VII).

$p_1=p\left(\frac{6.895\text{kPa}}{1\text{psi}}\right)$ (IX)

Substitute $25\text{psi}$ for $p$ in Equation (IX).

$\begin{array}{c}{p}_1=\left(25\text{psi}\right)\left(\frac{6.895\text{kPa}}{1\text{psi}}\right)\\ =172.375\text{kPa}\end{array}$

Thus, pressure in $\text{SI}$ unit is $172.375\text{kPa}$.

To determine

The section modulus in $\text{SI}$ unit.

Answer to Problem 30P

The section modulus in $\text{SI}$ unit is $196.64{\text{cm}}^3$.

Explanation of Solution

Write the expression for section modulus.

$Z_1=Z\left(n_4\right)$ (X)

Here, the section modulus in $\text{SI}$ unit is $Z_1$, the section modulus in ${\text{in}}^3$ is $Z$ and conversion factor for section modulus is $n_4$

Write the expression for conversion factor.

$n_4=\left(\frac{{\left(2.54\text{cm}\right)}^3}{{\left(1\text{in}\right)}^3}\right)$ (XI)

Conclusion:

Substitute $\left(\frac{{\left(2.54\text{cm}\right)}^3}{{\left(1\text{in}\right)}^3}\right)$ for $n_4$ in Equation (X).

$Z_1=Z\left(\frac{{\left(2.54\text{cm}\right)}^3}{{\left(1\text{in}\right)}^3}\right)$ (XII)

Substitute $12{\text{in}}^3$ for $Z$ in Equation (XII).

$\begin{array}{c}{Z}_1=\left(12{\text{in}}^3\right)\left(\frac{{\left(2.54\text{cm}\right)}^3}{{\left(1\text{in}\right)}^3}\right)\\ =196.6447{\text{cm}}^3\\ \approx 196.64{\text{cm}}^3\end{array}$

Thus, the section modulus in $\text{SI}$ unit is $196.64{\text{cm}}^3$.

To determine

The unit weight in $\text{SI}$ unit.

Answer to Problem 30P

The unit weight in $\text{SI}$ unit is $36.43\text{N}/\text{m}$.

Explanation of Solution

Write the expression for unit weight.

$w_1=w\left(n_5\right)$ (XIII)

Here, the unit weight in $\text{SI}$ unit is $w_1$, the unit weight in ${\text{cm}}^4$ is $w$ and conversion factor for unit weight is $n_5$.

Write the expression for conversion factor.

$n_5=\left(\frac{175.127\text{N}/\text{m}}{1\text{lbf}/\text{in}}\right)$ (XIV)

Conclusion:

Substitute $\left(\frac{175.127\text{N}/\text{m}}{1\text{lbf}/\text{in}}\right)$ for $n_5$ in Equation (XIII).

$w_1=w\left(\frac{175.127\text{N}/\text{m}}{1\text{lbf}/\text{in}}\right)$ (XV)

Substitute $0.208\text{lbf}/\text{in}$ for $w$ in Equation (XV).

$\begin{array}{c}{w}_1=\left(0.208\text{lbf}/\text{in}\right)\left(\frac{175.127\text{N}/\text{m}}{1\text{lbf}/\text{in}}\right)\\ =36.4264\text{N}/\text{m}\\ \approx 36.43\text{N}/\text{m}\end{array}$

Thus, the unit weight in $\text{SI}$ unit is $36.43\text{N}/\text{m}$.

To determine

The deflection in $\text{SI}$ unit.

Answer to Problem 30P

The deflection in $\text{SI}$ unit is $0.048\text{mm}$.

Explanation of Solution

Write the expression for deflection.

${\delta }_1=\delta \left(n_6\right)$ (XVI)

Here, the deflection in $\text{SI}$ unit is ${\delta }_1$, the deflection in $\text{in}$ is $\delta$ and conversion factor for deflection is $n_6$.

Write the expression for conversion factor.

$n_6=\left(\frac{25.4\text{mm}}{1\text{in}}\right)$ (XVII)

Conclusion:

Substitute $\left(\frac{25.4\text{mm}}{1\text{in}}\right)$ for $n_6$ in Equation (XVI).

${\delta }_1=\delta \left(\frac{25.4\text{mm}}{1\text{in}}\right)$ (XVIII)

Substitute $0.00189\text{in}$ for $\delta$ in Equation (XVIII).

$\begin{array}{c}{\delta }_1=\left(0.00189\text{in}\right)\left(\frac{25.4\text{mm}}{1\text{in}}\right)\\ =0.0480\text{mm}\\ \approx 0.048\text{mm}\end{array}$

Thus, the deflection in $\text{SI}$ unit is $0.048\text{mm}$.

To determine

The velocity in $\text{SI}$ unit.

Answer to Problem 30P

The velocity in $\text{SI}$ unit is $6.096\text{m}/\text{s}$.

Explanation of Solution

Write the expression for velocity.

$v_1=v\left(n_7\right)$ (XIX)

Here, the velocity in $\text{SI}$ unit is $v_1$, the velocity in $\text{ft}/\text{min}$ is $v$ and conversion factor for velocity is $n_7$.

Write the expression for conversion factor.

$n_7=\left(\frac{0.00508\text{m}/\text{s}}{1\text{ft}/\text{min}}\right)$ (XX)

Conclusion:

Substitute $\left(\frac{0.00508\text{m}/\text{s}}{1\text{ft}/\text{min}}\right)$ for $n_7$ in Equation (XIX).

$v_1=v\left(\frac{0.00508\text{m}/\text{s}}{1\text{ft}/\text{min}}\right)$ (XXI)

Substitute $1200\text{ft}/\text{min}$ for $v$ in Equation (XXI).

$\begin{array}{c}{v}_1=\left(1200\text{ft}/\text{min}\right)\left(\frac{0.00508\text{m}/\text{s}}{1\text{ft}/\text{min}}\right)\\ =6.096\text{m}/\text{s}\end{array}$

Thus, the velocity in $\text{SI}$ unit is $6.096\text{m}/\text{s}$.

To determine

The unit strain in $\text{SI}$ unit.

Answer to Problem 30P

The unit strain in $\text{SI}$ unit is $0.00215\text{mm}/\text{mm}$.

Explanation of Solution

Write the expression for unit strain.

${\epsilon }_1=\epsilon \left(n_8\right)$ (XXII)

Here, the unit strain in $\text{SI}$ unit is ${\epsilon }_1$, the unit strain in $\text{in}/\text{in}$ is $\epsilon$ and conversion factor for unit strain is $n_8$.

Write the expression for conversion factor.

$n_8=\left(\frac{\left(\frac{25.4\text{mm}}{25.4\text{mm}}\right)}{\left(\frac{\text{1}\text{in}}{\text{1}\text{in}}\right)}\right)$ (XXIII)

Conclusion:

Substitute $\left(\frac{\left(\frac{25.4\text{mm}}{25.4\text{mm}}\right)}{\left(\frac{\text{1}\text{in}}{\text{1}\text{in}}\right)}\right)$ for $n_8$ in Equation (XXII).

${\epsilon }_1=\epsilon \left(\frac{\left(\frac{25.4\text{mm}}{25.4\text{mm}}\right)}{\left(\frac{\text{1}\text{in}}{\text{1}\text{in}}\right)}\right)$ (XXIV)

Substitute $0.00215\text{in}/\text{in}$ for $\epsilon$ in Equation (XXIV).

$\begin{array}{c}{\epsilon }_1=\left(0.00215\text{in}/\text{in}\right)\left(\frac{\left(\frac{25.4\text{mm}}{25.4\text{mm}}\right)}{\left(\frac{\text{1}\text{in}}{\text{1}\text{in}}\right)}\right)\\ =0.00215\text{mm}/\text{mm}\end{array}$

Thus, the unit strain in $\text{SI}$ unit is $0.00215\text{mm}/\text{mm}$.

To determine

The volume in $\text{SI}$ unit.

Answer to Problem 30P

The volume in $\text{SI}$ unit is $29.988\times {10}^6{\text{mm}}^3$.

Explanation of Solution

Write the expression for volume.

$V_1=V\left(n_8\right)$ (XXV)

Here, the volume in $\text{SI}$ unit is $V_1$, the volume in ${\text{in}}^3$ is $V$ and conversion factor for volume is $n_8$.

Write the expression for conversion factor.

$n_8=\left(\frac{{\left(25.4\text{mm}\right)}^3}{{\left(1\text{in}\right)}^3}\right)$ (XXVI)

Conclusion:

Substitute $\left(\frac{{\left(25.4\text{mm}\right)}^3}{{\left(1\text{in}\right)}^3}\right)$ for $n_8$ in Equation (XXV).

$V_1=V\left(\frac{{\left(25.4\text{mm}\right)}^3}{{\left(1\text{in}\right)}^3}\right)$ (XXVII)

Substitute $1830{\text{in}}^3$ for $V$ in Equation (XXVII).

$\begin{array}{c}{V}_1=\left(1830{\text{in}}^3\right)\left(\frac{{\left(25.4\text{mm}\right)}^3}{{\left(1\text{in}\right)}^3}\right)\\ =\left(1830{\text{in}}^3\right)\left(\frac{\left(16387.064{\text{mm}}^3\right)}{{\left(1\text{in}\right)}^3}\right)\\ =29.9883\times {10}^6{\text{mm}}^3\\ \approx 29.988\times {10}^6{\text{mm}}^3\end{array}$

Thus, the volume in $\text{SI}$ unit is $29.988\times {10}^6{\text{mm}}^3$.