Chapter 8, Problem 8.3P

(a)

To determine

The new arm radius.

Answer to Problem 8.3P

  $\text{0}\text{.00638}\text{m}$ would be the new radius for the arm.

Explanation of Solution

Given:

The radius of [resent control arm is $\text{1}\text{cm}$ .

The length of the present control arm is $\text{30}\text{cm}$ .

The maximum control force exerted on the arm is $\text{7}{\text{.85×10}}^{\text{4}}\text{N}$ .

The least ductility provided to sufficient fracture is $15\%$

Formula used:

The expression for the weight of the material is given by

  $W_1=\rho \left(\pi r^{2}l\right)$ …… (I)

Here, $W_1$ is the weight of the material, $r$ is the radius, $\rho$ is the density of low alloy steel and $l$ is the length.

The expression for the new area is given as,

  $A_{\text{n}}=\frac{F}{\left(I\right){\sigma }_Y}$ …… (II)

Here, $A_{\text{n}}$ is the new area, $F$ is force, $I$ is the percentage of metal yield stress and ${\sigma }_Y$ is the yield stress

The expression for arm radius is given as

  $r^2=\frac{A_{\text{n}}}{\pi }$ …… (III)

Here, $A_{\text{n}}$ is a new area.

Calculation:

The weight of material is calculated as,

Substitute $7.85\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ , $\text{30}\text{cm}$ for $l$ and $\text{1}\text{cm}$ for $A$ in equation (I).

  $\begin{array}{c}{W}_1=7.85\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\left(\pi \times \left(1\text{cm}\times \text{0}\text{.01}\text{m}/{\text{cm}}^{\text{2}}\right)\times \left(30\text{cm×0}\text{.01}\text{m}/\text{cm}\right)\right)\\ =\left(7.85\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.424\times {10}^{-3}{\text{m}}^{\text{2}}/\text{cm}\right)\\ =0.7398\text{kg}\left(\frac{{10}^3\times {10}^{-3}{\text{m}}^3}{{\text{m}}^{\text{3}}}\right)\end{array}$

The new area is calculated as,

Substitute $50\%$ for $I$ , $\text{7}{\text{.85×10}}^{\text{4}}\text{N}$ for $F$ and $\text{1225}\text{MPa}$ for ${\sigma }_{\text{Y}}$ in equation (II).

  $\begin{array}{c}{A}_{\text{n}}=\frac{\text{7}{\text{.85×10}}^{\text{4}}\text{N}}{\left(50\%\right)\text{1225}\text{MPa}}\\ =\frac{\text{7}{\text{.85×10}}^{\text{4}}\text{N}}{\left(50\%\right)\text{1225M}\text{Pa}\times \left(\frac{{10}^6\text{Pa}}{\text{1}\text{MPa}}\right)}\\ =1.28\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$

The arm radius is calculated as,

Substitute $1.28\times {10}^{-4}{\text{m}}^{\text{2}}$ for $A_{\text{n}}$ in equation (III).

  $\begin{array}{c}{r}^2=\frac{1.28\times {10}^{-4}{\text{m}}^{\text{2}}}{\pi }\\ =4.0734\times {10}^{-4}{\text{m}}^{\text{2}}\\ r=\sqrt{4..0734\times {10}^{-4}{\text{m}}^{\text{2}}}\\ =0.00638\text{m}\end{array}$

Conclusion:

Therefore, the new arm radius is $\text{0}\text{.00638}\text{m}$ .

(b)

To determine

The weight reduction that is provided by the new steel.

Answer to Problem 8.3P

Areduction of $0.4386\text{kg}$ in the weight occurs by using the new steel.

Explanation of Solution

Formula used:

The expression for new weight of the material is given by,

  $W_2=\rho \left(A_{\text{n}}l\right)$ …… (IV)

Here, $A_n$ is new area, $\rho$ is the density of low alloy steel and $l$ is the length.

Calculation:

The weight of the material is calculated as,

Substitute $7.85\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ , $\text{30}\text{cm}$ for $l$ and $1.28\times {10}^{-4}{\text{m}}^{\text{2}}$ for $A_{\text{n}}$ in equation(IV).

  $\begin{array}{c}{W}_2=7.85\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\left(1.28\times {10}^{-4}{\text{m}}^{\text{2}}\times \text{30}\text{cm}\times 0.01\text{m}/\text{cm}\right)\\ =7.85\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\times 3.84\times {10}^{-5}{\text{m}}^{\text{3}}\\ =0.3014\text{}\text{kg}\end{array}$

The weight saving is calculated as,

  $W=W_1-W_2$ …… (V)

Substitute $0.7398\text{kg}$ for $W_1$ and $0.3014\text{kg}$ for in equation (V)

  $\begin{array}{c}W=0.7398\text{kg}-0.3014\text{kg}\\ =0.4386\text{kg}\end{array}$

Conclusion:

Therefore, the weight reduction that is provided by this new steel is $0.4386\text{kg}$ .