Chapter 6.13, Problem 80SEP

  The material for a rod of cross-sectional area 2.70 in.² and length 75.0 in. must be selected such that under an axial load of 120.000.0 lb. it will not yield and the elongation in the bar will remain below 0.105 in. (a) Provide a list of at least three different metals that would satisfy these conditions. (b) Narrow the list down if cost is an issue, (c) Narrow the list down if corrosion is an issue. Use Appendix I for properties and cost of common alloys only.

To determine

The list of three different metals which satisfy the given condition.

Explanation of Solution

Write the expression for the stress acting in the rod.

  $\sigma =\frac{F}{A_0}$                                                                                                                    …… (I)

Here, applied force is $F$, and area of the rod is$A_0$.

Write the expression for the engineering strain as given as follows:

  $\epsilon =\frac{\Delta l}{l}$                                                                                                                    …… (II)

Here, change in length is $\Delta l$, and initial length is l.

Write the expression for the Hook’s law as given as follows:

  $\sigma =E\epsilon$

  $E=\frac{\sigma }{\epsilon }$                                                                                                                   …… (III)

Here, stress is$\sigma$, young’s modulus is $E$, and engineering strain is $\epsilon$.

Conclusion:

Substitute 120,000 lb for $F$ and $2.7{\text{in}}^2$ for $A_0$ in equation (I).

  $\begin{array}{c}\sigma =\frac{120,000\text{ lb}}{2.7{\text{in}}^2}\\ =44444\frac{\text{lb}}{{\text{in}}^{\text{2}}}\times \frac{1\text{ksi}}{1000{\text{lb/in}}^2}\\ =44.44\text{ksi}\end{array}$

Substitute 0.105 in for $\Delta l$ and 75 in for l in equation (II).

  $\begin{array}{c}\epsilon =\frac{0.105\text{in}}{75\text{in}}\\ =0.0014\text{in/in}\end{array}$

The given rod is loaded with in the elastic region so hook’s law can be applied for obtaining the stress and strain values.

Substitute $44.44\text{ksi}$ for $\sigma$ and $0.0014\text{in/in}$ for $\epsilon$ in equation (III).

  $\begin{array}{c}E=\frac{44.4\text{ ksi}}{0.0014\text{in/in}}\\ =31714.3\text{ksi}\times \frac{1\text{ GPa}}{145.038\text{ksi}}\\ =218.66\text{GPa}\\ \approx \text{200}\text{GPa}\end{array}$

The material that needs to satisfy the given specification should possess at least 200 GPa as Young’s modulus.

Refer the Appendix I, “5. Room temperature modulus of elasticity” obtain the following metals whose Young’s modulus values are equal to or more than 200 GPa.

1. 1. Low alloy steels (Alloy 1006 cold drawn)
2. 2. Stainless steels (Alloy 405 cold rolled annealed)
3. 3. Miscellaneous alloys (Hayness 25 cold rolled and annealed)

To determine

Choose the material based on the cost of the material.

Explanation of Solution

Refer the Appendix I “16. Cost and relative cost of some selected materials” obtain the cost of metals that are chosen in part (a) whose young’s modulus values are more than 218.66 GPa .

1. 1. Low alloy steels (Alloy 1006 cold drawn) – 2.91 US $/lb
2. 2. Stainless steels (Alloy 405 cold rolled annealed) – 394.79 US $/lb
3. 3. Miscellaneous alloys (Hayness 25 cold rolled and annealed) – 40US $/lb

If cost is an issue, the best choice is Low alloy steels (Alloy 1006 cold drawn).

To determine

Choose the material based on the cost of the corrosion.

Explanation of Solution

Refer part (a).

The materials are,

1. 1. Low alloy steels (Alloy 1006 cold drawn)
2. 2. Stainless steels (Alloy 405 cold rolled annealed)
3. 3. Miscellaneous alloys (Hayness 25 cold rolled and annealed)

Residual stresses are induced when the material is cold worked. This residual stresses can cause corrosion, with proper annealing heat treatment residual stresses can be removed. Hence, comparing the stainless steels with miscellaneous alloys, the stainless steels will be the good choice and prevents the corrosion. Stainless steel alloys are readily available in the market.