Chapter 8, Multiple Constraints and Conflicting Objectives Video Solutions, Materials Selection in Mechanical Design

Problem 1

Multiple constraints: a light, stiff, strong tie (Fig. E8.1).
A tie, of length $L$ loaded in tension, is to support a load $F$, at minimum weight without failing (implying a constraint on strength) or extending elastically by more than $\delta$ (implying a constraint on stiffness, $F / \delta$ ). The table summarizes the requirements.
a. Follow the method of this chapter to establish two performance equations for the mass, one for each constraint, from which two material indices and one coupling equation linking them are derived. Show that the two indices and the coupling equation are (in order):
$$
M_{1}=\frac{\rho}{E} \quad \text { and } \quad M_{2}=\frac{\rho}{\sigma_{\gamma}} \text { and }\left(\frac{\rho}{\sigma_{\gamma}}\right)=\frac{L}{\delta}\left(\frac{\rho}{E}\right)
$$
b. Use these and the material chart of Fig. E8.2, which has the indices as axes, to identify candidate materials for the tie (1) when the permitted elastic strain is $\delta$ $/ L=10^{-3}$ and (2) when $\delta / L=10^{-2}$.

Khoobchandra Agrawal

Numerade Educator

02:30

Problem 2

Multiple constraints: a cheap column that must not buckle or crush (Fig. E8.3). The best choice of material for a light strong column depends on its aspect ratio: the ratio of its height $H$ to its diameter $D$. This is because short, fat columns fail by crushing; tall slender columns buckle instead. Derive two performance equations for the material cost of a column of solid circular
section and specified height $H$, designed to support a load $F$ large compared to its self-load, the first using the constraints that the column must not crush, the second that it must not buckle. The table summarizes the needs.
a. Proceed as follows
1. Write an expression for the material cost of the column - its mass times its cost per unit mass, $C_{m}$.
2. Express the two constraints as equations and use them to substitute for the free variable, $D$, to find the cost of the column that will just support the load without failing by either mechanism
3. Identify the material indices $M_{1}$ and $M_{2}$ that enter the two equations for the mass, showing that they are
$$
M_{1}=\left(\frac{C_{m} \rho}{\sigma_{c}}\right) \quad \text { and } \quad M_{2}=\left[\frac{C_{m} \rho}{E^{1 / 2}}\right]
$$
b. Data for six possible candidates for the column are listed in the table with this problem. Use these to identify candidate materials when $F=10^{5} \mathrm{~N}$ and $H=3 \mathrm{~m}$. Ceramics are admissible here, because they have high strength in compression. Data for candidate materials for the column
$$
\begin{array}{|l|l|l|l|l|}
\hline \text { Material Density } \rho\left(\mathrm{kg} / \mathbf{m}^{3}\right) & \text { Cost/kg } \mathcal{C}_{\text {is }} \text { (\$/kg) Modulus } E \text { (MPa) Compression strength } \sigma_{t} \text { (MPa) } \\
\hline \text { Wood (spruce) } & 700 & 0.5 & 10,000 & 25 \\
\hline \text { Brick } & 2100 & 0.35 & 22,000 & 95 \\
\hline \text { Granite } & 2600 & 0.6 & 20,000 & 150 \\
\hline \text { Concrete } & 2300 & 0.08 & 20,000 & 13 \\
\hline \text { Cast iron } & 7150 & 0.25 & 130,000 & 200 \\
\hline \text { Structural steel } & 7850 & 0.4 & 210,000 & 300 \\
\hline \text { Al-alloy 6061 } & 2700 & 1.2 & 69,000 & 150 \\
\hline
\end{array}
$$

M Hassan Anwar

Numerade Educator

02:17

Problem 3

Fig. E8.4 shows a material chart with the two indices of Exercise E8.2 as axes. Identify and plot coupling lines for selecting materials for a column with
$F=10^{6} \mathrm{~N}$ and $H=3 \mathrm{~m}$ (the same conditions as above), and for a second column with $F=10^{3} \mathrm{~N}$ and $H=20 \mathrm{~m}$.

James Kiss

Numerade Educator

00:48

Problem 4

Approximate values for exchange constants (1). In the US a typical family car covers 120,000 miles over its life and delivers, on average 25 miles per US gallon (6.6 miles per litre). Petrol in the US costs $\$ 0.63$ per litre, so the life cost for fuel is $\$ 11,455$. The typical family car, when loaded, weighs $2315 \mathrm{~kg}$. If the petrol consumption is directly proportional to the weight, what is the value of reducing the weight of the car by $1 \mathrm{~kg}$ (Fig. E8.5)?

Aadit Sharma

Numerade Educator

06:14

Problem 5

Approximate values for exchange constants (2). The makers of the car shown in Fig. $8.5$ plan to market it in Europe. If the car covers 120,000 miles $(193,000 \mathrm{~km})$ over its life and consumes on average $9.5$ litres per $100 \mathrm{~km}$. Petrol in the Europe costs $€ 1.3$ per litre ( $\$ 1.5$ per litre), so the life cost for fuel is $\$ 27,500$. The car, when loaded, weighs $2315 \mathrm{~kg}$. If the petrol consumption is directly proportional to the weight, what is the value of reducing the weight of the car by $1 \mathrm{~kg}$ ?

Trinity Steen

Numerade Educator

04:22

Problem 5

Approximate values for exchange constants (3). A Routemaster - a London bus - weighs 12 tonnes when empty, 18 tonnes with full. It does, on average, $6.6$ miles $(11 \mathrm{~km})$ per imperial gallon of diesel, which cost $£ 5$ per imperial gallon. The bus travels an average of $57,000 \mathrm{~km}$ per year and has a life of 20 years. Assuming that the average load factor is $50 \%$ (meaning that the bus weighs 15 tonnes on average) and that the fuel consumption is directly proportional to the weight, what is the value of weight saving for a Routemaster? (Data for Routemaster from London Transport.) (Fig. E8.6).

Raymond Matshanda

Numerade Educator

08:33

Problem 7

Conflicting objectives: an air cylinder for a truck (Fig. E8.7). Trucks rely on compressed air for braking and other power-actuated systems. The air is stored in one or a cluster of cylindrical pressure tanks like that shown here (length $L$, diameter $2 R$, hemispherical ends). Most are made of low carbon steel, and they are heavy. The task: to explore the potential of alternative materials for lighter air tanks, recognizing that there must be a trade-off between mass and cost - if it is too expensive, the truck owner will not want it even if it is lighter. The table summarizes the design requirements.
Show that the mass and material cost of the tank described in relative to one made of low carbon steel are given by
$$
\frac{m}{m_{o}}=\left(\frac{\rho}{\sigma_{y}}\right)\left(\frac{\sigma_{1, o}}{\rho_{\rho}}\right) \text { and } \frac{\mathrm{C}}{\mathrm{C}_{\sigma}}=\left(\frac{\mathrm{C}_{m} \rho}{\sigma_{y}}\right)\left(\frac{\sigma_{y \rho}}{C_{m \rho} P_{\rho}}\right)
$$
where $\rho$ is the density, $\sigma_{y}$ the yield strength and $C_{m}$ the cost per kg of the material, and the subscript ' $o$ ' indicates values for mild steel.

Eric Mockensturm

Numerade Educator

07:19

Problem 8

Conflicting objectives: an air cylinder for a truck-material choice. Explore the trade-off between relative cost and relative mass for the air cylinder of Exercise E8.7, considering the replacement of a mild steel tank with one made, first, of low alloy steel, and, second, one made of filamentwound CFRP. The relevant material properties in the table below. Define a relative penalty function $$
Z^{*}=\alpha^{*} \frac{m}{m_{0}}+\frac{C}{C_{0}}
$$
where $a^{\circ}$ is a relative exchange constant, and evaluate $Z$ for $a^{\circ}=1$ and for $a^{\circ}=100$.

Averell Hause

Carnegie Mellon University

04:16

Problem 9

Conflicting objectives: an air cylinder for a truck-the graphical method. Fig. E8.8, below, is a chart with axes of $\mathrm{m} / \mathrm{m}_{o}$ and $\mathrm{C} / \mathrm{C}_{o}$ derived in Exercise E8.7 Mild steel (here labelled 'Low carbon steel') lies at the coordinates (1,1). Sketch a trade-off surface - remember that it is simply a lower envelope to the data, with no mathematical function associated with it. Then plot contours of $Z^{*}$ (see Exercise E8.8) that are approximately tangent to the trade-off surface for $\alpha^{*}=1$ and for $\alpha^{*}=100$. They are plots of the linear relationship $$
\alpha^{*} \frac{m}{m_{0}}=-\frac{1}{\alpha^{*}} \frac{C}{C_{0}}
$$

Abigail Martyr

Numerade Educator

05:02

Problem 10

Conflicting objectives: low-cost insulating walls for freezers and refrigerator trucks (Fig. E8.9). Freezers and refrigerated trucks have panelwalls that provide thermal insulation, and at the same time are stiff, strong and light (stiffness to suppress vibration, strength to tolerate rough usage). To achieve this, the panels are usually of sandwich construction, with two skins of steel, aluminium or GFRP (providing the strength) separated by, and bonded to, a low density insulating core. In choosing the core we seek to minimize thermal conductivity, $\lambda$, and at the same time to maximize stiffness because this allows thinner steel faces, and thus a lighter panel, while still maintaining the overall panel stiffness. The table summarizes the design requirements.

Banhishikha Sinha

Numerade Educator