ME 499/699 Materials Selection

Homework -1 Solutions

1. Using Materials Selection Charts

(a) Using the E-ρ chart identify metals with both E > 100 GPa and E/ρ > 23 GPa/(Mg/m3). Solution

• Stage-1: Tree stage to select metals • Stage-2: Graph stage E-ρ chart to isolate metals with E>100GPa • Stage-3: Graph stage E-ρ chart

• Draw line with slope 1 • Edit Stage-3, and modify Selection. Enter X=1 Mg/m3 and Y=23 GPa. If the plot has X axis in kg/m3 then enter X=1000 kg/m3.

• Resultant selection is: Name Identity Cast iron, ductile (nodular) Commercially pure titanium High carbon steel Low alloy steel Low carbon steel Medium carbon steel Nickel Nickel-based superalloys Nickel-chromium alloys Stainless steel Titanium alloys

Three classes – steels, titanium alloys and nickel alloys

(b) Using the E-ρ chart establish whether woods have a higher specific stiffness E/ρ than epoxies.

Solution

• Certain wood products have E/ρ greater than epoxies Name Identity Bamboo BA Hardwood: oak, across grain Hardwood: oak, along grain Plywood Softwood: pine, across grain Softwood: pine, along grain

(c) Do titanium alloys have a higher or lower specific strength (strength/density, σ f / ρ) than tungsten alloys? This is important when you want strength at low weight (landing gear of aircraft, mountain bikes).

Solution

• Titanium alloys have a higher specific strength

2. Translating design requirements:

For the problems below, the best material for minimizing cost is to be selected. Translate the design requirements into the four steps – function, constraints, objectives and free variables. Assign numerical values for the requirements where possible.

(a) A material is required to manufacture office scissors. Paper is an abrasive material, and scissors sometimes encounter hard obstacles like staples.

Solution

• To resist abrasive wear the scissors must have blades of high hardness. In cutting, they will sooner or later encounter a staple or other hard obstruction that would chip a brittle blade – some toughness is required. These two parameters help reduce wear, but there are other factors that influence it, so it is sensible to specify good wear resistance. Finally, the scissors must be formed – if the handles are integral with the blades, they must be forged or stamped from sheet, requiring the ability to be processed in this way. • The design requirements can be classified into the four steps of material selection

Function • Scissors Constraints • High hardness • Adequate toughness: K1c > 15MPa.m0.5

• Good wear resistance • Able to be forged Objective • Minimize material cost Free variables • Choice of material

(b) A material is required for the windings of an electric air-furnace capable of temperatures

up to 1000ºC. Think out what attributes a material must have if it is to be made into windings and function properly in a furnace.

Solution

• If the material is to be used as windings it must be able to be drawn to wire and wound to a coil, requiring ductility. It must conduct electricity and be able to operate at 1000oC in air.

Function • High temperature furnace winding Constraints • Maximum service temperature, Tmax > 1000º C • Able to be rolled or drawn to wire • Good electrical conductor • Some ductility so that it can be wound, ε f > 2% • Good resistance to oxidation at elevated temperature Objective • Minimize material cost Free variables • Choice of material

3. Derivation of material indices:

The problems below deal with cantilever beams with a square cross-section and fixed length. Derive the material index M for case.

(a) A cantilever beam of given length L and fixed square cross-section (of side t) is loaded at its end by a load F. In order to minimize deflection, show that the material index to be maximized is M = E, where E is Young's modulus (neglect self-weight).

Solution

Function • End loaded cantilever beam Constraints • Length L is specified • Cross-section t x t is specified • End load F is specified Objective • Minimize deflection δ Free variables • Choice of material

The deflection of the end of an end-loaded cantilever beam is (See Appendix A.3)

( ) ⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛=

==

=

EtLF

tbhI

EIFL

14

1212

3

4

3

43

3

δ

δ

The last equation expresses the deflection as the product of the “loading”, “geometry” and “material” indices. To minimize deflection, the material index to be minimized is (1/E) or the material index to be maximized is

EM =

(b) A cantilever beam of given length L and fixed square cross-section (of side t) deflects

under its own weight (w per unit length). In order to minimize deflection, show that the material index to be maximized is M = E/ρ, where E is Young's modulus and ρ is the density.

Solution

Function • Self loaded cantilever beam Constraints • Length L is specified • Cross-section t x t is specified Objective • Minimize deflection δ Free variables • Choice of material

The beam is subject to a self load per unit length of

2gtw ρ= The deflection of the end of a cantilever beam subject to a distributed load of f is

( ) ( ) ⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛===

EtLg

tELgt

EIwL ρρδ 2

4

4

424

23

12/88

In order to minimize deflection δ, the material index to be maximized is

ρEM =

(c) A cantilever beam of given length L and square cross-section (i.e. size is not given)

deflects under its own weight (w per unit length). Show that for the lightest beam that does not deflect more than a given value δ, the material index to be maximized is M = E/ρ2, where E is Young's modulus and ρ is the density.

Solution

Function • End loaded cantilever beam Constraints • Length L is specified • maximum deflection δ is specified Objective • Minimize mass Free variables • Choice of material

• Choice of cross section A=t2

The mass of the beam, which is to be minimized, is

2Ltm ρ=

The deflection of a beam under self loading from part (b) above is:

( )

( ) ⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛=

ELgt

EtLg

ρδ

ρδ

42

2

4

23

23

Substituting for t2 into the mass equation gives

( )

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛=

ELg

ELgLm

25

4

23

23

ρδ

ρδ

ρ

In order to minimize mass m, the material index to be maximized is

2ρEM =