Question

Materials selection for light springs is shown in Figure 3. A family of lines of slope 2 link materials with equal values of Specific modulus specific strength Technical ceramics Melals and polymers yield sirength, ey Ceramics. glasses: modulus of rupture, MOR Elastomers: tensile tear strength, Composites: tensile failure, o Diamond 8,c SigN Al203 10-1 Composites Yield before 10 Siica glass Soda gass CFRP bucking Nontechnical ceramics Stone Brick Ti alloy CD 10-2 nMetals Wood ng alloys Search area Yield strein of Lead alloys PMMA Polymers 010 Natural materials Design guide lines 0 、PE 010 Foa 0-4er Leatner Cork oans Poiyuretharfe before yiel 10-5」 10-2 10-4 Silicones Elastomers MFA, 6 70-3 10-1 Specific strength, ơhp (MPa/(kg/m3)) Figure 3. A schematic E/p-OjP chart showing guide lines for the material index M2-0f2/pE for materials for light springs. Metals are disadvantageous to other materials because of their high densities Composites and elastomers are excellent One is shown at the value of M2 = 1 kJ/kg. Metals are less attractive than composites and much less attractive than elastomers because of their high density. You can store approximately 18 times more elastic energy, per unit weight in a rubber band than in the best spring steel. Candidates are listed in the Table 3 Write down the candidate materials in Table 3, which are suitable to the value of M2 and the description. Note that candidate materials were denoted by blue circles in Figure 3 and they were described above Table 3. Candidate Materials for Efficient Light Springs M2 of/E (kJ/kg) Description 0.9 2.6 3.9 6.5 0.4 0.9 0.3 0.7 1.3 2.1 Material Expensive than steel, corrosion resistant, but expensive Better than steel; expensive Poor, because of high density On a weight basis, this material makes good springs As good as steel, but with a high loss factor Outstanding; 20 times better than spring steel; but with high loss factor 20-50 Remarks Additional considerations should be given to the selection of a material for a spring. Springs for vehicle suspensions must sustain fatigue and corrosion. Engine valve springs must withstand high temperatures Polymers have a relatively high loss factor and dissipate energy when they vibrate unlike hardened metals. Polymers are not suitable for springs that carry a steady load for long periods of time, because they creep

Answer 1

First of all, lets list down the materials whose M₂ value is to be found out.

CFRP

Steel

Ti Alloys

Wood

PA

Silicon

Polyurethane

1. CFRP:

E = 70 GPa Strength = 600 MPa

Therefore, M₂ = 600² / 70 x 10³ = 5.142 kJ/kg

2. Steel:

E = 180 GPa Strength = 860 MPa

Therefore, M₂ = 860² / 180 x 10³ = 4.10 kJ/kg

3. Ti Alloys:

E = 102.7 GPa Strength = 590 MPa

Therefore, M₂ = 590² / 102.7 x 10³ = 3.38 kJ/kg

4. Wood:

E = 13.9 GPa Strength = 2.7 MPa

Therefore, M₂ = 2.7² / 13.9 x 10³ = 5.24 x 10^-4 kJ/kg

5. PA:

E = 2.5 GPa Strength = 85 MPa

Therefore, M₂ = 85² / 2.5 x 10³ = 2.89 kJ/kg

6. Silicon:

E = 180 GPa Strength = 180 MPa

Therefore, M₂ = 180² / 180 x 10³ = 0.18 kJ/kg

7. Polyurethane:

E = 151.4 MPa Strength = 4.5 MPa

Therefore, M₂ = 4.5² / 151.4 = 0.133 kJ/kg

Material	M2 value
	0.9 - 2.6
CFRP, Steel, Ti Alloys, Wood, PA	3.9 - 6.5
	0.4 - 0.9
Silicon, Polyurethane	0.3 - 0.7
	1.3 - 2.1
	20 - 50