Answer
For AISI 1018 CD steel,
specific strength = 47.42 kN-m/kg
For 2011-T6 aluminum,
specific strength = 62.3 kN-m/kg
For Ti-6 Al-4 V titanium,
specific strength = 187 kN-m/kg
For ASTM No. 40 cast iron,
specific strength = 40.7 kN-m/kg
Work Step by Step
a) for AISI 1018 CD steel,
from table A-20, S_{y} = 370 MPa = 370 * 1000 kPa
from table A-5,
unit weight w of carbon steel = 76.5 kN/cubic-metre
Dividing by gravitational acceleration g = 9.81 $\frac{m}{s^{2}}$,
we get density p = 76.5/9.81 k-kg/$m^{3}$
Multiply by 1000, we get p = 76.5*102 kg/$m^{3}$
$\frac{Sy}{p}$ = $\frac{370*1000}{76.5*102}$ = 47.42 kN-m/kg
b) for 2011-T6 aluminum,
from table A-22, S_{y} = 169 MPa = 169 * 1000 kPa
from table A-5,
unit weight w = 26.6 kN/cubic-metre
we get density p = 26.6*102 kg/$m^{3}$
$\frac{Sy}{p}$ = $\frac{169*1000}{26.6*102}$ = 62.3 kN-m/kg
c) for Ti-6Al-4V titanium,
from table A-24c, S_{y} = 830 MPa = 830 * 1000 kPa
from table A-5,
unit weight w of titanium = 43.4 kN/cubic-metre
we get density p = 43.4*102 kg/$m^{3}$
$\frac{Sy}{p}$ = $\frac{830*1000}{43.4*102}$ = 187 kN-m/kg
d) for ASTM No. 40 cast iron,
For this material, we do not have yield strength, hence we use ultimate tensile strength Sut
from table A-24 a, S_{ut} = 42.5 kpsi = 42.5*6.89 MPa = 292.8 MPa = 292.8 * 1000 kPa
from table A-5,
unit weight w = 70.6 kN/cubic-metre
we get density p = 70.6*102 kg/$m^{3}$
$\frac{S_{ut}}{p}$ = $\frac{292.8*1000}{70.6*102}$ = 40.7 kN-m/kg
