Are Problem The Feature Box in this chapter introduced spider silk as an impress

Question

Are Problem The Feature Box in this chapter introduced spider silk as an impressive natural material with a specific strength greater than structural steel. Given a tensile strength of a particular spider silk as 1.2 x 103 MPa with a density of 1.30 Mg/m3, calculate its specific strength and compare that result with the value for 1040 steel in Table 1 Table 1 Specific Strengths (Strength/Density) specific strength Group Material Noncomposites 1040 steela 9.9 x 106 (0.39 x 106) 16.9 x 106 (0 67 x 106) 2048 plate aluminuma

Explanation / Answer

Specify strength is obtained as yield strength/density

From given data ss for spider silk = (1.2*10^3*10*6 Pa)/(1.3*10^6 g/m^3)

=923.0769 Nm/gram

To compare this with given values in table we need to convert units to mm

Since mass*gravity = force we can divide this term by gravitational constant to get units as meter

1kg * g = 1N

We are using grams in denominator so 1g* g = 1/1000 N

Also we need to convert 1m to 1000 mm

Finally we get specific strength as 94.095505 *10^6 mm

Which is greater than both materials given in table

With respect to steela spider silk is 9.505 times better

With respect to aluminuma spider silk is 5.5678 times better

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