Question

Young's Modulus (Y) 20 X 1010 N/m2
Shear Modulus (S) 8.1 X 1010 N/m2
Bulk Modulus (B) 16 X 1010 N/m2


The table to the above represents various properties of steel. You have steel wire 4.9 meters in length that stretches 0.16 cm when subjected to a force of 400 N.

What would the diameter of the wire be if you wanted the wire to stretch 0.06 cm less when subjected to this same force?

Answer 1

You want the amount of stretching to decrease from 0.16 cm to 0.10 cm. That is 5/8 of the orginal stetching. To make that happen, the stress nust be decreased by a factor of 5/8. Since the load is the same, the area must increase by a factor of 8/5. The diamenter must increase by a factor sqrt(8/5) = 1.265

Now we have to figure out the original wire diameter -- the one that stretched 0.16 cm.

The wire that stretched 0.16 cm had an area A given by

F/A = Y * (deltaL)/L = 20*10^10*0.16*10^-2/4.9 = 6.53*10^7 N/m^2

A = 6.13*10^-6 m^2
D = sqrt[4A/pi} = 2.79*10^-3 m = 2.79 mm

The new wire diamneter is 1.265 times 2.79 mm, or about 3.5 mm