Question

A hanging wire made out of titanium with diameter 0.080 cm is initially 2.3 m long. When a 78 kg mass is hung from it, the wire stretches an amount 2.89 cm. A mole of titanium has a mass of 47.9 grams, and its density is 4.51 g/cm3. Based on these experimental measurements, what is the Young's modulus for titanium?

Find the effective spring stiffness of one interatomic bond in titanium.

Answer 1

Young's modulus = (tensile stress)/(strain)

In this case the strain is
(delta L)/L = 2.89cm/230 cm
= 1.257*10^-2

and the stress is

sigma = (78*9.8 N)/[(pi/4)*(8^10^-4m)^2] = 764.4 N/5.027*10^-7 m^2)
= 1.52*10^9 N/m^2

So E = 1.21*10^11 N/m^2
= 121 GPa

A spring stiffness for an individual Ti-Ti interatomic pair can be estimated by dividing
(Tensile force)/(area occupied by one molecule)
by (stretch per intermolecular molecule pair).

You will need a characteristic intermolecular distance or diameter for Ti atoms in the solid. Call it d. Get that from the number density of Ti atoms, n.
d = n^(-1/3)

n = [4.51 g/cm^3/(47.9g/mole)]*6.02*10^23 atom/mole = 5.56*10^22 atom/cm^3
n^-1/3 = d = 2.6*10^-8 cm
= 2.6*10^-10 m
spring stiffness = k
=(Tension/d^2)/(strain*d)
= E/d = 1.21*10^11/2.6*10^-10
= 4.6*10^20 N/m