Question

You are responsible for examining one of the biomaterials from a newly discovered organism. You make a test specimen 20 mm long, 1 mm high and 1 mm wide and apply a tensile force along its long axis. You find that 4.0 N is required to increase its length by 20%. The material is linearly elastic and isotropic. What is the Young's modulus of the material? How much strain energy was stored during your test?

Answer 1

Given

organism dimensions are

l = 20 mm , h = 1 mm , width w = 1 mm

the force required to get 20 % increase in its length is F = 4 N

the change in lenght about 20 % of its length is  

20% of 20 mm is

20*20/100 = 4 mm

the change in length is dl = (20+4)-20 = 4 mm

the are of cross section is A = w*h = 1*1 mm^2 = 1*10^-6 m^2

now the young's modulus is Y = F*l/A*e

Y = Tensile stress /Strain

Y = (F/A)/(dl/l)

Y = F*l/A*dl

Y = (4*20*10^-3)/(10^-6*4*10^-3)

Y = 20000000 = 20*10^6

the strain energy stored is  

U = p^2*L/(2*Y*A)

P= F/A

U = ((4/1*10^-6)(20*10^-3)/(2*20*10^6*1*10^-6)) = 2*10^-9 J