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A solid cylinder of cortical bone has a length of 500mm, diameter of 2cm and a Young’s Modulus of 17.4GPa. Determine the spring constant ‘k’

User Boxdog
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1 Answer

4 votes

Answer:

The spring constant is
1.09*10^(9)\ N/m

Step-by-step explanation:

Given that,

length = 500 mm

Diameter = 2 cm

Young's modulus = 17.4 GPa

We need to calculate the young's modulus

Using formula of young's modulus


Y=((F)/(A))/((\Delta l)/(l))....(I)


Y=(Fl)/(\Delta l A)

From hook's law


F=kx


k=(F)/(x)


F=k*\Delta l....(II)

Put the value of F in equation


Y=(k*\Delta l* l)/(\Delta l A)


Y=(kl)/(A)

We need to calculate the spring constant


k = (YA)/(l)....(II)

We need to calculate the area of cylinder

Using formula of area of cylinder


A=2\pi* r* l

Put the value into the formula


A=2\pi* 1*10^(-2)*500*10^(-3)


A=0.0314\ m^2

Put the value of A in (II)


k=(1.74*10^(10)*0.0314)/(500*10^(-3))


k=1.09*10^(9)\ N/m

Hence, The spring constant is
1.09*10^(9)\ N/m

User Jasan
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