Elasticity/Sample final6

Sample Final Exam Problem 6

Two long cylinders are in contact as shown in the figure below. Both cylinders are made of the same material which has a Young's modulus of 10 GPa and a Poisson's ratio 0.20. The smaller cylinder has a radius of 4 cm while the outer one has a radius of 10 cm. What is the width of the region of contact of the two cylinders under the action of a force of 1 kN per unit length of the cylinders.

Contact between two cylinders

Solution

The area of contact per unit length is given by

$a={\sqrt {{\frac {Pr_{1}r_{2}}{\pi (r_{1}+r_{2})}}\left({\frac {\kappa +1}{\mu }}\right)}}$

For plane strain

$\kappa =3-4\nu$

Therefore, for the material of the cylinders

$\kappa =3-(4)(0.20)=2.2~~;~~\mu ={\frac {E}{2(1+\nu )}}={\frac {10}{2(1+0.20)}}=4.2~{\text{GPa}}$

Since the outer cylinder contains the inner one, the radius of curvature can be considered to be negative. Therefore,

$r_{1}=4~{\text{cm}}=0.04~{\text{m}}~~;~~r_{2}=-10~{\text{cm}}=-0.1~{\text{m}}~~;~~$

The area of contact per unit length of the cylinders is

$a={\sqrt {{\frac {(1)(10^{3})(0.04)(-0.1)}{\pi (0.04-0.1)}}\left({\frac {2.2+1}{(4.2)(10^{9})}}\right)}}=0.13~{\text{mm}}$

The width of the region of contact is

${a=0.13~{\text{mm}}}$

Elasticity