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.

Solution
The area of contact per unit length is given by

a = \sqrt{\frac{P r_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}         }

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